Category: Class 11

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Tamilnadu Samacheer Kalvi 11th English Solutions Supplementary Chapter 5 The Singing Lesson

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Warm up

Question 1.
What are all the factors that influence our moods?
Human beings are bundles of emotions, A small angry look from a friend, a scolding from a teacher one admires, a taunt from mom could influence our moods the whole day.

Question 2.
How do you behave under the spells of different moods?
Answer:
When good things keep happening, we are happy. If we don’t get an easy question paper or the expected questions don’t appear we feel quite upset. If a close friend becomes angry, instead of analysing what caused it, we feel dejected. When centum is the goal, even 99% of marks disappoints us.

Question 3.
Do you think it is important not to be swayed by every passing mood?
Answer:
Yes, we should not be swayed by every passing emotions. But we are the sum total of our experiences. What ever bitter sweet experiences that occur do influence us. Latest researches . say that the food we eat, weather, clothes, the colour around us and punishing work schedules can adversely affect our moods. It would be ideal if we don’t allow external circumstances to influence our equanimity of mind and the ability to stay focused on our goals.

 

Question 4.
Suggest some ways by which we can maintain a calm temperament under all circumstances.
Answer:
Early morning walks, meditation and the practice of treating both success and failure, joy and sorrow with the same composure will naturally increase our life span on this planet. Listening to good music and reading good books, not only text books, can drastically reduce unpleasant stress. Thus we can maintain a calm temperament at all occasions.

Samacheer Kalvi 11th English The Singing Lesson Textual Questions

A. Based on your understanding of the story, answer the following questions in about 30 – 50 words each.

Question 1.
What was the knife that Miss Meadows carried with her?
Answer:
Miss. Meadows had received a letter from Mr. Basil calling off the marriage. She deemed it a kind of personal failure. Her anger and disappointed became despair. She carried cold despair buried deep in her heart like a wicked knife.

Question 2.
What kind of relationship existed between Miss Meadows and the Science Mistress?
Answer:
Both hated each other. Science Miss gave her a “Sugary smile” concealing her hostility. But Miss. Meadows enquiries responded to her deception always with a cold grimace.

Question 3.
Why was Miss Meadows upset and dejected?
Answer:
Mr. Basil had written a disquieting letter after his engagement with her. He had claimed that he was not a “Marrying man”. The thought of marriage gave him a feeling of disgust. He had struck out the word disgust and replaced it with ‘regret’ to lessen the hurt. So, Miss. Meadows was upset and dejected.

Question 4.
How would Miss Meadows usually treat Mary? How did her behaviour towards the girl change that day?
Answer:
Usually she would receive the flower from her favourite pupil Mary Beazley. She would tuck it in her belt with great tenderness and give a smile to her. The music class would start with a joyful note.

Question 5.
Why had Miss Meadows chosen ‘A Lament’ as the lesson that particular day?
Answer:
Miss. Meadows had chosen “A lament” as a lesson for that particular day. She was only in a mood to lament her broken engagement and shattered dreams. The choice of the music lesson reflected her real mood of dejection and despondency.

Question 6.
What brought agony to the girls during the music lessons?
Answer:
During the music lesson, Miss. Meadows did not show any warmth. She was icy cold and mechanical in her instructions. Children could easily realize that Miss. Meadow was in a wax. Miss. Meadow totally ignored the chrysanthemum from her favourite pupil. She also did not respond to her greeting. This sent tremors across the class. Young ones quickly understood the unstated message that music Miss was in one of her worst moods confirming their guess, she gave them “a lament” to practice.

The lesson was devoid of any warmth and joy. The lyrics of the song was so gloomy that children entered the world of unnatural agony and despair.

Question 7.
Bring out the substance of Basil’s letter to Miss Meadows.
Answer:
The content of Basil’s letter read, “I feel more and more strongly that our marriage would be a mistake. Not that I don’t love you. I love you as much as it is possible for me to love any woman but, truth to tell, I have come to the conclusion that I am not a marrying man and the idea of settling down fills me with nothing but – the word “disgust” was scratched out and “regret” written over the top.

Question 8.
Why did Miss. Wyatt summon Miss. Meadows to her room?
Answer:
A telegram addressed to Miss. Meadows was received at the school’s office. The Head mistress could not fathom the content of the Telegram. Believing that the telegram must be a harbinger of a tragedy, the Headniistress Miss. Wyatt summoned Miss. Meadows to her room.

Question 9.
How did Miss. Meadows express her joy, when she returned to the music class?
Answer:
Miss. Meadows changed the song for the children. She asked them to sing a joyful song beginning with flowers o’er laden. When she found that some children were still stuck up in the despondent mood, she reprimanded them. She told the girls, don’t look so doleful. It ought to sound warm, joyful and eager. And this time Miss. Meadow’s voice dominated the voices of all the little angels in her classroom. It was deep and glowing with a cheerful expression.

Question 10.
Briefly explain the cause of Miss Meadows’ joy at the end.
Answer:
Contrary to the expectation of Miss. Wyatt, the telegram was from Basil. It was an apology and reconciliatory in nature. In the telegram, her fiance had asked her to ignore the letter written when he must have been mad. A few hours before, she was the embodiment of disappointment and self-pity. The telegram had restored her joy. She could gather the pieces of her shattered dreams and hopes and built them anew. She was so happy that Miss. Wyatt’s warnings fell into deaf ears.

Vocabulary

A. Note the following words from the story. They all refer to different ways of walking. Find out their meanings and use each of them in meaningful sentences of your own. Refer a thesaurus and add a few more to the list.

(a) trod (b) fluttered (c) hurried (d) skipped (e) strode (f) sped

(a) trod – walked, stepped, strode, went
(b) fluttered – hovered, danced, flitted, flapped, oscillated, twitched, vibrated, flickered
(c) hurried – went fast, hastened, sped, charged, sprinted, chased, scampered, galloped, scrambled.
(d) skipped – capered, bobbed, bounded, jumped, leapt, gambolled,’frisked, romped
(e) strode – marched, trod, paced, stalked, dashed, ran, flaunted, joggled, tramped, rushed,zoomed
(f) sped – hurried, rushed, zipped, spurred, hurtled, sailed, hastened, quickened

B. Complete the mind map given below and write a brief summary of the story in your own words.

Answer:

(i) a letter from Basil breaking the engagement
(ii) irritable
(iii) sing a lament
(iv) Basil as a knife piercing her heart
(v) deserted her
(vi) of mpod
(vii) Head mistress Miss. Wyatt
(viii) the letter

A brief summary:
Miss Meadows was upset over a letter from Basil breaking the engagement. She remains gloomy and irritable in class. She taxes the students making them sing a lament. She thinks of the letter from Basil as a knife piercing her heart. Basil seems to have deserted her. Suddenly she is called by Head mistress Miss. Wyatt. The headmistress gives her a telegram which asks her to forget the letter. Miss Meadows feels happy and returns to the class with vigour and good cheer.

C. Answer the following questions in a paragraph of about 150 words each.

Question 1.
Describe Miss Meadows’ mood before and after receiving the telegram. How did it affect her class?
Answer:
Miss. Meadow was heart-broken. The letter written by Basil had pierced her heart and she was bleeding. Her hatred and anger became a knife and she carried it with her. Her icy cold response to science Miss demonstrates it. She is least bothered about the tender feelings of young children who look at her face all time for a friendly nod or smile of approval. Her favourite pupil Mary Beazley is baffled at her treatment of the chrysanthemum she had brought with so much love. The choice of the song “A lament” perfectly jells well with her worst mood. She is in fact in her heart lamenting over the loss of love, trust and future hopes. She is unnecessarily severe with young children forcing them to redo the singing which drives them to despair, pain and tears they manage to stifle.

After she receives the telegram from Basil apologizing for his insane letter, her mood changes to joy. She takes the chrysanthemum and keeps it close to her lips to conceal her blush. She goads the children to sing a song of joy congratulating some one for success. She persuades them to show warmth in their voices. Her warm and lively voice dominates the tremulous voices of the young ones. The young ones now realize that Miss Meadow who was in a wax earlier is now in her elements.

“My moods don’t just swing – they bounce, pivot, recoil, rebound, oscillate, fluctuate and occasionally PIROUETTE. ”

 

Question 2.
‘The only difference between a good day and a bad day is your attitude.’Relate this to a real life experience you have had. Share your thoughts in class.
Answer:
It is true that attitude makes or mars things. I was waiting for my Std X public exam resuls. I had put in 10 hours of work everyday. I was an average boy but I hoped to join the top five any time. The results were scheduled to be out the next morning. I have no internet connectivity at home. But one of my friends had it. His name is Murali. His father had come to my home. When I was out, he had told my parents I had failed in one subject. When I returned home, I found both my parents upset. I asked them what was wrong. They said it was a bad day. I asked them to tell me what was in their mind. They asked me to take things as they come by.

I didn’t understand. My dad told philosophically that failure is a stepping stone to success. Still they were not open with me. When I went out for tea, the local newspapers carried Std X public exam results. To my joy, I had passed with a first class. When I broke the news, my parents disclosed what had upset them. If they had just not reacted to the wrong information, their natural joy would not have been robbed. Now that they knew the information, they were over joyed. I just wondered at the human capacity to go down in misery and bounce back to extreme joy just by a turn of events. The best course of action to be happy all the time would be to. retain the key to our happiness and never give up on our joy to any external event.

“Don’t take it on yourself to repay a wrong. Trust the Lord and He will make it right.”

Question 3.
You are busy getting ready for school. You receive a Whats App message from your best friend, saying that he/she is very upset over the fight you had yesterday and doesn’t want to talk to you any more. This distresses you as she sounds very firm. However, today is a big day at school with two tests lined up. What will be your state of mind? How will you handle this situation?
Answer:
I always remember an anecdote. Kannadasan has recounted this anecdote. A temple elephant was proceeding to the temple. It’s mahout had washed him and applied sandal paste and holy ash on his forehead. Passerby greeted him like a God. As he was walking majestically, he was followed by a she elephant. A pig crossing the male elephant told its wife, “You see how the elephant was scared of me and gave way”. Overhearing the arrogant words of the pig, “The ‘she-elephant’ asked the ‘male elephant’ if it was true. The gentle animal smiled wisely and said, “I always focus on my goal.

We are on our way to a holy place. Even if I stamp on the pig by mistake, he would die. But I need to return to the tank for another wash.” “In life we need to avoid confrontation to ensure continuous progress in the chosen path. When I am a student, academics is quite important. Friendship is also important.

If a friend gets upset and if she really values my friendship, I can always say sorry and bring her around after the examination. If she is pig-headed and refuses to give up arrogance or anger, I will tell her I shall pray for her and move on. I will definitely find some one worthy of my true friendship. In reality, true friends, can’t be angry for long with each other. Realizing the value of true friends .

I’ll send a message wishing her the very best of luck for her exam and promise to sort out the issue in the evening. Nothing needs to be taken as a permanent failure in relationship or even in examination. I would like to remember the Chinese proverb “One can’t help birds of sorrow hover over one’s head. But one can prevent them from building nest in one’s head”.

“Never leave a true relationship for a few faults. Nobody is perfect; Nobody is correct. In the end… Affection is always greater than perfection. ”

Additional Question

Question 1.
Attempt a character sketch of Miss. Meadows.
Answer:
Miss. Meadows is a 30 year old lady. Mr. Basil a 25 year old young man gets engaged to her. When she is cherishing the dreams of a happy married life, a letter lands on her heart like a bombshell. It shatters her dreams. It pierces her heart. Being a sensitive lady, she feels her heart is bleeding. The contents of the letter keep haunting her memory. She is unable to focus in her music classes. The miraculous engagement was almost broken. The words, “our marriage could be a mistake” leaves her bleeding. She interpreted the scornful glance of science miss . as if she had known about the “break”. He had mentioned that he was not a marrying man”.

She wondered how she would react to the disclosure of the shattered engagement to the colleagues and the villagers. She even harboured the idea of leaving her job and go into hiding somewhere. As she is in a glooming mood, she doesn’t respond to the offer of chrysanthemum with a warm smile and thanks. She gives a ‘lament’ for practice. It is only when she receives a telegram of apology from her fiance her mood gets lifted. She flits on the wings of hope and sings a joyous song along with her students. She is a perfect example of ordinary mortals who are early hurt and quickly bounce back to hopeful life as well.

“Please don’t expect me to always be good, kind and loving. There are times when I will be cold, thoughtless and hard to understand.”

Additional Questions

I. Answer the following choosing from the options give below.

Question 1.
Miss. Meadows hugging the __________ stared in hatred at the science mistress.
(a) baton
(b) books
(c) bite
(d) knife
Answer:
(d) knife

Question 2.
Everything about Miss. Meadows was sweet, pale like __________
(a) money
(b) honey
(c) flower
(d) rose
Answer:
(b) honey

Question 3.
Science mistress was good at showing a __________ smile whenever she came across Miss. Meadows.
(a) honey
(b) sugary
(c) feigned
(d) deceptive
Answer:
(b) sugary

Question 4.
The story is set in __________ season.
(a) winter
(b) spring
(c) autumn
(d) rainy
Answer:
(c) autumn

Question 5.
One could witness an excitement in the school __________
(a) drowsy
(b) insipid
(c) gleeful
(d) sensational
Answer:
(c) gleeful

Question 6.
Everyday the presentation of a beautiful flower to Miss. Meadows by __________ her favourite has pupil become a ritual.
(a) Muriel
(b) Mr. Basil
(c) Rosy
(d) Mary Beazley
Answer:
(d) Mary Beazley

Question 7.
The little children in the music were thinking __________ is in a wax.
(a) Rowdy
(b) Meady
(c) Rosy
(d) Miss. Wyatt
Answer:
(b) Meady

Question 8.
The letter from Mr. Basil had __________ Miss. Meadows heart.
(a) gladdened
(b) soothed
(c) pierced
(d) embalmed
Answer:
(c) pierced

Question 9.
Mary’s __________ was totally ignored by Miss. Meadows.
(a) Rose
(b) Lilly
(c) Chrysanthemum
(d) apple
Answer:
(c) Chrysanthemum

Question 10.
The song chosen for practice in the music class was a __________
(a) love song
(b) lament
(c) joyous song
(d) hymn
Answer:
(b) lament

 

Question 11.
The rejection of the flower was a __________ moment in Mary Beazley’S school life.
(a) memorable
(b) unforgettable
(c) staggering
(d) astonishing
Answer:
(c) staggering

Question 12.
Nothing could be more __________ than the lament.
(a) important.
(b) beautiful
(c) tragic
(d) fabulous
Answer:
(c) tragic

Question 13.
The last time Mr. Basil had come to see Miss. Meadow, he had worn a __________ in his buttonhole.
(a) diamond
(b) pearl
(c) rose
(d) chrysanthemum
Answer:
(c) rose

Question 14.
Mr. Basil could not refuse the headmaster’s wife’s invitation for a __________ because he couldn’t afford to be unpopular.
(a) party
(b) lecture
(c) card game
(d) dinner
Answer:
(d) dinner

Question 15.
Basil had written to Miss. Meadows that their marriage would be a __________
(a) boon
(b) mistake
(c) marvel
(d) wonder
Answer:
(b) mistake

Question 16.
Mr. Basil’s previous letter was about a __________ book-case.
(a) sandal
(b) teak
(c) neem
(d) fumed-oak
Answer:
(d) fumed-oak

Question 17.
The tiny one who clung to the lament wriggled like __________ caught on a line.
(a) dogs
(b) cats
(c) elephant
(d) fishes
Answer:
(d) fishes

Question 18.
Miss. Meadows compliment with a strange, stony tone positively __________ the younger girls.
(a) encouraged
(b) boosted
(c) frightened
(d) pleased
Answer:

Question 19.
Miss. Meadows went on recalling the struck out word __________ in his letter.
(a) ‘pale’
(b) ‘white’
(c) ‘pleasing’
(d) ‘disgust’
Answer:
(d) ‘disgust’

Question 20.
Miss __________ was the school headmistress.
(a) Glory
(b) Mary
(c) Victoria
(d) Wyatt
Answer:
(d) Wyatt

Question 21.
Mrs. Wyatt really hoped for a news about a __________ through the telegram.
(a) marriage
(b) tragedy
(c) comedy
(d) practical joke
Answer:
(b) tragedy

Question 22.
Miss. Wyatt had sent for Miss. Meadow because __________ a had been received at the school office.
(a) letter
(b) money order
(c) book parcel
(d) telegram
Answer:
(d) telegram

Question 23.
On hearing the news of a telegram Miss. Meadows thought that __________ must have I committed suicide.
(a) Ryan
(b) Mary
(c) Mr. Basil
(d) Osborne
Answer:
(c) Mr. Basil

Question 24.
The telegram received the stress of Miss. Meadows but added to that of Miss __________
(a) Rose
(b) Beadle
(c) Mary
(d) Wyatt
Answer:
(d) Wyatt

Question 25.
Miss. Meadows had to struggle to lift the little girl from __________ spirit to a cheerful mood.
(a) doleful
(b) hateful
(c) disdainful
(d) cheerful
Answer:
(a) doleful

Question 26.
In order to indicate that she was her normal self again she took the __________ and held it to her lips.
(a) baton
(b) knife
(e) book
(d) chrysanthemum
Answer:
(d) chrysanthemum

 

Question 27.
The song after receiving the telegram sounded __________ , joyful and eager.
(a) cold
(b) distasteful
(c) mournful
(d) warm
Answer:
(d) warm

Question 28.
The fateful letter made Miss. Meadows even think of __________ her job and go into hiding.
(a) taking
(b) building
(c) resigning
(d) up scaling
Answer:
(c) resigning

Question 29.
After her brief visit to Miss. Wyatt’s room, dominated the students and it was glowing with __________
(a) pain
(b) bliss
(c) expression
(d) depression
Answer:
(c) expression

Question 30.
Miss Wyatt learnt that the telegram was Miss. Meadows’ __________
(a) dad
(b) fiance
(c) brother
(d) correspondent
Answer:
(b) fiance

II. Rearrange the sentences

Question 1.
(a) She did not say thanks with warmth to her favourite pupil for the flower.
(b) Most of the children realized with alarm that Miss. Meady was in a wax.
(c) Miss. Meadows looked dejected.
(d) Meadows, responding with a grimace went away and started her music.
(e) Science mistress greeted her with a sugary smile.
Answers:
(c) Miss. Meadows looked dejected.
(e) Science mistress greeted her with a sugary smile.
(d) Meadows, responding with a grimace went away and started her music.
(a) She did not say thanks with warmth to her favourite pupil for the flower
(b) Most of the children realized with alarm that Miss. Meady was in a wax.

Question 2.
(a) She was called to Miss. Wyatt’s room and given a telegram.
(b) Miss. Meadows, reflecting her despair, asked the children to sing “a lament”
(c) Miss. Meadows was upset over her fiance’s letter calling off the marriage
(d) After seeing the contents of the telegram, Miss. Meadows almost flew back to her class on the wings of hope and gave the class a joyful song to sing.
(e) Basil seemed to have deserted her.
Answers:
(c) Miss. Meadows was upset over her fiance’s letter calling off the marriage
(e) Basil seemed to have deserted her.
(b) Miss. Meadows, reflecting her despair, asked the children to sing “a lament”
(a) She was called to Miss. Wyatt’s room and given a telegram.
(d) After seeing the contents of the telegram, Miss. Meadows almost flew back to her class on the wings of hope and gave the class a joyful song to sing.

III. Identify the speaker

1. “Isn’t it cold? It might be winter” – science mistress to Miss. Meadows
2. “It is rather sharp – Miss Meadows to science mistress
3. You look frozen – Science mistress to Miss. Meadows
4. Oh! not quite as bad as that – Miss. Meadows to science mistress
5. ‘sh – sh! gives’ – Mary Beazley to fellow students
6. Silence, please! Immediately” – Miss Meadows to her student
7. “Good morning Miss. Meadows” – Mary Beasley to music miss
8. “Thank you Mary, How very nice! Turn to page 32” – Miss Meadows to Mary Beasley
9. Page fourteen, please mark the accent – Miss. Meadows to the music class children
10. What could have possessed him to write such a letter? – Monologue from Miss. Meadows
11. The third line should be one crescendo – Miss. Meadows to her class
12. “Away’ you must begin to die – to fade – until the listening ear” is nothing more than a whisper – Miss Meadows, to her music class students
13. Well, Monica, what is it? – Miss Meadows to Monica
14. Miss. Wyatt wants to see you in the mistress’s room. – Miss Monica to Miss. Meadows
15. “I shall met you in honour to talk quietly while I am away,” – Miss to her students in music class
16. “I sent for you just now because this telegram has come for you” – Miss Wyatt to Miss Meadows
17. “A telegram for me, Miss. Wyatt” – Miss Meadows to Miss Wyatt
18. “I hope it’s not bad news” – Miss Wyatt to Miss. Meadows
19. “I do hope, it’s nothing serious” – Miss Meadows to Miss Wyatt
20. Oh, no, thank you, Miss. Wyatt – Miss. Meadows to Miss Wyatt
21. You’ve fifteen minutes more of your class … Miss Meadows, haven’t you? – Miss Wyatt to Miss Meadows.
22. ”It’s from my fiance saying that… “ – Miss Meadows to Miss. Wyatt.

IV. Reading comprehension.

1. With despair – cold, sharp despair – buried deep in her heart like a wicked knife, Miss Meadows, in cap and gown and carrying a little baton, trod the cold corridors that led to the music hall. Girls of all ages, rosy from the air, and bubbling over with that gleeful excitement that comes from running to school on a fine autumn morning, hurried, skipped, fluttered by; from the hollow classrooms came a quick drumming of voices; a bell rang; a voice like a bird cried, “Muriel.”

And then there came from the staircase a tremendous knock-knock-knocking. Someone had dropped her dumb bells. The Science Mistress stopped Miss Meadows. “Good morning,” she cried, in her sweet, affected drawl. “Isn’t it cold? It might be winter.” Miss Meadows, hugging the knife, stared in hatred at the Science Mistress. Everything about her was sweet, pale, like honey. You would not have been surprised to see a bee caught in the tangles of that yellow hair. “It is rather sharp,” said Miss Meadows, grimly. The other smiled her sugary smile.

Question (a)
What is a baton used for?
Answer:
A baton is a long stick used for conducting a music orchestra.

Question (b)
What was the wicked knife?
Answer:
Sharp despair was the wicked knife.

Question (c)
Why was Miss. Meadows in a state of despair?
Answer:
Mr. Basic had written a letter calling of their engagement. So, Miss. Meadows was in a state of despair.

Question (d)
Why was the greeting of science mistress affected?
Science mistress did not have real feelings for Miss. Meadows. It was out of courtesy that she offered a sugary smile to Miss. Meadows and asked after the weather.

2. Forms Four, Five, and Six were assembled in the music hall. The noise was deafening. On the platform, by the piano, stood Mary Beazley, Miss Meadows’ favourite, who played accompaniments. She was turning the music stool. When she saw Miss Meadows, she gave a loud, warning “Sh-sh! Girls!” and Miss Meadows, her hands thrust in her sleeves, the baton under her arm, strode down the centre aisle, mounted the steps, turned sharply, seized the brass music stand, planted it in front of her, and gave two sharp taps with her baton for silence.

“Silence, please! Immediately!” and, looking at nobody, her glance swept over that sea of coloured flannel blouses, with bobbing pink faces and hands, quivering butterfly hair-bows, and music-books outspread. She knew perfectly well what they were thinking. “Meady is in a wax.” Well, let them think it! Her eyelids quivered; she tossed her head, defying them. What could the thoughts of those creatures matter to someone who stood there bleeding to death, pierced to the heart, by such a letter

Question (a)
Who was Miss. Meadows favourite pupil?
Answer:
Mary Beazley was Miss. Meadows’favourite pupil.

 

Question (b)
How did Mary alert the fellow students?
Answer:
On seeing Miss. Meadows, Mary alerted the fellow students saying,” sh – sh! girls”.

Question (c)
How did Miss. Meadows silence the music class?
Answer:
There was a brass music stand. Miss. Meadows gave two sharp taps with her baton to silence the music class.

Question (d)
According to Miss. Meadows, what were the children thinking about her on that bad day?
Answer:
Children were thinking that Miss. Meady is in a wax”.

Question (e)
What did the thoughts of children not affect Miss. Meadows?
Answer:
The letter from Mr. Basil had pierced her heart. She was bleeding to death. In such a state, Miss. Meadows could not possibly think about what the children were thinking about her.

The Singing Lesson About the Author

Kathleen Mansfield Murry (1888 – 1923) was a New Zealand short story writer who wrote under the pen-name Katherine Mansfield. She left New Zealand at the age of 19 and settled in the United Kingdom where she gained the friendship of great writers such as 0.11. Lawrence and Virginia Woolf. Bliss and The Garden Party were collections of short stories written by her. She wrote many poems and her collected letters were a great success.

The Singing Lesson Summary

This story depicts the fact that human moods are often influenced by experiences good or bad. The story revolves around the swings of mood experienced by Miss. Meadows and how it affected her work directly. Miss. Meadows, a music teacher aged 30, is engaged to 25 years old Basil. It was a huge surprise to everyone including the science teacher whom Miss. Meadows hates with all her heart. Suddenly she receives a disheartening letter from Basil that he is not a ‘marrying type of man’. The very idea of marriage gives him a feeling of “disgust”. But out of courtesy he had struck down the word disgust and written “regret”. After reading the letter Miss. Meady became gloomy. She had a feeling that her engagement was broken and it would soon come to the knowledge of everyone.

She would be a laughing stock. She will have to resign her job and go into hiding somewhere. This feeling of despair and disappointment had hurt her so much that she did not even accept the chrysanthemum from her favourite student Mary Beazley. A Chinese proverb says, “One cannot prevent. Under normal circumstances, she would have tucked the flower in her belt and returned a beaming smile to her favourite pupil. Miss. Meady made the children sing a lament creating an atmosphere of icy gloom befitting an occasion of mourning. She was very severe with young ones who didn’t evidence considerable pain and expression in her voice. A relief came to the drudgery of lament in the form of a telegram. Monica informed Miss. Meadows to meet the Headmistress. Miss. Meadows wondered if Mr. Basil had committed suicide. Her hands flew out in anxiety to collect the telegram from Miss. Wyatt.

Miss. Wyatt hoped it was bad news but out of politeness said, “I hope it is not bad news.” Miss. Meadows tore open the telegram. To her great relief it read, “Pay no attention to letter. Must have been mad, bought hat-stand today-Basil”. To answer the anxious query of Miss. Wyatt, Miss. Meadows blushed and said it was from her fiance. This happy turn of events upset Miss. Wyatt who disallowed telegrams on happy occasions. She reminded her of the 15 minutes left of her music class. Miss. Meadows ran all the way back to the music class. It appeared that she was flying on the wings of hope, love and joy. She picked up the deserted yellow chrysanthemum and held it to her lips to hide her smile.

Her mood suddenly switched over boundless joy. It reflected in the music lesson. She asked her pupils to turn to page 32 and sing the most joyous song the children had ever practised. Those who were blowing their nose from feigned sorrow and rigidity of the music, couldn’t suddenly move on to a joyful note. Miss. Meadow chided them for lack of feeling and expression.

Conclusion: The telegram restores her hopes, joy and faith in future, Many of us become prisoners of circumstances and over react to pinpricks in life. One should learn to be composed and take all kinds of information with a pinch of salt. Birds of sorrow hovering over one’s head. But one can prevent them from building nests in one’s head.

Textual:
accompaniments – music played to support an instrument, voice or group
aisle – a passage between rows of seats
drawl – slow, lazy way of talking
fiance – a man to whom one is engaged to be married
forte – a musical tone played loudly
grimace – expression of disgust on a person’s face of a music orchestra
tangles – a contused mass, twisted

 

Additional:
chided -rebuked
digust – revulsion
disallow – refuse
gloom – dejection
joyful – cheerful
joyous – happy
lament – expression of grief
regret – feel sad
relief – feeling of relaxation from tension
suicide – killing oneself
upset – pained

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Tamilnadu Samacheer Kalvi 11th English Speech Writing

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(i) You are Arun/Akila. You have to deliver a speech on the topic ‘Education Gives One Power’. You have jotted down the following notes:

 

Education trains mind—sharpens skill and abilities—Education: a source of power— improve self—be independent—earn money—ignorance to knowledge—removes superstition—develops a free spirit—important for women: gives them freedom from social ills—independent—responsible.
Write your speech in 150-200 words.
Answer:

Education Gives Power

Respected Principal, teachers and friends!

Education provides us knowledge. It trains our mind and sharpens our skills and abilities. Education refines our tastes and temperaments and builds our thought process. Vocational courses help young boys to earn and learn together. They provide means of earning livelihood and open the route to employment. Professional courses, as is evident from the name itself, equip us for adopting various professions. Some of these highly skilled professionals seek placements or jobs in esteemed companies and business concerns. Thus education is important for our survival. Decent living is impossible without good income or high salary.

 

Education improves the quality of our life and frees us of superstition, foolish, meaningless mind-blocks and rituals. If women are educated the whole family benefits as the food is hygienically prepared, children are healthy, well-mannered and disciplined. Education gives us power over our environment. We can control the situation and shape our destiny. Education spreads awareness among people and gives them freedom from social ills. It makes people independent by providing them means to earn their living. They become responsible citizens and realise their rights and duties. In short, education gives one power.

Thank you.

(ii) Shweta has to deliver a speech in the morning assembly on the topic: ‘The Generation Gap is Destroying Family Life’.
Write her speech in about 150-200 words.
Answer:
The Generation Gap Is Destroying Family Life

Respected Chairman, honorable judges, members of the staff and my dear friends!

 

I stand before you to speak my mind on ‘The Generation Gap is Destroying Family Life’. There are many points to support this contention.

The generation gap has caused a chasm between the old and the young. There is a clash of ideas and ideals, tastes, way of thinking and life-style. The young hanker after luxury, comfort and material happiness. They believe in full enjoyment and complete freedom. The elders insist on moral and spiritual values. They advocate renunciation and control on desires. The children think that their parents are slave to customs.

The generation gap is evident in the behaviour and manners of the two classes. The elders believe in strict obedience to a superior authority. They want that children must respect their elders and leam to be polite. The children pine for freedom. They resent all sorts of curbs. They demand freedom of expression. They want to have a say in family affairs. They insist that their views must be given proper consideration. This leads to bickerings, heart-burning and . tension in the family.

 

The elders insist on discipline and strict compliance of their orders. They advocate smiling acceptance of punishment meted out for breach of discipline. The young insist on unrestrained freedom and frankness. They advocate freedom of self-expression and are unwilling to tolerate any interference in their personal affairs—career, love-affair or marriage. Thus there is a wide gap between the two generations. This difference in their way of thinking and behavior is destroying family life.

Hence, I conclude that ‘The generation gap is destroying family life’.

Thank you.

 

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Tamilnadu Samacheer Kalvi 11th English Debate Writing

Check out the topics covered in Writing Debate Writing Questions and Answers before you start your preparation. Improve your grammar skills with the help of Samacheer Kalvi 11th English Book Solutions Questions and Answers pdf links. The solutions for Tamilnadu State Board 11th English Textbook are prepared by the English experts. So, if you follow Tamilnadu State Board Solutions 11th English Textbook Solutions you can cover all the topics in Writing Debate Writing Questons and Answers. This helps to improve your communication skills.

 

(i) You are chosen for representing your school at the regional level inter-school debate contest. Prepare a debate for the same on the topic given below:
‘Newspapers ought to contain more news and fewer advertisements
Answer:

Newspapers Ought to Contain More News and Fewer Advertisements
(by Pradeepa)

Respected Chairperson, August faculty and dear friends,

It is painful to see that journalists have lost all ethics and professionalism today. They adopt diverse techniques to increase the revenue of their paper. They try to sensationalise the news to win more readers. However, even the newspaper having the largest circulation is not self-dependent. It must have plenty of commercial advertisements to meet its running costs. In fact, it is these ads which provide the owners the requisite funds. Thus, advertisements cannot be ruled out. They are a necessary evil and must stay.

The owners and editors of the newspapers must also realise their responsibility to the nation. The press is the strongest pillar of democracy. It creates sensible public opinion in favor of good policies and criticises the wrong policies. The newspapers must maintain a balance between news and advertisements. A common man buys a newspaper for news. It would be better that the editors publish a classified advertisement supplement twice or thrice a week and save the general reader from the proliferation of advertisements while scanning the pages for views and news.

 

(ii) Prepare a debate on the topic ‘Corruption’.
Answer:
Corruption
Yesterday we had a debate about corruption. Of course, everybody was against it, particularly in Third World countries where it undermines economic progress and social and political institutions. But when it comes to business some people weren’t so sure. Corruption is often hidden in the form of fees or taxes or commissions. If these are needed to help get the business instead of your competitor, then a little flexibility is needed.

Fortunately, I was faced with this very early in my business career. I told their Lordships a little cautionary tale. My company developed some new technology in the textile industry and the way it could be used in many countries was through licensing. On one occasion, I had to convince an official to give us a permit. In his bookcase were copies of Dickens, Shakespeare, the Oxford English Dictionary and so I complimented him on his interest in English literature. He looked at me and handed me a copy of Dombey and Son. I opened it up and inside was a cutout which exactly fitted American dollar bills.

As I said in the debate, I am not sure whether it was my distaste for corruption, my uncertainty about the law, my concern for our reputation or my love of books, but this deal never did go through.

 

However, we did sell that technology in 23 other countries and later several people told me that they bought our license instead of others exactly because we did not pay bribes. This gave them more confidence that our technology would work.

It was the cash flow from these licenses that helped the early business grow. It also convinced me that a strong ethical culture is good for a business.

This is why we should stop watering down the Bribery Act and bring it into force now.

(iii) Impact of Demonetization on Common Man
Answer:
At the stroke of the hour on midnight of 9th November 2016, India lost 86% of its monetary base. The print, electronic and social media has been praising Prime Minister’s masterstroke by which he has reportedly destroyed the base of corruption in India. In this single move, the Government has attempted to tackle all the three issues affecting the economy, /.<?., a parallel economy, counterfeit currency in circulation and terror financing. There is no doubt that Prime Minister has pulled out a major coop and substantially enhanced his reputation as a strong leader.

