Category: Class 11

Students can Download Tamil Nadu 11th English Model Question Paper 1 Pdf, Tamil Nadu 11th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th English Model Question Paper 1

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3:00 Hours
Maximum Marks: 90

PART – I

I. Answer all the questions. [20 × 1 = 20]
Choose the correct synonym for the underlined words from the options given.

Question 1.
A peaceful pallor spread on his face.
(a) paleness
(b) rosiness
(c) ruddy
(d) rejuvenation
Answer:
(a) paleness

Question 2.
It was indeed a sensational bid.
(a) astounded
(b) ordinary
(c) sensitive
(d) exciting
Answer:
(a) astounded

Question 3.
I had a sumptuous meal to sate my hunger.
(a) sadden
(b) suppress
(c) satisfy
(d) struggle
Answer:
(c) satisfy

Choose the correct antonym for the underlined words from the options given.

Question 4.
The vessels dropped and scattered all over.
(a) speckled
(b) sprinkled
(c) gathered
(d) grew
Answer:
(c) gathered

Question 5.
There a curious smothering noise from my friend.
(a) suppressed
(b) expressed
(c) smoothened
(d) suffocated
Answer:
(b) expressed

Question 6.
Her face was wrinkled and weird.
(a) smooth
(b) creased
(c) lined
(d) crippled
Answer:
(a) smooth

Question 7.
Choose the clipped form of “percolate”.
(a) perk
(b) per
(c) perc
(d) colate
Answer:
(a) perk

Question 8.
Choose the right definition for the given term “plagiarist”.
(a) A student of bees
(b) One who accumulates; one who collects
(c) One who purloins the words, writings, or ideas of another, and passes them off as his own; a literary thief
(d) One who is tenacious of a strict adherence to official formalities
Answer:
(c) One who purloins the words, writings, or ideas of another, and passes them off as his own; a literary thief

Question 9.
Choose the meaning of the idiom ‘Beat around the bush’.
(a) Indirectly talking about an issue
(b) Do or say something exactly right
(c) To go to bed
(d) Without any hesitation; instantly
Answer:
(a) Indirectly talking about an issue

Question 10.
Choose the meaning of the foreign word in the sentence.
I’d like to learn taekwondo though it might not be the best option for me.
(a) Japanese language
(b) Chinese preparation
(c) Floor dancing
(d) Kick fist martial art
Answer:
(d) Kick fist martial art

Question 11.
Choose the word from the options given to form a compound word with “goat”.
(a) male
(b) female
(c) scape
(d) major
Answer:
(c) scape

Question 12.
Form a new word by adding a suitable suffix to the root word, “hero”?
(a) ity
(b) ism
(c) ness
(d) ish
Answer:
(b) ism

Question 13.
Choose the expanded form of “NSC”.
(a) National Savings Certificate
(b) National Service Certificate
(c) National Savings Career
(d) National Service Certificate
Answer:
(a) National Savings Certificate

Question 14.
The correct syllabification of the word “psychology” is…………..
(a) psych-ol-ogy
(b) psy-chol-ogy
(c) ps-ycho-logy
(d) psy-chol-o-gy
Answer:
(d) psy-chol-o-gy

Question 15.
A craze for establishing banks is known as …………..
(a) Etymology
(b) Stampomania
(c) Eulogomania
(d) bancomania
Answer:
(d) bancomania

Question 16.
Fill in the blank with the suitable preposition.
Ramesh went……….. Tom’s place to settle the bills.
(a) to
(b) on
(c) in
(d) from
Answer:
(a) to

Question 17.
Add a suitable question tag to the following statement.
He never fails in his duty,………..?
(a) isn’t he
(b) won’t-he
(c) will he
(d) does he
Answer:
(d) does he

Question 18.
Substitute the underlined word with the appropriate polite alternative.
Sagar’s cat had to be killed because it had cancer.
(a) put in
(b) put down
(c) put up
(d) put out
Answer:
(b) put down

Question 19.
Substitute the phrasal verb in the sentence with a single word.
However hard she tried, she could not figure out the meaning of the proverb.
(a) appreciate
(b) condemn
(c) understand
(d) faint
Answer:
(c) understand

Question 20.
Fill in the blank with a suitable relative pronoun
The children ……… shouted in the street are not from our school.
(a) which
(b) whose
(c) who
(d) that
Answer:
(c) who

PART – II

II. Answer any seven of the following: [7 × 2 = 14]
(i) Read the following sets of poetic lines and answer any four of the following. [4 × 2 = 8]

Question 21.
“I have learned to wear many faces like dresses”
(а) State the figure of speech in the above line.
(b) Who does the term ‘I’ refer to?
Answer:
(a) Simile
(b) 1 refers to the poet, Gabriel Okara.

Question 22.
“ When swollen eye meets gnarled fist
When snaps the knee, and cracks the wrist”
(a) What is the figure of speech employed in the second line.
(b) Why are the eyes swollen?
Answer:
(a) Onomatopoeia
(b) The eyes are swollen as they were injured by the opponent.

Question 23.
“Or the greenhouse glass is broken, and the trellis past repair
Ay, there’s the wonder of the thing! Macavity’s not there!”
(a) What is a greenhouse and a trellis?
(b) What is a wonder after the greenhouse glass is broken?
Answer:
(a) Greenhouse is a glass building in which plants that need protection from cold weather are grown. Trellis is a framework of light wooden or metal bars used as a support for fruit trees or creeper.
(b) Macavity is to be found nowhere when one finds out that the greenhouse glass is broken.

Question 24.
“Let’s talk of graves, of worms, and epitaphs,
Make dust our paper, and with rainy eyes
Write sorrow on the bosom of the earth.”
(a) What do you understand by rainy eyes?
(b) Identify the figure of speech in the first line.
Answer:
(a) Eyes shedding tears is said to be rainy eyes.
(b) Metaphor

Question 25.
“And I must think, do all I can,
That there was pleasure there…”
(a) What did the poet notice about the twigs?
(b) What was the poet’s thought about then?
Answer:
(a) The budding twigs spread out their fan to catch the breezy air.
(b) The poet thought the twigs were experiencing the j oy of their contact with the breezy air.

Question 26.
“The height you reach is not that we care;”
(a) How different is the poet’s perception of success?
(b) How does the poet use the word ‘height’?
Answer:
(a) Those who reach great heights in terms of scholarship, wealth and positions are not deemed great. Those who have competence and merit alone are respected.
(b) In terms of mountain climbing, climbing Everest is said to be the highest achievement.
But even scaling a small hillock is an achievement. Every human effort to succeed needs to be appreciated.

(ii) Do as directed (any three) [3 × 2 = 6]

Question 27.
Rewrite the following dialogue in reported form.
Kiran : Will you exchange the defective torch I had bought from you yesterday?
Shopkeeper : Do you have the receipt with you?
Answer:
Kiran asked the shopkeeper if he would exchange the defective torch he had bought from him the day before. The shopkeeper enquired if Kiran had the receipt with him.

Question 28.
Rewrite the following sentence in its passive form.
She had already cooked the food.
Answer:
The food had already been cooked by her.

Question 29.
Convert the following complex sentence to a compound sentence.
I called for Arjun who came at once.
Answer:
I called for Aijun and he came at once.

Question 30.
Hurry up. You will miss the flight. (Combine using ‘Unless’)
Answer:
Unless you hurry up, you will miss the flight.

PART – III

III. Answer any seven of the following: [7 × 3 = 21]
(i) Explain any two of the following with Reference to the Context: [2 × 3 = 6]

Question 31.
“Have I not reason to lament What Man has made of Man?”
Answer:
Reference: These lines are from the poem “Lines Written in Early Spring” written by William Wordsworth.

Context: Amidst happy nature, William Wordsworth couldn’t help feeling sad. At that occasion, he said these words.

Explanation: The mixed feelings of happiness and sadness is well brought out in these lines. He was inspired by a small woodland grove, a landscape of beauty. He came upon this spot when walking near Alford village. While sensing the blissful mood and happiness of birds, plants, creepers and the murmuring brook, he juxtaposed what humans did to their kind in Napoleonic wars.

Question 32.
“I am just glad as glad can he
That I am not them, that they are not me. ”
Answer:
Reference: The poet Ogden Nash says these words in the poem “Confession of a Bom Spectator’.

Context: While discussing about the athletes he admires, the poet says these words.

Explanation: The poet was a bom spectator. Right from his boyhood, he had seen boys aspire for sports championships. He had wondered at their ability to specialize in horse riding, to play hockey or basketball. He had seen young ones trying to play center in the football or be a tackle or offender in a game like kabaddi. But he has been absolutely glad that he is not them and they are not him.

Question 33.
We deem it our duty and mission in life,
To bless and praise the deserving ones
Answer:
Reference: These lines are from the poem “Everest is not the Only Peak” written by Kulothungan.

Context: The poet says these words highlighting the virtues of unsung heroes.

Explanation: The unsung heroes adhere to their ethical principles in life. Even if they have not reached great heights. They consider it their duty and mission to identify deserving people with natural talents and appreciate them.

(ii) Answer any two of the following questions briefly: [2 × 3 = 6]

Question 34.
Why was the author left with his grandmother in the village?
Answer:
The author was left with his grandmother in the village because his parents had to go to the city for work.

Question 35.
What difficulty did Mary Kom experience while eating Chinese preparation?
Answer:
Once Mary Kom and her team mates were given chopsticks to eat their food in China. Other friends, asked for spoons and managed. But Mary Kom ended up using both her hands to hold the chopsticks to pick up the food and push it into her mouth. She managed the complex work and satisfied her hunger.

Question 36.
What were the contents of Bryson’s bag?
Answer:
The contents of the bag were frequent flyer card, newspaper cuttings, loose papers, tobacco pipe, magazines, passport, English money and film.

(iii) Answer any three of the following: [3 × 3 = 9]

Question 37.
Re-arrange the shuffled words and frame into meaningful sentences, (change to pie chart/graph or table)

1. pillars/there/human life/man/of/woman/and/are/two
2. uppermost/soil layer/is/earth/the/of/the
3. it/palnts/which/supports/food/provide/all/living things/to/planet/on/this

Answer:

1. There are two pillars of human life – man and woman.
2. Soil is the uppermost layer of the earth.
3. It supports plants which provide food to all living things on this planet.

Question 38.
Describe the process of pitching a tent.
Answer:

1. Select a location free of debris and an area that is as level as possible for your camp site. Lay down your footprint or ground cloth.
2. Position the tent over the footprint with the doors facing away from the wind for the best ventilation.
3. Lay out the poles and assemble them.
4. Insert the tent poles and secure it to take up the tension in the poles.
5. Pull the tent upright.

Question 39.
Expand the following news headlines:

1. Scientists Develop Synthetic Cornea.
2. World Leaders Meet at Geneva.
3. ATMs Without Security Guards to Close.

Answer:

1. A newly developed implant made of plastic may soon offer patients the chance to see again as Scientists from Germany have developed synthetic corneas.
2. German Chancellor Angela Merkel and US President Donald Trump will converge upon the snowy Swiss town of Davos this week for the World Economic Forum’s Annual Meeting.
3. The Prime Minister Shri Narendra Modi has requested all ATM’S without proper security to be closed down.

Question 40.
Complete the proverbs choosing the suitable words given in brackets.

1. A stitch in ………. saves nine, (tear, time, line)
2. A thing begun is ……….. done, (half, full, not)
3. Beauty is in the ………… of the beholder, (imagination, mind, eye)

Answer:

1. time
2. half
3. eye

PART – IV

IV. Answer the following: [7 × 5 = 35]

Question 41.
Describe Mary Kom’s personal experiences during her first International Championship match from the time of selection to winning the medal.
Answer:
Mary Kom flew to Pennsylvania, USA to compete under 48 kg World Women’s Boxing Championship in 2001. On landing, she rushed to the sports arena and weighed. She was lucky as she could rest enough to face her opponent the following day. As soon as she won her opponent in the first match, she gained enormous confidence. Her fear of facing new opponents in the ring vanished completely. While her team-mates went on losing one after another, she went on to reach the finals.

She was even hopeful of winning gold. She had defeated Nadia Hokmi of Poland by RSC. She also defeated Jamie Behai of Canada by 21-9. The greatest disadvantage just before finals was that she lost her appetite. She was not accustomed to American food. However hard she tried, she could not eat. She lost her weight. She was just 46 kg before the finals. This cost her, her long cherished dream of gold medal and she lost to Hula Sahin of Turkey 13-5. She won her silver but was very sad. But the biggest thing she took home from Pennsylvania was not the medal but the conviction that she could take on any boxer in the world.

[OR]

Trace the thoughts that went on in the mind of the narrator when picture after picture was put up and sold at the auction.
Answer:
The author was enthusiastically participating in the bid at Christie with very little money on him. He sailed smooth for a long time raising the stakes on many paintings and carefully staying behind other competitors. It was fun watching till he got trapped in a net, set by his own tongue. When one particular painting was offered for 4000 guineas, the bidders maintained an uncomfortable silence when the author heard himself foolishly saying “and fifty”. The auctioneer banged the hammer finalizing the deal in the narrator’s favour.

It was then the narrator realized with alarm that he had no money on him. Suddenly he lost interest in fun- bidding. He started thinking fast for a way out of the tight comer he had created for himself. Many small and big paintings were offered and sold out fast. The Barbizon pictures were selling fast like hot cakes for 2000 to 3000 guineas. The author was running over the names of friends, relatives and even money lenders who might bail him out of the tight comer. He even speculated on the possibility of confessing his poverty to the staff of Christie and request them to put up the picture again for sale. Such a genuine mistake could have been rectified at the early stages of auction.

As he had enthusiastically participated in the bid for many paintings, the auctioneers wouldn’t buy his justification for the “genuine mistake”. As bidders stood in a queue to hand in their cheques/cash to collect their paintings, the narrator stood deliberately at the end. He never felt such a fool or had colder feet all his life.

Question 42.
According to the poet what contributes most to the injuries sustained by the athletes?
Answer:
According to the poet, zealous athletes play so rough that they do not even consider one another’s feelings in their dealings with other players. The players are mostly goaded by prize money or glory from the media’s light on them. They maim each other as they romp. Cracking vertebrae and spines don’t stop the rough players. Most of the players don’t have sportsmanship.

They don’t treat success and failure equally. In order to get the light of fame on their face, they are ready to permanently disable a rival player too. The crazy desire for championship titles and the light of fame on them leads them to ignore swollen eyes, snapping of knee joints or cracking of wrists. In short, the poet believes the apathy of zealous players and obvious indifference to the pain and debilitating injury contribute most to the injuries sustained by athletes.

[OR]

Explain how ‘Lines Written in Early Spring’ stresses the fact that Nature is meant for Man’s joy and pleasure to be preserved by Man.
Answer:
Nature’s holy plan is to offer joy and peace to all forms of life on earth. He firmly believes that man is meant to spend his days blissfully taking part in the vitality and joy surrounding him. The poet believes that the harmonious, peaceful and happy co-existence of birds, plants, trees and brooks soothes the troubled mind of man. But he remembered the depravity of man which was evident in Napoleonic wars. He was fed up with man’s capacity to destroy innocent lives and property.

The poet is unhappy with unnatural aspects of industrial revolution, the misery caused by wars, greedy and aggressive behaviour causing suffering in humans. Man has lost his sensitivity to listen to the joyful lessons of nature and has gone to the extent of denuding the forest which really sustains life on earth. The poet stands to reason that nature functions on God’s plan but the man changes the holy plan wrecking the natural joy of human life.

