Category: Class 11

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Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 4 With the Photographer

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With The Photographer Question Answer Warm Up

Photographs capture memorable moments. They fill us with nostalgia. Discuss the following questions.

With The Photographer By Stephen Leacock Question And Answers Question 1.
What are the occasions when photographs are taken?

e.g. birthday parties

Answer:

e.g. birthday parties	Weddings
Convocation	Functions
Annual day	Sports day
Tour	Meetings

With The Photographer Summary Question 2.
Why are photographs taken?

e.g. to freeze our favourite moments

Answer:

e.g. to freeze our favourite moments	to cherish memorable moments
To record best things	to freeze beauty
To document function	to lodge complaints with evidence

Samacheer Kalvi 11th English With the Photographer Textual Questions

1. Based on your understanding of the story, answer the following questions in two or three sentences each.

With The Photographer Essay In English Question (a)
Why did the author go to the photo studio?
Answer:
The author wanted his photograph taken. He wanted to leave it behind with his friends and relatives after his death. So, he went to a photographer.

With The Photographer Paragraph Question (b)
Describe the photographer.
Answer:
The photographer was a drooping’man in a gray suit. He had dim eyes as a natural scientist.

With The Photographer Question (c)
Bring out the significance of what Leacock was reading at the photographer’s.
Answer:
The poet had to wait. To kill the time, he read Ladies companion for 1912 and the girls’ magazine for 1902 and the infants’journal for 1888. He read the old magazines which were meant for people who lived long ago.

Question (d)
Why did Leacock assume that the photographer was praying?
Answer:
The photographer hid himself behind the camera under a cloth. He did not move or say anything for quile’sometime. So, the author assumed that the photographer was praying.

Question (e)
How did the inner room get light?
Answer:
There was a frosted window. A sheet of factory cotton hung against it. A beam of sunlight filtered through it. Thus the studio got the needed light.

Question (f)
Why did the photographer take a longtime to photograph Leacock?
Answer:
The photographer wanted to prove his skills in photography. Somehow, the author’s face didn’t appear to be good enough for a photo. So, he gave many instructions like “open the mouth, close it, droop your ear, roll your eyes, turn your face up” and went on correcting the pose. This took a lot of time.

Question (g)
What angered the author?
Answer:
The author’s facial features and body was severely criticized by the photographer. He gave plenty of instructions. He himself held the author’s face and twisted it. He said, “I don’t like the head … The ears are bad”. He was asked to expand the lungs, contract the waist, turn the face upward. All this added to the author’s annoyance. He became really angry.

Question (h)
Why did the photographer feel happy after taking the photograph?
Answer:
The photographer felt happy after taking the photograph. The author was angry and was about to get up from his place. The animation on his face was caught in the photograph.

Question (i)
Why did Leacock visit the studio on Saturday?
Answer:
The photographer had asked the author to come and collect the proof on Saturday. So, the author went to the studio on Saturday.

Question (J)
How did the author react on seeing his photograph?
Answer:
The author was upset to see his photo. It carried little likeness of him. He even asked the photographer if it was himself on the photo.

Question (k)
What changes had the photographer effected on Leacock’s face in the picture?
Answer:
The photographer had retouched the author’s eyes. His eyebrows were removed and put in new ones. The mouth was adjusted a little. The author had a consolation that his ears were recognizable. The photographer declared his intention of removing it and fixing them anew.

Question (l)
What was the human side to the photographer?
Answer:
The photographer stated, “I think the face would be better three – quarters full. The author was glad to find that he had such a human side to him.

 

Question (m)
Why was the photographer proud to receive Leacock on Saturday?
Answer:
While handing over the proof of the photo, there was a certain pride in the photographer’s manner. He believed that with his technical expertise, he had corrected all the ugly features and made the photo presentable. So, he was naturally proud.

Question (n)
What was the only similarity between Leacock’s face and his photograph?
Answer:
The author’s ears were almost the same in the photograph.

2. Based on your understanding of the lesson, complete the sentences given below to make a summary of the story ‘With the Photographer’ in a paragraph.

(a) The narrator went to the photographer to
(b) The photographer made the author wait for
(c) While waiting in the studio the narrator kept reading
(d) The photographer told him to
(e) The narrator got angry because
(f) The photographer was pleased after
(g) He was called on Saturday to
(h) On seeing the photograph the narrator
(i) The photographer had made changes
(j) The photograph did not look like
(k) The narrator was frustrated as
(l) He left the studio saying

Answers:

(a) take a photograph of himself
(b) an hour
(c) old useless magazines
(d) sit down
(e) the photographer repeatedly said his face was wrong
(f) developing the negative
(g) see the proof
(h) was annoyed
(i) in the author’s face
(j) himself
(k) he could not share it with his friends
(l) that it was a worthless bauble for him

3. Answer the following questions in a paragraph of 100 -150 words each:

Question (a)
Stephen Leacock’s visit to the photo studio turns out to be an annoying experience for him. Discuss citing relevant instances from the story.
Answer:
The author had to wait for an hour and read the magazines like “The Ladies Companion, Girls’ Magazine and the Infants’ Journal. He had a disturbing feeling that he had done an unwarrantable thing in breaking in on the photographer’s privacy and his scientific pursuits with a face like his. After studying his face for sometime from behind the camera, he said, the face was wrong. He commented that the face would be better three quarters full. Then he held the author’s face making him believe that he was going to kiss it. He twisted the author’s face as far as it would go. He said that he didn’t like the head. He asked him to open the mouth a little and then close it. Then he said that the ears was bad. He suggested that he should droop them a little more. He asked him to roll his eyes under the lids.

He asked him to turn his face upward a little and keep his hand on the knee. He instructed the author to hump the neck and contract the waist and also twist the hip. It was the last straw when he said that he didn’t quite like his face for, it was just a trifle too full. These numerous instructions and cynical comments about the features of his face annoyed him. He exploded with anger saying that he had lived with the same face for forty years. He even wanted to leave the place without taking the photograph. When he was about to get up, the photographer clicked the button. The photographer looked pleased. He said that he had caught the author in a moment of animation. Thus the experience with the photographer was really annoying.

“Once you start to dislike someone, Everything they do begins to annoy you. ”

Question (b)
“To me it is but a worthless bauble.”Why did the photographer’s touch of technical expertise appear a worthless bauble to Leacock?
Answer:
The author wanted to have himself photographed. The purpose was to leave behind the image or picture of his likeness to his family and relatives. It would remind them of him after he is dead. The photo could reconcile his absence or loss to them. But the phototgrapher had retouched the eyebrows, eyes, mouth etc. All these features did n6t resemble the author. In a depressed mood, the author said his ears were almost the same.

But the photographer said he could completely replace his ears using a new technique. When the author saw the photo it was technically sound but when it came to likeness, it was a disaster. The purpose of taking the photo was lost. The poet asked the photographer to do all sorts of corrections and keep it with himself and for his friends as a technically sound photo. But for himself it was a worthless bauble. The author broke into tears and left the studio.

“God spent a long time crafting you, and God doesn’t make worthless things. ”

Additional Questions

Question (a)
Attempt a character sketch of the photographer.
Answer:
The photographer was a drooping man in gray coat. He appeared to be a scientist making a study on the faces of people who visit his studio. He wore a serious face all the time. He always looked at the faces of people through his camera. He disapproved of the features of people which did not conform to his estimate of perfect looks. He was proud of his expertise in using various editing skills. He boasted of his skill of removing eyebrows and refixing them artistically. He could correct the mouth, nose and eyes of photos.

Infact, he had corrected every feature of the author’s face except his ears which carried some resemblance to the original. Even that he offered to replace with better looking ears using sophisticated technique.

“To a photographer, photography is to place Head, Heart and Eye along the same line of sight. ”

I. Choose the correct answer from the following:

Question 1.
The photographer was a ____ man in gray suit.
(a) cheerful
(b) drooping
(c) drowsy
(d) flamboyant
Answer:
(b) drooping

Question 2.
The author has to wait for ____ to have his photo taken.
(a) two hours
(b) 15 minutes
(c) 45 minutes
(d) an hour
Answer:
(a) two hours

Question 3.
The author visited a ____ to have his photo taken.
(a) Jeweller’s
(b) dentist
(c) X-ray centre
(d) Studio
Answer:
(d) Studio

Question 4.
Stephen Leacock wanted a ____ taken to leave behind with his friends and relatives.
(a) portrait
(b) photo
(c) will
(d) video
Answer:
(b) photo

Question 5.
The photographer rolled a machine into the centre of the room. The machine was an old ____
(a) megaphone
(b) Radio
(c) TV
(d) camera
Answer:
(d) camera

Question 6.
The cameraman was apparently ____ for light and air.
(a) calm
(b) composed
(c) frantic
(d) foolish
Answer:
(c) frantic

 

Question 7.
The author believed that he had eroded into the ____ of the photographer.
(a) life
(b) business
(c) privacy
(d) publicity
Answer:
(c) privacy

Question 8.
Initially the photographer spent just a ____ behind the camera.
(a) minute
(b) second
(c) rupee
(d) pound
Answer:
(b) second

Question 9.
The photographer said “The face is quite ____ ”.
(a) right
(b) impressive
(c) wrong
(d) handsome
Answer:
(c) wrong

Question 10.
The photographer commented, “The face would be better ____ full”.
(a) one quarter
(b) three – quarter
(c) two quarter
(d) four quarter
Answer:
(b) three – quarter

Question 11.
The author closed his eyes when the photographer held his head in his hands. He thought the photographer was going to ____ him.
(a) touch
(b) kiss
(c) kick
(d) hug
Answer:
(b) kiss

Question 12.
The photographer twisted the author’s ____ as far as it would go.
(a) hand
(b) face
(c) leg
(d) ear
Answer:
(b) face

Question 13.
The photographer asked the author to drop his ____ a little.
(a) shoulders
(b) hands
(c) ears
(d) eyes
Answer:
(c) ears

Question 14.
The photographer instructed the author to expand his ____
(a) eyes
(b) ears
(c) hands
(d) lungs
Answer:
(d) lungs

Question 15.
Inspite of making many corrections in the position, the photographer found author’s face just a ____ too full.
(a) little
(b) lot
(c) a trifle
(d) a lot
Answers
(c) a trifle

II. Identify the speakers.

1. “I want my photograph taken.” – Stephen Leacock
2. “Sit there, and wait”. – The photographer
3. “The face is quite wrong”. – The photographer
4. “I know….. I have always known it”. – Stephen Leacock

III. Spoken by Stephen Leacock to photographer.

1. I’m sure it would.
2. Stop….. This is my face. It is not yours.
3. I’ve lived with it for forty years
4. I know its faults.
5. I know’ tis my face. I know it wasn’t made for me, but it’s my face, the only one I have.
6. Is it me?
7. The eyes don’t look very much like mine.
8. Fine…. but my eyebrows are not like that?
9. Oh, you don’t, don’t you?
10. What about the mouth?
11. The ears though strike me as a good likeness they’re like mine.
12. Heaven gave it to me, humble thought the gift may have been.
13. “To me it is, but a worthless bauble”,
14. What I wanted is no longer done.
15. “Go on, then with your brutal work”

IV. Spoken by the photographer to Stephen Leacock:

1. I don’t like the head.
2. Open the mouth a little.
3. Close it.
4. The ears are bad drop them a little more.
5. Put the hands on the knees.
6. “I think… that I caught the features just in a moment of animation.”
7. No I don’t care for it. I like to get the hair clear back’to the superficies and make out a new brow line.
8. It’s adjusted a little, yours is too low. I found I couldn’t use it.
9. “But I can fix that all right in the print. We have a process now – the sulphide for removing ears entirely. I’ll see if
10. Yes,…. it’s you.
11. Oh, no, “ I’ve retouched them. They came out splendidly, don’t they?
12. The eye brows are removed. We have a process now – the delphide for putting in new ones.
13. I don’t like the hair low on the skull. .

V. Rearrange the sentences logically:

Question 1.
(a) The photographer told him to sit.
(b) The photographer repeatedly said that his face was wrong.
(c) The author went to the photographer to have his photo taken.
(d) The photographer made him wait for an hour.
(e) While waiting in the studio, the narrator kept reading old magazines.

Answer:

(c) The author went to the photographer to have his photo taken.
(d) The photographer made him wait for an hour.
(e) While waiting in the studio, the narrator kept reading old magazines.
(a) The photographer told him to sit.
(b) The photographer repeatedly said that his face was wrong.

Question 2.
(a) He said that he didn’t like the head.
(b) He gave several instructions and kept on commenting about the author’s ears, eyes and face.
(c) The photographer said that the author’s face would be better three quarters full.
(d) The author got angry and said he had lived with the same face for forty years and knew its fault.
(e) He took the author’s face in his hand and twisted it sideways.

Answers:

(c) The photographer said that the author’s face would be better three quarter full.
(e) He took the author’s face in his hands and twisted it sideways.
(a) He said that he didn’t like the head.
(b) He gave several instructions and kept on commenting about the author’s ears, eyes and the face.
(d) The author got angry and said he lived with the same face for forty years and knew its faults.

Question 3.
(a) The photographer offered to remove the ears and fix them anew with a new teclmique.
(b) The ears alone resembled those of the author.
(c) The photographer asked the author to come on Saturday to see the proof.
(d) On seeing the photograph, the author was annoyed.
(e) The photographer had made several changes in the author’s face.

Answers:

(c) The photographer asked the author to come on Saturday to see the proof.
(d) On seeing the photograph, the author was annoyed.
(e) The photographer had made several changes in the author’s face.
(b) The ears alone in the photo resembled those of the author.
(a) The photographer offered to remove the ears and fix them anew with a new technique

Question 4.
(a) Calling it a worthless bauble, he left the studio in tears.
(b) As the photographer had made several changes, he could not share it with his friends.
(c) Stephen Leacock wanted to have his photo taken.
(d) He wanted his family and friends to remember him with the photo.
(e) The photographer, after making him wait for an hour and criticising his looks, took his photo.

Answers:

(c) Stephen Leacock wanted to have his photo taken.
(d) He wanted his family and friends to remember him with the photo after his death.
(e) The photographer, after making him wait for an hour and criticising his looks took his photo.
(b) As the photographer had made several changes, he could not share it with his friends.
(a) Calling it a worthless bauble, he left the studio in tears.

VI. Read the following paragraph and answer the questions given below.

1. “I want my photograph taken,” I said. The photographer looked at me without enthusiasm. He was a drooping man in a gray suit, with the dim eye of a natural scientist. But there is no need, to describe him. Everybody knows what a photographer is like.
“Sit there,” he said, “and wait.”

I waited an hour. I read the Ladies Companion for 1912, the Girls Magazine for 1902 and the Infants Journal for 1888. I began to see that I had done an unwarrantable thing in breaking in on the privacy of this man’s scientific pursuits with a face like mine. After an hour the photographer opened the inner door.
Come in,” he said severely.

I went into the studio.
“Sit down,” said the photographer.

Question (a)
Who wanted to have his photograph taken?
Answer:
Stephen Leacock wanted to have his photo taken.

Question (b)
What was the reaction of the photographer when the speaker disclosed his intention of visiting the studio?
Answer:
The photographer looked at the speaker without any enthusiasm and asked him to sit and wait.

Question (c)
How long did the author wait?
Answer:
He waited for an hour.

Question (d)
What did the author realized while waiting at the studio?
Answer:
The author realized that he had done an unnarratable thing in breaking in on the privacy of the photographer’s scientific pursuits with a face like his.

Question (e)
Why did the photographer make people wait for a long time before taking a photograph?
Answer:
In older times, it took enormous time to load a negative into the big camera. They needed time to fix the camera in the right angle and ensured the appropriate lighting for taking the photo. So, the photographers made people wait for a long time.

2. “The face is quite wrong,” he said. “I know,” I answered quietly; “I have always known it.” He sighed. “I think,” he said, “the face would be better three-quarters full.” “I’m sure it would,” I said enthusiastically, for I was glad to find that the man had such a human side to him. “So would yours. In fact,” I continued, “how many faces one sees that are apparently hard, narrow, limited, but the minute you get them three-quarters full they get wide, large, almost boundless in-”

But the photographer had ceased to listen. He came over and took my head in his hands and twisted it sideways. I thought he meant to kiss me, and I closed my eyes. But I was wrong.

Question (a)
What statement disturbed the author?
Answer:
The photographer said, “The face is quite wrong”. This statement disturbed the author.

Question (b)
What observation cheered up the author?
Answer:
The photographer said, “The face would be better three quarters full”. This cheered up the author.

Question (c)
How did the author respond to the photographer’s observation?
Answer:
The author, in a bid to please the photographer said that one comes across hard, narrow limited faces which become all right when they are three quarters full.

Question (d)
Why did the photographer sigh?
Answer:
The photographer had a feeling that the author’s face was not good for a photo. So, he 1 sighed.

Question (e)
How was the author wrong?
Answer:
When the photographer held the author’s head in his hand, he thought that he was going to kiss him. But he was wrong.

3. “I don’t like the head,” he said. Then he went back to the machine and took another look. “Open the mouth a little,” he said. I started to do so. “Close it,” he added quickly. Then he looked again.“The ears are bad,” he said; “droop them a little more. Thank you. Now the eyes. Roll them in under the lids. Put the hands on the knees, please, and turn the face just a little upward. Yes, that’s better. Now just expand the lungs! So! And hump the neck-that’s it-and just contract ! the waist-ha!-and twist the hip up toward the elbow-now! I still don’t quite like the face, it’s just a trifle too full, but-” I swung myself round on the stool. “Stop,” I said with emotion but, ‘ I think, with dignity.

Question (a)
What did the photographer not like?
Answer:
The photographer did not like the author’s head.

Question (b)
What did the photographer do after going back the machine?
Answer:
He asked the author to open the mouth a little and close it.

Question (c)
What did the photographer tell about the ears?
Answer:
The photographer told the author that the ears were bad.

Question (d)
What did the photographer ask the author to do with?
Answer:
The photographer asked the author to roll his eyes under the lids.

Question (e)
What was the photographer’s final comment comments after suggesting various changes in the posture?
Answer:
The photographer said, “I still don’t like the face. Its just a trifle too full.

4. “This face is my face. It is not yours, it is mine. I’ve lived with it for forty years and I know its faults. I know it’s out of drawing. I know it wasn’t made for me, but it’s my face, the only one I have-” I was conscious of a break in my voice but I went on-”such as it is, I’ve learned

( to love it. And this is my mouth, not yours. These ears are mine, and if your machine is too narrow-” Here I started to rise from the seat.)

Snick! The photographer had pulled a string. The photograph was taken. I could see the machine still staggering from the shock. “I think,” said the photographer, pursing his lips in a pleased smile, “that I caught the features just in a moment of animation.” “So!” I said bitingly,- “features, eh? You didn’t think I could animate them, I suppose? But let me see the picture.’’ “Oh, there’s nothing to see yet,” he said, “I have to develop the negative first. Come back on Saturday and I’ll let you see a proof of it.”

 

Question (a)
Why was there a break in the author’s voice?
Answer:
The photographer had given a lot of instructions and made insulting remarks about his face. He could not possibly allow a photographer to insult a face with which the author had lived for forty years. So, there was a break in his voice protesting against his rude remarks.

Question (b)
How did the author retaliate the photographer’s rudeness?
Answer:
The author said his camera was too narrow to take a proper photograph.

Question (c)
Why did the photographer take the snap when author refused to be photographed?
Answer:
The photographer found the author animated and caught his features in animation.

Question (d)
How did the author reveal his ignorance of the art of photography?
Answer:
Soon after the photo was taken, the author asked for the printout of the photo. In those days, developing a photo involved a long and cumbersome process.

Question (e)
When did the photographer ask the author to see a proof of the photo?
Answer:
The photographer wanted the author to come to collect the proof on Saturday.

5. “Fine,” I said, “but surely my eyebrows are not like that?” “No,” said the photographer, with a momentary glance at my face, “the eyebrows are removed. We have a process now-the Delphide-for putting in new ones. You’ll notice here where we’ve applied it to carry the hair away from the brow. I don’t like the hair low on the skull.”

“Oh, you don’t, don’t you?” I said. “No,” he went on, “I don’t care for it. I like to get the hair clear back to the superficies and make out a new brow line.” “What about the mouth?” I said with a bitterness that was lost on the photographer, “is that mine?” “It’s adjusted a little,” he said, “yours is too low. I found I couldn’t use it.”

Question (a)
What was the only consolation to the author in the proof given to him?
Answer:
The ears in the photo appeared to be like the author’s own. The rest of the facial features had been edited by the photographer. It was the only consolation.

Question (b)
Why did the author express scorn while talking to the proud photographer?
Answer:
The photographer could not accept the ears resembling those of the author. He offered to remove them completely and fix them anew with the process called sulphide. This infuriated the author so much that he expressed scorn in the words he spoke to the photographer.

Question (c)
What was the purpose of the author’s visit to the studio?
Answer:
The author visited with the purpose of taking a photo, a picture which would look like him.

Question (d)
What did the author want to do with the photograph he wanted?
Answer:
The author wanted to leave his photo with his friends who might keep it after his death to reconcile them to his loss.

Question (e)
Why did the author treat the photograph as a bauble?
Answer:
The photographer had used all his craftsmanship to correct the eyebrows, mouth and all the. facial features except the ears. The author was shocked to see the photo dissimilar to his look. Except his ears all looked different. The photographer offered to replace his ears using sulphide process. So, the author treated the photo as a worthless bauble and asked the photographer to retain it.

With the Photographer About the author.

Stephen Leacocok (1869-1944) was a Canadian teacher, political scientist, writer and a humorist. He was the best known English- speaking humorist in the world in the years 1915-1925. He is popular for his light humour and criticism of people’s follies. In his honour “The Stephen Leacock Memorial Medal for Humour’ has been named. His stories and novels earned him a world wide reputation. He was awarded the “Royal Society of Canada’s Lome Pierce Medal in 1937. He is still remembered for his satire and witty criticism in his prose writings.

With the Photographer Summary

Leacock goes to a photographer to get himself photographed. He waits for an hour. He is called into the inner room. The photographer is a serious man. He is obviously not satisfied with Leacock’s face. He says his face is wrong. He says it should be three quarters full. Leacock talks about different kinds of faces and agrees with the photgrapher’s view. Meanwhile, the photographer withdraws himself behind the camera under a covered cloth. He comes close to Leacock. He holds his face tenderly. Leacock closes his eyes thinking he is going to kiss his face. But the photographer turns his face in many directions to suit the angle from which he is going to shoot him with his static camera. He gives many directions to Leacock as to how he should pose.

Infact his instructiohs, “close your mouth, droop your ears, turn your face, expand you lungs, raise your hip” annoy Leacock. He is really confused and frightened. He becomes impatient and scolds him for finding fault with his face. It was after all the only face he had lived with for forty years. He is photographed when he is angry and about to get up. The photographer is pleased because he took the photo-when he was animated.

He asks Leacock to come again on Saturday to see the proof of his photograph. To the great annoyance of the author, the photographer claimed to have edited his eyebrows and mouth and wanted to edit his ears using some sophisticated techniques. Leacock tells the photographer that he wanted a photo of his likeness so that family members and friends could see the photo and remember him after his death. He is so angry that he asks the photographer to keep the corrected photo and left the studio in tears of humiliation.

With the Photographer Glossary

Textual:
animation – excitement
bauble – a thing of no value / trifle
beckoned – called
boundless – limitless
ceased – stopped
depict – show, give a picture of
drooping – bending
emboss – cause to bulge out
frantic – mad, desperate
grave – serious
pursuits – quest
reconcile – to comfort and heal
staggering – shaking or vibrating
super ficies – surface / outer face
trifle – bit
unwarrantable – illegal, wrongful
withering scorn – disapproving hatred

 

Additional:
animated – lively
annoyance – irritation
apparently – obviously
beam – a line of light
conscious – be aware of
contract (u) – shrink
dim – dull
droop – bend
expand – stretch / widen
frightened – got afraid
frosted – dimmed by frost
hooked – bent
humiliation – feeling of being hurt
likeness – similarity
quietly – silently
sophisticated – modem
tenderly – gently
unfolded – opened

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Samacheer Kalvi 11th Commerce Solutions Chapter 5 Hindu Undivided Family and Partnership

Get the Questions and Answers, in Tamilnadu State Board 11th Commerce Solutions for Chapter 5 Hindu Undivided Family and Partnership. Learn the concepts of 11th Commerce Chapter-Wise by referring to the Tamilnadu State Board Solutions for Chapter 5 Hindu Undivided Family and Partnership Questions and Answers. Hence we suggest the students to Download Samacheer Kalvi 11th Commerce Book Solutions Questions and Answers pdf to enhance your knowledge.

Samacheer Kalvi 11th Commerce Hindu Undivided Family and Partnership Textbook Exercise Questions and Answers

I. Choose the Correct Answer

11th Commerce 5th Lesson Questions And Answers Question 1.
The firm of Hindu Undivided Family is managed by whom?
(a) Owner
(b) Karta
(c) Manager
(d) Partner
Answer:
(b) Karta

Samacheer Kalvi Guru 11th Commerce Question 2.
In the firm of Hindu Undivided Family, how one gets the membership?
(a) By Agreement
(b) By Birth
(c) By Investing Capital
(d) By Managing
Answer:
(b) By Birth

Hindu Undivided Family Business Examples Question 3.
The members in the joint Hindu family are known as ………………..
(a) Karta
(b) Coparceners
(c) Generations
(d) Partners
Answer:
(b) Coparceners

Question 4.
Only the male members in the family get the right of inheritance by birth as ………………..
(a) Hindu law
(b) Mitakshara Law
(c) Dayabhaga law
(d) None of these
Answer:
(b) Mitakshara Law

Question 5.
partnership is formed by ………………..
(a) Agreement
(b) relationship among partners
(c) The direction of government
(d) Friendship
Answer:
(a) Agreement

Question 6.
Registration of partnership is ………………..
(a) Compulsory
(b) Optional
(c) Not necessary
(d) None of the above
Answer:
(b) Optional

Question 7.
A temporary partnership which is formed to complete a specific job doing a specified period of time is called ………………..
(a) Partnership – at – will
(b) Particular partnership
(c) Limited Partnership
(d) Joint Venture
Answer:
(d) Joint Venture

Question 8.
The partnership deed is also called ………………..
(a) Articles of Association
(b) Articles of Partnership
(c) Partnership Act
(d) Partnership
Answer:
(b) Articles of Partnership

Question 9.
A partnership is registered with ………………..
(a) Registrar of Companies
(b) Registrar of Co – operatives
(c) Registrar of Firms
(d) District Collector
Answer:
(c) Registrar of Firms

II. Very Short Answer Questions

Question 1.
Who is called KARTA?
Answer:
All the affairs of a Joint Hindu Family are controlled and managed by one person who is known as ‘Karta’ or ‘Manager’.

Question 2.
What are the two schools of Hindu law?
Answer:

1. Dayabhaga and
2. Mitakshara

Question 3.
Who is a called a Partner?
Answer:
The persons who enter into partnership are individually called ‘Partners’.

Question 4.
Who is a Sleeping partner?
Answer:
Such a partner contributes capital and shares in the profits or losses of the firm but does not take part in the management of the business.

Question 5.
Who is a Minor?
Answer:
Under the Indian Majority Act, person who has not completed 18 years of age is a minor.

Question 6.
How many types of Dissolution?
Answer:

1. Without order of court.
2. By order of court.

III. Short Answer Questions

Question 1.
What is the meaning of Joint Hindu Family Business?
Answer:
The Joint Hindu Family Business is a distinct form of organisation peculiar to India. Joint Hindu Family Firm is created by the operation of law. It does not have any separate and distinct legal entity from that of its members.

Question 2.
Write any 3 features of HUF?
Answer:

1. Governed by Hindu Law : The business of the Joint Hindu Family is controlled and managed under the Hindu law.
2. Membership by Birth : The membership of the family can be acquired only by birth. As soon as a male child is born in the family, that child becomes a member.
3. Liability : Except the Karta, the liability of all other members is limited to their shares in the business.

Question 3.
Explain the nature of liability of Karta.
Answer:
The Karta is not only liable to the extent of his share in the business but his separate property is equally attachable and amount of debt can be recovered from his personal property.

Question 4.
What is the meaning of Coparceners?
Answer:
The members of the Joint Hindu Family business are called Coparceners. It is regulated by the provisions of Hindu Law. According to Hindu Succession Act, 1956, a Coparcener will have a share in the Coparcenaries property after the death of the Co – parcener.

Question 5.
Define Partnership?
Answer:
According to Prof. Haney, “The relations which exist between persons, competent to make contracts, who agree to carry on a lawful business in common with a view to private gain”.

Question 6.
What is the minimum and maximum number of members in the partnership concern?
Answer:
Since partnership is the outcome of an agreement, the minimum number of persons required to form a partnership is two. Maximum is restricted to 10 in the case of banking business and to 20 in all other cases.

Question 7.
What is the meaning of Partnership Deed?
Answer:
Though a partnership agreement need not necessarily be in writing, it is important to have a written agreement in order to avoid misunderstandings; it is desirable to have a written agreement. A carefully drafted partnership deed helps in ironing out differences which may develop among partners and in ensuring smooth running of the partnership business. It should be properly stamped and registered.

Question 8.
Who is called a Secret partner?
Answer:
A secret partner is one whose association is not known to the general public. Other than this distinct feature, he is like rest of the partners in all respects.

Question 9.
What is meant by Joint and Several Liability?
Answer:
Every partner is jointly and severally liable for all acts of the firm. It means that in case the assets are inadequate for meeting the claims of creditors, even their personal properties should be made available. The creditors can recover their claims from all the partners.

IV. Long Answer Questions

Question 1.
What is the implied authority of Karta?
Answer:
1. In a joint family firm, only Karta has the implied authority to enter into a contract for debts and pledge the property of the firm for the ordinary purpose of the business of the firm.

2. The Karta is the senior most male member of the family. The members of the family have full faith and confidence in Karta. Only Karta is entitled to deal with outsiders. But other members can deal with outsiders only with the permission of Karta.