 

Views and Counterviews:

The idea of demonetization is good but it has to be taken into consideration that most of the black money is kept in the form of.land, buildings or gold or kept abroad. What is in cash constitutes only 4% of the total amount of black money on which taxes are not being paid.

Out of this, a lot of money is in circulation in an everyday transaction like if someone is building a house; the bill is not paid through banks for sand, bricks, etc. This money goes into the other systems though it has been drawn from bank. These things will come under control with this step.

Small farmers, sellers, merchants, daily wage labourers and traders are suffering because of lack of proper planning, intelligence and foresight such as recaliberation of ATM machines. There was need to pile up enough 100 Rupee notes and other smaller denomination notes in the market before taking this step. It is being said by critics that this step was taken only to bolster the image of the Prime Minister as he has been unable to deliver on GDP growth, inflation and bringing the black money from abroad.

Demonetization is an established practice in monetary policy to tackle black money. The Prime Minister has explained why this is a financial surgical strike. It was meant to be suddenly implemented. In the past, demonetization has taken place twice but it fails because the idea is to tackle the black money existing in circulation. This is not to tackle corruption per se or the Government is not saying that 100% corruption will be tackled. If announcement and time would have been given, this step might not have been successful in controlling black money and counterfeit currency in circulation coming from Pakistan, Nepal or other countries.

 

People are facing problems because the limit of withdrawal has not been kept at a higher level. If this would have been kept at a higher level, there were chances that the recycling of black money might begin. The ideal money in circulation has to come to the banking channels.

It is also being said that what is being attempted is replacement of currency and not demonetization itself which was unnecessary. This is a terrible setback for the international standing of the Indian economy. At this time, the economy is struggling with slowdown. There is demand sluggishness in the economy leading to practically no private sector investment and stagnant industrial growth. If we look at the farm sector, this is the harvest time. Farmers generally deal in cash and India is also largely a cash economy. The cash transactions in this economy are far more than the total number of electronic transactions done on a daily basis. In the tribal heartland of the country, the poor people through middlemen are getting their currencies exchanged for Rs 300 or Rs 400 because of lack of proper information which is hitting them.

The stock of the black economy that constitutes a major part of the GDP is significant. Even if 50% of this amount is withdrawn, the kind of relief that RBI will get on its liabilities and the sort of deposits commercial banks will get will lead to a rise in the deposit and later on there will be decrease in lending rates plus fiscal deficit. The black money in circulation is like a steroid in the economy which keeps the demand going and gives a feeling that everything is working well. The problem is that investment is not taking place in the economy and the rate of growth of capital formation is down. The only way to bring this up is to divert more funds into investments which will happen when the cost of capital comes down.

 

Conclusion:
So far, it can be said that this is a historical step and should be supported by all. One should look at the bigger picture which will definitely fetch results in the long term. This is what the people have been asking for a long time which has finally happened.

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Writing Debate Writing Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

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Tamilnadu Samacheer Kalvi 11th English Speaking Group Discussion

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(i) Intelligent Quotient (IQ) Versus Emotional Quotient (EQ)

RAJ:
According to my opinion, those who have a good IQ level will always have the ability to reason out things and find an amicable solution to any problem that he/she faces in life. Even in today’s world, when a young aspirant is made to face an interview with a board of panelist, one of the key methods to finalise the situation is a problem solving abilities a candidate possesses. Nowadays, your mental appearance is more important than your physical appearance. Though eloquence in communication was given weight age, the skill and ability to convince others is much more important than flamboyant language. Presence of mind, common sense and practicality is more important than bookish knowledge. Therefore people with more IQ is needed than people with better EQ.

 

Chandran:
According to my understanding, I strongly contradict the views of Raj. It is the character that brings you huge success and I think good character, good personalities and good leaders are mostly driven by emotions.

Emotional quotient and intelligence quotient are undoubtedly the most important aspects of a person’s life and can make a huge difference between an average life and a successful life. To be successful, one may want to have a great IQ but, that success will not last long if your EQ is not appropriate. But on the contrary, if you have a good EQ and a little less IQ, then the chances of you becoming successful to carry on a successful career increases.

LEELA:
Everyone is equal. All should be recognised. One needs to be balanced with all qualities just as a balanced diet is needed for healthy living. It is illogical for learned people to say that IQ is much more needed than EQ or vice versa. I feel everyone is unique and is important. There is always something that one can contribute to. It needs to be balanced. Team play is vital at such a situation. A person with better IQ/EQ has his own workspace and methodology he contributes to the work differently as a person with different IQ/EQ. One who has a high IQ level can work smartly to do even the toughest tasks easily and hence procure popularity and wealth. Likewise, one who has better EQ can develop the same popularity and also leadership qualities. He/She can withstand adverse condition and overcome them.

 

Let me detail you with examples. To manage a team, a person of high EQ is needed whereas someone who needs to plan tasks and perform it in the best possible way, a person with high IQ should be there.

A person with higher IQ may work on any area and be extremely successful but may lack good social circle. This will surely lead to depression or emotional setbacks.

A person with higher EQ may win a large social circle of friends and supporters who are a big pillar of support at difficult times but due to their lack of technical abilities that needs better IQ, will turn out to be failure.

Hence, I want to reiterate that a person with an average IQ level and an equal amount of EQ level will be the best to survive in the present society. He can work anywhere happily with a fair wealth and fair emotional stability. He would be the most satisfied person.

Therefore a balanced IQ & EQ is the most desirable to be successful.

 

RAJ:
But I strongly feel that IQ is more important than EQ. Emotions are the biggest barrier to decision-making abilities. If we think about what others say and starts taking into consideration of others opinion, we get messed up and hence lack in making the right decision. Aren’t we living in a world where everything changes in an instant. So, we need someone who is confident of themselves and not get carried away by emotions. Hence I would prefer IQ over EQ,

Chandran:
I strongly opine that you consider the weight age of EQ more than IQ because if you . are perfect in EQ then you can manage with less IQ. For instance, if you have the qualities of EQ such as good personality skills, a better way of communication, and good attitude you can face any competitive situation. On the contrary, although you have strong IQ but poor EQ, then no one will take you seriously or you will be less focused leading to a few people joining you. So overall conclusion is that you don’t need to have a balanced or same level of IQ or EQ. If you are the best in EQ, then you can easily survive in this competitive era with less IQ.

 

LEELA:
There is no point in Mr. Raj arguing for IQ and Mr.Chandran taking sides with those who have better EQ, IQ and EQ have their own importance in our life and both should be in balanced proportion. I feel IQ makes us ambitious but EQ put limits on these ambitions. High IQ and Low EQ are the traits that can be observed in terrorists. So today, our topic is IQ vs EQ. I stay strong in both needed for a successful person. EQ is important for our social . life. AS we live in a society, we need good EQ levels for our personal and social behaviour and character to be well accepted in the society. Whereas, better IQ is surely the need for enhancing our knowledge to take the right decisions in this society and be a successful person. A balance between both is the need of the hour for both act as the two wheels in our journey of a successful life cycle.

(ii) To Live or Not

Kiruba:
We live in a world where most of us have a pet at home. When they fall sick, we rush them to the hospital. When our heart melts to see an animal or a bird suffer because of sickness, how will our conscience allow us to kill a fellow human being just because he is incurably ill? Moreover, he has done no harm to the society and his illness is not because of his fault.Therefore, we must provide him proper treatment and allow him to live as long as nature and God has willed his life span to be in this world.

 

Suganthi:
Exactly. Well said. God has gifted us life. So, he alone has the right to take it back. No human being has the right to interfere in His decisions or will for us in this world. Every man/woman has the right to live as long as God intends him/her to lead a life in this world.

Therefore, the illogical reasoning that a man suffering from an incurable disease, should be put to death is inexcusable beyond reason and a thought to be condemned.

Babu:
This world is governed by Darwin’s survival of the fittest principle. Darwin drew parallels between his own economic theories and his biological ones: “This survival of the fittest”, which he had sought to express in mechanical terms, is that which he has also called ‘natural selection’, or the preservation of favoured races in the struggle for life. An incurably diseased person is weak and has no value whatsoever to the society. More so, he has no means to live. Therefore, it would be in the fitness of things to kill him even against his wish. Therefore I bring to your kind notice that it is the natural selection to kill a person who cannot live befitting the needs of the society.

 

Kiruba:
Now a days, new vistas of progress have been opened in medical sciences and alternative ‘ medicine like Acupuncture, Acupressure, Reiki Pranik healing, Touch therapy, Herbal therapy, Diet therapy, etc. hold a ray of hope for the so called incurably diseased persons. So, why snatch life from them? When the kith and kin are ready to try all the means for survival, why should one interfere and speak things agaainst their survival. Miracles do happen and ‘ prayers are surely answered. Nothing wrong in giving a chance for someone fighting and struggling to live. We have no right to speak about putting another person to death.

Babu:
An incurably diseased person is the cause of constant worry to his family. His demands are unending and not withstanding the best possible attention, care and treatment given to him, he always remains dissatisfied and disgruntled. This adversely affects the peace of mind and comfort of the family members and all those who are trying to help them. Many a times, the physical worry and mental agony faced by the caretakers fall a prey to many other difficult situations and even fall sick being in an unhealthy environment for weeks and months. Therefore, the best way out of such a situation is to put an end to his life.

 

Suganthi:
It is not always the case that incurably diseased persons spread contagious diseases as some might argue. Even in those rare cases where it may be true, these persons are not real health hazards because it is medically established now that all incurable diseases are not contagious. However, as a precautionary measure, we should open separate hospitals or isolation wards for persons suffering from incurable contagious diseases and thus quarantine them. Isn’t that a better solution than thinking of putting an end to their life? To me, it’s a murder under legal permits.

Babu:
But these days we are burdened with the responsibility of reducing our rapidly increasing population. The many diseased persons constitute a good part of it. Even otherwise their contribution to society is a burden. It would be justified and reasonable not to allow them to drag on their agonizing life. Killing an incurably diseased person will put an end to research work in medical science. As I have already stated, there are many hospitals and many doctors who start experimenting with diseased persons. Even otherwise, suffering people have been the subject of research work for those who are into higher education. Thereby, its appropriate to free them from pain and suffering!

 

Kiruba:
Such an act only points out to the degenerated values in this society. It is almost compared to that of an individual throwing an appliance when it is old and doesn’t yield good results. Isn’t there a difference between living and non-living? How can a living human being be treated like a material object? What has happened to the values that students were taught or imbibed in their life? Such a thinking is only an assertion of degenerated moral values.

Conclusion:
India is a land of values. We are a secular society. We believe in respecting each other. Let us not be carried away by degenerated moral values and learn to respect one another and help them to live in peace without a feeling of guilt.

 

The main aim is to share the knowledge and help the students of 11th English to secure the best score in their final exams. Use the concepts of Samacheer Kalvi 11th English Book Solutions Speaking Group Discussion Questions and Answers in Real time to enhance your skills. If you have any doubts you can post your comments in the comment section, We will clarify your doubts as soon as possible without any delay.

For those looking for help on 11th Physics can use the Tamilnadu State Board Solutions for 11th Physics Chapter 10 Oscillations prevailing for free of cost.

Download the Samacheer Kalvi 11th Physics Book Solutions Questions and Answers Notes Pdf, for all Chapter Wise through the direct links available and take your preparation to the next level. The Tamilnadu State Board Solutions for 11th Physics Chapter 10 Oscillations Questions and Answers covers all the topics and subtopics within it. Practice using these Samacheer Kalvi 11th Physics Book Solutions Questions and Answers for Chapter 10 Oscillations PDF and test your preparation level and bridge the knowledge gap accordingly.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 10 Oscillations

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Samacheer Kalvi 11th Physics Oscillations Textual Evaluation Solved

Samacheer Kalvi 11th Physics Oscillations Multiple Choice Questions

Question 1.
In a simple harmonic oscillation, the acceleration against displacement for one complete oscillation will be [model NSEP 2000-01 ]
(a) an ellipse
(b) a circle
(c) a parabola
(d) a straight line
Answer:
(d) a straight line

Question 2.
A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3s. The time period is
(a) 15s
(b) 6s
(c) 12s
(d) 9s
Answer:
(c) 12s
Hint:
Time period of Oscillation = 2 × (time taken to go from A to B + the next time taken to return at B)
= 2 × (3 + 3)
= 2 × 6
Time period = 12s

Question 3.
The length of a second’s pendulum on the surface of the Earth is 0.9 m. The length of the same pendulum on surface of planet X such that the acceleration of the planet X is n times greater than the Earth
(a) 0.9 n
(b) \(\frac{0.9}{n} \mathrm{m}\)
(c) 0.9 n²m
(d) \(\frac{0.9}{n^{2}}\)
Answer:
(a) 0.9 n
Hint:
Second’s pendulum on the surface of planet,
Time period, T = 2 sec

Question 4.
A simple pendulum is suspended from the roof of a school bus which moves in a horizontal direction with an acceleration a, then the time period is …..

Answer:
\(\mathrm{T} \propto \frac{1}{g^{2}+a^{2}}\)

Question 5.
Two bodies A and B whose masses are in the ratio 1:2 are suspended from two separate massless springs of force constants k_A and k_B respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is

Answer:

Hint:
The maximum velocity in the Oscillation is given as v_max = Aω
The ratio of maximum velocity, v_A: v_B = 1 : 2

Question 6.
A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is

Answer:
(b) \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{\sqrt{2}}\)
Hint:
Spring constant of spring depends upon number of coil in the spring! When you cut the spring then the number of coil remain half of the original, so
k’ = 2k
Hint:
According to law of accelaration, the time period of simple pendulum

Question 5.
Two bodies A and B whose masses are in the ratio 1 : 2 are suspended from two separate massless springs of force constants k_A and k_B respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is

Answer:
(b) \(\sqrt{\frac{k_{\mathrm{B}}}{8 k_{\mathrm{A}}}}\)
Hint:
The maximum velocity in the Oscillation is given as v_max = Aω
The ratio of maximum velocity, v_A : v_B = 1 : 2

Question 6.
A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is

Answer:
(b) \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{\sqrt{2}}\)
Hint:
Spring constant of spring depends upon number of coil in the spring! When you cut the spring then the number of coil remain half of the original, so
k’ = 2k

Question 7.
The time period for small vertical oscillations of block of mass m when the masses of the pulleys are negligible and spring constant k₁ and k₂ is ….

Answer:

Question 8.
A simple pendulum has a time period T₁. When its point of suspension is moved vertically upwards according as y = kt², where y is vertical distance covered and k = 1 ms^-2, its time period becomes T₂. Then [IIT 2005]

Answer:
(c) \(\frac{6}{5}\)
Hint:

Question 9.
An ideal spring of spring constant k, is suspended from the ceiling of a room and a block of mass M is fastened to its lower end. If the block is released when the spring is un-stretched, then the maximum extension in the spring is : [IIT 2002]

Answer:
(c) \(2 \frac{\mathrm{Mg}}{k}\)
Hint:
Work by gravity + Work by spring = Change in kinetic energy

Question 10.
A pendulum is hung in a very high building oscillates to and fro motion freely like a simple harmonic oscillator. If the acceleration of the bob is 16 ms-2 at a distance of 4 m from the mean position, then the time period is [NEET 2018 model]
(a) 2s
(b) 1s
(c) 2πs
(d) πs
Answer:
(d) πs
Hint:
Acceleration, a = 16 m/s²
displacement, y = 4m
According to SHM,

Question 11.
A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom, the period of oscillation will
(a) first increase and then decrease
(b) first decrease and then increase
(c) increase continuously
(d) decrease continuously
Answer:
(a) first increase and then decrease

Question 12.
The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are ………

Answer:
(c) \(\mathrm{kg} \mathrm{s}^{-1}\)

Question 13.
When a damped harmonic oscillator completes 100 oscillations, its amplitude is reduced to \(\frac{1}{3}\) of its initial value. What will be its amplitude when it completes 200 oscillations?

Answer:
(d) \(\frac{1}{9}\)
Hint:
In damped vibration, amplitude at any instant t is

Question 14.
Which of the following differential equations represents a damped harmonic oscillator?

Answer:

Question 15.
If the inertial mass and gravitational mass of the simple pendulum of length / are not equal, then the time period of the simple pendulum is …..

Answer:

Samacheer Kalvi 11th Physics Oscillations Short Answer Questions

Question 1.
What is meant by periodic and non-periodic motion? Give any two examples, for each motion.
Answer:
Periodic motion: Any motion which repeats itself in a fixed time interval is known as periodic motion. Examples: Hands in pendulum clock, swing of a cradle.
Non-Periodic motion: Any motion which does not repeat itself after a regular interval of time is known as non-periodic motion. Example: Occurrence of Earth quake, eruption of volcano.

Question 2.
What is meant by force constant of a spring?
Answer:
The force constant (or) spring factor is defined as the restoring force produced per unit displacement.

Question 3.
Define time period of simple harmonic motion.
Answer:
Time period: The time period is defined as the time taken by a particle to complete one oscillation. It is usually denoted by T. For one complete revolution, the time taken is t = T, therefore,

Question 4.
Define frequency of simple harmonic motion.
Answer:
The number of oscillations produced by the particle per second is called frequency. It is denoted by f. SI unit for frequency is s^-1 or hertz (Hz).
Mathematically, frequency is related to time period by f = \(\frac{1}{\mathrm{T}}\)

Question 5.
What is an epoch?
Answer:
The phase of a vibrating particle corresponding to time t = 0 is called initial phase or epoch. At, t = φ, φ = φ₀.
The constant φ₀ is called initial phase or epoch. It tells about the initial state of motion of the vibrating particle.

Question 6.
Write short notes on two springs connected in series.
Answer:
Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position the phase of the vibrating particle. At time t = 0 s (initial time), the phase φ = φ₀ is called epoch (initial phase) where φ₀ is called the angle of epoch.

Question 7.
Write short notes on two springs connected in parallel.
Answer:
When two or more springs are connected in parallel, we can replace (by removing) all these springs with an equivalent spring (effective spring) whose net effect is same as if all the springs are in parallel connection.

Question 8.
Write down the time period of simple pendulum.
Answer:
The angular frequency of this oscillator (natural frequency of this system) is

The frequency of oscillations is

and time period of oscillations is

Question 9.
State the laws of simple pendulum?
Answer:
Law of length: For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.

Law of acceleration: For a fixed length, the time period of a simple pendulum is inversely proportional to square root of acceleration due to gravity.

Question 10.
Write down the equation of time period for linear harmonic oscillator.
Answer:
From Newton’s second law, we can write the equation for the particle executing simple harmonic motion

Comparing the equation with simple harmonic motion equation, we get

which means the angular frequency or natural frequency of the oscillator is

The frequency of the oscillation is

and the time period of the oscillation is

Question 11.
What is meant by free oscillation?
Answer:
When the oscillator is allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.

Question 12.
Explain damped oscillation. Give an example.
Answer:
The oscillations in which the amplitude decreases gradually with the passage of time are called damped Oscillations.
Example:

1. The oscillations of a pendulum or pendulum oscillating inside an oil filled container.
2. Electromagnetic oscillations in a tank circuit.
3. Oscillations in a dead beat and ballistic galvanometers.

Question 13.
Define forced oscillation. Give an example.
Answer:
The body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Sound boards of stringed instruments.

Question 14.
What is meant by maintained oscillation? Give an example.
Answer:
While playing in swing, the oscillations will stop after a few cycles, this is due to damping. To avoid damping we have to supply a push to sustain oscillations. By supplying energy from an external source, the amplitude of the oscillation can be made constant. Such vibrations, are known as maintained vibrations.
Example: The vibration of a tuning fork getting energy from a battery or from external power supply.

Question 15.
Explain resonance. Give an example.
Answer:
The frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known as resonance vibrations. Example: The breaking of glass due to sound

Samacheer Kalvi 11th Physics Oscillations Long Answer Questions

Question 1.
What is meant by simple harmonic oscillation? Give examples and explain why every simple harmonic motion is a periodic motion whereas the converse need not be true.
Answer:
Simple harmonic motion is a special type of oscillatory motion in which the acceleration or force on the particle is directly proportional to its displacement from a fixed point and is always directed towards that fixed point. In one dimensional case, let x be the .displacement of the particle and a_x be the acceleration of the particle, then


where b is a constant which measures acceleration per unit displacement and dimensionally it is equal to T^-2.
By multiplying by mass of the particle on both sides of equation (1) and from Newton’s second law, the force is
F_x = -kx …(3)
where k is a force constant which is defined as force per unit length. The negative sign indicates that displacement
and force (or acceleration) are in opposite directions. This means that when the displacement of the particle is taken towards right of equilibrium position (x takes positive value), the force (or acceleration) will point towards equilibrium (towards left) and similarly, when the displacement of the particle is taken towards left of equilibrium position (x takes negative value), the force (or acceleration) will point towards equilibrium (towards right). This type of force is known as restoring force because it always directs the particle executing simple harmonic motion to restore to its original (equilibrium or mean) position. This force (restoring force) is central and attractive whose center of attraction is the equilibrium position.
In order to represent in two or three dimensions, we can write using vector notation

where \(\vec{r}\) is the displacement of the particle from the chosen origin. Note that the force and displacement have a linear relationship. This means that the exponent of force \(\overrightarrow{\mathrm{F}}\) and the exponent of displacement \(\vec{r}\) are unity. The sketch between cause (magnitude of force | \(\overrightarrow{\mathbf{F}}\) |) and effect (magnitude of displacement | \(\vec{r}\) |) is a straight line passing through second and fourth quadrant.
By measuring slope \(\frac{1}{k}\), one can find the numerical value of force constant k.

Question 2.
Describe Simple Harmonic Motion as a projection of uniform circular motion.
Answer:
The projection of uniform circular motion on a diameter of SHM
Consider a particle of mass m moving with uniform speed v along the circumference of a circle whose radius is r in anti-clockwise direction (as shown in figure). Let us assume that the origin of the coordinate system coincides with the center O of the circle. If ω is the angular velocity of the particle and θ the angular displacement of the particle at any instant of time t, then θ = ωt. By projecting the uniform circular motion on its diameter gives a simple harmonic motion. This means that we can associate a map (or a relationship) between uniform circular (or revolution) motion to vibratory motion. Conversely, any vibratory motion or revolution can be mapped to uniform circular motion. In other words, these two motions are similar in nature.

Let us first project the position of a particle moving on a circle, on to its vertical diameter or on to a line parallel to vertical diameter as shown in figure. Similarly, we can do it for horizontal axis or a line parallel to horizontal axis.

As a specific example, consider a spring mass system (or oscillation of pendulum). When the spring moves up and down (or pendulum moves to and fro), the motion of the mass or bob is mapped to points on the circular motion.
Thus, if a particle undergoes uniform circular motion then the projection of the particle on the diameter of the circle (or on a line parallel to the diameter) traces straight line motion which is simple harmonic in nature. The circle is known as reference circle of the simple harmonic motion. The simple harmonic motion can also be defined as the motion of the projection of a particle on any diameter of a circle of reference.

Question 3.
What is meant by angular harmonic oscillation? Compute the time period of angular harmonic oscillation.
Answer:
Time period and frequency of angular SHM:
When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation. The point at which the resultant torque acting on the body is taken to be zero is called mean position. If the body is displaced from the mean position, then the resultant torque acts such that it is proportional to the angular displacement and this torque has a tendency to bring the body towards the mean position.

Let \(\vec{\theta}\) be the angular displacement of the body and the resultant torque \(\vec{\tau}\) acting on the body is

\(\kappa\) is the restoring torsion constant, which is torque per unit angular displacement. If I is the
moment of inertia of the body and a is the angular acceleration then

Question 4.
Write down the difference between simple harmonic motion and angular simple harmonic motion.
Answer:
Comparision of simple harmonic motion and angular harmonic motion

Question 5.
Discuss the simple pendulum in detail.
Answer:
Simple pendulum: A pendulum is a mechanical system which exhibits periodic motion. It has a bob with mass m suspended by a long string (assumed to be massless and inextensible string) and the other end is fixed on a stand. At equilibrium, the pendulum does not oscillate and hangs vertically downward. Such a position is known as mean position or equilibrium position. When a pendulum is displaced through a small displacement from its equilibrium position and released, the bob of the pendulum executes to and fro motion. Let l be the length of the pendulum which is taken as the distance between the point of suspension and the centre of gravity of the bob. Two forces act on the bob of the pendulum at any displaced position.

(i) The gravitational force acting on the body (\(\overrightarrow{\mathrm{F}}=m \vec{g}\)) which acts vertically downwards.
(ii) The tension in the string \(\overrightarrow{\mathrm{T}}\) which acts along the string to the point of suspension. Resolving the gravitational force into its components:
(a) Normal component: The component along the string but in opposition to the direction of tension, F_as = mg cos θ
(b) Tangential component: The component perpendicular to the string i.e., along tangential direction of arc of swing, F_ps = mg sin θ
Therefore, The normal component of the force is, along the string,

From the figure, we can observe that the tangential component W_ps of the gravitational force always points towards the equilibrium position i.e., the direction in which it always points opposite to the direction of displacement of the bob from the mean position. Hence, in this case, the tangential force is nothing but the restoring force. Applying Newton’s second law along tangential direction, we have

where, s is the position of bob which is measured along the arc. Expressing arc length in terms of angular displacement i.e.,

Because of the presence of sin 0 in the above differential equation, it is a non-linear differential equation (Here, homogeneous second order). Assume “the small oscillation approximation”, sin θ ≈ θ, the above differential equation becomes linear differential equation

This is the well known oscillatory differential equation. Therefore, the angular frequency of this oscillator (natural frequency of this system) is

Question 6.
Explain the horizontal oscillations of a spring.
Answer:
Horizontal oscillations of a spring-mass system: Consider a system containing a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in figure. Let x₀ be the equilibrium position or mean position of mass m when it is left undisturbed. Suppose the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position Let F be the restoring force (due to stretching of the spring) which is proportional to the amount of displacement of block. For one dimensional motion, mathematically, we have

where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. Notice that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity). This is not always true; in case if we apply a very large stretching force, then the amplitude
of oscillations becomes very large (which means, force is proportional to displacement containing higher powers of x) and
therefore, the oscillation of the system ìs not linear and hence, it is called non-linear oscillation. We restrict ourselves only to linear oscillations throughout our discussions, which means Hooke’s law is valid (force and displacement have a linear relationship).

From Newton’s second law, we can write the equation for the particle executing simple harmonic motion

Comparing the equation with simple harmonic motion equation, we get

The frequency of the oscillation is
>
and the time period of the oscillation is

Question 7.
Describe the vertical oscillations of a spring.
Answer:
Vertical oscillations of a spring: Let us consider a massless spring with stiff ness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring elongates by a length l. Let F₁ be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free-body diagram for this system as shown in figure. When the system is under equilibrium,
F₁ + mg = 0 …(1)
But the spring elongates by small displacement /, therefore,

Substituting equation (2) in equation (1), we get
-kl + mg = 0

Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y +1) is


The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as

Question 8.
Write short notes on the oscillations of liquid column in U-tube.
Answer:
Oscillation of liquid in a U-tube:
Consider a U-shaped glass tube which consists of two open arms with uniform cross sectional area A. Let us pour a non- viscous uniform incompressible liquid of density ρ in the U-shaped tube to a height h as shown in the figure. If the liquid and tube are not disturbed then the liquid surface will be in equilibrium position O.

It means the pressure as measured at any point on the liquid is the same and also at the surface on the arm (edge of the tube on either side), which balances with the atmospheric pressure. Due to this the level of liquid in each arm will be the same. By blowing air one can provide sufficient force in one arm, and the liquid gets disturbed from equilibrium position O, which means, the pressure at blown arm is higher than the other arm. This creates difference in pressure which will cause the liquid to oscillate for a very short duration of time about the mean or equilibrium position and finally comes to rest. Time period of the oscillation is

Question 9.
Discuss in detail the energy in simple harmonic motion.
Answer:
Energy in simple harmonic motion:
(a) Expression for Potential Energy
For the simple harmonic motion, the force and the displacement are related by Hooke’s law
\(\overrightarrow{\mathrm{F}}=-k \vec{r}\)

Since force is a vector quantity, in three dimensions it has three components.
Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case
F = — kx …… (1)
The work done by the conservative force field is independent of path. The potential energy U can be calculated from the following expression.

Comparing (1) and (2), we get

This work done by the force F during a small displacement dx stores as potential energy

where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation y = A sift ωt, we get
x = A sin ωt

This variation of U is shown in figure.

(b) Expression for Kinetic Energy
Kinetic energy

Since the particle is executing simple harmonic motion, from equation

This variation with time is shown in figure.

(c) Expression for Total Energy
Total energy is the sum of kinetic energy and potential energy
E = KE + U …(11)


Alternatively, from equation (5) and equation (10), we get the total energy as

which gives the law of conservation of total energy.
Thus the amplitude of simple harmonic oscillator, can be expressed in terms of total energy.

Question 10.
Explain in detail the four different types of oscillations.
Answer:
Free oscillations: When the oscillator is* allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration. In this case, the amplitude, frequency and the energy of the vibrating object remains constant.

Examples:

1. Vibration of a tuning fork.
2. Vibration in a stretched string.
3. Oscillation of a simple pendulum.
4. Oscillations of a spring-mass system.

Damped oscillations: During the oscillation of a simple pendulum (in previous case), we have assumed that the amplitude of the oscillation is constant and also the total energy of the oscillator is constant. But in reality, in a medium, due to the presence of friction and air drag, the amplitude of oscillation decreases as time progresses. It implies that the oscillation is not sustained and the energy of the SHM decreases gradually indicating the loss of energy.

The energy lost is absorbed by the surrounding medium. This type of oscillatory motion is known as damped oscillation. In other words, if an oscillator moves in a resistive medium, its amplitude goes on decreasing and the energy of the oscillator is used to do work against the resistive medium. The motion of the oscillator is said to be damped and in this case, the resistive force (or damping force) is proportional to the velocity of the oscillator.

Examples:

1. The oscillations of a pendulum (including air friction) or pendulum oscillating inside an oil filled container.
2. Electromagnetic oscillations in a tank circuit.
3. Oscillations in a dead beat and ballistic galvanometers.

Maintained oscillations: While playing in swing, the oscillations will stop aft er a few cycles, this is due to damping. To avoid damping we have to supply a push to sustain oscillations. By supplying energy from an external source, the amplitude of the oscillation can be made constant. Such vibrations are known as maintained vibrations.
Example: The vibration of a tuning fork getting energy from a battery or from external power supply.
Forced oscillations: Any oscillator driven by an external periodic agency to overcome the damping is known as forced oscillator or driven oscillator. In this type of vibration, the body executing vibration initially vibrates with its natural frequency and due to the presence of external periodic force, the body later vibrates with the frequency of the applied periodic force. Such vibrations are known as forced vibrations.
Example: Sound boards of stringed instruments.

Samacheer Kalvi 11th Physics Oscillations Numerical Problems

Question 1.
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is


Answer:
Oscillations of a particle dropped in a tunnel along the diameter of the earth.
Consider earth to be a sphere of radius R and centre O. A straight tunnel is dug along the diameter of the earth. Let ‘g’ be the value of acceleration due to gravity at the surface of the earth.
Suppose a body of mass ‘m’ is dropped into the tunnel and it is at point R i.e., at a depth d below the surface of the earth at any instant.
If g’ is acceleration due to gravity at P.

If y is distance of the body from the centre of the earth, then

Negative sign indicates that the force acts in the opposite direction of displacement.

Question 2.
Calculate the time period of the oscillation of a particle of mass m moving in the potential defined as
Answer:

Question 3.
Consider a simple pendulum of length l = 0.9 m which is properly placed on a trolley rolling down on a inclined plane which is at θ = 45° with the horizontal. Assuming that the inclined plane is frictionless, calculate the time period of oscillation of the simple pendulum.
Answer:
Length of the pendulum l = 0.9 m
Inclined angle θ = 45°

Question 4.
A piece of wood of mass m is floating erect in a liquid whose density is p. If it is slightly pressed down and released, then executes simple harmonic motion. Show that its time period of oscillation is
Answer:

Question 5.
Consider two simple harmonic motion along x and y-axis having same frequencies but different amplitudes as x = A sin (ωt + φ) (along x axis) and y = B sin ωt (along y axis).

Note: when a particle is subjected to two simple harmonic motion at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
Answer:
(a) \(y=\frac{\mathrm{B}}{\mathrm{A}} x\), equation is a straight line passing through origin with positive slope.
(b) \(y=-\frac{B}{A} x\), equation is a straight line passing through origin with negative slope.
(c) \(\frac{x^{2}}{\mathrm{A}^{2}}+\frac{y^{2}}{\mathrm{B}^{2}}=1\), equation is an ellipse whose center is origin. A2 B2
(d) \(x^{2}+y^{2}=\mathrm{A}^{2}\), equation is a circle whose center is origin.
(e) \(\frac{x^{2}}{\mathrm{A}^{2}}+\frac{y^{2}}{\mathrm{B}^{2}}-\frac{2 x y}{\mathrm{AB}} \frac{1}{\sqrt{2}}=\frac{1}{2}\), equation is an ellipse which (oblique ellipse which means tilted ellipse)

Question 6.
Show that for a particle executing simple harmonic motion
(a) the average value of kinetic energy is equal to the average value of potential energy.
(b) average potential energy = average kinetic energy = \(\frac{1}{2}\) (total energy)

Suppose a particle of mass m executes SHM of period T. The displacement of the particles at any instant t is given by y = A sin ωt

Question 7.
Compute the time period for the following system if the block of mass m is slightly displaced vertically down from Its equilibrium position and then released. Assume that the pulley is light and smooth, strings and springs are light.
Answer:

Case (a): Pulley is fixed rigidly here. When the mass displace by y and the spring will also stretch by y. Therefore, F = T = ky

Case (b): Mass displace by y, pulley also displaces by y. T = 4ky

Samacheer Kalvi 11th Physics Oscillations Additional Questions 

Question 1.
The total energy of a particle vibrating in SHM is proportional to the square of its ……..
(a) velocity
(b) acceleration
(c) amplitude
(d) none of these
Answer:
(a) veloctiy

Question 2.
In order to double the period of a simple pendulum ……….
(a) its length should doubled
(b) its length should be quadrupled.
(c) the mass of its bob should be doubled.
(d) the mass of its bob should be quadrupled.
Answer:
(b) its length should be quadrupled.