Question 43.
Write an essay of about 150 words by developing the following hints.
C.V. Burgess-master craftsman-reveals-few names-first patient Joe-wife Emily- surgical room-Emily apprehensive-two children-Dorothea-Dentist hospital becomes play area-snobbish woman-whole play resolves-dramatic irony of patients’ guess- the dentists’ room -opening the tool cabinet-The groaning noise-vexation of Emily- Joe add to the dramatic irony-nurse moves about-feigned seriousness-the fact of the misplacement of key-which adds comic.
Answer:
C.V. Burgess is a master craftsman who reveals only a’few names. The first patient Joe and his wife Emily are the most dominant characters. Joe is inside the surgical room. Emily is anxious about the husband. Among the two children the dramatist uses only the girl’s name Dorothea and the Dentist hospital becomes a play area for Dorothea and the little boy who claim the same magazine for reading. The snobbish woman who goes on showing her photo album gives us an impression if she came to see the doctor or to show her photos. The whole play revolves around the dramatic irony of patients’ guess as to what happened inside the dentists’ room and what really happened.

The pliers, hack saw and the huge hammer were taken inside the dentist’s room only for opening the tool cabinet. But the patients wondered how these would’be used in dental surgery. The groaning noise from inside the dentist’s rooms and the vexation of Emily Joe add to the fear of the patients waiting. A few women patients leave the waiting room scared of subjecting themselves to the torture of having their bad teeth extracted with carpentry tools. The nurse moves about with all feigned seriousness without disclosing the fact of the misplacement of key.

[OR]

Narrator – wants – photograph – photographer wait for an hour – comments – angry – called on Saturday – proof – Narrator shocked – photograph – not like him – worthless bauble.
Answer:
‘With the Photographer’ by Stephen Leacock is narrated in the first person. The narrator while sitting in the photographer’s studio begins, to read some magazines and sees how other people look and the narrator begins to feel insecure about his appearance. It is also noticeable that the photographer takes a dislike to his face judging it to be wrong.

What should have been a simple process of taking a photograph becomes something of a nightmare for the narrator. How confident the narrator becomes is noticeable when he returns to the photographer’s studio the following Saturday. He realises that the photograph that has been taken of him looks nothing like him. This angers the narrator as he was simply looking for a photograph that would show his likeness. He accepts that he may not be to everybody’s liking when it comes to his physical appearance but is angered by the changes made. The photographer has retouched the photograph so much that the narrator does not recognise himself.

The end of the story is also interesting as the reader realises that it is just a worthless bauble when he begins to cry. He has been judged solely by his appearance by the photographer whose job was to simply take a life like photograph.

Question 44.
Write a summary or Make notes of the following passage.
Goa is a vibrant, living entity and more than just a geographical sunspot on the western coast of India. Famous for its silver sands and mesmeric sunsets, its recorded history datesback to the 3rd century B.C. It is blessed with marvelous weather, fabulous beaches, picturesque hill-top forts, gracious people and awe inspiring cathedrals. Arombol, 50 km north of Panaji, is a unique beach, which is both rocky and sandy.

It was a sweet water pond right near the seashore that’s very pleasant to bask in. Goa not only has almost 120 km long silver beaches but also offers long, wide and picturesque rivers and scenic lakes. Aqua sports hold great attraction for tourists to Goa. Some of the most popular aqua sports are swift rides on water scooters and speedboats at the Bay of Doha Paula. The best time to visit Goa, the ideal Serene Beach Resort is during the relatively cool winter months between late September and mid-March.
Answer:
Summary
No. of words given in the original passage: 160
No. of words to be written in the summary: 160/3 = 51 ± 5
Rough Draft
Goa is in the western coast of India. It is famous for silver sands, sunsets, weather, beaches, forts, gracious people and eathedrals. Arombol is a rocky and sandy beach with a small water pond with sweet water. Aqua sports like swift rides on water scooters and speedboats are available at the Bay of Dona Paula. It is good to visit Goa between late September and mid March.

Fair Draft
Goa
Goa is in the western coast of India. It is famous for silver sands, sunsets, weather, beaches, forts, gracious people and cathedrals. Arombol is a rocky and sandy beach with a small water pond with sweet water. Aqua sports like swift rides on water scooters and speedboats are available at the Bay of Dona Paula. It is good to visit Goa between late September and mid March.

No. of words in the summary: 67

[OR]

Note-making
Title: Goa
Answer:
Location:
western coast of India

Tourist attraction:
silver sands, sunsets, weather, beaches, forts, gracious people and cathedrals Arombol beach – sweet water pond Aqua sports held at at the Bay of Dona Paula

Best time to visit:
between late September and mid-March

Question 45.
Read the following advertisement and respond to it with a resume/bio-data/CV considering yourself fulfilling the conditions specified:
[Write XXXX for your name and YYYY for your address]
Wanted:
Applications are invited for the post of a Typist in a reputed school of Madurai. The . candidate must have at least 5 years of experience. The applicant must have a pleasant personality. He/she should be good in English. Attractive salary. Interested candidates should apply to The Principal, AKS International, Indirapuram, Madurai within 10 days with detailed resume.

14th January, 2020

From
Savitha
2, Gandhi street
Madurai
To
The Correspondent
AKS International School
Indirapuram
Madurai
Respected Sir/Madam,
Sub: Application for the post of a Typist
With reference to your advertisement dated 8th January 2019,1 hereby wish to apply for the post of a typist. I have rich experience and can communicate well with a pleasing personality. If given an opportunity, I will satisfy my superiors to the best of my ability.
Please find enclosed my resume for your kind perusal.

Yours sincerely,
Savitha
Address on the Envelope To
The Correspondent
AKS International School
Indirapuram
Madurai

Resume:
Name : Savitha Rajeev
Date Of Birth : 8th May, 1991
Marital Status : Married
Husband’s Name : Mr. Rajeev
Address For Communication : Yyy
Contact Number – Mobile: 9988776655
Residence : 22445566
Mother Tongue : Tamil
Language Known : English and Tamil
Educational Background:

Professional Experience:

Hobbies : Photograph, Gardening, Reading, Travelling.
Expected Salary : 27,000/ per month
Salary Drawn : 23,000/ per month
Reference : 1. Mr. Ravi (Manager -no. 9998887777
2. Mrs. Rani (Manager-Raj Enterprises) 9900000222

Declaration
I hereby declare that the above given information is true to my knowledge.
Station: YYY
Date. 14.01.2020

Savitha
Signature of the Applicant

[OR]

Write an essay in about 150 words on ‘Convincing the public not to discriminate against those with AIDS’.
Answer:
Abolish discrimination against those with AIDS
People living with HIV infection and AIDS should have the same basic rights and responsibilities as those which apply to all citizens of the country. They have the same rights to liberty and autonomy, security of the person and to freedom of movement as the rest of the population. No restrictions should be placed on the free movement of HIV-infected people, and they may not be segregated, isolated or quarantined in prisons, schools, hospitals or elsewhere merely because of their HIV-positive status.

Only a sexual relationship without proper precautions will be a risk factor. Therefore they should be accepted whole heartedly without any discrimination. People with HIV infection or AIDS are entitled to the right to make their own decisions about any matter that affects marriage and child-bearing – although counselling about the consequences of their decisions should be provided. Due to the fear of isolation, ignorance, denial, and discrimination, people will allow HIV to develop into AIDS, further decreasing life expectancy, since the body’s immune system function will have been significantly lowered.

Along with family bonds and intimate relationships, a spiritual relationship is also strained. Fear and vulnerability included fear of punishment from God, fear of being discovered as HIV/ AIDS-positive and fear of the future and death. They should be given the love, warmth, psycho-social and emotional support.

Question 46.
Read the following sentences, spot the errors and rewrite the sentences correctly.
(a) People of diverse cultures lives in India.
(b) There is many people who exhibit unity on diversity.
(c) Water is needed to irrigating the fields.
(d) We live in times when incomes is rising.
(e) The opening ceremony for the 2016 Summer Olympic Games took place in Maracan stadium.
Answer:
(a) People of diverse cultures lives in India.
(b) There are many people who exhibit unity on diversity.
(c) Water is needed to irrigate the fields.
(d) We live in times when incomes are rising.
(e) The opening ceremony of the 2016 Summer Olympic Games took place in Maracana stadium.

[OR]

Fill in the blanks appropriately.
(a) They could ……….. untie the ………. (knot / not)
(.b) The world ………….. avoid war in the larger interest of human race. (Fill in with a modal verb)
(c) Every student …………. respect the national symbols, (use semi-modal)
(d) ……….. you have the hall ticket you can enter the examination hall. (Use a suitable link word).
Answer:
(a) not, knot
(b) should
(c) must
(d) If

Question 47.
Identify each of the following sentences with the fields given below.
(a) The flight was cancelled due to fog.
(b) Spicy food can cause acidity in the stomach.
(c) Meena stumbled upon a chance to practice running a race.
(d) The company has recommended a dividend of 75 percent.
(e) The monitor is not working. Get it repaired.
(Sports, Weather, Commerce, Nutrition and Dietetics, Computer)
Answer:
(a) Weather
(b) Nutrition and Dietetics
(c) Sports
(d) Commerce
(e) Computer

[OR]

Read the following passage carefully and answer the questions that follow.
For millions of people in India, river Ganga is the most sacred river. It is considered as mother and goddess. It is also a lifeline to millions of Indians who live along its course and depend on it for their daily needs. River Ganga is the third largest river in the world by the amount of water that flows through it. It is the longest river in India. The river water of Ganga is used for irrigation, transportation and fishing. The Gangetic plain formed by river Ganga is one of the most fertile lands on earth.

This is why almost 10% of the world population lives here and earns its livelihood. The Ganga, in India is the most worshipped body of water. The irony here is that in spite of being the most worshipped river, it is also the dirtiest one. It carries some metals thrown out by tanneries, waste produced by industries and urban waste from different cities. All this has made river Ganga the fifth most polluted river in the world. Another major reason that adds to the Ganga river pollution is the coal based power plants on its banks which burn tons of coal every year and produce tons of fly ash. This ash mixed with domestic waste water is released, in the river.

This bad situation calls for an urgent need to make efforts to reduce pollution and revive river Ganga. To achieve these objectives, Government of India has started a programme named “Namami Ganga Programme”. The main pillars of this programme are sewage treatment, river surface cleaning, afforestation, riverfront development and public awareness. The importance of the success of ‘Namami Gange Programme’ can be seen through the following lines; “If Ganga dies, India dies. If Ganga thrives, India thrives. No Ganga, No India.”
Questions.

1. For whom is river Ganga a lifeline?
2. For what purpose is the Ganga river water used?
3. Most people in India consider the Gan^a as
4. What are the pollutants that make the river dirty?
5. Write any two pillars of the “Namami Gange”?

Answer:

1. River Ganga is a lifeline for millions of Indians.
2. The Ganga river water used for irrigation, transportation and fishing.
3. Goddess
4. Metals thrown out by tanneries, waste produced by industries and urban waste from different cities makes the river dirty.
5. The pillars of this programme are sewage treatment, river surface cleaning, afforestation, riverfront development and public awareness.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 3 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
40 ml of methane is completely burnt using 80ml of oxygen at room temperature. The volume of gas left after cooling to room temperature. The volume of gas left after cooling to room temperature is …………………….
(a) 40 ml of CO₂ gas
(b) 40 ml of CO₂ gas and 80 ml of H₂O gas
(c) 60 ml of CO₂ gas and 60 ml H₂O gas
(d) 120 ml of CO₂ gas
Solution:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
Answer:
(a) 40 ml of CO₂ gas

Question 2.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1,0
(b) 3, 1,-1, +\(\frac{1}{2}\)
(c) 3, 2, +1, –\(\frac{1}{2}\)
(d) 3, 0, 0, +\(\frac{1}{2}\)
Solution:
3p_x electron; n = 3 (main shell)
for p_x orbital, l = 1, m = -1, s = \(\frac{1}{2}\)
Answer:
(b) 3, 1,-1, +\(\frac{1}{2}\)

Question 3.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.
Answer:
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.

Question 4.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 5.
Lithium shows diagonal relationship with ………………………
(a) Sodium
(b) Magnesium
(c) Calcium
(d) Ahuninium
Answer:
(b) Magnesium

Question 6.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\) \((\frac { \partial V }{ \partial T } )\)
(a) T
(b) 1/T
(c) P
(d) None of these
Solution:

Answer:
(b) 1/T

Question 7.
Heat of combustion is always …………………….
(a) Positive
(b) Negative
(c) Zero
(d) Either positive or negative
Answer:
(b) Negative

Question 8.
If in a mixture where Q = K, then what happens?
(a) The reaction shift towards products
(b) The reaction shift towards reactants
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) Nothing happens
Answer:
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 9.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.
Answer:
(c) CO₂

Question 10.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, + 1
(b) + 1, 0, -1
(c) – 2, 0, + 2
(d) 0, 0, 0
Solution:

Formal charge of O_A/O_B = N_V – (N_e + \(\frac { N_{ b } }{ 2 } \)) = 6 – (4 + \(\frac{4}{2}\)) = 6 – 6 = 0
Formal charge of C = 4 – (0 + \(\frac{8}{2}\)) = 4 – 4 = 0
Answer:
(d) 0, 0, 0

Question 11.
In the hydrocarbon CH₃ – CH₃ – CH = CH – CH₂ – C = CH the state of hybridisation of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) None of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 12.
Enzyme present in apple is ………………………..
(a) Polyphenol oxidase
(b) Polyphenol reductase
(c) Polyphenol
(d) Polyphenol hydrolase
Answer:
(a) Polyphenol oxidase

Question 13.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ………………………….
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 14.
In Finkelstein reaction, the mechanism followed is ……………………….
(a) S_N1
(b) E₁
(c) E₂
(d) S_N2
Answer:
(d) S_N2

Question 15.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Define Avogadro Number?
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 × 10²³.

Question 17.
Explain the meaning of the symbol 4P. Write all the four quantum numbers for these electrons?
Answer:
4f²: It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac{1}{2}\), –\(\frac{1}{2}\)

Question 18.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”?
Answer:
No, It is not correct. Thq accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
What are the uses of calcium hydroxide?
Calcium hydroxide is used
Answer:

1. In the preparation of mortar, a building material.
2. In white wash due to its disinfectant nature.
3. In glass making and tanning industry.
4. For the preparation of bleaching powder and for the purification of sugar.

Question 20.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride?
Answer:
Given:
∆H_f = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 kJ^-1 mol^-1
T_f = ?
∆S_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}\); T_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}}\)
T_f = \(\frac{30400 \mathrm{J} \mathrm{mol}^{-1}}{28.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}\) = 1070.4 K

Question 21.
How is a gas-solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
Example: In carbonate beverages the following equilibrium exists.
CO₂(g) ⇄ CO₂ (solution).

Question 22.
What type of hybridisations are possible in the following geometeries?

1. Octahedral
2. Tetrahedral
3. Square planar

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 23.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

1. A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
2. The tube is heated till reaction ceases and become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
3. The contents of the dist is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 24.
Complete the reactions and identify the products?

Answer:

PART – III

Answer any six questions in which question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
The balanced equation for a reaction is given below 2x + 3y → 41 + m
When 8 moles of x react with 15 moles of y, then

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2x + 3y → 41 + m
1. 2x reacts with 3y to give products.
8x reacts with 15y means, y is the excess because 8 moles of x should react with 4 × 3y = 12y moles of y to give products.
In this reaction 15y moles are used.
Therefore, 3 moles of y is excess and x is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 × 41 i.e. 161 and 4m as product.
8x+ 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left is 3 moles of y.

Question 26.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium oxide has very strong ionic bonds as compared to magnesium fluoride.
2. Mg²⁺ and O^2- have charges of +2 and -2, respectively.
3. Oxygen ion is smaller than fluoride ion.
4. The smaller the ionic radii, the smaller the bond length in MgO and the bond is stronger than MgF₂.
5. Due to more strong bond nature in MgO, it has high melting point than MgF₂.