3. Except the Karta, the liability of all other members is limited to their shares in the business. The Karta is not only liable to the extent of his share in the business but his separate property is equally attachable and amount of debt can be recovered from his personal property.

Question 2.
Can a minor be admitted in the Joint Hindu Family business – Why?
Answer:
In a partnership, minor cannot become co – partner though he may be admitted to the benefit of partnership. In a Joint Hindu Family firm even a new born baby can be a partner. The membership of the family can be acquired only by birth. As soon as the male child is born in the family, that child becomes the member.

Question 3.
What are the contents of Partnership Deed?

Answer:

1. Name : Name of the Firm.
2. Nature of Business : Nature of the proposed business to be carried on by the partners.
3. Duration of Partnership : Duration of the partnership business whether it is to be run for a fixed period of time or whether it is to be dissolved after completing a particular venture.
4. Capital Contribution : The capital is to be contributed by the partners. It must be remembered that capital contribution is not necessary to become a partner for, one contribute his organising power, business acumen, managerial skill etc., instead of capital.
5. Withdrawal from the Firm : The amount that can be withdrawn from the firm by each partner.
6. Profit/Loss Sharing : The ratio in which the profits or losses are to be shared. If the profit sharing ratio is not specified in the deed, all the partners must share the profits and bear the losses equally.
7. Interest on Capital : Whether any interest is to be allowed on capital and if so. the rate of interest.
8. Rate of Interest on Drawing : Rate of interest on drawings, if any.
9. Loan from Partners : Whether loans can be accepted from the partners and if so the rate of interest payable thereon.
10. Account Keeping : Maintenance of accounts and audit.
11. Salary and Commission to Partners : Amount of salary or commission payable to partners for their services. (Unless this is specifically provided, no partner is entitled to any salary).
12. Retirement : Matters relating to retirement of a partner. The arrangement to be made for paying out the amount due to a retired or deceased partner must also be stated.
13. Goodwill Valuation : Method of valuing goodwill on the admission, death or retirement of a partner.
14. Distribution of Responsibility : Distribution of managerial responsibilities. The work that is entrusted to each partner is better stated in the deed itself.
15. Dissolution Procedure : Procedure for dissolution of the firm and the mode of settlement of accounts thereafter.
16. Arbitration of Dispute : Arbitration in case of disputes among partners. The deed should provide the method for settling disputes or difference of opinion. This clause will avoid costly litigations.

Question 4.
Explain the types of dissolution of partnership firm.
Answer:
Dissolution of Partnership is different from the dissolution of partnership firm. It is due to the fact that when the jurally relation present between all partners, comes to an end, it is known as dissolution of firm, however, when any one of the partners become incapacitated, then the partnership between the concerned partner and other partners of the firm, comes to an end, but the firm may continue to operate, if other partners desire so.

Question 5.
Write any three differences between Dissolution of Partnership and Dissolution of Firm?
Answer:
Dissolution of Partnership:

• In the case of dissolution of partnership, only one or more of the partners terminate their connections with the firm.
• Dissolution of partnership may or may not bring the business of the firm to an end.
• In dissolution of partnership, the business will continue even after dissolution.

Dissolution of Firm:

• Whereas all the partners terminate their connections with the firms in the case of dissolution of firm.
• But dissolution of the firm brings the business of the firm to an end.
• But business cannot be continued in the case of dissolution of firm.

Question 6.
Write the procedure for Registration of a Firing Procedure for registration:
Answer:
A statement should be prepared stating the following particulars.

1. Name of The firm.
2. The principal place of business.
3. Name of other places where the firm carried on business.
4. Names and addresses of all the partners.
5. The date on which .each partner joined the firm.
6. The duration of the firm.

This statement signed by all the partners should be produced to the Registrar of Firms along with the necessary registration fee of Rs.3. Any change in the above particulars must be communicated to the Registrar within 14 days of such alteration.

Samacheer Kalvi 11th Commerce Hindu Undivided Family and Partnership Additional Questions and Answers

I. Choose the Correct Answer:

Question 1.
……………. is that form of business organisation which is ow’ned and controlled by a single individual
(a) Sole trading concern
(b) Partnership firm
(c) Joint Hindu family business
(d) Joint stock companies
Answer:
(a) Sole trading concern

Question 2.
……………… is known as individual entrepreneurship.
(a) Partnership
(b) Sole trader
(c) Joint stock company
(d) Co – operative
Answer:
(b) Sole trader

Question 3.
When his business assets are not sufficient to pay off the business debts, he has to pay from his personal property.
(a) Unlimited Liability
(b) Flexibility
(c) Small capital
(d) Limited Liability
Answer:
(a) Unlimited Liability

Question 4.
“He receives all the profits and risks all of his property in the success or failure of the enterprise”- was said by ………………..
(a) Wheeler
(b) J.L. Hansen
(c) H. Haney
(d) O.R. Krishnasamy
Answer:
(a) Wheeler

Question 5.
Which of the following is under non – corporate enterprise?
(a) Government
(b) Co – operative
(c) Company
(d) Sole trading concern
Answer:
(d) Sole trading concern

II. Very Short Answer Questions

Question 1.
Write any two examples of Joint Hindu Family business run in India?
Answer:

1. Reliance Industries
2. Tata Consultancy Services.

Question 2.
Mention any four kinds of Partners?
Answer:

1. Active partner
2. Sleeping partner
3. Nominal partner
4. Partner in profits only

Question 3.
What is partnership firm?
Answer:
The persons who enter into partnership are collectively known as ‘Firm’.

Question 4.
Write any two types of dissolution through court?
Answer:

1. When a partner becomes of unsound mind.
2. Permanent incapacity observed in its formation, management or in its closure.

III. Short Answer Questions

Question 1.
What is unlimited liability?
Answer:
The liability of a sole trader is unlimited. Since, apart from his business assets, even his private properties are also available for satisfying the claims of creditors. Hence, creditors may give more loans because they can get back the loan from the personal properties of sole traders.

Question 2.
What is Particular partnership?
Answer:
When a partnership is formed to carry on a particular venture or a business of temporary nature, it is called particular partnership. In other words, it comes to an end on the completion of the particular venture.

Question 3.
What is meant by Mitakshara Law?
Answer:
According to Mitakshara law, only the male members in the family get the right of inheritance by birth. It is applied throughout India except Assam and West Bengal.

Question 4.
What is meant by Dayabhaga Law?
Answer:
According to Dayabhaga Law, the right of property develops on the Coparceners by succession and not by birth. The share in the property is not fluctuating on the basis of births and deaths. The share are specified prior to partition. The coparceners can alienate their share of property given without the concern of their coparceners.

IV. Long Answer Questions

Question 1.
What are the rights of a partner? (Any five)
Answer:
1. Right to take part in business : Every partner has a right to take part in the management of the business.

2. Right to be consultant : Every partner has the right to be consulted in all the matters concerning the firm. The decision of the majority will prevail in all the routine matters.

3. Right of access to books, record and document : Every partner has the right of access to all records and books of accounts, and to examine and copy them.

4. Right to share profit : Every partner is entitled to share the profits in the agreed ratio. If no profit – sharing ratio is specified in the deed, they must be shared equally.

5. Right to receive interest : A partner has the right to receive interest on loans advanced by him to the firm at the agreed rate, and where.no rate is stipulated, interest @ 6% p.a. allowed.

Question 2.
What are the circumstances under which a partnership firm is dissolved? (any five)
Answer:
1. By agreement or mutual consent : A firm may be dissolved when all the partners agree to close the affairs of the firm. Just as a partnership is created by contract, it can also be terminated by contract.

2. By insolvency of all the partners but one : If any of the partners adjudged an insolvent »(or if all the partners become insolvent) it is necessary to dissolve the firm.

3. Business becoming unlawful : When the business carried on by the partnership becomes illegal, the partnership firm is automatically dissolved.

4. By notice of dissolution : In the case of partnership at will when any partner gives in writing to all the other partners indicating his intension to dissolve the firm, the firm will be dissolved.

5. Continued loss : If the business of the firm cannot be continued expect at a loss, the same may be dissolved by the court on application by a partner.

Case Study

Question a.
A father had self acquired agricultural land. He transferred the said land in the name of his three sons. The revenue records reflect the names of the three sons with 1/3rd share against each name. Father died recently. However physical partition of the said land amongst the three brothers has not been done as they have mutually decided against it. Eldest son has started managing the land since father’s demise. Is the land in question ancestral property of the three brothers? Can the three brothers claim to a HUF? If yes, then since when are they HUF – after father’s demise or since the date land transferred in their names?
Answer:
Yes, the land is ancestral property of the three brothers. It is their father’s land. The three brothers can claim to a HUF. After their father’s death, they can get the land transferred in their names.

Question b.
Draw a family tree diagram as you think. Just imagine you are running a business under the Joint Hindu Family system.
Answer:

For Future Learning

Question a.
Raman with members of his extended family established a Joint Hindu Family business of Handicrafts. Raman being the head of family controlled the business as ‘Karta’. He had authority to take all decisions for the business. Many times, he sold goods for cash without informing other members of the family business. This resulted in lesser profits. He also sold one of the family properties and gave money to his daughter as a wedding gift. What values did Karta ignore in the above case?
Answer:

1. Unlimited liability
2. No consultation with family members.
3. Quick decision making.
4. Hasty decision.

Question b.
Palani is an Electronics Engineer. He has met two businessmen who wish to enter into a partnership with him for the manufacture of tape – recorders. They are prepared to make the investment and offer a fourth share in profits to Palani. Would you have any special words of advice for Palani?
Answer:
Palani is a working partner. He knows the field of product manufacturing of tape – recorders. So he is an expert in the business.

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Students can Download Accountancy Chapter 8 Bank Reconciliation Statement Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

Samacheer Kalvi 11th Accountancy Bank Reconciliation Statement Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Bank Reconciliation Statement Questions And Answers Pdf Question 1.
A bank reconciliation statement is prepared by ………………
(a) Bank
(b) Business
(c) Debtor to the business
(d) Creditor to the business
Answer:
(b) Business

11th Accountancy Chapter 8 Book Back Answers Question 2.
A bank reconciliation statement is prepared with the help of ………………
(a) Bank statement
(b) Cash book
(c) Bank statement and bank column of the cash book
(d) Petty cash book
Answer:
(c) Bank statement and bank column of the cash book

Accountancy Class 11 Chapter 8 Solutions Question 3.
Debit balance in the bank column of the cash book means ………………
(a) Credit balance as per bank statement
(b) Debit balance as per bank statement
(c) Overdraft as per cash book
(d) None of the above
Answer:
(a) Credit balance as per bank statement

Bank Reconciliation Statement Problems And Solutions Pdf Question 4.
A bank statement is a copy of ………………
(a) Cash column of the cash book
(b) Bank column of the cash book
(c) A customer’s account in the bank’s book
(d) Cheques issued by the business
Answer:
(c) A customer’s account in the bank’s book

Bank Reconciliation Statement Questions And Answers Question 5.
A bank reconciliation statement is prepared to know the causes for the difference between:
(a) The balance as per the cash column of the cash book and bank column of the cash book
(b) The balance as per the cash column of the cash book and bank statement
(c) The balance as per the bank column of the cash book and the bank statement
(d) The balance as per petty cash book and the cash book
Answer:
(c) The balance as per the bank column of the cash book and the bank statement

Bank Reconciliation Questions For Class 11 Question 6.
When money is withdrawn from bank, the bank ………………
(a) Credits customer’s account
(b) Debits customer’s account
(c) Debits and credits customer’s account
(d) None of these
Answer:
(b) Debits customer’s account

Bank Reconciliation Statement Class 11 Question 7.
Which of the following is not the salient feature of bank reconciliation statement?
(a) Any undue delay in the clearance of cheques will be shown up by the reconciliation
(b) Reconciliation statement will discourage the accountant of the bank from embezzlement
(c) It helps in finding the actual position of the bank balance
(d) Reconciliation statement is prepared only at the end of the accounting period
Answer:
(d) Reconciliation statement is prepared only at the end of the accounting period

Bank Reconciliation Statement Class 11 Solutions Question 8.
Balance as per cash book is ₹ 2,000. Bank charge of ₹ 50 debited by the bank is not yet shown in the cash book. What is the bank statement balance now?
(a) ₹ 1,950 credit balance
(a) ₹ 1,950 credit balance
(b) ₹ 1,950 debit balance
(c) ₹ 2,050 debit balance
(d) ₹ 2,050 credit balance
Answer:
(a) ₹ 1,950 credit balance

Bank Reconciliation Statement Questions Question 9.
Balance as per bank statement is ₹ 1,000. Cheque deposited, but not yet credited by the bank is ₹ 2, 000. What is the balance as per bank column of the cash book?
(a) ₹ 3,000 overdraft
(b) ₹ 3,000 favourable
(c) ₹ 1,000 overdraft
(d) ₹ 1,000 favourable
Answer:
(b) ₹ 3,000 favourable

Bank Reconciliation Questions Question 10.
Which one of the following is not a timing difference?
(a) Cheque deposited but not yet credited
(b) Cheque issued but not yet presented for payment
(c) Amount directly paid into the bank
(d) Wrong debit in the cash book
Answer:
(d) Wrong debit in the cash book

II. Very Short Answer Questions

Bank Reconciliation Statement Answers Question 1.
What is meant by bank overdraft?
Answer:
It is not possible to have unfavourable cash balance in the cash book. But, it is possible to have unfavourable balance in the bank account. When the business
is not having sufficient money in its bank account, it can borrow money from the bank. As a result of this, amount is overdrawn from bank.

Accounting Chapter 8 Bank Reconciliation Question 2.
What is bank reconciliation statement?
Answer:
If every entry in the cash book matches with the bank statement, then bank balance will be the same in both the records. But, practically it may not be possible. When the balances do not agree with each other, the need for preparing a statement to explain the causes arises. This statement is called bank reconciliation statement (BRS).

Chapter 8 Bank Reconciliation Question 3.
State any two causes of disagreement between the balance as per bank column of cash book and bank statement.
Answer:
(a) Cheques issued but not yet presented for payment.
(b) Cheques deposited into bank but not yet credited.

Bank Reconciliation Statement Solutions Question 4.
Give any two expenses which may be paid by the banker as per standing instruction.
Answer:
Insurance premium, loan instalment, etc., paid as per standing instructions.

Bank Reconciliation Solutions Question 5.
Substitute the following statements with one word/phrase
(a) A copy of customer’s account issued by the bank.
(b) Debit balance as per bank statement.
(c) Statement showing the causes of disagreement between the balance as per cash book and balance as per bank statement.
Answer:
(a) Pass book
(b) Pass book favourable
(c) (1) Timing difference, (2) Errors in recording

Question 6.
Do you agree on the following statements? Write “yes” if you agree, and write “no” if you disagree.
(a) Bank reconciliation statement is prepared by the banker.
(b) Adjusting the cash book before preparing the bank reconciliation statement is compulsory.
(c) Credit balance as per bank statement is an overdraft.
(d) Bank charges debited by the bank increases the balance as per bank statement.
(e) Bank reconciliation statement is prepared to identify the causes of differences between balance as per bank column of the cash book and balance as per cash column of the cash book.
Answer:
(a) No
(b) No
(c) No
(d) No
(e) Yes

III. Short Answer Questions

Question 1.
Give any three reasons for preparing bank reconciliation statement.
Answer:
The main reasons for preparing bank reconciliation statement are:

1. To identify the reasons for the difference between the bank balance as per the cash book and bank balance as per bank statement.
2. To identify the delay in the clearance of cheques.
3. To ascertain the correct balance of bank column of cash book.

Question 2.
What is meant by the term “cheque not yet presented?”
Answer:
When the cheques are issued by the business, it is immediately entered on the credit side of the cash book by the business. But, this may not be entered in the bank statement on the same day. It will be entered in the bank statement only after it is presented with the bank.

Question 3.
Explain why does money deposited into bank appear on the debit side of the cash book, but on the credit side of the bank statement?
Answer:
When the cheques are deposited into bank, the amount is debited in the cash book on the same day. But, these may not be shown in the bank pass book on the same day because these will be entered in the bank statement only after the collection of the cheques.

Question 4.
What will be the effect of interest charged by the bank, if the balance is an overdraft?
Answer:
The bank has to cover the cost of running the customer’s account. So debit is given to the account of the business towards bank charges. Also, if the business had taken any loan or overdrawn, interest has to be paid by the business. These entries for bank charges and interest are made in the bank statement. But, the entry is made in the cash book only when the bank statement is received by the business. Till then, the Cash book shows more balance than bank statement.

Question 5.
State the timing differences in BRS with examples.
Answer:
The timing differences in BRS are:
(a) cheques issued but not yet presented for payment
(b) cheques deposited into bank but not yet credited
(c) bank charges and interest on loan and overdraft
(d) interest and dividends collected by the bank
(e) dishonour of cheques and bills
(f) amount paid by parties directly into the bank
(g) payment made directly by the bank to others
(h) bills collected by the bank on behalf of its customer

IV. Exercises

Question 1.
From the following particulars prepare a bank reconciliation statement of Jayakumar as on 31^st December, 2016. (3 Marks)
(a) Balance as per cash book ₹ 7,130
(b) Cheque deposited but not cleared ₹ 1,000
(c) A customer has deposited ₹ 800 into the bank directly
Answer:
Bank reconciliation statement of Jayakumar as on 31^st December, 2016

Question 2.
From the following particulars of Kamakshi traders, prepare a bank reconciliation statement as on 31^st March, 2018. (3 Marks)
(a) Debit balance as per cash book ₹ 10,500
(b) Cheque deposited into bank amounting to ₹ 5,500 credited by bank, but entered twice in the cash book
(c) Cheques issued and presented for payment amounting to ₹ 7,000 omitted in the cash book
(d) Cheque book charges debited by the bank ₹ 200 not recorded in the cash book.
(e) Cash of ₹ 1,000 deposited by a customer of the business in cash deposit machine not recorded in the cash book.
Answer:
Bank reconciliation statement of Kamakshi as on 31^st March, 2018

Question 3.
From the following information, prepare bank reconciliation statement to find out the bank statement balance as on 31^st December, 2017. (5 Marks)

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 4.
On 31^st March, 2017, Anand’s cash book showed a balance of ₹ 1,12,500. Prepare bank reconciliation statement. (5 Marks)
(a) He had issued cheques amounting to ₹ 23,000 on 28.3.2017, of which cheques amounting to ₹ 9,000 have so far been presented for payment.
(b) A cheque for ₹ 6,300 deposited into bank on 27.3.2017, but the bank credited the same only on 5th April 2017.
(c) He had also received a cheque for ₹ 12,000 which, although entered by him in the cash book, was not deposited in the bank.
(d) Wrong credit given by the bank on 30th March 2017 for ₹ 2,000.
(e) On 30th March 2017, a bill already discounted with the bank for ₹ 3,000 was dishonoured, but no entry was made in the cash book.
(f) Interest on debentures of ₹ 700 was received by the bank directly.
(g) Cash sales of ₹ 4,000 wrongly entered in the bank column of the cash book.
Answer:
Bank reconciliation statement of Anand as on 31^st March, 2017

Question 5.
From the following particulars of Siva and Company, prepare a bank reconciliation statement as on 31^st December, 2017. (2 Marks)
(a) Credit balance as per cash book ₹ 12,000.
(b) A cheque of ₹ 1,200 issued and presented for payment to the bank, wrongly credited in the cash book as ₹ 2,100.
(c) Debit side of bank statement was undercast by ₹ 100.
Answer:
Bank reconciliation statement of Siva and Company as on 31^st December, 2017

Question 6.
From the following particulars of Raheem traders, prepare a bank reconciliation statement as on 31^st March, 2018. (3 Marks)
(a) Overdraft as per cash book ₹ 2,500
(b) Debit side of cash book was undercast by ₹ 700
(c) Amount received by bank through RTGS amounting to ₹ 2,00,000, omitted in the cash book.
(d) Two cheques issued for ₹ 1,800 and ₹ 2,000 on 29^th March 2018. Only the second cheque is presented for payment.
(e) Insurance premium on car for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
Answer:
Bank reconciliation statement of Raheem as on 31^st March, 2018

Question 7.
From the following information, prepare bank reconciliation statement as on 31^st December, 2017 to find out the balance as per bank statement. (5 Marks)

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 8.
Prepare bank reconciliation statement from the following data. (5 Marks)

Answer:

Question 9.
From the following particulars of Veera traders, prepare a bank reconciliation statement as on 31^st December, 2017. (2 Marks)
(a) Credit balance as per bank statement ₹ 6,000
(b) Amount received by bank through NEFT for ₹ 3,500, entered twice in the cash book.
(c) Cheque dishonoured amounting to ₹ 2,500, not entered in cash book.
Answer:
Bank reconciliation statement of Veera Traders as on 31^st December, 2017

Question 10.
Prepare bank reconciliation statement from the following data and find out the balance as per cash book as on 31^st March, 2018. (3 Marks)

Answer:
Bank reconciliation statement as on 31^st March, 2018

Question 11.
Ascertain the cash book balance from the following particulars as on 31^st December, 2017: (5 Marks)

1. Credit balance as per bank statement ₹ 2,500
2. Bank charges of ₹ 60 have not been entered in the cash book
3. Cheque deposited on 28^th December 2017 for ₹ 1,000 was not yet credited by the bank
4. Cheque issued on 24^th December 2017 for ₹ 700, not yet presented for payment
5. A dividend of ₹ 400 collected by the bank directly but not entered in the cash book
6. A cheque of ₹ 600 had been dishonoured, but no entry was made in the cash book
7. Interest on term loan ₹ 1,200 debited by bank but not accounted in cash book
8. No entry had been made in the cash book for a trade subscription of ₹ 500 paid vide banker’s order on 23^rd December 2014

Answer:
Bank reconciliation statement as on 31^st December, 2017

Question 12.
From the following particulars of Raja traders, prepare a bank reconciliation statement as on 31^st January, 2018. (5 Marks)
(a) Balance as per bank statement ₹ 5,000
(b) Cheques amounting to ₹ 800 had been recorded in the cash book as having been deposited into the bank on 25^th January 2018, but were entered in the bank statement on 2^nd February 2018.
(c) Amount received by bank through NEFT amounting to ₹ 3,000, omitted in the cash book.
(d) Two cheques issued for ₹ 3,000 and ₹ 2,000 on 29^th March 2018. Only the first cheque is presented for payment.
(e) Insurance premium on motor vehicles for ₹ 1,000 paid by the bank as per standing instruction not recorded in the cash book.
(f) Credit side of cash book was undercast by ₹ 700
(g) Subsidy received directly by the bank from the state government amounting to ₹ 10,000, not entered in cash book.
Answer:
Bank reconciliation statement of Raja Traders as on 31^st January, 2018

Question 13.
From the following particulars of Simon traders, prepare a bank reconciliation statement as on 31^st March, 2018. (2 Marks)
(a) Debit balance as per bank statement ₹ 2,500
(b) Cheques deposited amounting to ₹ 10,000, not yet credited by bank.
(c) Payment through net banking for ₹ 2,000, omitted in the cash book
Answer:
Bank reconciliation statement of Simon Traders as on 31^st March, 2018

Question 14.
From the following particulars, ascertain the cash book balance as on 31^st December, 2016.

1. Overdraft balance as per bank statement ₹ 1,26,640 (3 Marks)
2. Interest on overdraft entered in the bank statement, but not yet recorded in cash book ₹ 3,200
3. Bank charges entered in bank statement, but not found in cash book ₹ 600
4. Cheques issued, but not yet presented for payment ₹ 23,360
5. Cheques deposited into the bank but not yet credited ₹ 43,400
6. Interest on investment collected by the bank ₹ 24,000

Answer:
Bank reconciliation statement as on 31^st December, 2016

Question 15.
From the following particulars of John traders, prepare a bank reconciliation statement as on 31^st March, 2018. (5 Marks)
(a) Bank overdraft as per bank statement ₹ 4,000
(b) Cheques amounting to ₹ 2,000 had been recorded in the cash book as having been deposited into the bank on 26^th March 2018, but were entered in the bank statement on 4^th April 2018.
(c) Amount received by bank through cash deposit machine amounting to ₹ 5,000, omitted in the cash book.
(d) Amount of ₹ 3,000 wrongly debited to John traders account by the bank, for which no details are available.
(e) Bills for collection credited by the bank till 29th March 2017 amounting to ₹ 4,000, but no advice received by John traders.
(f) Electricity charges made through net banking for ₹ 900 was wrongly entered in cash column of the cash book instead of bank column.
(g) Cash sales wrongly recorded in the bank column of the cash book for ₹ 4,000.
Answer:
Bank reconciliation statement of John Traders as on 31^st March, 2018

Question 16.
Prepare bank reconciliation statement from the following data. (3 Marks)

Answer:
Bank reconciliation statement

Question 17.
Prepare bank reconciliation statement as on 31^st March, 2017 from the following extracts of cash book and bank statement.

Bank Statement

Answer:

Question 18.
A trader received his bank statement on 31^st December, 2017 which showed an overdraft balance of ₹ 12,000. On the same day, his cash book showed a debit balance of ₹ 2,000.
Analyse the following transactions. Choose the possible causes and prepare a bank reconciliation statement to show the causes of differences.
(a) Cheque deposited for ₹ 2,000 on 21^st December, 2017. Bank credited the same on 26^th December, 2017.
(b) Cheque issued for payment on 26^th December, 2017 amounting to ₹ 2,500, not yet presented until 31^st, December, 2017.
(c) Bank charges amounting to ₹ 200 not yet entered in the cash book.
(d) Online payment for ₹ 1,500 entered twice in the cash book.
(e) Cheque deposited amounting to ₹ 1,000, but omitted in the cash book. The same cheque was dishonoured by bank, but not yet entered in cash book.
(f) Cheque deposited, not yet credited by bank amounting to ₹ 17,800.
Answer:
Bank reconciliation statement as on 31^st December, 2017

Note: Transactions (a) and (c) have been entered in both cash book and bank statement also. So we need not false in to accounts.

Textbook Case Study Solved

Magesh, an enthusiastic young entrepreneur, started a business on 1^st  December, 2017. He opened a current account with a nationalised bank for his business transaction. In the same bank, he maintains his personal savings bank account too. He did not find time to maintain his cash book. So he appointed a person called Dinesh to take care of bank transactions. But that person was inexperienced.

Question 1.
On 1^st December, 2017, the opening balance as per cash book and bank record was the same. On 2^nd December, Magesh issued a cheque for ₹ 2,000 to a supplier, but the same was entered in the credit side of the cash book as ₹ 200.
Answer:
Credit balance as per bank statement is ₹ 19,700.

Question 2.
On 3^rd, December, Magesh issued his savings bank account cheque for his personal expenses amounting to ₹ 2,500, but Dinesh assumed this as current account cheque and the same was entered in the cash book as drawings.
Answer:
Over casting of debit side of bank column of the cash book is ₹ 1800.

Question 3.
Dinesh was asked to deposit cash of ₹ 1,000 in cash deposit machine in order to make a payment to one of the business’ supplier. He credited the same in the bank column of the cash book.
Answer:
Wrong debit in cash book is ₹ 2,500.

Question 4.
On 15^th December, one of his customers made online payment to Magesh’s current account, amounting to ₹ 1,000. There was no entry in the cash book for this.
Answer:
Wrong credit in cash book is ₹ 1,000.
Instead of personal bank account he can open business bank account (i.e.) current account.

Question 5.
Dinesh received his salary in cash for ₹ 5,000. He credited this amount in the bank column of cash book.
Answer:
Online payment no recorded in cash book is ₹ 1,000.

Question 6.
Bank made payment on 23^rd December, amounting to ₹ 2,500, as per standing instruction. But, there is no entry in the cash book for the same.
Answer:
Wrong credit in the cash book of bank column is ₹ 5,000.

Question 7.
On 31^st, December 2017, Magesh received a bank statement from his bank, which showed a credit balance of ₹ 19,700. He instructed Dinesh to check the statement with the cash book. On comparing both, Dinesh found that the cash book showed a balance of ₹ 14,500. He was puzzled. He needs your help to reconcile the balances.
Answer:
Insurance premium paid but not entered in cash book is ₹ 2,500.
Bank reconciliation cash book as on 31^st December 2017

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Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 4 Cost and Revenue Analysis

Tamilnadu State Board Solutions for 11th Economics Chapter 4 Cost and Revenue Analysis Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Cost and Revenue Analysis Text Book Back Questions and Answers

Part – A

11th Economics Chapter 4 Book Back Answers Multiple Choice Questions

11th Economics Chapter 4 Question 1.
Cost refers to _______
(a) price
(b) value
(c) fixed cost
(d) cost of production
Answer:
(d) cost of production

Cost And Revenue In Economics Class 11 Question 2.
Cost functions are also known as …………………….. function.
(a) Production
(b) Investment
(c) Demand
(d) Consumption
Answer:
(a) Production

Cost And Revenue In Economics Question 3.
Money cost is also known as _______ cost.
(a) explicit
(b) implicit
(c) social
(d) real
Answer:
(a) explicit

Cost And Revenue Analysis Question 4.
Explicit cost plus implicit cost denote ………………… cost.
(a) Social
(b) Economic
(c) Money
(d) Fixed
Answer:
(b) Economic

Samacheer Kalvi Guru 11 Economics Question 5.
Explicit costs are termed as
(a) out of pocket expenses
(b) social cost
(c) real cost
(d) sunk cost
Answer:
(a) out of pocket expenses

Cost And Revenue Analysis In Economics Question 6.
The costs of self – owned resources are termed as ……………………… cost.
(a) Real
(b) Explicit
(c) Money
(d) Implicit
Answer:
(d) Implicit

Cost And Revenue Questions Question 7.
The cost that remains constant at all levels of output is _______ cost.
(a) fixed
(b) variable
(c) real
(d) social
Answer:
(a) fixed

Cost And Revenue Analysis In Economics Pdf Question 8.
Identify the formula for estimating average variable cost.
(a) TC/Q
(b) TVC/Q
(c) TFC/Q
(d) TAC/Q
Answer:
(b) TVC/Q

Samacheer Kalvi Guru 11th Economics Question 9.
The cost incurred by producing one more unit of output is _______ cost.
(a) variable
(b) fixed
(c) marginal
(d) total
Answer:
(c) marginal

Question 10.
The cost that varies with the level of output is termed as …………………. cost.
(a) Money
(b) Variable cost
(c) Total cost
(d) Fixed cost
Answer:
(b) Variable cost

Question 11.
Wage is an example for _______ cost of the production.
(a) fixed
(b) variable
(c) marginal
(d) opportunity
Answer:
(b) variable

Question 12.
The cost per unit of output is denoted by …………………… cost.
(a) Average
(b) Marginal
(c) Variable
(d) Total
Answer:
(a) Average

Question 13.
Identify the formula of estimating average cost.
(a) AVC/Q
(b) TC/Q
(c) TVC/Q
(d) AFC/Q
Answer:
(b) TC/Q

Question 14.
Final total cost where TFC = 100 and TVC = 125.
(a) 125
(b) 175
(c) 225
(d) 325
Answer:
(c) 225

Question 15.
Long-run average cost curve is also called as _______ curve.
(a) demand
(b) planning
(c) production
(d) sales
Answer:
(b) planning

Question 16.
Revenue received from the sale of products is known as …………………….. revenue.
(a) Profit
(b) Total revenue
(c) Average
(d) Marginal
Answer:
(b) Total revenue

Question 17.
Revenue received from the sale of an additional unit is termed as _______ revenue.
(a) profit
(b) average
(c) marginal
(d) total
Answer:
(c) marginal

Question 18.
Marginal revenue is the addition made to the ……………………….
(a) Total sales
(b) Total revenue
(c) Total production
(d) Total cost
Answer:
(b) Total revenue

Question 19.
When price remains constant, AR will be _______ MR.
(a) equal to
(b) greater than
(c) less than
(d) not related to
Answer:
(a) equal to

Question 20.
A bookseller sold 40 books with a price of ₹10 each. The total revenue of the sellers is ₹ …………………….
(a) 100
(b) 200
(c) 300
(d) 400
Answer:
(d) 400

Part – B

Answer the following questions in one or two sentences

Question 21.
Define cost.
Answer:
Cost refers to the total expenses incurred in the production of a commodity.