Question 3.
A simple harmonic oscillator has amplitude A and time period T. Its maximum speed is …….

Answer:
(d) \(\frac{2 \pi \mathrm{A}}{\mathrm{T}}\)
Solution:

Question 4.
A simple harmonic oscillator has a period of 0.01 s and an amplitude of 0.2 m. The magnitude of the velocity in m/s at the centre of oscillation is ……..
(a) 20π
(b) 40π
(c) 60π
(d) 80π
Answer:
(b) 40π
Solution:
Velocity is maximum at the centre of oscillation and is given by

Question 5.
A particle is executing SHM. Then the graph of acceleration as a function of displacement is ……..
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(a) straight line
Solution:
In SHM, F ∝ y ⇒ a ∝ y; Thus the graph is a straight line.

Question 6.
A particle is executing SHM. Then the graph of velocity as a function of displacement is …….
(a) straight line
(b) circle
(c) ellipse
(d) hyperbola
Answer:
(c) ellipse
Solution:

Question 7.
The amplitude of a vibrating body situated in a resisting medium ………
(a) decreases linearly with time
(b) decrease exponentially with time
(c) decreases with time in some other manner
(d) remains constant with time
Answer:
(b) decreases exponentially with time

Question 8.
The frequency of a vibrating body situated in air …….
(a) is the same as its natural frequency
(b) is higher than its natural frequency
(c) is lower than its natural frequency
(d) can have any value
Answer:
(c) is lower than its natural frequency

Question 9.
The equation represents the equation of motion for a ……. vibration
(a) free
(b) damped
(c) forced
(d) resonant
Answer:
(b) damped

Question 10.
The displacement equation of an oscillator is y = 5 sin (0.2 7 πt + 0.5 π) in SI units. The time period of oscillation is
(a) 10 s
(b) 1 s
(c) 0.2 s
(d) 0.5 s
Answer:
(a) 10 s
Solution:
Comparing with the standard equation y = A sin (ωt + φ)

Question 11.
A loaded spring vibrates with a period T. The spring is divided into four equal parts and the same load is suspended from one as these parts. The new time period is ………

Answer:
(b) \(\frac{\mathrm{T}}{2}\)
Solution:
Let the force constant of the spring be k. Then T = \(2 \pi \sqrt{\frac{m}{k}}\)
If the spring is divided into four equal parts, then the force constant of each part will be 4k.

Question 12.
The vertical extension in a light spring by a weight of 1 kg, in equilibrium is 9.8 cm. The period of oscillation of the spring, in seconds, will be ……

Answer:
(a) \(\frac{2 \pi}{10}\)
Solution:

Question 13.
A particle executing SHM has an acceleration of 64 cm/s² with its displacement is 4 cm. Its time period, in seconds is ……

Answer:
(a) \(\frac{\pi}{2}\)
Solution:

Question 14.
A body executes SHM with an amplitude A. Its energy is half kinetic and half potential when the displacement is ……..

Answer:
\(\frac{\mathrm{A}}{\sqrt{2}}\)
Solution:

Question 15.
The maximum displacement of a particle executing SHM is 1 cm and the maximum acceleration is (1.57)² cm/s². Its time period is …..
(a) 0.25s
(b) 4.0s
(c) 1.57s
(d) 3.14s
Answer:
(b) 4.0s
Solution:

Question 16.
The velocity of a particle, undergoing SHM is v at the mean position. If its amplitude is doubled, the velocity at the mean position will be …….
(a) 2 v
(b) 3 v
(c) \(2 \sqrt{2} v\)
(d) 4v
Answer:
(a) 2 v
Solution:

Question 17.
A girl is swinging on a swing in the sitting position. How will the period of swing be affected if she stands up?
(a) The period will now be shorter
(b) The period will now be longer
(c) The period will remain unchanged
(d) The period may become longer or shorter depending upon the height of the girl
Answer:
(a) The period will now be shorter
Solution:
The effective value of I will decrease. Therefore, the time period will be shorter.

Question 18.
The equation of SHM of a particle is > where k is a positive constant. The time period of motion is given by …….

Answer:
(a) \(\frac{2 \pi}{\sqrt{k}}\)
Solution:
Here k is same as ω²

Question 19.
The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minutes will be \(\frac{1}{x}\) time the original, where x is ………
(a) 2 × 3
(b) 2³
(c) 3²
(d) 3 × 2²
Answer:
(b) 2³
Solution:
The amplitude decreases exponentially with time and becomes half in 1 minute.
Amplitude after 3 minutes = \(\left(\frac{1}{2}\right)^{3}\) of the original value. Thus, x = 2³

Question 20.
When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is ……..

Answer:
(c) \(\frac{a}{2}\)
Solution:

Question 21.
A massless spring, having force constant k, oscillates with a frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from one of the parts. The frequency of oscillation will now be
For a simple pendulum the graph between length and time period will be …….
(a) n
(b) \(n \sqrt{2}\)
(c) \(\frac{n}{\sqrt{2}}\)
(d) 2n
Answer:
(a) n
Solution:

Question 22.
For a simple pendulum the graph between the length and time period will be a ……….
(a) hyperbola
(b) Parabola
(c) Straight line
(d) none of these
Answer:
(b) Parabola

Question 23.
A particle is executing simple harmonic motion given by The velocity of the particle when its displacement is 3 units is …….

Answer:
(a) 16 units
Solution:

Question 24.
When a particle oscillates simple harmonically, its potential energy varies periodically. If the frequency of oscillation of the particle is n, the frequency of potential energy variation is ……
(a) \(\frac{n}{2}\)
(b) n
(c) 2n
(d) 4n
Answer:
(c) 2n

Question 25.
A particle, moving along the x-axis, executes simple harmonic motion when the force acting on it is given by (A and k are positive constants.) …….
(a) – Akx
(b) A cos (kx)
(c) A exp (- kx)
(d) Akx
Answer:
(a) – Akx

Question 26.
The motion of a particle is expressed by the equation a = -bx, where x is the displacement from the mean position, a is the acceleration and b is a constant. The periodic time is ………

Answer:
(b) \(\frac{2 \pi}{\sqrt{b}}\)
Solution:

Question 27.
The angular velocity and the amplitude of a simple pendulum are ω and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is ……….

Solution:
(d) \(\frac{a^{2}-x^{2}}{x^{2}}\)
Solution:

Question 28.
A particle is oscillating according to the equation x = 5 cos (0.5 π t) where t is in seconds. The particle moves from the position of equilibrium to the position of maximum displacement in time ………
(a) 1 s
(b) 2 s
(c) 0.5 s
(d) 4 s
Answer:
(a) 1 s
Solution:

Time taken to move from the position of equilibrium to the position of maximum displacement is t = \(\frac{T}{4}\) = 1s

Question 29.
A seconds pendulum is placed in a space laboratory orbiting around the Earth at a height 3R from the Earth’s surface where R is the radius of the Earth. The time period of the pendulum will be ……..
(a) zero
(b) 2/3s
(c) 4 s
(d) infinite
Answer:
(d) infinite
Solution:

Question 30.
A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?

Answer:
(a) \(\frac{n}{2}\)
Solution:

Question 31.
Two simple pendulums of lengths 0.5m and 2.0m respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has completed …….. oscillations.
(a) 5
(b) 3
(c) 1
(d) 2
Answer:
(d) 2
Solution:
The time period of the shorter pendulum is half that of the longer pendulum. Therefore, the pendulums will again be in phase ( at the mean position). When the shorter pendulum has completed 2 oscillations.

Question 32.
A body is executing simple harmonic motion with an angular frequency 2 rod/sec. The velocity of the body at 20mm displacement, when the amplitude of motion is 60mm, is
(a) 90mm/s
(b) 113mm/s
(c) 118 mm/s
(d) 131 mm/s
Answer:
(b) 113 mm/s
Solution:

Question 33.
If the displacement of a particle executing SHM, is given by y = 0.30 sin (220t + 0.64) in metre, then the frequency and the maximum velocity of the particle are (t is in seconds)
(a) 35 Hz, 66m/s
(b) 45 Hz, 66 m/s
(c) 58 Hz, 113 m/s
(d) 35 Hz, 132 m/s
Answer:
(a) 35 Hz, 66 m/s
Solution:

Question 34.
The kinetic energy of a particle, executing SHM, is 16 J when it is at its mean position. If the amplitude of oscillations is 25 cm, and the mass of the particle is 5.12 kg, the time period of its oscillation is …….
(a) π/5 s
(b) 2π s
(c) 20π s
(d) 5π s
Answer:
(a) π/5 s
Solution:

Question 35.
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x) = kx² .Where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Answer:
(b) independent of a
Solution:
Since V(x) = Kx², the motion is simple harmonic. In SHM, the time period is independent of the amplitude of oscillation.

Question 36.
The amplitude of a damped oscillation reduces to one third of its original value a₀ in 20s. The amplitude of such oscillation after a period of 40s will be ……..
(a) a₀/9
(b) a₀/6
(c) a₀/2
(d) a₀/27
Answer:
(a) a₀/9
Solution:
In the first 20s, the amplitude reduces to one-third of the original value, i.e., to a₀/3, In the next 20s, it will reduce to one-third of the reduced value, i.e., to a₀/9.

Question 37.
Masses m_A and m_B hanging from the ends of strings of lengths l_A and l_B are executing. Simple harmonic motions. If their frequencies are related as f_A= 2f_B, then ……

Answer:
(c) l_A = l_B/4 regardless of masses.
Solution:

Question 38.
Two simple harmonic motions act on a particle. These harmonic motions are x = A cos (ωt + δ); y = A cos (ωt + α) When the resulting motion is …….
(a) A circle and the actual motion is clockwise
(b) an ellipse and the actual motion is counter clockwise
(c) a ellipse and the actual motion is clockwise
(d) a circle and the actual motion is counter clockwise
Answer:
(d) a circle and the actual motion is counter clockwise

Solution:
Squaring and adding the two equations, We get x² + y² = A². This is an equation of a circle. Hence the resultant motion is circular. The motion is counter clockwise.

Question 39.
A metal bob is suspended from a coiled spring. When set into vertical vibrations on the earth. It oscillates up and down with frequency f If the same experiment is carried out in a satellite circling the Earth the frequency of vibration will be …….
(a) f
(b) zero
(c) infinite
(d) depend on the distance of the satellite from the earth.
Answer:
(a) f
Solution:
The frequency of oscillation of a mass spring system depends only on the mass and the spring constant.

Question 40.
In forced oscillations of a particle, the amplitude is maximum for a frequency ω₁ of the force, while the energy is maximum for a frequency ω₂ of the force. Then ……
(a) ω₁ < ω₂
(b) ω₁ > ω₂ when damping is small and ω₁ > ω₂ when damping is large.
(c) ω₁ > ω₂
(d) ω₁ = ω₂
Answer:
(d) ω₁ = ω₂

Question 41.
Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?
(a) When v is maximum, a is maximum
(b) Value of a is zero, whatever may be the value of v
(c) When v is zero, a is zero
(d) When v is maximum, a is zero
Answer:
(d) When v is maximum, a is zero
Solution:

Question 42.
The function sin² (ωt) represents
(a) a simple harmonic motion with a period π/ω
(b) a simple harmonic motion with a period 2π/ω
(c) a periodic, but not simple harmonic motion with a period π/ω
(d) a periodic, but not simple harmonic motion with a period 2π/ω
Answer:
(a) a simple harmonic motion with a period π/ω

Question 43.
A particle executing simple harmonic motion has a kinetic energy K₀ cos² ωt. The maximum
values of the potential energy and the total energy are, respectively ….
(a) k₀/2 and k₀
(b) k₀ and 2k₀
(c) k₀ and k₀
(d) 0 and 2k₀
Answer:
(c) k₀ and k₀

Question 44.
A particle executing simple harmonic motion of amplitude 31.4 cm/s. The frequency of its oscillation is ………
(a) 3 Hz
(b) 4 Hz
(c) 2 Hz
(d) 1 Hz
Answer:
(d) 1 Hz

Question 45.
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is ……..
(a) 0.5π
(b) π
(c) 0.707π
(d) zero
Answer:
(a) 0.5π

Question 46.
Which one of the following equations of motion represents simple harmonic motion?
(a) Acceleration = – k₀x + k₁x²
(b) Acceleration = – k(x + a)
(c) Acceleration = k(x + a)
(d) Acceleration = kx
Answer:
(b) Acceleration = k(x + a)

Question 47.
Which of the following functions represent SHM?

(a) I and III
(b) I and II
(c) only I
(d) I, II and III
Answer:
(a) I and III

Question 48.
Two simple harmonic motions of angular frequencies 100 and 1000 rad/s have the same displacement amplitude. The ratio of their maximum accelerations is ……….
(a) 1 : 10
(b) 1 : 10²
(c) 1 : 10³
(d) 1 : 10⁴
Answer:
(b) 1 : 10
Solution:
The magnitude of the maximum acceleration is given by

Question 49.
The period of oscillation of a simple pendulum is T in a stationary lift. If the lift moves upwards with an acceleration of 8g, the period will ……..
(a) remain the same
(b) decrease by T/2
(c) increase by T/3
(d) none of these
Answer:
(c) increase by T/3
Solution:
Thus, the new time period is T/3. Hence the correct option is (d).

Question 50.
A simple harmonic oscillator consist of a particle of mass m and an ideal spring with spring constant k The particle oscillates with a time period T. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be …….

Answer:
(b) \(\frac{\mathrm{T}}{\sqrt{2}}\)
Solution:

If the spring is cut into two equal parts, the force constant of each part becomes 2k. Therefore,

Question 51.
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is …….
(a) 3 Hz
(b) 4 Hz
(c) 2 Hz
(d) 1 Hz
Answer:
(d) 1 Hz
Solution:

Samacheer Kalvi 11th Physics Oscillations 2 Marks Questions

Question 1.
What is meant by Oscillatory motion?
Answer:
When an object or a particle moves back and forth repeatedly for some duration of time, its – motion is said to be oscillatory (or vibratory).

Question 2.
What is meant by simple harmonic motion (SHM)?
Answer:
A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.

Question 3.
What is meant by displacement in SHM?
Answer:
The distance travelled by the vibrating particle at any instant of time t from its mean position is known as displacement.
y = A sin ωt
The maximum displacement from the mean position is known as amplitude (A) of the vibrating particle.

Question 4.
Define velocity in SHM.
Answer:
Velocity: The rate of change of displacement is velocity.

Question 5.
Define acceleration in SHM.
Answer:
Acceleration: The rate of change of velocity is acceleration.

Question 6.
What is meant by phase in SHM?
Answer:
Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position.

Question 7.
What is meant by angular oscillation?
Answer:
When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation.

Question 8.
Define Amplitude.
Answer:
The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude.

Samacheer Kalvi 11th Physics Oscillations 3 Marks Questions

Question 1.
Derive the expression for resultant spring constant when two springs having constant k₁ and k₂ are connected in series.
Answer:
Let x₁ and x₂ be the elongation of springs from their equilibrium position (un-stretched position) due to applied force F. Then, the net displacement of the mass point is

Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as

Suppose we have n springs connected in series, the effective spring constant in series is

If all spring constants are identical i.e., k₁ = k₂ = … = k_n = k then

This means that the effective spring constant reduces by the factor n. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,
>
Then the ratio of compressed distance or elongated distance x₁ and x₂ is

The elasticity potential energy stored in first and second springs are respectively. Then their ratio is

Question 2.
Derive the expression for resultant spring constant when two springs having constant k₁ and k₂ are connected in parallel.
Answer:
k₁ and k₂ attached to a mass m as shown in figure. The results can be generalized to any number of springs in parallel.

Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is
F = – k_pX …(1)
where k_p is called effective spring constant.
Let the first spring be elongated by a displacement x due to force F₁ and second spring be elongated by the same displacement x due to force F₂, then the net force
F = – k₁x – k₂x …..(2)
Equating equations (2) and (1), we get
k_p = k₁ + k₂…..(2)
Generalizing, for n springs connected in parallel,

If all spring constants are identical i.e., k₁ = kk₂ = … = k_n = k then

This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

Samacheer Kalvi 11th Physics Oscillations Numerical Problems

Question 1.
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum oscillator. The acceleration of the bob of the pendulum is 20 ms^-2 at a distance of 5 m from the mean position. To find the time period of oscillation.
Answer:
Given; a = 20ms^-2 ; y = 5 m a = ω²y

Question 2.
The acceleration due to gravity on the surface of the moon is 1.7 ms^-2. What is the time period of simple pendulum on the moon if its time period on the earth is 3.5 s? Give g on Earth = 9.8 ms^-2
Answer:

Question 3.
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.
Answer:

Question 4.
A body of mass m is attached to lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. Find the value of m is kg.
Answer:

Question 5.
Two simple harmonic motions are represented by the equations:

What is the ratio of their amplitudes ?
Answer:

Question 6.
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 Nm^-1. The block is pulled to a distance x = 10 cm from its equilibrium position at t = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Answer:

Question 7.
A 5 kg collar is attached to a spring of force constant 500 Nm^-1. It slides without friction on a horizontal rod as shown in figure. The collar is displaced from its equilibrium position by 10.0 cm and released.
Calculate:
(i) the period of oscillation
(ii) the maximum speed, and
(iii) the maximum acceleration of the collar.
Answer:

Question 8.
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer:
When 0.02 kg mass is added, the spring streches by 7 cm

When 0.02 kg mass ia removed, the period of vibration will be

Question 9.
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?
Answer:

Question 10.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is 5 cm.
Answer:
Here A = 5 cm, T = 0.2s
Velocity and acceleration at any displacement x are given by

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Samacheer Kalvi 11th Physics Kinetic theory of Gases Textual Evaluation Solved

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Multiple Choice Questions

Question 1.
A particle of mass m is moving with speed u in a direction which makes 60° with respect to x axis. It undergoes elastic collision with the wall. What is the change in momentum in x and y direction?

Answer:
\(\Delta p_{x}=-2 m u, \Delta p_{y}=0\)
Solution:
The change in momentum of the molecule in x direction
\(\Delta p_{x}\) = Final momentum – Initial momentum ,
= After collision – Before collision
= – mu – mu = – 2mu
The change in momentum of the molecule in Y direction \(\Delta p_{y}\) = 0

Question 2.
A sample of ideal gas is at equilibrium. Which of the following quantity is zero?
(a) rms speed
(b) average speed
(c) average velocity
(d) most probable speed
Answer:
(c) average velocity

Question 3.
An ideal gas is maintained at constant pressure. If the temperature of an ideal gas increases
from 100K to 1000K then the rms speed of the gas molecules
(a) increases by 5 times
(b) increases by \(\sqrt{10}\) times
(c) remains same
(d) increases by 7 times
Answer:
(b) increases by \(\sqrt{10}\) times
Solution:

Question 4.
Two identically sized rooms A and B are.connected by an open door. If the room A is air conditioned such that its temperature is 4° lesser than room B, which room has more air in it?
(a) Room A
(b) RoomB
(c) Both room has same air
(d) Cannot be determined
Answer:
(a) Room A

Question 5.
The average translational kinetic energy of gas molecules depends on ……
(a) number of moles and T
(b) only on T
(c) P and T
(d) P only
Answer:
(a) number of moles and T

Question 6.
If the internal energy of an ideal gas U and volume V are doubled then the pressure ……
(a) doubles
(b) remains same
(c) halves
(d) quadruples
Answer:
(b) remains same
Solution:

Question 7.
The ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\) for a gas mixture consisting of 8 g of helium and 16 g of oxygen is …… [Physics Olympiad – 2005]
(a) 23/15
(b) 15/23
(c) 27/17
(d) 17/27
Answer:
(c) 27/17
Solution:

Question 8.
A container has one mole of monoatomic ideal gas. Each molecule has /degrees of freedom.
What is the ratio of \(\gamma=\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) = ?

Answer:
(d) \(\frac{f+2}{f}\)
Solution:

Question 9.
If the temperature and pressure of a gas is doubled the mean free path of the gas molecules
(a) remains same
(b) doubled
(c) tripled
(d) quadruples
Answer:
(a) remains same
Solution:
Mean free path of the molecule

Question 10.
Which of the following shows the correct relationship between the pressure and density of an ideal gas at constant temperature?

Answer:

Question 11.
A sample of gas consists of µ₁ moles of monoatomic molecules, µ₂ moles of diatomic molecules and µ₃ moles of linear triatomic molecules. The gas is kept at high temperature. What is the total number of degrees of freedom?
(a) [3µ₁ + 7(µ₂ + µ₃)] N_A
(b) [3µ₁ + 7µ₂ + 6µ₃] N_A
(c) [7µ₁ + 3(µ₂ + µ₃)] N_A
(d) [3µ₁ + 6(µ₂ + µ₃)] N_A
Answer:
(a) [3µ₁ + 7(µ₂ + µ₃)] N_A

Question 12.
If S_P and S_V denote the specific heats of nitrogen gas per unit mass at constant pressure and constant volume respectively, then …….. [JEE 2007]
(a) S_P – S_V = 28 R
(b) S_P – S_V = R/28
(c) S_P – S_V = R/14
(d) S_P – S_V = R
Answer:
(b) S_P – S_V = R/28

Question 13.
Which of the following gases will have least rms speed at a given temperature?
(a) Hydrogen
(b) Nitrogen
(c) Oxygen
(d) Carbon dioxide
Answer:
(d) Carbon dioxide

Question 14.
For a given gas molecule at a fixed temperature, the area under the Maxwell-Boltzmann distribution curve is equal to

Answer:
(a) \(\frac{\mathrm{PV}}{k T}\)

Question 15.
The following graph represents the pressure versus number density for ideal gas at two different temperatures T₁ and T₂. The graph implies …….
(a) T₁ = T₂
(b) T₁ > T₂
(c) T₁ < T₂
(d) Cannot be determined
Answer:
(b) T₁ > T₂

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Short Answer Questions

Question 1.
What is the microscopic origin of pressure?
Answer:
According to the kinetic theory of a gases, the pressure exerted by the molecules depends on
(i) Number density n = \(\frac{N}{V}\)
(ii) Mass of the molecule
(iii) Mean square speed

Question 2.
What is the microscopic origin of temperature?
Answer:
The average K.E per molecule \(\overline{\mathrm{KE}}=\epsilon=\frac{3}{2} k \mathrm{T}\)
The equation implies that the temperature of a gas is a measure of the average translational K.E. per molecule of the gas.

Question 3.
Why moon has no atmosphere?
Answer:
Moon has no atmosphere. The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

Question 4.
Write the expression for rms speed, average speed and most probable speed of a gas
molecule.
Answer:

Question 5.
What is the relation between the average kinetic energy and pressure?
Answer:
The internal energy of the gas U = \(\frac{3}{2} \mathrm{N} k \mathrm{T}\)
The above equation can also be written as U = \(\frac{3}{2} \mathrm{PV}\)
since PV = NkT

From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density \(\left(u=\frac{U}{V}\right)\)
Writing pressure in terms of mean kinetic energy density

where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H. S of equation (2) by 2, we get

From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 6.
Define the term degrees of freedom.
Answer:
The minimum number of independent coordinates needed to specify the position and configuration of a thermo-dynamical system in space is called the degree of freedom of the system.

Question 7.
State the law of equipartition of energy.
Answer:
According to kinetic theory, the average kinetic energy of system of molecules in thermal equilibrium at temperature T is uniformly distributed to all degrees of freedom (x or y or z
directions of motion) so that each degree of freedom will get \(\frac{1}{2}\) kT of energy. This is called law of equipartition of energy.

Question 8.
Define mean free path and write down its expression.
Answer:
The average distance travelled by the molecule bfetween collisions is called mean free path (λ).

Question 9.
Deduce Chailes’ law based on kinetic theory.
Answer:
Charles’ law: From PV = \(\frac{2}{3} U\). For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that

This is Charles’ law.

Question 10.
Deduce Boyle’s law based on kinetic theory.
Answer:
Boyle’s law: From PV = \(\frac{2}{3} U\)
But the internal energy of an ideal gas is equal to N times the average kinetic energy (ε) of each molecule. U = Nε. For a fixed temperature, the average translational kinetic energy ε will remain constant. It implies that

Thus PV = constant Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle’s law.

Question 11.
Deduce Avogadro’s law based on kinetic theory.
Answer:
Avogadro’s law: This law states that at constant temperature and pressure, equal volumes of all gases contain the same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases,

At the same temperature, average kinetic energy per molecule is the same for two gases.

Dividing the equation (1) by (2) we get N₁ = N₂
This is Avogadro’s law. It is sometimes referred to as Avogadro’s hypothesis or Avogadro’s Principle.

Question 12.
List the factors affecting the mean free path.
Answer:

1. Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several metre away than smell of cold food.
2. Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 13.
What is the reason for Brownian motion?
Answer:
According to kinetic theory, any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Long Answer Questions

Question 1.
Write down the postulates of kinetic theory of gases.
Answer:

1. All the molecules of a gas are identical, elastic spheres.
2. The molecules of different gases are different.
3. The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
4. The molecules of a gas are in a state of continuous random motion.
5. The molecules collide with one another and also with the walls of the container.
6. These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
7. Between two successive collisions, a molecule moves with uniform velocity.
8. The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
9. The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
10. These molecules obey Newton’s laws of motion even though they move randomly.

Question 2.
Derive the expression of pressure exerted by the gas on the walls of the container.
Answer:
Expression for pressure exerted by a gas: Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.


The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.
A molecule of mass m moving with a velocity \(\vec{v}\) having components (v_x, v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (- v_x, v_y, v_z).
The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = – mv_x
The change in momentum of the molecule in x direction = Final momentum – initial momentum = – mv_x – mv_x = – 2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2mv_x
The number of molecules hitting the right side wall in a small interval of time At.
The molecules within the distance of v_x∆t from the right side wall and moving towards’ the right will hit the wall in the time interval ∆t. The number of molecules that will hit the right side wall in a time interval At is equal to the product of volume (Av_x∆t) and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume \(\frac{\mathrm{N}}{\mathrm{V}}\).
We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.
The number of molecules that hit the right side wall in a time interval ∆t

In the same interval of time At, the total momentum transferred by the molecules

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

Pressure, P = force divided by the area of the wall

Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term \(v_{x}^{2}\) by the average \(\overline{v_{x}^{2}}\) in equation (4).

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{z}^{2}}\) . The mean square speed is written as

Question 3.
Explain in detail the kinetic interpretation of temperature.
Answer:
To understand the microscopic origin Of temperature in the same way, from pressure exerted by a gas,

Comparing the equation (1) with ideal gas equation PV = NkT,

Multiply the above equation by \(\frac{3}{2}\) on both sides,

R.H.S. of the equation (3) is called average kinetic energy of a single molecule (\(\overline{\mathrm{KE}}\)).
The average kinetic energy per molecule

Equation (3) implies that the temperature, of a gas is a measure of the average translatiohal kinetic energy per molecule of the gas.
Equation (4) is a very important result from kinetic theory of gas. We can infer the following from this equation.
(i) The average kinetic energy of the molecule is directly proportional to absolute temperature of the gas. The equation (3) gives the connection between the macroscopic world (temperature) to microscopic world (motion of molecules).
(ii) The average kinetic energy of each molecule depends only on temperature of the gas hot on mass of the molecule. In other words, if the temperature of an ideal gas is measured using thermometer, the average kinetic energy of each molecule can be calculated without seeing the molecule through naked eye.
By multiplying the total number of gas molecules with average kinetic energy of each molecule, the internal energy of the gas is obtained.

Here, we understand that the internal energy of an ideal gas depends only on absolute temperature and is independent of pressure and volume.

Question 4.
Describe the total degrees of freedom for monoatomie molecule, diatomic molecule and triatomic molecule,
Monoatomie molecule: A monoatomie molecule by virtue of its nature has only three translational degrees of freedom.
Therefore f = 3
Example: Helium, Neon, Argon
Diatomic molecule: There are two cases.
(i) At Normal temperature: A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction. Physically the molecule can be regarded as a system of two point masses fixed at the ends of a massless elastic spring. The center of mass lies in the center of the diatomic molecule. So, the motion of the center of mass requires three translational degrees of freedom (figure a). In addition, the diatomic molecule can rotate about three mutually perpendicular axes (figure b). But the moment of inertia about its own axis of rotation is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom.
f = 5

2. At High Temperature: At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy] (figure c ). So totally there are seven degrees of freedom.
f = 7
Examples: Hydrogen, Nitrogen, Oxygen.

Triatomic molecules: There are two cases.
Linear triatomic molecule:
In this type, two atoms lie on either side of the central atom. Linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to diatomic molecule except there is an additional atom at the center. At normal temperature, linear triatomic molecule will have five degrees of freedom. At high temperature it has two additional vibrational degrees of freedom. So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide.

Non-linear triatomic molecule: In this case, the three atoms lie at the vertices of a triangle.
It has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. The total degrees of freedom.
f = 6
Example: Water, Sulphurdioxide.

Question 5.
Derive the ratio of two specific heat capacities of monoatomic, diatomic and triatomic molecules.
Answer:
Application of law of equipartition energy in specific heat of a gas, Meyer’s relation C_P – C_V = R connects the two specific heats for one mole of an ideal gas.
Equipartition law of energy is used to calculate the value of C_P – C_V and the ratio between them \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\). Here γ is called adiabatic exponent.


Note that the C_V and C_P are higher for diatomic molecules than the mono atomic molecules. It implies that to increase the temperature of diatomic gas molecules by 1°C it require more heat energy than monoatomic molecules.

(iii) Triatomic molecule
(a) Linear molecule:

(b) Non-linear molecule:

Note that according to kinetic theory model of gases the specific heat capacity at constant volume and constant pressure are independent of temperature. But in reality it is not sure.
The specific heat capacity varies with the temperature.

Question 6.
Explain in detail the Maxwell Boltzmann distribution function.
Answer:
Maxwell-Boltzmann: In speed distribution function
Consider an atmosphere, the air molecules are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed. In the previously we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of 5 ms^-1 to 10 ms^-1 or 10 ms^-1 to 15 ms^-1 etc. In general our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell’s speed distribution function.

The above expression is graphically shown as follows:

From the above figure, it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The ms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three.
(i) The area under the graph will give the total number of gas molecules in the system
(ii) Figure 2 shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

Question 7.
Derive the expression for mean free path of the gas.
Answer:
Expression for mean free path
We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves along a straight path with uniform velocity. This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume. Assume that only one molecule is in motion and all others are at rest.
If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, consider the molecule to move in an imaginary cylinder of volume πd²vt. It collides with any molecule whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd²vtn. The total path length divided by the number of collisions in time t is the mean free path.

Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion. So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path

The equation (2) implies that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases so it decreases the distance travelled by the molecule before collisions.
Case 1: Rearranging the equation (2) using ‘m’ (mass of the molecule)

But mn = mass per unit volume = ρ (density of the gas)

The equation (4) implies the following:
(i) Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several meter away than smell of cold food.
(ii) Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.

Question 8.
Describe the Brownian motion.
Answer:
Brownian motion is due to the bombardment of Brownian motion suspended particles by molecules of the surrounding fluid. But during 19th century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.

According to kinetic theory, any particle suspended – in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:

1. Brownian motion increases with increasing temperature.
2. Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Numerical Problems

Question 1.
A fresh air is composed of nitrogen N₂ (78%) and oxygen O₂ (21%). Find the rms speed of N₂ and O₂ at 20°C.
Answer:
Absolute temperature T = 20°C + 273 = 293K
Gas constant R = 8.32 J mol^-1 K^-1

Question 2.
If the rms speed of methane gas in the Jupiter’s atmosphere is 471.8 m s^-1, show that the surface temperature of Jupiter is sub-zero.
Answer:
RMS speed of methane gas (v_rms) = 471.8 ms^-1
Molar mass of methane gas (M) = 16.04 g per mol

Question 3.
Calculate the temperature at which the rms velocity of a gas triples its value at S.T.P. (Standard temperature T₁ = 273 K)
Answer:
At STP temperature T₁ = 273 K

Question 4.
A gas is at temperature 80°C and pressure \(5 \times 10^{-10} \mathrm{N} \mathrm{m}^{-2}\). What is the number of molecules per m³ if Boltzmann’s constant is \(1.38 \times 10^{-23} \mathrm{J} \mathrm{K}^{-1}\).
Answer:

Question 5.
From kinetic theory of gases, show that Moon cannot have an atmosphere (Assume k = 1.38 × 10^-23 J K^-1 Temperature T = 0°C = 273 K).
Answer:

Question 6.
If 10²⁰ oxygen molecules per second strike 4 cm² of wall at an angle of 30° with the normal when moving at a speed of 2 × 10³ ms^-1, find the pressure exerted on the wall.
(mass of 1 atom = 2.67 × 10^-26 kg).
Answer:

Momentum normal to the wall at angle 30°

Question 7.
During an adiabatic process, the pressure of a mixture of monoatomic and diatomic gases is found to be proportional to the cube of the temperature. Find the value of
Answer:

Question 8.
Calculate the mean free path of air molecules at STP. The diameter of N₂ and O₂ is about 3 × 10^-10 m.
Answer:

Question 9.
A gas made of a mixture of 2 moles of oxygen and 4 moles of argon at temperature T. Calculate the energy of the gas in terms of RT. Neglect the vibrational modes.
Answer:
For two moles of diatomic nitrogen with no vibrational mode,

For four mole of monatomic argon,

Total energy of the gas, U = U₁ + U₂ = 5 RT + 6 RT
U = 11 RT

Question 10.
Estimate the total number of air molecules in a room of capacity 25 m³ at a temperature of 27°C.
Answer:

The number of molecules in the room,

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Multiple Choice Questions

I. Choose the correct answer from the following:

Question 1.
Oxygen and hydrogen gases are at the same temperature the ratio of the average K.E of an oxygen molecule and that of a hydrogen molecule is
(a) 16
(b) 4
(c) 1
(d) \(\frac{1}{4}\)
Answer:
(c) 1
Solution:
P₁V₁ = P₂V₂, if T is constant.