Question 27.
A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction
2KClO_3(s) → 2KCl + 3O₂
The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:

Question 28.
List the characteristics of entropy?
Characteristics of entropy:
Answer:

• Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
• In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
• Entropy is defined as “for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
  \(\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T}\)
• If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy,
• The change in entropy of a process represented by ∆S and is given by the equation,
  ∆S_sys = S_f – S_i
• If S_f > S_i, ∆S is positive, the reaction is spontaneous and reversible.
  If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
• Unit of entropy: SI unit of entropy is J K^-1.

Question 29.
Derive the values of K_C and K_P for the synthesis of HI.
Answer:
H₂(g) + I₂(g) ⇄ 2HI(g)
Let us consider the formation of HI in which V moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’.
Let ‘x’ moles of each of H₂ and I₂ react together to form 2x moles of HI.

Applying mass of action

Caluculation of K_P: K_P = K_C.RT^∆ng
Here ∆n_g = n_p – n_r = 2 -2 = 0
Hence, K_P = K_C
K_P = \(\frac{4 x^{2}}{(a-x)(b-x)}\)

Question 30.
Describe the classification of organic compounds based on their structure?
Answer:
Classification of organic compounds based on the structure

Question 31.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical, carbocation and carbanion?

Answer:

Question 32.
An alkyl halide with molecular formula C₆H₁₂Br on dehydrohalogenation gave two isomeric alkenes X and,Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide?
Answer:

1. C₆ H₁₃ Br is 3 – Bromo – 4 – methylpentane.
2. 3 – Bromo – 4 – methylpentane on dehydration give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methylpentane

Question 33.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

1. With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin, cataract, sunburn, skin cancer etc.
2. By killing many of the phytoplanktons, it can damage the fish productivity.
3. Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) Balance the following equations by ion electron method?
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
(II) Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?

[OR]

(b)
(I) How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn²⁺ (z = 25) and argon (z = 18)?
(II) Explain about the significance of de Broglie equation?
Answer:
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H₂O to balance oxygen atoms

Add Hcl to balance hydrogen atoms
KMnO₄ + 5e^– + 8 HCl → MnCl₂ + 4H₂O …………………. (4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H₃BO₃ = (1 × 3) + (11 × 1) + (16 × 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 × 0.543
= 33.66 g

[OR]

(b) (I) Fe → Fe³⁺ + 3e^–
Fe (Z = 26)
Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25) Electronic configuration is
1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 35 (a).
(I) Mention any two anomalous properties of second period elements?
(II) Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason?

[OR]

(b)
(I) How do you expect the metallic hydrides to be useful for hydrogen storage?
(II) Write a note about ortho water and para water?
Answer:
(a) (I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, \(\mathbf{r}_{\mathrm{Na}^{+}}\), > \(\mathbf{r}_{\mathrm{Mg}^{2+}}\), > \(\mathbf{r}_{\mathrm{Al}^{3+}}\).

[OR]

(b)
(I) In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

(II)
1. Water exists in space in the interstellar clouds, in proto-planetary disks, in the comets and icy satellites on the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho – H₂O and para – H₂O, in which the directions are antiparallel.

3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para – H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para – H₂O (OPR = 2.5) than on Earth.

Question 36 (a).
Derive the values of critical constants from the Van der Waals constants?

[OR]

(b) Derive the values of K_p and K_C for dissociation of PCl₅?
Answer:
(a) Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT for 1 mole.
From this equation, the values of critical constant P_C, V_C and T_C are derived in terms of a and b the Vander Waals constants.

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point, all the three values of V are equal to the critical volume Vc. i.e. V = V_C„
i.e; V = V_C
V – V_C = 0 ………………… (5)
(V – V_C)³ ………………….. (6)
(V³ – 3V_CV²) + 3V_C²V – V_C³ …………………. (7)
As the equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

Substituting the values of V_C and P_C in equation (9)

Critical constants a and b can be calculated rising Vander Waals constants as follows:

(b) Consider that V moles of PCl₅is taken in container of volume‘V’
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

Applying law of mass action

K_p caluculation: K_p = K_C. \(\mathrm{RT}^{\mathrm{An}}\); ∆n_g = 2 – 1 = 1
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where ‘n’ is the total number of moles at equilibrium
n = a – x + x + x = a + x

Question 37 (a).
(I) Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement?
(II) Explain how non-ideal solutions shows positive deviation from Raoult’s law?

[OR]

(b)
(I) How will you distinguish between electrophiles and nucleophiles?
(II) Complete the following reactions and identify the products?
(a)
(b)
Answer:
(a) (I) When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.
(II)

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alco¬hol and water-water interaction).
3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure
   predicted by Raoult’s law.
5. Here, the mixing process is endothermic i.e., DHmixing > 0 and there will be a slight increase in volume (DVmixing > 0)
6. 

(b)
(I)
Electrophiles:

1. They are electron deficient.
2. They are cations.
3. They are lewis acids.
4. Accept an electron pair.
5. Attack on electron rich sites.

Nucleophiles:

1. They are electron rich.
2. They are anions.
3. They are lewis bases.
4. Donate an electron pair.
5. Attack on electron deficient sites.

(II)
(a)
(b)

Question 38 (a).
(I) Is it possible to prepare methane by Kolbe’s electrolytic method?
(II) Explain how 2-butyne reacts with
(a) Lindlar’s catalyst
(b) Sodium in liquid ammonia.

[OR]

(b) (I) Discuss the aromatic nucleophilic substitution reactions of chlorobenzene?
(II) CCl₄ > CHCl₃ > CH₂C₁₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason?
Answer:
(a) (I) Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.
(II) (a) 2-butyne reacts with Lindlar’s catalyst: 2-butyne can be reduced to cis – 2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis – 2 – butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans – 2 – butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans – 2 – butene.

[OR]

(b) (I) Aromatic nucleophilic substitution reactions:
Dow’s process:

1.

2.

3.

(II) The boiling point of chloro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is ……………………
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Solution:

The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3/4.4 = 0.075 mol
Answer:
(c) 0.075

Question 2.
Two electrons occupying the same orbital are distinguished by …………………
(a) Azimuthal quantum number
(b) Spin quantum number
(c) Magnetic quantum number
(d) Orbital quantum number
Solution:
Spin quantum number For the first electron ms = + \(\frac{1}{2}\)
For the second electron ms = – \(\frac{1}{2}\)
Answer:
(b) Spin quantum number

Question 3.
Statement – 1: Ionization enthalpy of N is greater than that of O.
Statement – II: N has exactly half filled electronic configuration which is more stable than electronic configuration of O.
(a) Statement – I is wrong but statement – II is correct
(b) Statement – I is correct but statement – II is wrong.
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.
(d) Statement – I and II are correct but statement – II is not the correct explanation of statement – I.
Answer:
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.

Question 4.
Water gas is …………………….
(a) H₂O_(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 5.
Among the following the least thermally stable is ……………………
(a) K₂CO₃
(b) Na₂CO₃
(c) BaCO₃
(d) Li₂CO₃
Li₂CO₃ is least stable.
Answer:
(d) Li₂CO₃

Question 6.
Match the following.

Answer:

Question 7.
C(diamond) → C(graphite), ∆H = -ve, this indicates that …………………
(a) Graphite is more stable than diamond
(b) Graphite has more energy than diamond
(c) Both are equally stable
(d) Stability cannot be predicted
Answer:
(a) Graphite is more stable than diamond

Question 8.
In the equilibrium, 2A(g) ⇄ 2B(g) + C₂(g)
the equilibrium concentrations of A, B and C₂ at 400 K are 1 × 10^-4M, 2.0 × 10^-3M, 1.5 × 10⁴M respectively. The value of K_C for the equilibrium at 400 K is ……………………..
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 × 10^-2
Solution:
[A] = 1 × 10^-4M; [B] = 2 × 10^-3M; [C] = 1.5 × 10^-4M
2A(g) ⇄ 2B(g) + C₂(g)

= 6.0 × 10^-2 = 0.06
Answer:
(a) 0.06

Question 9.
Which of the following is a non-aqueous solution?
(a) Salt solution
(b) Sugar solution
(c) Br₂ in CCl₄
(d) Ethanol dissolved in water
Answer:
(c) Br₂ in CCl₄

Question 10.
Which of the following molecule does not exist due to its zero bond order?
(a) \(\mathrm{H}_{2}^{-}\)
(b) \(\mathrm{He}_{2}^{+}\)
(c) He₂
(d) \(\mathrm{H}_{2}^{+}\)
Answer:
(c) He₂

Question 11.
Which of the following is optically active?
(a) 3 – Chloropentane
(b) 2 – Chloropropane
(c) Meso – tartaric acid
(d) Glucose
Answer:
(d) Glucose

Question 12.
Which of the following represent a set of nucleophiles?
(a) BF₃, H₂O, NH^2-
(b) AlCl₃, BF₃, NH₃
(c) CN^–, RCH₂^–, ROH
(d) H⁺, RNH₃⁺, CCl₂
Answer:
(c) CN^–, RCH₂^–, ROH

Question 13.
Propyne on passing through red hot iron tube gives ……………………
(a)
(b)
(c)
(d) one of these
Answer:
(a)

Question 14.
Consider the following statements:

(I) E₂ reaction is a bimolecular elimination reaction of second order
(II) E₂ reaction takes place in two steps.
(III) E₂ reaction generally Jakes place in primary alkyl halides.

Which of the above statements is/are not correct?
(a) (I) only
(b) (II) only
(c) (III) only
(d) (I) & (III)
Answer:
(II) E₂ reaction takes place in two steps.

Question 15.
Photo chemical smog formed in congested metropolitan cities mainly consists of ………………….
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

PART – II

Answer any six questions in which question No. 23 is compulsory. [6 × 2 = 12]

Question 16.
Why interstitial hydrides have a lower density than the parent metal?
Answer:

1. d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
2. Hydrogen atom being small in size occupy some in the metallic lattice producing distortion without any change in its type.
3. The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.

Question 17.
Prove that calcium oxide is a basic oxide?
Answer:
Calcium oxide is a basic oxide. It combines with acidic oxides at high temperature.

1. 
2. 

Question 18.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means?
Answer:

1. The mathematical relationship between the volume of a gas and the number of moles is V ∝ n
2. \(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \) Constant, where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂
   and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of the moles of the gas.

Question 19.
Why pressure has no effect on the synthesis of HI?
Answer:
When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = 0.

Question 20.
Draw the lewis structure of PCl₅ and SF₆
Answer:

Question 21.
How are naphthalene and camphor purified?
Answer:

1. Naphthalene, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to be purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resulting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 22.
How will you convert ethyl chloride into
(I) Ethane
(II) n – butane
Answer:
(I) Conversion of ethyl chloride into ethane:

(II) Conversion of ethyl chloride into n – butane:
Wurtz reaction:

Question 23.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle, why?
Answer:
(I) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene), it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

(II) With the use of 1 % ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl₂ gas into non poisonous diethyl carbonate.
COCl₂ + 2C₂H₅OH → CO(OC₂H₅)₂ + 2HCl.

Question 24.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

1. Classical smog is caused by coal-smoke and fog.
2. It occurs in cold humid climate.
3. The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
4. Chemically it is reducing in nature because of high concentration of SO₂ and so it is also called reducing smog.
5. It is primarily responsible for acid rain.
6. It also causes bronchial irritation.

Photochemical smog:

1. Photochemical smog is cause by photochemical oxidants.
2. It occurs in warm, dry and sunny climate.
3. The chemical composition is the mixture of NO₂ and O₃ gases.
4. Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃ and so it is also called oxidising smog.
5. It causes irritation to eyes, skin and lungs and increase the chances of asthma.
6. It causes corrosion of metals, stones and painted surfaces.

PART – III

Answer any six questions in which question No. 28 is compulsory. [6 × 3 = 18]

Question 25.
An ice cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:

1. In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
2. Liquid water at 0°C has the density as 999.82 kg/cm³. Maximum density is attained by water only at 4°C as 1000 kg/cm³.
3. When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called anomalous expansion of water.
4. The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water,
5. At 0°C, ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 26.
Write the chemical equations for the reactions involved in Solvay process of preparation of sodium carbonate?
Answer:
Solvay process:
The Solvay process is represented by the below chemical equations:

Question 27.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if
(a) It Is compressed to a smaller volume at constant temperature
(b) The temperature is raised while keeping the volume constant
(c) More gas is introduced into the same volume and at the same temperature
Answer:
(a) If a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) If more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. If the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 28.
Calculate \(\Delta \mathrm{H}_{\mathrm{r}}^{0}\) for the reaction
CO₂(g) + H₂(g) → CO(g) + H₂O (g)
given that \(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) for CO₂(g), CO(g) and H₂O(g) are – 393.5, – 111.31 and – 242 kJ mol^-1 respectively.
Answer:
Given:
\(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) CO₂ = -393.5 KJ mol^-1

Question 29.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic?
Answer:
(I) Electronic configuration of O atom is 1s² 2s² 2p⁴

(II) Electronic configuration of O₂ molecule is

(III) Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \) = \(\frac{10-6}{2}\) = 2

(IV) Molecule has two unpaired electrons, hence it is paramagnetic.

Question 30.
Give the principle involved in the estimation of halogen in an organic compound by Carius method?
Answer:
Estimation of halogens: Carius method
(I) A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.

(II) C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

(III) The precipate AgX is filtered, washed, dried and weighted.

(IV) From the mass of AgX and the mass of organic compound taken, the percentage of halogen are caluculated.
(V)

Question 31.
What polymerisation? Explain with suitable example?
Answer:
A polymer is a larga molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation, a few examples are:

Question 32.
Compare \(\mathbf{S}_{\mathrm{N}^{1}}\) and \(\mathbf{S}_{\mathrm{N}^{2}}\) reaction mechanisms?
Answer:

Question 33.
From where does ozone come in the photochemical smog?
Answer:
(I) Photochemical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.

(II) Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃. So it is also called oxidising smog.

(III) Photochemical smog is formed by following reactions:
N₂ + O₂ 2NO
2NO + O₂ 2NO₂

(O) + O₂ O₃
O₃ + NO NO₂ + (O)

(IV) NO and O₃ are strong oxidising agents and they can react with unbumt hydrocarbons in polluted air to form formaldehyde, acrolein and PAN.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
(I) An atom of an element contains 35 electrons and 45 neutrons. Deduce

1. The number of protons
2. The electronic configuration for the element
3. All the four quantum numbers for the last electron

(II) How many unpaired electrons are present in the ground state of Fe²⁺ (z = 26), Mn²⁺ (z = 25) and argon (z=18)?

[OR]

(b)
(I) Explain why hydrogen is not placed with the halogen in the periodic table.
(II) Complete the following reactions.
Al₄C₃ + D₂O → ?
CaC₂ + D₂O → ?
Mg₃N₂, + D₂O → ?
Ca₃P₂ + D₂O → ?
Answer:
(a) (I) An element X contains 35 electrons and 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m = +1, s = –\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^–
Fe (Z = 26) Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

[OR]

(b) (I)

1. Hydrogen resembles alkali metals as well as halogens.
2. Hydrogen resembles more alkali metals than halogens.
3. Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
4. In most of its compounds hydrogen exists in +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

(II)

Question 35 (a).
(I) Why alkafr metals have high chemical reactivity? How this changes along the group?
(II) Distinguish between alkali metals and alkaline earth metals?