Question 22.
Define cost function?
Answer:
The functional relationship between cost and output is expressed as ‘Cost Function’.
A Cost Function may be written as
C = f (Q)
Eg: TC = Q³ – 18Q² + 91Q + 12
Where, C = Cost, and Q = quantity of output. Cost functions are derived functions because they are derived from Production Functions.

Question 23.
What do you mean by fixed cost?
Answer:
Fixed cost does not change with the change in the quantity of output. The expenses on fixed factors remain unchanged irrespective of the level of output and these expenses are called fixed cost.

Question 24.
Define Revenue?
Answer:
The amount of money that a producer receives in exchange for the sale of goods is known as revenue. In short, revenue means sales revenue. It is the amount received by a firm from the sale of a given quantity of a commodity at the prevailing price in the market.

Question 25.
Explicit Cost – Define.
Answer:
Explicit cost refers to the actual expenditures of the firm to purchase or hire the inputs.

Question 26.
Give the definition for ‘Real Cost’?
Answer:
Real Cost refers to the payment made to compensate for the efforts and sacrifices of all factor owners for their services in production. Real Cost includes the efforts and sacrifices of landlords in the use of land, capitalists to save and invest, and workers in foregoing leisure. Real costs are considered pains and sacrifices of labour as the real cost of production.

Question 27.
What is meant by Sunk cost?
Answer:
A cost incurred in the past and cannot be recovered in the future is called a sunk cost.

Part – C

Answer the following questions in One Paragraph

Question 28.
Distinguish between fixed cost and variable cost.
Answer:
Fixed Cost:

1. Does not change with the change in the quantity of output.
2. It is also called as ‘Supplementary cost’ or ‘Overhead cost’.
   (Eg.) Rent of the factory, Permanent worker’s salary.

Variable Cost:

1. These costs vary with the level of output
2. It is also called as ‘Prime cost’, ‘Special cost’ or ‘Direct cost’.
   (Eg.) Cost of raw materials, Temporary worker’s salary.

Question 29.
State the differences between money cost and real cost.
Answer:
Money Cost:

1. It is the total money expenses incurred by a firm in producing a commodity.
2. It includes the cost of raw materials, payments of wages and salaries, rent, interest, etc.,
3. It is also called as prime cost or direct cost or nominal cost.

Real Cost:

1. It refers to the payment made to compensate for the efforts and sacrifices of all factor owners for their services in production.
2. It includes the efforts and sacrifices of factors of production.
3. Landlords effort is the use of land, capitalists to save and invest and workers in foregoing leisure.

Question 30.
Distinguish between explicit cost and implicit cost.
Answer:
Explicit Cost:

1. Payment made to others for the purchase of factors of production.
2. It includes wages, payment for raw material, rent, interest, expenditure on transport and advertisement.
3. It is also called as accounting cost or out of pocket cost or money cost.

Implicit Cost:

1. Payment made to the use of resources that the firm already owns.
2. Cash payment is not made for the use of producer’s own land, building, machinery and other factors of production.
3. Implicit cost is also called as imputed cost or book cost.

Question 31.
Define opportunity cost and provide an example?
Answer:

1. Opportunity cost refers to the cost of the next best alternative use. In other words, it is the value of the next best alternative foregone.
2. For example, a farmer can cultivate both paddy and sugarcane in farmland.
3. If he cultivates paddy, the opportunity cost of paddy output is the amount of sugarcane output given up.
4. Opportunity Cost is also called as “Alternative Cost” or Transfer cost.

Question 32.
Stale the relationship between AC and MC.
Answer:
There is a unique relationship between the AC and MC curves.

1. When AC is falling, MC lies below AC.
2. When AC becomes constant, MC also becomes equal to it.
3. When AC starts increasing, MC lies above the AC.
4. MC curve always cuts AC at its minimum point from below.

Question 33.
Write a short note on Marginal Revenue?
Answer:

1. Marginal Revenue [MR] is the addition to the total revenue by the sale of an additional unit of a commodity.
2. MR can be found out by dividing change in total revenue by the change in quantity sold out.
3. MR = ∆TR/∆Q where MR denotes Marginal Revenue, ∆TR denotes change in Total Revenue and ∆Q denotes change in total quantity.
4. The other method of estimating MR is:

MR = TR_n – TR_n-1, (or) TR_n+1 – TR_n
Where, MR denotes Marginal Revenue,
TR_n denotes total revenue of n^th item,
TR_n-1 denotes Total Revenue of n – 1^th item and
TR_n+1 denotes Total Revenue of n + 1^th item.
If TR = PQ,
MR = dTR/dQ = P, which is equal to AR.

Question 34.
Discuss the Long run cost curves with a suitable diagram.
Answer:
In the long run all factors of production become variable. 1 he existing size of the firm can be increased. There are neither fixed inputs nor fixed costs in the long run.

LAC = LTC/Q
LAC – Long-run average cost LTC – Long-run total cost
Q – denotes the quantity of output.
The LAC curve is derived from short run average cost curves. It is the locus of points denoting the least cost curve of producing the corresponding output.
The LAC curve is called as ‘Planning curve’ or ‘Envelope curve’

Part – D
Answer the following questions in about a page

Question 35.
If total cost =10+Q³, find out AC, AVC, TFC, AFC when Q = 5.
Answer:
TC = 10 + Q³
AC = \(\frac { TC }{ Q } \)
AC = \(\frac{10+Q^{3}}{Q}\)
If Q=5, Q = 5 × 5 × 5 = 125
AC = \(\frac { 10 + 125 }{ 5 } \) = \(\frac { 135 }{ 5 } \) = 27

AVC:
TC = 10 + Q³
TC = TFC + TVC
TVC = Q³
AVC = \(\frac { TVC }{ Q } \)
= \(\frac{Q^{3}}{Q} = Q\)²
If Q = 5, then AVC = 5²
AVC = 25

TFC:
TC = 10 + Q³
TC = TFC + TVC
TFC = 10

AFC:
AFC = \(\frac { TFC }{ Q } \)
= \(\frac { 10 }{ 5 } \)
AFC = 2

Question 36.
Discuss the short run cost curves with suitable diagram :
Answer:
1. Total fixed cost: TFC
All payments for the fixed factors of production are known as total fixed cost. It does not change with output.

2. Total variable cost: TVC
All payments to the variable factors of production is called as total variable cost. As output increases TVC also increases.

3. Total cost curves: TC
Total cost means the sum total of all payments made in the production. It is the total cost of production.
TC = TFC + TVC.

4. Average fixed cost: AFC
It refers to the fixed cost per unit of output.
AFC = \(\frac { TFC }{ Q } \)

5. Average variable cost: AVC
It refers to the total variable cost per unit of output.
AVC = \(\frac { TFC }{ Q } \)

6. Average cost:
It refers to the total cost per unit of output.
AC = \(\frac { TC }{ Q } \) (or) AC = AFC + AVC

1. ATC curve is also a ‘U’ shaped curve.
2. Initially ATC declines, reaches a minimum and rises beyond the optimum output.
3. The ‘U’ shape of the AC reflects the law of the variable proportions.

Marginal Cost:
Marginal cost is the additional made to the total cost by producer one extra unit of output

Question 37.
Bring out the relationship between AR and MR curves under various price conditions.
Answer:
If a firm is able to sell additional units at the same price then AR and MR will be constant. If the firm sells its additional units only by reducing the price then both AR and MR will fall and be different.

Constant AR and MR: (at fixed price)
If price remains constant, MR also remain constant and coincide with AR. Under perfect competition as the price is constant, AR is equal to MR and their shape will be straight line horizontal to X axis.
TR – AR – MR – Constant price

Declining AR and MR: (at declining price)

When a firm sells large quantities at lower prices both AR and MR will fall. But the fall in MR will be more steeper than the fall in the AR and MR lies below AR.

The MR curve divides the distance between AR curve and Y axis into two equal parts. The decline in AR need not be a straight line or linear. If the prices are declining with the increase in quantity sold, the AR can be non-linear may be concave or convex to the origin

AR,TR,MR declining price

Samacheer Kalvi 11th Economics Cost and Revenue Analysis Additional Questions and Answers

Part – A

Choose the best options

Question 1.
Real Costis ……………………
(a) Pain and sacrifice
(b) Subjective concept
(c) Efforts and foregoing leisure
(d) All the above
Answer:
(d) All the above

Question 2.
Average fixed cost is obtained by _______
(a) TC / Q
(b) TFC / Q
(c) TVC / Q
(d) None of the above
Answer:
(b) TFC / Q

Question 3.
The Marginal Cost curve is ……………………..
(a) V-shaped
(b) Upward
(c) Downward
(d) U – shaped
Answer:
(d) U – shaped

Question 4.
Total fixed cost + Total variable cost is?
(a) AC-MC
(b) TC-AC
(c) TC
(d) None
Answer:
(c) TC

Question 5.
What is break-even point?
(a) No profit no loss point
(b) No profit
(c) No loss
(d) Profit – point
Answer:
(a) No profit no loss point

Question 6.
Break-Even point is _______
(a) Total cost and total revenue
(b) Average revenue and financial revenue
(c) No profit – no loss point
(d) All the above
Answer:
(c) No profit – no loss point

Question 7.
What is the other name for “Opportunity Cost”?
(a) Real Cost
(b) Money Cost
(c) Economic Cost
(d) Social Cost
Answer:
(a) Real Cost

Question 8.
Average variable cost is _______
(a) TFC / Q
(b) TVC / Q
(c) TC / Q
(d) None
Answer:
(b) TVC / Q

Question 9.
…………………….. revenue means price of the product.
(a) Total
(b) Marginal
(c) Profit
(d) Average
Answer:
(d) Average

Question 10.
Cost function is the
(a) Relationship between total cost and output
(b) Relationship between revenue and cost
(c) Relationship between wages and interest d. None of the above
(d) None of this above
Answer:
(a) Relationship between total cost and output

Question 11.
Match the following

(a) 1 – (iv), 2 – (i), 3 – (iii), 4 – (ii)
(b) 1 – (iv), 2 – (i), 3 – (ii), 4 – (iii)
(c) 1 – (ii), 2 – (iii), 3 – (iv), 4 – (i)
(d) 1 – (iii), 2 – (iv), 3 – (i), 4 – (ii)
Answer:
(b) 1 – (iv), 2 – (i), 3 – (ii), 4 – (iii)

Match the following and choose the answer using the codes given below

Question 1.

(a) 1 2 3 4
(b) 3 4 1 2
(c) 2 3 4 1
(d) 4 3 1 2
Answer:
(b) 3 4 1 2

Question 2.

(a) 3 4 2 1
(b) 1 2 3 4
(c) 2 3 4 1
(d) 4 3 1 2
Answer:
(a) 3 4 2 1

Question 3.

(a) 2 3 4 1
(b) 3 4 2 1
(c) 1 2 3 4
(d) 4 3 2 1
Answer:
(d) 4 3 2 1

Question 4.

(a) 2 3 4 1
(b) 2 1 4 3
(c) 4 3 2 1
(d) 2 4 1 3
Answer:
(b) 2 1 4 3

Choose the correct statement

Question 5.
(a) In the long run, all the factors are fixed
(b) The LAC curve is called an envelope curve.
(c) LAC is equal to the long-run total cost
(d) LAC curve cannot be derived from short-run cost curves
Answer:
(b) The LAC curve is called as an envelope curve.

Question 6.
(a) Revenue means the price of the product.
(b) Marginal revenue is equal to the price of the product.
(c) Revenue means sales revenue d. Average revenue is the total income of the firm.
(d) Average revenue is the total income of the firm
Answer:
(c) Revenue means sales revenue d. Average revenue is the total income of the firm.

Choose the incorrect pair

Question 7.

Answer:
(d) MR is infinity (iv) TR is decreasing

Question 8.

Answer:
(a) ATC  – (1) TC_n – TC_n-1

Analyze the reason for the following

Question 9.
Assertion (A) : If AR remains constant MR is also constant.
Reason (R) : MR is the addition made to the TR by the sale of an additional unit of a commodity.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is true (R) is false.
Answer:
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 10.
Assertion (A) : Real cost refers to the payment made to compensate the efforts and sacrifice of all factor owners.
Reason (R) : Adam Smith regarded pain and sacrifice of labour as real cost of production.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Choose the incorrect statement

Question 11.
(a) When AC is falling, MC lies below AC.
(b) When AC becomes constant, MC also equal to it.
(c) When AC starts increasing, MC lies above the AC.
(d) MC never cuts AC curve.
Answer:
(d) MC never cuts AC curve.

Question 12.
(a) MR is equal to zero and TR is decreasing
(b) AR and MR curve depends upon the elasticity of AR curve
(c) When price elasticity is greater than one, MR is positive
(d) When price elasticity is less than one, MR is negative
Answer:
(a) MR is equal to zero and TR is decreasing

Choose the odd one out

Question 13.
(a) Money cost
(b) Total variable
(c) Real cost
(d) prime cost
Answer:
(b) Total variable

Question 14.
(a) When P = 3, Q = 8 then TR = 24
(b) When P = Q = 1 then TR = 10
(c) When P = 4, Q = 6 then TR = 27
(d) When P = 5, Q = 7 then TR = 35
Answer:
(c) When P = 4, Q = 6 then TR = 27

Fill in the blanks with the suitable option given below

Question 15.
Economics profit is ______
(a) TR – TC
(b) TC – TR
(c) AC – MC
(d) None
Answer:
(a) TR – TC

Question 16.
Cost function is the ________
(a) Relationship between total cost and output
(b) Relationship between revenue and cost
(c) Relationship between wages and interest
(d) None of the above
Answer:
(a) Relationship between total cost and output

Question 17.
Break – even point is _________
(a) Total cost and total revenue
(b) Average revenue and financial revenue
(c) No profit – No loss point
(d) All the above
Answer:
(c) No profit – No loss point

Choose the best option

Question 18.
Average fixed cost is obtained by
(a) TC / Q
(b) TVC / Q
(c) AC / Q
(d) TFC / Q
Answer:
(d) TFC / Q

Question 19.
Long-run average cost curve can also be called as _____
(a) Planning curve
(b) Envelope curve
(c) Boat-shaped curve
(d) All the above
Answer:
(d) All the above

Question 20.
Total fixed cost + Total variable cost is _____
(a) AC – MC
(b) TC
(c) TC – AC
(d) None
Answer:
(b) TC

Part – B

Answer the following questions in one or two sentences

Question 1.
Give the definition for Economic Cost?
Answer:

1. Economic cost refers to all payments made to the resources owned and purchased or hired by the firm in order to ensure their regular supply to the process of production. It is the summation of explicit and implicit costs.
2. Economic Cost is relevant to calculate the normal profit and thereby the economic profit of a firm.

Question 2.
What is the economic cost?
The economic cost is the summation of explicit and implicit costs.

Question 3.
What is the Average Variable Cost?
Answer:
The average variable cost refers to the total variable cost per unit of output. It is obtained by dividing total variable cost (TVC) by the quantity of output [Q], AVC = TVC/Q, where AVC denotes Average variable cost, TVC denotes total variable cost and Q denotes the quantity of output.

Question 4.
What is a floating cost?
Answer:
Floating cost refers to all expenses that are directly associated with business activities but not with asset creation.

Question 5.
What are variable costs?
Answer:
Variable cost vary with the level of output. It is also called as prime cost, special cost or direct cost.

Question 6.
What is total revenue?
Answer:
Total revenue is the amount of income received by the firm from the sale of its products.

Question 7.
What is marginal revenue?
Answer:
Marginal revenue is the addition to the total revenue by the sale of an additional unit of a commodity.
MR = TR_n – TR_n-1

Part – C

Answer the following questions in One Paragraph

Question 1.
How can you calculate average fixed cost ?
Answer:
Average fixed cost is the fixed cost per unit of output. It is obtained by dividing the total fixed cost by the quantity of output.
AFC = \(\frac { TFC }{ Q } \)
(Eg.) If TFC is 100;
Q = 10 Find AFC
AFC = \(\frac { TFC }{ Q } \)
= \(\frac { 1000 }{ 10 } \)
AFC = 100.

Question 2.
Define the Prime Cost?
Answer:

1. All costs that vary with output, together with the cost of administration are known as Prime Cost.
2. Prime Cost = Variable Costs + Costs of Administration.

Question 3.
Write a note on average revenue.
Answer:
Average revenue is the revenue per unit of the commodity sold. It is calculated by dividing the total revenue (TR) by the number of units sold (Q).
AR = \(\frac { TR }{ Q } \)
If TR = PQ
AR = \(\frac { PQ }{ Q } \) = P
AR = P
AR – Average revenue, TR – Total revenue, Q – quantity of unit sold.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3

11th Maths Exercise 2.3 Question 1.
Represent the following inequalities in the interval notation:
Solution:

⇒ x ∈ [-1, 4)

[] closed interval, end points are included
() ➝ open interval
end points are excluded

(ii) x ≤ 5 and x ≥ -3[i] x ≤ 5 and x ≥ -3
Solution:

x ∈ [-3, 5)

(iii) x < -1 or x < 3
Solution:

x ∈ (-∞, -1) or x ∈ (-∞, 3)

(iv) – 2x > 0 or 3x – 4 < 11
Solution:
-2x > 0 ⇒ 2x < 0 ⇒ x < 0
x ∈ (-∞, 0)
3x – 4 < 11
⇒ 3x – 4 + 4 < 11 + 4

11th Maths Exercise 2.3 Answers Question 2.
Solve 23x < 100 when
(i) x is a natural number,
(ii) x is an integer.
Solution:
23x <100

(i.e.,) x > 4.3
(i) x = 1, 2, 3, 4 (x ∈ N)
(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4 (x ∈ Z)

11th Maths Chapter 2 Solution Samacheer Kalvi Question 3.
Solve -2x ≥ 9 when
(i) x is a real number,
(ii) x is an integer,
(iii) x is a natural number.
Solution:
-2x > 9 ⇒ 2x ≤ -9

(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4
(iii) x = 1, 2, 3, 4

Samacheer Kalvi 11th Maths Solution Question 4.

Solution:

(ii)

Solution:

Samacheer Kalvi 11th Maths Example Sums Question 5.
To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored 84, 87, 95, 91 in first four subjects, what is the minimum mark one scored in the fifth subject to get A grade in the course?
Solution:
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87+95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93

Samacheer Kalvi 11th Maths Solutions Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Solution:
12% solution of acid in 600 l ⇒ 600 × \(\frac{12}{100}\) = 72 l of acid
15% of 600 l ⇒ 600 × \(\frac{15}{100}\) = 90 l
18% of 600 l ⇒ 600 × \(\frac{18}{100}\) = 108 l
Let x litres of 18% acid solution be added

(600 + x)15 ≥ 7200 + 30x
9000+ 15x ≥ 7200 + 30x
1800 ≥ 15x
x ≤ 120
Let x litres of 18% acid solution be added

10800 + 18 ≤ 7200 + 30x
3600 ≤ 12x
x > 300
The solution is 120 ≤ x > 300

Samacheer Kalvi Class 11 Maths Solutions Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.
Solution:
Let the two numbers be x and x + 2
x + x + 2 < 40
⇒ 2x < 38

⇒ x< 19 and x > 10
so x = 11 ⇒ x + 2 = 13
x = 13 ⇒ x + 2 = 15
x = 15 ⇒ x + 2 = 17
When x = 17 ⇒ x + 2 = 19
So the possible pairs are (11, 13), (13, 15), (15, 17), (17, 19)

Samacheer Kalvi 11 Maths Solutions Question 8.
A model rocket is launched from the ground. The height h of the rocket after t seconds from lift off is given by h(t) = -5t² + 100t; 0 ≤ r ≤ 20. At what time the rocket is 495 feet above the ground?
Solution:
h(t) = -5t² + 1oot
at t = 0, h(0) = 0
at t = 1, h(1) = -5 + 100 = 95
at t = 2, h(2) = -20 + 200 = 180
at t =3, h(3) = -45 + 300 = 255
at t = 4, h(4) = -80 + 400 = 320
at t = 5, h(5) = -125 + 500 = 375
at t = 6, h(6) = – 180 + 600 = 420
at t = 7, h(7) = -245 + 700 = 455
at t = 8, h(8) = – 320 + 800 = 480
at t = 9, h(9) = -405 + 900 = 495
So, at 9 secs, the rocket is 495 feet above the ground.

Samacheer Kalvi 11th Maths Answers Question 9.
A Plumber can be paid according to the following schemes: In the first scheme he will be paid Rs. 500 plus Rs.70 per hour, and in the second scheme he will be paid Rs. 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
Solution:
I scheme with x hr
500 + (x- 1) 70 = 500 + 70x – 70
= 430 + 70x
II scheme with x hours
120x
Here I > II
⇒ 430 + 70x > 120x
⇒ 120x – 70x < 430
50x < 430
\(\frac{50 x}{50}<\frac{430}{50}\)
x < 8.6 (i.e.) when x is less than 9 hrs the first scheme gives better wages.

Samacheer Kalvi Guru 11th Maths Question 10.
A and B are working on similar jobs but their annual salaries differ by more than Rs 6000. If B earns Rs. 27000 per month, then what are the possibilities of A’s salary per month?
Solution:
A’s monthly salary = ₹ x
B’s monthly salary = ₹ 27000
Their annual salaries differ by ₹ 6000
A’s salary – 27000 > 6000
A’s salary > ₹ 33000
B’s salary – A’s salary > 6000
27000 – A’s salary > 6000
A’s salary < ₹ 21000
A’s monthly salary will be lesser than ₹ 21,000 or greater than ₹ 33,000

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3 Additional Questions

11th Maths Solutions Samacheer Kalvi Question 1.

Solution:

Multiplying both sides by 30, we get 15x ≥ 2(4x – 1) ⇒ 15x ≥ 8x – 2⇒ 15x – 8x ≥ -2

11th Maths Samacheer Kalvi Solutions Question 2.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x be the marks obtained by Ravi in the third test.

⇒ 145 + x ≥ 180 ⇒ x >180 – 145
⇒ x ≥ 35
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note. A minimum of 35 marks.
⇒ Marks greater than or equal to 35.

Samacheerkalvi.Guru 11th Maths Question 3.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get Grade ‘A’ in the course.
Solution:
Let x be the marks obtained by Sunita in the fifth examination. Then,

⇒ 368 + x ≥ 450 ⇒ x ≥ 450 – 368
⇒ x ≥ 82
Thus, Sunita must obtain marks greater than or equal to 82,
i. e., a minimum of 82 marks.

Samacheer Kalvi Guru 11th Maths Solution Question 4.
Find the pairs of ceonsecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2 .
According to the given conditions.
x < 10, x + 2 < 10 and x + (x + 2) > 11
⇒ x < 10, x < 8

From (1) and (2), we get 9
\(\frac{9}{2}\) < x < 8
Also, x is an odd positive integer. x can take values 5 and 7.
So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)

Samacheer Kalvi 11th Maths Question 5.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let x be the smaller of the two consecutive even positive integers, then the other is x + 2. According to the given conditions.

Also, x is an even positive integer.
x can take the values 6, 8 and 10.
So, the required possible pairs will be (x, x + 2) = (6, 8), (8, 10), (10, 12)

Maths Solution Class 11 Samacheer Kalvi Question 6.
Forensic Scientists use h = 61.4 + 2.3F to predict the height h in centimetres for a female whose thigh bone (femur) measures F cm. If the height of the female lies between 160 to 170 cm find the range of values for the length of the thigh bone?
Solution:
Given h = 61.4 + 2.3 F
Given h = 160 ⇒ 160 = 61.4 + 2.3 F
⇒ 2.3 F = 160 – 61.4 = 98.6

Given h = 170 ⇒ 170 = 61.4 + 2.3 F
⇒ 170 – 61.4 = 2.3 F
2.3F = 108.6

So the ranges of values are 42.87 < x < 47.23

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

11th Maths Exercise 4.4 Question 1.
By the principle of mathematical induction, prove that, for n ≥ 1

Solution:

∴ P(k+ 1) is true.
Thus P(K) is true ⇒ (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

11th Maths Chapter 4 Exercise 4.4 Question 2.
By the principle of mathematical induction, prove that, for n > 1

Solution:

∴ P(1) is true
Let P(n) be true for n = k

∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(k) is true for all n ∈ N.

Exercise 4.4 Class 11 Question 3.
Prove that the sum of the first n non-zero even numbers is n² + n.
Solution:
Let P(n) : 2 + 4 + 6 +…+2n = n² + n, ∀ n ∈ N

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

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

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

Ex 4.4 Class 11 Question 4.
By the principle of Mathematical induction, prove that, for n ≥ 1.

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n ∈ N

Exercise 4.4 Maths Class 11 Solutions Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.

Exercise 4.4 Maths Class 11 Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.

Ex 4.4 Class 11 Maths Question 7.
Using the Mathematical induction, show that for any natural number n

Solution:


∴ P(k + 1) is true
Thus p(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction,
p(n) is true for all n ∈ z

4.4 Maths Class 11 Question 8.
Using the Mathematical induction, show that for any natural number n,

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

Exercise 4.4 Class 11 Pdf Question 9.
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1
Solution:
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t_{(k + 1)}
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! [1 + k + 1] – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
= (k + 1 + 1)! – 1
∴ P(k + 1) is true ⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.

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

Chapter 4 Class 11 Maths Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1,

Solution:

Chapter 4 Maths Class 11 Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Solution:
Let P(n) : n³ – 7n + 3
Step 1:
P(1) = (1)³ – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).

Step 2:
P(k) : k³ – 7k + 3 = 3λ. Let it be true
⇒ k³ = 3λ + 7k – 3

Step 3:
P(k + 1) = (k + 1)³ – 7(k + 1) + 3
= k³ + 1 + 3k² + 3k – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= (3λ + 7k – 3) + 3k² – 4k – 3 (from Step 2)
= 3k² + 3k + 3λ – 6
= 3(k² + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).

10th Maths Exercise 4.4 11th Sum Question 13.
Use induction to prove that 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.
Solution:
P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9 = 5² + 24 – 9
= 25 + 24 – 9 = 40 ÷ by 20
So P(1) is true
Assume that the given statement is true for n = k
(i.e) 5^{k + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9
= 25 + 24 – 9
So P(1) is true
To prove P(k + 1) is true
P(k + 1) = 5^{k + 1 + 1} + 4 × 6^{k + 1 + 1} – 9
= 5 × 5 ^{k + 1} + 4 × 6 × 6^k – 9
= 5[20C + 9 – 4 × 6^k] + 24 × 6^k – 9 [from(1)]
= 100C + 45 – 206^k + 246^k – 9
= 100C + 46^k + 36
= 100C + 4(9 + 6^k)
Now for k = 1 ⇒ 4(9 + 6^k) = 4(9 + 6)
= 4 × 15 = 60 ÷ by 20 .
for k = 2 = 4(9 + 6²) = 4 × 45 = 180 ÷ 20
So by the principle of mathematical induction 4(9 + 6^k) is ÷ by 20
Now 100C is ÷ by 20.
So 100C + 4(9 + 6^k) is ÷ by 20
⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.

Samacheer Kalvi 11th Maths Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5, is divisible by 9, for all natural numbers n.
Solution:
P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9
P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5
= 10 + 48 + 5 = 63 ÷ by 9
So P(1) is true. Assume that P(k) is true
(i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9
(i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer)
⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ……(1)
To prove P(k + 1) is true.
Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5
= 10 × 10^k + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5
= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5
= 90C – 45 – 18 × 4^{k + 2}
= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9
So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

Question 15.
Prove that using the Mathematical induction

Solution:

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

Question 1.
Prove by induction the inequality (1 + x)ⁿ ≥ 1 + nx, whenever x is positive and n is a positive integer.
Solution:
P(n) : (1 +x)ⁿ ≥ 1 +nx
P(1): (1 + x)¹ ≥ 1 + x
⇒ 1 + x ≥ 1 + x, which is true.
Hence, P(1) is true.
Let P(k) be true
(i.e.) (1 + x)^k ≥ 1 + kx
We have to prove that P(k + 1) is true.
(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x
Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true]
Multiplying both sides by (1 + x), we get
(1 + x)^k(1 + x) ≥ (1 + kx)(1 + x)
⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx²
⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ….. (1)
Now, 1 + (k + 1) x + kx² ≥ 1 + (k + 1)x …… (2)
[∵ kx² > 0]
From (1) and (2), we get
(1 + x)^{k + 1} ≥ 1 + (k + 1)x
∴ P(k + 1) is true if P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values, of n.