Question 2.
According to the kinetic theory of gases,
(a) the pressure of a gas is proportional to the rms speed of the molecules.
(b) the rms speed of the molecules of a gas is proportional to the absolute temperature.
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
(d) the pressure of a gas is proportional to the square root of the rms speed of the molecules.
Answer:
(c) the rms speed of the molecules of a gas is proportional to the square root of the absolute temperature.
Solution:
The rms speed of the molecules of a gas is proportional to the square root of the absolute temperature. .

Question 3.
Pressure exerted by a perfect gas is equal to
(a) mean K.E. per unit volume.
(b) half of mean K.E. per unit volume.
(c) one-third of mean K.E. per unit volume.
(d) two-third of mean K.E. per unit volume.
Answer:
(d) two-third of mean K.E. per unit volume.
Solution: P = \(\frac{2}{3}\) K.E. (average K.E. of the gas per unit volume).

Question 4.
The temperature of an ideal gas is increased from 27°C to 927°C. The root mean square speed of its molecules becomes.
(a) 3 times
(b) double
(c) 4 times
(d) 6 times
Answer:
(b) double
1200
Solution:

Question 5.
Two gases are enclosed in a container at constant temperature. One of the gases, which is diatomic, has relative molecular mass eight times the other, which is monoatomic. The ratio pf the rms speed of the molecules of the monoatomic gas to that of the molecules of the diatomic gas is
(a) 8
(b) 4
(c) 2\(\sqrt{2}\)
(d) 2
Answer:
(c) 2\(\sqrt{2}\)
Solution:

Question 6.
If the absolute temperature of a gas is increased 3 times the rms velocity of the molecules will be ………
(a) 3 times
(b) 9 times
(c) 73 times
(d) 76 times
Answer:
(c) 73 times

Question 7.
At a given temperature which of the following gases possesses maximum rms velocity of molecules?
(a) H₂
(b) O₂
(c) N₂
(d) CO₂
Answer:
(a) H₂
Solution:

Question 8.
Two vessels have equal volumes. One of them contains hydrogen at one atmosphere and the other helium at two atmospheres. If both the samples are at the same temperature, the rms velocity of the hydrogen molecules is …..
(a) equal to that of the helium molecules
(b) twice that of the helium molecules
(c) half that of the helium molecules
(d) \(\sqrt{2}\) times that of the helium molecules
Answer:
(d) \(\sqrt{2}\) times that of the helium molecules
Solution:

Question 9.
A gas is enclosed in a container which is then placed on a fast moving train. The temperature of the gas …..
(a) rises
(b) remains unchanged
(c) falls
(d) becomes unsteady
Answer:
(c) falls

Question 10.
The mean translational K.E. of a perfect gas molecule at absolute temperature T is (K is Boltzmann constant)

Answer:
(c) \(\frac{3}{2} k \mathrm{T}\)

Question 11.
Ajar has mixture of hydrogen and Oxygen gases in the ratio 1 : 5. The ratio of mean kinetic energies of hydrogen and Oxygen molecules is ……..
(a) 1 : 5
(b) 5 : 1
(c) 1 : 1
(d) 1 : 25
Answer:
(c) 1 : 1
Solution:
The mean kinetic energy depends only on the temperature.

Question 12.
The pressure exerted on the walls of the container by a gas is due to the fact that the gas molecules ……..
(a) lose their K.E
(b) Stick to the walls
(c) are accelerated towards the walls
(d) change their momenta due to collision with the walls.
Answer:
(d) change their momenta due to collision with the walls.

Question 13.
Pressure exerted by a gas is …….
(a) independent of the density of the gas
(b) inversely proportional to the density of the gas
(c) directly proportional to the density of the gas
(d) directly proportional to the square of the density of the gas.
Answer:
(c) directly proportional to the density of the gas

Question 14.
Four molecules have speed 2 km/s, 3 km/s, 4 km/s and 5 km/s. The rms speed of these molecules in km/s is …….

Answer:
(a) \(\sqrt{\frac{27}{2}}\)
Solution:

Question 15.
A real gas behaves as an ideal gas at …….
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(a) low pressure and high temperature

Question 16.
The kinetic theory of gases breaks down most at ………
(a) low pressure and high temperature
(b) high pressure and low temperature
(c) low pressure and low temperature
(d) high pressure and high temperature.
Answer:
(b) high pressure and low temperature

Question 17.
Two different ideal gases are enclosed in two different vessels at the same pressure. If ρ₁ and ρ₂ are their densities and v₁ and v₂ their rms speeds, respectively then ____ is equal to ……

Answer:
(a) \(\sqrt{\frac{\rho_{2}}{\rho_{1}}}\)
Solution:

Question 18.
A cylinder of capacity 20 litres is filled with hydrogen gas. The total average K.E. of translatory motion of its molecules is 1.5 × 10⁵ J. The pressure of hydrogen in the cylinder is …….

(d) 5 × 10⁶ Nm^-2
Solution:

Question 19.
The molecular weights of oxygen and hydrogen are 32 and 2, respectively. The rms velocities of their molecules at a given temperatures, will be in the ratio
(a) 4 : 1
(b) 1 : 4
(e) 1 : 16
(6) 16 : 1
Answer:
(b) 1 : 4
Solution:

Question 20.
The average energy of a molecules of a monoatomic gas at temperature T is (K Boltzmann
constant)
(a) \(\frac{1}{2}\)kT
(b) kT
(c) \(\frac{3}{2}\)kT
(d) \(\frac{5}{2}\)kT
Answer:
(c) \(\frac{3}{2}\)kT

Question 21.
The temperature at which the molecules of nitrogen will have the same rms velocity as the
molecules of oxygen at 127° C is
(a) 77°C
(b) 350°C
(c) 273°C
(d) 457°C
Answer:
(a) 77°C
Solution:
Let the required temperature be T. Then
>

Question 22.
The temperature of a gas is raised from 27°C to 927°C. The root mean square speed of its molecules …….

(b) gets halved
(c) remains the same
(d) gets doubled
Answer:
(d) gets doubled
Solution:

Question 23.
The temperature at which the K.E of a gas molecules is double its value at 27°C is
(a) 54°C
(b) 300 K
(c) 327°C
(d) 108°C
Answer:
(c) 327°C
Solution:

Question 24.
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root mean square velocity of the gas molecules is v, at 480 K it becomes
(a) 4v
(b) 2v
(c) \(\frac{v}{2}\)
(d) \(\frac{v}{4}\)
Answer:
(b) 2v
Solution:

Question 25.
The average translational K.E. of O₂(molar mass 32) molecules at a particular temperature is 0.048 eV. The translational K.E. of N₂ (molar mass 28) molecules in eV at the same temperature is ……….
(a) 0.0015
(b) 0.003
(c) 0.048
(d) 0.768
Answer:
(c) 0.048
Solution:
Average translational K.E. of a gas molecule = \(\frac{3}{2}\)kT.
It is independent of molecular mass.

Question 26.
The K.E. of one mole of a gas at normal temperature and pressure is ……..

Solution:
(d) 3.4 × 10³J
Solution:

Question 27.
The average K.E. of a hydrogen gas molecule at STP will be (Boltzmann constant K_B= 1.38 × 10^-23 JK^-1) ………
(a) 0.186 × 10^-28 J
(b) 0.372 × 10^-20 J
(c) 0.56 × 10^-20 J
(d) 5.6 × 10^-20 J
Answer:
(c) 0.56 × 10^-20 J
Solution:

Question 28.
The rms speed of the particles of fume of mass 5 × 10^-17 kg executing Brownian motion in air at STP is ……
(a) 1.5 ms^-1
(b) 3.0 ms^-1
(c) 1.5 cm s^-1
(d) 3.0 cm s^-1
Answer:
(c) 1.5 cm s^-1
Solution:

Question 29.
To what temperature should the hydrogen at room temperature (27°C) be heated at constant pressure so that the RMS velocity of its molecules becomes double its previous value?
(a) 1200°C
(b) 927°C
(c) 600°C
(d) 108°C
Answer:
(b) 927°C
Solution:

Question 30.
A vessel contains oxygen at 400 K. Another similar vessel contains an equal mass of hydrogen at 300K. The ratio of the rms speeds of molecules of hydrogen and oxygen is

Answer:
(d) \(2 \sqrt{3}\)
Solution:

Question 31.
A chamber contains a mixture of helium gas (He) and hydrogen gas (H₂). The ratio of the root- mean-square speeds of the molecules of He and H₂ is ……..
(a) 2
(b) \(\sqrt{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{\sqrt{2}}\)
Solution:

Question 32.
On colliding with the walls in a closed container, the ideal gas molecules.
(a) transfer momentum to the walls
(b) lose momentum completely
(c) move with smaller speeds
(d) perform Brownian motion.
Answer:
(a) transfer momentum to the walls

Question 33.
The speeds of 5 molecules of a gas (in arbitrary units) are as follows: 2, 3, 4, 5, 6 The root mean square speed for these molecule is
(a) 2.91
(b) 3.52
(c) 4.00
(d) 4.24
Answer:
(d) 4.24
Solution:

Question 34.
At absolute zero temperature, the K.E. of the molecules becomes ………
(a) zero
(b) maximum
(c) minimum
(d) none of these
Answer:
(a) zero

Question 35.
If the rms speed of the molecules of a gas is 1000 ms^-1 the average speed of the molecule is
(a) 1000 ms^-1
(b) 922 ms^-1
(c) 780 ms^-1
(d) 849 ms^-1
Answer:
(b) 922 ms^-1

Question 36.
The gas having average molecular speed four times that of SO₂ (molecular mass 64) is
(a) He (molecular mass 4)
(b) O₂ (molecular mass 32)
(c) H₂ (molecular mass 2)
(d) CH₄ (molecular mass 16)
Answer:
(a) He (molecular mass 4)
Solution:

Samacheer Kalvi 11th Physics Kinetic Theory of Gases 2 Marks Questions

Question 1.
State Avogadro’s law.
Answer:
It states that equal volumes of all gases under similar conditions of temperature and pressure, contain equal number of molecules.

Question 2.
Define root mean square speed (v_rms). Write down its equations.
Answer:
Root mean square speed is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by

Question 3.
Define Avogadro’s number.
Answer:
It is defined as the number of particles present in one mole of the substance. It is denoted by N_A

Question 4.
Define Average speed. Write it equation.
Answer:
Average speed is defined as the mean (or) average of all the speeds of molecules.

Question 5.
Define most probable speed of the gas. Write its expressions.
Answer:
It is defined as the speed acquired by most of the molecules of the gas.

Question 6.

Answer:

Question 7.
What is the reason “No hydrogen in Earth’s atmosphere” ?
Answer:
As the root mean square speed of hydrogen is much less than that of nitrogen, it easily escapes from the earth’s atmosphere.
In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters
many disastrous consequences.

Question 8.
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.

Question 9.
What does the universal gas constant R signify? Give its value.
Answer:
The universal gas constant R signifies the workdone by (or on) a gas per mole per kelvin. Its value is R = 8.314 J mol^-1 K^-1

Question 10.
What is Boltzmann’s constant? Give its value.
Answer:
Boltzmann’s constant is defined as the gas constant per molecule.

Question 11.
When do the real gases obey more correctly the gas equation: PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas. At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

Samacheer Kalvi 11th Physics Kinetic Theory of Gases Additional Numerical Problems

Question 1.
If the rms speed of hydrogen molecules at 300 K is 1930 ms^-1. Then what is the rms speed of oxygen molecules
Answer:

Question 2.
The rms velocity of the molecules in a sample of helium is 5/7 times that of molecules in a sample of hydrogen. If the temperature of hydrogen is 0°C. Then, what is the temperature of helium?
Answer:

Question 3.
A cylinder of fixed capacity 44.8 litres contain helium gas at standard pressure and temperature. What is the amount of heat needed to raise the temperature of the gas by 15°C? (R = 8.31 J mol^-1 K^-1)
Answer:
Volume of 1 mole of He at STP = 22.4 litres
Total volume of He at STP = 44.8 litres

Molar specific heat of He (monoatomic gas) at constant volume,

Question 4.
An insulated container containing monoatomic gas of molar mass m is moving with a velocity v₀. If the container is suddenly stopped. Find the change in temperature.
Answer:
Suppose the container has n moles of the monoatomic gas. Then the loss in K.E. of the gas

If the temperature of the gas changes by ∆T, then heat gained by the gas,

Question 5.
Estimate the total number of molecules inclusive of oxygen, nitrogen, water vapour and other constituents in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atmospheric pressure. (k_B = 1.38 × 10^-23 JK^-1)
Solution:

Question 6.
Estimate the average energy of a helium atom at
(i) room temperature (27°C)
(ii) the temperature on the surface of the sun (6000 K) and
(iii) the temperature of 10⁷ K. (k_B = 1.38 × 10^-23JK^-1)
Answer:

Question 7.
The molecules of a given mass of a gas have rms velocity of 200 ms 1 at 27°C and 1.0 × 10s Nm^-2 pressure. When the temperature and pressure of the gas are respectively. 127 °C and 0.05 × 10s Nm^-2.
Find the rms velocity of its molecules in ms^-1
Answer:

Question 8.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Mass of oxygen molecule (m) = 2.76 × 10^-26 kg
Boltzmann’s constant (k_B) = 1.38 × 10^-23 JK^-1)
Answer:

Question 9.
The temperature of a gas is raised from 27°C to 927°C. What is the rms molecular speed?
Answer:

Question 10.
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. Find the ratio \(\frac{\mathbf{C}_{\mathbf{P}}}{\mathbf{C}_{\mathbf{V}}}\) of the
mixture.
Answer:

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Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved

Samacheer Kalvi 11th Physics Heat and Thermodynamics Multiple Choice Questions
Question 1.
In hot summer after a bath, the body’s …….
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer:
(a) internal energy decreases

Question 2.
The graph between volume and temperature in Charles’ law is …….
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer:
(c) a straight line

Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is …..
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer:
(b) adiabatic

Question 4.
An ideal gas passes from one equilibrium state (P₁, V₁, T₁, N) to another equilibrium state
(2P₁, 3V₁, T₂, N). Then ………

Answer:
(b) \(\mathrm{T}_{1}=\frac{\mathrm{T}_{2}}{6}\)
Solution:
From ideal gas equation, PV = NkT
One equilibrium state (P₁, V₁, T₁, N)
Another equilibrium state (P₂, V₂, T₂, N)
P₂ = 2P₁, and V₂ = 3V₁

Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase
(a) mass
(b) weight
(c) center of mass
(d) moment of inertia
Answer:
(d) moment of inertia

Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process …….
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0 Answer: (a) Q > 0, W > 0

Question 7.
When you exercise in the morning, by considering your body as thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U < 0, W < 0
(d) ∆U = 0, W > 0
Answer:
(b) ∆U < 0, W > 0

Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0 (c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
Answer:
(c) ∆U > 0, Q > 0

Question 9.
An ideal gas is taken from (P_i, V_i) to (P_f, V_f) in three different ways. Identify the process in which the work done on the gas the most.

(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer:
(b) Process B

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A ➝ B ➝ C ➝ D is shown below. (Processes D ➝ A and B ➝ C are adiabatic)

The corresponding PV diagram for the process is (all figures are schematic)

Answer:

Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer:
(a) 8280 K
Solution:
According to Wien’s displacement law,

Question 12.
Identify the state variables given here?
(a) Q, T, W
(b) P, T, U
(c) Q, W
(d) P, T, Q
Answer:
(b) P, T, U

Question 13.
In an isochoric process, we have ……
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
Answer:
(a) W = 0

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is ….. [NEET 2018]
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(c) 26.8%
Solution:
The freezing point of water = 0°C = 273 K
Boiling point of water = 100°C = 373 K

Question 15.
An ideal refrigerator has a freezer at temperature -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is …..
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer:
(c) 40.2°C
Solution:

Samacheer Kalvi 11th Physics Heat and Thermodynamics Short Answer Questions

Question 1.
‘An object contains more heat’- is it a right statement? If not why?
Answer:
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at higher temperature to an object lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.

Question 2.
Obtain an ideal gas law from Boyle’s and Charles’ law.
Answer:
Boyle’s law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.

Question 3.
Define one mole.
Answer:
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).

Question 4.
Define specific heat capacity and give its unit.
Answer:
Specific heat capacity of a substance is defined as the amount of heat energy required into raise the temperature of 1 kg of a substance by 1 Kelvin or 1°C

The SI unit for specific heat capacity is J kg^-1 K^-1

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by IK or 1°C.

Question 6.
What is a thermal expansion?
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Question 7.
Give the expressions for linear, area and volume thermal expansions.
Answer:
Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T.

Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as

Volume Expansion: For a small change in temperature ∆T the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.

Question 8.
Define latent heat capacity. Give its unit.
Answer:
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material.

Question 9.
State Stefan-Boltzmann law.
Answer:
Stefan Boltzmann law states that, the total amount of heat radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.

Question 10.
What is Wien’s law?
Answer:
Wien’s law states that, the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.

Question 11.
Define thermal conductivity. Give its unit.
Answer:
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady state conditions is known as thermal conductivity of a material.

Question 12.
What is a black body?
Answer:
A black body is one which neither reflects nor transmits but absorbs whole of the heat radiation incident on it.
The absorptive power of a perfect black body is unity.

Question 13.
What is a thermodynamic system? Give examples.
Answer:
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V) and Temperature (T). The remaining part of the universe is called surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas and radiation.

Question 14.
What are the different types of thermodynamic systems?
Answer:

1. Open system can exchange both matter and energy with the environment.
2. Closed system exchange energy but not matter with the environment.
3. Isolated system can exchange neither energy nor matter with the environment.

Question 15.
What is meant by ‘thermal equilibrium’?
Answer:
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

Question 16.
What is mean by state variable? Give example.
Answer:
In thermodynamics, the state of a thermodynamic system is represented by a set of variables called thermodynamic variables.
Examples: Pressure, temperature, volume and internal energy etc.

Question 17.
What are intensive and extensive variables? Give examples.
Answer:
Extensive variable depends on the size or mass of the system.
Example : Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.

Question 18.
What is an equation of state? Give an example.
Answer:
Equation of state: The equation which connects the state variables in a specific manner is called equation of state. An ideal gas obeys the equation PV = NkT at thermodynamic equilibrium.
For example, if we push the piston of a gas container, the volume of the gas will decrease but pressure will increase or if heat is supplied to the gas, its temperature will increase, pressure and volume of the gas may also increase.

Question 19.
State Zeroth law of thermodynamics.
Answer:
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Question 20.
Define the internal energy of the system.
Answer:
The internal energy of a thermodynamic system is the sum of kinetic and potential energies of all the molecules of the system with respect to the center of mass of the system.

Question 21.
Are internal energy and heat energy the same? Explain.
Answer:
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”

Question 22.
Define one calorie.
Answer:
One calorie is defined as the amount of heat energy needed to raise the temperature of one gram of water by one degree Celsius at a pressure of one atmosphere.

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
Joule essentially converted mechanical energy to internal energy. In his experiment potential energy is converted to rotational kinetic energy of paddle wheel and this rotational kinetic energy is converted to internal energy of water.

Question 24.
State the first law of thermodynamics.
Answer:
This law states that ‘Change in internal energy (∆U) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.

Question 25.
Can we measure the temperature of the object by touching it?
Answer:
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.

Question 26.
Give the sign convention for Q and W.
Answer:
System gains heat Q is positive
System loses heat Q is negative
Work done on the system W is negative
Work done by the system W is positive

Question 27.
Define the quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout.

Question 28.
Give the expression for work done by the gas.
Answer:
When a gas expands against pressure, it does work on the surroundings. The work done in expansion for volume V₁ to V₂ is given by

If the pressure remains constant during expansion,
Then W = P(V₂ – V₁) = P∆V
If the volume remains constant, then W = 0.
If there is no external pressure, then no work is done. For example, when a gas expands freely in vaccum, no work is done by it.

Question 29.
What is PV diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer:
It implies that to increase the temperature of the gas at constant volume requires less heat than increasing the temperature of the gas at constant pressure. In other words s is always greater than s_v.

Question 31.
Give the equation of state for an isothermal process.
Answer:
It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is PV = µRT

Question 32.
Give an expression for work done in an isothermal process.
Answer:

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT

Question 34.
Apply first law for
(a) an isothermal
(b) adiabatic
(c) isobaric processes.
Answer:
(a) For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W
(b) This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process. For an adiabatic process, the first law becomes ∆U = W.
(c) The first law of thermodynamics for isobaric process is given by
∆U = Q – P∆Y

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic process is given by

Here γ is called adibatic exponent
(γ = C_p/C_γ) which depends on the nature of the gas

Question 36.
Give an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by


We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin.

Question 37.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 38.
Draw the PV diagram for:
(a) Isothermal process
(b) Adiabatic process
(c) isobaric process
(d) Isochoric process
(a) Isothermal process:
Answer:
(a) Isothermal process

(b) Adiabatic process:

(c) isobaric process:

(d) Isochoric process:

Question 39.
What is a cyclic process?
Answer:
This is a thermodynamic process in which the thermodynamic system returns to its initial state after undergoing a series of changes. Since the system comes back to the initial state, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. From the first law of thermodynamics, the net heat transferred to the system is equal to work done by the gas.
Q_net = Q_in – Q_out = W (for a Cyclic Process)

Question 40.
What is meant by a reversible and irreversible processes?
Answer:
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.

Question 41.
State Clausius form of the second law of thermodynamics.
Answer:
“Heat always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.

Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
Answer:
Kelvin-Planck statement: It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work. This implies that no heat engine in the universe can have 100% efficiency.

Question 43.
Define heat engine.
Answer:
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Question 44.
What are processes involves in a Carnot engine?
Answer:
There are four processes involves in a carnot engine:

1. source
2. sink
3. insulating stand
4. working substance

Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
Answer:
According to first law of thermodynamics work can be completely converted into heat. Since the system comes back to the initial stage, the change in the internal energy is zero. In cyclic process, heat can flow in to system and heat flow out of the system. The net heat transferred to the system is equal to work done by the gas.
Q_net = Q_in = Q_out = W (for 3 Cyclic Process)

Question 46.
State the second law of thermodynamics in terms of entropy.
Answer:
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.

Question 47.
Why does heat flow from a hot object to a cold object?
Answer:
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics.

Question 48.
Define the coefficient of performance.
Answer:
It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Long Answer Questions

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
Meaning of heat: When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating. Due to flow of heat sometimes the temperature of the body will increase or sometimes it may not increase.

Meaning of work: When you rub your hands against each other the temperature of the hands increases. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now if you place your hands on the cheek, the temperature of the cheek increases. This is because the hands are at higher temperature than the cheek. In the above example, the temperature of hands is increased due to work and temperature of the cheek is increased due to heat transfer from the hands to the chin. It is shown in the Figure. By doing work on the system, the temperature in the system will increase and sometimes may not. Like heat, work is also not a quantity and through the work energy is transferred to the system. So we cannot use the word ‘the object contains more work’ or ‘less work’.
Either the system can transfer energy to the surrounding by doing work on surrounding or the surrounding may transfer energy to the system by doing work on the system. For the transfer of energy from one body to another body through the process of work, they need not be at different temperatures.

Question 2.
Discuss the ideal gas laws.
Answer:
Boyle’s law: For a given gas at low pressure (density) kept in a container of volume V, experiments revealed the following information.
When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume \(\mathrm{P} \propto \frac{1}{\mathrm{V}}\)
Charles’ law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature V ∝ T.
By combining these two equations we have
PV = CT. Here C is a positive constant.
We can infer that C is proportional to the number of particles in the gas container by considering the following argument. If we take two containers of same type of gas with same volume V, same pressure P and same temperature T, then the gas in each container obeys the above equation. PV = CT. If the two containers of gas is considered as a single system, then the pressure and temperature of this combined system will be same but volume will be twice and number of particles will also be double.

For this combined system, V becomes 2V, so C should also double to match with the ideal gas equation \(\frac{P(2 V)}{T}=2 C\). It implies that C must depend on the number of particles in the gas and also should have the dimension of \(\left[\frac{\mathrm{pv}}{\mathrm{T}}\right]=\mathrm{JK}^{-1}\). So we can write the constant C as k times the number of particles N.
Here k is the Boltzmann constant (1.381 × 10^-23 JK^-1) and it is found to be a universal constant. So the ideal gas law can be stated as follows
PV = NkT …(1)
The equation (1) can also be expressed in terms of mole.
Suppose if a gas contains p mole of particles then the total number of particles can be written as
N = µN_A …… (2)
where N_A is Avogadro number (6.023 × 10²³ mol^-1)
Substituting for N from equation (2), the equation (1) becomes PV = µN_AkT. Here N_Ak = R called universal gas constant and its value is 8.314 J /mol. K
So the ideal gas law can be written for µ mole of gas as
PV = μRT …(3)
This is called the equation of state for an ideal gas. It relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 3.
Explain in detail the thermal expansion.
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.
All three states of matter (solid, liquid and gas) expand when heated. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. The relative change in the size of solids is small. Railway tracks are. given small gaps so that in the summer, the tracks expand and do not buckle. Railroad tracks and bridges have expansion joints to allow them to expand and contract freely with temperature changes.

Liquids, have less intermolecular forces than solids and hence they expand more than solids. This is the principle behind the mercury thermometers.
In the case of gas molecules, the intermolecular forces are almost negligible and hence they expand much more than solids. For example in hot air balloons when gas particles get heated, they expand and take up more space.
The increase in dimension of a body due to the increase in its temperature is called thermal expansion.
The expansion in length is called linear expansion. Similarly the expansion in area is termed as area expansion and the expansion in volume is tenned as volume expansion.

Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\left(\frac{\Delta \mathrm{L}}{\mathrm{L}_{0}}\right)\) is directly proportional to ∆T

Where, α_L = coefficient of linear expansion,
∆L = Change in length,
L = Original length,
∆T = Change in temperature.
Area Expansion: For a small change in temperature ∆T the fractional change in area \(\left(\frac{\Delta \mathrm{A}}{\mathrm{A}_{0}}\right)\) of a substance is directly proportional to ∆T and it can be written as

Where, α_A = coefficient of area expansion.
∆A = Change in area,
A = Original area,
∆T = Change in temperature.
Volume Expansion: For a small change in temperature AT the fractional change in volume \(\left(\frac{\Delta \mathrm{V}}{\mathrm{V}_{0}}\right)\) of a substance is directly proportional to ∆T.

Where, α_v = coefficient of volume expansion,
∆V = Change in volume,
V = Original volume,
∆T = Change in temperature,
Unit of coefficient of linear, area and volumetric expansion of solids is °C^-1 or K^-1

Question 4.
Describe the anomalous expansion of water: How is it helpful in our lives?
Answer:
Anomalous expansion of water : Liquids expand on heating and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as it is cooled from room temperature, until it reach 4°C. Below 4°C the volume increases and so the density decreases. This means that the water has a maximum density at 4°C. This behavior of water is called anomalous expansion of water.

In cold countries during the winter season, the surface of the lakes will be at lower temperature than the bottom as shown in the Figure. Since the solid water (ice) has lower density than its liquid form, below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). This is due to the anomalous expansion of water. As the water in lakes and ponds freeze only at the top the species living in the lakes will be safe at the bottom.

Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
Calorimetry: Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, the heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be Anomalous expansion of water in lakes mathematically expressed as


Heat gained or lost is measured with a calorimeter. Usually the calorimeter is an insulated container of water. A sample is heated at high temperature (T₁) and immersed into water at room temperature (T₂) in the calorimeter. After some time both sample and water reach a final equilibrium temperature T_f. Since the calorimeter is insulated, heat given by the water.

Q_gain = – Q_lost
Note the sign convention. The heat lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity

Here S₁ and s₂ specific heat capacity of hot sample and water respectively.

Question 6.
Discuss various modes of heat transfer.
Answer:
There are three modes of heat transfer: Conduction, Convection and Radiation.

Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.
Convection: Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Boiling water in a cooking pot is an example of convection. Water at the bottom of the pot receives more heat. Due to heating, the water expands and the density of water decreases at the bottom. Due to this decrease in density, molecules rise to the top. At the same time the molecules at the top receive less heat and become denser and come to the bottom of the pot. This process goes on continuously. The back and forth movement of molecules is called convection current.
To keep the room warm, we use room heater. The air molecules near the heater will heat up and expand. As they expand, the density of air molecules will decrease and rise up while the higher density cold air will come down. This circulation of air molecules is called convection current.
Radiation: When we keep our hands near the hot stove we feel the heat even though our hands are not touching the hot stove. Here heat transferred from the hot stove to our hands is in the form of radiation. We receive energy from the sun in the form of radiations. These radiations travel through vaccum and reach the Earth. It is the peculiar character of radiation which requires no medium to transfer energy from one object to another. The conduction or convection requires medium to transfer the heat.
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from hot stove.

Question 7.
Explain in detail Newton’s law of cooling.
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,

Dividing both sides of equation (2) by dt

Where a is some positive constant. From equation (3) and (4)


Integrating equation (5) on both sides,

Where b₁, is the constant of integration. Taking exponential both sides, we get

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Answer:
Wien’s law and Vision:
The Sun is approximately taken as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation of Wien’s law.

It is the wavelength at which maximum intensity is 508 nm. Since the Sun’s temperature is around 5700 K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part, of the spectrum. It is shown in Figure.

The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible not in infrared or X-ray ranges in the spectrum.
Suppose if humans had evolved in a planet near the star Sirius (9940K), then they would have had the ability to see the Ultraviolet rays!

Question 9.
Discuss the:
(a) thermal equilibrium
(b) mechanical equilibrium
(c) chemical equilibrium
(d) thermodynamic equilibrium.
Answer:
(a) Thermal equilibrium: When a hot cup of coffee is kept in the room, heat flows from coffee to the surrounding air. After sometime the coffee reaches the same temperature as the surrounding air and there will be no heat flow from coffee to air or air to coffee. It implies that the coffee and surrounding air are in thermal equilibrium with each other. Two systems are said to be in thermal equilibrium with each other if they arc at the same temperature, which Mechanical equilibrium will not change with time.

(b) Mechanical equilibrium: Consider a gas container with piston. When some mass is placed on the piston, it will move downward due to downward gravitational force and after certain humps and jumps the piston will come to rest at a new position. When the downward gravitational force given by the piston is balanced by the upward force exerted by the gas, the system is said to be in mechanical equilibrium. A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.
(c) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.
(d) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Answer:
Joule’s mechanical equivalent of heat: The temperature of an object can be increased by heating it or by doing some work on it.
In the eighteenth century, James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa.
In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy equal to 2 mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.

This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C, 4.186 J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J
This is called Joule’s mechanical equivalent of heat.

Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Answer:
Work done in volume changes: Consider a gas contained in the cylinder fitted with a movable piston. Suppose the gas is expanded quasi-statically by pushing the piston by a small distance dx. Since the expansion occurs quasi-statically the pressure, temperature and internal energy will have unique values at every instant.

The small work done by the gas on the piston
dW = Fdx …(1)
The force exerted by the gas on the piston F = PA.
Here A is area of the piston and P is pressure exerted by the gas on the piston.
Equation (1) can be rewritten as Work done by the gas
dW = PA dx …(2)
But Adx = dV= change in volume during this expansion process.
So the small work done by the gas during the expansion is given by
dW = PdV ….(3)
dV is positive since the volume is increased. Here, dW is positive.
In general the work done by the gas by increasing the volume from V_i to V_f is given by

Suppose if the work is done on the system, then V_i > V_f. Then, W is negative.
Note here the pressure P is inside the integral in equation (4). It implies that while the system is doing work, the pressure need not be constant. To evaluate the integration we need to first express the pressure as a function of volume and temperature using the equation of state.

Question 12.
Derive Mayer’s relation for an ideal gas.
Answer:
Mayer’s relation: Consider p mole of an ideal gas in a container with volume V, pressure P and temperature T.
When the gas is heated at constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If C_v is the molar specific heat capacity at constant volume, from equation.

Suppose the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = µC_PdT …(3)
If W is the workdone by the gas in this process, then
W = P dV …(4)
But from the first law of thermodynamics,
Q = dU + W …. (5)
Substituting equations (2), (3) and (4) in (5), we get,

For mole of ideal gas, the equation of state is given by

Since the pressure is constant, dP = 0

This relation is called Mayer’s relation It implies that the molar specific heat capacity of an ideal gas at constant pressure is greater than molar specific heat capacity at constant volume. The relation shows that specific heat at constant pressure (s_p) is always greater than specific heat at constant volume (s_v).

Question 13.
Explain in detail the isothermal process.
Answer:
Isothermal process: It is a process in which the temperature remains constant but the pressure and volume of a thermodynamic system will change. The ideal gas equation is
PV = µRT
Here, T is constant for this process So the equation of state for isothermal process is given by
PV= Constant …(1)
This implies that if the gas goes from one equilibrium state (P₁, V₁) to another equilibrium state (P₂, V₂) the following relation holds for this process
P₁V₁ = P₂V₂ …(2)
Since PV = constant, P is inversely proportional to \(v\left(P \propto \frac{1}{V}\right)\). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm. We know that for an ideal gas the internal energy is a function of temperature only. For an isothermal process since temperature is constant, the internal energy is also constant. This implies that dU or ∆U = 0.
For an isothermal process, the first law of thermodynamics can be written as,
Q = W …(3)


From equation (3), we infer that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat to the system, the temperature will increase. For isothermal process this is not true. The isothermal compression takes place when the piston of the cylinder is pushed. This will increase the internal energy which will flow out of the system through thermal contact.

Question 14.
Derive the work done in an isothermal process.
Answer:
Work done in an isothermal process: Consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (P_i, V_i) to the final state (P_f, V_f). We can calculate the work done by the gas during this process. The work done by the gas,

As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid. Writing pressure in terms of volume and temperature,

Substituting equation (2) in (1) we get

In equation (3), we take uRT out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get

As a result the work done on the gas in an isothermal compression is negative.
In the PV diagram the work done during the isothermal expansion is equal to the area under the graph.

Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas which turns out to be the area with a negative sign.

Question 15.
Explain in detail an adiabatic process.
Answer:
Adiabatic process: This is a process in which no heat flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. So the pressure, volume and temperature of the system may change in an adiabatic process.
For an adiabatic process, the first law becomes ∆U = W.
This implies that the work is done by the gas at the expense of internal energy or work is done on the system which increases its internal energy.