[OR]

(b)
(I) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude?
(II) Explain the graphical representation of Charles’ law?
Answer:
(a) (I) Alkali metals exhibit high chemical reactivity because of their low ionization enthalpy and their larger size.
The reactivity of alkali metals increases from Li to Cs, since the value of ionization energy decreases down the group (Li to Cs). All the alkali metals are highly reactive towards the more electronegative elements such as oxygen and halogens.

(II)
Alkali Metals:

1. Alkali metals are soft.
2. They have a single electron in the valence shell and their electronic configuration is [noble gas] ns¹.
3. They have low melting points.
4. Hydroxides are strongly basic.
5. Carbonates do not decompose.
6. Nitrates give corresponding nitrites and oxygen as products.
7. They show +1 oxidation states.
8. Their carbonates are soluble in water except Li₂CO₃.
9. Except Li, alkali metals do not form complex compounds.

Alkaline earth metals:

1. Alkaline earth metals are hard.
2. They have two electrons in the valence shell and their electronic configuration is [noble gas] ns².
3. They have relatively high melting points.
4. Hydroxides are less basic.
5. Carbonates decompose to form oxide, when heated to high temperatures.
6. Nitrates give corresponding oxides, nitrogen dioxide and oxygen as products.
7. They show +2 oxidation states.
8. Their carbonates are insoluble in water.
9. They can form complex compounds.

[OR]

(b)
(I) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease.
As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

(II)

1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3. i.e. P₁ < P₂ < P₃ < P₄ < P₅. When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 36 (a).
(I) Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
(II) Derive the relationship between standard free energy (∆G°) and equilibrium constant (K_eq)

[OR]

(b)
(I) 2.56g of Sulphur is dissolved in 100 g of carbon disulphide. The solution boils at 319. 692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS₂ is 319. 450K. Given that K_b for CS₂ = 2.42 K kg mol^-1
(II) Show that the sum of mole fraction of a solution is equal to one?
Answer:
(a) (I) A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

(II)

1. In a reversible process, system is at all times in perfect equilibrium with its surroundings.

2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.

3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.

4. It is possible only if at equilibrium, the free energy of a systepi is minimum.

5. Lets consider a general equilibrium reaction,
A + B ⇄ C + D
The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.
∆G = ∆G° +RTIn Q ………………. (1)
where Q is reaction quotient and is defined as the ratio of concentrajion of the products to the concentration of the reactants under non-equilibrium condition.

6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes, ∆G° =-RTln K_eq ……………… (2)
This equation is known as Van’t Hoff equation.
∆G° = -2.303 RTlogKeq ………………. (3)
We also know that,
∆G° = ∆H° -T∆S° = – RT In K …………….. (4)

[OR]

W₂ = 2.56 g; W₁ = 100 g
T = 319.692; K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K
M₂ = image 28
M₂ = 256 g mol^-1
Molecular mass of sulphur in solution = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac{256}{2}\) = 8
Hence, molecular formula of sulphur is S₈.

(II) Consider a solution containing two components A and B whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A
and n_B respectively.

Question 37 (a).
(I) Explain about the procedure and calculation behind the carius method of estimation of sulphur?
(II) What is the difference between distillation, distillation under reduced pressure and steam distillation?

[OR]

(b)
(I) An organic compound (A) of a molecular formula C₂H₄ which is a simple alkene. A reacts with dil H₂SO₄ to give B. A again reacts with Cl₂ to give C. Identify A, B and C and write the equations.
(II) Why chloro acetic acid is stronger acid than acetic acid?
Answer:
(a) (I) Carius method

• Procedure: A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and heated for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape.
• The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker. H₂SO₄ formed is converted to BaSO₄ (white ppt.) The precipitate is filtered, washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(II) Calculation:
Mass of organic compound = Wg
233 g of BaSO₄ contains 32 g of sulphur
Percentage of sulphur = (\(\frac{32}{233}\) × \(\frac{x}{w}\) × 100)%

(II) Distillation is used in case of volatile liquid mixed with a non-volatile impurities.
Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

[OR]

(b) (I)

1. C₂H₄ is CH₂ = CH₂ is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dil H₂SO₄ to give ethanol (B)

3. Ethylene (A) reacts with Cl₂ to give 1, 2 dichloro ethane (C)

(II) Chloro acetic acid: image 31
Chloro acetic acid has Cl – group and it has high electronegativity and shows -I effect. Therefore Cl – atom to facilitate the dissociation of O – H bond very fastly. Whereas in the case of acetic acid, has CH₃ group and it shows +1 effect, therefore dissociation of O – H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 38 (a).
(I) Write a chemical reaction useful to prepare the following:

1. Freon – 12 from carbon tetrachloride.
2. Carbon tetrachloride from carbon disulphide.

(II) What are ambident nucleophiles? Explain with an example.

[OR]

(I) Write about hydrosphere (or) Why Earth is called as Blue planet?
(II) Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Freon-12 from carbon tetrachloride:
Freon-12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This reaction is called “Swarts reaction.”

2. Carbon tetrachloride from carbon disulphide:
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl₃ as catalyst to give carbon tetrachloride.

(II) Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:

It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

[OR]

(b)

1. Hydrosphere include all types of water sources like oceans, seas, rivers, lakes, streams, underground water, polar ice – caps, clouds etc.
2. It covers about 75% of the earth’s surface. Hence earth is called as Blue planet:

(II) Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as

1. Insecticides
2. Fungicides and
3. Herbicides.

1. Insecticides:
Insecticides like DDT, BHC, Aldrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish..

2. Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

3. Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Students can Download Tamil Nadu 11th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Maths Model Question Paper 5 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 2:30 Hours
Maximum Marks: 90

PART – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If n((A × B) ∩ (A × C)) = 8 and n(B ∩ C) = 2 then n(A) = …………………
(a) 6
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 2.
The value of log₃ \(\frac{1}{81}\) is ……………….
(a) -2
(b) -8
(c) -4
(d) -9
Answer:
(c) -4

Question 3.
The value of log₃ 11 log₁₁ 13 log₁₃ 15 log₁₅ 27 log₂₇ 81 is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is …………………..
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Answer:
(c) 0

Question 5.
If tanα and tan β are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……………………..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the value of a is …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r= 35 then n = …………………..
(a) 1
(b) 6
(c) 5
(d) 4
Answer:
(a) 1

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = …………………….
(a) -2
(b) \(\frac{1}{2}\)
(c) – \(\frac{1}{2}\)
(d) 2
Answer:
(d) 2

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = …………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …………………..
(a) (7, 3)
(b) (4, 1)
(c) (1,-1)
(d) (3, 4)
Answer:
(b) (4, 1)

Question 11.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B is a symmetric matrix
(b) AB is a symmetric matrix
(c) (AB) = (BA)^T
(d) A^TB = AB^T
Answer:
(d) A^TB = AB^T

Question 12.
If [3 -1 2] B = [5, 6] then the order of B is ………………….
(a) 3 × 2
(b) 2 × 3
(c) 3 × 1
(d) 1 × 1
Answer:
(a) 3 × 2

Question 13.
1f \(\underset { x\rightarrow 0 }{ lim } \) \(\frac{sin px}{tan 3x}\) = 4 then the value of p is …………………….
(a) 6
(b) 9
(c) 12
(d) 4
Answer:
(c) 12

Question 14.
For \(\vec { a } \) = \(\vec { i } \) + \(\vec { j } \) – 2\(\vec { k } \), \(\vec { b } \) = \(\vec { i } \) + 2\(\vec { j } \) + \(\vec { k } \) and \(\vec { c } \) = \(\vec { i } \) – 2\(\vec { j } \) + 2\(\vec { k } \) the unit vector parallal to is \(\vec { a } \)
+ \(\vec { b } \) + \(\vec { c } \) is ……………………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}-\vec{j}+\vec{k}}{\sqrt{6}}\)
Answer:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀ x with respect to log_x 10 is …………………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac { x^{ 2 } }{ 100 } \)
Answer:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e^x+5logx) is …………………..
(a) e^xx⁴(x + 5)
(b) e^xx(x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Answer:
(a) e^xx⁴(x + 5)

Question 17.
If f(x) = x tan^-1x thenf'(1) = ……………………..
(a) 1 + \(\frac { \pi }{ 4 } \)
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)
(c) \(\frac{1}{2}\) – \(\frac { \pi }{ 4 } \)
(d) 2
Answer:
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)

Question 18.
∫cosec xdx = …………………….
(a) log tan \(\frac{x}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) log (cosecx – cot x) + c
(d) all of them
Answer:
(d) all of them

Question 19.
Ten coins are tossed; The probability of getting atleast 8 heads is …………………..
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is …………………..
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Answer:
(a) \(\frac{19}{33}\)

PART – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A?
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}\) = \(\frac{1+sinθ}{cosθ}\)
Answer:

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
No. of non-collincar points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.
∴ No. of Tnangle formed ¹⁵C₃
= \(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\) = 455

Question 24.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1?
Answer:
Givc sum of the intercepts = 1
⇒ when x intercept a then y intercept = 1 – a
Equation of the line is \(\frac{x}{a}\) + \(\frac{y}{1-a}\) = 1
The line passes through (8, 3) ⇒ \(\frac{8}{a}\) + \(\frac{3}{1-a}\) = 1
(i.e) 8 (1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 6a + 8 =0
(a – 2)(a – 4) = 0 ⇒a = 2 or 4
1. When a = 2 equation of the line is \(\frac{x}{2}\) + \(\frac{y}{-2}\) = 1 (i.e) \(\frac{x}{2}\) – y = 1 ⇒ x – 2y = 2
2.When a = 4 equation of the line is \(\frac{x}{4}\) + \(\frac{y}{1-4}\) = 1 (i.e) \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1 ⇒ 3x – 4y = 12

Question 25.
Find the values of p, q, r, & s if \(\left[\begin{array}{ccc}
p^{2}-1 & 0 & -31-q^{3} \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]\)
Answer:
When two matrices (of some order) are equal then their correspondings entries are equal.

Question 26.
Find |\(\vec { a } \) × \(\vec { b } \)| where \(\vec { a } \) = 3\(\vec { i } \) + 4\(\vec { j } \) and \(\vec { b } \) = \(\vec { i } \) + \(\vec { j } \) + \(\vec { k } \)
Answer:

Question 27.
At the given point x₀ discover whether the given function is continous or discontinous citing the reasons for your answer?
Answer:

Question 28.
Evaluate y = xe^-x2
Answer:

Question 29.
Evaluate ∫\(\frac{1}{x logx}\) dx?
Answer:

Question 30.
Evaluate [((256)^-1/2)^-1/4]³
Answer:

PART – III

III. Answer any seven questions. Question No. 40 Is compulsory. [7 × 3 = 21]

Question 31.
Find the largest possible domain for the real valued functionsf defined by f(x) = \(\sqrt { x^{ 2 }-5x+6 } \)

Question 32.
Show that tan 75° + cot 75° = 4?

Question 33.
There are 10 bulbs in a room. Each one of them can be operated independently. Find the number of ways in which the room can be illuminated?

Question 34.
Find the \(\sqrt [ 3 ]{ 126 } \) approximately to two decimal places?

Question 35.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = O and parallel to the line 3x + 4y = 7?

Question 36.
If \(\left|\begin{array}{ccc}
a & b & a \alpha+b \\
b & c & b \alpha+c \\
a \alpha+b & b \alpha+c & 0
\end{array}\right|\) = 0 prove that a, b, c are in G.P. or a is a root of ax² + 2bx + c =0

Question 37.
Evaluate \(\underset { x\rightarrow 3 }{ lim } \)\(\frac { x^{ 2 }-9 }{ x-3 } \) ¡f it exists by finding f(3^–) and f(3⁺)

Question 38.
Find the derivative of tan^-1 (1 + x²) with respect x² + x + 1?

Question 39.
Evaluate: ∫x⁵e^{x2 dx}

Question 40.
Prove that the line segment joining the mid points of the adjacent sides of a quadrilateral from parllelogram?

PART – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) Graph the functions f(x) = x³ and g(x) = \(\sqrt [ 3 ]{ x } \) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

[OR]

(b) 1f x = -2 is one root of x³ – x² – 17x = 22 then find the other roots of the equation?

Question 42.
(a) lf A + B + C= it prove that cosA + cos B + cosC = 1 + 4 sin (\(\frac{A}{2}\)) sin (\(\frac{B}{2}\) sin (\(\frac{C}{2}\))

[OR]

(b) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) show that A² – 4A – 5I = O

Question 43.
(a) If ⁿ⁺¹C₈ : ^(n-3)P₄ = 57 : 16, find the value of n?

[OR]

(b) If the letters of the word IITJEE arc permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE?

Question 44.
(a) The line \(\frac{x}{a}\) + \(\frac{y}{b}\)= 1 moves in such a way that \(\frac { 1 }{ a^{ 2 } } \) + \(\frac { 1 }{ b^{ 2 } } \) = \(\frac { 1 }{ c^{ 2 } } \) where c is a constant. Find the locus of the foot of the perpendicular from the origin on the given line?

[OR]

(b) Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them?

Question 45 (a).
Prove that \(\left|\begin{array}{lll}
1 & x^{2} & x^{3} \\
1 & y^{2} & y^{3} \\
1 & z^{2} & z^{3}
\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)

(b) Evaluate \(\underset { x\rightarrow \infty }{ lim } \) \(\frac{3}{x-2}\) – \(\frac{2 x+11}{x^{2}+x-6}\)

Question 46.
(a) Evaluate \(\frac{1}{6 x-7-x^{2}}\)

(b) Evaluate ∫e^tan-1x (\(\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \)) dx

Question 47 (a).
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

[OR]

(b) Firm manufactures PVC pipes in three plants viz. X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,

1. find the probability that the selected pipe is a defective one.
2. if the selected pipe ¡s a defective, then what is the probability that it was produced by plant Y?