Question 2.
3²ⁿ – 1 is divisible by 8.
Solution:
P(n) = 3²ⁿ – 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1 – 1 = 9 – 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k) = 3^2k – 1 is divisible by 8 ….. (1)
Now, P(k + 1) = 3^{(2k + 2)} – 1 = 3^2k.3² – 1
= 3²(3^2k – 1) + 8
Now, 3^2k – 1 is divisible by 9. [Using (1)]
∴ 3² (3^2k – 1) + 8 is also divisible by 8.
Hence, 3²ⁿ – 1 is divisible by 8 ∀ n E N

Question 3.
Prove by the principle of mathematical induction if x and y are any two distinct integers, then xⁿ – yⁿ is divisible by x – y. [OR]
xⁿ – yⁿ is divisible by x – y, where x – y ≠ 0.
Solution:
Let the given statement be P(n).
(i.e.) P(n): xⁿ – yⁿ = M(x – y), x – y ≠ 0

Step I.
When n = 1,
xⁿ – yⁿ = x – y = M(x – y) ….(1)
⇒ P(1) is true.

Step II.
Assume that P(k) is true.
(i.e.) x^k – y^k = M(x – y), x – y ≠ 0
We shall now show that P(k + 1) is true
Now, x^{k + 1} – y^{k + 1} = x^{k + 1} – x^ky + x^{k + 1}y – y^{k + 1}
= x^k(x – y) + y(x^k – y^k)
= x^k(x – y) + yM(x – y) [Usng ….. (1)]
= (x – y)(x^k – yM)
∴ By the principle of mathematical induction, P(n) is true for all n ∈ N

Question 4.
Prove by the principle of mathematical induction that for every natural number n, 3^{2n + 2} – 8n – 9 is divisible by 8.
Solution:
Let P(n): 3^{2n + 2} – 8n – 9 is divisible by 8.
Then, P(1): 3^{2.1 + 2} – 8.1 – 9 is divisible by 8.
(i.e.) 3⁴ – 8 – 9 is divisible by 8 or 81 – 8 – 9 is divisible by 8
(or) 64 is divisible by 8, which is true.
Suppose P(k) is true, then
P(k) : 3^{2k + 2} – 8k – 9 is divisible by 8
(i.e.) 3^{2k + 2} – 8k – 9 = 8m, where m ∈ N (or)
3^{2k + 2} = 8m + 8k + 9
P(k + 1) is the statement given by, …(1)
P(k + 1) : 3^{2(k + 1) + 2} – 8(k + 1) – 9

∴ P(k + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 5.
Use the principle of mathematical induction to prove that for every natural number n.

Solution:
Let P(n) be the given statement, i.e.

⇒ P(1) is true.
We note than P(n) is true for n = 1.
Assume that P(k) is true

Now, we shall prove that P(k + 1) is true whenever P(k) is true. We have,

∴ P(k + 1) is also true whenever P(k) is true
Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.

Question 6.
n³ – n is divisible by 6, for each natural number n ≥ 2.
Solution:
Let P(n) : n³ – n

Step 1 :
P(2): 2³ – 2 = 6 which is divisible by 6. So it is true for P(2).

Step 2 :
P(A): k³ – k = 6λ. Let it is be true for k ≥ 2
⇒ k³ = 6λ + k …(i)

Step 3 :
P(k + 1) = (k + 1)³ – (k + 1)
= k³ + 1 + 3k² + 3k – k – 1 = k³ – k + 3(k² + k)
= k³ – k + 3(k² + k) = 6λ + k – k + 3(k² + k)
= 6λ + 3(k² + k) [from (i)]
We know that 3(k² + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.

Question 7.
For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Solution:
Let P(n) : 7ⁿ – 2ⁿ

Step 1 :
P(1) : 7¹ – 2¹ = 5λ which is divisible by 5. So it is true for P(1).

Step 2 :
P(k): 7^k – 2^k = 5λ. Let it be true for P(k)

Step 3 :
P(k + 1) = 7^{k + 1} – 2^{k + 1}

So, it is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Question 8.
n² < 2ⁿ, for all natural numbers n ≥ 5.
Solution:
Let P(n) : n² < 2ⁿ for all natural numbers, n ≥ 5

Step 1 :
P(5) : 1⁵ < 2⁵ ⇒ 1 < 32 which is true for P(5)

Step 2 :
P(k): k² < 2k. Let it be true for k ∈ N

Step 3 :
P(k + 1): (k + 1)² < 2^{k + 1}
From Step 2, we get k² < 2^k
⇒ k² < 2^{k + 1} < 2k + 2k + 1


From eqn. (i) and (ii), we get (k + 1)² < 2^{k + 1}
Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.

Question 9.
In 2n < (n + 2)! for all natural number n.
Solution:
Let P(n) : 2n < (n + 2)! for all k ∈ N.

Hence, P(k + 1) is true whenever P(k) is true.

Question 10.
1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N.
Solution:
Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N

Step 1:
P(1) : 1 = 1(2.1 – 1) = 1 which is true for P(1)

Step 2:
P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1). Let it be true.

Step 3:
P(k + 1) : 1 + 5 + 9 + … + (4k – 3) = k(4k + 1)
= k(2k – 1) + (4k + 1) = 2k² – k + 4k + 1
= 2k² + 3k + 1 = 2k² + 2k + k + 1
= 2k(k + 1) + 1 (k + 1) = (2k + 1)(k + 1)
= (k+ 1) (2k + 2 – 1) = (k + 1) [2(k + 1) – 1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 5 Motion of System of Particles and Rigid Bodies

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Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Textual Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

11th Physics Chapter 5 Book Back Answers Question 1.
The center of mass of a system of particles does not depend upon,
(a) position of particles
(b) relative distance between particles
(c) masses of particles
(d) force acting on particle
Answer:
(d) force acting on particle

11th Physics 5th Chapter Book Back Answers Question 2.
A couple produces, [AIPMT 1997, AIEEE 2004]
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion [AIPMT 1997]
Answer:
(a) pure rotation

11th Physics Lesson 5 Book Back Answers Question 3.
A particle is moving with a constant velocity along a line parallel to positive X – axis. The magnitude of its angular momentum with respect to the origin is –
(a) zero
(b) increasing with x
(c) decreasing with x
(d) remaining constant [IIT 2002]
Answer:
(d) remaining constant

Samacheer Kalvi 11th Physics Question 4.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 m s^-2
(d) 25 m s^-2
[NEET 2017]
Answer:
(b) 25 rad s^-2

11th Physics Samacheer Kalvi Question 5.
A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia,
(a) increases
(b) decreases
(c) remains constant
(d) depends on direction of rotation. [IIT 1998]
Answer:
(a) increases

Samacheer Kalvi Guru 11th Physics Question 6.
A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes,
(a) L
(b) L / 2
(c) 2 L
(d) L / 2 [AFMC 1998, AIPMT 2015]
Answer:
(d) L / 2

Samacheer Kalvi Physics 11th Question 7.
A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about –
(a) the center point of the circle.
(b) the point on the circumference of the circle
(c) any point inside the circle.
(d) any point outside the circle. [IIT 2003]
Answer:
(a) the center point of the circle.

Samacheer Kalvi 11th Physics Solution Book Question 8.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along –
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) tangent to the path [AIPMT 2012]
Answer:
(a) a line perpendicular to the plane of rotation

Samacheerkalvi.Guru 11th Physics Question 9.
Two discs of same moment of inertia rotating about their regular axis passing through center and perpendicular to the plane of disc with angular velocities ω₁ and ω₁. They are brought in to contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is-
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²
(b) I(ω₁ – ω₂)ω₂
(c) \(\frac {1}{8}\) I(ω₁ – ω₂)ω²
(d) \(\frac {1}{2}\) I(ω₁ – ω₂)ω²
Answer:
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²

11 Physics Samacheer Solutions Question 10.
A disc of moment of inertia I_a is rotating in a horizontal plane about its symmetry axis with a constant angular speed to. Another disc initially at rest of moment of inertia I_b is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is-

Answer:

Samacheer Kalvi Class 11 Physics Solutions Question 11.
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle 0 without slipping and slipping down the incline without rolling is –
(a) 5 : 7
(b) 2 : 3
(c) 2 : 5
(d) 7 : 5
[AIPMT 2014]
Answer:
(a) 5 : 7

Samacheer Kalvi Guru 11 Physics Question 12.
From a disc of radius R a mass M, a circular hole of diameter R, whose rim passes through the center is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through it?
(a) 15MR²/32
(b) 13MR²/32
(c) 11MR²/32
(d) 9MR²/32 [NEET 2016]
Answer:
(b) 13MR²/32

Samacheer Kalvi 11 Physics Question 13.
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is,
(a) \(\sqrt{\frac{4}{3} g h}\)
(b) \(\sqrt{\frac{10}{7} g h}\)
(c) \(\sqrt{2gh}\)
(d) \(\sqrt{\frac{1}{2} g h}\)
Answer:
(a) \(\sqrt{\frac{4}{3} g h}\)

Physics Class 11 Samacheer Kalvi Question 14.
The speed of the center of a wheel rolling on a horizontal surface is vQ. A point on the rim in level with the center will be moving at a speed of speed of,
(a) zero
(b) v₀
(c) \(\sqrt{2}\)v₀
(d) 2 v₀
[PMT 1992, PMT 2003, IIT 2004]
Answer:
(c) \(\sqrt{2}\)v₀

Samacheer Kalvi 11th Physics Solution Question 15.
A round object of mass m and radius r rolls down without slipping along an inclined plane. The fractional force,
(a) dissipates kinetic energy as heat.
(b) decreases the rotational motion.
(c) decreases the rotational and transnational motion ,
(d) converts transnational energy into rotational energy [PMT 2005]
Answer:
(d) converts transnational energy into rotational energy

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions

Samacheer Kalvi.Guru 11th Physics Question 1.
Define center of mass.
Answer:
The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

11th Physics Samacheer Kalvi Solution Question 2.
Find out the center of mass for the given geometrical structures.
(a) Equilateral triangle
(b) Cylinder
(c) Square
Answer:

(a) For equilateral triangle, center of mass lies at its centro-id.
(b) For cylinder, center of mass lies at its geometrical center.
(c) For square, center of mass lies at the point where the diagonals meet.

11 Samacheer Physics Solutions Question 3.
Define torque and mention its unit.
Answer:
Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is,
\(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\)

11th Physics Samacheer Solutions Question 4.
What are the conditions in which force cannot produce torque?
Answer:
The forces intersect (or) passing through the axis of rotation cannot produce torque as the perpendicular distance between the forces is 0 i.e. r = 0.
∴ \(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\) = 0

Question 5.
Give any two examples of torque in day – to – day life.
Answer:

• The opening and closing of a door about the hinges.
• Turning of a nut using a wrench.

Question 6.
What is the relation between torque and angular momentum?
Answer:
We have the expression for magnitude of angular momentum of a rigid body as, L = I ω. The expression for magnitude of torque on a rigid body is, τ = I α.
We can further write the expression for torque as,
τ = I\(\frac {dω}{dt}\) (∴ α = \(\frac {dω}{dt}\))
Where, ω is angular velocity and α is angular acceleration. We can also write equation,
τ = \(\frac {d(Iω)}{dt}\)
τ = \(\frac {dL}{dt}\)

Question 7.
What is equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium where both its linear momentum and angular momentum remain constant.

Question 8.
How do you distinguish between stable and unstable equilibrium?
Answer:
Stable Kquilibrium:

• The body tries to come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly higher if disturbed from equilibrium.
• Potential energy of the body is minimum and it increases if disturbed.

Unstable Equilibrium:

• The body cannot come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly lower if disturbed from equilibrium.
• Potential energy of the body is not minimum and it decreases if disturbed.

Question 9.
Define couple.
Answer:
A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

Question 10.
State principle of moments.
Answer:
Principle of moment states that when an object is in equilibrium the sum of the anticlockwise moments about a point is equal to the sum of the clockwise moments.

Question 11.
Define center of gravity.
Answer:
The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

Question 12.
Mention any two physical significance of moment of inertia
Answer:
Moment of inertia for point mass,
I = \(m_{i} r_{i}^{2}\)
Moment of inertia for bulk object,
I = ∑\(m_{i} r_{i}^{2}\)

Question 13.
What is radius of gyration?
Answer:
The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

Question 14.
State conservation of angular momentum.
Answer:
The law of conservation of angular momentum states that when no external torque acts on the body the net angular momentum of a rotating rigid body remains constant.

Question 15.
What are the rotational equivalents for the physical quantities, (i) mass and (ii) force?
Answer:
The rotational equivalents for (i) mass and (ii) force are moment of inertia and torque respectively.

Question 16.
What is the condition for pure rolling?
Answer:
In pure rolling, there is no relative motion of the point of contact with the surface when the rolling object speeds up or shows down. It must accelerate or decelerate respectively.

Question 17.
What is the difference between sliding and slipping?
Sliding:

• Velocity of center of mass is greater than Rω i.e. V_CM > Rω.
• Velocity of transnational motion is greater than velocity of rotational motion.
• Resultant velocity acts in the forward direction.

Slipping:

• Velocity of center of mass is lesser than Rω. i.e. V_CM < Rω
• Velocity of translation motion is lesser than velocity of rotational motion.
• Resultant velocity acts in the backward direction.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions

Question 1.
Explain the types of equilibrium with suitable examples.
Answer:

• Transnational motion – A book resting on a table.
• Rotational equilibrium – A body moves in a circular path with constant velocity.
• Static equilibrium – A wall – hanging, hanging on the wall.
• Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
• Stable equilibrium – A table on the floor
• Unstable equilibrium – A pencil standing on its tip.
• Neutral equilibrium – A dice rolling on a game board.

Question 2.
Explain the method to find the center of gravity of a irregularly shaped lamina.
Answer:
There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R as shown in figure, the vertical lines I PP’, QQ’, RR’ all pass through the center of gravity. Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.

Determination of center of gravity of plane lamina by suspending

Question 3.
Explain why a cyclist bends while negotiating a curve road? Arrive at the expression for angle of bending for a given velocity.
Answer:
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X – axis and the vertical line through O as Z – axis as shown in Figure.

The system as a frame is rotating about Z – axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be \(\frac{m v^{2}}{r}\) This force will act through the center of gravity. The forces acting on the system are,

• gravitational force (mg)
• normal force (N)
• frictional force (f)
• centrifugal force (\(\frac{m v^{2}}{r}\)).

As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure.
For rotational equilibrium,
τ_net = 0
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is I BC which causes an (\(\frac{m v^{2}}{r}\) BC) Which causes an anticlockwise turn that is taken as positive.


While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle 0 from vertical given by the above expression to stay in equilibrium (i.e. to avoid a fall).

Question 4.
Derive the expression for moment of inertia of a rod about its center and perpendicular to the rod.
Answer:
Let us consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is, dI = (dm) x²

As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, λ = \(\frac {M}{l}\)
The (dm) mass of the infinitesimally small length as, dm = λ dx = \(\frac {M}{l}\) dx
The moment of inertia (I) of the entire rod can be found by integrating dl,

As the mass is distributed on either side of the origin, the limits for integration are taken from to – l/2 to l/2.

Question 5.
Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.
Answer:
Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in figure.
The moment of inertia (dl) of this small mass (dm) is,
dI = (dm)R²
The length of the ring is its circumference (2πR). As the mass is uniformly distributed, the mass per unit length (λ) is,
λ = \(\frac {mass}{lengh}\) = \(\frac {M}{2πR}\)
The mass (dm) of the infinitesimally small length is,
dm = λ dx = \(\frac {M}{2πR}\) dx
Now, the moment of inertia (I) of the entire ring is,
To cover the entire length of the ring, the limits of integration are taken from 0 to 2πR.

Question 6.
Derive the expression for moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.
Answer:
Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,
dI = (dm)R²
As the mass is uniformly distributed, the mass per unit area (σ) is σ = \(\frac {mass}{area}\) = \(\frac{M}{\pi R^{2}}\)
The mass of the infinitesimally small ring is,
dm = σ 2πr dr = \(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\) 2πr dr
where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness), dm = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r dr
dI = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r³ dr
The moment of inertia (I) of the entire disc is,

Question 7.
Discuss conservation of angular momentum with example.
Answer:
When no external torque acts on the body, the net angular momentum of a rotating rigid body remains constant. This is known as law of conservation of angular momentum.
τ = \(\frac {dL}{dt}\)
If τ = 0 then, L = constant.
As the angular momentum is L = Iω, the conservation of angular momentum could further be written for initial and final situations as,
I_iω_i = I_iω_i (or) Iω = constant
The above equations say that if I increases ω will decrease and vice – versa to keep the angular momentum constant.

There are several situations where the principle of conservation of angular momentum is applicable. One striking example is an ice dancer as shown in Figure A. The dancer spins slowly when the hands are stretched out and spins faster when the hands are brought close to the body.

Stretching of hands away from body increases moment of inertia, thus the angular velocity decreases resulting in slower spin. When the hands are brought close to the body, the moment of inertia decreases, and thus the angular velocity increases resulting in faster spin. A diver while in air as in Figure B curls the body close to decrease the moment of inertia, which in turn helps to increase the number of somersaults in air.

Question 8.
State and prove parallel axis theorem.
Answer:
Parallel axis theorem:
Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is – given by the relation,
I = I_C + M d²
Let us consider a rigid body as shown in figure. Its moment of inertia about an axis AB passing through the center of mass is I_C. DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C. For this, let us consider a point mass m on the body at position x from its center of mass.

The moment of inertia of the point mass about the axis DE is, m (x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
I = ∑ m (x + d)²
This equation could further be written as,
I = ∑ m(x² + d² + 2xd)
1= ∑ (mx² + md² + 2 dmx)
l = ∑ mx² + md² + 2d ∑ mx
Here, ∑ mx² is the moment of inertia of the body about the center of mass. Hence,I_C = ∑ mx²
The term, ∑ mx = 0 because, x can take positive and negative values with respect to the axis AB. The summation (∑ mx) will be zero.
Thus, I = I_C + ∑ m d² = I_C + (∑m) d²
Here, ∑ m is the entire mass M of the object (∑ m = M).
I = I_C + Md²

Question 9.
State and prove perpendicular axis theorem.
Answer:
Perpendicular axis theorem:
This perpendicular axis theorem holds good only for plane laminar objects. The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y – axes lie in the plane and Z – axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are I_X and I_Y respectively – and I_Z is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_Z = I_X + I_Y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y – axes lie on the plane and Z – axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z – axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z – axis as, I_Z = ∑ mr²
Here, r² = x² + y²
Then, I_Z = ∑ m (x² + y²)
I_Z = ∑ m x² + ∑ m y²
In the above expression, the term ∑ m x² is the moment of inertia of the body about the Y-axis and similarly the term ∑ m y²is the moment of inertia about X- axis. Thus,
I_X = ∑ m y² and I_Y = ∑ m x²
Substituting in the equation for Iz gives,
I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

Question 10.
Discuss rolling on inclined plane and arrive at the expression for the acceleration.
Answer:
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

For transnational motion, mg sin θ is the supporting force and f is the opposing force, mg sin θ f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sin 0 cannot cause torque as it passes through it but the frictional force f can set torque of Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK² we get,
Rf = mK² \(\frac {a}{R}\); f = ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
Now equation becomes,
mg sin θ – ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = ma
mg sin θ = ma + ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
a \(\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = g sin θ
After rewriting it for acceleration, we get,
a = \(\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\)
We can also find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane.
v² = u² + 2as. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = \(\frac {h}{sin θ}\)
By taking square root,

The time taken for rolling down the incline could also be written from first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,
t = \(\frac {v}{a}\)

The equation suggests that for a given incline, the object with the least value of radius of gyration K will reach the bottom of the incline first.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Conceptual Questions

Question 1.
When a tree is cut, the cut is made on the side facing the direction in which the tree is required to fall. Why?
Answer:
A cut on the tree is made on the side facing the direction in which the tree is required to fall because that side will no longer be supported by the normal force from the bottom, therefore the gravitational force tries to rotate it. So the torque given by the gravity to the tree makes the tree fall on the side as anticipated.

Question 2.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his center of gravity will go away from the body. It affects the balance, to avoid this he bends. By which center of gravity will realign within the body again. So balance is maintained.

Question 3.
Why is it much easier to balance a meter scale on your finger tip than balancing on a match stick?
Answer:
A meter scale is larger then a match stick. So the center of gravity for meter scale is higher than a matchstick when we keep it vertically. It is easier to balance the object whose center of gravity is higher than the object whose centro of gravity is lower. So, it is hard to balance a match stick than a meter scale.

Question 4.
Two identical water bottles one empty and the other filled with water are allowed to roll down an inclined plane. Which one of them reaches the bottom first? Explain your answer.
Answer:
Mass of the empty water bottle mostly concentrated on its surface. So moment of inertia of empty water bottle is more than the bottle filled with water. As we know, moment of inertia is inversely proportional to angular velocity. Therefore, the bottle filled with water whirls with greater speed and reaches the ground first.

Question 5.
Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of same angular momentum, compare the moment of inertia using the graph.
Answer:
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac { 1 }{ 2 }\) Iω².
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

It resembles with y = Kx². If angular momentum is same for two objects, kinetic energy is inversely proportional to moment of inertia.
Moment of inertia of the object whose kinetic energy is lesser will have greater magnitude.

Question 6.
Three identical solid spheres move down through three inclined planes A, B and C all same dimensions. A is without friction, B is undergoing pure rolling and C is rolling with slipping. Compare the kinetic energies E_A, E_B and E_C at the bottom.
Answer:
Even though, the three identical solid spheres of same dimensions move down through three different inclined plane, according to the law of conservation of energy, the potential energy possessed by these three solid spheres will be converted into kinetic energies. So the kinetic energies E_A, E_B and E_C are equal at the bottom.

Question 7.
Give an example to show that the following statement is false. Any two forces acting on a body can be combined into single force that would have same effect.
Answer:
A single force i.e. resultant of two forces acting on a body depends upon the angle between them also. The simple example for this is if two forces 5 N and 5 N acting on the object in the opposite direction, the single resultant force acting on the body is zero. But, if two forces acting on the object along the same direction, then the resultant i.e. the single force is 5 + 5 = 10 N. Hence the given statement “any two forces acting on a body can be combined into single force that would leave same effect” is wrong.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numerical Problems

Question 1.
A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total energy of the disc when rolling along a horizontal table with a velocity of 20 cm s^-2.
Answer:
Given,
Mass of the disc = 100 g = 100 x 10^-3 kg = \(\frac { 1 }{ 10 }\)kg
Velocity of disc = 20 cm s^-1 = 20 x 10^-2 ms^-1 = 0.2 ms^-1

Question 2.
A particle of mass 5 units is moving with a uniform speed of v = \(3 \sqrt{2}\) units in the XOY plane along the line y = x + 4. Find the magnitude of angular momentum.
Answer:
Given,
Mass = 5 units
Speed = v = \(3 \sqrt{2}\) units
Y = X + 4
Angular momentum = L = m(\(\bar{r} \times \bar{v}\))
= m(x\(\hat{i}\) +y\(\hat{j}\))x(v\(\hat{i}\) + v\(\hat{j}\)) = m[xv\(\hat{k}\)-vy\(\hat{k}\)] = m[xv\(\hat{k}\)– v(x + 4)\(\hat{k}\)]
L = -mv\(\hat{k}\) = -4 x 5 x \(3 \sqrt{2}\)\(\hat{k}\) = – 60\(\sqrt{2}\)\(\hat{k}\)
L = 60\(\sqrt{2}\) units.

Question 3.
A fly wheel rotates with a uniform angular acceleration. If its angular velocity increases from 20π rad/s to 40π rad/s in 10 seconds, find the number of rotations in that period.
Answer:
Given,
Initial angular velocity ω₀ = 20 π rad/s
Final angular velocity ω = 40 π rad/s
Time t = 10 s
Solution:
Angular acceleration α = \(\frac{\omega-\omega_{0}}{t}\) = \(\frac {40π – 20π }{ 10 }\)
α = 2π rad/s²
According to equation of motion for rotational motion

The number of rotations = n = \(\frac {θ}{ 2π }\)
n = \(\frac {300π}{ 2π }\) = 150 rotations.

Question 4.
A uniform rod of mass m and length / makes a constant angle 0 with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this gravity is.
Answer:
Moment of inertia of the rod about the axis which is passing through its center of gravity is


Moment of inertia of a uniform rod of mass m and length l about one axis which passes through one end of the rod

Question 5.
Two particles P and Q of mass 1 kg and 3 kg respectively start moving towards each other from rest under mutual attraction. What is the velocity of their center of mass?
Answer:
Given,
Mass of particle P = 1 kg Mass of particle Q = 3 kg
Solution:
Particles P and Q forms a system. Here no external force is acting on the system,

We know that M = \(\frac {d}{dt}\) (V_CM ) = f
It means that, C.M. of an isolated system remains at rest when no external force is acting and internal forces do not change its center of mass.

Question 6.
Find the moment of inertia of a hydrogen molecule about an axis passing through its center of mass and perpendicular to the inter-atomic axis.
Given: mass of hydrogen atom 1.7 x 10²⁷kg and inter atomic distance is equal to 4 x 10^-10m.
Answer:
Given,
Inter-atomic distance : 4 x 10^-10 m
Mass of H₂ atom : 1.7 x 10^-27 kg
Moment of inertia of H₂ =

Question 7.
On the edge of a wall, we build a brick tower that only holds because of the bricks’ own weight. Our goal is to build a stable tower whose overhang d is greater than the length l of a single brick. What is the minimum number of bricks you need?
(Hint: Find the center of mass for each brick and add.)
Answer:
Given:

Length of the brick = l
Length of the overhang = d
The mono of bricks can be decided only by using the concept of position of center of gravity. The first brick is in contact with the ground and it will not fall over.
Let one end of brick 2 is coinciding with the center of brick 1 i.e. x = 0.
∴ The position of n brick is
x_n = (n – 1) \(\frac {L}{4}\)
The center of gravity is in the midway between the center of brick 2 and the center of brick n.
position of G =

brick tower will fall when G >\(\frac {L}{4}\) it shows that n > 4.