The adiabatic process can be achieved by the following methods:
(i) Thermally insulating the system from surroundings so that no heat flows into or out of the system; for example, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.
(ii) If the process occurs so quickly that there is no time to exchange heat with surroundings even though there is no thermal insulation. A few examples are shown in Figure.
The equation of state for an adiabatic process is given by

Here γ is called adiabatic exponent (γ = C_p/C_v) which depends on the nature of the gas. The equation (1) implies that if the gas goes from an equilibrium state (P_i, V_i) to another equilibrium state (P_f, V_f) adiabatically then it satisfies the relation

The PV diagram for an adiabatic process is also called adiabat. But actually the adiabatic curve is steeper than isothermal curve.
We can also rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac{\mu \mathrm{RT}}{\mathrm{V}}\). Substituting this equation in the equation (1), we have

Note here that is another constant. So it can be written as

The equation implies that if the gas goes from an initial equilibrium state (T_i, V_i) to final equilibrium state (T_f, V_f) adiabatically then it satisfies the relation

The equation of state for adiabatic process can also be written in terms of T and P as

Question 16.
Derive the work done in an adiabatic process.
Answer:
Work done in an adiabatic process: Consider µ moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston of cross sectional area A is fitted in the cylinder.
Let W be the work done when the system goes from the initial state (P_i, V_i, T_i) to the final state (P_f, V_f, T_f) adiabatically.


By assuming that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PV_γ = constant (or)


In adiabatic expansion, work is done by the gas. i.e., W_adia is positive. As T_i > T_f the gas
cools during adiabatic expansion.
In adiabatic compression, work is done on the gas. i.e., W_adia is negative. As T_i< T_f the
temperature of the gas increases during adiabatic compression.

To differentiate between isothermal and adiabatic curves in PV diagram, the adiabatic curve is drawn along with isothermal curve for T_f and T_i. Note that adiabatic curve is steeper than isothermal curve. This is because γ > 1 always.

Question 17.
Explain the isobaric process and derive the work done in this process.
Isobaric process: This is a thermodynamic process that occurs at constant pressure. Even though pressure is constant in this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have

In an isobaric process the temperature is directly proportional to volume.
V ∝ T (Isobaric process) …. (2)
This implies that for a isobaric process, the V-T graph is a straight line passing through the origin.
If a gas goes from a state (V_i, T_i) to (V_f, T_f) at constant pressure, then the system satisfies the following equation

Examples for Isobaric process:
(i) When the gas is heated and pushes the piston so that it exerts a force equivalent to atmospheric pressure plus the force due to gravity then this process is isobaric.
(ii) Most of the cooking processes in our kitchen are isobaric processes. When the food is cooked in an open vessel, the pressure above the food is always at atmospheric pressure. The PV diagram for an isobaric process is a horizontal line parallel to volume axis.
Figure (a) represents isobaric process where volume decreases figure
(b) represents isobaric process where volume increases.

The work done in an isobaric process: Work done by the gas

In an isobaric process, the pressure is constant, so P comes out of the integral,

Where ∆V denotes change in the volume. If ∆V is negative, W is also negative. This implies that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas equation.

The equation (6) can also be rewritten using the ideal gas equation.
From ideal gas equation

Substituting this in equation (6) we get

In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the above diagram is equal to the work done by the gas.
The first law of thermodynamics for isobaric process is given by

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process: This is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables.
The pressure – volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.

The equation of state for an isochoric process is given by


We can infer that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (P_i, T_i) to (P_f, T_f) at constant volume, then the system satisfies the following equation

For an isochoric processes, ∆V = 0 and W = 0. Then the first law becomes
∆U = Q …(3)
Implying that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.
Suppose a system loses heat to the surroundings through conducting walls by keeping the volume constant, then its internal energy decreases. As a result the temperature decreases; the pressure also decreases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
Limitations of first law of thermodynamics: The first law of thermodynamics explains well the inter convertibility of heat and work. But it does not indicate the direction of change.
For example,
(a) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law, it is possible for the energy to flow from hot object to cold object or from cold object to hot object. But in nature the direction of heat flow is always from higher temperature to lower temperature.
(b) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted to the kinetic energy of the car. So the first law is not sufficient to explain many of natural phenomena.

Question 20.
Explain the heat engine and obtain its efficiency.
Answer:
Heat Engine: In the modem technological world, the role of automobile engines plays a vital role in for transportation. In motor bikes and cars there are engines which take in petrol or diesel as input, and do work by rotating wheels. Most of these automobile engines have efficiency not greater than 40%. The second law of thermodynamics puts a fundamental restriction on efficiency of engines. Therefore understanding heat engines is very important.

Reservoir: It is defined as a thermodynamic system which has very large heat capacity. By taking in heat from reservoir or giving heat to reservoir, the reservoir’s temperature does not change.
Example: Pouring a tumbler of hot water in to lake will not increase the temperature of the lake. Here the lake can be treated as a reservoir.
When a hot cup of coffee attains equilibrium with the open atmosphere, the temperature of the atmosphere will not appreciably change. The atmosphere can be taken as a reservoir.
We can define heat engine as follows: Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.
A heat engine has three parts:
(a) Hot reservoir
(b) Working substance
(c) Cold reservoir A Schematic diagram for heat engine is given below:
1. Hot reservoir (or) Source: It supplies heat to the engine. It is always maintained at a high temperature T_H.
2. Working substance: It is a substance like gas or water, which converts the heat supplied into Work.
3. Cold reservoir (or) Sink: The heat engine ejects some amount of heat (Q_L) in to cold reservoir after it doing work. It is always maintained at a low temperature T_L.
The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.
The efficiency of the heat engine is defined as the ratio of the work done (out put) to the heat absorbed (input) in one cyclic process.
Let the working substance absorb heat Q_H units from the source and reject Q_L units to the sink after doing work W units.
We can write. Input heat = Work done + ejected heat
Q_H = W + Q_L
W = Q_H – Q_L
Then the efficiency of heat engine

Note here that Q_H, Q_L and W all are taken as positive, a sign convention followed in this expression.
Since Q_L < Q_H, the efficiency (η) always less than 1. This implies that heat absorbed is not completely converted into work. The second law of thermodynamics placed fundamental restrictions on converting heat completely into work.

Question 21.
Explain in detail Carnot heat engine.
Answer:
In the year 1824 a young French engineer Sadi Carnot proved that a certain reversible engine operated in cycle between hot and cold reservoir can have maximum efficiency. This engine is called Carnot engine.
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine has four parts which are given below.
(i) Source: It is the source of heat maintained at constant high temperature T_H. Any amount of heat can be extracted from it, without changing its temperature.

(ii) Sink: It is a cold body maintained at a constant low temperature T_L. It can absorb any amount of heat.
(iii) Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
(iv) Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
Efficiency of a Carnot engine: Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.

Applying isothermal conditions, we get,

Here we omit the negative sign. Since we are interested in only the amount of heat (Q_L) ejected into the sink, we have

Substituting equation (5) in (4), we get

Note : T_L and T_H should be expressed in Kelvin scale.
Important results:
1. η is always less than 1 because T_L is less than T_H. This implies the efficiency cannot be 100%.
2. The efficiency of the Carnot’s engine is independent of the working substance. It depends only on the temperatures of the source and the sink. The greater the difference between the two temperatures, higher the efficiency.
3. When T_H = T_L the efficiency η = 0. No engine can work having source and sink at the same temperature.

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
The quantity \(\frac{Q}{T}\) is called entropy. It is a very important thermodynamic property of a system.

It is also a state variable. \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) is the entropy received by the Carnot engine from hot reservoir is entropy given out by the Carnot engine to the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. But for all practical engines 1 ike diesel and petrol engines which are not reversible engines, they satisfy the relation \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}>\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\)
In fact we can reformulate the second law of thermodynamics as follows “For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process should occur.
Entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics. Entropy is also called ‘measure of disorder’. All natural process occur such that the disorder should always increases.
Consider a bottle with a gas inside. When the gas molecules are inside the bottle it has less disorder. Once it spreads into the entire room it leads to more disorder. In other words when the gas is inside the bottle the entropy is less and once the gas spreads into entire room, the entropy increases. From the second law of thermodynamics, entropy always increases. If the air molecules go back in to the bottle, the entropy should decrease, which is not allowed by the second law of thermodynamics. The same explanation applies to a drop of ink diffusing into water. Once the drop of ink spreads, its entropy is increased. The diffused ink can never become a drop again. So the natural processes occur in such a way that entropy should increase for all irreversible process.

Question 24.
Explain in detail the working of a refrigerator.
Answer:
Refrigerator: A refrigerator is a Carnot’s engine working in the reverse order.
Working Principle: The working substance (gas) absorbs a quantity of heat Q_L from the cold body (sink) at a lower temperature T_L. A certain amount of work W is done on the working substance by the compressor and a quantity of heat Q_H is rejected to the hot body (source) ie, the atmosphere at T_H. When you stand beneath of refrigerator, you can feel warmth air. From the first law of thermodynamics, we have
Q_L + W = Q_H ….. (1)
As a result the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.
Coefficient of performance (COP) (β): COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

Inferences.
1. The greater the COP, the better is the condition of the refrigerator. A typical refrigerator has COP around 5 to 6.
2. Lesser the difference in the temperatures of the cooling chamber and the atmosphere, higher is the COP of a refrigerator.
3. In the refrigerator the heat is taken from cold object to hot object by doing external work. Without external work heat cannot flow from cold object to hot object. It is not a violation of second law of thermodynamics, because the heat is ejected to surrounding air and total entropy of (refrigerator + surrounding) is always increased.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.

The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa
Answer:

Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:
T = – 53°C = 220 K
P = 0.9 × 10³ Pa
V = 1 m³

Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₂ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
Let T be the equilibrium temperature and let n₁ and n₂ be the number of moles in vessels 1 and 2 respectively. As there is no loss of energy,

Substituting n₁ and n₂ values and solving, we get

Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion α_L is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its center and perpendicular to an axis of the rod.
Answer:
Moment of inertia of a uniform rod of mass and length l about its perpendicular bisector. Moment of inertia of the rod

Increase in length of the rod when temperature is increased by ∆T, is given by
L’ = L(1 + α_L∆T)
New moment of inertia of the rod

Question 5.
Draw the TP diagram (P – x axis, T – y axis), VT(T – x axis, V – y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) Isobaric process
Answer:

Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:
For ideal gas equation of state
PV = nRT
P₁ = 500 kPa, T₁ = 25°C = 25 + 273 = 298K, P₂ = 520 kPa, T₂ = ?
Expansion of tyre is negligible (V_constant)

Question 7.
Normal human body of the temperature is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body).
Answer:
Normal human body temperature (T) = 98.6°F
Convert Fahrenheit into Kelvin,

So, T = 98.6° F = 310 K
From Wien’s displacement law

Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Answer:
From equation for adiabatic process,

Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
Answer:

Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.

Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x – direction and V is taken along y – direction. Analyze the nature of heat exchange in each process.
Answer:
Process 1 to 2 = increase in volume. So heat must be added.
Process 2 to 3 = Volume remains constant. Increase in temperature.
The given heat is used to increase the internal energy.

Process 3 to 1 : Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system.
It is an isobaric compression where the work is done on the system.

Question 11.
An ideal gas is taken in a cyclic process as shown in the figure.
Calculate
(a) work done by the gas.
(b) work done on the gas
(c) Net work done in the process
Answer:
(a) Work done by the gas (along AB)


(c) Net work done in the process = Area under the curve AB
= Rectangle area + triangle area

Question 12.
For a given ideal gas 6 × 10⁵J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m³ at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change ¡n internal energy of the gas
(c) graph this process ¡n PV and TV diagram.
Answer:
Heat energy supplied to gas Q = 6 × 10⁵J
Change in volume ∆V = (6 – 4) = 2 m³
1 atm = 1.0 13 × 10⁵ Nm^-2
(a) Work done by the gas W = P × ∆V = 1.013 × 10⁵ × 2 = 2.026 × 10⁵
W = 202.6kJ1
(b) Change in internal energy of the gas

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency,
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) by increasing the temperature of the hot reservoir from 300°C to 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Heat engine operates at initial temperature = 100°C + 273 = 373 K
Final temperature = 300°C + 273 = 573 K
At melting point = 273 K

(a) By decreasing the cold reservoir, efficiency

(b) By increasing the temperature of hot reservoir, efficiency,

Method (a) More efficiency than method (b).

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained at a temperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Efficiency of Carnot engine (η₁) = 45% = 0.45
Initial intake temperature (T₁) = 327°C = 600 K
New efficiency (η₂) = 60% = 0.6
Efficiency of Carnot engine is given by

T₁ is temperature of source ; T₂ is temperature of sink

Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer:

Samacheer Kalvi 11th Physics Heat and Thermodynamics Textual Evaluation Solved Additional Questions Solved

I. Choose the correct answer from the following:

Question 1.
The coefficient of volume expansion of a solid is x times the coefficient of superficial expansion. Then x is
(a) 1.5
(b) 2
(c) 2.5
(d) 3
Answer:
(a) 1.5

Question 2.
A solid metal ball has a spherical cavity. If the ball is heated, the volume of the cavity will
(a) increase
(b) decrease
(c) remain unaffected
(d) remain unaffected but the shape of the cavity will change.
Answer:
(a) increase

Question 3.
A metal sheet with a circular hole is heated. The hole will
(a) contract
(b) expand
(c) remain unaffected
(d) contract or expand depending on the value of the linear expansion coefficient.
Answer:
(b) expand

Question 4.
The length of a metal rod at 0°C is 0.5m. When it is heated, its length increases by 2.7 mm. The final temperature of the rod is (coefficient of linear expansion of the metal = 90 × 10⁶/°C) ……
(a) 20°C
(b) 30°C
(c) 40°C
(d) 60°C
Answer:
(d) 60°C
Solution:
l_t = l₀(1 + αt)

Question 5.
A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically with iron strip towards left and copper strip towards right. If this bimetal is heated, it will
(a) remain straight
(b) bend towards right
(c) bend towards left
(d) bend forward
Answer:
(c) bend towards left
Solution:
Since α_copper > α_iron, the bimetal will bend towards iron, i.e., towards left.

Question 6.
When water is heated from 0°C to 10°C, its volume …..
(a) decreases
(b) increases
(c) first increase and then decrease
(d) first decreases and then increases.
Answer:
(d) first decreases and then increases.

Question 7.
A block of wood is floating on water at 0°C with a certain volume V above water level. The temperature of water is slowly raised to 20°C. How does the volume V change with the rise of temperature?
(a) remain unchanged
(b) decrease continuously
(c) decrease till 4°C and then increase
(d) increase till 4°C and then decrease.
Answer:
(a) increase till 4°C and then decrease.
Solution:
The density of water increases from 0° to 4°C and then decreases. Therefore, the volume V of the block above water level will increase till 4°C and then decrease.

Question 8.
An iron tyre is to be fitted on a wooden wheel 0.1m in diameter. The diameter of the tyre is 6 mm smaller than that of the wheel. The tyre should be heated by a temperature of (coefficient of volume expansion of iron is 3.6 × 10^-5/°C)
(a) 167°C
(b) 334°C
(c) 500°C
(d) 1000°C
Answer:
(a) 167°C
Solution:

Question 9.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm². The force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C is (coefficient of linear expansion of steel is 10 51°C and Young’s modulus of steel is 2 × 10¹⁰ N/m²) …….
(a) 40 N
(b) 80 N
(c) 120 N
(d) 160 N
Answer:
(d) 160N
Solution:
The required force is given by

Question 10.
Which of the following will make the volume of an ideal gas four times?
(a) double the absolute temperature and double the pressure.
(b) Halve the absolute temperature and double the pressure.
(c) Quarter the absolute temperature at constant pressure.
(d) Quarter the pressure at constant temperature.
Answer:
(d) Quarter the pressure at constant temperature.

Question 11.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas becomes.
(a) 54° C
(b) 150 K
(c) 327° C
(d) 327 K
Answer:
(c) 327°C
Solution:

Question 12.
An air bubble doubles in radius on rising from the bottom of a lake to its surface. If the atmospheric pressure is equal to that of a column of water of height H, the depth of lake is
(a) H
(b) 2H
(c) 7H
(d) 8H
Answer:
(c) 7H
Solution:
Since the radius becomes double, the volume becomes eight times. Therefore, according to Boyle’s law, the pressure becomes one-eighth. Now, the pressure at the surface Hρg. Therefore pressure at the bottom must be 8 Hρg. Hence the depth of the lake is 7H.

Question 13.
The mass of 1 litre of helium under a pressure of 2 atm and at a temperature of 27°C is
(a) 0.16 g
(b) 0.32 g
(c) 0.48 g
(d) 0.64 g
Answer:
(b) 0.32 g

Question 14.
Pressure exerted by a perfect gas is equal to …….
(a) mean kinetic energy per unit volume
(b) half of mean kinetic energy per unit volume
(c) one-third of mean kinetic energy per unit volume
(d) two-thirds of mean kinetic energy per unit volume
Answer:
(d) two-thirds of mean kinetic energy per unit volume

Question 15.
Two vessels A and B contain the same ideal gas. The volume of B is twice that of A, the pressure in B is twice that in A and the temperature of B is twice that of A. The ratio of the number of gas molecules in A and B is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(a) 1 : 2

Question 16.
According to Boyle’s law, PV = C when the temperature of the gas remains constant. The value of C depends on
(a) temperature of the gas
(b) nature of the gas
(c) quantity of the gas
(d) both temperature and quantity of the gas.
Answer:
(d) both temperature and quantity of the gas.

Question 17.
The pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1°C. The initial temperature was
(a) 250 K
(b) 250°C
(c) 25 K
(d) 25°C
Answer:
(c) 25 K

Question 18.
A temperature difference of 25°C is equivalent to a temperature difference of
(a) 25°F
(b) 45°F
(c) 67°F
(d) 77°F
Answer:
(b) 45°F

Question 19.
A temperature at which both the Fahrenheit and the centigrade scales have the same value is
(a) 40°
(b) -40°
(c) 20
(d) – 20°
Answer:
(b) – 40°
Solution: Let the required temperature be t. then, \(\frac{t}{5}=\frac{t-32}{9} \Rightarrow t=-40^{\circ}\)

Question 20.
If the temperature of patient is 40°C, his temperature on the Fahrenheit scale will be
(a) 72°F
(b) 96°F
(c) 100°F
(d) 104°F
Answer:
(d) 104°F

Question 21.
The correct value of 0°C on the Kelvin scale is ……..
(a) 273.15 K
(b) 272.85 K
(c) 273 K
(d) 273.2 K
Answer:
(a) 273.15 K

Question 22.
When a gas in a closed vessel was heated so as to increase its temperature by 5°C, there occurred an increase of 1% in its pressure, the original temperature of the gas was …….
(a) 50°C
(b) 227°C
(c) 273°C
(d) 500°C
Answer:
(b) 227°C

Question 23.
A perfect gas at 27°C is heated at constant pressure so as to double its volume. The temperature of the gas will be …….
(a) 600°C
(b) 54°C
(c) 327°C
(d) 300°C
Answer:
(c) 327°C

Question 24.
Temperature can be expressed as a derived quantity in terms of ……
(a) length and mass
(b) mass and time
(c) length, mass and time
(d) none of these
Answer:
(d) none of these

Question 25.
The equation of state corresponding to 8 g of O₂ is
(a) PV = 8RT
(b) PV = RT/4
(c) PV = RT
(d) PV = RT/2
Answer:
(b) PV = RT/4
Solution:
8g of O₂ is 1/4 of a mole of O₂, which is 32g. Thus, the required equation of state is PV = \(\frac{1}{4} \mathbf{R} \mathbf{T}\).

Question 26.
At a given volume and temperature, the pressure of a gas ….
(a) varies inversely as its mass
(b) varies inversely as the square of its mass
(c) varies linearly as its mass
(d) is independent of its mass
Answer:
(c) varies linearly as its mass

Question 27.
Oxygen boils at -183°C. This temperature in Fahrenheit scale is
(a) -215.7°
(b) – 297.4°
(c) -310.6°
(d) – 373.2°
Answer:
(b) – 297.4°
Solution:

Question 28.
A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the decrease in temperature as registered by the centigrade thermometer?
(a) 80°
(b) 60°
(c) 40°
(d) 30°
Answer:
(c) 40°
Solution:

Question 29.
The change in temperature of a body is 50°C. The change on the kelvin scale is …….
(a) 50K
(b) 323K
(c) 70K
(d) 30K
Answer:
(a) 50K

Question 30.
Mercury thermometers can be used to measure temperature up to ……..
(a) 260°C
(b) 100°C
(c) 360°C
(d) 500°C
Answer:
(c) 360°C

Question 31.
For an ideal gas the inter particle interaction is ……..
(a) attractive
(b) repulsive
(c) very large
(d) zero
Answer:
(d) zero

Question 32.
Device used to measure very high temperature is …….
(a) Pyrometer
(b) Thermometer
(c) Bolometer
(d) calorimeter
Answer:
(a) Pyrometer

Question 33.
Two metal rods A and B are having their initial lengths in the ratio 2 : 3, and coefficients of linear expansion in the ratio 3 : 4. When they are heated through same temperature difference,the ratio of their linear expansions is ……
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 3
Answer:
(a) 1 : 2

Question 34.
Boyles’ law is applicable in ……
(a) isochoric process
(b) isothermal process
(c) isobaric process
(d) both (a) and (b)
Answer:
(b) isothermal process

Question 35.
A rod, when heated from 0°C to 50°C, expands by 1.0 mm. Another rod, twice as long as the first at 0°C and of the same material, is heated from 0°C to 25°C. The second rod will expand by ……
(a) 0.5 mm
(b) 1.0 mm
(c) 2.0 mm
(d) 4.0 mm
Answer:
(b) 1.0 mm

Question 36.
A container contains hydrogen gas at pressure P and temperature T. Another identical container contains helium gas at pressure 2P and temperature T/2. The ratio of the number of molecules of the two gases is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
Answer:
(a) 1 : 4
Solution:
The ratio of the number of molecules is same as the ratio of the number of moles.
Now n = \(\frac{P V}{R T}\)

Question 37.
Density of water is maximum at the temperature of
(a) 32°F
(b) 39.2°F
(c) 42°F
(d) 40°F
Answer:
(b) 39.2°F
Solution:
The density of water is maximum at 4°C. Let F be the corresponding temperature on the Fahrenheit scale. Then using the equation

Question 38.
The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, is (R is the gas constant) …….
(a) PV = (5/32) RT
(b) PV = 5RT
(c) PV = (5/2)RT
(d) PV = (5/16)RT
Answer:
(a) PV = (5/32) RT
Solution:
Number of moles, n = \(\frac{5}{32}\)

Question 39.
A bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown. The metal X has a higher coefficient of expansion compared to that for metal Y. When the bimetallic strip is placed in a cold bath?

(a) it will bend towards the right
(b) it will bend towards the left
(b) it will not bend but shrink
(d) it will neither bend nor shrink
Answer:
(b) it will bend towards the left.
Solution:
In cold bath, the metal X will contract more than the metal Y. Therefore, the strip will bend towards the left.

Question 40.
An ideal gas is expanding such that PT² = constant the coefficient of volume expansion of the gas is

Answer:
\(\frac{3}{\mathrm{T}}\)
Solution:
Coefficient of volume expansion \(\alpha_{\mathrm{v}}=\frac{d \mathrm{V}}{\mathrm{v} d \mathrm{T}}\)
PT² = K and PV = nRT

Question 41.
A metallic solid sphere is rotating about its diameter as axis of rotation. If the temperature is increased by 200°C, the percentage increase in its moment of inertia is : (coefficient of linear expansion of the metal = 10^-5/°C)
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4
Answer:
(d) 0.4

Question 42.
The difference between volume and pressure coefficients of an ideal gas is ……

Answer:
(d) Zero

Question 43.
Which of the following instruments is used in the measurement of temperatures above 2000°C?
(a) Gas thermometer
(b) Pyrometer
(c) Bolometer
(d) Thermo-electric Pile
Answer:
(b) Pyrometer

Question 44.
At 0°C, Pressure measured by barometer is 760 mm. What will be the pressure at 100°C?
(a) 760
(b) 730
(c) 780
(d) none of these
Answer:
(d) none of these

Question 45.
The temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale?
(a) 73°W
(b) 117°W
(c) 200°W
(d) 139°W
Answer:
(b) 117°W
Solution:
Let t be the required temperature. Then,

Question 46.
Two balloons are filled, one with pure He gas and the other with air. If the pressure and temperature in both the balloons are same the number of molecules per unit volume is
(a) more in the He filled balloon
(b) more in the air filled balloon.
(c) same in both the balloon.
(d) in the ratio 1 : 4.
Answer:
(c) same in both the balloons.
Solution: Assuming ideal gas behavior, the number of moles per unit volume is \(\frac{n}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\) Since P and T are same in both the balloon, \(\frac{n}{V}\) is also same in both.

Question 47.
Pressure of an ideal gas is increased by keeping temperature constant. What is the effect on the kinetic energy of molecules?
(a) increase
(b) no change
(c) decrease
(d) can’t be determined
Answer:
(b) no change
Solution:
K.E of an ideal gas depends only on the temperature. Hence, it remains the same.

Question 48.
One mole of gas occupies a volume of 200 ml. at 100 mm pressure. What is the volume occupied by two moles of this gas at 400 mm pressure and at same temperature?
(a) 50 ml
(b) 100 ml
(c) 200 ml
(d) 400 ml
Answer:
(b) 100 ml
Solution:

⇒ Volume of 2 moles of the gas at 400 mm pressure = 2 × 50 = 100 ml

Question 49.
There is a change in length when a 33000 N tensile force is applied on a steel rod of area of cress-section 10^-3 m². The change of temperature required to produce the same elongation, if the steel rod is heated, is (modulus of elasticity of steel = 3 × 10¹¹ N/m², coefficient of linear expansion of steel = 1.1 × 10^-5/°C) …….
(a) 20°C
(b) 15°C
(c) 10°C
(d) 0°C
Answer:
(c) 10°C
Solution:

Question 50.
In the given (V – T) diagram, what is the relation between pressures P₁ and P₂?
(a) P₂ = P₁
(b) P₂ > P₁
(c) P₂ < P₁
(d) cannot be predicted
Answer:
(c) P₂ < P₁

Question 51.
Boiling water is changing into steam. Under this condition the specific heat of water is
(a) zero
(b) one
(c) infinite
(d) less than one
Answer:
(c) infinite
Solution:
In order to change boiling water into steam, heat has to be given but there is no increase of temperature. Therefore, under this condition the specific heat of water is infinite.

Question 52.
The first law of thermodynamics is concerned with the conservation of ……
(a) number of molecules
(b) energy
(c) number of moles
(d) temperature
Answer:
(b) energy

Question 53.
The gas law \(\frac{P V}{T}\) = constant is true for
(a) isothermal changes only
(b) adiabatic changes only
(c) both isothermal and adiabatic changes
(d) neither isothermal nor adiabatic changes
Answer:
(c) both isothermal and adiabatic changes

Question 54.
The pressure-temperature relationship for an ideal gas undergoing adiabatic change is …..

Answer:
(a) \(\mathrm{p}^{1-\gamma} \mathrm{T}^{\gamma}\) = constant

Question 55.
For a certain gas the ratio of specific heats is given to be γ = 1.5. For this gas ……..
(a) C_v = 3R
(b) C_P = 3R
(c) C_V = 5R
(d) C_P = 5R
Answer:
(b) C_P = 3R
Solution:

Question 56.
Cooking takes longest time ………
(a) at the sea level
(b) at Shimla
(c) at mount Everest (if tried)
(d) in a submarine 100 m below the surface of water.
Answer:
(c) at mount Everest (if tried)

Question 57.
A closed bottle containing water at room temperature is taken to the moon and then the lid is opened. The water will ……
(a) freeze
(b) boil
(c) decompose into hydrogen and oxygen
(d) not change at all.
Answer:
(b) boil
Solution:
There is no atmosphere on the moon and so there is no pressure

Quarter 58.
A gas receives an amount of heat equal to 110 joules and performs 40 joules of work. The change in the internal energy of the gas is …….
(a) 70 J
(b) 150 J
(c) 110 J
(d) 40 J
Answer:
(a) 70 J

Question 59.
For a mono-atomic gas, the molar specific heat at constant pressure divided by the molar gas constant R, is equal to ……
(a) 2.5
(b) 1.5
(c) 5.0
(d) 3.5
Answer:
(a) 2.5

Question 60.
Heat capacity of a substance is infinite. It means …….
(a) infinite heat is given out
(b) infinite heat is taken in
(c) no change in temperature whether heat is taken in or given out
(d) all of these
Answer:
(c) no change in temperature whether heat is taken in or given out

Question 61.
We consider a thermodynamic system. If ∆U represent the increase in its energy and W the work done by the system, which of the following statements is true?
(a) ∆U = – W in an isothermal process
(b) ∆U = – W in an adiabatic process
(c) ∆U = W in an isothermal process
(d) ∆U = W in an adiabatic process
Answer:
(b) ∆U = – W in an adiabatic process
Solution:
According to the first law of thermodynamics ∆Q = AU + W
In an adiabatic process, ∆Q = 0. Therefore, ∆U = – W

Question 62.
The first operation involved in a carnot cycle is
(a) isothermal expansion
(b) adiabatic expansion
(c) isothermal compression
(d) adiabatic compression
Answer:
(a) isothermal expansion

Question 63.
During an adiabatic process, if the pressure of an ideal gas is proportional to the cube of its temperature, the ratio \(\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}\) is ….

Answer:
(d) \(\frac{3}{2}\)
Solution:

Question 64.
In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas ……..
(a) The temperature will decrease.
(b) the volume will decrease.
(c) the pressure will remain constant.
(d) the temperature will increase.
Answer:
(a) The temperature will decrease.
Solution:
According to the first law of thermodynamics, the internal energy decreases. Hence the temperature will decrease.

Question 65.
In a carnot heat engine 8000J of heat is absorbed from a source at 400 K and 6500 J of heat is rejected to the sink. The temperature of the sink is …..
(a) 325 K
(b) 100 K
(c) 200 K
(d) 273 K
Answer:
(a) 325 K

Question 66.
2 Kg of water of 60°C is mixed with 1 kg of water at 30°C kept in a vessel of heat capacity 220 J K^-1. The specific heat of water is 4200 J Kg-1K ‘. Then the final temperature is nearly.
(a) 35°C
(b) 45°C
(c) 55°C
(d) 50°C
Answer:
(d) 50°C
Solution:
According to the principle of calorimetry,

Question 67.
A carnot engine absorbs heat at 127°C and rejects heat at 87°C. The efficiency of engine is
(a) 10%
(b) 30%
(c) 50%
(d) 70%
Answer:
(a) 10%
Solution:

Question 68.
The first law of thermodynamics confirms the law of ……
(a) conservation of momentum
(b) conservation of energy
(c) flow of heat in a particular direction
(d) conservation of heat energy and mechanical energy
Answer:
(b) conservation of energy

Question 69.
The internal energy of an ideal gas depends on
(a) only pressure
(b) only volume
(c) only temperature
(d) none of these
Answer:
(c) only temperature

Question 70.
An ideal gas heat engine operators in a carnot’s cycle between 227°C and 127°C. It absorbs 6 × 10⁴J at high temperature. The amount of heat converted into work is
(a) 2.4 × 10⁴ J
(b) 4.8 × 10⁴J
(c) 1.2 × 10⁴J
(d) 6 × 10⁴J
Answer:
(c) 1.2 × 10⁴J

Question 71.
In an isochoric process
(a) ∆U = ∆Q
(b) ∆Q = ∆W
(c) ∆U = ∆W
(d) None of these
Answer:
(a) ∆U = ∆Q
Solution:
In an isochoric process, volume remains constant. Therefore, no work is done by or on the system. So, ∆W = 0 Hence ∆U = ∆Q.

Question 72.
The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio specific heats at constant pressure to that at constant volume is ………
(a) 8/7
(b) 5/7
(c) 9/7
(d) 7/5
Answer:
(d) 7/5
Solution:

Question 73.
If energy dQ is supplied to a gas isochorically, increase in internal energy is dU. Then ….
(a) dQ > dU
(b) dQ < dU
(c) dQ = dU
(d) dQ = -dU
Answer:
(c) dQ = dU

Question 74.
A diatomic ideal gas is used in a camot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency of the engine is ……
(a) 0.25
(b) 0.5
(c) 0.75
(d) 0.99
Answer:
(c) 0.75

Question 75.
The mechanical equivalent of heat J is:
(a) a constant
(b) a physical quantity
(c) a conversion factor
(d) none of the above
Answer:
(c) a conversion factor

Question 76.
Which of the following process is reversible?
(a) transfer of heat by radiation
(b) Transfer of heat by conduction
(c) Electrical heating of nichrome wire
(d) Isothermal compression
Answer:
(d) Isothermal compression

Question 77.
An ideal gas heat engine operates in camot cycle between 227°C and 127°C. It absorbs 6 × 10⁴ cal of heat converted to work is ……
(a) 1.2 × 10⁴ cal
(b) 4.8 × 10⁴ cal
(c) 6 × 10⁴ cal
(d) 2.4 × 10⁴ cal
Answer:
(a) 1.2 × 10⁴ cal

Question 78.
Ten moles of an ideal gas at constant temperature 600 K is compressed from 100 l to 10 L. The work done in the process is …..
(a) 4.11 × 10⁴ J
(b) -4.11 × 10⁴ J
(c) 11.4 × 10⁴ J
(d) – 11.4 × 10⁴ J
Answer:
(d) -11.4 × 10⁴ J
Solution:
The process is isothermal. The work done is,

Question 79.
A gas is compressed at a constant pressure of 50 N/m² from a volume 4 m³. Energy of 100 J is then added to the gas by heating. Its internal energy is …..
(a) increased by 400 J
(b) increased by 200 J
(c) increased by 100 J
(d) decreased by 200 J
Answer:
(a) increased by 400 J
Solution:
∆U = ∆Q – ∆W = ∆Q – P∆V = 100 – 50 (4 – 10) = 400 J

Question 80.
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then ……
(a) W = 0
(b) Q = W = O
(c) E = 0
(d) Q = 0
Answer:
(c) E = 0
Solution:
In a cyclic process, a system starts in one state and comes back to the same state. Therefore, the change in internal energy is zero.