Students can Download Tamil Nadu 11th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ………………..
(a) 0
(b) 2 m/s
(c) 4 m/s
(d) 2π m/s
Hint:
Displacement of the cyclist in half revolution is
d = diameter of the circular track
i.e., d= 80 m
Time taken, t = 40 s
Average velocity, V =  = \(\frac{80}{40}\)
V = 2 m/s
Answer:
(b) 2 m/s

Question 2.
A wheel has angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 seconds it has rotated through an angle of (in radian) ………………..
(a) 10
(b) 12
(c) 4
(d) 6
Answer:
(a) 10

Question 3.
If the origin of co-ordinate system lies at the centre of mass. The sum of the moments of the masses of the system about the centre of mass is …………………
(a) May be greater than zero
(b) May be less than zero
(c) May be equal to zero
(d) Always zero
Answer:
(d) Always zero

Question 4.
Dimensional formula for co-efficient of viscousity
(a) ML^-2 T^-2
(b) ML^-2 T^-1
(c) ML^-1 T^-1
(d) M^-1 L^-1 T^-1
Answer:
(c) ML^-1 T^-1

Question 5.
Action and reaction …………………
(a) Acts on same object
(b) Acts on two different objects
(c) Have resultant not zero
(d) Acts on the same direction
Answer:
(b) Acts on two different objects

Question 6.
A spring is stretched by applying load to its free end. The strain produced in the spring is …………………
(a) Volumetric
(b) Shear
(c) Longitudinal
(d) Longitudinal and shear
Answer:
(d) Longitudinal and shear

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 ms^-2
(d) 25 ms^-2

Hint:
τ = F × r
Iα = F × r
MR² × α = 30 × \(\frac{40}{100}\); \(\frac { 3\times 40\times 40\times \alpha }{ 100\times 100 } \) = 12
\(\frac { 3\times 16\times \alpha }{ 100 } \) = 12; α = 25 rad/s²
Answer:
(b) 25 rad s^-2

Question 8.
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (E is total energy) ………………
(a) \(\frac{2}{3}\) E
(b) \(\frac{1}{3}\) E
(c) \(\frac{1}{4}\) E
(d) \(\frac{1}{2}\) E
Hint:
PE = \(\frac{1}{2}\) kx²
⇒ \(P E_{V_{2}}=\frac{1}{2} K\left(\frac{A}{2}\right)^{2}=\frac{1}{4}\left(\frac{1}{2} K A^{2}\right)\)
Answer:
(c) \(\frac{1}{4}\) E

Question 9.
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/s and 7.5 m/s² respectively. Amplitude of the oscillation is ………………
(a) 0.36
(b) 0.28
(c) 0.61
(d) 0.53
Hint:
x = A sin ωt
∴ a = \(\frac { d^{ 2 }x }{ dt^{ 2 } } \) = -Aω² sinωt
∴Maximum acceleration |a_max| = Aω²
Now Aω² = 7.5
A = \(\frac { 7.5 }{ \omega ^{ 2 } } \) = \(\frac { 7.5 }{ (3.5)^{ 2 } } \) = 0.61
Answer:
(c) 0.61

Question 10.
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become ………………….
(a) \(\frac{n}{4}\)
(b) \(\sqrt{2n}\)
(c) n
(d) \(\frac { n }{ \sqrt { 2 } } \)
Hint:

Answer:
(c) n

Question 11.
The theory of refrigerator is based on …………….
(a) Joule-Thomson effect
(b) Newton’s particle theory
(c) Joule’s effect
(d) None of the above
Answer:
(d) None of the above

Question 12.
Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is ………………..
(a) 54 cal
(b) 60 cal
(c) 546 cal
(d) 600 cal
Hint:
Workdone (W) = – p.dv = nRT
= 0.1 × (0.2 cal) × (273 + 27) = 0.1 × 2 × 300
W = 60 cal
Answer:
(b) 60 cal

Question 13.
When a lift is moving upwards with acceleration a, then time period of simple pendulum in it will ………………..
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)
(b) 2π\(\sqrt { \frac { g+a }{ l } } \)
(c) \(\frac{1}{2π}\)\(\sqrt { \frac { 1 }{ g+a } } \)
(d) \(\frac{1}{2π}\)\(\sqrt { \frac { g+a }{ l } } \)
Answer:
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)

Question 14.
A disc is rotating with angular speed ω. If a child sits on it, what is conserved?
(a) Linear momentum
(b) Angular momentum
(c) Kinetic energy
(d) Potential energy
Answer:
(b) Angular momentum

Question 15.
The vectors \(\vec { A } \) and \(\vec { B } \) are such that |\(\vec { A } \) + \(\vec { B } \)| = |\(\vec { A } \) – \(\vec { B } \)|. The angle between the two vector is ………………..
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Hint:
The angle between two vector is always 90°.
Answer:
(d) 90°

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Velocity – time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement – time graph for the motion of the object?
Answer:
The given graph shows that the velocity of the object is constant. That is, the velocity of the object is not changing, so the acceleration of the object is zero. Since the acceleration of an object is given by


Displacement – time graph for the motion of the object is shown in the figure above.

Question 17.
Can a body subjected to a uniform acceleration always move in a straight line?
Answer:
It will be a straight line in one dimensional motion but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform acceleration.

Question 18.
Calculate the viscous force on a ball of radius 1mm moving through a liquid of viscosity 0.2 Nsm^-2 at a speed of 0.07 ms^–
Answer:
Radius of the ball (a) = 1mm = 1 × 10^-3m
Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2
Speed of the ball (v) = 0.07 ms^-1
According to Stoke’s law
Viscous force F = 6 π η av
= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07
= 0.26376 × 10^-3 = 2.64 × 10^-4N

Question 19.
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10ms^-2)
Answer:
Given data: F = 30 N, load (m) 2 kg; height = 10m, g = 10 ms^-2
Gravitational forcc F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J

Question 20.
Why are shockers used in automobiles like car?
Answer:
In the event of jump or jerk, the lime of action of force increases. Since the product of force aid time is constant in a given situation. therefore the force decreases.

Question 21.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighted 250 N on the surface?
Answer:
As g_d = g (1 – \(\frac{d}{R}\)) ⇒ mg_d = mg(1 – \(\frac{d}{R}\))
Here d = \(\frac{R}{2}\)
∴mg_d = (250) × (1 – \(\frac { R/2 }{ R } \)) = 250 × \(\frac{1}{2}\) = 125N

Question 22.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\vec { A } \) and \(\vec { B } \) are perpendicular to each other than their scalar product \(\vec { A } \).\(\vec { B } \) = O because cos 90° = 0. Then the vectors \(\vec { A } \) and \(\vec { B } \) are said to be mutually orthogonal.

Question 23.
An air bubble of radius r in water is at a depth of h below the water surface at some instant. If P is atmospheric pressure and d and T are the density and surface tension of ater respectively. Calculate the pressure P inside the bubble?
Answer:
Excess ot pressure inside the air bubble in water = \(\frac{2T}{r}\)
∴ Total pressure inside the air bubble
= atmospheric pressure + pressure due to liquid column + Excess pressure due to surface tension
= P + hρg + \(\frac{2T}{r}\)

Question 24.
Define beats?
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second.
n = |f₁ – f₂| per second.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Define centripetal acceleration and give any two examples?
Answer:
The acceleration that is directed towards the centre of the circle along the radius and perpendicular to the velocity of the particle is known as centripetal or radial or normal acceleration.
Example:-

1. In the case of planets revolving round the Sun or the moon revolving round the earth, the centripetal force is provided by the gravitational force of attraction between them.
2. For an electron revolving round the nucleus in a circular path, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Question 26.
Write any six properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors \(\vec { A } \) and \(\vec { B } \) may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \). But, \(\vec { A } \) × \(\vec { B } \) = – \(\vec { B } \) × \(\vec { A } \) .
Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin θ = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e., θ = 0° or θ = 180°
(\(\vec { A } \) × \(\vec { B } \))_min = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0° \(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

(VI) The self-vector products of unit vectors are thus zero.
\(\hat { i } \) × \(\hat { i } \) = \(\hat { j } \) × \(\hat { j } \) = \(\hat { k } \) × \(\hat { k } \) = 0

Question 27.
Show that the pressure of the gas is equal to two third of mean kinetic energy per unit volume?
Answer:
The internal energy of the gas is given by
U = \(\frac{3}{2}\) NkT
The above equation can also be written as
U = \(\frac{3}{2}\) PV
Since PV = NkT
P = \(\frac{2}{3}\) \(\frac{U}{V}\) = \(\frac{2}{3}\) u
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 (u = \(\frac{U}{V}\))
Writing pressure in terms of mean kinetic energy density using equation.
P = \(\frac{1}{3}\) nm\(\overline { V^{ 2 } } \) = \(\frac{1}{3}\) ρ\(\overline { V^{ 2 } } \)
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
P = \(\frac{2}{3}\)(\(\frac{ρ}{2}\) \(\overline { V^{ 2 } } \))
P = \(\frac{2}{3}\) \(\overline { KE } \)
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 28.
Derive an expression for gravitational potential energy?
Answer:
The gravitational tbrce 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
dW = \(\vec { F } \)_ext . d\(\vec { r } \) ……………….. (1)
The work is done against the gravitational force, therefore,
\(\vec { F } \)_ext = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \) ………………. (2)
Substituting equation (2) in (1), we get
dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).d\(\vec { r } \) ………………… (3)
d\(\vec { r } \) = dr \(\hat { r } \) ⇒ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).(dr \(\hat { r } \))
\(\hat { r } \).\(\hat { r } \) = 1 (Since both are unit vectors)
∴ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \) dr …………….. (4)
Thus the total work done for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses and m₁ and m₂ when the separation between them are r and r’ respectively.

Question 29.
A satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by satellite due to Earth’s gravitational force?
Answer:
g’ = g(1 – \(\frac { 2h }{ R_{ e } } \))
= g(\(\frac { 1-2\times 1600\times 10^{ 3 } }{ 6400\times 10^{ 3 } } \)) = g(1 – \(\frac{2}{4}\))
g’ = g (1 – \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = g (1- \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = \(\frac{8}{2}\) (or) g’ = \(\frac{9.8}{2}\) = 4.9ms^-2

Question 30.
Explain the v ariation of a g with latitude?
Answer:
When an object is on the surface fo 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 mωR’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g = g – ω²R 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 31.
A bullet of mass 50g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms^-2
Answer:
m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁ The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\)
v = \(\frac{0.05 u_{1}+(0.45 \times 0)}{(0.05+0.45)}\) = \(\frac{0.05}{0.50}\)u₁
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
v = \(\sqrt{2gh}\)
v = \(\sqrt{2 \times 10 \times 1.8}\) = \(\sqrt{36}\)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
6 = \(\frac{0.05}{0.50}\)u₁ or u₁ = \(\frac{0.05}{0.50}\) × 6 = 10 × 6
u₁ = 60ms^-1

Question 32.
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 33.
Calculate how many times more intense is 90 dB sound compared to 40 dB sound?
Answer:
Given Data:
L = log \(\frac { I }{ I_{ 0 } } \) = log I – log I₀
We get 90 dB = 9 B = log I₁ – log I₀ ……………….. (1)
40 dB = 4 B = logI₂ – logI₀ ………………….. (2)
Subtract (2) from (1)
50 dB = 5B = log I₁ – logI₂
5 = log₁₀ (\(\frac{I_{1}}{I_{2}}\))
\(\frac{I_{1}}{I_{2}}\) = 10⁵

PART – IV

Answer all the questions. [ 5 × 5 = 25]

Question 34 (a)
Obtain an expression for the time period T of a simple pendulum. The time period T depends on

1. Mass of the bob(m)
2. Length of the pendulum (l)
3. Acceleration due to gravity (g) at the place where the pendulum is suspended, (constant k = 2π)

Answer:
Example:
An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon

1. Mass m of the bob
2. Length l of the pendulum and
3. Acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is k = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c ⇒ a = 0, b = 1/2, and c = -1/2
From the above equation
T = \(\mathrm{k} \mathrm{m}^{0} l^{1}=g^{-1}-2\)
T = \(k\left(\frac{1}{g}\right)^{1}\) = \(k \sqrt{1 / g}\)
Experimentally k = 2π, hence
T = \(2 \pi \sqrt{1 / g}\)

[OR]

(b) Obtain an expression for the escape speed in detail?
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
\(\mathrm{E}_{i}=\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) ………………. (1)
where, M_E is the mass of the Earth and R_E the radius of the Earth.
The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) 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
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 same initial speed to escape Earth’s gravity.

Question 35 (a).
Derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
\(\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………….. (1)
Total kinetic energy after collision
KE_f = \(\frac{1}{2}\) (m₁ + m₂)v² …………….. (2)
Then the loss of kinetic energy is
Loss of KE, ∆Q = KE_f – KE_i = \(\frac{1}{2}\) (m₁ + m₂)v² – \(\frac{1}{2}\) m₁ u₁² – \(\frac{1}{2}\) m₂ u₂² ………………. (3)
Substituting equation v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) in equation (3), and on simplying (expand v by using the algebra) (a + b)² = a² + b² + 2ab, we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

(b) A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. Calculate the initial velocity of the shell?
Answer:
Given Data :
m = 200 gm = 0.2 kg; M = 4 kg.
Energy generated = 1.05 KJ = 1.05 × 10³ J
According to law of Conservation of linear momentum
mv = Mv’
∴v’ = (\(\frac{m}{M}\)) v
Total K.E of the gun and bullet

[OR]

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

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²

(d) Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about

1. The diameter of the disc
2. The axis, tangent to the disc and parallel to its diameter
3. The axis through the centre of the disc and perpendicular to its plane

Answer:
1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m
Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR²
Id = \(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m²

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc
= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = \(\frac{5}{4}\) × 0.5 × 1
= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc
= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²

Question 36 (a).
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it?
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum.

When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force \(\vec { F } \)₁₂ on particle 2 and particle 2 exerts an exactly equal and opposite force \(\vec { F } \)₁₂ on particle 1 according to Newton’s third law.
\(\vec { F } \)₂₁ = –\(\vec { F } \)₁₂ …………… (1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as
\(\vec { F } \)₁₂ = \(\frac{d \bar{p}_{1}}{d t}\) and \(\vec { F } \)₂₁ = \(\frac{d \vec{p}_{2}}{d t}\) ……………… (2)

Here \(\vec { P } \)₁ is the momentum of particle 1 which changes due to the force \(\vec { F } \)₁₂ exerted by particle 2. Further \(\vec { P } \)₂ is the momentum of particle 2. This changes due to \(\vec { F } \)₂₁ exerted by particle 1.
Substitute equation (2) in equation (1)

It implies that \(\vec { P } \)₁ + \(\vec { P } \)₂ = (constant vector always)

\(\vec { P } \)₁ + \(\vec { P } \)₂ is the total linear momentum of the two particles ( \(\vec { P } \)_tot = \(\vec { P } \)₁ + \(\vec { P } \)₂). It is also called as total linear momentum of the system. Flere, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system ( \(\vec { P } \)_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec { P } \)₁ and \(\vec { P } \)₂ can vary,
in such a way that \(\vec { P } \)₁ + \(\vec { P } \)₂ is a constant vector.

The forces \(\vec { F } \)₁₂ and \(\vec { F } \)₁₂ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

To find the recoil velocity of a gun when a bullet is fired from it:
Consider the firing of a gun. Here the system is Gun + bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { P } \)₁ be the momentum of the bullet and \(\vec { P } \)₂ the momentum of the gun before firing. Since initially both are at rest.

\(\vec { P } \)₁ = 0, \(\vec { P } \)₂ = 0.

Total momentum before firing the gun is zero, \(\vec { P } \)₁ + \(\vec { P } \)₂ = 0

According to the law of conservation of linear momentum, total linear momemtum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { P } \)₁ + \(\vec { P } \)₂. To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { P } \)₂ to \(\vec { P } \)₂. Due to the conservation of linear momentum, \(\vec { P } \)₁+ \(\vec { P } \)₂‘= 0. It implies that \(\vec { P } \)₁‘ = –\(\vec { P } \)₂ the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum \(\vec { P } \)₂. It is called ‘recoil momentum’. This is an example of conservation of total linear momentum.

[OR]

(b) 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;, the initial total energy of the object is
\(E_{i}=\frac{1}{2} M v_{i}^{2}-\frac{G M M_{E}}{R_{E}}\) ………………. (1)
where, M_E is the mass of the Earth and RE the radius of the Earth. The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) 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 = 0 …………….. (2)

Substituting (1) in (2) we get,
\(\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{F}}}=0\)
\(\frac{1}{2} \mathrm{M} v_{i}^{2}=\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) …………….. (3)

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 same initial speed to escape Earth’s gravity.

Question 37 (a).
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.
\(\frac{dQ}{dt}\) ∝ (T – T_s) …………….. (1)

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 of T in time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt
\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) = -a(T – T_s) ……………….. (4)

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
\(\mathrm{T}=\mathrm{T}_{\mathrm{s}}+b_{2} e^{\frac{-a}{m s}}\) ……………. (6)
Here b₂ = _e^b1 = Constant

[OR]

(b) Derive an expression for pressure exerted by the gas on the wall 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 in 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 sanie 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 bêfore 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 ∆t.

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 &. The number of molecules that will hit the right side wall in a time interval ∆t 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 ii is number of molecules per
unit volume \(\frac{N}{V}\) 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.