Question 8.
The 747 boing plane is landing at a speed of 70 m s^-1. Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?
Answer:
The types of the plane will leave a skid mark if the speed of the types in contact with ground is lesser than the velocity of the plane. The condition for this is –
v > ω
(When the type attained an angular velocity of V/R)
The types will stop the skidding and starts the rolling.
The forces acting on the wheel after the plane touches down are,
N – P Normal force W – weight
The wheel is not accelerating means
N = ω
The torque about the center of the wheel is
τ = RF = µωR
The angular acceleration is

According to equation of motion, time taken to stop the skidding by the wheel is,

The six mark will have a length of
l = vt = 70 x 0.03 = 2.1 m
Note:
The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is 68.63 m. The normal force experienced by plane through its 16 wheels is ω = 232 KN.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Additional Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The changes produced by the deforming forces in a rigid body are –
(a) very large
(b) infinity
(c) negligibly small
(d) small
Answer:
(c) negligibly small

Question 2.
When a rigid body moves all particles that constitute the body follows-
(a) same path
(b) different paths
(c) either same or different path
(d) circular path
Answer:
(b) different path

Question 3.
For bodies of regular shape and uniform mass distribution, the center of mass is at –
(a) the comers
(b) inside the objects
(c) the point where the diagonals meet
(d) the geometric center
Answer:
(d) the geometric center

Question 4.
For square and rectangular objects center of mass lies at –
(a) the point where the diagonals meet
(b) at the comers
(c) on the center surface
(d) any point
Answer:
(a) the point where the diagonals meet

Question 5.
Center of mass may lie –
(a) within the body
(b) outside the body
(c) both (a) and (b)
(d) only at the center
Answer:
(c) both (a) and (b)

Question 6.
The dimension of point mass is –
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 7.
The motion of center of mass of a system of two particles is unaffected by their internal forces –
(a) irrespective of the actual directions of the internal forces
(b) only if they are along the line joining the particles
(c) only if acts perpendicular to each other
(d) only if acting opposite
Answer:
(a) irrespective of the actual directions of the internal forces

Question 8.
A circular plate of diameter 10 cm is kept in contact with a square plate of side 10 cm. The density of the material and the thickness are same everywhere. The center of mass of the system will be
(a) inside the circular plate
(b) inside the square plate
(c) At the point of contact
(d) outside the system
Answer:
(6) inside the square plate

Question 9.
The center of mass of a system of particles does not depend on
(a) masses of particles
(b) position of the particles
(c) distribution of masses
(d) forces acting on the particles
Answer:
(d) forces acting on the particles

Question 10.
The center of mass of a solid cone along the line from the center of the base to the vertex is at –
(a) \(\frac { 1 }{ 2 }\) th of its height
(b) \(\frac { 1 }{ 3 }\) of its height
(c) \(\frac { 1 }{ 4 }\) th of its height
(d) \(\frac { 1 }{ 5 }\) th of its height
Answer:
(d) \(\frac { 1 }{ 5 }\) th of its height

Question 11.
All the particles of a body are situated at a distance of X from origin. The distance of the center of mass from the origin is –
(a) ≥ r
(b) ≤ r
(c) = r
(d) > r

Question
A free falling body breaks into three parts of unequal masses. The center of mass of the three parts taken together shifts horizontally towards –
(a) heavier piece
(b) lighter piece
(c) does not shift horizontally
(d) depends on vertical velocity
Answer:
(c) does not shift horizontally

Question 13.
The distance between the center of carbon and oxygen atoms in the gas molecule is 1.13 A. The center of mass of the molecule relative to oxygen atom is –
(a) 0.602 Å
(b) 0.527 Å
(c) 1.13 Å
(d) 0.565 Å
Answer:
(b) 0.527 Å
Given,
Inter atomic distance = 1.13 Å
Mass of carbon atom = 14
Mass of oxygen atom = 16
Let C.M. of molecule lies at a distance of X from oxygen atom-
i.e. m₁r₁ = m₂r₂
16 X = 14(1.13 – X)
30 X = 15.82
X = 0.527 Å

Question 14.
The unit of position vector of center of mass is-
(a) kg
(b) kg m²
(c) m
(d) m²
Answer:
(c) m

Question 15.
The sum of moments of masses of all the particles in a system about the center of mass is-
(a) minimum
(b) maximum
(c) zero
(d) infinity
Answer:
(c) zero

Question 16.
The motion of center of mass depends on-
(a) external forces acting on it
(b) internal forces acting within it
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) external forces acting on it

Question 17.
Two particles P and Q move towards with each other from rest with the velocities of 10 ms^-1 and 20 ms^-1 under the mutual force of attraction. The velocity of center of mass is-
(a) 15 ms^-1
(b) 20 ms^-1
(c) 30 ms^-1
(d) zero
Answer:
(d) zero

Question 18.
The reduced mass of the system of two particles of masses 2 m and 4 m will be –
(a) 2 m
(b) \(\frac {2 }{ 3 }\)y m
(c) \(\frac {3}{ 2 }\)y m
(d) \(\frac { 4 }{ 3 }\)m
Answer:
(d) \(\frac { 4 }{ 3 }\)m

Question 19.
The motion of the center of mass of a system consists of many particles describes its –
(a) rotational motion
(b) vibratory motion
(c) oscillatory motion
(d) translator y motion
Answer:
(c) oscillatory motion

Question 20.
The position of center of mass can be written in the vector form as –

Answer:

Question 21.
The positions of two masses m₁ and m₂ are x₁ and x₂. The position of center of mass is –

Answer:

Question 22.
In a two particle system, one particle lies at origin another one lies at a distance of X. Then the position of center of mass of these particles of equal mass is –

Answer:
(a) \(\frac {X}{2}\)

Question 23.
Principle of moments is –

Answer:

Question 24.
Infinitesimal quantity means –
(a) collective particles
(b) extremely small
(c) nothing
(d) extremely larger
Answer:
(b) extremely small

Question 25.
In the absence of external forces the center of mass will be in a state of –
(a) rest
(b) uniform motion
(c) may be at rest or in uniform motion
(d) vibration
Answer:
(c) may be at rest or in uniform motion

Question 26.
The activity of the force to produce rotational motion in a body is called as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 27.
The moment of the external applied force about a point or axis of rotation is known as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 28.
Torque is given as –
(a) \(\vec{r}\) . \(\vec{F}\)
(b) \(\vec{r}\) x \(\vec{F}\)
(c) \(\vec{F}\) x \(\vec{r}\)
(d) r F cos θ
Answer:
(b) \(\vec{r}\) x \(\vec{F}\)

Question 29.
The magnitude of torque is –
(a) rF sin θ
(b) rF cos θ
(c) rF tan θ
(d) rF
Answer:
(a) rF sin θ

Question 30.
The direction of torque ácts –
(a) along \(\vec{F}\)
(b) along \(\vec{r}\) & \(\vec{F}\)
(c) Perpendicular to \(\vec{r}\)
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)
Answer:
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)

Question 31.
The unit of torque is –
(a) is
(b) Nm^-2
(c) Nm
(d) Js^-1
Answer:
(c) Nm

Question 32.
The direction of torque is found using –
(a) left hand rule
(b) right hand rule
(c) palm rule
(d) screw rule
Answer:
(b) right hand rule

Question 33.
if the direction of torque is out of the paper then the rotation produced by the torque is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 34.
If the direction of the torque is inward the paper then the rotation is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 35.
if \(\vec{r}\) and \(\vec{F}\) are parallel or anti parallel, then the torque is –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(a) zero

Question 36.
The maximum possible value of torque is –
(a) zero
(b) infinity
(c) \(\vec{r}\) + \(\vec{F}\)
(d) rF
Answer:
(d) rF

Question 37.
The relation between torque and angular acceleration is –
(a) \(\vec{τ}\) = \(\frac{1}{\alpha}\)
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)
(c) \(\vec{α}\) = I \(\vec{τ}\)
(d) \(\vec{τ}\) = \(\frac{\vec{\alpha}}{\mathrm{I}}\)
Answer:
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)

Question 38.
Angular momentum is –
(a) \(\vec{P}\) x \(\vec{r}\)
(b) \(\vec{r}\) x \(\vec{P}\)
(c) \(\overrightarrow{\frac{r}{\vec{p}}}\)
(d) \(\vec{r}\) . \(\vec{P}\)
Answer:
(b) \(\vec{r}\) x \(\vec{P}\)

Question 39.
The magnitude of angular momentum is given by –
(a) rp
(b) rp sin θ
(c) rp cos θ
(d) rp tan θ
Answer:
(b) rp sin θ

Question 40.
Angular momentum is associated with –
(a) rotational motion
(b) linear motion
(c) both (a) and (b)
(d) circular motion only
Answer:
(c) both (a) and (b)

Question 41.
Angular momentum acts perpendicular to –
(a) \(\vec{r}\)
(b) \(\vec{P}\)
(c) both \(\vec{r}\) and \(\vec{P}\)
(d) plane of the paper
Answer:
(c) both \(\vec{r}\) and \(\vec{P}\)

Question 42.
Angular momentum is given by –
(a) \(\frac {I}{ω}\)
(b) τω
(c) Iω
(d) \(\frac {ωI}{2}\)
Answer:
(c) Iω

Question 43.
The rate of change of angular momentum is –
(a) Torque
(b) angular velocity
(c) centripetal force
(d) centrifugal force
Answer:
(a) Torque

Question 44.
The forces acting on a body when it is at rest –
(a) is gravitational force
(b) Normal force
(c) both gravitational as well as normal force
(d) No force is acting
Answer:
(c) both gravitational as well as normal force

Question 45.
The net force acting on a body when it is at rest is –
(a) gravitational force
(b) Normal force
(c) Sum of gravitational and normal force
(d) zero
Answer:
(d) zero

Question 46.
If net force acting on a body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) both (a) and (b)
(d) none
Answer:
(a) transnational equilibrium

Question 47.
If the net torque acting on the body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(b) rotational equilibrium

Question 48.
when the net force and net torque acts on the body is zero then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(d) none

Question 49.
When the net force and net torque acts on the body is zero then the body is in –
(a) static equilibrium
(b) Dynamic equilibrium
(c) both (a) and (b)
(d) transnational equilibrium
Answer:
(c) both (a) and (b)

Question 50.
When two equal and opposite forces acting on the body at two different points, it may give –
(a) net force
(b) torque
(c) stable equilibrium
(d) none
Answer:
(b) torque

Question 51.
The torque in rotational motion is analogous to in transnational motion –
(a) linear momentum
(b) mass
(c) couple
(d) force
Answer:
(d) force

Question 52.
Which of the following example does not constitute a couple?
(a) steering a car
(b) turning a pen cap
(c) ball rolls on the floor
(d) closing the door
Answer:
(c) ball rolls on the floor

Questioner 53.
If the linear momentum and angular momentum are zero, then the object is said to be in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(d) all the above

Question 54.
When the body is disturbed, the potential energy remains same, then the body is in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(c) neutral equilibrium

Question 55
The point where the entire weight of the body acts is called as –
(a) center of mass
(b) center of gravity
(c) both (a) and (b)
(d) pivot
Answer:
(b) center of gravity

Question 56.
The forces acting on a cyclist negotiating a circular Level road is /are –
(a) gravitational force
(b) centrifugal force
(c) frictional force
(d) all the above
Answer:
(d) all the above

Question 57.
While negotiating a circular level road a cyclist has to bend by an angle θ from vertical to stay in an equilibrium is-
(a) \(\tan \theta=\frac{r g}{r^{2}}\)
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)
(c) θ = \(\sin ^{-1}\left(\frac{r g}{r^{2}}\right)\)
(d) zero
Answer:
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)

Question 58.
Moment of inertia for point masses –
(a) m²r
(b) rw²
(c) mr²
(d) zero
Answer:
(c) mr²

Question 59.
Moment of inertia for bulk object –
(a) rm²
(b) rw²
(c) \(m_{i} r_{i}^{2}\)
(d) \(\Sigma m_{i} r_{i}^{2}\)
Answer:
(d) \(\Sigma m_{i} r_{i}^{2}\)

Question 60.
For rotational motion, moment of inertia is a measure of –
(a) transnational inertia
(b) mass
(c) rotational inertia
(d) invariable quantity
Answer:
(c) rotational inertia

Question 61.
Unit of moment of inertia –
(a) kgm
(b) mkg^-2
(c) kgm²
(d) kgm^-1
Answer:
(c) kgm²

Question 62.
Dimensional formula for moment of inertia is –
(a) [ML^-2]
(b) [M²L^-1]
(c) [M^-2]
(d) [ML²]
Answer:
(d) [ML²]

Question 63.
Moment of inertia of a body is a –
(a) variable quantity
(b) invariable quantity
(c) constant quantity
(d) measure of torque
Answer:
(a) variable quantity

Question 64.
Moment of inertia of a thin uniform rod about an axis passing through the center of mass and perpendicular to the length is –
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(b) \(\frac { 1 }{ 12 }\)Ml²

Question 65.
Moment of inertia ofa thin uniform rod about an axis passing through one end and perpendicular to the length is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(a) \(\frac { 1 }{ 3 }\)Ml²

Question 66.
Moment of inertia of a thin uniform rectangular sheet about an axis passing through the center of mass and perpendicular to the plane of the sheet is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )

Question 67.
Moment of inertia of a thin uniform ring about an axis passing through the center of gravity and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\)MR²
(d) \(\frac { 3 }{ 2 }\)MR²
Answer:
(a) MR²

Question 68.
Moment of inertia of a thin uniform ring about an axis passing through the center and lying on the plane (along diameter) is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 69.
Moment of inertia of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 70.
Moment of inertia of a thin uniform disc about an axis passing through the center lying on the plane (along diameter is)
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(d) \(\frac { 1 }{ 4 }\) MR²

Question 71.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(a) MR²

Question 72.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)

Question 73.
Moment of inertia of a uniform solid cylinder about an axis passing through the center and along the axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 74.
Moment of inertia of a uniform solid cylinder about as axis passing perpendicular to the length and passing through the center is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)

Question 75.
Moment of inertia of a thin hollow sphere about an axis passing through the center along its diameter is
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(a) \(\frac { 2 }{ 3 }\)MR²

Question 76.
Moment of inertia of a thin hollow sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(b) \(\frac { 5 }{ 3 }\)MR²

Question 77.
torment of inertia of a uniform solid sphere about an axis passing through the center along its diameter is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(d) \(\frac { 2 }{ 5 }\)MR²

Question 78.
Moment of inertia of a uniform solid sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(c) \(\frac { 7 }{ 5 }\)MR²

Question 79.
The ratio of K²/R² of a thin uniform ring about an axis passing through the center and perpendicular to the plane is-
(a) 1
(b) 2
(c) \(\frac { 7 }{ 5 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(a) 1

Question 80.
The ratio of K²/ R² of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) 1
(b) 2
(c) \(\frac {1}{ 2 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(c) \(\frac {1}{ 2 }\)

Question 81.
When no external torque acts on the body, the net angular momentum of a rotating body.
(a) increases
(b) decreases
(c) increases or decreases
(d) remains constant
Answer:
(d) remains constant

Question 82.
Moment of inertia of a body is proportional to –
(a) ω
(b) \(\frac { 1 }{ ω }\)
(c) ω²
(d) \(\frac{1}{\omega^{2}}\)
Answer:
(b) \(\frac { 1 }{ ω }\)

Question 83.
When the hands are brought closer to the body, the angular velocity of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 84.
When the hands are stretched out from the body, the moment of inertia of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 85.
The work done by the torque is –
(a) F. ds
(b) F. dθ
(c) τ dθ
(d) r.dθ
Answer:
(c) τ dθ

Question 86.
Rotational Kinetic energy of a body is –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\) Iω²
(c) \(\frac { 1 }{ 2 }\)Iv²
(d) \(\frac { 1 }{ 2 }\)mω²
Answer:
(b) \(\frac { 1 }{ 2 }\) Iω²

Question 87.
Rotational kinetic energy is given by –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\)Iv²
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)
(d) \(\frac{2 \mathrm{I}}{\mathrm{L}^{2}}\)
Answer:
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)

Question 88.
If E is a rotational kinetic energy then angular momentum is-
(a) \(\sqrt{2 \mathrm{IE}}\)
(b) \(\frac{\mathrm{E}^{2}}{2 \mathrm{I}}\)
(c) \(\frac{2 \mathrm{I}}{\mathrm{E}^{2}}\)
(d) \(\frac{E}{I^{2} \omega^{2}}\)
Answer:
(a) \(\sqrt{2 \mathrm{IE}}\)

Question 89.
The product of torque acting on a body and angular velocity is –
(a) Energy
(b) power
(c) work done
(d) kinetic energy
Answer:
(b) power

Question 90.
The work done per unit time in rotational motion is given by –
(a) \(\vec{F}\) .v
(b) \(\frac {dθ}{dt}\)
(c) τ ω
(d) I ω
Answer:
(c) τ ω

Question 91.
While rolling, the path of center of mass of an object is –
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line

Question 92.
In pure rolling, the velocity of the point of the rolling object which comes in contact with the surface is –
(a) maximum
(b) minimum
(c) zero
(d) 2 V_CM
Answer:
(c) zero

Question 93.
In pure rolling velocity of center of mass is equal to –
(a) zero
(b) Rω
(c) \(\frac { ω }{ R }\)
(d) \(\frac { R }{ ω }\)
Answer:
(b) Rω

Question 94.
In pure rolling, rotational velocity of points at its edges is equal to-
(a) Rω
(b) velocity of center of mass
(c) transnational velocity
(d) all the above
Answer:
(a) Rω

Question 95.
Sliding of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(c) V_trans > V_rot

Question 96.
Sliding of the object occurs while –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(d) V_CM > Rω

Question 97.
Slipping of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(a) V_trans < V_rot

Question 98.
Slipping of the object occurs when –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(c) V_CM < Rω

Question 99.
In sliding, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(a) forward direction

Question 100.
In slipping, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(b) backward direction

Question 101.
When a solid sphere is undergoing pure rolling, the ratio of transnational kinetic energy to rotational kinetic – energy is –
(a) 2 : 5
(b) 5 : 2
(c) 1 : 5
(d) 5 : 1
Answer:
(b) 5 : 2

Question 102.
Time taken by the rolling object in inclined plane to reach its bottom is –

Answer:

Question 103.
The velocity of the rolling object on inclined plane at the bottom of inclined plane is –

Answer:

Question 104.
Moment of inertia of an annular disc about an axis passing through the centre and perpendicular to the plane of disc is –

Answer:

Question 105.
Moment of inertia of a cube about an axis passing through the center of mass and perpendicular to face is –
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)
(b) \(\frac {1}{3}\) Ma²
(c) \(\frac {Ma}{6}\)
(d) \(\frac{\mathrm{Ma}^{2}}{12}\)
Answer:
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)

Question 106.
Moment of inertia of a rectangular plane sheet about an axis passing through center of mass and perpendicular to side b in its plane is –
(a) \(\frac{\mathrm{Ml}^{2}}{12}\)
(b) \(\frac{\mathrm{Ma}^{2}}{12}\)
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)
(d) \(\frac{\mathrm{Ml}^{2}}{6}\)
Answer:
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)

Question 107.
Rotational kinetic energy can be calculated by using –
(a) \(\frac{1}{2}\) I ω²
(b) \(\frac{\mathrm{L}^{2}}{2I}\)
(c) \(\frac{1}{2}\) Lω
(d) all the above
Answer:
(b) \(\frac{\mathrm{L}^{2}}{2I}\) )

Question 108.
The radius of gyration of a solid sphere of radius r about a certain axis is r. The distance of that axis from the center of the sphere is –
(a) \(\frac{2}{5}\)r
(b) \(\sqrt{\frac{2}{5}}\)r
(c) \(\sqrt{0.6r}\)
(d) \(\sqrt{\frac{5}{3}}\)
Answer:
(c) \(\sqrt{0.6r}\)
From parallel axis theorem
I = I_G + Md²
mr² = \(\frac{2}{5}\) mr² + md²
d = \(\sqrt{\frac{3}{5}}\)r = \(\sqrt{0.6r}\)

Question 109.
A wheel is rotating with angular velocity 2 rad/s. It is subjected to a uniform angular acceleration 2 rad/s² then the angular velocity after 10 s is
(a) 12 rad/s
(b) 20 rad/s
(c) 22 rad/s
(d) 120 rad/s
Answer:
(c) 22 rad/s
ω = ω₀ + αt
Here ω₀ = 2 rad/s,
α = 2 rad/s2
ω = 10 s
ω = 2 + 2 x 10 = 22 rad/s

Question 110.
Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_b > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively,
then,
(a) L_B > L_A
(b) L_A > L_B
(c) L_A = \(\frac{L_{B}}{2}\)
(d) L_A = 2L_B
Answer:
(a) L_B > L_A

Question 111.
Three identical particles lie in x, y plane. The (x, y) coordinates of their positions are (3, 2), (1, 1), (5, 3) respectively. The (x, y) coordinates of the center of mass are –
(a) (a, b)
(b) (1, 2)
(c) (3, 2)
(d) (2, 1)
Answer:
(c) The X and Y coordinates of the center of mass are

Question 112.
A solid cylinder of mass 3 kg and radius 10 cm is rotating about its axis with a frequency of 20/π. The rotational kinetic energy of the cylinder
(a) 10 π J
(b) 12 J
(c) \(\frac{6 \times 10^{2}}{\pi}\) J
(d) 3 J
Answer:
(b) 12 J
Given,
M = 3 kg
R = 0.1 m
v = 20 / π
Angular frequency ω = 2πv = \(\frac{2π x 20}{π}\) = 40 rad/s^-1
Moment of inertia of the cylinder about its axis = I = \(\frac{1}{2}\) mR² = \(\frac{1}{2}\) x 3 x (0.1)² = 0.015 kg m²
K.E. = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) x 0.015 x (40)² = 12 J

Question 113.
A circular disc is rolling down in an inclined plane without slipping. The percentage of rotational energy in its total energy is
(a) 66.61%
(b) 33.33%
(c) 22.22%
(d) 50%
Answer:
(b) 33.33%
Rotational K.E. = \(\frac{1}{2}\)Iω² \(\frac{1}{2}\)(\(\frac{1}{2}\)MR²)ω² = \(\frac{1}{4}\) MR²ω²
Transnational K.E. = \(\frac{1}{2}\)MV² = \(\frac{1}{2}\)M(Rω)² = \(\frac{1}{2}\) MR²ω²
Total kinetic energy = E_rot + E_trans = \(\frac{1}{4}\) MR²ω²\(\frac{1}{2}\)M(Rω)² = \(\frac{3}{4}\) MR²ω²
% of E_rot = \(\frac{E_{\text {rot }}}{E_{\text {Tot }}}\) x 100% = 33.33%

Question 114.
A sphere rolls down in an inclined plane without slipping. The percentage of transnational energy in its total energy is
(a) 29.6%
(b) 33.4%
(c) 71.4%
(d) 50%
Answer:
(c) 71.4%
Rotational K.e. E_rot =

Question 115.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is –
(a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(c) According to law of conservation of linear momentum
MV = (M + M) V_CM
V_CM = \(\frac{MV}{M + M}\) = \(\frac{10 × 10}{10 + 4}\) = 10 ms^-1

Question 116.
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is –
(a) \(\frac{L}{4}\)
(b) \(\frac{L}{2}\)
(c) L
(d) 2L
Answer:
(a) \(\frac{L}{4}\)
We know that
angular momentum L = Mr²
Here, m and co are constants L α r²
If r becomes \(\frac{r}{2}\) angular momentum becomes \(\frac{1}{4}\) th of its initial value.

Question 117.
The moment of inertia of a thin uniform ring of mass 1 kg and radius 20 cm rotating about the axis passing through the center and perpendicular to the plane of the ring is –
(a) 4 x 10^-2 kg m²
(b) 1 x 10^-2 kg m²
(c) 20 x 10^-2 kg m²
(d) 10 x 10^-2 kg m²
Answer:
(b) Moment of inertia I = MR² = 1 x (10 x 10^-2)² = 1 x 10^-2 kg m².

Question 118.
A solid sphere is rolling down in the inclined plane, from rest without slipping. The angle of inclination with horizontal is 30°. The linear acceleration of the sphere is –
(a) 28 ms^-2
(b) 3.9 ms^-2
(c) \(\frac{25}{7}\)ms^-2
(d) \(\frac{1}{20}\)ms^-2
Answer:
(c) \(\frac{25}{7}\)ms^-2
We know that,a =

Question 119.
An electron is revolving in an orbit of radius 2 A with a speed of 4 x 10⁵ m /s. The angular momentum of the electron is [Me = 9 x 10^-31 kg]
(a) 2 x 10^-35 kg m² s^-1
(b) 72 x 10^-36 kg m² s^-1
(c) 7.2 x 10^-34 kg m² s^-1
(d) 0.72 x 10^-37 kg m² s^-1
Answer:
(b) Angular momentum L = mV x r = 9 x 10^-31 x 4 x 10⁵ x 2 x 10^-10 = 72 x 10^-36kg m² s^-1

Question 120.
A raw egg and hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the eggs to stop is –
(a) =1
(b) < 1
(c) > 1
(d) none of these
Answer:
(d) When a raw egg spins, the fluid inside comes towards its side.
∴ “1” will increase in – turn it decreases ω. Therefore it takes lesser time than boiled egg.
∴\(\frac {time fìr raw egg}{time for boiled egg}\) < 1

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (1 Mark)

Question 1.
What is a rigid body?
Answer:
A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Question 2.
When an object will have procession? Give one example.
Answer:
The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a procession.
Example:
The spinning top when it is about to come to rest.

Question 3.
Define angular momentum. Give an expression for it.
Answer:
The angular momentum of a point mass is defined as the moment of its linear momentum.
\(\vec{L}\) = \(\vec{r}\) x \(\vec{p}\) or L = rp sin θ

Question 4.
When an angular momentum of the object will be zero?
Answer:
If the straight path of the particle passes through the origin, then the angular momentum is zero, which is also a constant.

Question 5.
When an object be in-mechanical equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

Question 6.
Derive an expression for the power delivered by torque.
Answer:
Power delivered is the work done per unit time. IF we differentiate the expression for work done with respect to time, we get the instantaneous
power (P).
p = \(\frac{dw}{dt}\) = τ \(\frac{dθ}{dt}\)
p = τ dω

Question 7.
A boy sits near the edge of revolving circular disc

1. What will be the change in the motion of a disc?
2. If the boy starts moving from edge to the center of the disc, what will happen?

Answer:

1. As we know L = Iω = constant if the boy sits on the edge of revolving disc, its I will be increased in turn it reduces angular velocity.
2. If the boy starts moving towards the center of the disc, its I will decrease in turn that increases its angular velocity.

Question 8.
Are moment of inertia and radius of gyration of a body constant quantities?
Answer:
No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis..

Question 9.
A cat is able to land on its feet after a fall. Which principle of physics is being used? Explain.
Answer:
A cat is able to land on its feet after a fall. This is based on law of conservation of angular ~ momentum. When the cat is about to fall, it curls its body to decrease the moment of inertia and increase its angular velocity. When it lands it stretches out its limbs. By which it increases its moment of inertia and inturn it decreases its angular velocity. Hence, the cat lands safety.

Question 10.
About which axis a uniform cube will have minimum moment of inertia ?
Answer:
It will be about an axis passing through the center of the cube and connecting the opposite comers.

Question 11.
State the principle of moments of rotational equilibrium.
Answer:
∑ =\(\bar{\tau}\) = 0

Question 12.
Write down the moment of inertia of a disc of radius R and mass m about an axis in its plane at a distance R / 2 from its center.
Answer:
\(\frac { 1 }{ 2 }\) MR²

Question 13.
Can the couple acting on a rigid body produce translator motion ?
Answer:
No. It can produce only rotatory motion.

Question 14.
Which component of linear momentum does not contribute to angular momentum?
Answer:
Radial Component.

Question 15.
A system is in stable equilibrium. What can we say about its potential energy ?
Answer:
PE. is minimum.

Question 16.
Is radius of gyration a constant quantity ?
Answer:
No, it changes with the position of axis of rotation.

Question 17.
Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter?
Answer:
Sphere of smaller density will have larger moment of inertia.

Question 18.
The moment of inertia of two rotating bodies A and B are I_A and I_B (I_A > I_B) and their angular momenta are equal. Which one has a greater kinetic energy ?
Answer:
K = \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) ⇒ K_A > K_A

Question 19.
A particle moves on a circular path with decreasing speed. What happens to its angular momentum?
Answer:
As \(\vec{L}\) = \(\vec{r}\) x m\(\vec{v}\) i.e., \(\vec{L}\) magnitude decreases but direction remains constant.

Question 20.
What is the value of instantaneous speed of the point of contact during pure rolling ?
Answer:
Zero.

Question 21.
Which physical quantity is conserved when a planet revolves around the sun ?
Answer:
Angular momentum of planet.

Question 22.
What is the value of torque on the planet due to the gravitational force of sun ?
Answer:
Zero.

Question 23.
If no external torque acts on a body, will its angular velocity be constant ?
Answer:
No.

Question 24.
Why there are two propellers in a helicopter ?
Answer:
Due to conservation of angular momentum.

Question 25.
A child sits stationary at one end of a long trolley moving uniformly with speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, then what is the effect of the speed of the centre of mass of the (trolley + child) system ?
Answer:
No change in speed of system as no external force is working.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (2 Marks)

Question 26.
State the factors on which the moment of inertia of a body depends.
Answer:

• Mass of body
• Size and shape of body
• Mass distribution w.r.t. axis of rotation
• Position and orientation of rotational axis

Question 27.
On what factors does radius of gyration of body depend?
Answer:
Mass distribution.

Question 28.
Why the speed of whirl wind in a Tornado is alarmingly high?
Answer:
In this, air from nearly regions get concentrated in a small space, so I decreases considerably. Since Iω = constant so ω increases so high.

Question 29.
Can a body be in equilibrium while in motion? If yes, give an example.
Answer:
Yes, if body has no linear and angular acceleration then a body in uniform straight line of motion will be in equilibrium.

Question 30.
There is a stick half of which is wooden and half is of steel, (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why?
Answer:
I (first case) > 1 (Second case)
∴ τ r = l α
⇒ α (first case) < α (second case)

Question 31.
If earth contracts to half of its present radius what would be the length of the day at equator?
Answer:

Question 32.
An internal force cannot change the state of motion of center of mass of a body. Flow does the internal force of the brakes bring a vehicle to rest?
Answer:
In this case the force which bring the vehicle to rest is friction, and it is an external force.

Question 33.
When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium.
Answer:
For translation equilibrium
∑ F_ext  = 0
For rotational equilibrium
∑ \(\overline{\mathrm{τ}}\)_ext  = 0

Question 34.
How will you distinguish between a hard boiled egg and a raw egg by spinning it on a table top’
Answer:
For same external torque, angular acceleration of raw egg will be small than that of Hard boiled egg.

Question 35.
Equal torques are applied on a cylinder and a sphere. Both have same mass and radius. Cylinder rotates about its axis and sphere rotates about one of its diameter. Which will acquire greater speed and why?
Answer:
τ = I α α = \(\frac { τ }{ I }\)
α in cylinder, α_C = \(\frac{\tau}{I_{C}}\)
α in sphere, α_S = \(\frac{\tau}{I_{S}}\)

Question 36.
In which condition a body lying in gravitational field is in stable equilibrium?
Answer:
When vertical line through center of gravity passes through the base of the body.

Question 37.
Give the physical significance of moment of inertia. Explain the need of fly wheel in Engine.
Answer:
It plays the same role in rotatory motion as the mass does in translator y motion.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (3 Marks)

Question 38.
Three mass point m₁, m₂, m₃ are located at the vertices of equilateral A of side ‘a’. What is the moment of inertia of system about an axis along the altitude of A passing through mi?
Answer:

Question 39.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any linear push) on a perfectly friction less table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure. Will the disc roll?
Answer:

For A V_A = R ω₀ in forward direction
For B = V_B = R ω₀ in backward direction R
For C, V_C = \(\frac {R}{2}\) ω₀ in forward direction disc will not roll.

Question 40.
Find the torque of a force 7\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) about the origin which acts on a particle whose position vector is \(\hat{j}\) +\(\hat{j}\) – \(\hat{j}\)
Answer:

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numericals

Question 41.
Three masses 3 kg, 4 kg and 5 kg are located at the comers of an equilateral triangle of side 1 m. Locate the center of mass of the system.
Answer:
(x,y) = (0.54 m, 0.36 m)

Question 42.
Two particles mass 100 g and 300 g at a given time have velocities 10\(\hat{j}\) – 7\(\hat{j}\) – 3\(\hat{j}\) and 7\(\hat{i}\) – 9 \(\hat{j}\) + 6\(\hat{k}\) ms^-1 respectively. Determine velocity of center of mass.
Answer:
Velocity of center of mass = \(\frac{31 \hat{i}-34 \hat{j}+15 \hat{k}}{2}\) ms^-1

Question 43.
From a uniform disc of radius R, a circular disc of radius R / 2 is cut out. The center of the hole is at R / 2 from the center of original disc. Locate the center of gravity of the resultant flat body.
Answer:
Center of mass of resulting portion lies at R/6 from the center of the original disc in a direction opposite to the center of the cut out portion.

Question 44.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds,

1. What is its angular acceleration (assume the acceleration to be uniform)
2. How many revolutions does the wheel make during this time ?

Answer:
a = 4π rad s^-2
n = 576

Question 45.
A meter stick is balanced on a knife edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm, what is the mass of the meter stick?
Answer:
m = 66.0 gm.