Question 81.
A carnot engine takes heat from a reservoir at 627°C and rejects heat to a sink at 27°C. Its efficiency is
(a) 3/5
(b) 1/3
(c) 2/3
(d) 200/209
Answer:
(c) 2/3

Question 82.
A carnot engine operates with source at 127°C and sink at 27°C. If the source supplies 40 KJ of heat energy, the work done by the engine is
(a) 1 KJ
(b) 4 KJ
(c) 10 KJ
(d) 30 KJ
Answer:
(c) 10 KJ
Solution:

Question 83.
The ratio of the specific heat \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\) in term of degrees of freedom (n) is given by:

Answer:
(b) \(\left(1+\frac{2}{n}\right)\)
Solution:

Question 84.
The heat required to increase the temperature of 4 moles of a mono-atomic ideal gas from 273 K to 473 K constant volume is ….
(a) 200 R
(b) 400 R
(c) 800 R
(d) 1200 R
Answer:
(d) 1200 R
Solution:

Question 85.
The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is – 20°C, the temperature of the surroundings to which it rejects heat is ……
(a) 21°C
(b) 31°C
(c) 41°C
(d) 11°C
Answer:
(b) 31°C

II. Write brief answer to the following questions:

Question 1.
What is meant by ‘heat’?
Answer:
When an object at higher temperature is placed in contact with another object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat.

Question 2.
What is meant by ‘temperature’? Give its unit.
Answer:
Temperature is the degree of hotness or coolness of a body. Hotter the body higher is its temperature. The temperature will determine the direction of heat flow when two bodies are in thermal contact. The SI unit of temperature is kelvin (K).

Question 3.
Define Avogadro’s number N_A?
Answer:
The Avogadro’s number N_A is defined as the number of carbon atoms contained in exactly 12g of ¹²C

Question 4.
Define heat capacity (or) thermal capacity?
Answer:
The heat capacity of a body is defined as the amount of heat required to raise its temperature through one degree.

Question 5.
What is meant by ‘Triple point of a substance’?
Answer:
The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid and solid) of that substance coexist in thermodynamic equilibrium.

Question 6.
What do you meant by change of state of a substance?
Answer:
The transition of a substance from one state to another by heating or cooling it is called change of state.

Question 7.
Define coefficient of linear expansion. Give its unit.
Answer:
The coefficient of linear expansion of the material of a solid rod is defined as the increase in length per unit original length per degree rise in its temperature.
The unit of α_L is °C^-1 (or KT^-1)

Question 8.
Define coefficient of area expansion? Give its unit.
Answer:
The coefficient of area expansion of a metal sheet is defined as the increase in its surface area per unit original surface area per degree rise in its temperature.
The unit of α_A is °Cα^-1 (or) KT^-1

Question 9.
Define coefficient of Volume expansion? Give its unit.
Answer:
The coefficient of Volume expansion of a substance is defined as the increase in volume per unit original volume per degree rise in its temperature.
The unit of α_v is °C^-1 (or) KT^-1

Question 10.
What is meant by conduction?
Answer:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Question 11.
What is meant by Convection?
Answer:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Question 12.
What is meant by Radiation? Give example.
Answer:
Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example:
1. Solar energy from the Sun.
2. Radiation from room heater.

Question 13.
State Newton’s Law of cooling.
Answer:
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

Question 14.
What do you meant by absolute zero of temperature?
Answer:
The lowest temperature of 0 K at which a gas is supposed to have zero volume (and zero pressure) and at which entire molecular motion stops is called absolute zero of temperature.

Question 15.
State Prevost theory of heat exchange.
Answer:
Prevost theory states that all bodies emit thermal radiation at all temperatures above absolute zero irrespective of the nature of the surroundings.

Question 16.
What is meant by ‘Mechanical equilibrium’?
Answer:
A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermo dynamic system or on the surrounding by thermodynamic system.

Question 17.
What is meant by ‘chemical equilibrium’?
Answer:
Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.

Question 18.
What is meant by ‘thermodynamic equilibrium’.
Answer:
In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 19.
Briefly explain how such a quasi-static process can be carried out.
Answer:
Quasi-static process: Consider a system of an ideal gas kept in a cylinder of volume V at pressure P and temperature T. When the piston attached to the cylinder moves outward the volume of the gas will change. As a result the temperature and pressure will also change because all three variables P,T and V are related by the equation of state PV = NkT. If a block of some mass is kept on the piston, it will suddenly push the piston downward. The pressure near the piston will be larger than other parts of the system. It implies that the gas is in non-equilibrium state. We cannot determine pressure, temperature or internal energy of the system until it reaches another equilibrium state. But if the piston is pushed very slowly such that at every stage it is still in equilibrium with surroundings, we can use the equation of state to calculate the internal energy, pressure or temperature. This kind of process is called quasi-static process.
A quasi-static process is an infinitely slow process in which the system changes its variables (P,V,T) so slowly such that it remains in thermal, mechanical and chemical equilibrium with its surroundings throughout. By this infinite, slow variation, the system is always almost close to equilibrium state.

Question 20.
Define specific heat capacity at constant pressure.
Answer:
Specific heat capacity at constant pressure (s_P): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the pressure constant is called specific heat capacity of at constant pressure.

Question 21.
Define specific heat capacity at constant volume.
Answer:
Specific heat capacity at constant volume (s_V): The amount of heat energy required to raise the temperature of one kg of a substance by 1 K or 1°C by keeping the volume constant is called specific heat capacity at constant volume.

Question 22.
Define molar specific heat capacity at constant volume.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1 °C at constant volume is called molar specific heat capacity at constant volume.

Question 23.
Define molar specific heat capacity at constant pressure.
Answer:
The amount of heat required to rise the temperature of one mole of a substance by 1K or 1°C at constant pressure is called molar specific heat capacity at constant pressure.

Question 24.
What is a isobaric process?
Answer:
It is a thermodynamic process which occurs at a constant pressure.

Question 25.
What is a isochoric process?
Answer:
It is a thermodynamic process which occurs at a constant volume.

Samacheer Kalvi 11th Physics Heat and Thermodynamics Numerical Problems

Question 1.
Copper wire of length l increase in length by 1% when heated from temperature T₁ to T₂. Find the percentage change in area when a copper plate of dimensions 2l × l is heated from T₁ to T₂
Answer:

Question 2.
A steel rod of length 25 cm has a cross-sectional area of 0.8 cm². Find the force required to stretch this rod by the same amount as the expansion produced by heating it through 10°C. (coefficient of linear expansion of steel is 10^-5/°C and Young’s modulus of steel is 2 × 10¹⁰Nm^-2)
Answer:

Question 3.
The temperature at the bottom of a 40 m deap lake is 12°C and that at the surface is 35°C. An air bubble of volume 1.0 cm3 rises from the bottom to the surface. Find its volume, (atmospheric pressure = 10 m of water)
Answer:
Let P₁, V₁ and T₁ be the pressure, bubble volume and absolute temperature at the bottom of the lake and let P₂, V₂ and T₂ be the corresponding quantities at the surface. Then

Question 4.
Ajar ‘A’ is filled with an ideal gas characterised by parameters P, V and T and another jar B is filled with an ideal gas with parameters 2P, \(\frac{\mathrm{V}}{4}\) and 2T. Find the ratio of the number of molecules in jar A and B.
Answer:

Question 5.
By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature?
Answer:
According to Boyle’s law,

Question 6.
A flask is filled with 13g of an ideal gas at 27°C and its temperature is raised to 52°C. Find the mass of the gas that hias to be released to maintain the temperature of the gas in the flask at 52°C and the pressure is same as the initial pressure.
Answer:
Let n be the initial number of moles and ri be the final number of moles. Since pressure and volume remain the same, we have

Question 7.
A rod of metal-1 of length 50.0 cm elongates by 0.10 cm when it is heated from 0°C to 100°C. Another rod of metal-2 of length 80.0 cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50.0 cm, made by welding pieces of rod 1 rad and 2 rad placed end to end, elongates by 0.03 cm when it is heated from 0°C to 50°C. Then what is the length of metal-1 in the third rod at 0°C?
Answer:
For metal-1, For metal-2,

Let the lengths of metal-1 and metal- 2 in the third rod at 0°C be l₁ and l₂, respectively.

Question 8.
A balloon is filled at 27°C and 1 atm pressure by 500 m³ He. Then find the volume of He at -3°C and 0.5 mm Hg pressure.
Answer:

Question 9.
During an experiment an ideal gas is found to obey an additional law VP² = constant. The gas is initially at temperature T and Volume V. Find the resulting temperature when it expands to volume 2V.
Answer:

Question 10.
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weights of A and B is:
Answer:
According to the ideal gas equation for one mole,

Where ρ is the density and M is the molecular weight of the gas.

Question 11.
For a gas the difference between the two specific heats is 4150 J/Kg K. What is the specific heat of the gas at constant volume if the ratio of specific heat is 1.4?
Answer:

Question 12.
A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming γ(= C_P/C_V) to be 1.5, find the new pressure of the gas.
Answer:
Let the initial volume be V. After isothermal expansion, pressure = \(\frac{P}{4}\), volume = 4V
Let P’ is the pressure after adiabatic compression. The volume then is V. Therefore,

Question 13.
A carnot engine operating between temperature T₁ and T₂ has efficiency \(\frac{1}{6}\). When T₂ is lowered by 62 K, its efficiency increases to \(\frac{1}{3}\). Then find the values of T₁ and T₂.
Answer:

Question 14.
A perfect gas goes from state A to state B by absorbing 8 × 10⁵ J of heat and doing 6.5 × 10⁵ J of external work. It is now transferred between the same two states in another process in which it absorbs 10s J of heat. In second process. Find the work done in the second process.
Answer:
According to the first law of thermodynamics
∆U = ∆Q – ∆W
In the first process ∆U = 8 × 10⁵ – 6.5 × 10⁵ = 1.5 × 10⁵ J
Now ∆U, being a state function, remains the same in the second process,
∆W = ∆Q – ∆U = 1 × 10⁵ – 1.5 × 10⁵
∆W = – 0.5 × 10⁵J
The negative sign shows that work is done on the gas.

Question 15.
A carnot engine, having efficiency of \(\eta=\frac{1}{10}\) as heat engine, is used as a refrigerator. If the work done on the system is 10 J, then find the amount of energy absorbed from the reservoir at lower temperature.
Answer:
As heat engine, let Q₁ be the heat energy absorbed from the source and Q₂ be the energy rejected to the sink. Then

From equation (1) and (2),

Question 16.
An ideal gas compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
Answer:
Work done on the gas = Area under the curve

W_adiabatic > W_isothermal > W_isobaric
W_isochoric is obviously zero because in an isochoric process there is no change in Volume.

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Samacheer Kalvi 11th Physics Properties of Matter Textual Evaluation Solved

Samacheer Kalvi 11th Physics Properties of Matter Multiple Choice Questions
Question 1.
Consider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they are stretched by the same load then the stress on Y is ……….
(a) equal to that on X
(b) thrice that on X
(c) nine times that on X
(d) Half that on X
Answer:
(c) nine times that on X
Solution:

Question 2.
If a wire is streched to double of its original length, then the strain in the wire is ………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4
Solution:

In this case Young’s modulus is equal to stress

Question 3.
The load-elongation graph of three wires of the same material are shown in figure. Which of the following wire is the thickest?
(a) wire 1
(b) wire 2
(c) wire 3
(d) all of them have same thickness
Answer:
(a) wire 1
Solution:

Wire 1 is the thickest compared to other wires.
Because the elongation of the wire 1 is minimum.

Question 4.
For a given material, the rigidity modulus is \(\left(\frac{1}{3}\right)^{r d}\) of Young’s modulus. Its Poisson’s ratio is
(a) 0
(b) 0.25
(c) 0.3
(d) 0.5
Answer:
(d) 0.5
Solution:
The relationship of Poisson’s ratio, rigidity and Young’s modulus is

Question 5.
A small sphere of radius 2 cm falls from rest in a viscous liquid. Heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to ……… [NEET model 2018]
(a) 2²
(b) 2³
(c) 2 ⁴
(d) 2⁵
Answer:
(d) 2⁵
Solution:
Rate of heat produced

Question 6.
Two wires are made of the same material and have the same volume. The area of cross sections
of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount? (NEET model 2018)
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(b) 4
Solution:
Since the two wires have same volume
Length of wire 1 = l, Area of wire 1 = A
Length of wire 2 = \(\frac{l}{2}\), Area of wire 2 = 2A

Question 7.
With an increase in temperature, the viscosity of liquid and gas, respectively will
(a) increase and increase
(b) increase and decrease
(c) decrease and increase
(d) decrease and decrease
Answer:
(c) decrease and increase

Question 8.
The Young’s modulus for a perfect rigid body is ……..
(a) 0
(b) 1
(c) 0.5
(d) infinity
Answer:
(d) infinity

Question 9.
Which of the following is not a scalar?
(a) viscosity
(b) surface tension
(c) pressure
(d) stress
Answer:
(d) stress

Question 10.
If the temperature of the wire is increased, then the Young’s modulus will …….
(a) remain the same
(b) decrease
(c) increase rapidly
(d) increase by very a small amount
Answer:
(b) decrease
Solution:
As temperature is increased, the strain i.e., change dimensions of the body is increased, as a result, the stiffness of the material is reduced. Which causes decrease in magnitude of modulus of elasticity.

Question 11.
Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is ∆l. If Y represents the Young’s modulus, then which of the following graphs is a straight line? [NEET 2014 model]
(a) ∆l versus V
(b) ∆l versus Y
(c) ∆l versus F
(d) ∆l versus \(\frac{1}{l}\)
Answer:
(c) ∆l versus F
Solution:
From Young’s modulus,

Question 12.
A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

Answer:

Question 13.
The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(a) length = 200 cm, diameter = 0.5 mm
(b) length = 200 cm, diameter = 1 mm
(c) length = 200 cm, diameter = 2 mm
(d) length = 200 cm, diameter = 3 mm
Answer:
(a) length = 200 cm, diameter = 0.5 mm
Solution:
From Young’s modulus,

Question 14.
The wettability of a surface by a liquid depends primarily on
(a) viscosity
(b) surface tension
(c) density
(d) angle of contact between the surface and the liquid
Answer:
(d) angle of contact between the surface and the liquid

Question 15.
In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 ms^-1 at a point where the diameter of the pipe is 20 cm. The velocity of water 1.5 (ms^-1) at a point where the diameter of the pipe is.
(a) 8
(b) 16
(c) 24
(d) 32
Answer:
(b) 16
Solution:
From equation of continuity,

Samacheer Kalvi 11th Physics Properties of Matter Short Answer Questions

Question 1.
Define stress and strain.
Answer:
Stress: The restoring force per unit area of a deformed body is known as stress.
Strain: Strain produced in a body is defined as the ratio of change in size of a body to the original size.

Question 2.
State Hooke’s law of elasticity.
Answer:
The stress is proportional to the strain in the elastic limit.

Question 3.
Define Poisson’s ratio.
Answer:
Poisson’s ratio, which is defined as the ratio of relative contraction (lateral strain) to relative
expansion (longitudinal strain). It is denoted by the symbol μ.

Question 4.
Explain elasticity using intermolecular forces.
Answer:
Elastic behaviour of solid. In a solid, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to the neighbouring molecules. When deforming force is applied on a body so that its length increases, then the molecules of the body go far apart.

Question 5.
Which one of these is more elastic, steel or rubber? Why?
Answer:
Steel is more elastic than rubber. If an equal stress is applied to both steel and rubber, the steel produces less strain. So the Young’s modulus is higher for steel than rubber. The object which has higher young’s modulus is more elastic.

Question 6.
A spring balance shows wrong readings after using for a long time. Why?
Answer:
When a spring balance has been used for a long time, it develops an elastic fatigue, the spring of such a balance take longer time to recover its original configuration and therefore it does not give correct measurement.

Question 7.
What is the effect of temperature on elasticity?
Answer:
As the temperature of substance increases, its elasticity decreases.

Question 8.
Write down the expression for the elastic potential energy of a stretched wire.
Answer:
The work done in stretching the wire by dl,
dW = F.dl
The total work done in stretching the wire from 0 to l is

From Young’s modulus of elasticity, force becomes,

Substituting equation (2) in (1) we get,

This work done is known as the elastic potential energy of a stretched wire.

Question 9.
State Pascal’s law in fluids.
Answer:
If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Question 10.
State Archimedes principle.
Answer:
If states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.

Question 11.
What do you mean by upthrust or buoyancy?
Answer:
The upward force exerted by a fluid that opposes the weight of an immersed object in a fluid is called upthrust or buoyant force.

Question 12.
State the law of floatation.
Answer:
The law of floatation states that a body will float in a liquid if the weight of the liquid displaced by the immersed part of the body equals the weight of the body.

Question 13.
Define coefficient of viscosity of a liquid.
Answer:
The coefficient of viscosity of a liquid is the viscous force acting tangentially per unit area of a liquid layer having a unit velocity gradient in a direction perpendicular to the direction of flow of the liquid.

Question 14.
Distinguish between streamlined flow and turbulent flow.
Answer:

Streamlined flow	Turbulent Flow
When a liquid flow such that each particle of the liquid passing a point moves along the same path and has the same velocity as its predecessor than the flow of liquids is said to be streamlined flow.	During the flow of fluid, when the critical velocity is exceeded by the moving fluid, the motion becomes turbulent.

Question 15.
What is Reynold’s number? Give its significance.
Answer:
It is a dimensionless number which determines the nature of the flow of fluid through a pipe. Reynold’s number is given by,

Question 16.
Define terminal velocity.
Answer:
The maximum constant velocity acquired by a body while falling freely through a viscous medium is called the terminal velocity V_T.

Question 17.
Write down the expression for the Stoke’s force and explain the symbols involved in it.
Answer:
The viscous force F acting on a spherical body of radius r depends directly on:
(i) radius (r) of the sphere
(ii) velocity (v) of the sphere and
(iii) coefficient of viscosity η of the liquid

On solving, we get x = 1, y = 1 and z = 1. Therefore, F = kηrv
Experimentally, Stoke found that the value of k = 6π
F = 6πηrv This relation is known as Stoke’s law.

Question 18.
State Bernoulli’s theorem.
Answer:
It states that the sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined, irrotational flow remains constant along
a streamline.

Question 19.
What are the energies possessed by a liquid? Write down their equations.
Answer:
A liquid in motion possesses following three types of energy:
(i) Kinetic energy: It is the energy possessed by a liquid by virtue of its motion.

(ii) Potential energy: It is the energy possessed by a liquid by virtue of its height above the ground level.
P.E. = mgh

(iii) Pressure energy: It is the energy possessed by a liquid by virtue of its pressure. Pressure energy per unit mass = \(\frac{\mathbf{P}}{m / v}\)

Question 20.
Two streamlines cannot cross each other. Why?
Answer:
If two streamlines cross each other, there will be two directions of flow at the point of intersection which is impossible.

Question 21.
Define surface tension of a liquid. Mention its S.I. Unit and dimension.
Answer:
The surface of a liquid is defined as the, force per unit length of a liquid (or) the energy per unit area of the surface of a liquid. T = \(\frac{\mathbf{F}}{l}\)
SI unit and dimensions of T are Nm^-1 and MT^-2

Question 22.
How is surface tension related to surface energy?
Answer:
The surface energy per unit area of a surface is numerically equal to the surface tension.

Question 23.
Define angle of contact for a given pair of solid and liquid.
Answer:
The angle between tangents drawn at the point of contact to the liquid surface and solid surface inside the liquid is called the angle of contact for a pair of solid and liquid. It is denoted by θ.

Question 24.
Distinguish between cohesive and adhesive forces.
Answer:
The force between the like molecules which holds the liquid together is called ‘cohesive force’. When the liquid is in contact with a solid, the molecules of the these solid and liquid will experience an attractive force which is called‘adhesive force’.

Question 25.
What are the factors affecting the surface tension of a liquid?
Answer:

1. The presence of any contamination or impurities.
2. The presence of dissolved substances.
3. Electrification
4. Temperature

Question 26.
What happens to the pressure inside a soap bubble when air is blown into it?
Answer:
When air is blown into a soap bubble, the pressure inside a bubble is decreased P = \(\frac{4 \mathrm{T}}{\mathrm{R}}\)

Question 27.
What do you mean by capillarity or capillary action?
Answer:
In a liquid whose angle of contact with solid is less than 90°, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Question 28.
A drop of oil placed on the surface of water spreads out. But a drop of water placed on oil contracts to a spherical shape. Why?
Answer:
The surface tension of the water is more than that of oil. Therefore, when oil is poured over water, the greater value of surface tension of water pulls oil in all directions, and as such it spreads on the water. On the other hand, when water is poured over oil, it does not spread over it because the surface tension of oil being less than that of water, it is not able to pull water over it.

Question 29.
State the principle and usage of venturimeter.
Answer:
This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem.

Samacheer Kalvi 11th Physics Properties of Matter Long Answer Questions

Question 1.
State Hooke’s law and verify it with the help of an experiment.
Answer:
Hooke’s law is for a small deformation, when the stress and strain are proportional to each other. It can be verified in a simple way by stretching a thin straight wife (stretches like spring) of length L and uniform cross sectional area A suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire.

The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ∆L. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ∆L on the Y-axis.
Therefore ∆L = (slope)F
Multiplying and dividing by volume,
V = AL


Comparing with stress and strain equations,
σ ∝ ε
i.e., the stress is proportional to the strain in the elastic limit.

Stress-strain profile curve: The stress versus strain profile is a plot in which stress and strain are noted for each load and a graph is drawn taking strain along the X-axis and stress along the Y-axis. The elastic characteristics of the materials can be analyzed from the stress-strain profile.

(a) Portion OA: In this region, stress is very small such that stress is proportional to strain, which means Hooke’s law is valid. The point A is called limit of proportionality because above this point Hooke’s law is not valid. The slope of the line OA gives the Young’s modulus of the wire.
(b) Portion AB: This region is reached if the stress is increased by a very small amount. In this region, stress is not proportional to the strain. But once the stretching force is removed, the wire will regain its original length. This behaviour ends at point B and hence, the point B is known as yield point (elastic limit). The elastic hebaviour of the material (here wire) in stress-strain curve is OAB.
(c) Portion BC: If the wire is stretched beyond the point B (elastic limit), stress increases and the wire will not regain its original length after the removal of stretching force.
(d) Portion CD: With further increase in stress (beyond the point C), the strain increases rapidly and reaches the point D. Beyond D, the strain increases even when the load is removed and breaks (ruptures) at the point E. Therefore, the maximum stress (here D) beyond which the wire breaks is called breaking stress or tensile strength. The corresponding point D is known as fracture point. The region BCDE represents the plastic behaviour of the material of the wire.

Question 2.
Explain the different types of modulus of elasticity.
Answer:
From Hooke’s law, the stress in a body is proportional to the corresponding strain, provided the deformation is very small. Here we shall define the elastic modulus of a given material. There are three types of elastic modulus.
(a) Young’s modulus
(b) Rigidity modulus (or Shear modulus)
(c) Bulk modulus
Young’s Modulus: When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young’s modules.

S.I. unit of Young modulus is Nm^-2 or pascal.

Bulk modulus: Bulk modulus is defined as the ratio of volume stress to the volume strain.

The negative sign indicates when pressure is applied on the body, its volume decreases. Further, the equation implies that a material can be easily compressed if it has a small value of bulk modulus. In other words, bulk modulus measures the resistance of solids to change in their volume. For an example, we know that gases can be easily compressed than solids, which means, gas has a small value of bulk modulus compared to solids. The S.I. unit of K is the same as that of pressure i.e., N m^{– 2} or Pa (pascal).
The rigidity modulus or shear modulus: The rigidity modulus is defined as the ratio of the shearing stress to shearing strain,

Further, the above implies, that a material can be easily twisted if it has small value of rigidity modulus. For example, consider a wire, when it is twisted through an angle θ, a restoring torque is developed, that is
\(\tau \propto \theta\)
This means that for a larger torque, wire will twist by a larger amount (angle of shear θ is large). Since, rigidity modulus is inversely proportional to angle of shear, the modulus of rigidity is small. The S.I. unit of ηR is the same as that of pressure i.e., N m^-2 or pascal.

Question 3.
Derive an expression for the elastic energy stored per unit volume of a wire.
Answer:
When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension 1 and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.
The work done in stretching the wire by dl, dW = Fdl
The total work done in stretching the wire from 0 to l is

From Young’s modulus of elasticity

Substituting equation (2) in equation (1), we get

Since, l is the dummy variable in the integration, we can change l to l’ (not in limits), therefore

Energy per unit volume is called energy density

Question 4.
Derive an equation for the total pressure at a depth ‘A’ below the liquid surface.
Answer:
In order to understand the increase in pressure with depth below the water surface, consider a water sample of cross sectional area in the form of a cylinder. Let h₁ and h₂ be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively. Let F₁ be the force acting downwards on level 1 and F₂ be the force acting upwards on level 2, such that, F₁ = P₁ A and F₂ = P₂A. Let us assume the mass of the sample to be m and under equilibrium condition, the total upward force (F₂) is balanced by the total downward force (F₁ + mg), in other words, the gravitational force will act downward which is being exactly balanced by the difference between the force F₂ – F₁.
F₂ – F₁ = mg …… (1)
where m is the mass of the water available in the sample element. Let ρ be the density of the water then, the mass of water available in the sample element is
m = ρV = ρA(h₂ – h₁)
V = A(h₂ – h₁)
Hence, gravitational force, F_G = ρA (h₂ – h₁)g
On substituting the W value in equation (1)
F₂ = F₁ + mg ⇒ P₂A = P₁A + ρA(h₂ – h₁)g
Cancelling out A on both sides,

P₂ = P₁ + ρ(h₂ – h₁ )g …(2)

If we choose the level 1 at the surface of the liquid (i.e., air-water interface) and the level 2 at a depth ‘h’ below the surface, then the value of h₁ becomes zero (h₁ = 0) and in turn P₁ assumes the value of atmospheric pressure (say P_a). In addition, the pressure (P₂) at a depth becomes P. Substituting these values in equation, we get
P = P_a + ρgh …… (3)
which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where Pa is the atmospheric pressure which is equal to 1.013 × 10⁵ Pa.
If the atmospheric pressure is neglected or ignored then P = ρgh ….. (4)
for a given liquid, ρ is fixed and g is also constant, then the pressure due to the fluid column is directly proportional to vertical distance or height of the fluid column. This implies, the height of the fluid column is more important to decide the pressure and not the cross sectional or base area or even the shape of the container.

Let us consider three vessels of different shapes A, B and C as shown in figure. These vessels are connected at the bottom by a horizontal pipe. When they are filled with a liquid (say water), it occupies the same level even though the vessels hold different amounts of water. It is true because the liquid at the bottom of each section of the vessel experiences the same pressure.

Question 5.
State and prove Pascal’s law in fluids.
Answer:
Statement of Pascal’s law: If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Application of Pascal’s law: Hydraulic lift: A practical application of Pascal’s law is the hydraulic lift which is used to lift a heavy load with a small force. It is a force Hydraulic lift multiplier. In consists of two cylinders A and B connected to each other by a horizontal pipe, filled with a liquid. They are fitted with frictionless pistons of cross sectional areas A₁ and A₂ (A₂ > A₁). Suppose a downward force F is applied on the smaller piston, the pressure of the liquid under this piston increase to But according to Pascal’s law, this increased pressure P is transmitted undiminished in all directions. So a pressure is exerted on piston B. Upward force on piston B is

Therefore by changing the force on the smaller piston A, the force on the piston B has been increased by the factor \(\frac{\mathrm{A}_{2}}{\mathrm{A}_{1}}\) and this factor is called the mechanical advantage of the lift.

Question 6.
State and prove Archimedes principle.
Answer:
Archimedes Principle: It states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.
Upthrust (or) buoyant force = Weight of liquid displaced
Proof: Consider a body of height ‘h’ lying inside a liquid of density ρ, at a depth x below the free surface of the liquid. Area of cross section of the body is ‘a’. The forces on the sides of the body cancel out.
Pressure at the upper face of the body, P₁ = xρg
Pressure at the lower face of the body, P₂ = (x + h)ρg
Thrust acting on the upper face of the body is F₁ = P₁a = xρga acting vertically downwards,
Thrust acting on the lower face of the body is F₂ = P₂a = (x + h) ρga acting vertically upwards.
The resultant force (F₂ – F₁) is acting on the body direction and is called upthrust (U).
U = F₂ – F₁ = (x + h) ρga – xρga = ahρg
But ah = V, Volume of the body = Volume of liquid

U = Vρg = Mg
i.e., Upthrust or buoyant force = Weight of liquid displaced.
This proves the Archimedes principle.

Question 7.
Derive the expression for the terminal velocity of a sphere moving in a high viscous fluid using stokes force.
Answer:
Expression for terminal velocity: Consider a sphere of radius r which falls freely through a
highly viscous liquid of coefficient of viscosity η. Let the density of the material of the sphere be ρ and the density of the fluid be σ.
Gravitational force acting on the sphere,

Here, it should be noted that the terminal speed of the sphere is directly proportional to the
square of its radius. If a is greater than ρ, then the term (ρ – σ) becomes negative leading to a
negative terminal velocity.

Question 8.
Derive Poiseuille’s formula for the volume of a liquid flowing per second through a pipe
under streamlined flow.
Answer:
Consider a liquid flowing steadily through a horizontal capillary tube. Let v = \(\left(\frac{\mathrm{v}}{t}\right)\) be the volume of the liquid flowing out per second through a capillary tube. It depends on (1) coefficient of viscosity (η) of the liquid, (2) radius of the tube (r), and (3) the pressure gradient \(\left(\frac{P}{l}\right)\).

Substituting in equation (1)

So, equating the powers of M, L and T on both sides, we get
a + c = 0, -a + b – 2c = 3, and – a – 2c = – 1
We have three unknowns a, b and c. We have three equations, on solving, we get
a = – 1, b = 4 and c = 1
Therefore, equation (1) becomes,

Experimentally, the value of k is shown to be \(\frac{\pi}{8}\), we have

The above equation is known as Poiseuille’s equation for the flow of liquid through a narrow tube or a capillary tube. This relation holds good for the fluids whose velocities are lesser than the critical velocity (v_c).

Question 9.
Obtain an expression for the excess of pressure inside a
(i) liquid drop
(ii) liquid bubble
(iii) air bubble.
Answer:
1. Excess of pressure inside air bubble in a liquid:
Consider an air bubble of radius R inside a liquid having surface tension T. Let P₁ and P₂ be the pressures outside and inside the air bubble, respectively.
Now, the excess pressure inside the air bubble is ∆P = P₁ – P₂
In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,

(i) The force due to surface tension acting towards right around the rim of length 2πR is F_T = 2πRT
(ii) The force due to outside pressure P, is to the right acting across a cross sectional area of πR² is \(\mathrm{P}_{1} \pi \mathrm{R}^{2}\)
(iii) The force due to pressure P₂ inside the bubble, acting to the left is \(\mathrm{F}_{\mathrm{P}_{2}}=\mathrm{P}_{2} \pi \mathrm{R}^{2}\)
As the air bubble is in equilibrium under the action of these forces, \(\mathrm{F}_{\mathrm{P}_{2}}=\mathrm{F}_{\mathrm{T}}+\mathrm{F}_{\mathrm{P}_{1}}\)
Excess pressure is ∆P = P₂ – P₁ = \(\frac{2 \mathrm{T}}{\mathrm{R}}\)

2. Excess pressure inside a soap bubble: Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Therefore, the force on the soap bubble due to surface tension is 2 × 2πRT. The various forces acting on the soap bubble are,

(i) Force due to surface tension F_T = 4πRT towards right.

3. Excess pressure inside the liquid drop: Consider a liquid drop of radius R and the surface tension of the liquid is T.
The various forces acting on the liquid drop are:

Question 10.
What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method.
Answer:
In a liquid whose angle of contact with solid is less than 90°, suffers capillary rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Practical application of capillarity
(i) Due to capillary action, oil rises in the cotton within an earthen lamp. Likewise, sap raises from the roots of a plant to its leaves and branches.
(ii) Absorption of ink by a blotting paper.
(iii) Capillary action is also essential for the tear fluid from the eye to drain constantly.
(iv) Cotton dresses are preferred in summer because cotton dresses have fine pores which act as capillaries for sweat.
Surface Tension by capillary rise method: The pressure difference across a curved liquid-air interface is the basic factor behind the rising up of water in a narrow tube (influence of gravity is ignored). The capillary rise is more dominant in the case of very fine tubes. But this phenomenon is the outcome of the force of surface tension. In order to arrive a relation between the capillary rise (h) and surface tension (T), consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.
The surface tension force F_T, acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, it resolved into two components
(i) Horizontal component T sin θ and
(ii) Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus Total upward force = (T cos θ) (2πr) = 2πrT cos θ
where θ is the angle of contact, r is the radius of the tube. Let ρ be the density of water and h be the height to which the liquid rises inside the tube. Then,

The upward force supports the weight of the liquid column above the free surface, therefore,

If the capillary is a very fine tube of radius (i.e., radius is very small) then \(\frac{r}{3}\) can be neglected when it is compared to the height h. Therefore,

This implies that the capillary rise (h) is inversely proportional to the radius (r) of the tube, i. e., the smaller the radius of the tube greater will be the capillarity.

Question 11.
Obtain an equation of continuity for a flow of fluid on the basis of conservation of mass.
Answer:
The mass flow rate through a pipe, it is necessary to assume that the flow of fluid is steady, the flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant with respect to time.

Under this condition, the path taken by the fluid particle is a streamline.
Consider a pipe AB of varying cross sectional area a₁ and a₂ such that a₁ > a₂. A non-viscous -and incompressible liquid flows steadily through the pipe, with velocities v₁ and v₂ in area a₁ and a₂, respectively.
Let m₁ be the mass of fluid flowing through section A in time ∆t, m₁ = (a₁v₁∆t)ρ
Let m₂ be the mass of fluid flowing through section B in time ∆t, m₂ = (a₂v₂∆t)ρ
For an incompressible liquid, mass is conserved m₁ = m₂

which is called the equation of continuity and it is a statement of conservation of mass in the flow of fluids.
In general, a v = constant, which means that the volume flux or flow rate remains constant throughout the pipe. In other words, the smaller the cross section, greater will be the velocity of the fluid.