Te no.of molecules that hit the right side wall in a time interval ∆t

= \(\frac{n}{2} \mathrm{A} v_{x} \Delta t\) ……………….. (1)
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ……………….. (2)

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)

F = \(\frac{\Delta p}{\Delta t}=n m \mathrm{A} v_{x}^{2}\) ……………….. (3)
Pressure P = force divided by the area of the wall

P = \(\frac{F}{A}\) = nmv²_x …………………. (4)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v²_x by the average \(\bar{v}_{x}^{2}\) in equation (4)

P = nm\(\bar{v}_{x}^{2}\) ………………… (5)

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, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as

\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3}\)nm \(\bar{v}^{2}\) or P = \(\frac{1}{3}\) \(\frac{N}{V}\) m\(\bar{v}^{2}\) as [n = \(\frac{N}{V}\)] ……………… (6)

Question 38 (a).
Explain with graphs the difference between work done by a constant force and by a variable force. Arrive at an expression for power and velocity. Give some examples for the same?
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ………………. (1)
The total Work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation.

dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,
W = \(\int_{r_{i}}^{r_{f}} d \mathrm{W}=\int_{r_{i}}^{r_{f}} \mathrm{F} \cos \theta d r\) ………………. (4)

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Expression for power and velocity

The work done by a force \(\vec { F } \) for a displacement \(\bar { dr } \) is
W = ∫\(\vec { F } \).\(\vec { dr } \) ……………. (1)
Left hand side of the equation (1) can be written as
W = ∫dW = ∫\(\frac{dW}{dt}\) (multiplied and divided by dt) ………………… (2)
Since, velocity is \(\vec { v } \) = \(\frac{d \vec{r}}{d t}\); \(\vec { dr } \) = \(\vec { v } \) dt. Right hand side of the equation (I) can be written as

Substituting equation (2) and equation (3) in equation (1), we get

This relation is true for any arbitrary value of di. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{dW}{dt}\) – \(\vec { F } \).\(\vec { v } \) = 0 Or \(\frac{dW}{dt}\) = \(\vec { F } \).\(\vec { v } \)
Hence power P = \(\vec { F } \).\(\vec { v } \)

[OR]

(b) 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.
\(\frac{dQ}{dt}\) ∝ (T – T_s)
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,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt

\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s law of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) -a(T – T_s) …………………. (4)
Where a is a positive constant.
From equation (3) and (4)
– a(T – T_s) = ms \(\frac{dT}{dt}\)
\(\frac{d T}{T-T_{s}}\) = -a\(\frac{a}{ms}\) dt ………………. (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
\(\mathrm{T}=\mathrm{T}_{s}+b_{2} e^{\frac{-a}{m s}}\) ………………… (6)
Here b₂ = e^b1 = constant

Students can Download Tamil Nadu 11th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 2 English Medium

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
   These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be ………………..
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Hint:
Volume of the sphere, V = \(\frac{4}{3}\) πr³
\(\frac{∆V}{V}\) × 100 = 3 × (\(\frac{∆r}{r}\) × 100) = 3 × 2%
\(\frac{∆V}{V}\) × 100 = 6%
Answer:
(d) 6%

Question 2.
A ball is dropped from a building. It takes 4s to reach the ground. The height of the building is (use g= 10 m/s²)
(a) 20 m
(b) 40 m
(c) 80 m
(d) 75 m
Hint:
s = ut + \(\frac{1}{2}\) at² s = h, g = a, u = 0
h = \(\frac{1}{2}\) gt²
h = \(\frac{1}{2}\) × 10 × (4)²; h = 80m
Answer:
(c) 80 m

Question 3.
For inelastic collision between two spherical rigid bodies ……………………
(a) the total kinetic energy is conserved
(b) the total mechanical energy is not conserved
(c) the linear momentum is not conserved
(d) the linear momentum is conserved
Answer:
(d) the linear momentum is conserved

Question 4.
Two rods OA and OB of equal length and mass are lying on xy plane as shown in figure. Let I_x, I_y and I_z be the moments of inertia of the the rods about x, y and z axis respectively, then …………………….
(a) I_x = I_y > I_z
(b) I_x > I_y > I_z
(c) I_x = I_y < I_z
(d) I_z > I_y > I_x

Hint:
I_x = I_y = 2\(\left[\frac{\mathrm{M} l^{2}}{3} \sin ^{2} 45^{\circ}\right]\) = \(\frac { ml^{ 2 } }{ 3 }\)
I_z = \(\left[\frac{m l^{2}}{3}\right]\) = \(\frac { 2ml^{ 2 } }{ 3 }\)
Answer:
(c) I_x = I_y < I_z

Question 5.
The motion of a rocket is based on the principle of conservation of ………………….
(a) Linear momentum
(b) Mass
(c) Angular momentum
(d) Kinetic energy
Answer:
(a) Linear momentum

Question 6.
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 7.
A given glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown. If the tube is rotated with a constant angular velocity ω, then …………………
(a) Water levels in both sections A and B go up
(b) Water level in section A goes up and that in B comes down
(c) Water level in section B goes up and that in A comes down
(d) Water level remain same in both
Answer:
(a) Water levels in both sections A and B go up

Question 8.
The efficiency of a heat engine working between the freezing point and boiling point of water is …………………..
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Hint:
Freezing point of water T_L = 0°C = 273K
Boiling point of water T_H – 100°C = 373K
∴ Efficiency, η = 1 – \(\frac { T_{ L } }{ T_{ H } } \) = 1- \(\frac{273}{373}\) = 0.2861; η = 26.8%.
Answer:
(c) 26.8%

Question 9.
Two waves represented by the following equation are travelling in the same medium
y₁ = 5 sin 2π (75 t – 0.25 x), y₂ = 10 sin 2π (150 – 0.25 x)
The intensity ratio of the two waves is ………………….
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Hint:
I ∝ A² ⇒ \(\frac { I_{ 1 } }{ I_{ 2 } }\) = (\(\frac { A_{ 1 } }{ A_{ 2 } }\))² = (\(\frac{5}{2}\))² = \(\frac{1}{4}\)
Answer:
(b) 1 : 4

Question 10.
A man pushes a wall and fails to displace it. He does …………………..
(a) Negative work
(b) Positive but not maximum work
(c) No work at all
(d) Maximum work
Answer:
(c) No work at all

Question 11.
A car moving on a horizontal road may be thrown out of the road in taking a turn …………………
(а) By the gravitational force
(b) Due to lack of sufficient centripetal force
(c) Due to rolling frictional force between tyre and road
(d) Due to the reaction of the ground
Answer:
(b) Due to lack of sufficient centripetal force

Question 12.
The volume of a gas expands by 0.25 m³ at a constant pressure of 10³ N/m, the workdone is equal to ……………………..
(a) 250 W
(b) 2.5 W
(c) 250 N
(d) 250 J
Hint:
Workdone = P. ∆V = 10³ × 0.25 = 250 J
Answer:
(d) 250 J

Question 13.
When three springs of spring constants k₁, k₂, k₃ connected in parallel, then the resultant spring constant is ……………………
(a) K = k₁ + k₂ + k₃
(b) \(\frac{1}{K}\) = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(c) K = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(d) K = k₁ – k₂ – k₃
Answer:
(a) K = k₁ + k₂ + k₃

Question 14.
The distance of the two planets from Sun are 10¹³ and 10¹² m respectively. The ratio of time period of the planets is …………………..
(a) 100
(b) \(\frac { 1 }{ \sqrt { 10 } } \)
(c) \(\sqrt{10}\)
(d) 10\(\sqrt{10}\)
Hint:
According to Kepler’s third law of planetary motion.
\(\frac { T_{ 1 } }{ T_{ 2 } } \) = \(\sqrt{\frac{R_{1}^{3}}{R_{2}^{3}}}\) = \(\sqrt{\frac{\left(10^{13}\right)^{3}}{\left(10^{12}\right)^{3}}}\) = \(\sqrt{\frac{10^{39}}{10^{36}}}\) = \(\sqrt{10^{3}}\) = 10\(\sqrt{10}\)
Answer:
(d) 10\(\sqrt{10}\)

Question 15.
The dimensional formula of planck’s constant is ………………..
(a) [M L² T^-1]
(b) [M L² T^-3]
(c) [M L T^-1]
(d) [M L³ T^-3]
Answer:
(a) [M L² T^-1]

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A particle is moving along a circular track of radius lm with uniform speed. What is the ratio of the distance covered and the displacement in half revolution?
Answer:
Distance covered by a particle = π × 1 = πm
Displacement covered by a particle = 2 × 1 = 2m
Ratio between distance and displacement

Question 17.
Give one argument in favour of the fact that frictional force is a non-conservative force?
Answer:
The direction of the frictional force is opposite to the direction of motion. When a body is moved, say from A to B and then back to A, work is required to be done both during forward and backward motion. So, the net workidone in a round trip is not zero. Hence, the frictional force is a non-conservative force.

Question Question 18.
Why does a gas not have a unique value of specific heat?
Answer:
This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating.

Question 19.
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 min. Calculate the velocity of river water in km/h?
Answer:
Resultant velocity = \(\frac{1km}{(15/60)h}\) = 4 km/h
If v is velocity of river, then v² + 4² = 5² ⇒ v = \(\sqrt{2-16}\) =3 km/h

Question 20.
In a dark room would you be able to tell whether a given note had been produced by a Piano or a Violin?
Answer:
Yes, in a dark room we can easily identify a sound produced by a Piano or a Violin by using the knowledge of timber or quality of sound. The two sources even though having the same intensity and fundamental frequency will be associated with different number of overtones of different relative intensities. These overtones combine and produce different sounds which enables us to identify them.

Question 21.
What is mean by P – V 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 22.
Why does a parachute descend slowly?
Answer:
The surface area of a parachute is much larger as compared to the surface area of stone. So, the air resistance in the case of a parachute is much larger than in the case of a stone. This explains as to why parachute descends slowly.

Question 23.
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 24.
Write a note on reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Write the rules for determining significant figure?
Answer:
Rules for counting significant figures:

Question 26.
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 27.
How do you classify the physical quantities on the basis of dimension?
Answer:
(I) Dimensional variables:
Physical quantities, which possess dimensions and have variable values are called dimensional variables. Examples are length, velocity, and acceleration etc.

(II) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.

(III) Dimensional Constant:
Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck’s constant etc.

(IV) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are n, e, numbers etc.

Question 28.
State 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.
T ∝ \(\sqrt{l}\)

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.
T ∝ \(\frac { 1 }{ \sqrt { g } } \)

Question 29.
Explain super position principle for 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
\(\overrightarrow{\mathrm{E}}_{\text {total }}=\overrightarrow{\mathrm{E}}_{1}+\overrightarrow{\mathrm{E}}_{2}+\ldots \overrightarrow{\mathrm{E}}_{n}=-\frac{\mathrm{G} m_{1}}{r_{1}^{2}} \hat{r}_{1}-\frac{\mathrm{G} m_{2}}{r_{2}^{2}} \hat{r}_{2}-\ldots \cdot \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)
\(\overrightarrow{\mathrm{E}}_{\mathrm{total}}=-\sum_{i=1}^{n} \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)

Question 30.
Write a note on static friction?
Answer:
Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
0 ≤ f_s ≤ µ_sN
where, µ_s – coefficient of static friction
N – Normal force then, equation shows that f_s can take any value from 0 & µ_sN. If object is at rest, when no external force acts on it then f_s = 0. If object is at rest, also external force acts on it, then f_s = F_ext
But still the static friction f_s is less than µ_sN when object begins to slide, the static friction (f_s) acting on the object attains maximum.

Question 31.
A small metal ball falls in liquid with a terminal velocity of V. If a ball of radius twice of first ball but same mass falls through a same medium, calculate the terminal velocity with which it falls?
Answer:
Given v = \(\frac{2 r^{2} \rho g}{9 \eta}\)
mass = \(\frac{4}{3}\) πr³ ρ = \(\frac{4}{3}\)π(2r)³ ρ¹ or ρ¹ = \(\frac{ρ}{8}\)
Terminal velocity of second ball is
v¹ = \(\frac{2(2 r)^{2}\left(\frac{\rho}{8}\right) g}{9 \eta}\) = \(\frac{v}{2}\)
v¹ = \(\frac{v}{2}\)

Question 32.
Derive an expression for co-efficient of performance of refrigerator?
Answer:
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.
COP = β = \(\frac { Q_{ L } }{ W } \) …………………… (1)
From the equation Q_L + W = Q_H
β = \(\frac{Q_{L}}{Q_{H}-Q_{L}}\)
β = \(\frac{1}{\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{L}}}-1}\) ………………. (2)
But we know that \(\frac { Q_{ H } }{ Q_{ L } } \) = \(\frac { T_{ H } }{ T_{ L } } \)
Substituting this equation (1) we get β = \(\frac{1}{\frac{T_{H}}{T_{L}}-1}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}\)

Question 33.
Derive an expression for Laplace’s correction?
Answer:
Laplace’s correction:
In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat.

Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………… (4)

where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume.

Differentiating equation (4) on both the sides, we get
\(\mathrm{V}^{\gamma} d \mathrm{P}+\mathrm{P}\left(\gamma \mathrm{V}^{\gamma-1} d \mathrm{V}\right)=0\)
or
\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………… (5)

where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (5) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_a = \(\sqrt{\frac{B_{A}}{\rho}}=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\gamma v_{T}}\)

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_a = (\(\sqrt{1.4}\)) (280m s^-1) = 331.30 m s^-1, which is very much closer to experimental data.

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34 (a).
Explain different types of error?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows

(i) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(ii) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(iii) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(iv) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(v) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer. Who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs.

For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.

If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ………………. a_n. The arthematic mean is
\(a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots a_{n}}{n}\) (or) \(a_{m}=\frac{1}{n} \sum_{i=1}^{i=n} a_{i}\)

[OR]

(b) By using equations of motion, derive an expression for range and maximum height reached by the object thrown at an oblique angle θ with respect to the horizontal direction?
Answer:

This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.

(Oblique projectile):
Examples:

1. Water ejected out of a hose pipe held obliquely.
2. Cannot fired in a battle ground.

Consider an object thrown with initial velocity at an angle 0 with the horizontal.
Then,
\(\vec { u } \) = u_x\(\hat { i } \) + u_y\(\hat { j } \)
where u_x = u cos θ is the horizontal component and u_y = u sin θ the vertical component of velocity.

Since the acceleration due to gravity is in the direction opposite to the direction of vertical component u_y, this component will gradually reduce to zero at the maximum height of the projectile.

At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (u_x = u cos θ) remains the same till the object reaches the ground.

Hence after the time f, the velocity along horizontal motion v_x = u_x + a_xt = u_x = u cos θ.
The horizontal distance travelled by projectile in time t is s_x = u_xt + \(\frac{1}{2}\)a_xt²
Here, s_x = x, u_x = u cos θ, a_x = 0.

Thus, x = u cos θ.t or t = \(\frac{x}{u cosθ}\) …………….. (1)
Next, for the vertical motion v_y= u_y + a_yt
Here u_y = u sin θ, ay = – g (acceleration due to gravity acts opposite to the motion).
Thus, v_y = u sin θ – gt
The vertical distance travelled by the projectile in the same time t is
Here, s_y = y, u_y = u sin θ, a_x = – g Then,
y = u sin θ t – \(\frac{1}{2}\)gt² ……………….. (2)
Subsitute the value of t from equation (i) and equation (ii), we have
y = \(u \sin \theta \frac{x}{u \cos \theta}-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\)
y = \(x \tan \theta-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\) ………………… (3)
Thus the path followed by the projectile is an inverted parabola.