Question 46.
A solid sphere is rolling op a friction less plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.
Answer:

Question 47.
Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centers and perpendicular to their planes.
Answer:
2 : 1

Question 48.
Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the center), and rotating with angular speed col and ω₂ are brought into contact face to face with their axes of rotation coincident,

1. What is the angular speed of the two – disc system ?
2. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₂.

Answer:

1. Let co be the angular speed of the two-disc system. Then by conservation of angular momentum
   
2. Initial K.E. of the two discs.
   

Hence there is a loss of rotational K.E. which appears as heat.
When the two discs are brought together, work is done against friction between the two discs.

Question 49.
In the HCL molecule, the separating between the nuclei of the two atoms is about 1.27 Å (1Å = 10^-10m). Find the approximate location of the CM of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in all its nucleus.
Answer:
As shown in Fig. suppose the H nucleus is located at the origin. Then,
x₁ = 0, x₂ = 1.27 Å, m₁ = 1, m₂ = 35.5
The position of the CM of HCl molecule is


Thus the CM of HCl is located on the line joining H and Cl nuclei at a distance of 1.235 Å from the H nucleus.

Question 50.
A child stands at the center of turn table with his two arms out stretched. The turn table is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/3 times the initial value?

1. Assume that the turn table rotates without friction
2. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation.

How do you account for this increase in kinetic energy ?
Answer:
Here ω = 40 rpm, I₂ = \(\frac { 1 }{ 2 }\) I₁
By the principle of conservation of angular momentum,
I₁ω₁ = I_2ω2 or I₁ x 4o = \(\frac {2}{5}\) I₁ ω₁ or ω₂ = 100 rpm
(ii) Initial kinetic energy of rotation –

new kinetic energy of rotation –

Thus the child’s new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position.

Question 51.
To maintain a rotor at a uniform angular speed of 200 rad s^-1 an engine needs to transmit a torque of 180 N m. What is the power required by the engine? Assume that the engine is 100% efficient.
Here ω = 200 rad s^-1, τ = 180 N m
Power, P = τω = 180 x 200 = 36,000 W = 36 kW.

Question 52.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front and back wheel.
Answer:

For transnational equilibrium of car
N_F + N_B = W = 1800 x 9.8 = 17640 N
For rotational equilibrium of car
1.05 N_F = 0.75 N_B
1.05 N_F = 0.75(17640 – N_F )
1.8 N_F = 13230
N_F = 13230 / 1.8 = 7350 N
N_B = 17640 – 7350 = 10290 N
Force on each front wheel = \(\frac {7350}{ 2 }\) = 3675 N
Force on each back wheel = \(\frac {10290}{ 2 }\) = 5145 N

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions (5 Marks)

Question 1.
Derive an expression for center of mass for distributed point masses.
Answer:
A point mass is a hypothetical point particle which has nonzero mass and no size or shape. To find the center of mass for a collection of n point masses, say,m₁ , m₂, m₃ ….. m_n  we have to first choose an origin and an appropriate coordinate system as shown in Figure. Let, x₁, x₂, x₃ …….. x_n be the X – coordinates of the positions of these point masses in the X direction from the origin.
The equation for the X coordinate of the center of mass is,

where, ∑ mi ¡s the total mass M of all the particles. ( ∑ mi = M).Hence,

Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in figure.

Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (x_CM, y_CM z_CM). in general, the position of center of mass can be written in a vector form as,

where, is the position vector of the center of mass and \(\vec{r}_{i}\) = x_i \(\hat{j}\) + y_i\(\hat{j}\) + z_i\(\hat{k}\) is the position vector of the distributed point mass; where, \(\hat{i}\), \(\hat{j}\), and \(\hat{j}\) are the unit vectors along X, Y and Z-axis respectively.

Question 2.
Discuss the center of mass of two point masses with pictorial representation.
Answer:
With the equations for center of mass, let us find the center of mass of two point masses m₁ and m₂, which are at positions x₁ and x₂ respectively on the X – axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.

(1) When the masses are on positive X-axis:
The origin is taken arbitrarily so that the masses m₁ and m₂ are at positions x₁ and x₂ on the positive X-axis as shown in figure (a). The center of mass will also be on the positive X- axis at x_CM as given by the equation,
\(x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\)

(2) When the origin coincides with any one of the masses:
The calculation could be minimized if the origin of the coordinate system is made to coincide with any one of the masses as shown in figure (b). When the origin coincides with the point
mass m₁, its position x₁ is zero, (i.e. x₁ = 0). Then,
\(x_{\mathrm{CM}}=\frac{m_{1}(0)+m_{2} x_{2}}{m_{1}+m_{2}}\)
The equation further simplifies as,
x_CM = \(\frac{m_{2} x_{2}}{m_{1}+m_{2}}\)

(3) When the origin coincides with the center of mass itself:
If the origin of the coordinate system is made to coincide with the center of mass, then, x_CM = O and the mass rn1 is found to be on the negative X- axis as shown in figure (c). Hence, its position x1 is negative, (i.e. – x₁).

The equation given above is known as principle of moments.

Question 3.
Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.
Answer:
Let us consider a rigid body rotating with angular velocity ω about an axis as shown in figure. Every particle of the body will have the same angular velocity ω and different tangential velocities v based on its positions from the axis of rotation. Let us choose a particle of mass m_i situated at distance r_i from the axis of rotation. It has a tangential velocity v_i given by the relation, v_i = r_i ω. The kinetic energy KE_i. of the particle is,
KE_i = \(\frac{1}{2} m_{i} v_{i}^{2}\)
Writing the expression with the angular velocity,
KE = \(\frac{1}{2}\) m_i(r_iω)² = \(\frac{1}{2} m_{i} r_{i}^{2}\)ω²

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
KE = \(\frac{1}{2}\left(\sum m_{i} r_{i}^{2}\right)\)ω²
where, the term ∑ m_ir_i^r is the moment of inertia I of the whole body. ∑ m_ir_i^r
Hence, the expression for KE of the rigid body in rotational motion is –
KE = \(\frac{1}{2}\) Iω²
This is analogous to the expression for kinetic energy in transnational motion.
KE = \(\frac{1}{2}\) Mv²

Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}\) Iω²
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

Question 4.
Discuss how the rolling is the combination of transnational and rotational and also be possibilities of velocity of different points in pure rolling.
Answer:
The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc. While rolling, the point undergoes transnational motion along with its center of mass and rotational motion with respect to its center of mass.

Combination of Translation and Rotation:
We will now see how these transnational and rotational motions arc related in rolling. If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2πR (its circumference). One would agree that not only the center of mass. but all the points Of l the disc are displaced by the same 2πR after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the transnational and rotational motion. Especially the point on the edge undergoes a path of a cyclonic as shown in the figure.

As the center of mass takes only a straight line path. its velocity v_CM is only transnational velocity v_TRANS (v_CM = v_TRANS). All the other points have two velocities. One is the transnational velocity v_TRANS (which is also the velocity of center of mass) and the other is the rotational velocity v_ROT (v_ROT = rω). Here, r ¡s the distance of the point from the center of mass and o is the angular velocity. The rotational velocity v_ROT is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is v. This resultant velocity y is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all’the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.
As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = o). For example, in figure, at the point of contact, v_TRANS is forward (to right) and v_ROT is backwards (to the left).

That implies that, v_TRANS and v_ROT are equal in magnitude and opposite in direction (v = v_TRANS – v_ROT = 0). Hence, we conclude that in pure rolling, for all the points on the edge, the magnitudes of v_TRANS and v_ROT are equal (v_TRANS = v_ROT) As v_TRANS = v_CM and v_ROT = Rω, in pure rolling we have,
v_CM = Rω

We should remember the special feature of the above equation. In rotational motion, as per the relation v = rω, the center point will not have any velocity as r is zero. But in rolling motion, it suggests that the center point has a velocity v_CM given by above equation v_CM – Rω. For the topmost point, the two velocities v_TRANS and v_ROT are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity v is the sum of these two velocities, v = v_TRANS + v_ROT In other form, v = 2 v_CM as shown in figure below.

Question 5.
Derive an expression for kinetic energy in pure rolling.
Answer:
As pure is the combination of transnational and rotational motion, we can write the total kinetic energy (KE) as the sum of kinetic energy due to transnational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT).
KE = KE_TRANS + KE_ROT ………(i)
If the mass of the rolling object is M, the velocity of center of mass is v_CM, its moment of inertia about center of mass is I_CM and angular velocity is ω, then

With center of mass as reference:
The moment of inertia (I_CM) of a rolling object about the center of mass is, I_CM = MK² and v_CM = Rω. Here, K is radius of gyration.

With point of contact as reference:
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as o, then,
KE = \(\frac {1}{2}\) I₀ω²

Here, I₀ is the moment of inertia of the object about the point of contact. By parallel axis
theorem, I₀ = I_CM + MK² Further we can write, I₀ MK² + MR². With v_CM = Rω or ω = \(\frac{v_{\mathrm{CM}}}{\mathrm{R}}\)

As the two equations (v) and (vi) are the same, it ¡s once again confirmed that the pure tolling problems could be solved by considering the motion as any one of the following two cases.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.

Question 6.

1. Can a body in translator y motion have angular momentum? Explain.
2. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string?

Answer:
(1) Yes, a body in translatory motion shall have angular momentum unless fixed point about which angular momentum is taken lies on the line of motion of body
\(|\overrightarrow{\mathrm{L}}|\) = rp sin θ
= 0 only when θ = O° or 180°

(2) MI of stone I = ml² (l – length of string) l is large, a is very small
τ = Iα
α = \(\frac {τ}{I}\) = \(\frac{\tau}{m l^{2}}\)
if l is large a is very small.
∴ more difficult to revolve.

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 2 Kingdom Animalia

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Samacheer Kalvi 11th Bio Zoology Kingdom Animalia Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Samacheer Kalvi Guru 11th Zoology Question 1.
The symmetry exhibited in cnidarians is –
(a) Radial
(b) Bilateral
(c) Pentamerous radial
(d) Asymmetrical
Answer:
(a) Radial

Class 11 Zoology Chapter 2 Notes Question 2.
Sea anemone belongs to phylum –
(a) Protozoa
(b) Porifera
(c) Coelenterata
(d) Echinodermata
Answer:
(c) Coelenterata

Samacheer Kalvi Guru 11th Biology Question 3.
The excretory cells that are found in platyhelminthes are –
(a) Protonephridia
(b) Flame cells
(c) Solenocytes
(d) All of these
Answer:
(d) All of these

Samacheer Kalvi 11 Zoology Solutions Question 4.
In which of the following organisms, self-fertilization is seen?
(a) Fish
(b) Round worm
(c) Earthworm
(d) Liver fluke
Answer:
(d) Liver fluke

Samacheer Kalvi 11th Zoology Solutions Question 5.
Nephridia of Earthworms are performing the same functions as –
(a) Gills of prawn
(b) Flame cells of Planaria
(c) Trachea of insects
(d) Nematoblasts of Hydra
Answer:
(b) Flame cells of Planaria

Zoology Class 11 Chapter 2 Question 6.
Which of the following animals has a true coelom?
(a) Ascaris
(b) Pheretima
(c) Sycon
(d) Taenia solium
Answer:
(b) Pheretima

Kingdom Zoology Question 7.
Metameric segmentation is the main feature of –
(a) Annelida
(b) Echinodermata
(c) Arthropoda
(d) Coelenterata
Answer:
(a) Annelida

11th Zoology Solutions Question 8.
In Pheretima locomotion occurs with help of –
(a) circular muscles
(b) longitudinal muscles and setae
(c) circular, longitudinal muscles and setae
(d) parapodia
Answer:
(c) circular, longitudinal muscles and setae

Samacheer Kalvi Guru 11th Bio Zoology Question 9.
Which of the following have the highest number of species in nature?
(a) Insects
(b) Birds
(c) Angiosperms
(d) Fungi
Answer:
(a) Insects

Kingdom Animalia Pdf Question 10.
Which of the following is a crustacean?
(a) Prawn
(b) Snail
(c) Sea anemone
(d) Hydra
Answer:
(a) Prawn

Zoology Kingdom Question 11.
The respiratory pigment in cockroach is –
(a) Haemoglobin
(b) Haemocyanin
(c) Haemoerythrin
(d) None of the above
Answer:
(c) Haemoerythrin

11th Zoology Samacheer Kalvi Question 12.
Exoskeleton of which phylum consists of chitinous cuticle?
(a) Annelida
(b) Porifera
(c) Arthropoda
(d) Echinodermata
Answer:
(a) Annelida

Question 13.
Lateral line sense organs occur in –
(a) Salamander
(b) Frog
(c) Water snake
(d) Fish
Answer:
(d) Fish

Question 14.
The limbless amphibian is –
(a) Icthyophis
(b) Hyla
(c) Rana
(d) Salamander
Answer:
(a) Icthyophis

Question 15.
Four chambered heart is present in –
(a) Lizard
(b) Snake
(c) Scorpion
(d) Crocodile
Answer:
(d) Crocodile

Question 16.
Which of the following is not correctly paired?
(a) Humans – Ureotelic
(b) Birds – Uricotelic
(c) Lizards – Uncotelic
(d) Whale – Ammonotelic
Answer:
(d) Whale – Ammonotelic

Question 17.
Which of the following is an egg laying mammal?
(a) Deiphinus
(b) Macropus
(c) Ornitho rhynchus
(d) Equus
Answer:
(c) Ornitho rhynchus

Question 18.
Pneumatic bones are seen in –
(a) Mammalia
(b) Aves
(c) Reptilia
(d) Sponges
Answer:
(b) Aves

Question 19.

Match the following columns and select the correct option.

Column -1	Column – II
(P) Pila	(i) Devil fish
(q) Dentalium	(ii) Chiton
(r) Chaetopleura	(iii) Apple snail
(s) Octopus	(iv) Tusk shell

(a) p – (ii), q – (i), r – (iii), s – (iv)
(b) p – (iii), q – (iv), r – (ii), s – (i)
(c) p – (ii), q – (iv), r – (i), s – (iii)
(d) p – (i), q – (ii), r – (iii), s – (iv)
Answer:
(b) p – (iii), q – (iv), r – (ii), s – (i)

Question 20.
In which of the following phyla, the adult shows radial symmetry but the larva shows bilateral symmetry?
(a) Mollusca
(b) Echinodermata
(c) Arthropoda
(d) Annelida
Answer:
(b) Echinodermata

Question 21.
Which of the following is correctly matched?
(a) Physalia – Portugese man of war
(b) Pennatula – Sea fan
(c) Adamsia – Sea pen
(d) Gorgonia – Sea anemone
Answer:
(a) Physalla – Portugese man of war

Question 22.
Why are spongin and spicutes important to a sponge?
Answer:
Spongin and spicules provide support and supports the soft body parts of the sponges. The spicules give the sponges rigidity and form to the sponges

Question 23.
What are the four characteristics common to most animals?
Answer:
The characteristics common to most animals are the arrangement of cell layers.

• The levels of organization.
• Nature of coelom.
• The presence or absence of segmentation and notochord.
• Organization of the organ system.

Question 24.
list the features that all vertebrates show at some point in their deelopnwnt.
Answer:
All vertebrates possess notochord during embryonic stay. li is repLaced by vertebra) column. All vertebrates possess pained appendages such as fins or lunits. Skin is covered by protective skeleton comprising of scales. fiathcrs hairs, claws, nails etc. Respiration is aerobic through gills, skin. buccopharyngeal cavity’ and lungs. All vertebrates have a muscular heart with two, three or four chambers and kidneys for excretion and osmoregulation.

Question 25.
Compare closed and opened circulatory system.
Answer:
Closed circulatory system:

• The circulation in which blood is present inside the blood vessels is called closed circulatory’ system
• It is found in higher organisms, e.g. annelids, cephalochordates and vertebrates.

Open circulatory system:

• The circulation in which blood remains filled in tissue spaces due to the absence of blood vessels is called open circulatory system
• It is found in lower organisms. e.g. arthropods, molluscs and echinoderms.

Question 26.
Compare schizocoelom with enterocoeloni.
Schizocoelom:

• The coelom which is formed by splitting of Mesoderm is called schizocoelom.
• It is found in lower invertebrates like annelids, arthropods and molluscs.

Enterocoeloni:

• The coelom which is formed from the Mesodermal pouches of archenteron is called enteroceolem.
• It is found in echinoderms, hemichordates and chordates.

Question 27.
Identify the structure that the archenteron becomes in a developing animal.
Answer:
The archenteron becomes the cavity of the digestive tract.

Question 28.
Observe the animal below and answer the following questions
(a) Identify the animal
(b) What type of symmetry does this animal exhibit?
(c) Is this animal Cephalized?
(d) How many germ layers does this animal have?
(e) How many openings does this animal’s digestive system have?
(f) Does this animal have neurons?

Answer:
(a) Sea anemone (Adarnasia)
(b) Radial symmetry
(c) No
(d) Two (ectoderm and endoderm)
(e) One
(f) No.

Question 29.
Choose the term that does not belong in the following group and explain why it does not belong?
Answer:

• Notochord, cephalisation, dorsal nerve cord and radial symmetry.
• Notochord, cephalisation and dorsal nerve cord are the characteristic features of chordates. The radial symmetry is not a characteristic feature of chordate. It is the feature of cnidarian and adult echinoderms. Hence it does not belong to the group.

Question 30.
Why flat worms are called acoelomates?
Answer:
Flat worms are called acoelomate because they do not possess a body cavity.

Question 31.
What are flame cells?
Answer:
Flame cells are the specialized excretory cells in flat worms. They help in excretion and osmoregulation.

Question 32.
Concept Mapping – Use the following terms to create a concept map that shows the major characteristic features of the phylum nematoda: Round worms, pseudocoelomates, digestive tract, cuticle, parasite and sexual dimorphism.
Answer:

Question 33.
In which phyla is the larva trochopore found?
Answer:
Annelida.

Question 34.
Which of the chordate characteristics do tunicates retain as adults?
Answer:
Ventral and tabular heart. Respiration is through gill slits.

Question 35.
List the characteristic features that distinguish cartilaginous fishes with living jawless fishes.
Answer:
Cartilaginous Fishes:

• These have powerful jaws which help in predation.
• These are free living predatory fishes.
• These are advanced over jawless fishes.
• These have a tough skin covered by dermal
• Respiration is by lamelliform gills without operculum.
• Paired fins are present.
• Fertilization is internal
• Viviparous e.g. Scoliodon

Living Jawless Fishes:

• They do not have jaws and the mouth is circular and suctorial.
• These are ectoparasites on fishes.
• These are primitive over cartilaginous fishes.
• The skin is soft and devoid of scales.
• These have six to fifteen pair of gills slits.
• Paired fins are absent.
• Fertilization is external
• Oviparous, e.g. Lamprey

Question 36.
List three features that characterise bony fishes.
Answer:

• These fishes have bony endoskeleton.
• The skin is covered by ganoid, cycloid or ctenoid scales.
• Gills are covered by an operculum.
• They are ammonotelic.
• They have mesonephric kidneys.
• External fertilization is seen.

Question 37.
List the functions of air bladder in fishes.
Answer:

• Air bladder helps in gaseous exchange.
• It helps in maintaining buoyancy.

Question 38.
Write the characteristics that contributes to the success of reptiles on land.
Answer:

• The characteristics that contribute to the success of reptiles on land are as follows:
• The presence of dry and comified skin with epidermal scales or scutes which prevent the loss of water.
• The presence of metanephric kidney.
• They are uricotelic (they excrete uric acid to prevent the loss of water).

Question 39.
List the unique features of bird’s endoskeleton.
Answer:

• The endoskeleton of birds is bony
• The long bones are hollow with air cavities (pneumatic)
• The body is covered by feathers.

Question 40.
Could the number of eggs or young ones produced by an oviparous and viviparous female be equal? Why?
Answer:
No. The number of eggs or young ones produced by an oviparous and viviparous female cannot be equal. When the oviparous animals lay eggs in the external environment or in the medium, the chance of survival and successful development into the adults are not certain. But in case of viviparous animals, young ones are nurtured by the adult animals. Hence, oviparous animals lay more eggs if they are fertilized in the medium or in water.

Entrance Examination Questions Solved
Choose the correct answer
Question 1.
Classification of sponges is primarily based on the …………. (JCECE-2003)
(a) body organization
(b) body plan
(c) skeleton
(d) canal system
Answer:
(c) skeleton

Question 2.
Symmetry is cnidaria is …………. (AMU-2009)
(a) radial
(b) bilateral
(c) pentamerous
(d) spherical
Answer:
(a) radial

Question 3.
Cavity of coelenterates is called …………. (BHU-2008)
(a) coelenteron
(b) coelom
(c) cavity
(d) none of these
Answer:
(a) coelenteron

Question 4.
Sea anemone belongs to phylum …………. (BCECE-2005)
(a) protozoa
(b) porifera
(c) coelenterata
(d) echinodermata
Answer:
(c) coelenterata

Question 5.
Medusa is the reproductive organ of …………. (BHU-2008)
(a) Hydra
(b) Aurelia
(c) Obelia
(d) Sea anemone
Answer:
(b) Aurelia

Question 6.
The excretory cells, that are found in platyhelminthes …………. (J & K CET-2007)
(a) Protonephridia
(b) Flame cells
(c) Solenocytes
(d) All of these
Answer:
(b) Flame cells

Question 7.
In which of the following organisms, self-fertilization is seen? (CCET-2007)
(a) Fish
(b) Round worm
(c) Earthworm
(d) Liver fluke
Answer:
(d) Liver fluke

Question 8.
Nephridia of Earthworms are performing same function as …………. (J & KCET-2003)
(a) gills of prawn
(b) flame cells of Planaria
(c) trachea of insects
(d) nematoblasts of Hydra
Answer:
(b) flame cells of Planaria

Question 9.
Phylum of Taenia solium is …………. (BCECE-2004)
(a) Aschelminthes
(b) Annelids
(c) Platylelminthes
(d) Mollusca
Answer:
(c) Platylelminthes

Question 10.
Ascaris is found in …………. (RPMT-2004)
(a) body cavity
(b) lymph nodes
(c) tissue
(d) alimentary canal
Answer:
(d) alimentary canal

Question 11.
Which of the following animals has a true coelom? (J & K CET-2007)
(a) Ascaris
(b) Pheretima
(c) Sycon
(d) Taenia solium
Answer:
(b) Pheretima

Question 12.
Metameric segmentation is the main feature of ………….
(a) Annelida
(b) Echinodermata
(c) Arthropoda
(d) Coelenterata
Answer:
(a) Annelida

Question 13.
Body cavity lined by mesoderm is called …………. (J & T CET-2005)
(a) coelenteron
(b) pseudocoel
(c) coelom
(d) blastocoel
Answer:
(c) coelom

Question 14.
Which of the following have the highest number of species in nature? (AIPMT-2011)
(a) Insects
(b) Birds
(c) Angiosperms
(d) Fungi
Answer:
(a) Insects

Question 15.
Which of the following is a crustacean? (Guj-CET-2011)
(a) Prawn
(b) Snail
(c) Sea anemone
(d) Hydra
Answer:
(a) Prawn

Question 16.
The respiratory pigment present in cockroach is …………. (OJEE-2010)
(a) Haemoglobin
(b) Haemocyanin
(c) Oxyhaemoglobin
(d) None of these
Answer:
(d) None of these

Question 17.
Book lungs are respiratory organs in …………. (AMU-2008)
(a) Insects
(b) Arachnids
(c) Molluscans
(d) Echinoderms
Answer:
(b) Arachnids

Question 18.
The excretory organ in cockroach is …………. (Kerala-CEE-2007)
(a) Malpighian corpuscle
(b) Malpighian tubules
(c) Green gland
(d) Metanephridia
Answer:
(b) Malpighian tubules

Question 19.
Exoskeleton of which phylum consists of chitinous cuticle? (J & KCET-2007)
(a) Annelida
(b) Porifera
(c) Arthropoda
(d) Echinodermata
Answer:
(c) Arthropoda

Question 20.
In cockroach, vision is due to …………. (PMET-2005)
(a) one compound eye
(b) two compound eyes
(c) two simple eyes
(d) two compound and two simple eyes.
Answer:
(b) two compound eyes

Question 21.
Which of the following respires through gills? (J & K CET-2005)
(a) Whale
(b) Turtle
(c) Frog
(d) Prawns
Answer:
(d) Prawns

Question 22.
Animals active at night are called …………. (J & K CET-2004)
(a) diurnal
(b) nocturnal
(c) parasites
(d) nocto – diumal
Answer:
(b) nocturnal

Question 23.
Salient features of Arthropoda is ……….. (RPMT-2003)
(a) aquatic and free living
(b) chitinous exoskeleton and jointed appendages
(c) radulla
(d) none of those
Answer:
(b) chitinous exoskeleton and jointed appendages

Question 24.
The second largest number of species containing phylum in the animal kingdom is ……….. (J & K CET-2008)
(a) Annelida
(b) Arthropoda
(c) Mollusca
(d) Chordata
Answer:
(c) Mollusca

Question 25.
Mollusca is ………… (JCECE-2006)
(a) Triploblastic, acoelomate
(b) Triploblastic, coelomate
(c) Diploblastic, acoelomate
(d) Diploblastic, coelomate
Answer:
(b) Triploblastic, coelomate

Question 26.
Tube feet are the locomotory organs of ………….
(a) Platyhelminthes
(b) Echinodermata
(c) Mollusca
(d) Arthropoda
Answer:
(b) Echinodermata

Question 27.
Given below are four matchings of a animal and its kind of respiratory organ …………. (PMT 2003)
(A) Silver fish – Trachea
(B) Scorpion – Book lung
(C) Sea squirt – Pharyngeal gills
(D) Dolphin – Skin
The correct matchings are
(a) A and B
(b) A,B and C
(c) B and D
(d) C and D
Answer:
(b) A,B and C

Question 28.
Which one of the following is a matching pair of an animal and a certain phenomenon it exhibits? (PMT 2003)
(a) Pheretima – Sexual dimorphism
(b) Rana – Complete metamorphosis
(c) Chameleon – Mimicry
(d) Taenia – Polymorphism
Answer:
(b) Rana – Complete metamorphosis

Question 29.
Two common characters found in centipede, cockroach and crab …………. (PMT 2006)
(a) book lungs and antennae
(b) compound eyes and anal cerci
(c) joint legs and chitinous exoskeleton
(d) green gland and tracheae
Answer:
(c) joint legs and chitinous exoskeleton

Question 30.
Which one of the following groups of animals is bilaterally symmetrical and triploblastic? (PMT 2009)
(a) Aschelminthes (round worms)
(b) Ctenophores
(c) Sponges
(d) Coelenterates (cnidarians)
Answer:
(a) Aschelminthes (round worms)

Question 31.
Which one feature is common to leech, cockroach and scorpion? (AIIMS 2004)
(a) Nephridia
(b) Ventral nerve cord
(c) Cephalization
(d) Antennae
Answer:
(b) Ventral nerve cord

Question 32.
Which one of the following features is common in silverfish, scorpion, dragonfly and prawn?
(a) Three pairs of legs and segmented body
(b) Chitinous cuticle and two pairs of antennae
(c) Jointed appendages and chitinous exoskeleton
(d) Cephalothorax and trachea
Answer:
(c) Jointed appendages and chitinous exoskeleton

Question 33.
Peripatus is known as a connecting link, because it has the characters of both …………. (BHU 1993)
(a) Fishes & amphibians
(b) Reptiles & birds
(c) Aves & fishes
(d) Arthropoda & annelids
Answer:
(d) Arthropoda & annelids

Question 34.
Osphradium of Pila globosa is ………… (BHU 1994, 2000, 2007)
(a) thermoreceptor
(b) Pheretima
(c) chemoreceptor
(d) tangoreceptor
Answer:
(c) chemoreceptor

Question 35.
Green glands present in some arthropods help in …………. (BHU 1998, 2007)
(a) respiration
(b) excretion
(c) digestion
(d) none of these
Answer:
(b) excretion

Question 36.
Squid, cuttle fish and Octopus belongs to class of …………. (BHU 1998, 2001)
(a) decapoda
(b) scaphopoda
(c) cephalopoda
(d) apods
Answer:
(c) cephalopoda

Question 39.
The canal system is a characteristic feature of …………. (BHU 1999, 2002)
(a) sponges
(b) echinoderms
(c) helminthes
(d) coelenterates
Answer:
(a) sponges

Question 40.
Malpighian tubules are …………. (BHU 2006)
(a) excretory organs of insects
(b) excretory organs of frog
(c) respiratory organs of insects
(d) endocrine glands of insects
Answer:
(a) excretory organs of insects

Question 41.
Caterpillar and maggot are …………. (BHU 2007)
(a) larvae
(b) nymphs
(c) adults
(d) pupa
Answer:
(a) larvae

Question 42.
Excretory organ of platyhelminthes is …………. (BHU 2008)
(a) gills
(b) flame cells
(c) nephridia
(d) trachea
Answer:
(b) flame cells

Question 43.
Water vascular system is a characteristic of …………. (BHU 2008)
(a) ctenophore
(b) annelid
(c) echinodermata
(d) arthropoda
Answer:
(c) echinodermata

Question 44.
Tube feet are the characteristic structures of …………. (DPMT 1993, 2008)
(a) jellyfish
(b) starfish
(c) cuttlefish
(d) crayfish
Answer:
(b) starfish

Question 45.
Hormone, which helps in metamorphosis in insects is …………. (DPMT 1996)
(a) pheromone
(b) ecdysone
(c) thyroxine
(d) all of these
Answer:
(b) ecdysone

Question 46.
The muscles associated with the heart of insects are …………. (DPMT 1996, 2006)
(a) alary
(b) striped
(c) radial
(d) pericardial
Answer:
(a) alary

Question 47.
Which of the following organisms is pseudocoelomate? (DPMT 2001, 2006)
(a) Hookworm
(b) Liver fluke
(c) Jelly fish
(d) Leech
Answer:
(a) Hookworm

Question 48.
Which of the following is not reported to have any fresh water forms? (DPMT 2003)
(a) Mollusca
(b) Sponges
(c) Coelenterates
(d) Echinoderms
Answer:
(d) Echinoderms