Question 12.
State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.
Answer:
Bernoulli’s theorem: According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,


This is known as Bernoulli’s equation.
Proof: Let us consider a flow of liquid through a Flow of liquid through a pipe AB
pipe AB. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.
Let the force exerted by the liquid at A is F_A = P_Aa_A
Distance travelled by the liquid in time t is d = v_At

since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is

Similarly, let a_B, v_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at E_B, we get

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then

Question 13.
Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube.
Answer:
Venturimeter : This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli’s theorem. It consists of two wider tubes A and A’ (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A manometer in the form of U-tube is also attached between the wide and narrow tubes. The manometer contains a liquid of density ‘ρ_m’.
Let P₁ be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density ‘ρ’ flows from the pipe with speed ‘v₁’ and into the narrow region, its speed increases to ‘v₂’. According to the Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid pressure P₂ at the narrow region of the tube B. Therefore, the pressure difference between the tubes A and B‘ is noted by measuring the height difference (∆P = P₁ – P₂) between the surfaces of the manometer liquid.
From the equation of continuity, we can say that Av₁ = av₂ which means that

From the above equation, the pressure difference

The volume of the liquid flowing out per second is

Samacheer Kalvi 11th Physics Properties of Matter Numerical Problems

Question 1.
A capillary of diameter dmm is dipped in water such that the water rises to a height of 30 mm. If the radius of the capillary is made of its previous value, then compute the
height up to which water will rise in the new capillary?
Answer:

Question 2.
A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of 4 × 10⁵ N is applied at the other end. If the rigidity modulus of the cylinder is 6 × 10¹⁰ Nm^-2 then, calculate the twist produced in the cylinder.
Answer:

Question 3.
A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.
Answer:
From the excess pressure inside a soap bubble

Here the two bubbles having the same pressure and temperature. So the radius of the combined bubbles,

Question 4.
A block of Ag of mass x kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of Ag is 10 and tension in the string is 37.12 N then compute the mass of Ag block.
Answer:
From the terminal velocity condition, F_G – U = F F = T, m = x

Question 5.
The reading of pressure meter attached with a closed pipe is 5 × 10⁵ Nm^{– 2}. On opening the valve of the pipe, the reading of the pressure meter is 4.5 × 10⁵ Nm^-2. Calculate the speed of the water flowing in the pipe.
Answer:
Using Bernoulli’s equation

Here initial velocity V₁ = 0 and density of water p = 1000 kg m^-3

Samacheer Kalvi 11th Physics Properties of Matter Conceptual Questions

Question 1.
Why coffee runs up into a sugar lump (a small cube of sugar) when one corner of the sugar lump is held in the liquid?
Answer:
Dip the comer of a sugar cube in coffee, and get the whole cube coffee-flavoured due to “capillary action”.

Question 2.
Why two holes are made to empty an oil tin?
Answer:
When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside.

Question 3.
We can cut vegetables easily with a sharp knife as compared to a blunt knife. Why?
Answer:
The area of a sharp edge is much less than the area of a blunt edge. For the same total force, the effective force per unit area is more for the sharp edge than the blunt edge. Hence, a sharp knife cuts easily than a blunt knife.

Question 4.
Why the passengers are advised to remove the ink from their pens while going up in an aeroplane?
Answer:
We know that atmospheric pressure decreases with height. Since ink inside the pen is filled at the atmospheric pressure existing on the surface of Earth, it tends to come out to equalise the pressure. This can spoil the clothes of the passengers, so they are advised to remove the ink from the pen.

Question 5.
We use straw to suck soft drinks, why?
Answer:
When we suck through the straw, the pressure inside the straw becomes less than the atmospheric pressure. Due to the pressure difference, the soft drink rises in the straw and we are able to take the soft drink easily.

Samacheer Kalvi 11th Physics Properties of Matter Additional Questions

I. Choose the correct answer from the following:

Question 1.
The force required to stretch a steel wire 1 cm² in cross section to double its length is (given Y = 2 × 10¹¹ Nm^-2) ……….

Answer:
(b) 2 × 10⁷ N
Solution:

Question 2.
The fractional change in volume per unit increase in pressure is called ……..
(a) Pressure co-efficient
(b) Volume co-efficient
(c) Bulk modulus
(d) Compressibility
Answer:
(d) Compressibility

Question 3.
The modulus of rigidity of a liquid is
(a) zero
(b) 1
(c) infinite
(d) none of these
Answer:
(a) zero

Question 4.
The Young’s modulus of a wire of length L and radius r is Y. If the length is reduced to \(\frac{\mathrm{L}}{2}\) and radius to \(\frac{\mathrm{r}}{2}\), its Young’s modulus will be …….
(a) \(\frac{\mathrm{Y}}{2}\)
(b) Y
(c) 2Y
(d) 4Y
Answer:
(b) Y
Solution:
The Young’s modulus is a property of the material. So, it remains the same.

Question 5.
A spherical ball contracts in volume by 0.01% when subjected to a normal uniform pressure of 100 atmospheres. The bulk modulus of the material of the ball in dynes/cm² is …….
(a) 1 × 10¹²
(b) 10 × 10¹²
(c) 100 × 10¹²
(d) 2 × 10¹¹
Answer:
(a) 1 × 10¹²
Solution:

Question 6.
The value of Poisson’s ratio lies between
(a) 0 and 1
(b) – 0.5 and 1
(c) 0 and 0.5
(d) – 1 and 1
Answer:
(c) 0 and 0.5

Question 7.
Poisson’s ratio cannot have the value ……
(a) 0.1
(b) 0.2
(c) 0.5
(d) 0.7
Answer:
(d) 0.7

Question 8.
The bulk modulus for an incompressible fluid is ………
(a) zero
(b) 1
(c) ∞
(d) between 0 and 1
Answer:
(c) ∞

Question 9.
The breaking stress of a wire depends upon
(a) length of the wire
(b) material of the wire
(c) radius of the wire
(d) shape of the cross-section
Answer:
(b) material of the wire

Question 10.
Shearing stress causes change in ……
(a) length
(b) breadth
(c) shape
(d) volume
Answer:
(c) shape

Question 11.
A certain force increases the length of a wire by 1 mm. The force required to increases its length by 2 mm is …….
(a) 2F
(b) 4F
(c) 8F
(d) 16F
Answer:
(a) 2F
Solution:
∆l ∝ F

Question 12.
Two wires of same material, having cross-sectional areas in the ratio 1 : 2 and lengths in the ratio 1 : 4 are stretched by the same force. The ratio of the stresses in the wires will be …….
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(b) 2 : 1
Solution:

Question 13.
If the tension on a wire is removed at once, then ……..
(a) it will break
(b) its temperature will reduce
(c) there will be no change in its temperature
(d) its temperature will increase
Answer:
(d) its temperature will increase
Solution:
When tension is applied, the distance between the atoms of the wire increases, there by increasing the potential energy of the wire. When the tension is removed, the potential energy decreases. This energy is converted into heat energy. So the temperature of the wire increases.

Question 14.
In steel, the Young’s modulus and the strain at the breaking point are 2 × 10¹¹ Nm^-2 and 0.15 respectively. The stress at the breaking point for steel is therefore ………
(a) 2 × 10⁸ Nm^-2
(b) 3 × 10¹⁰ Nm^-2
(c) 3 × 10¹²Nm^-2
(d) None of these
Answer:
(b) 3 × 10¹⁰ Nm^-2
Solution:
Breaking stress = Breaking strain × Young’s modulus
= 0.15 × 2 × 10¹¹ = 3 × 10¹⁰ Nm^-2 .

Question 15.
The pressure in a liquid at a given depth below the surface ……….
(a) is always exerted downward
(b) is the same in all directions
(c) equals the total weight of liquids above that depth
(d) depends upon the amount of liquid below that depth
Answer:
(b) is the same in all directions

Question 16.
The pressure at the bottom of a liquid tank does not depend on ………
(a) acceleration due to gravity
(b) density of the liquid
(c) height of the liquid
(d) area of the liquid surface
Answer:
(d) area of the liquid surface

Question 17.
The pressure of the Earth’s atmosphere at sea level is due to the ……
(a) gravitational attraction of the Earth for the atmosphere
(b) evaporation of water from the seas and oceans
(c) fact that most living things constantly breathe air
(d) heating of the atmosphere by the Sun
Answer:
(a) gravitational attraction of the Earth for the atmosphere

Question 18.
The operating principle of a hydraulic press is …….
(a) Pascal’s Law
(b) Archimedes principle
(c) Newton’s law of gravitation
(d) Boyle’s law
Answer:
(a) Pascal’s Law

Question 19.
A floating body always displaces its own ……..
(a) mass of liquid
(b) volume of liquid
(c) weight of liquid
(d) none of these
Answer:
(c) weight of liquid

Question 20.
The pressure in a water tap at the base of a building is 3 × 10⁶ dynes/cm² and on its top it is 1.6 × 10⁶ dynes/cm². The height of the building is approximately
(a) 7 m
(b) 14 m
(c) 70 m
(d) 140 m
Answer:
(b) 14 m
Solution:

Question 21.
The weight of a body in air is 100 N. How much will it weight in water, if it displaces 400 cc of water?
(a) 90 N
(b) 94 N
(c) 98 N
(d) None of these
Answer:
(d) None of these
Solution:
Upthrust = Weight of water displaced = 0.4 × 9.8 = 3.92 N
Apparent weight = 100 – 3.92 = 96.08 N

Question 22.
A body is floating in a liquid with \(\frac{1}{5}\) of its volume outside the liquid. If the relative density of the body is 0.9, that of the liquid is ……..

Answer:

Question 23.
A boat having length 3m and breadth 2m is floating on a lake. It sinks by 1 cm when a man gets on it. The mass of the man is ………
(a) 60 kg
(b) 55 kg
(c) 65 kg
(d) 70 kg
Answer:
(a) 60kg
Solution:
Weight of man = Weight of additional water displaced
mg = 3 × 2 × 0.01 × 103 × g m = 60 kg

Question 24.
A bird weighs 2 kg and is inside an airtight cage of 1 kg. If its starts to fly, then what is the weight of the bird and cage assembly?
(a) 3 kg
(b) 2 kg
(c) 1 kg
(d) none of these
Answer:
(a) 3 kg
Solution:
When the bird flies, the upthrust on it is equal and opposite to the down thrust on the cage. Therefore, the weight of the assembly remains unchanged.

Question 25.
Two light balls are suspended as shown in the figure. When a stream of air passes through the space between them, the distance between the balls will ……..

(a) increase
(b) decrease
(c) remain the same
(d) may increase or decrease depending on the speed of air
Answer:
(b) decrease
Solution:
When the speed of the air between the balls increases, then according to Bernoulli’s theorem, the pressure in this region decreases. Therefore, the balls will be pushed towards each other by the air pressure in the outer region.

Question 26.
The rate of leak from a hole in a tank is …….
(a) independent of its height from the bottom
(b) more if situated near the bottom
(c) more if situated near its top
(d) more at midway between top and bottom
Answer:
(b) more if situated near the bottom

Question 27.
When a fluid passes through the constricted part of a pipe, its ………
(a) velocity and pressure decrease
(b) velocity and pressure increase
(c) velocity decreases and pressure increases
(d) velocity increases and pressure decreases
Answer:
(d) velocity increases and pressure decreases

Question 28.
Bernoulli’s principle does not explain ……….
(a) curved path of a spinning ball
(b) surviving of a fish in a lake
(c) working of a paint sprayer
(d) automatic blowing off the roofs of houses during blizzard in hilly areas
Answer:
(b) surviving of a fish in a lake

Question 29.
An ideal liquid flows through a horizontal tube of variable diameter. The pressure is lowest where the ……..
(a) velocity is highest
(b) velocity is lowest
(c) diameter is largest
(d) none of these
Answer:
(a) velocity is highest

Question 30.
Bernoulli’s equation is applicable in the case of ……….
(a) streamlined flow of compressible fluids
(b) streamlined flow of incompressible fluids
(c) turbulent flow of compressible fluids
(d) turbulent flow of incompressible fluids
Answer:
(b) streamlined flow of incompressible fluids

Question 31.
Bernoulli’s theorem is based on the conservation of …….
(a) mass
(b) momentum
(c) energy
(d) all of the above
Answer:
(c) energy

Question 32.
Bernoulli’s theorem is applicable to ………
(a) flow of liquids
(b) viscocity
(c) surface tension
(d) static fluid pressure
Answer:
(a) flow of liquids

Question 33.
The working of an atomiser depends on ….
(a) Bernoulli’s principle
(b) Boyle’s law
(c) Archimedes principle
(d) Pascal’s law
Answer:
(a) Bernoulli’s principle

Question 34.
‘Dynamic lift’ is related to ……
(a) Bernoulli’s principle
(b) Archimedes principle
(c) Equation of continuity
(d) Pascal’s law
Answer:
(a) Bernoulli’s principle

Question 35.
A gale blows over a house. The force due to the gale on the roof is ………
(a) in the downward direction
(b) zero
(c) in the upward direction
(d) horizontal
Answer:
(c) in the upward direction

Question 36.
If a stream of air is blown under one of the pans of a physical balance in equilibrium, then the
pan will ……..
(a) go up
(b) go down
(c) not be affected
(d) go up or down depending on the velocity of the stream
Answer:
(b) go down

Question 37.
Water venturimeter works on the principle of ……….
(a) Newton’s third law of motion
(b) Stokes’s formula
(c) Bernoulli’s theorem
(d) Hooke’s law
Answer:
(c) Bernoulli’s theorem

Question 38.
Aeroplanes are made to run on runway before take-off because it ………
(a) decreases friction
(b) decreases viscous drag of air
(c) decreases atmospheric pressure
(d) provides required life to the aeroplane
Answer:
(d) provides required life to the aeroplane

Question 39.
When the terminal velocity is reached, the acceleration of a body moving through a viscous
medium is …….
(a) zero
(b) positive
(c) negative
(d) either (b) or (c) depending upon other factors.
Answer:
(a) zero

Question 40.
If a raindrop with a mass of 0.05 g falls with constant velocity, the retarding force of atmospheric
friction is (neglect density of air) …….
(a) zero
(b) 49 dynes
(c) 490 dynes
(d) none of these
Answer:
(b) 49 dynes
Solution:
Since the rain drop is falling with constant velocity, the retarding upward force is equal to its weight in magnitude, F = 0.05 × 980 = 49 dynes

Question 41.
If temperature rises, the coefficient of viscosity of a liquid …….
(a) decreases
(b) increases
(c) remains unchanged
(d) increases for some liquids and decreases for others
Answer:
(a) decreases

Question 42.
The velocity of a rain drop attains constant value because of ……..
(a) surface tension
(b) upthrust of air
(c) viscous force exerted by air
(d) air currents
Answer:
(c) viscous force exerted by air

Question 43.
With increase in temperature the viscosity of
(a) a gas decreases and a liquid increases
(b) a gas increases and a liquid decreases
(c) both gases and liquids decrease
(d) both gases and liquids increase
Answer:
(b) a gas increases and a liquid decreases

Question 44.
Two small spheres of radii r and 2r fall through a viscous liquid with the same constant speed.
The viscous forces experienced by them are in the ratio ……..
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(a) 1 : 2
Solution:

Question 45.
Viscosity is the property of liquids by virtue of which they ……
(a) oppose the relative motion of its parts
(b) push neighbouring molecules
(c) attract other molecules
(d) become conducting
Answer:
(a) oppose the relative motion of its parts

Question 46.
Streamlined flow is more likely for liquids with ………
(a) high density and low viscosity
(b) low density and high viscosity
(c) high density and high viscosity
(d) low density and low viscosity
Answer:
(b) low density and high viscosity

Question 47.
The dimensional formula of coefficient of viscosity is ……

Answer:
(d) ML^-1 T^-1

Question 48.
A good lubricant should have …….
(a) high viscosity
(b) low viscosity
(c) moderate viscosity
(d) high density
Answer:
(a) high viscosity

Question 49.
If a liquid wets a solid surface, the angle of contact is …….
(a) 0°
(b) 90°
(c) less than 90°
(d) greater than 90°
Answer:
(c) less than 90°

Question 50.
When some detergent is added to water, the surface tension ……..
(a) remains unaffected
(b) increases
(c) decreases
(d) may increase or decrease
Answer:
(c) decreases

Question 51.
Rain drops are spherical because of …….
(a) gravitational force
(b) surface tension
(c) low viscosity of water
(d) air resistance
Answer:
(b) surface tension

Question 52.
At critical temperature the surface tension of a liquid …….
(a) is zero
(b) is the same as that at any other temperature
(c) is infinity
(d) cannot be determined
Answer:
(a) is zero

Question 53.
A liquid will not wet the surface of a solid if the angle of contact is …..
(a) acute
(b) obtuse
(c) zero
(d) \(\frac{\pi}{2}\)
Answer:
(b) obtuse

Question 54.
The surface tension of soap solution is 25 × 10^-3 Nm^-1. The excess pressure inside a soap bubble of diameter 1 cm is ………
(a) 5 Pa
(b) 10 Pa
(c) 20 Pa
(d) None of these
Answer:
(c) 20 Pa
Solution:

Question 55.
Surface tension does not depend on ……..
(a) nature of the liquid
(b) temperature of the liquid
(c) atmospheric pressure
(d) presence of impurities
Answer:
(c) atmospheric pressure

Question 56.
Meniscus of mercury in a capillary is ……
(a) concave
(b) convex
(c) plane
(d) cylindrical
Answer:
(b) convex

Question 57.
The potential energy of a molecule on the surface of a liquid compared to that of a molecules
inside the liquid is …….
(a) smaller
(b) the same
(c) greater
(d) zero
Answer:
(c) greater

Question 58.
At which of the following temperatures, the value of surface tension of water is minimum?
(a) 4°C
(b) 25°C
(c) 50°C
(d) 15°C
Answer:
(d) 15°C

Question 59.
If the surface tension of water is 0.06 Nm^-1 then the capillary rise in a tube of diameter 1 mm is (angle of contact = 0°) ……..
(a) 1.22 cm
(b) 2.44 cm
(c) 3.12 cm
(d) 3.86 cm
Answer:
(b) 2.44 cm
Solution:

Question 60.
The surface tension phenomenon is the result of the tendency of a system to ………
(a) conserve energy
(b) conserve volume
(c) keep potential energy minimum
(d) keep surface area minimum
Answer:
(c) keep potential energy minimum

Samacheer Kalvi 11th Physics Properties of Matter 2 Mark Questions

II. Write brief answer to the following questions:

Question 1.
What is meant by ‘Mean positions of atoms’?
Answer:
The interaction between the atoms, they position themselves at a particular interatomic distance. This position of atoms in this bound condition is called their mean positions.

Question 2.
What is meant by ‘Evaporation’?
Answer:
When a liquid is heated at constant pressure to its boiling point or when the pressure is reduced at a constant temperature it will convert to a gas. This process of a liquid changing to a gas is called evaporation.

Question 3.
What are the physical states of matter?
Answer:
Solid, liquid, gas, plasma, Bose-Einstein condensates, quark-gluon plasmas and hot plasma these are the physical states of matter.

Question 4.
Define elasticity? Give its example.
Answer:
A body regains its original shape and size after the removal of deforming force, it is said be elastic and the property is called elasticity. Example: Rubber, metals, steet ropes

Question 5.
Define deforming force?
Answer:
The force which changes the size or shape of a body is called a deforming force.

Question 6.
Define ‘Plasticity’?
Answer:
If a body does not regain its original shape and size after removal of the deforming force, it is said to be a plastic body and the property is called plasticity.
Example : Glass.

Question 7.
Explain the classification of longitudinal stress?
Answer:
Longitudinal stress can be classified into two types, tensile stress and compressive stress. Tensile stress: Internal forces on the two sides of ∆A may pull each other, i.e., it is stretched by equal and opposite forces. Then, the longitudinal stress is called tensile stress. Compressive stress: When forces acting on the two sides of ∆A push each other, ∆A is pushed by equal and opposite forces at the two ends. In this case, ∆A is said to be under compression. Then, the longitudinal stress is called compressive stress.

Question 8.
Explain the classification of longitudinal strain?
Answer:
Longitudinal strain can be classified into two types: –

1. Tensile strain: If the length is increased from its natural length then it is known as tensile
   strain.
2. Compressive strain: If the length is decreased from its natural length then it is known as compressive strain.

Question 9.
Define elastic limit?
Answer:
Elastic limit: The maximum stress within which the body regains its original size and shape after the removal of deforming force is called the elastic limit.

Question 10.
What is meant by “Breaking stress or tensile strength”?
Answer:
The maximum stress ulitimate stress point beyond which the wire breaks is called breaking stress or tensile strength.

Question 11.
Define compressibility?
Answer:
The reciprocal of the bulk modulus is called compressibility. It is defined as the fractional change in volume per unit increase in pressure.

Question 12.
Define relative density (or) specific gravity?
Answer:
The relative density of a substance is defined as the ratio of the density of a substance to the density of water at 4°C. It is a dimensionless positive scalar quantity.

Question 13.
What is atmospheric pressure?
Answer:
The pressure exerted by the atmosphere at sea level is called atmospheric pressure. The atmospheric pressure at sea level is 1.013 × 10⁵ Nm^-2 (or) Pa

Question 14.
Explain hydrostatic paradox with suitable example.
Answer:
Hydrostatic paradox: The pressure exerted by a liquid column depends only on the height of the liquid column and not on the shape of the containing vessel:
Let us consider three vessels A, B and C of different shapes. These vessels are connected at the bottom by a horizontal pipe. When they are filled a liquid (say water), it occupies the same level even though the vessels hold different amount of water. It is true because the liquid at the bottom of each section of the vessel experiences the same pressure.

Question 15.
Write the examples of floating bodies?
Answer:

1. A person can swim in sea water more easily than in river water.
2. Ice floats on water.
3. The ship is made of steel but its interior is made hollow by giving it a concave shape.

Question 16.
Define ‘Viscosity’?
Answer:
Viscosity is defined as ‘the property of a fluid to oppose the relative motion between its layers’.

Question 17.
Define tube of flow.
Answer:
A bundle of streamlines having the same velocity over any cross section perpendicular to the
direction of flow then such bundle is called a ‘tube of flow’.

Question 18.
Write down the applications of viscosity?
Answer:
(i) The oil used as a lubricant for heavy machinery parts should have a high viscous coefficient. To select a suitable lubricant, we should know its viscosity and how it varies with temperature. [Note: As temperature increases, the viscosity of the liquid decreases]. Also, it helps to choose oils with low viscosity used in car engines (light machinery).
(ii) The highly viscous liquid is used to damp the motion of some instruments and is used as brake oil in hydraulic brakes.
(iii) Blood circulation through arteries and veins depends upon the viscosity of fluids.
(iv) Millikan conducted the oil drop experiment to determine the charge of an electron. He used the knowledge of viscosity to determine the charge.

Question 19.
What is meant by‘Molecular range’?
Answer:
It is the maximum distance upto which a molecule can exert force of attraction on another molecule. It is of the order of 10^-9 m for solids and liquids.

Question 20.
What is sphere of influence?
Answer:
It is a sphere drawn around a particular molecule as centre and molecular range as radius.

Samacheer Kalvi 11th Physics Properties of Matter Numerical Questions

Question 1.
Two steel wires of lengths 1 m and 2 m have diameters 1 mm and 2 mm respectively. If they are stretched by forces of 40 N and 80 N respectively, find the ratio of their elongations.
Solution:

Question 2.
A wire of length 2m and cross-sectional area 2 × 10^-6 m² is made of a material of Young’s modulus 2 × 10¹¹ Nm^-2. What is the work done in stretching it through 0.1 mm.
Answer:

Question 3.
A wire is stretched by 0.01 m when it is stretched by a certain force. Another wire of the same material but double the length and double the diameter is stretched by the same force. What is the elongation in metres?
Answer:

Question 4.
Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by same force. Calculate the ratio of workdone in stretching them.
Answer:

Question 5.
The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10^-11 Pa^-1 and density of water is 10³ kg m^-3. What fractional compression of water will be obtained at the bottom of the ocean?
Answer:

Question 6.
An iceberg of density 900 kgm^-3 is floating in water of density 1000 kgm^-3. What is the percentage of volume of iceberg outside the water?
Answer:
Fraction of volume inside water = Relative density of the body

Fraction of volume outside water = 1 – 0.9 = 0.1
Percentage of volume outside water = 0.1 × 100 = 10%

Question 7.
A cubical copper block has each side 2.0 cm. It is suspended by a string and submerged in oil of density 820 kg m^-3. Calculate the tension in the string.
(density of copper = 8920 kg m^-3)
Answer:
Tension = true weight – upthrust due to oil

Question 8.
By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density = 13.6 g cm^-2). Using the straw, he can drink water from a glass upto a maximum depth of.
Answer:
Pressure difference between atmosphere and lungs
∆P = 760 – 750 = 10 mm = 1 cm of Hg
Maximum depth upto which the student can suck water from a glass = 1 × 13.6 = 13.6 cm

Question 9.
A sphere made of a material of specific gravity 8 has a concentric spherical cavity and just sinks in water. Calculate the ratio of the radius of the cavity to that of the outer radius of the sphere.
Answer:
Let the radius of the sphere be R and that of the cavity be r. Then, since the sphere just sinks in water,

Question 10.
A solid floats in liquid A with half its volume immersed and in liquid B with \(\frac{2}{3}\) of its volume immersed. The densities of the liquids A and B are in the ratio.
Answer:

Question 11.
An ideal liquid is flowing in a cylindrical tube of internal diameter 4 cm with a velocity of 5 ms^-1. If this tube is connected to another tube of internal diameter 2 cm, then the velocity of the liquid in the second tube will be?
Answer:
According to the equation of continuity

Question 12.
The cylindrical tube of a spray pump has radius R, one end of which has ‘n’ fine holes, each of radius r. If the speed of the liquid in the tube is V, what is the speed of the ejection of the liquid through the holes.
Answer:
Let the required speed be v’.
According to equation of continuity a’v’ = av

Question 13.
What is the pressure on a swimmer 10 m below the surface of a lake?
Answer:

Question 14.
A square metal plate of 10 cm side moves parallel to another plate with a velocity of 10 cm s^-1, both plates immersed in water. If the viscous force is 200 dyne and viscosity of water is 0.01 poise. What is their distance apart?
Answer:

Question 15.
Find the terminal velocity of a steel ball 2 mm in diameter falling through glycerine. Relative density of steel is 8, relative density of glycerine is 1.3 and viscosity of glycerine is 8.3 poise.
Answer:

Question 16.
The terminal velocity of a copper ball of radius 2 mm falling through a tank of oil at 20°C is 6.5 cm s^-1. Compute the viscosity of the oil at 20°C. Density of oil = 1.5 × 10³ kg m ^-3, density of copper is 8.9 × 10³ kg m^-3.
Answer:

Question 17.
Water is flowing in a pipe of radius 1.5 cm with an average velocity of 15 cm s^-1. What is the nature of flow? Given coefficient of viscosity of water is 10^-3 kg m^-1s^-1 and its density
Answer:

As R_c > 3000, so the flow is turbulent.

Question 18.
Water is flowing with a speed of 2 ms^-1 in a horizontal pipe with cross- sectional area decreasing from 2 × 10^-2 m^-2 to 0.01 m² at pressure 4 × 10⁴ Pa. What will be the pressure at small cross-section?
Answer:

Question 19.
Calculate the height to which water will rise in capillary tube of 1.5 mm diameter. Surface tension of water is 7.4 × 10^-3 Nm^-1.
Solution:

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Samacheer Kalvi 11th Physics Gravitation Textual Evaluation Solved

Samacheer Kalvi 11th Physics Gravitation Multiple Choice Questions

Question 1.
The linear momentum and position vector of the planet is perpendicular to each other at
(a) perihelion and aphelion
(b) at all points
(c) only at parihelion
(d) no point
Answer:
(a) perihelion and aphelion

Question 2.
If the masses of the Earth and Sun suddenly double, the gravitational force between them will
(a) remain the same
(b) increase 2 times
(c) increase 4 times
(d) decrease 2 times
Answer:
(c) increase 4 times
Solution:
The gravitation force of attraction is given by F =
If the masses are doubled then the force will be

The gravitational force between them will be increase 4 times.

Question 3.
A planet moving along an elliptical orbit is closest to the Sun at distance r₁ and farthest away at a distance of r₂ If V₁ and v₂ are linear speeds at these points respectively. Then the ratio \(\frac{v_{1}}{v_{2}}\) is ……. [NEET 2016]
Answer:
(a) \(\frac{r_{2}}{r_{1}}\)
Solution:
According to the Law of conservation of angular momentum

Question 4.
The time period of a satellite orbiting Earth in a circular orbit is independent of ……..
(a) Radius of the orbit
(b) The mass of the satellite
(c) Both the mass and radius of the orbit
(d) Neither the mass nor the radius of its orbit
Answer:
(b) The mass of the satellite

Question 5.
If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be
(a) 64.5
(b) 1032
(c) 182.5
(d) 730
Answer:
(b) 1032
Solution:
By Kepler’s law

If the distance between the Earth and Sun were to be doubled from its present Value

Question 6.
According to Kepler’s second law, the radial vector to a planet from the Sun sweeps out equal areas in equal intervals of time. This law is a consequence of
(a) conservation of linear momentum
(b) conservation of angular momentum
(c) conservation of energy
(d) conservation of kinetic energy
Answer:
(b) conservation of angular momentum

Question 7.
The gravitational potential energy of the Moon with respect to Earth is
(a) always positive
(b) always negative
(c) can be positive or negative
(d) always zero.
Answer:
(b) always negative

Question 8.
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A, K_B and K_C respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then ….. [NEET 2018]

(a) K_A > K_B > K_C
(b) K_B < K_A < K_C
(c) K_A < K_B < K_C
(d) K_B > K_A > K_C
Answer:
(a) K_A > K_B > K_C
Solution:
The kinetic energy of a planet becomes

So, kinetic energy is inversly proportional to the orbital distance. Where the distance will be closer E_K is larger and the distance will be larger E_K is less.

Question 9.
The work done by the Sun’s gravitational force on the Earth is ….
(a) always zero
(b) always positive
(c) can be positive or negative
(d) always negative
Answer:
(c) can be positive or negative

Question 10.
If the mass and radius of the Earth are both doubled, then the acceleration due to gravity g1
(a) remains same
(b) \(\frac{g}{2}\)
(c) 2g
(d) 4g
Answer:
(b) \(\frac{g}{2}\)
Solution:
Acceleration due to gravity g’ = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
If the mass and radius of the Earth are both doubled

Question 11.
The magnitude of the Sun’s gravitational field as experienced by Earth is …..
(a) same over the year
(b) decreases in the month of January and increases in the month of July
(c) decreases in the month of July and increases in the month of January
(d) increases during day time and decreases during night time.
Answer:
(c) decreases in the month of July and increases in the month of January

Question 12.
If a person moves from Chennai to Trichy, his weight …..
(a) increases
(b) decreases
(c) remains same
(d) increases and then decreases
Answer:
(b) decreases

Question 13.
An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If
the lift is in free fall, the reading in the spring scale is ……..
(a) 98 N
(b) zero
(c) 49 N
(d) 9.8 N
Answer:
(b) zero
Solution:
The lift is in freefall. It and its contents will experience apparent weightlessness just like astronauts.
The spring balance reading will change from 100 N to zero.

Question 14.
If the acceleration due to gravity becomes 4 times its original value, then escape speed
(a) remains same
(b) 2 times of original value
(c) becomes halved
(d) 4 times of original value
Answer:
(b) 2 times of original value
Solution:
Escape speed 4 times of ‘g’

Question 15.
The kinetic energy of the satellite orbiting around the Earth is
(a) equal to potential energy
(b) less than potential energy
(c) greater than kinetic energy
(d) zero
Answer:
(b) less than potential energy

Samacheer Kalvi 11th Physics Gravitation Short Answer Questions

Question 1.
State Kepler’s three laws.
Answer:
1. Law of Orbits: Each planet revolves moves around the Sun in an elliptical orbit with the Sun at one of the foci of the ellipse.
2. Law of area: The radial vector line joining the Sun to a planet sweeps equal areas in equal intervals of time.
3. Law of period: The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.

Question 2.
State Newton’s Universal law of gravitation.
Answer:
Newton’s law of gravitation: States that the gravitational force between two masses is directly proportional to product of masses and inversely proportional to square of the distance between the masses.
In the mathematical form, it can be written as,

Question 3.
Will the angular momentum of a planet be conserved? Justify your answer.
Answer:
The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

Question 4.
Define the gravitational field. Give its unit.
The gravitational field intensity \(\overrightarrow{\mathrm{E}}_{1}\) at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg^-1.

Question 5.
What is meant by superposition of gravitational field?
Answer:
The total gravitational field at a point due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of gravitational fields.

Question 6.
Define gravitational potential energy.
Answer:
Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

Question 7.
Is potential energy the property of a single object? Justify.
Answer:
There is no potential energy for a single object. The gravitational potential energy depends upon the two masses and the distance between them.

Question 8.
Define gravitational potential.
Answer:
The gravitational potential is defined as the amount of work required to bring unit mass from infinity to that point.

Question 9.
What is the difference between gravitational potential and gravitational potential energy?
Answer:

Gravitational Potential	Gravitational potential Energy
The amount of work required to bring unit mass from infinity to that point in point an gravitational field.	The work done by an external agent in moving the body from infinity to that in an gravitational field.
The unit of V(r) is J kg-1	The unit of U(r) is J (or) joule.

Question 10.
What is meant by escape speed in the case of the Earth?
Answer:
The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as, v_e = \(\sqrt{2 g R_{E}}\).

Question 11.
Why is the energy of a satellite (or any other planet) negative?
Answer:
The energy of satellite is negative. Because the energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Question 12.
What are geostationary and polar satellites?
Answer:
Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the earth.

Question 13.
Define weight.
Answer:
The weight of an object \(\overrightarrow{\mathrm{W}}\) is defined as the downward force whose magnitude W is equal to that of upward force that must be applied to the object to hold it at rest or at constant velocity relative to the earth. The direction of weight is in the direction of gravitational force. So the magnitude of weight of an object is denoted as, W = N = mg.