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y = u sin θ, a = -g, s = h_max, and at the maximum height v_y = 0
Hence, (0)² = u² sin² θ = 2gh_max
or
h_max = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) …………….. (4)

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight.
This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\) a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f.
Then 0 = u sinθ T_f – \(\frac{1}{2}\)gT^2_f
T_f = 2u \(\frac{sin θ}{g}\) …………………. (5)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write

Range R = Horizontal component of velocity x time of flight = u cos θ × T_f = \(\vec{r}_{1} \times \vec{r}_{2}\)

The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum,
sin 2θ = 1. This implies 2θ = π/2
or θ = \(\frac{π}{4}\)
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range is given by.
R_max = \(\frac { u^{ 2 } }{ g }\) ……………….. (6)

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows: Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let 0 be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between R and A. Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

(I) Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.
From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴ AN = B cos θ and sin θ = \(\frac{BN}{B}\) ∴BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B² sin² θ
⇒ R² = A² + B ² (cos² θ + sin² θ) + 2AB cos θ
⇒ R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

(II) Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
IF \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA + AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1(\(\frac{B \sin \theta}{A+B \cos \theta}\))

[OR]

(b) Arrive at an expression for velocity of objects in one dimensional elastic collision?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁u₁ + m₂v₁ ………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) ………………. (2)
Further,

Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ……………….. (3)
After simplifying and rearranging terms,
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
m₁(u₁ + v₁) = m₂(v₂ + u₂) (v₂ – u₂) …………………. (4)
Dividing equation (4) by (2) gives,
\(\frac{m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)}{m_{1}\left(u_{1}-v_{1}\right)}=\frac{m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)}{m_{2}\left(v_{2}-u_{2}\right)}\)
u₁ + v₁ = v₂ + u₁
Rearranging, u₁ – u₂ = v₂ – v₁ ………………… (5)
Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………. (6)
or v₂ = u₁ + v₁ – u₁ ………………. (7)

To find the final velocities v₁ and v₂:
Substituting equation (7) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – 2a₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ = m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂) u₁ + 2m₂u₂ = (m₁ + m₂)v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂ ……………. (8)
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ ………………. (9)

Question 36 (a).
Discuss the variation of g with change in altitude and depth?
Answer:
When an object is on the surface fo 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 objecf would have been mg. Elowever, the object experiences an additional centrifugal force due to spinning of the Earth.

This centrifugal force is given by mω²R’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g’ = g – ω²2 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.

Variation of g with depth:
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

g’ = \(\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}_{\mathrm{e}}-d\right)^{2}}\) ………………. (1)
Here M’ is the mass of the Earth of radius (R_e – d) Assuming the density of the earth ρ be constant,
ρ = \(\frac{M}{V}\) …………………. (2)

where M is the mass of the Earth and V its volume, Thus,

Here also g’ < g. As depth increases, 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.

[OR]

(b) 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 figure (1), 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 rms speed, average speed and most probable speed are indicated in the figure (1). It can be seen that the rms speed is greatest among the three.

1. The area under the graph will give the total number of gas molecules in the system.
2. 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 37 (a).
What is meant by simple harmonic motion?
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.

(b) 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? Given g on Earth = 9.8 ms^-2?
Answer:

(c) A man with wrist watch on his hand falls from the top of the tower. Does the watch , give correct time during the free fall?
Answer:
Yes. Because the working of a wrist watch does pot depend on gravity at that place but depends on spring action.

[OR]

(d) State Wien’s law?
Answer:
When the animals feed cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

(e) 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, \(\frac{F-32}{180}\) = \(\frac{K-273}{100}\)
So, T = 98.6°F = 310K
From Wien’s displacement law
Maximum wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9348 × 10^-9 m
λ_max = 9348 nm (at 98.6°F)
During high fever, human body temperature
T = 104°F = 313K
Peak wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9259 × 10^-9 m
λ_max = 9259 (at 104°F)

(f) Animals curl into a ball, when they feel very cold. Why?
Answer:
When animals feel cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

Question 38 (a).
Explain the horizontal oscillations of 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 x₀. 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.
F ∝ x
F = -kx ……………… (1)

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 is 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.

\(m \frac{d^{2} x}{d t^{2}}=-k x\) ……………….. (1)
\(\frac{d^{2} x}{d t^{2}}=-\frac{k}{m} x\) ……………….. (2)
Comparing the equation with simple harmonic motion equation, we get
ω² = \(\frac{k}{m}\) ………………….. (3)

which means the angular frequency or natural frequency of the oscillator is
ω = \(\sqrt{\frac{k}{m}} \mathrm{rad} s^{-1}\) ……………….. (4)

The frequency of the oscillation is
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \mathrm{Hertz}\) …………………… (5)
and the time period of the oscillation is
T = \(\frac{1}{f}\) = 2π \(\sqrt{m/k}\) seconds …………………. (6)

[OR]

(b) 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 capillar 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:

1. 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.
2. Absorption of ink by a blotting paper.
3. Capillary action is also essential for the tear fluid from the eye to drain constantly.
4. 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

1. Horizontal component T sin θ and
2. Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus.

Total upward force = (T cos θ) (2nr) = 2nrT 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,
T = \(\frac{r \rho g h}{2 \cos \theta}\)
Liquid rises through a height h
h = \(\frac{2 \mathrm{T} \cos \theta}{r \rho g} \Rightarrow h \alpha \frac{1}{r}\)
This implies that the capillary rise (h) is inversely proportional to the radius (r) of the tube, i.e., the smaller the radius of thd tube greater will be the capillarity.

Subject Matter Experts at SamacheerKalvi.Guide have created Tamil Nadu State Board Samacheer Kalvi 11th English Book Answers Solutions Guide Pdf Free Download are part of Samacheer Kalvi 11th Books Solutions.

Let us look at these TN State Board New Syllabus Samacheer Kalvi 11th Std English Guide Pdf of Text Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, and revise our understanding of the subject.

Students can also read Tamil Nadu Samacheer Kalvi 11th English Model Question Papers 2020-2021 English & Tamil Medium.

Samacheer Kalvi 11th English Book Solutions Answers Guide

Samacheer Kalvi 11th English Book Back Answers

Samacheer Kalvi 11th English Book Solutions Prose

• Chapter 1 The Portrait of a Lady
• Chapter 2 The Queen of Boxing
• Chapter 3 Forgetting
• Chapter 4 Tight Corners
• Chapter 5 The Convocation Address
• Chapter 6 The Accidental Tourist

Samacheer Kalvi 11th English Book Solutions Poem

• Chapter 1 Once Upon A Time
• Chapter 2 Confessions of a Born Spectator
• Chapter 3 Lines Written in Early Spring
• Chapter 4 Macavity – The Mystery Cat
• Chapter 5 Everest is Not The Only Peak
• Chapter 6 The Hollow Crown

Samacheer Kalvi 11th English Book Solutions Supplementary

• Chapter 1 After Twenty Years
• Chapter 2 A Shot in the Dark
• Chapter 3 The First Patient
• Chapter 4 With the Photographer
• Chapter 5 The Singing Lesson
• Chapter 6 The Never Never Nest

We hope these Tamilnadu State Board Samacheer Kalvi Class 11th English Book Solutions Answers Guide Pdf Free Download will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding TN State Board New Syllabus Samacheer Kalvi 11th Standard English Guide Pdf of Text Book Back Questions and Answers, Chapter Wise Important Questions, Study Material, Question Bank, Notes, drop a comment below and we will get back to you as soon as possible.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 14 Computerised Accounting Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 14 Computerised Accounting

11th Accountancy Guide Computerised Accounting Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
n accounting, computer is commonly used in the following areas:
a) Recording of business transactions
b) Payroll accounting
c) Stores accounting
d) All the above
Answer:
d) All the above

Question 2.
Customised accounting software is suitable for ________.
a) Small, conventional business
b) Large, medium business
c) Large, typical business
d) None of the above
Answer:
b) Large, medium business

Question 3.
Which one is not a component of computer system?
a) Input unit
b) Output unit
c) Data
d) Central Processing Unit
Answer:
c) Data

Question 4.
An example of output device is ________.
a) Mouse
b) Printer
c) Scanner
d) Keyboard
Answer:
b) Printer

Question 5.
One of the limitations of computerised accounting system is ________.
a) System failure
b) Accuracy
c) Versatility
d) Storage
Answer:
a) System failure

Question 6.
Expand CAS ________.
a) Common Application Software
b) Computerised Accounting System
c) Centralised Accounting System
d) Certified Accounting System
Answer:
b) Computerised Accounting System

Question 7.
Which one of the following is not a method of codification of accounts?
a) Access codes
b) Sequential codes
c) Block codes
d) Mnemonic code
Answer:
a) Access codes

Question 8.
TALLY is an example of ________.
a) Tailor-made accounting software
b) Ready-made accounting software
c) In-built accounting software
d) Customised accounting software
Answer:
b) Ready-made accounting software

Question 9.
People who write codes and programmes are called as ________.
a) System analysts
b) System designers
c) System operators
d) System programmers
Answer:
d) System programmers

Question 10.
Accounting software is an example of ________.
a) System software
b) Application software
c) Utility software
d) Operating software
Answer:
b) Application software

II. Very Short Answer Questions

Question 1.
What is a computer?
Answer:

1. A computer can be described as an electronic device designed to accept raw data as input, processes them and produces meaningful information as output.
2. It has the ability to perform arithmetic and logical operations as per given set of instructions called program.

Question 2.
What is CAS?
Answer:

1. Computerised accounting system refers to the system of maintaining accounts using computers.
2. It involves the processing of accounting transactions through the use of hardware and software in order to keep and produce accounting records and reports.
3. Computerised accounting system takes accounting transactions as inputs that are processed through accounting software.

Question 3.
What is hardware?
Answer:

1. The physical components of a computer constitute its hardware.
2. Hardware consists of input devices and output devices that make a complete computer system.
3. Examples of input devices are keyboard, optical scanner, mouse, joystick, touch screen and stylus which are used to feed data into the computer.
4. Output devices such as monitor and printer are media to get the output from the computer.

Question 4.
What is meant by software?
Answer:
A set of programs that form an interface between the hardware and the user of a computer system are referred to as software.

Question 5.
What is accounting software?
Answer:
Accounting software describes a type of application software that records and processes accounting transactions within functional modules such as accounts payable, accounts receivable, journal, general ledger, payroll, and trial balance. It functions as an accounting information system

Question 6.
Name any two accounting packages.
Answer:

1. Readymade software
2. Customised software
3. Tailormade software

Question 7.
Give any two examples of readymade software.
Answer:

1. Tally ERP
2. Profit Books

Question 8.
What is coding?
Answer:

1. Coding refers to creating computer programming code.
2. The process of assigning something for classification or identification is known as coding.

Question 9.
What is grouping of accounts?
Answer:

1. Each minor head in accounting have number of sub-heads.
2. After classification of accounts into various groups namely, major, minor and sub-heads and allotting codes to each account these are programmed into the computer system.
3. A proper codification requires a systematic grouping of accounts.
4. The major groups or heads could be Assets, Liabilities, Revenues and Expenses.
5. The sub-groups or minor heads could be capital, non-current liabilities, current assets, sales and so on.

Question 10.
What are mnemonic codes?
Answer:

1. A mnemonic is a term, symbol or name used to define or specify a computing function.
2. Assembly language also uses a mnemonic to represent machine operation, or opcode.
3. Example are SJ – Sales Journals ; HQ – Head Quarters.

III. Short Answer Questions

Question 1.
What are the various types of accounting software?
Answer:
1. Readymade software:

• These packages are standardised or readymade packages which can be used by the business enterprises immediately on procurement. These packages are used by small and conventional business enterprises.
• Cost of installation and maintenance is very low. Training cost is negligible and sometimes the vendor provides free of cost training.
• This software’s are used by those enterprises where financial transactions are simple, uniform and routine in nature. Few examples of such type of software are Tally, Busy, Marg, and Profit books.

2. Customised software:

• Many a time, it is not possible that ready-to-use packages suit the requirements of the business enterprise.
• In such circumstances, customised packages may help the business enterprise for fulfilling their requirements. Customised packages can be modified according to the need of the enterprise.
• For example, software can record attendance of the employees and on the requirement of the customer it can also count the absence of employees in a month, etc.

3. Tailor made software:

• Large enterprises have their own way of functioning.
• For effective management information system, varied and specific information is frequently required by many users which may not be needed in case of small or medium scale enterprises.
• In such enterprises, depending upon their functioning, need based software’s known as tailored packages are installed.
• The cost of these packages is very high and specific training for using these packages is also required.

Question 2.
Mention any three limitations of computerised accounting system.
Answer:
1. Heavy cost of installation – Computer hardware needs replacement and software needs to be updated from time to time with the availability of newer versions.

2. Cost of training – To ensure effective and efficient use of computerised system of accounting, newer versions of hardware and software are to be introduced. These require special training and hence, cost is incurred to train the staff personnel.

3. Fear of unemployment – On account of the introduction of computerised accounting system, the employees feel insecure that they may lose employment and show less interest in computer related work.

4. Disruption of work – When computerised system is introduced, the existing process of accounting and other works are interrupted. This results in certain changes in the working environment.

5. System failure – The danger of a system crashing due to some failure in hardware can lead to subsequent interruption of work. This is more when no back-up is made.

Question 3.
State the various types of coding methods.
Answer:
Following are the three methods of codification.
(a) Sequential codes – In sequential code, numbers and/or letters are assigned in consecutive order. These codes are applied primarily to source documents such as cheques, invoices, etc. A sequential code can facilitate document search.
For example:

• Code – Accounts
• CL001 – ABC LTD
• CL002 – XYZ LTD
• CL003 – SCERT

(b) Block codes – In a block code, a range of numbers is partitioned into a desired number of sub-ranges and each sub- range is allotted to a specific group. In most of the cases of block codes, numbers within a sub – range follow sequential coding scheme, i.e., the numbers increase consecutively.
For example:

• Code       – Dealer type
• 100 – 199 – Small pumps
• 200 – 299 – Medium pumps
• 300 – 399 – Pipes
•  400 – 499 – Motors

(c) Mnemonic codes – A mnemonic code consists of alphabets or abbreviations as symbols to codify a piece of information.
For example:

• Code – Information
• SJ – Sales Journals
• HQ – Head Quarters

Question 4.
List out the various reports generated by computerised accounting system.
Answer:
Computerised accounting system takes accounting transactions as inputs that are processed through accounting software to generate the following reports:

1. Day books /Journals
2. Trading account
3. Ledger
4. Profit and loss account
5. Trial balance
6. Balance sheet, etc.

Question 5.
Tate the input and output devices of a computer system,
Answer:
1. Input devices – Examples of input devices are keyboard, optical scanner, mouse, joystick, touch screen and stylus which are used to feed data into the computer.