Question 49.
Pseudocoelom is not found in ………… (DPMT 2004)
(a) Ascaris
(b) Ancylostoma
(c) Fasciola
(d) none of these
Answer:
(c) Fasciola

Question 50.
Animals devoid of respiratory, excretory and circulatory organs belong to phylum ………… (DPMT 2004)
(a) echinodermata
(b) platyhelminthes
(c) porifera
(d) mollusca
Answer:
(c) porifera

Question 51.
Cilia of gills of bivalve molluscs help in …………. (DPMT 2005)
(a) protection
(b) respiration
(c) excretion
(d) feeding
Answer:
(b) respiration

Question 52.
All flat worms differ from all round worms in having …………. (DPMT 2009)
(a) triploblastic body
(b) solid mesoderm
(c) bilateral symmetry
(d) metamorphosis in the life history
Answer:
(b) solid mesoderm

Question 53.
Parthenogenesis can be seen in …………. (UPCPMT 1995)
(a) frog
(b) honey bee
(c) moth
(d) all of these
Answer:
(b) honey bee

Question 54.
The endocrine gland of insects, which secretes the juvenile hormone, is …………. (UP-CPMT 1995)
(a) corpora allata
(b) corpora albicans
(c) corpora myecaena
(d) all of these
Answer:
(a) corpora allata

Question 55.
Malpighian tubules are …………. (UP – CPMT 1996, 2008)
(a) excretory organs of insects
(b) respiratory organs of insects
(c) excretory organs of frog
(d) endocrine glands of insects
Answer:
(a) excretory organs of insects

Question 56.
In mollusca, eye is present over a stalk called …………. (UP-CPMT 2000, 2007)
(a) osphradium
(b) ostracum
(c) ommatophore
(d) operculum
Answer:
(c) ommatophore

Question 57.
Which of the following symmetries is found in adult sea anemone? (UP – CPMT 2004)
(a) Radial
(b) Biradial
(c) Bilateral
(d) Spherical
Answer:
(a) Radial

Question 58.
Feeding in sponges takes place through …………. (UP-CPMT 2005)
(a) choanocytes
(b) nurse cells
(c) ostia
(d) osculum
Answer:
(a) choanocytes

Question 59.
Osphradium is meant for …………. (UP-CPMT 2005)
(a) excretion
(b) nutrition
(c) selection and rejection of food
(d) grinding of food
Answer;
(c) selection and rejection of food

Question 60.
Excretory product of spider is …………. (UPCPMT 2007)
(a) uric acid
(b) ammonia
(c) guanine
(d) none of these
Answer:
(c) guanine

Question 61.
Which of the following is not the character of Taenia soliuml …………. (UPCPMT 2007)
(a) Polysis
(b) Proglottid
(c) Metamerism
(d) Strobila
Answer:
(c) Metamerism

Question 62.
Daphnia is commonly known as …………. (UP-CPMT 2007)
(a) clam shrimp
(b) fairy shrimp
(c) water fleas
(d) tadpole shrimp
Answer:
(c) water fleas

Question 63.
Wuchereria is found in …………. (UP-CPMT 2007)
(a) lymph nodes
(b) lungs
(c) eye
(d) gonds
Answer:
(a) lymph nodes

Question 64.
“Turbellarians” are free living …………. (UP-CPMT 2008)
(a) flatworms
(b) trematodes
(c) nematodes
(d) cesrtodes
Answer:
(a) flatworms

Question 65.
Polyp phase is absent in …………. (UP-CPMT 2008)
(a) Physalia
(b) Obselia
(c) Hydra
(d) Aurelia
Answer:
(d) Aurelia

Question 66.
Animals having pseudocoelomate and triploblastic nature are present in phyla …………. (UP-CPMT 2008).
(a) annelida
(b) arthropoda
(c) aschelminthes
(d) platyhelminthes
Answer:
(c) aschelminthes

Question 67.
Primitive nervous system is formed in …………. (UP-CPMT 2009)
(a) sponge
(b) cnidaria (coelenterate)
(c) echinodermata
(d) annelida
Answer:
(b) cnidaria (coelenterate)

Question 68.
Tissues are absent in the body of …………. (UP-CPMT 2009)
(a) sponge
(b) annelida
(c) platyhelminthes
(d) arthropoda
Answer:
(a) sponge

Question 69.
Linmulus belongs to class ………….
(a) onychophora
(b) insect
(c) merostomata
(d) Crustacea
Answer:
(c) merostomata

Question 70.
Ambulacral system is mainly useful for ………….
(a) locomotion
(b) feeding
(c) circulation
(d) defence
Answer:
(b) feeding

Question 71.
Which of the following is a excretory organ in mollusca?
(a) Keber’s organ
(b) nephridia
(c) Malpighian organ
(d) Flame cells
Answer:
(a) Keber’s organ

Question 72.
Mouth parts of housefly are ………….
(a) Piercing and sucking type
(b) Biting and sucking type
(c) Sponging and sucking type
(d) biting and chewing type
Answer:
(c) Sponging and sucking type

Question 73.
Anus is absent in
(a) Periplaneta
(b) Unio
(c) Fasciola
(d) Pheretima
Answer:
(c) Fasciola

Question 74.
Asymmetry in gastropoda is due to ………….
(a) twisting
(b) torsion
(c) coiling
(d) none of these
Answer:
(b) torsion

Question 75.
The pigment haemocyanin is found in ………….
(a) mollusca
(b) chordate
(c) echinodermata
(d) annelida
Answer:
(a) mollusca

Question 76.
The development of adult characteristics in a moulting insect is promoted by ………….
(a) pheromone
(b) thyroxine
(c) juvenile hormone
(d) ecdysone
Answer:
(d) ecdysone

Question 77.
If you are given an insect, a spider, a Peripatus and a crab, based on which character you can identify an arachnid from others?
(a) one pair of legs
(b) sense organs
(c) four pairs of legs
(d) number of wings
Answer:
(c) four pairs of legs

Question 78.
Choanocytes perform ………….
(a) reproduction
(b) nutrition
(c) discretion of spicules
(d) excretion
Answer:
(b) nutrition

Question 79.
Common characteristics of cockroach, housefly and mosquito are ………….
(a) one pair each of wings and halters
(b) three pairs of legs and one pair of developed wings
(c) two pair of legs and two compound eye
(d) compound and simple eyes
Answer:
(a) one pair each of wings and halters

Question 80.
The secondary host of Taenia is ………….
(a) snail
(b) pig
(c) man
(d) dog
Answer:
(b) pig

Question 81.
The exoskeleton of insect is made up of ………….
(a) pectin
(b) lignin
(c) chitin
(d) suberin
Answer:
(c) chitin

Question 82.
Collar cells are found in ………….
(a) aschelminthes
(b) cnidaria
(c) arthropoda
(d) sponges
Answer:
(d) sponges

Question 83.
Ommatidia are the units that constitute the compound eyes in …………. (AMU 1995).
(a) Fish ‘
(b) Insects
(c) Mammals
(d) Birds
Answer:
(b) Insects

Question 84.
Which of the following animals possesses ink gland?(AMU 2003)
(a) Blue Whale
(b) Scorpion
(c) Sea Urchin
(d) Cuttle Fish
Answer:
(d) Cuttle Fish

Question 85.
Comb plates are present in …………. (AMU 2004)
(a) echinoderms
(b) ctenophores
(c) annelids
(d) molluscs
Answer:
(b) ctenophores

Question 86.
Which of the following does not belong to phylum cnidaria?(AMU 2004)
(a) Sea pen
(b) Sea lily
(c) Sea-fan
(d) Sea anemone
Answer:
(b) Sea lily

Question 87.
Protonephridia are the excretory structures present in …………. (AMU 2005)
(a) Planaria
(b) Roundworm
(c) Tapeworm
(d) Prawn
Answer:
(a) Planaria

Question 88.
Which of the following is not an annelid? (AMU 2007)
(a) Leech
(b) Earthworm
(c) Sea mouse
(d) Sea cucumbers
Answer:
(d) Sea cucumbers

Question 89.
Blood worms are the larvae of …………. (AMU 2007)
(a) Hirudinaria
(b) Chironomus
(c) Limulus
(d) Daphnia
Answer:
(b) Chironomus

Question 90.
Pick the odd pair …………. (AMU 2008)
(a) Porifera : spicules
(b) Scyphozoan : coral reef
(c) Nematode : pseudocoelomate
(d) Cestoda : proglottid
Answer:
(b) Scyphozoan : coral reef

Question 91.
Insect metamorphosis having larval stage is called …………. (AFMC 1994)
(a) Incomplete metamorphosis
(b) Retrogressive metamorphosis
(c) Heteromorphosis
(d) Complete metamorphosis
Answer:
(d) Complete metamorphosis

Question 92.
Which of the following is not an insect? (AFMC 1996)
(a) Cockroach
(b) Spider
(c) Mosquito
(d) Bedbug
Answer:
(b) Spider

Question 93.
Which of the following enters intestine by penetrating through skin? (AFMC 2003)
(a) Hook worm
(b) Ascaris
(c) Pin worm
(d) Filarialworm
Answer:
(a) Hook worm

Question 94.
In nemathelminthes the coelom is not lined by peritoneum is …………. (AFMC 2004)
(a) acoelom
(b) pseudocoelom
(c) enterocoelom
(d) haemocoel
Answer:
(b) pseudocoelom

Question 95.
Leech secretes which of the following anticoagulant? (AFMC 2004)
(a) Hirudin
(b) Heparin
(c) Serotonin
(d) Histamine
Answer:
(a) Hirudin

Question 96.
Canal system in porifera is not concerned with …………. (AFMC 2005)
(a) respiration
(b) nutrition
(c) sexual reproduction
(d) none of these
Answer:
(c) sexual reproduction

Question 97.
Johnston’s organ is present in …………. (AFMC 2007)
(a) antenna of insect
(b) head of cockroach
(c) abdomen of housefly
(d) abdomen of spider
Answer:
(a) antenna of insect

Question 98.
Which of the following is not an arachnid? (AFMC 2007)
(a) Spider
(b) Itchmite
(c) Louse
(d) Tick
Answer:
(c) Louse

Question 99.
Fasciola hepatica is …………. (AFMC 2007)
(a) hermaphrodite self fertilizing
(b) hermaphrodite, cross fertilizing
(c) unisexual
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 100.
Match the excretory organs listed under column I with the animals given under column II. Choose the answer which gives the correct combination of alphabets of the column.

Column I	Column II
A. Nephridia	p. Hydra
B. Malpighian tubules	q. Leech
C. protonephridia	r. Shark
D. kidneys	s. Round worms
	t. Cockroach

(a) A – q; B -1; C – s; D – r
(b) A = s ; B= q; C= p; D= t
(c) A -1; B – q; C – s; D – r .
(d) A = q; B = s ; C =t; D= p
Answer:
(a) A – q; B -1; C- s; D – r

Question 101.
Entomology is concerned with the study of ………….
(a) formation and properties of soil
(b) agricultural practices
(c) various aspects of human life
(d) various aspects of insects.
Answer:
(d) various aspects of insects.

Question 102.
Which phylum of the animal kingdom is exclusively marine? (Orissa 2003, 2006)
(a) Porifera
(b) Arthropoda
(c) Echinodermata
(d) Molluscs
Answer:
(c) Echinodermata

Question 103.
Study of ticks and mites is ………….
(a) Acarology
(b) Entomology
(c) Malacology
(d) Carcinology
Answer:
(a) Acarology

Question 104.
Larva of mosquito is ………….
(a) maggot
(b) caterpillar
(c) grub
(d) none of these
Answer:
(d) none of these

Question 105.
Transparent hairs on catkins and caterpillars function to?
(a) Trap heat
(b) Trap moisture
(c) Reflect light
(d) Drink water
Answer:
(b) Trap moisture

Question 106.
Which of the following traits is not the characteristic of echinodermata?
(a) Water vascular system
(b) Trochophore larva
(c) Aristotle’s lantern
(d) Radial and indeterminate cleavage
Ans.
(b) Trochophore larva

Question 107.
Which of the following is pseudocoelomate?
(a) Nematode
(b) Chordate
(c) Echinodermata
(d) Arthropoda
Answer:
(a) Nematode

Question 108.
Which is not correct for sponges?
(a) Internal fertilization
(b) External fertilization
(c) Gemmule formation
(d) Gametes are formed from epidemial cells.
Answer:
(b) External fertilization

Question 109.
Triploblastic, schizocoelic and unsegmented soft bodied animals belongs to the phylum ……….. (J&K 1998)
(a) annelid
(b) mollusca
(c) nemathelminthes
(d) none of the above
Answer:
(b) mollusca

Question 110.
Which one of the following animals belongs to the phylum cnidaria? (J&K 1998)
(a) Silver fish
(b) Squid
(c) Jelly fish
(d) Echidna
Answer:
(c) Jelly fish

Question 111.
Palaemon (prawn) is a …………. (J & K 2000)
(a) fish
(b) insect
(c) soft shell mollusca
(d) crustacean
Answer:
(d) crustacean

Question 112.
Tapeworm occurs as a parasite in …………. (J&K 2001)
(a) liver
(b) stomach
(c) intestine
(d) all of these
Answer:
(c) intestine

Question 113.
What distinguishes an insect from crustacean? (J&K 2002. 2005)
(a) number of eyes
(b) arrangement of nerve cords
(c) number of appendages
(d) presence of wings.
Answer:
(c) number of appendages

Question 114.
Leeches are usually …………. (J&k 2005)
(a) herbivorous
(b) insectivorous
(c) carnivorous
(d) sanguvorous
Answer:
(d) sanguvorous

Question 115.
Wuchereria bancrofti is a common filarial worm. It belongs to the phylum …………. (J&K 2007)
(a) Platyhelminthes
(b) Nemathelminthes
(c) Annelid
(d) Coelenterate
Answer:
(b) Nemathelminthes

Question 116.
The dioecius animal is …………. (J&K 2008)
(a) Liver fluke
(b) Aurelia
(c) Tapeworm
(d) Earthworm
Answer:
(b) Aurelia

Question 117.
Malpighian tiihuies remove excretory products from ………….
(a) Mouth
(b) Haemolymph
(c) Oesophagus
(d) Alimentary canal
Answer:
(b) Haemolymph

Question 118.
Which of the following cell type is capable of giving rise to other ccli types in sponges?
(a) Pinacocytes
(b) Archaeocytes
(c) Thesocytes
(d) Collencytes
Answer:
(b) Archaeocytes

Question 119.
The infective stage of Entarnoeba histolytica is ………….
(a) cyst
(b) spore
(c) egg
(d) trophozoite
Answer;
(d) trophozoite

Question 120.
Gonads of Obelia occur in ………….
(a) on blastocyst
(b) in hydrula stage
(c) radial canals of medusa
(d) bases of entacies of medusa
Answer:
(c) radial canals of medusa

Question 121.
Which one of the following features is common to leech, cockroach and scorpion?
(a) nephridia
(b) ventral nerve cord
(c) cephalization
(d) antennae
Answer:
(b) ventral nerve cord

Question 122.
Excretory organs of flatworms are ………….
(a) Malpighian tubules
(b) Neprons
(c) Protonephridia
(d) Nnepridia
Answer:
(c) Protonephridia

Question 123.
Sea cucumbers belong to class ………….
(a) Echinoidea
(b) Ilolothuroidea
(c) Ophiuroidea
(d) Asteroidean
Answer:
(b) Holothuroidea

Question 124.
One of the following is a very unique feature of the mammals …………. (PMT2004, DPMT 1996. 1998)
(a) Homeothermy
(b) Presence of diaphragam
(c) Four chambered heart
(d) Rib cage
Answer:
(b) Presence of diaphragam

Question 125.
Uricotelisum is found in …………. (PMT 2004)
(a) Mammals and birds
(b) Fishes and fresh water protozoans
(c) Birds, reptiles and insects
(d) Frogs and toads
Answer:
(c) Birds, reptiles and insects

Question 126.
Which one of the following characters is not typical of the class mammalian? (PMT 2004)
(a) Thecodont dentition
(b) Alveolar lungs
(C) Ten pairs of cranial nerves
(d) Seven cervical vertebrate
Answer:
(c) Ten pairs of cranial nerves

Question 127.
Which one of the following in birds, indicates their reptilian ancestry? (PMT 2008)
(a) Two special chambers crop and gizzard in their digestive tract
(b) Eggs with a calcareous shell
(c) Scales on their hind limbs
(d) Four – chambered heart
Answer:
(e) Scales on their hind limbs

Question 128.
Which one of the following pairs of animals comprises ‘Jawless fishes’? (PMT2009)
(a) Mackerals and rohu
(b) Lampreys and hag fishes
(c) Guppies and hag fishes
(d) Lampreys and eels
Answer:
(b) Lampreys and hag fishes

Question 129.
Camouflage of chameleon is associated with …………. (AIIMS 1995)
(a) Chromoplast
(b) Chromosome
(c) Chromatophore
(d) Chromomere
Answer:
(c) Chromatophore

Question 130.
In fast swimming fishes, propulsion is due to …………. (A1IMS 2000)
(a) Pelvic fin
(b) Pectoral fin
(c) Dorsal fin
(d) Caudal fin
Answer:
(d) Caudal fin

Question 131.
Body temperature of cold blooded animals …………. (AIIMS 2000)
(a) Is constant
(b) Fluctuates with surrounding temperature
(c) Becomes very low (a) times
(d) Is very cold
Answer:
(b) Fluctuates with surrounding temperature

Question 132.
Which of the following is an egg laying mammal? (AUMS 2001)
(a) Kangaroo
(b) Platypus
(c) Penguin
(d) Whale
Answer:
(b) Platypus

Question 133.
Which of the following are uricotelic animals? (AIIMS 2002)
(a) rohu and frog
(b) camel
(c) lizard and crow
(d) earthworm and eagle
Answer:
(c) lizard and crow

Question 134.
Which of the following does not come under the class mammals? (AIIMS 2007)
(a) flying fox
(b) hedgehog
(c) manatee
(d) lamprey
Answer:
(d) lamprey

Question 135.
Which of the following is concerned with the formation of urea in rabbit? (BHU 1994,2007)
(a) spleen
(b) kidney
(c) blood
(d) liver
Answer:
(d) liver

Question 136.
Lateral line is present in …………. (BHU 1996)
(a) dog fish
(b) jelly fish
(c) star fish
(d) none of these
Answer:
(a) dog fish

Question 137.
The largest and heaviest mammals in the world is …………. (BHU 1994)
(a) blue whale
(b) elephant
(c) lion
(d) tiger
Answer:
(a) blue whale

Question 138.
Ichthyophis is a member of …………. (AIIMS 1997)
(a) amphibian
(b) mollusca
(c) reptilian
(d) annelid
Answer:
(a) amphibian

Question 139.
Renal portal system is absent in …………. (AIIMS 1998,2008)
(a) reptiles
(b) amphibians
(c) reptiles and amphibians
(d) birds
Answer:
(b) amphibian

Question 140.
Bone marrow is absent in …………. (AIIMS 2000)
(a) reptilian
(b) amphibian
(c) fishes
(d) birds
Answer:
(d) birds

Question 141.
Urea is formed in which organ of rabbit? (AIIMS 2001)
(a) liver
(b) kidney
(c) spleen
(d) lung
Answer:
(a) liver

Question 142.
Which of the following is not classified as amphibian? (AIIMS 2003)
(a) frog
(b) salamander
(c) tortoise
(d) ichthiophis
Answer:
(c) tortoise

Question 143.
The excretory material of bony fish is …………. (AIIMS 2004)
(a) urea
(b) protein
(c) ammonia
(d) amino acid
Answer:
(c) ammonia

Question 144.
Limbless amphibians belong to the order …………. (AIIMS 2007)
(a) anura
(b) urodela
(c) gymnophiona
(d) lissamphibia
Answer:
(c) gymnophiona

Question 145.
Which of the following snakes is non-poisonous?…………. (AIIMS 2007)
(a) cobra
(b) krait
(c) viper
(d) python
Answer:
(d) python

Question 146.
Placoid scales are found in …………. (AIIMS 2008)
(a) reptilia
(b) bony fishes
(c) cartilaginous fishes
(d) amphibians
Answer:
(c) cartilaginous fishes

Question 147.
Which of the following is a correct sequence of decreasing order of number of species? (AIIMS 2008)
(a) aves, pisces, reptiles, amphibians, mammals
(b) pisces, aves, reptiles, mammals, amphibians
(c) pisces, mammals, reptiles, amphibians, aves
(d) amphibians, aves, pisces, mammals, reptiles
Answer:
(b) pisces, aves, reptiles, mammals, amphibians

Question 148.
Excretory organ in Balanoglossus are …………. (DPMT 1991, 2008)
(a) nephridia
(b) antennary gland
(c) collar cord
(d) proboscis gland
Answer:
(d) proboscis gland

Question 149.
Reptiles share which of the following character with birds and mammals? (DPMT 1994)
(a) Amnion
(b) Homeothermy
(c) Diaphragm
(d) Hippie
Answer:
(a) Amnion

Question 150.
Cowper’s gland is present in …………. (DPMT 1996)
(a) Frog
(b) Earthworm
(c) Rabbit
(d) Cockroach
Answer:
(c) Rabbit

Question 151.
Which of the following pairs belong to the category of cold blooded animals? (DPMT 1998)
(a) bat & rat
(b) snakes & birds
(c) frog & snakes
(d) birds & monkey
Answer:
(c) frog & snakes

Question 152.
The character of birds without exception is …………. (UP-CPMT 1995)
(a) omnivorous
(b) beak without teeth
(c) flying wings
(d) lay eggs with calcareous shells
Answer:
(b) beak without teeth

Question 153.
Quill feathers (a) the base of quill wings are called …………. (UP-CPMT 1995)
(a) remiges
(b) coverts
(c) barbules
(d) down feathers
Answer:
(a) remiges

Question 154.
Which of the following pair of organisms are uricotelic? (UP-CPMT 2000)
(a) cartilaginous fishes and mammals
(b) reptiles and mammals
(c) birds and insects
(d) bony fishes and lizards
Answer:
(c) birds and insects

Question 155.
In the urinogenital organs of rabbit which one of following part is present in male but not in female? (UP-CPMT 2005)
(a) Urethra
(b) Fallopian tube
(c) Vagina
(d) Vas deferens
Answer:
(d) Vas deferens

Question 156.
Which one of the following features is present in some stage of the life history of all chordates? (UP-CPMT 2000)
(a) Blood flowing forward in dorsal blood vessel
(b) Pharyngeal gill slits
(c) A ventral hollow nerve cord
(d) Heart lying dorsally
Answer:
(b) Pharyngeal gill slits

Question 157.
Thoracic cage in rabbit is made up of …………. (UP-CPMT 2006)
(a) Ribs, vertebral column & diaphragm
(b) Ribs, diaphragm & sternum
(c) Vertebral column, diaphragm & sternum
(d) Ribs, vertebral column & sternum
Answer:
(d) Ribs, vertebral column & sternum

Question 158.
Which of the following has exoskeleton of scales and paired copulatory organ or penis? (UP-CPMT 2007)
(a) Sharks
(b) Lizards
(c) Urodela
(d) Urochordata
Answer:
(b) Lizards

Question 159.
Laterally compressed tail is found in ………….
(a) Fresh water snakes
(b) Terrestrial snakes
(c) Marine non-poisonous snakes
(d) Marine poisonous snakes
Answer:
(d) Marine poisonous snakes

Question 160.
Which of the following is characteristic feature of fishes?
(a) Tail and venous heart
(b) Epidermal scales and tail
(c) Venous heart and gills
(d) Epidcrmal scales and gills
Answer:
(c) Venous heart and gills

Question 161.
Similarity between fish and tadpole is ………….
(a) Scales
(b) Legs
(c) Lateral line
(d) Fins
Answer:
(c) Lateral line

Question 162.
Four-chambered heart is present in ………….
(a) frog
(b) crocodile
(c) shark
(d) lizard
Answer:
(b) crocodile

Question 163.
Right aortic arch is present in ………….
(a) reptiles only
(b) mammals only
(c) birds only
(d) both birds and mammals
Answer:
(c) birds only

Question 164.
Kidney of adult reptiles are …………. (AMU 1996)
(a) mesonephric
(b) metanephric
(c) pronephnc
(d) both (a) and (b)
Answer:
(b) metanephric

Question 165.
Marine fishes drink sea water to …………. (AMU 2001)
(a) meet their body salt requirements
(b) compensate loss of water from their body
(c) flush out nitrogenous wastes from their body
(d) achieve all of the above
Answer:
(d) achieve all of the above

Question 166.
In which of the following fishes the males have brood pouch, where eggs laid by the female remain till they hatch? (AMU 2002)
(a) Lung fish
(b) Climbing perch
(c) Salmon
(d) Sea horse
Answer:
(d) Sea horse

Question 167.
Match the names of branches of science listed under column-I with the field study given under column-II choose the choice which gives the correct combination of the alphabets …………. (AMU 2000)

(a) A – s, B – p, C – q, D – r
(b) A – q, B – s , C – r, D – q
(c) A – s, B – q, C – p, D – r
(d) A – p, B – s, C – r, D – q
Answer:
(a) A – s, B – p, C – q, D – r

Question 168.
Identify the edible fresh water teleosts …………. (AMU 2001)
(a) Sharks
(b) Rays and skates
(c) Hilsa ilisha
(d) Catla catla
Answer:
(d) Catla catla

Question 169.
Turtles are …………. (AMU 2002)
(a) Pisces
(b) Reptiles
(c) Molluscans
(d) Arthropods
Answer:
(b) Reptiles

Question 170.
Harversian systems are found in the bones of …………. (AMU 2002)
(a) Pigeon
(b) Panther
(c) Pipe fish
(d) Python
Answer:
(b) Panther

Question 171.
Choose the correct combination of alphabets which matches the zoological names given under column I with their common names given under column-II

(a) A – F,B – G,C – E,D – H
(b) A – G,B – E,C – H,D – F
(c) A – F,B – E,C – H,D – G
(d) A – F,B – E,C – G,D – H
Answer:
(c) A – F,B – E,C – H,D – G

Question 172.
Which of the following statements is true? (AMU 2003)
(a) All chordates are vertebrates
(b) All vertebrates are chordates
(c) Invertebrates possess a tubular nerve cord
(d) Non-chordates have a vertebral column
Answer:
(b) All vertebrates are chordates

Question 173.
Choose the cat fish from the following. (AMU 2004)
(a) Cirrhina mriga / a
(b) Wa / logo atiti
(c) Lobeo rohita
(d) Catia calla
Answer:
(b) Wallago allu

Question 174.
A four chambered heart is not found in …………. (AMU 2004)
(a) Mammals
(b) Birds
(c) Snake
(d) Crocodile
Answer:
(c) Snake

Question 175.
Calotes versicolor is a …………. (AMU 1997)
(a) House lizard
(b) Rock lizard
(c) Garden lizard
(d) Flying lizard
Answer:
(c) Garden lizard

Question 176.
Scientific name of king cobra is …………. (AMU 2002)
(a) Naja naja
(b) Amphibians
(c) Naja Hannah
(d) Vipera russelli
Answer:
(c) Naja Hannah

Question 177.
Branch of zoology dealing with the study of amphibians and reptiles is called …………. (AMU 2003)
(a) Ichthyology
(b) Ornithology
(c) Herpetology
(d) Malacology
Answer:
(c) Herpetology

Question 178.
Adaptation of colour vision is found in …………. (AMU 2006)
(a) Mammals
(b) Aves
(c) Reptiles
(d) All of these
Answer:
(d) All of these

Question 179.
Epidemial scale is the characteristic feature of class reptilian, which of the following class is without epidermal scale? (AMU 2006)
(a) Fish
(b) Aves
(c) Mammals
(d) Amphibians
Answer:
(d) Amphibians

Question 180.
Duck-billed platypus is a connecting link between …………. (AMU 2007)
(a) Reptile and bird
(b) Living and non-living
(c) Reptile and mammal
(d) Echinodermata and chordate
Answer:
(c) Reptile and mammal

Question 181.
Which of the following is a egg laying mammal? (J&K 2005)
(a) Dolphin
(b) Platypus
(c) Whale
(d) Walrus
Answer:
(b) Platypus

Question 182.
In sharks, one of the following is absent …………. (J&K 2008)
(a) Claspers
(b) Placoid scales
(c) Cartilaginous endoskeleton
(d) Air bladder
Answer:
(d) Air bladder

Question 183.
Which one of the following animals belongs to cyclostomata? (J&K2008)
(a) Channa
(b) Loris
(c) Dodo
(d) Pertomyzon
Answer:
(d) Pertomyzon

Question 184.
Which of the following is dominant in desert?
(a) Lizard
(b) Tiger
(c) Leopard
(d) Hyla
Answer:
(a) Lizard

Question 185.
Two examples in which the nitrogenous wastes are excreted from body in the form of uric acid are ………….
(a) birds and lizards
(b) insects and bony fishes
(c) mammals and molluscs
(d) frogs and cartilaginous fishes
Answer:
(a) birds and lizards

Question 186.
The arrangement of ear ossicles in mammalian ear is ………….
(a) stapes, malleus, incus
(b) malleus, incus, stapes
(c) incus, malleus, stapes
(d) columella, inatleus, incus
Answer:
(b) malleus, incus, stapes

Question 187.
Snake has ………….
(a) movable eyelids
(b) immovable eyelids
(c) no cyclids
(d) eyelids in pouches
Answer:
(b) immovable eyelids

Question 188.
Which among these is correct combination of aquatic mammals? (NEET 2017)
(a) Dolphins, Seals, Trygon
(b) Whales, Dolphin, Seals
(c) Trygon, Whales, Seals
(d) Seals, Dolphin, Sharks
Answer:
(b) Whales, Dolphin, Seals

Question 189.
In case of poriferance, the spongocoel is lined with flagellated cells called …………. (NEET 2017)
(a) Oscula
(b) Coenocytes
(c) Mesenchymal cells
(d) Ostia
Answer:
(b) Coenocytes

Question 190.
Which is the National Aquatic animal of India? (NEET 2016)
(a) River Dolphin
(b) Blue whale
(c) Sca horse
(d) Gangetic shark
Answer:
(a) River Dolphin

Question 191.
An important characteristic that Hernichordates share with chordates is …………. (NEET2O17)
(a) Ventral tubular nerve chord
(b) Pharynx with gill slits
(c) Pharynx without gill slits
(d) Absence of notochord
Answer:
(a) Ventral tubular nerve chord

Samacheer Kalvi 11th Bio Zoology Kingdom Animalia  Additional Questions & Answers

Multiple Choice Question And Answer
Choose the correct answer

Question 1.
Which of the following has loose aggregates of cells without tissues?
(a) cnidarians
(b) flatworms
(c) sponges
(d) echinodenns
Answer:
(c) sponges

Question 2.
Which of the following has open type of circulation?
(a) frogs
(b) garden lizard
(c) man
(d) cockroach
Answer:
(d) cockroach

Question 3.
Which of the following is advantageous for the animals in locomotion, food capture etc.?
(a) asymmetrical
(b) radially symmetrical
(c) biradially symmetrical
(d) bilaterally symmetrical
Answer:
(d) bilaterally symmetrical

Question 4.
Which of the following restricts the free movement of internal organs?
(a) acoelom
(b) pseudocoelom
(c) schizocoelom
(d) enterocoelom
Answer:
(a) acoelom

Question 5.
Radiata include
(a) Diploblastic and bilaterally symmetrical animals
(b) Triploblastic and radially symmetrical animals
(c) Diploblastic and radially symmetrical animals
(d) Triploblastic and bilaterally symmetrical animals
Answer:
(c) Diploblastic and radially symmetrical animals.