Question 14.
Why is there no lunar eclipse and solar eclipse every month?
Answer:
If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so dining new Moon we can observe solar eclipse. But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

Question 15.
How will you prove that Earth itself is spinning?
Answer:
The Earth’s spinning motion can be proved by observing star’s position over a night. Due to Earth’s spinning motion, the stars in sky appear to move m circular motion about the pole star.

Samacheer Kalvi 11th Physics Gravitation Long Answer Questions

Question 1.
Discuss the important features of the law of gravitation.
Answer:
1. As the distance between two masses increases, the strength of the force tends to decrease because of inverse dependence on r². Physically it implies that the planet Uranus experiences less |F| gravitational force from the Sun than the Earth since Uranus is at larger distance from the Sun compared to the Earth.
2. The gravitational forces between two particles always constitute an action-reaction pair. It implies that the gravitational force exerted by the Sim on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of this reaction force is towards Earth.

3. The torque experienced by the Earth due to the gravitational force of the Sun is given by

. It implies that angular momentum \(\overrightarrow{\mathrm{L}}\) is a constant vector.

4. The expression has one inherent assumption that both M₁, and M₂ are treated as point masses. When it is said that Earth orbits around the Sun due to Sun’s gravitational force, we assumed Earth and Sun to be point masses.
5. Point masses holds even for small distance.
6. There is also another interesting result. Consider a hollow sphere of mass M. If we place another object of mass ‘m’ inside this hollow sphere the force experienced by this mass ‘m’ will be zero.

Question 2.
Explain how Newton arrived at his law of gravitation from Kepler’s third law.
Answer:
Newton law of gravitation states that a particle of mass M₁ attracts any other particle of mass M₂ in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square to the distance between them. In mathematical form, it can be written as:

where \(\hat{r}\) is the unit vector from M₁ towards M₂ as shown in given figure, and G is the Gravitational constant that has the value of 6.67 × 10^-11 N m² kg^-2, and r is the distance between the two masses M₁ and M₂.

In given figure the vector \(\overrightarrow{\mathrm{F}}\) denotes the gravitational force experienced by M₂ due to M₁. Here the negative sign indicates that the gravitational force is always attractive in nature and the direction of the force is along the line joining the two masses. In Cartesian coordinates, the square of the distance is expressed as r² = (x² + y² + z²)

Question 3.
Explain how Newton verified his law of gravitation.
Answer:
Newton inverse square law: Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the centre is


The velocity in terms of known quantities r and T, is

Here T is the time period of revolution of the planet. Substituting this value of v in equation (1) we get,

Substituting the value ‘a’ from (3) in Newton’s second law, F = ma, where ‘m’ is the mass of the planet.

By substituting equation (6) in the force expression, we can arrive at the law of gravitation.

Here negative sign implies that the force is attractive and it acts towards the center. But Newton strongly felt that according to his third law, if Earth is attracted by the Sun, then the Sim must also be attracted by the Earth with the same magnitude of force. So he felt that the Sun’s mass (M) should also occur explicitly in the expression for force (7). From this insight, he equated the constant 4π²k to GM which turned out to be the law of gravitation.

Again the negative sign in the above equation implies that the gravitational force is attractive.

Question 4.
Derive the expression for gravitational potential energy.
Answer:
The gravitational force is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂ are initially separated by a distance r’. Assuming m₁ to be fixed in its position, work must be done on m₂ to move the distance from r’ to r.

To move the mass m₂ through an infinitesimal displacement \(d \vec{r}\) from \(\vec{r}\) to \(\vec{r}\) + \(d \vec{r}\), work has to be done externally. This infinitesimal work is given by

The work is done against the gravitational force, therefore,

Substituting equation (2) in (1), we get

Thus the total work for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses m₁ and m₂ when the seperation between them are r and r’ respectively.

Case 1: If r < r’ : Since gravitational force is attractive, m₂ is attracted by m₁. Then m₂ can move from r’ to r without any external Work. Here work is done by the system spending its internal energy and hence the work done is said to be negative.
Case 2: If r > r’: Work has to be done against gravity to move the object from r’ to r. Therefore work is done on the body by external force and hence work done is positive.

Question 5.
Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U = mgh.
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy.
Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.

Here r = R_e + h, where R_e is the radius of the Earth, h is the height above the Earth’s surface


Substituting equation (4) and (5) we get,
U = -mg_e + mgh …….. (6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂, then the potential energy at h₁ is
U(h₁) = – mgR_e + mgh₁ …(7)
and the potential energy at h₂ is
U(h₂) = – mgR_e + mgh₂ …(8)
The potential energy difference between h₁ and h₂ is
U(h₂) – U(h₁) = mg(h₂ – h₁) …(9)
The term mgR_e in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

Question 6.
Explain in detail the idea of weightlessness using lift as an example.
Answer:
When a man is standing in the elevator, there are two forces acting on him.
1. Gravitational force which acts downward. If we take the vertical direction as positive y direction, the gravitational force acting on the man is \(\overrightarrow{\mathrm{F}}_{\mathrm{G}}=-m \hat{g} \hat{j}\)
2. The normal force exerted by floor on the man which acts vertically upward, \(\overrightarrow{\mathrm{N}}=\mathrm{N} \hat{j}\)
Weightlessness of freely falling bodies: Freely falling objects experience only gravitational force. As they fall freely, they are not in contact with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is equal to the acceleration due to the gravity of the Earth, i.e., (a = g)
Newton’s 2nd law acting on the man N = m(g – a)
a = g ∴ N = m(g – g) = 0.

Question 7.
Derive an expression for escape speed.
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i, the initial total energy of the object is

where, M_E is the mass of the Earth and R_E the radius of the Earth. The term is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0
According to the law of energy conservation,
E_i = E_f …. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the sarfte initial speed to escape Earth’s gravity.

Question 8.
Explain the variation of g with latitude.
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mω²R’.


where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
a_PQ = ω²R’ cos λ = ω²R cos² λ
Since R’ = R cos λ
Therefore, g’ = g – ω² R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 9.
Explain the variation of g with altitude from the Earth’s surface.
Answer:
Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is

We find that g’ < g. This means that as altitude h increases the acceleration due to gravity g decreases.

Question 10.
Explain the variation of g with depth from the Earth’s surface.
Answer:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points.
The part of the Earth which is above the radius (R_e – d) do not contribute to the acceleration. The result is proved earlier and is given as

Here M’ is the mass of the Earth of radius (R_e – d)
Assuming the density of Earth ρ to be constant, \(\rho=\frac{M}{V}\)
where M is the mass of the Earth and V its volume, thus,


Here also g’ < g. As depth increase, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

Question 11.
Derive the time period of satellite orbiting the Earth.
Answer:
Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to 2π(R_E + h) and time taken for it is the time period, T. Then,


Squaring both sides of the equation (2) we get

Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_E. Then,

By substituting the values of R_E = 6.4 × 10⁶ m and g = 9.8ms^-2, the orbital time period is obtained as T ≅ 85 minutes.

Question 12.
Derive an expression for energy of satellite.
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,

Here M_s – mass of the satellite, M_E -mass of the Earth, R_E – radius of the Earth.
The Kinetic energy of the satellite is

Here v is the orbital speed of the satellite and is equal to

Substituting the value of v in (2) the kinetic energy of the satellite becomes,

Therefore the total energy of the satellite is

The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
Note: As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Question 13.
Explain in detail the geostationary and polar satellites.
Answer:
The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Kepler’s third law is used to find then radius of the orbit.

Substituting for the time period (24 hours = 86400 seconds), mass, and radius of the Earth, h turns out to be 36,000 km. Such satellites are called “geo-stationary satellites”, since they appear to be stationary when seen from Earth.

India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication. Another type of satellite which is placed at a distance of 500 to 800 krn from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. A polar satellite covers a small strip of area from pole to pole during one revolution. In the next revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

Question 14.
Explain how geocentric theory is replaced by heliocentric theory using the idea of retrograde motion of planets.
Answer:
When the motion of the planets are observed in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called “retrograde motion of planets.

Careful observation for a period of a year clearly shows that Mars initially moves eastwards (February to June), then reverses its path and moves backwards (July, August, September). It changes it direction of motion once again and continues its forward motion (October onwards). In olden days, astronomers recorded the retrograde motion of all visible planets and tried to explain the motion. According to Aristotle, the other planets and the Sun move around the Earth in the circular orbits. If it was really a circular orbit it was not known how the planet could reverse its motion for a brief interval. To explain this retrograde motion, Ptolemy introduced the concept of “epicycle” in his geocentric model. According to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as “epicycle’. A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth. Essentially Ptolemy retained the Earth centric idea of Aristotle and added the epicycle motion to it.

But Ptolemy’s model became more and more complex as every planet was found to undergo retrograde motion. In the 15th century, the Polish astronomer Copernicus proposed the heliocentric model to explain this problem in a simpler manner. According to this model, the Sun is at the centre of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth.
Earth orbits around the Sun faster than Mars. Because of the relative motion between Mars and Earth, Mars appears to move backwards from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copernicus model. It was because of its simplicity, the heliocentric model slowly replaced the geocentric model.

Question 15.
Explain in detail the Eratosthenes method of finding the radius of Earth.
Answer:
Eratosthenes observed that during noon time of summer solstice the Sun’s rays cast no shadow in the city Syne which was located 500 miles away from Alexandria. At the same day and same time he found that in Alexandria the Sun’s rays made 7.2 degree with local vertical. He realized that this difference of 7.2 degree was due to the curvature of the Earth.

If S is the length of the arc between the cities of Syne and Alexandria, and if R is radius of Earth, then

1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436 km, which is amazingly close to the correct value of 6378 km.

Question 16.
Describe the measurement of Earth’s shadow (umbra) radius during total lunar eclipse.
Answer:
Lunar eclipse and measurements of shadow of Earth: On January 31, 2018 there was a total lunar eclipse which was observed from various place including Tamil Nadu. It is possible to measure the radius of shadow of the Earth at the point where the Moon crosses.

When the Moon is inside the umbra shadow, it appears red in colour. As soon as the Moon Schematic diagram of umbra disk radius exits from the umbra shadow, it appears in crescent shape.

By finding the apparent radii of the Earth’s umbra shadow and the Moon, the ratio of the these radii can be calculated.
The apparent radius of Earth’s umbra shadow = R_s = 13.2 cm
The apparent radius of the Moon = R_m = 5.15 cm

The radius of Moon R_m = 1737 km
The radius of the Earth’s umbra shadow is R_s = 2.56 × 1737 km ≅ 4446
The correct radius is 4610 km

The error will reduce if the pictures taken using a high quality telescope are used.

Samacheer Kalvi 11th Physics Gravitation Conceptual Questions

Question 1.
In the following what are the quantities which that are conserved?
(a) Linear momentum of planet
(b) Angular momentum of planet
(c) Total energy of planet
(d) Potential energy of a planet
Answer:
Since there is a net force acting on the planet, its velocity changes which means its linear momentum changes. In fact, the absolute value of linear momentum changes too since the planet’s speed is variable as it goes around in its elliptical orbit . So the linear momentum of planet is not conserved. But the angular momentum about the sun is conserved. Since the torque of gravitational force is zero.

The total mechanical energy remains constant for an isolated system objects that interact with conservative forces. So, the total energy of the system of planet is conserved. The single energy of the planet is not conserved.

Question 2.
The work done by Sun on Earth in one year will be
(a) Zero
(b) None zero
(c) positive
(d) negative
Answer:
(a) Zero

Question 3.
The work done by Sun on Earth at any finite interval of time is
(a) positive, negative or zero
(b) Strictly positive
(c) Strictly negative
(d) It is always zero
Answer:
(d) It is always zero
2 & 3. No work is done on the earth revolving around it in perfectly circular orbit. The earth revolves around the sun due to gravitational force of attraction between the sun and the earth. This force around the sun. This centripetal force is always perpendicular to the linear displacement.
Work done = W = F.d cos θ Since θ = 90°

Question 4.
If a comet suddenly hits the Moon and imparts energy which is more than the total energy of the Moon, what will happen?
Answer:
A comet with small velocity and high mass, doesn’t trigger the moon much. It just makes a circular shaped impact. The moon is ment for the protection for life on earth and to attain stability for the earth rotation. But a comet with large mass and with large velocity may destroy the moon completely or its impact makes the moon, go out of its orbit.

Question 5.
If the Earth’s pull on the Moon suddenly disappears, what will happen to the Moon?
Answer:
Basically, the moon would become the third planet, however the orbit may change. The moon is in orbit around the earth, and what happens will depend on just when the earth disappears. The moon travels around the earth at about 1 km/sec and the Earth – moon pair travel around the sun at about 30 km/sec. This extra 1 km/sec movement will result in an orbit change.

Question 6.
If the Earth has no tilt, what happens to the seasons of the Earth?
Answer:
If the earth weren’t tilted on its axis, there would be no seasons. And humanity would suffer.

Question 7.
A student was asked a question ‘why are there summer and winter for us? He replied as ‘since Earth is orbiting in an elliptical orbit, when the Earth is very faraway from .the Sun (aphelion) there will be winter, when the Earth is nearer to the Sun (perihelion) there will be winter}. Is this answer correct? If not, what is the correct explanation for the occurrence of summer and winter?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.

Question 8.
The following photographs are taken from the recent lunar eclipse which occurred on January 31, 2018. Is it possible to prove that Earth is a sphere from these photographs?
Answer:
Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse.

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
An unknown a planet orbits the Sun with distance twice the semi major axis distance of the Earth’s orbit. If the Earth’s time period is Tp what is the time period of this unknown planet?
Answer:
By Kepler’s 3rd law T² ∝ a³
Time period of unknown planet = T₂
Time period of Earth = T₁
Distance of unknown planet from the Sun = a₂
Distance of the Earth from the Sun = a₁

Question 2.
Assume that you are in another solar system and provided with the set of data given below consisting of the planets’ semi major axes and time periods. Can you infer the relation connecting semi major axis and time period?
Answer:

In a given datas tells us the relation connecting to the semi major axis is proportional to the two times of square of the time period.
a ∝ 2T²

Question 3.
If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them?
Answer:
By Newton’s law of gravitation

Here, the masses and mutual distance between the two objects are doubled .

There is no change in the gravitational force between them.

Question 4.
Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitational field is zero.
Answer:
Let the point be the position when the gravitational field is zero.,

The point P is at a distance \(\frac{r}{3}\) from mass ‘m’ and \(\frac{2r}{3}\) from mass ‘4m’

Question 5.
If the ratio of the orbital distance of two planets \(\frac{d_{1}}{d_{2}}\) = 2, what is the ratio of gravitational field experienced by these two planets?
Answer:
The gravitational field experienced by planets 1

The gravitational field experienced by planet 2

Question 6.
The moon I_o orbits Jupiter once in 1.769 days. The orbital radius of the Moon I₀ is 421700 km. Calculate the mass of Jupiter?
Answer:
Kepler’s third law is used to find the mass of the planet

Question 7.
If the angular momentum of a planet is given by \(\overrightarrow{\mathbf{L}}=5 t^{2} \hat{i}-6 \hat{t} \hat{j}+3 \hat{k}\). What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum?
Answer:
The torque experienced by the planet

Question 8.
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle.
Answer:
The net gravitational force = Centripetal force

Question 9.
Suppose unknowingly you wrote the universal gravitational constant value as G = 6.67 × 10¹¹ instead of the correct value G = 6.67 × 10^-11, what is the acceleration due to gravity g’ for this incorrect G? According to this new acceleration due to gravity, what will be your weight W?
Answer:
Data: Incorrect Gravitational constant G = 6.67 × 10¹¹ Nm² kg^-2
Mass of the Earth M_e = 5.972 × 10²⁴ kg
Radius of the earth R_e = 6371 km (or) 6371 × 10³ m

Question 10.
Calculate, the gravitational field at point O due to three masses m₁, m₂, and m₃ whose positions are given by the following figure. If the masses m₁ and m₂ are equal what is the change in gravitational field at the point O?
Answer:

(Negative sign indicates field acting along negative x direction)

Question 11.
What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million km. The mass of the Earth is 5.9 × 10²⁴ kg and mass of the Sun is 1.9 × 10³⁰ kg.
Answer:
Mass of the Earth M_E = 5.9 × 10²⁴ kg
Mass of the Sun M_S = 1.9 × 10³⁰ kg
Distance between the Sun and Earth
r = 150 million km; r = 150 × 10⁹ m
Gravitational constant G = 6.67 × 10^-11 Nm² kg^-2
The gravitational potential energy

Question 12.
Earth revolved around the Sun at 30 km s^-1. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.
Answer:
Mass of the Earth M_E = 5.9 × 10²⁴ kg
Speed of the Earth rovolves around the Sun

= 2.655 × 10³³ + (- 4.985 × 10³³)
= 2.655 × 10³³ – 4.985 × 10³³
E = -2.33 × 10³³ joule (or) J
‘-Ve’ implies that Earth is bounded with Sun.

Question 13.
An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy with that velocity should the object be thrown from Earth?
Answer:
An object is thrown up with an initial velocity is v_i, So Total energy of the object is

Now, the object reaches a height with a non-zero K.E.
K.E becomes infinity. P.E becomes zero.

Question 14.
Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small?
Answer:
Variation of g’ with depth

Variation of g’ with altitude

Question 15.
Calculate the change in g value in your district of Tamil Nadu. (Hint: Get the latitude of your district of Tamilnadu from the Google). What is the difference in g values at ‘ Chennai and Kanyakumari?
Answer:
Variation of ‘g’ value in the latitude to chennai

Period of revolution (T) = 1 day = 86400 sec
Radius of the Earth (R) = 6400 × 10³ m
Latitude of Chennai (λ) = 13° = 0.2268 rad

Variation of ‘g’ value in the latitude of Kanyakumari

Samacheer Kalvi 11th Physics Gravitation Additional Questions

I. Choose the correct answer from the following:

Question 1.
According to Kepler, planet move in
(a) Circular orbits around the Sun
(b) Elliptical orbits around the Sim with Sun at exact centre
(c) Straight lines with constant velocity ‘
(d) Elliptical orbits around the Sun with Sun at one of its foci.
Answer:
(d) Elliptical orbits around the Sun with Sun at one of its foci.

Question 2.
Kepler’s second law regarding constancy of aerial velocity of a planet is consequence of the law of conservation of ………
(a) energy
(b) angular momentum
(c) linear momentum
(d) None of these
Answer:
(b) angular momentum
Hint:

Question 3.
According to Kepler, the period of revolution of a planet (T) and its mean distance from the Sun (a) are related by the equation
(a) T³ a³ = constant
(b) T² a^-3 = constant
(c) Ta³ = constant
(d) T²a = constant
Answer:
(b) T² a^-3 = constant
Hint:

Question 4.
The period of Moon’s rotation around the Earth is nearly 29 days. If Moon’s mass were 2 fold
its present value and all other things remained unchanged the period of Moon’s rotation would be nearly ….. days.

Answer:
(d) 29
Hint:
Time period does not depends upon the mass of satellite.

Question 5.
The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4
Hint:

Question 6.
The radius of orbit of a planet is two times that of Earth. The time period of planet is years …….
(a) 4.2
(b) 2.8
(c) 5.6
(d) 8.4
Answer:
(b) 2.8
Hint:

Question 7.
A geostationary satellite orbits around the earth in a circular orbit of radius 3600 km the time period of a satellite orbiting a few hundred kilometers above the earth’s surface (R_E = 6400 km) will be approximately be …… hours.
(a) 1/2
(b) 1
(c) 2
(d) 4
Answer:
(c) 2
Hint:

Question 8.
What does not change in the field of central force?
(a) Potential energy
(b) kinetic energy
(c) linear momentum
(d) Angular momentum
Answer:
(d) Angular momentum
Hint:
For central force torque is zero.

Question 9.
A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is …… hours.

Answer:
(b) \(48 \sqrt{2}\)
Hint:

Question 10.
If the Earth is at one-fourth of its present distance from the sun the duration of year will be:
(a) half the present year
(b) one-eight the present year
(c) one-fourth the present year
(d) one-sixth the present year
Answer:
(b) one-eight the present year
Hint:

Question 11.
The Earth E moves in an elliptical orbit with the Sun S at one of the foci as shown in figure.
Its speed of motion will be maximum at a point ……..

(a) C
(b) A
(c) B
(d) D
Answer:
(b) A
Hint: Speed at the Earth will be maximum when its distance from the Sun is minimum because
mvr = constant

Question 12.
Rockets are launched in eastward direction to take advantage of …..
(a) the clear sky on eastern side
(b) Earth’s rotation
(c) the thinner atmosphere on this side
(d) Earth’s tilt
Answer:
(b) Earth’s rotation
Hint:
Because Earth rotation from west to east direction.

Question 13.
Two sphere of mass M₁ and M₂ are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be ……

Answer:
(a) F
Hint:
Gravitational force does not depend upon the medium.

Question 14.
Which of the following statement about the gravitational constant is true?
(a) It is a force
(b) It has same value in all system of unit
(c) It has not unit
(d) It depends on the value of the masses
Answer:
(a) It is a force

Question 15.
Energy required to move a body of mass ‘M’ from an orbit of radius 2R to 3R is …….

Answer:
(d) \(\frac{\mathrm{GMm}}{6 \mathrm{R}}\)
Hint:
Change in P.E. in displacing a body from r₁ and r₂ is given by:

Question 16.
The mass of the earth is 6 × 10²⁴ kg and that of the Moon is 7.4 × 10²² kg. The constant of gravitation G is 6.67 × 10^-11 Nm² kg^-2. The potential energy of the system is – 7.79 × 10²⁸J. The mean distance between the Earth and Moon is …… metre.
(a) 3.80 × 10⁸
(b) 3.37 × 10⁸
(c) 7.60 × 10⁸
(d) 1.90 × 10²
Answer:
(a) 3.80 × 10⁸
Hint:

Question 17.
What is the intensity of gravitational field at the center of spherical shell?
(a) \(\frac{\mathrm{G} m}{r^{2}}\)
(b) g
(c) zero
(d) None of these
Answer:
(c) zero

Question 18.
A body of mass m is taken from the Earth’s surface to a height equal to the radius R of the earth. If g is the acceleration to gravity at the surface of the Earth, then find the change in the potential energy of the body ……

Answer:
(b) \(\frac{1}{2} m g R\)
Hint:

Question 19.
A satellite is orbiting around the Earth in a circular orbit with velocity v. If m is the mass of the satellite, its total energy is ……

Answer:

Hint:
The total energy is negative of the kinetic energy.

Question 20.
Escape velocity of a body of 1 kg. On a planet is 100 ms^-1. Gravitational potential energy of the body at the planet is ……
(a) -5000 J
(b) – 1000 J
(c) – 2400 J
(d) 4000 J
Answer:
(a) – 5000 J

Question 21.
A particle falls towards earth from infinity. It’s velocity reaching the Earth would be ……

Answer:
(b) \(\sqrt{2 g \mathrm{R}}\)
Hint:
This should be equal to escape velocity is = \(\sqrt{2 g \mathrm{R}}\)

Question 22.
An artificial satellite is revolving round the Earth in a circular orbit, its velocity is half the escape velocity. Its height from the Earth surface is …. km.
(a) 6400
(b) 12800
(c) 3200
(d) 1600
Answer:
(a) 6400

Question 23.
The escape velocity of a body on the surface of the Earth is 11.2 km/s. If the mass of the Earth is increase to twice its present value and the radius of the earth becomes half, the escape velocity becomes = …… kms^-1
(a) 5.6
(b) 11.2
(c) 22.4
(d) 494.8
Answer:
(c) 22.4
Hint:

If M becomes double and R becomes half, then escape velocity becomes two times.

Question 24.
The velocity with which a projectile must be fired so that it escapes Earth’s gravitational does
not depend on ……..
(a) Mass of Earth
(b) Radius of the projectile’s orbit
(c) Mass of the projectile
(d) Gravitational constant
Answer:
(c) Mass of the projectile

Question 25.
The escape velocity for a body projected vertically upwards from the surface of Earth is 11 kms^-1. If the body is projected at an angle of 45° with the vertical, the escape velocity will be …. kms^-1

Answer:
(d) 11
Hint:
Escape velocity does not depends upon the angle of projection.

Question 26.
Two satellites of mass ml and m₂(m₁> m₂) are revolving round the earth in circular orbits of r₁ and r₂ (r₁ > r₂) respectively. Which of the following statement is true regarding their speeds v₁ and v₂ ……..

Answer:
(b) v₁ < v₂
Hint:

Question 27.
As astronaut orbiting the earth in a circular orbit 120 km above the surface of Earth, gently drops a spoon out of space-ship. The spoon will ……….
(a) fall vertically down to the Earth
(b) move towards the moon
(c) will move along with space-ship
(d) will move in an irregular way then fall down to Earth
Answer:
(c) will move along with space-ship
Hint:
The velocity of the spoon will be equal to the orbital velocity when dropped out of the space-ship

Question 28.
A satellite revolves around the Earth in an elliptical orbit. Its speed.
(a) is the same at all point in the orbit
(b) is greatest when it is closest to the Earth
(c) is greatest when it is farthest to the Earth
(d) goes on increasing or decreasing continuously depending upon the mass of the satellite. Answer:
(b) is greatest when it is closest to the Earth

Question 29.
A satellite is moving around the Earth with speed v in a circular orbit of radius r. If the orbit
radius is decreased by 1% its speed will …….
(a) increase by 1%
(b) increase by 0.5%
(c) decrease by 1%
(d) decrease by 0.5%
Answer:
(b) increase by 0.5%

Question 30.
Orbital velocity of an artificial satellite does not depend upon …….
(a) mass of Earth
(b) mass of satellite
(c) radius of Earth
(d) acceleration due to gravity
Answer:
(b) mass of satellite

Question 31.
The orbital speed of Jupiter is …….
(a) greater than the orbital speed of Earth
(b) less then the orbital speed of Earth
(c) zero
(d) equal to the orbital speed of Earth
Answer:
(b) less then the orbital speed of Earth
Hint:

Question 32.
As we go grom the equator to the poles, the value of g …..
(a) remains constant
(b) decreases
(c) increases
(d) decreases upto latitude of 45°
Answer:
(c) increases

Question 33.
The value of g on the Earth surface is 980 cm/sec². Its value at a height of 64 km from the Earth surface is ……. cms²
(a) 960.40
(b) 984.90
(c) 982.45
(d) 977.55
Answer:
(a) 960.40
Hint:

Question 34.
The Moon s radius is \(\frac{1}{4}\) that of earth and its mass is \(\frac{1}{80}\) times that of the Earth. If g represents the acceleration due to gravity on the surface of Earth, that on the surface of the Moon is ……

Answer:
(b) \(\frac{g}{5}\)
Hint:

Question 35.
If the density of small planet is that of the same as that of the earth while the radius of the
planet is 0.2 times that of the Earth, the gravitational acceleration on the surface for the planet is ……
(a) 0.2g
(b) 0.4g
(c) 2g
(d) 4g
Answer:
(a) 0.2g
Hint:

Question 36.
Assuming Earth to be a sphere of a uniform density, what is value of gravitational acceleration in mine 100 km below the Earth surface = ….. ms^-2
(a) 9.66
(b) 7.64
(c) 5.00
(d) 3.1
Answer:
(a) 9.66
Hint:

Question 37.
The radii of two planets are respectively R₁ and R₂ and their densities are respectively ρ₁ and ρ₂ the ratio of the accelerations due to gravity at their surface is ……

Answer:
(d) \(g_{1}: g_{2}=\mathrm{R}_{1} \rho_{1}: \mathrm{R}_{2} \rho_{2}\)

Question 38.
The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to:

Answer:
(c) dR
Hint:

Question 39.
The acceleration of a body due to the attraction of the Earth (radius R) at a distance 2R from the surface of the Earth is …….

Answer:
\(\frac{g}{9}\)
Hint:

Question 40.
If density of Earth increased 4 times and its radius becomes half of then out weight will be ……
(a) four times its present value
(b) doubled
(c) remains same
(d) halved
Answer:
(b) doubled
Hint:
g ∝ ρR

Question 41.
The radius of the Earth is 6400 km and g = 10 ms^-2 in order that a body of 5 kg weights zero at the equator, the angular speed of the Earth is ….. rad s^-1

Answer:
(c) \(\frac{1}{800}\)
Hint:

Question 42.
Weight of a body is maximum at …..
(a) Moon
(b) poles of Earth
(c) equator of Earth
(d) centre of Earth
Answer:
(b) poles of Earth

Question 43.
The weight of an astronaut, in an artificial satellite revolving around the Earth is:
(a) zero
(b) equal to that on the Earth
(c) more than that on Earth
(d) less than that on Earth
Answer:
(a) zero

Samacheer Kalvi 11th Physics Gravitation 2 Marks Questions

Question 1.
Distinguish between the terms gravitation and gravity.
Answer:
Gravitation: It is the force of attraction between any two bodies in the universe.
Gravity: It is the force of attraction between the earth and any object lying on or near its surface.

Question 2.
Why is G called the universal gravitational constant?
Answer:
The value of G does not depend on the nature and size of the bodies. It also does not depend on the nature of the medium between the two bodies. That is why G is called universal gravitational constant.

Question 3.
What is meant by the term free fall?
Answer:
The motion of a body under the influence of gravity alone is called a free fall.

Question 4.
What is meant by acceleration due to gravity? Is is a scalar or a vector?
Answer:
The acceleration produced in a freely falling body under the gravitational pull of the earth. It is a vector having direction towards the centre of the earth.

Question 5.
What do you mean by weight of a body? Is it a scalar or vector?
Answer:
Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by w = mg (or) \(\overrightarrow{\mathrm{W}}=m \vec{g}\)

Question 6.
Define orbital velocity.
Answer:
Orbital velocity is the velocity required to put the satellite into its orbit around the earth.

Question 7.
Give some uses of geostationary satellites.
Answer:

• In communicating radio, T.V and telephone signals across the world.
• In studying upper regions of the atmosphere.
• In forecasting weather.
• In deter ming the exact shape and dimensions of the earth.
• In studying solar radiations and cosmic rays.

Question 8.
Give the uses of polar satellites.
Answer:

• Polar satellites are uses in weather and environment monitoring.
• They are used in spying work for military purposes.
• They are used to study topography of Moon, Venus and Mars.

Samacheer Kalvi 11th Physics Gravitation Numerical Problems

Question 1.
A geo-stationary satellite is orbiting the Earth of a height of 6R above the surface of Earth R being the radius of the Earth calculate the time period of another satellite at a height of 2.5R from the surface of Earth.
Distance of satellite from the center are 7R and 3.5 R respectively.
Answer:

Question 2.
The time period of a satellite of Earth is 5 hours. If the separation between the Earth and the satellite is increased to four times the previous value, the new time period will become.
Answer:

Question 3.
The figure shows elliptical orbit of a planet ‘M’ about the Sun ‘S’, the shaded area SCD is twice the shaded area SAB. If t₁ is the time for the planet to move from C and D and t₂ is the time to move from A to B then.
Answer:

Question 4.
A satellite moves in a circle around the Earth, the radius of this circle is equal to one half of the radius of the Moon’s orbit. The satellite completes one revolution in…… lunar month.
Answer:
Time period of revolution of moon around the Earth T_m = 1 lunar month

Question 5.
Two identical solid copper spheres of radius R are placed in contact with each other. The gravitational force between them is proportional to.
Answer:


∴ The gravitational force between them is proportional to 4th power of radius.

Question 6.
Two satellites A and B of the same mass are revolving around the Earth in circular orbits such that the distance of B from the centre of the Earth is thrice as compared to the distance of A from the centre. What will be the ratio of centripetal force on B to that on A.
Answer:
The necessary centripetal force is provided by the gravitational force of attraction, for circular orbit

The ratio of centripetal force,

Question 7.
An infinite number of bodies, each of mass 2 kg, are situated on x-axis at distances lm, 2m, 4m, 8m from the origin. What will be the resultant gravitational potential due to this system at the origin.
Answer:
Gravitational potential at the origin is

Question 8.
A body of mass ‘m’ kg starts falling from a point 2R above the Earth’s surface. What is its K.E. When it has fallen to a point ‘R’ above the Earth’s surface.
Answer:

Question 9.
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to done against the gravitational force between them to take the particle is away from the sphere.
Answer:

So, the amount of work done to take the particle upto infinite will be 6.67 × 10^-10 J

Question 10.
The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms^-2 and 6400 km respectively. The required energy for this work will be.
Answer:
Potential energy U = – mgR_e + mgh
(The first term is independent of the height, so it can be taken to zero.)
W = U = mgh [h ≈ R]
= 1000 × 10 × 6400 × 10³ = 64 × 10⁹
W = 6.4 × 10¹⁰ J

Question 11.
If the mean radius of the Earth is R, its angular velocity is ω, and the acceleration due to gravity at the surface of the Earth is g, then what will be the cube of the radius of the orbit of a geostationary satellite.
Answer:
Let r be the radius of the geostationary orbit. Angular velocity of revolution of a geostationary satellite is same as the angular velocity of rotation of the Earth.

Question 12.
The escape velocity of a body from Earth’s surface is v_e. What will be the escape velocity of the same body from a height equal to 7R from Earth’s surface.
Answer:

Question 13.
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth’s surface. Find the mean density of the Earth.
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 14.
If the mass of Earth is 80 times of that of a planet and diameter is double that of planet and ‘g’ on the Earth is 9.8 ms-2. Calculate the value of ‘g’ on that planet?
Answer:

Question 15.
At what distance from the centre of Earth, the value acceleration due to gravity ‘g’ will be half that of the surface?
Answer:
According to acceleration due to gravity

Question 16.
A body weight 700 g on the surface of Earth. How much it weight on the surface of planet whose mass is \(\frac{1}{7}\) and radius is half that of the Earth.
Answer:
Acceleration due to gravity g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

Question 17.
An object weight 72 N on the Earth. What its weight at a height \(\frac{\mathbf{R}}{2}\) from Earth.
Answer:

Question 18.
A body weight 500 N on the surface of the Earth. How much would it weight half way below the surface of Earth.
Answer:
Weight on surface of Earth, mg = 500 N
Weight below the surface of Earth at d = \(\frac{\mathrm{R}}{2}\)
From variation of ‘g’ with depth

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