Question 6.
The minute pores on the body of sponges are called
(a) oseuliem
(b) ostia
(c) choanocytes
(d) spongocoel
Answer:
(b) ostia

Question 7.
Which of the following statements is correct?
(a) Polyp forms are free-living
(b) Medusa forms are sessile
(c) Medusa produces gametes
(d) Polyp reproduces sexually
Answer:
(c) Medusa produces gametes

Question 8.
Which of the following is the adaptation of flatworms for the endoparasitic mode of life?
(a) They are dorsoventrally flattened
(b) They have hooks, suckers or both
(c) Their body is not segmented
(d) They reproduce sexually
Answer:
(b) They have hooks, suckers or both

Question 9.
Sexual dimorphism is seen in –
(a) Sycon
(b) Hydra
(c) Liver flukes
(d) Ascaris
Answer:
(d) Ascaris

Question 10.
Which of the following shows metamerically segmented body?
(a) Aschelminthes
(b) Annelida
(c) Arthropoda
(d) Platyhelminthes
Answer:
(b) Annelida

Question 11.
Which of the following is the characteristic feature of the phylum Arthropoda?
(a) They have segmented legs
(b) They have collablasts for food capture
(c) They are end oparasites of animals
(d) They do not have chitinous exoskeleton
Answer:
(a) They have segmented legs

Question 12.
Which of the following is the rasping organ of molluscs found in the mouth?
(a) radula
(b) pallium
(c) misceral mass
(d) mantle
Answer;
(a) radula

Question 13.
Which of the following is bilaterally symmetrical in larval stages and radially symmetrical in adult?
(a) Molluscs
(b) Echinoderms
(c) Arthropods
(d) Annelids
Answer:
(b) Echinoderms

Question 14.
Which of the following has the anterior proboscis, collar and trunk?
(a) Ascidian
(b) Star fish
(c) Sea cucumber
(d) Balanoglossus
Answer:
(d) Balanoglossus

Question 15.
Urochordate means
(a) Chordates which have notochord in the head region
(b) Chordates which have notochord in the tail region of larval forms
(c) Chordates which have notochord in the tail region of adults
(d) Chordates which have no notochord
Answer:
(b) Chordates which have notochord in the tail region of larval forms.

Question 16.
Which of the following loses all the chordate characters in the adult stage?
(a) Cephalochordates
(b) Hemichordates
(c) Tunicates
(d) Chordates
Answer:
(c) Tunicates

Question 17.
Which of the following has cartilagenous endoskeleton with notochord?
(a) Exocoetus
(b) Labeo
(c) Hyla
(d) Scolidon
Answer:
(d) Scolidon

Question 18.
Which is the class of animals adapted for dual mode of life?
(a) Pisces
(b) Amphibia
(c) Reptilia
(d) Mammalia
Answer:
(b) Amphibia

Question 19.
Which of the following is the flight adaptation of birds?
(a) Pneumatic bones and strong flight muscles
(b) Homeothermic condition
(c) Migration to distant places
(d) Presence of homy covering on the beak
Answer:
(a) Pneumatic bones and strong flight muscles

Question 20.
Which of the following are truly terrestrial animals?
(a) Lung fishes
(b) Amphibians
(c) Mammals
(d) Reptiles
Answer:
(d) Reptiles

II Give Reasons
Question 1.
Closed type of circulation is advanced.
Answer:
Closed type of circulation is found in higher organisms like prochordates and vertebrates. The invertebrates have open type of circulation except annelids. The closed type of circulation is advanced because blood flows in blood vessels. There is a clear separation of oxygenated and deoxygenated blood.

Question 2.
Pseudocoelomates are more advanced than acoelomates.
Answer:
Acoelomates do not have body cavity. Their body is solid and hence the movement of internal organs is restricted. Pseudocoelomates have pseudocoelomic fluid in the pseudocoelom. It acts as a hydrostatic skeleton and allows free movement of visceral organs and circulation of nutrients.

Question 3.
Sponges are primitive animals.
Answer:
Sponges have cellular grade of organisation. Tissues are not formed. The cells are loosely arranged. The division of labour is found among the group of cells.

Question 4.
Bioluminence is advantageous to ctenophores.
Answer:
Ctenophores are exclusively marine. They emit light. It helps the animals in finding food, mate and escape from the predators.

Question 5.
Tape worm and liver fluke are not destroyed by the digestive juices secreted by human beings.
Answer:
The body of helminth parasites is covered by a thick covering called tegument. This protects the worms. They have hooks, suckers or both for attachment.

Question 6.
All vertebrates are chordates but all chordates are not vertebrates.
Answer:
All Vertebrates have notochord during embryonic development. Later it is replaced by the vertebral column. But in lower chordates like prochordates, vertebral column is not present. They have only notochord in the adult or larval stage. Hence it is said that all vertebrates are chordates but all chordates do not have vertebrate characters

Answer the following

Questions 1.
Distinguish invertebrates and chordates.
Answer:
Invertebrates:

• The major group of animals which do not have notochord or vertebral column are Invertebrates.
• These are lower animals.

Chordates:

• The major group of animals which have notochord or vertebral column are chordates.
• These are higher animals.

Question 2.
Distinguish between invertebrates and vertebrates.
Answer:
Invertebrates:

• The major group of animals which do not have notochord or vertebral column are Invertebrates.
• These are lower animals.

Vertebrates:

• The major group of animals which have vertebral column are vertebrates
• These are higher animals.

Question 3.
What are choanocytes?
Answer:
The inner layer of sponges have a special type of cells called choanocytes. These flagellated collar cells create and maintain water flow through the sponge. It helps in respiration and digestion.

Question 4.
Distinguish between open type of circulation and closed type of circulation.
Answer:
Open Type of circulation:

• The circulation in which blood remains filled in tissue spaces is known as open type of circulation.
• This is seen in lower organisms, e.g. arthropods, molluscs echinoderms and urochordates.

Closed type of circulation:

• The circulation in which blood flows inside the blood vessels is known as closed type of circulation.
• This is seen in higher organisms, e.g. Annelids, cephalochordates and vertebrates.

Question 5.
Distinguish between Diploblastic animals and triploblastic animals.
Answer:
Diploblastic animals:

•  The animals in which the cells are arranged in two embryonic layers, the ectoderm and endoderm are called diploblastic animals.
• These are lower organisms, e.g. Cnidaria and ctenophora

Triploblastic animals:

• The animals in which the cells are arranged in three embryonic layers, the ectoderm, mesoderm and endoderm are called triploblastic animals.
• These are higher organisms, e.g. Platyhelminthes to mammalia.

Question 6.
What are asymmetrical animals?
Answer:
The animals which lack a definite body plan and any plane passing through the center of the body does not divide them into two equal halves are known as asymmetrical animals, e.g. Sponges.

Question 7.
What is radial symmetry?
Answer:
When any plane passing through the central axis of the body divides an organism into two identical parts, it is called radial symmetry, e.g. Cnidarian.

Question 8.
What is bilateral symmetry?
Answer:
The symmetry in which the animals have two similar halves on either side of the central place is bilateral symmetry, e.g. Flatworms and annelids.

Question 9.
What is biradial symmetry?
Answer:
The symmetry in which the animals have two planes of symmetry, longitudinal and sagittal axis and longitudinal and transverse axis is biradial symmetry, e.g. Ctenophores.

Question 10.
What are the advantages of bilateral symmetrical animals?
Answer:
The bilaterally symmetrical animals can seek food, locate mates, escape from predators and move more efficiently. These animals have dorsal ventral sides and anterior, posterior ends, right and left sides. They exhibit cephalization with sense organs and brain at the anterior end of the animal.

Question 11.
What are acoelomates?
Answer:
The animals which do not possess a body cavity are called acoelomates. The body is solid without perivisceral cavity. These have restricted free movement of internal organs, e.g. Flatworms.

Question 12.
What are pseudocoelomates?
Answer:
The animals which have the body cavity that is not fully lined by the mesodermal epithelium are called pseudocoelomates. The pseudocoel is filled with pseudocoelomic fluid. It acts as a hydrostatic skeleton and allows free movement of visceral organs and circulation of nutrients e.g. Roundworms.

Question 13.
What are eucoelomates?
Answer:
Eucoelomates are the animals which have true coelom that develops with the mesoderm and is lined by mesodermal epithelium called peritonium.

Question 14.
Distinguish between schizocoelomates and enterocoelomates
Answer:
Schizocoelomates:
In schizocoelometes, the body cavity is formed by splitting of mesoderm, e.g. Annelids, arthropods and molluscs.

Enterocoelomates:
In enterocoelomates, the body cavity is formed from the mesodermal pouches of archenteron. e.g. Echinoderms, hemichordates and chordates.

Question 15.
Distinguish between parazoa and eunietazoa.
Answer:
Parazoa:
These include multicellular animals whose cells are loosely arranged without the formation of tissues or organs, e.g. Sponges

Eumetazoa:
These include multicellular animals with well defined tissues, organs and organ systems.

Question 16.
Distinguish between radiata and bilateria.
Answer:
Radiata:

• These include radially symmetrical animals.
• There are diploblastic e.g., Cnidarians and ctenophores

Bilateria:

• These include bilaterally symmetrical animals.
• There are triploblastic. e.g. Flat – worms

Question 17.
Distinguish between protostomia and deuterostomia.
Answer:
Protostomia:

• These include the eumetazoans in which embryonic blastopores develops into mouth.
• Acoelomata, pseudocoelomata and schizocoelomata are the three sub divisions of this division.

Deuterostomia:

• These include the cutnetazoans in which embryonic blastopore develops into anus.
• Enterocoelmata is the only one subdivision of this division.

Question 18.
What is canal system?
Answer:
The water transport system in sponges through which water enters through minute opores and goes out through the large opening called osculum. It helps in nutrition, circulation, respiration and excretion.

Question 19.
Distinguish between ostia and osculum.
Answer:
Ostia:

• The minute pores lining the body wall of sponges are called ostia.
• Water enters through ostia.

Osculum:

• The large opening in sponges is called osculum.
• Water goes out through osculum.

Question 20.
What spongocoel?
Answer:
The central cavity of the sponges is called spongocoel.

Question 21.
What are choanocytes?
Answer:
Choanocytes are the collar cells lining the spongocoel and the canals of sponges. These are helpful in creating water current in sponges.

Question 22.
Distinguish between asexual reproduction and sexual reproduction.
Answer:
Asexual Reproduction:

• The reproduction without involvement of gametes is called Asexual Reproduction.
• Zygote is not formed.

Sexual Reproduction:

• The reproduction with the involvement of gametes is called sexual reproduction.
• Zygote is formed by the process called fertilization.

Question 23.
Name the larvae of sponges.
Answer:
Parenchymula and amphiblastula.

Question 24.
What is indirect development?
Answer:
The development with different types of larval stages is called indirect development.

Question 25.
What is holozoic nutrition?
Answer:
The nutrition in which solid food materials are taken in by animals is called holozoic nutrition.

Question 26.
What are cnidocytes or cnidoblasts or nematocysts?
Answer:
The stinging cells found on the tentacles of cnidarians are called cnidocytes or cnidoblasts or nematocysts. They are useful for anchorage, defense and capturing prey.

Question 27.
What is coelenteron?
Answer:
The central visceral cavity of cnidarians is called coelenteron.

Question 28.
Distinguish between polyp and medusa.
Answer:
Polyp:

• The sessile body form of cnidarians is called polyp.
• It is the asexual generation

Medusa:

• The free living body form of cnidarians is called medusa.
• It is the asexual generation

Question 29.
What is metagenesis or Alternation of generation?
Answer:
The cnidarians exhibit sexual and asexual forms that alternate with each other. This is called metagenesis or Alternation of generation.

Question n 30.
Name the larva of cnidarians?
Answer:
Planula larva.

Question 31.
What are lasso cells or colloblasts?
Answer:
The special cells of ctenophores which helps in food capture are lasso cells or colloblasts.

Question 32.
Name the larva of ctenophores?
Answer:
Cydippid larva.

Question 33.
What are solanocytes?
Answer:
The specialized excretory cells of flatworms, flame cells are called solanocytes.

Question 34.
What are the larvae of flatworms?
Answer:
Miracidium, Sporocyst, redia, cercaria and metacercaria.

Question 35.
What is regeneration?
Answer:
The ability to regrow the lost parts is called regeneration, e.g. Planaria.

Question 36.
What is metamerism? –
Answer:
The body of annelids are divided into segments. This phenomenon is known as metamerism.

Question 37.
Name the respiratory pigments of annelids?
Answer:
Haemoglobin and chlorocruorin.

Question 38.
Name the larva of annelids.
Answer:
Trochophore larva.

Question 39.
What is moulting or ecdysis?
Answer:
The chitinous exoskeleton of arthropods is shed periodically. This process is known as moulting or ecdysis.

Question 40.
What are the respiratory organs of arthropods?
Answer:
Gills, Book gills, Book lungs and trachea.

Question 41.
Name the sensory organs of arthropods?
Answer:
Antennae, Simple and compound eyes and statocysts.

Question 42.
What are ctenidia?
Answer:
The feather like gills of molluscs are called ctenidia.

Question 43.
What is radula?
Answer:
The rasping organ found in the mouth of molluscs is called radula.

Question 44.
What is the function of ospharidium?
Answer:
Ospharidium are helpful to test the purity of water.

Question 45.
Name the respiratory pigment of molluscs.
Answer:
Haemocyanin, a copper containing pigment.

Question 46.
Name the larva of molluscs?
Answer:
Veliger larva.

Question 47.
What is water vascular system?
Answer:
The system which helps in nutrition and respiration in echinoderms is called water vascular system. Water enters into the body through special organs.

Question 48.
Name the larva of hemichordates?
Answer:
Tornaria larva.

Question 49.
What are urochordates?
Answer:
The chordates which have notochord only in the tail region of the larval stage are called urochordates e.g. Ascidian.

Question 50.
Distinguish between Agnatha and Gnathostomata.
Answer:
Agnatha:

• These include jawless fish-like aquatic vertebrates.
• They do not have paired appendages.

Gnathostomata:

• These include jawed vertebrates.
• They have paired appendages.

Question 51.
What are poikilothermic?
Answer:
The animals which change their body temperature according to the environment are called poikilothermic.

Question 52.
What is anadromous migration?
Answer:
The migration of marine fishes to fresh water body like rivers for spawning is known as anadromous migration.

Question 53.
Distinguish between oviparous and viviparous animals.
Answer:
Oviparous animals:

• The egg laying animals are known as oviparous animals.
• They lay eggs containing yolk for embryonic development e.g. birds.

Viviparous animals:

• The animals which give birth to young ones are called viviparous animals.
• The developing embryo derives nutrients from the parent, e.g. man

Question 54.
What are Ammonotelic animals?
Answer:
The animals which excrete ammonia dissolved in water are called ammonotelic animals. More water is spent, e.g. fishes.

Question 55.
What are ureotelic animals?
Answer:
The animals which excrete urea along with water are called ureotelic animals. Less water is spent e.g. man.

Question 56.
What are urecotelic animals?
Answer:
The animals which excrete uric acid in the form of pellets are called urecotelic animals. Very less water is spent e.g. birds.

Question 57.
Distinguish between hibernation and aestivation.
Answer:
Hibernation:

• The dormancy period for animals during winter is called hibernation.
• It is known as winter sleep.

Aestivation:

• The dormancy period for animals during summer is called Aestivation.
• It is know as summer sleep.

Question 58.
Distinguish between cleidoic eggs and non-cleidoic eggs.
Answer:
Cleidoic eggs:

• The eggs which have a thick and hard outermost shell are cleidoic eggs.
• This is a terrestrial adaptation, e.g., Reptiles and birds.

Non Cleidoic eggs:

• The eggs Which do not have a protective shell are non – Cleidoic eggs
• This is seen in aquatic animals, e.g., Fishes, amphibians.

Question 59.
What is rhamphotheca?
Answer:
The homy covering on the beak of birds is called rhamphotheca.

Question 60.
Name some flightless birds.
Answer:
Ostrich, kiwi and penguin.

Question 61.
Distinguish between poikilothermic and homeothermic.
Answer:
Poikilothermic:

• The animals which change their body temperature according to the environment are called poikilothermic animals.
• These cold blooded animals, e.g., fishes, amphibians and reptiles.

Homeothermic:

• The animals which maintain constant body temperature irrespective of environmental changes are called homeothermic animals.
• These are warm blooded animals, e.g., birds and mammals.

Question 62.
Explain various patterns of organisation in animals.
Answer:
Animals exhibit different patterns of organisation:
Cellular level of organisation :

• Cells are loosely arranged without the formation of tissues.
• There is division of labour among the cells, e.g., sponges.

Tissue level of organisation :

• Cells which perform similar function are grouped into tissues.
• The tissues perform a common function, e.g., cnidarians.

Organ level of organisation :
Different kinds of tissues aggregate to form an organ to perform a specific function e.g., flatworms and other hyper phyla.

Organ system level of organisation :

• The tissues are organised to form organs and organ systems.
• All the organ system function in a coordinated manner.

Question 63.
Explain symmetry in animals.
Answer:
Symmetry is the body arrangement in which parts lie on opposite side of the axis are identical. If the animals lack a definite body plane or irregular shaped and any plane passing through the center of the body does not divide them into two equal halves, these are known as asymmetrical, e.g., sponges, adult gastropods.

When any plane passing through the central axis of the body divides an organism into two equal parts, it is known as radial symmetry. They have a top and bottom side, e.g., cnidarians. Echinoderms have five planes of symmetry and show pentamerous radial symmetry. Animals which have two pairs of symmetrical sides are biradially symmetrical. Animal which have two similar halves on either side of the control plane show bilateral symmetry.

Question 64.
Classify animals based on coelom.
Answer:
The cavity between the body wall and the gut wall is called coelom. If the animals do not have coelom, they are called acoelomates. e.g., flatworms. In some animals, the body cavity is not fully lined by the mesodermal epithelium. The mesoderm is formed as scattered pouches between the ectoderm and endoderm. Such a body cavity is called a pseudocoel. The animals which have pseudocoel e.g. round worms.

If the coelom develops within the mesoderm and is lined by mesodermal epithelium it is called eucoelom. The animals which have true coelom are called eucoelomates. If the body cavity is formed by splitting of mesoderm, the animals are called schizocoelomates e.g., Annelids, arthropods and molluscs. If the body cavity is formed from the mesodermal ‘ pouches of archenteron, the animals are called enterocoelomate animals, e.g., echinoderms, hemichordates and chordates.

Question 65.
Classify the animal kingdom.
Answer:

Question 66.
Write the general characters of the phylum porifera.
Answer:

• They are aquatic, asymmetrical.
• They have pores all over the body.
• They are multicellular with cellular level of organisation. Tissues are not formed.
• They have canal system for circulation of water.
• They have skeleton made of calcareous or siliceous spicules.
• Nutrition is holozoic and digestion intracellular.
• Asexual reproduction by fragmentation and gemmule formation.
• Indirect development with parenchymula and amphiblastula larvae, e.g., Sycon and Spongilla.

Question 67.
Write the general characters of the phylum cnidaria.
Answer:

• The cnidaria are aquatic, radially symmetrical and diploblastic.
• The tentacles have stinging cells called cnidocytes or cnidoblasts or nematocysts.
• They exhibit tissue level of organisation.
• They have a central gastrovascular cavity called coelenderon.
• Digestion is by both extracellular and intracellular.
• Alternation of generation is seen in cnidarians which have polyp and medusa forms.
• Development is indirect with planula larva e.g. Physalia.

Question 68.
Write the general characters of the phylum ctenophora.
Answer:

• The animals are marine, diploblastic and radially symmetrical.
• They have eight external rows of ciliated comb plates which help in locomotion.
• Bioluminescence is seen.
• They lack nematocysts but have lasso cells which help in food capture.
• Digestion is by both extracellular and intracellular.
• Sexual reproduction is seen.
• Fertilization is external and development is indirect.
• Cydippid larva is seen, e.g., Pleurobrachia.

Question 69.
Write the general characters of flatworms.
Answer:

• The flatworms are flat, bilaterally symmetrical, triploblastic animals.
• These are acoelomates with organ system level of organisation.
• They are endoparasites. They have hooks or suckers or both.
• They show pseudosegmentation.
• Flame cells or solanocytes are the excretory cells.
• Sexes are not separate.
• Fertilization is internal. Development is indirect with many larval stages like miracidium sporocyst, redia, cercaria and metacercaria. e.g., Taenia and liver fluke.

Question 70.
Write the general characters of the phylum aschelminthes.
Answer:

• The body of these worms is circular.
• They are free living or parasite.
• They are triploblastic, pseudocoelomates with organ system level of organisation.
• Body is covered by cuticle.
• Digestive system is complete with mouth, pharynx and anus.
• Excretory system consists of rennet glands.
• Sexes are separate. Sexual dimorphism is seen.
• Fertilization is internal.
• Development may be direct or indirect, e.g., Ascaris.

Question 71.
Write the general characters of the phylum annelida.
Answer:

• They are aquatic or terrestrials, free living or parasitic.
• They are triploblastic, bilaterally symmetrical, schizocoelomates with organ system level of organisation.
• The body is metamerically segmented.
• Longitudinal and circular muscles help in locomotion.
• Closed type of circulation is seen.
• Respiratory pigments are present.
• Sexual reproduction is seen. Development is direct or indirect with a trochophore larva e.g., earthworm.

Question 72.
Write the general characters of the phylum arthropoda.
Answer:

• These are bilaterally symmetrical, triploblastic, schizocoelomate, segmented animals.
• They have organ system grade of organisation.
• They have jointed appendages.
• Body is covered by chitinous exoskeleton. Body is divided into head, thorax and abdomen.
• Body cavity is filled with colourless blood. It is called haemocoel.
• Respiratory organs are gills, book gills, book lungs, trachea.
• Open type circulation is seen.
• Sense organs are present.
• Fertilization is internal, e.g., Limulus and insects.

Question 73.
Write the general characters of the phylum mollusca.
Answer:

• Molluscs are terrestrial or aquatic with organ system level of organisation.
• They are triploblastic, bilaterally symmetrical, coelomate animals.
• Body is divided into head, foot and visceral hump.
• The digestive system is complete.
• Nephridia are the excretory organs.
• Open type of circulatory system is seen.
• Blood contains copper containing respiratory pigment called hemocyanin.
• They are oviparous.
• Development is indirect with a veliger larva, e.g., Pila and Octopus.

Question 74.
Write the general characters of the phylum echinodermata.
Answer:

• The adults are radially symmetrical but the larvae are bilaterally symmetrical.
• They have mesodermal endoskeleton of calcareous ossicles called spines.
• Water vascular system is present.
• Tube feet are the organs of locomotion, respiration and capture of food.
• The digestive system is complete with mouth on the ventral side and anus on the dorsal side.
• Excretory organ are absent.
• Open type of circulatory system is present.
• Reproduction is by sexual method.
• Fertilization is external.
• Indirect development with bilaterally symmetrical larval forms, e.g., starfish.

Question 75.
Write the general characters of hemichordata.
Answer:

• The Hemichordates have both invertebrate and vertebrate characters.
• They are worm like, tubiculous animals.
• They are bilaterally symmetrical, triploblastic coelomate animals with organ system level of organisation.
• The body is divided into proboscis, collar and trunk.
• They are ciliary feeders.
• Circulatory system is simple and open.
• Excretion is by a single proboscis gland or glomerulus situated in the proboscis.
• Sexes are separate.
• Fertilization is external.
• Development is indirect with tomaria larva, e.g., Balanoglossus.

Question 76.
Write the general characters of urochordates or tunicates.
Answer:

• They are marine, sessile, pelagic or free swimming.
• Body is unsegmented and covered by a test or tunic.
• Ault forms are sac like.
• Coelom is absent.
• Notochord is present only in the tail region of the larval stage.
• Heart is ventral and tubular.
• Nerve cord is present only in the larval stage.
• They are hermaphrodites and development is indirect with a free swimming tadpole larva.
• Retrogressive metamorphosis is seen e.g. Ascidia.

Question 77.
Write the general characters of cephalochordates.
Answer:

• They are marine found in shallow waters.
• They lead a burrowing mode of life.
• They are fish like with notochord, nerve cord and pharyngeal gill slits throughout their life.
• Closed type of circulatory system is seen without heart.
• Excretion is by protonephridia.
• Sexes are separate.
• Fertilization is external.
• Development is indirect with a larva e.g. Amphioxus.

Question 78.
Write the general character of the subphylum vertebrata.
Answer:

• The vertebrates have notochord during embryonic stage. In adult, it is replaced by vertebral column. ,
• They have paired appendages.
• The skin is covered by protective skeleton like scales, feathers, hairs, claws, nails etc.
• Respiration is by gills, skin, buccopharyngeal cavity and lungs.
• They have a ventral heart with two, or three or four chambers.

Question 79.
Classify the subphylum Vertebrata.
Answer:

Question 80.
Write the general characters of the class Cyclostomata.
Answer:

• They are primitive, poikilothermic, jawless aquatic vertebrates.
• Body is slender and eel-like with six to fifteen pairs of gill slips.
• Mouth is circular without jaws and suctorial.
• Heart is two chambered. Closed types of circulation is seen.
• No paired appendages.
• Cranium and vertebral column are cartilaginous, e.g. Petromyzon.

Question 81.
Write the general characters of the class chondrichthyes.
Answer:

• They are cartilaginous marine fishes. .
• Notochord is persistent throughout life.
• The skin is covered by dermal placoid. scales.
• Caudal fin is heterocercal.
• Jaws are powerful.
• Respiration is by lamelliform gills without operculum.
• Mesonephric kidneys are present.
• They have two chambered heart.
• They are ureotelic.
• They are poikilothermic and viviparous.
• Sexes are separate and fertilization is internal, e.g., Scoliodon.

Question 82.
Write the general characters of the class osteichthyes.
Answer:

• They are marine or freshwater fishes.
• They have bony endoskeleton.
• The body is spindle-shaped.
• The skin is covered by ganoid, cycloid or ctenoid scales.
• Respiration is by four pairs of gills covered by operculum.
• Air bladder is present which helps in gaseous exchange (lung fishes) for maintaining buoyancy.
• They have two chambered heart.
• They have mesonephric kidneys and ammonotelic.
• Lateral line sense organ is present.
• External fertilization is seen and most forms are oviparous e.g. Labeo and Catla.

Question 83.
Write the general characters of the class Amphibia.
Answer:
Amphibians are adapted to live both in water and on land.

• They are poikilothermic.
• The body is divisible into head and trunk.
• They have two pairs of limbs.
• The skin is smooth or rough, moist, pigmented and glandular.
• Eyes have eyelids.
• Respiration is by gills, lungs and through the skin.
• Heart is three chambered.
• Kidneys are mesonephric and ureotelic.
• Sexes are separate and fertilization is external.
• They are oviparous and development is indirect, e.g. Bufo and Rana.

Question 84.
Write the general characters of the class Reptilia.
Answer:

• They are mostly terrestrial.
• The body is covered by dry and comified skin with scales.
• They have three chambered heart.
• They are poikilothermic.
• They are oviparous and they lay cleidoic eggs.
• They have metanephric kidney and are uricotelic.
• Fertilization is internal, e.g. Chelone and Chameleon.

Question 85.
Write the general characters of the class Aves.
Answer:

• They have feathers on the body.
• The forelimbs are modified into wings. The hind limbs are adapted for walking, running, swimming and perching.
• The skin has oil gland.
• The exoskeleton consists of feathers, scales, claws on legs.
• The bones are pneumatic.
• Respiration is by lungs with air sacs.
• The heart is four chambered with right systemic arch.
• They are homeothermic.
• Migration and parental care are seen.
• Urinary bladder is absent.
• Females have only left ovary.
• They are oviparous. The eggs are cleidoic.

Question 86.
Write the general characters of the class Mammalia.
Answer:

• The body is covered by hairs.
• They have mammary glands.
• They have two pairs of limbs adapted for walking, running, climbing, burrowing, swimming and flying.
• The skin has sweat glands, scent glands and sebaceous glands.
• Exoskeleton includes homy epidermal horns, spines, claws, nails, hooves and bony dermal plates.
• They have thecodont, heterodont and diphyodont teeth.
• External ears or pinnae are present.
• Heart is four chambered with left systemic arch.
• RBCs are non-nucleated.
• They have metanephric kidneys and they are ureotelic.
• They are homeothermic.
• Sexes are separate and fertilization is internal e.g. Platypus, kangaroo, elephants and man.

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