2. Output devices – Examples Output devices such as monitor and printer are media to get the output from the computer.

11th Accountancy Guide Computerised Accounting Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
An example of input device is ________.
a) Mouse
b) Printer
c) Monitor
d) Headphone
Answer:
a) Mouse

Question 2.
The facts and figures that are fed into a computer for further processing are called ________.
a) Procedure
b) Connectivity
c) Data
d) Reliability
Answer:
c) Data

Question 3.
________ packages are used by small and conventional business enterprises.
a) Readymade software
b) Customised software
c) Tailor made software
d) None of these
Answer:
a) Readymade software

Question 4.
________ packages c n be modified according to the need of the enterprise.
a) Readymade software
b) Customised software
c) Tailor made software
d) None of these
Answer:
b) Customised software

Question 5.
________ packages are used by medium or large nature business enterprises.
a) Readymade software
b) Customised software
c) Tailor made software
d) None of these
Answer:
b) Customised software

Question 6.
________ is a step by step series of instructions to per! rm a specific function and achieve desired output.
a) Procedure
b) Connectivity
c) Data
d) Reliability
Answer:
a) Procedure

Question 7.
The physical components of a computer constitute it ________.
a) Hardware
b) Software
c) Data
d) Procedure
Answer:
a) Hardware

Question 8.
A set of tools and programs to manage the overall working of a computer using a defined set of hardware components is called an ________.
a) Programming software
b) Utility software
c) Application software
d) Operating system
Answer:
d) Operating system

Question 9.
________ are designed specifically for managing the computer device and its resources.
a) Programming software
b) Utility software
c) Application software
d) Operating system
Answer:
b) Utility software

Question 10.
________ is an identification mark.
a) Hardware
b) Software
c) Data
d) Code
Answer:
d) Code

Question 11.
________ codes are applied primarily to source documents such as cheques, invoices, etc.
a) Sequential codes
b) Block codes
c) Mnemonic codes
d) None of these
Answer:
a) Sequential codes

Question 12.
________ code can facilitate document search.
a) Mnemonic codes
b) Block codes
c) Sequential codes
d) None of these
Answer:
c) Sequential codes

Question 13.
________ code, a range of numbers is partitioned into a desired number of sub-ranges and each sub-range is allotted to a specific group.
a) Mnemonic codes
b) Block codes
c) Sequential codes
d) None of these
Answer:
b) Block codes

Question 14.
________ code consists of alphabets or abbreviations as symbols to codify a piece of information.
a) Mnemonic codes
b) Block codes
c) Sequential codes
d) None of these
Answer:
a) Mnemonic codes

Question 15.
________ consists of input devices and output devices that make a complete computer system.
a) Hardware
b) Software
c) Data
d) Code
Answer:
a) Hardware

Question 16.
How many formats are available white exporting a file?
a) 2
b) 3
c) 5
d) 7
Answer:
b) 3

Question 17.
Tally package was developed by ________.
a) Tally solutions
b) Microsoft
c) Apple Solutions
d) None of the above
Answer:
a) Tally solutions

Question 18.
Which device of computer operation dispenses with the use of the keyboard?
a) Joystick
b) Mouse
c) Light Pen
d) Touch pen
Answer:
b) Mouse

Question 19.
Which of the following device primarily used to provide hard copy?
a) CRT
b) Pen drive
c) Printer
d) Card Reader
Answer:
c) Printer

Question 20.
Which of the following produces high-quality output?
a) Impact Printer
b) Non-Impact Printer
c) Both (a) and (b)
d) one of the above
Answer:
b) Non-Impact Printer

Question 21.
Which of the following is not a hardware?
a) Printer
b) Scanner
c) Interpreter
d) All of the above
Answer:
c) Interpreter

Question 22.
The copimonly used input device is the ________.
a) Mouse
b) Monitor
c) Keyboard
d) None of the above
Answer:
c) Keyboard

Question 23.
The shortcut to use calculator is ________.
a) Ctrl + M
b) Ctrl + N
c) Ctrl + O
d) Ctrl + C
Answer:
b) Ctrl + N

Question 24.
Suspense account is grouped under ________.
a) Assets
b) Liabilities
c) Income
d) Expenses
Answer:
b) Liabilities

Question 25.
________ is a step by step series of instructions to perform a specific function and achieve desired output.
a) Procedure
b) Data
c) Connectivity
d) All the above
Answer:
a) Procedure

II. Short Answer Questions

Question 1.
What is Operating system?
Answer:
A set of tools and programs to manage the overall working of a computer using a defined set of hardware components is called an operating system. It is the interface between the user and the computer system.
Example: DOS, Windows, UBUNTU, Imac, etc.

Question 2.
What is Programming software?
Answer:
Special software to accept data and interpret them in the form of machine/assembly language under-standable by a computer.
Example: C, PASCAL, COBOL, etc.

Question 3.
What is Utility software?
Answer:
These are designed specifically for managing the computer device and its resources.
Example: File manager, Anti-virus software, etc.

Question 4.
Give any two examples of Application software?
Answer:

1. General purpose software.
2. Specific purpose software.

Question 5.
Give any two examples of System software?
Answer:

1. Operating system.
2. Programming software.
3. Utility software.

Question 6.
What is data?
Answer:
The facts and figures that are fed into a computer for further processing are called data. Data are raw input until the computer system interprets them using machine language, stores them in memory, classifies them for processing and produces results in conformance with the instructions given to it. Processed and useful data are called information which is used for decision making.

Question 7.
What are sequential codes? ‘
Answer:
In sequential code, numbers and/or letters are assigned in consecutive order. These codes are applied primarily to source documents such as cheques, invoices, etc. A sequential code can facilitate document search.
For example:

Question 8.
What are sequential codes?
Answer:
In a block code, a range of numbers is partitioned into a desired number of sub-ranges and each sub-range is allotted to a specific group. In most of the cases of block codes, numbers within a sub-range follow sequential coding scheme, i.e., the numbers increase consecutively.
For example:

Question 9.
What is General purpose software?
Answer:
This type of application can be used for a variety of tasks and not limited to one particular function.
Example: MS-Office.

Question 10.
What is Specific purpose software?
Answer:
This software is created to execute one specific task and they are customised to the needs of user.
Example:
Accounting software, payroll software, etc.

Question 11.
Explain the basic Features of computerised accounting system.
Answer:
i) Simple and integrated – CAS is designed to automate and integrate ail the business operations such as purchase, sales, finance, inventory and manufacturing. The CAS may be integrated with enhanced Management Information System (MIS), multi-lingual and data organisation capabilities to simplify all the business processes of the organisation easily and cost-effectively.

ii) Speed – It can perform functions at much higher speed than doing the same manually.

iii) Accuracy – Computers perform functions with high degree of accuracy. If hardware, software and input by people are proper, the computerised accounting system can assure of accurate outcome.

iv) Reliability – Computers are used to process large volumes of data and hence, data provided by it are reliable.

v) Versatility – Computer and accounting software have the ability to perform diverse tasks. For example, by simply recording accounting entries through accounting software, one can get trial balance, trading account, profit and loss account, balance sheet and diverse reports.

vi) Transparency – With computerised accounting, the organisation will have greater transparency of day-to-day business operations and access to the vital information.

vii) Scalability – CAS enables processing of any volume of data in tune with the change in the size of the business.

viii) On-line facility – CAS offers online facility to store and process transaction and data so as to retrieve information to generate and view financial reports in any part of the world.

Question 12.
Discuss the Advantages of Computerised Accounting System.
Answer:
i) Faster processing – Computers require far less time than human beings in performing a particular task. Therefore, accounting data are processed faster using a computerised accounting system.

ii) Accurate information – There is less space for error because only one account entry is needed for each transaction unlike repeated posting of the same accounting data in manual system.

iii) Reliability – Computer systems are immune to boredom, tiredness or fatigue. Therefore, these can perform repetitive functions effectively and are highly reliable.

iv) Easy availability of information – The data are easily available and can be communicated to different users at the same time.

v) Up-to-date information – Account balances will always be up to date since the records are automatically updated as and when accounting data are entered or stored.

vi) Efficiency – The computer based accounting system ensures better use of time and resources.

vii) Storage and retrieval – Computer based systems require a fractional amount of physical space as compared to the books of accounts in the form of journals, ledgers and accounting registers.

viii) Works as a motivator to employees – Employees using computer systems feel more valued as they are trained and specialised for the job.

Question 13.
Explain the Components of Computerised Accounting System.
Answer:
i) Hardware – The physical components of a computer constitute its hardware. Hardware consists of input devices and output devices that make a complete computer system. Examples of input devices are keyboard, optical scanner, mouse, joystick, touch screen and stylus which are used to feed data into the computer. Output devices such as monitor and printer are media to get the output from the computer.

ii) Software – A set of programs that form an interface between the hardware and the user of a computer system are referred to as software.

iii) People – The most important element of a computer system is its users. They are also called live-ware of the computer system.

iv) Procedure – Procedure is a step by step series of instructions to perform a specific function and achieve desired output.

v) Data – The facts and figures that are fed into a computer for further processing are called data. Data are raw input until the computer system interprets them using machine language, stores them in memory, classifies them for processing and produces results in conformance with the instructions given to it. Processed and useful data are called information which is used for decision making.

vi) Connectivity – When two or more computers are connected to each other, they can share information and resources such as sharing of files (data/music, etc), sharing of printer, sharing of facilities like the internet. This sharing is possible using wires, cables, satellite, infra-red, Bluetooth, microwave transmission, etc.

Question 14.
Differences between manual and computerised accounting system.
Answer:

Question 15.
Compare ready to use software, customised software and tailor made software.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 12 Final Accounts of Sole Proprietors – I Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 12 Final Accounts of Sole Proprietors – I

11th Accountancy Guide Final Accounts of Sole Proprietors – I Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
Closing Stock is an item of ________.
a) Fixed Asset
b) Current asset
c) Fictitious Asset
d) Intangible asset
Answer:
b) Current asset

Question 2.
Balance sheet is ________.
a) An account
b) A statement
c) Neither a statement nor an account
d) None of the above
Answer:
b) A statement

Question 3.
Net profit of the business increases the ________.
a) Drawings
b) Receivables
c) Debts
d) Capital
Answer:
d) Capital

Question 4.
Carriage inwards will be shown ________ .
a) In the trading account
b) In the profit and loss account
c) On the liabilities side
d) On the assets side
Answer:
a) In the trading account

Question 5.
Bank overdraft should be shown ________.
a) In the trading account
b) Profit and loss account
c) On the liabilities side
d) On the assets side
Answer:
c) On the liabilities side

Question 6.
Balance sheet shows of the business ________.
a) Profitability
b) Financial position
c) Sales
d) Purchases
Answer:
b) Financial position

Question 7.
Drawings appearing in the trial balance is ________.
a) Added to the purchases
b) Subtracted from the purchases
c) Added to the capital
d) Subtracted from the capital
Answer:
d) Subtracted from the capital

Question 8.
Salaries appearing in the trial balance is shown on the ________.
a) Debit side of trading account
b) Debit side of profit and loss account
c) Liabilities side of the balance sheet
d) Assets side of the balance sheet
Answer:
b) Debit side of profit and loss account

Question 9.
Current assets does not include ________.
a) Cash
b) Stock
c) Furniture
d) Prepaid expenses
Answer:
c) Furniture

Question 10.
Goodwill is classified as ________.
a) A current asset
b) A liquid asset
c) A tangible asset
d) An intangible asset
Answer:
d) An intangible asset

II. Very Short Answer Type Questions

Question 1.
Write a note on trading account.
Answer:

1. Trading refers to buying and selling of goods with the intention of making profit.
2. Trading account is a nominal account which shows the result of buying and selling of goods for an accounting period.
3. It is prepared to find out the difference between the revenue from sales and cost of goods sold.

Question 2.
What are wasting assets?
Answer:

1. When the asset is used regularly, it depreciates, eventually having little or no residual value.
2. During the period of depreciation, the asset is called a “wasting asset”.
3. Example, natural resources, such as gas and timber, are wasting assets that eventually are used and then have no remaining value.

Question 3.
What are fixed assets?
Answer:

1. Fixed assets are those assets which are acquired or constructed for continued use in the business and last for many years such as land and building, plant and machinery, motor vehicles, furniture etc.
2. It is classified into a. Tangible Assets and b. Intangible Assets.

Question 4.
What is meant by purchases returns?
Answer:

1. Purchases returns or returns outwards, are a normal part of business.
2. Goods may be returned to supplier if they carry defects or if they are not according to the specifications of the buyer.

Question 5.
Name any two direct expenses and indirect expenses. .
Answer:

1. Direct Expenses: Carriage inwards, Wages, Import Duty, and Royalty
2. Indirect Expenses: Office Expenses, Selling Expenses, Administrative Expenses

Question 6.
Mention any two differences between trial balance and balance sheet.
Answer:

Question 7.
What are the objectives of preparing trading account?
Answer:

1. Trading account provideds information about gross profit and gross loss.
2. It provides an opportunity to safeguard against possible losses.
3. It provides information about direct expenses and direct incomes.

Question 8.
What is the need for preparing profit and loss account?
Answer:

1. Ascertainment of net profit and net loss
2. To compare the profits
3. To have a control on expenses
4. It is used to prepare the balance sheet

III. Short Answer Questions

Question 1.
What are final accounts? What are its constituents?
Answer:

1. The business entities are interested in knowing periodically the results of business operations carried on and the financial soundness of the business.
2. In other words, they want to know the profitability and the financial position of the business.
3. These can be ascertained by preparing the final accounts or financial statements.
4. The final accounts are usually prepared at the end of the accounting period on the basis of balances of ledger accounts shown by the trial balance.

The final accounts or financial statements include the following:

1. Income Statement or Trading and Profit and Loss Account and
2. Position Statement or Balance Sheet.

The purposes of preparing the financial statements are:

1. To ascertain the financial performance of an enterprise and
2. To ascertain the financial position of an enterprise.
3. The income statement and balance sheet are prepared for these purposes respectively.
4. Income statement gives the manner in which the profit or loss for an accounting period is arrived at.
5. Hence, at the close of the accounting period, all nominal accounts (i.e. expenses, losses, revenues, gains, purchases, purchases returns, sales and sales returns) are to be closed by transferring to the income statement or trading and profit and loss account.

Question 2.
What is meant by closing entries? Why are they passed?
Answer:
1. Balances of all the nominal accounts are required to be closed on the last day of the accounting year to facilitate the preparation of trading and profit and loss account.

2. It is done by passing necessary closing entries in the journal proper.

3. Purchases have debit balance and a purchases return has credit balance.

4. At the end of the accounting year, the balance in purchases returns account is closed by transferring to purchases account.

5. Similarly, sales account has credit balance and sales returns have debit balance.

6. At the end of the accounting year, the balance in sales returns account is closed by transferring to sales account.

7. The various closing entries are as follows
(e.g.) for closing purchases returns account

Question 3.
What is meant by gross profit and net profit?
Answer:
Gross profit:

1. The difference between the totals of two sides of the trading account indicates either gross profit or gross loss.
2. If the total of the credit side is more, the difference represents gross profit.
3. On the other hand, if the total of the debit side is higher, the difference represents gross loss.
4. The gross profit or gross loss is transferred to profit and loss account.

Net profit:

1. After debiting indirect expenses and losses and crediting all indirect incomes and gains, if the total of the credit side of the profit and loss account exceeds the debit side, the difference is termed as net.
2. profit.
3. On the other hand, if the total in the debit side exceeds the credit side, the difference is termed as net loss. Net profit or net loss is transferred to the capital account.

Question 4.
“Balance sheet is not an account”- Explain.
Answer:
1. A balance sheet is a part of the final accounts. However, the balance sheet is a statement and not an account. It has no debit or credit sides and as such the words ‘To’ and ‘By’ are not used before the names of the accounts shown therein.

2. A balance sheet is a summary of the personal and real accounts, which have balances. Personal and real accounts having debit balances are shown on the right hand side known as assets side, whereas personal and real accounts having credit balances are shown on the left hand side known as liabilities side.

3. The totals of the two sides of the balance sheet must be equal. If the totals are not equal, it indicates existence of error. It must satisfy the accounting equation, i.e., Assets = Capital + Liabilities, following the dual aspect concept.

4. Balance sheet is prepared on a particular date and not for a fixed period. It discloses the financial position of a business on a particular date. It gives the balances only for the date on which it is prepared.

5. It shows the financial position of the business according to the going concern concept.

Question 5.
What are the advantages of preparing a balance sheet?
Answer:
1. The main purpose of preparing a balance sheet is to ascertain the true financial position of the business at a particular point of time.

2. It helps in comparing the cost of various assets of the business such as the amount of closing stock, amount due from debtors, amount of fictitious assets, etc.

Moreover as assets and liabilities of similar nature are grouped and presented in balance sheet, a comparative study of these assets and liabilities is facilitated. It helps in comparing the various liabilities of the business.

3. It helps in finding out the solvency position of the firm. The firm’s solvency position is favourable if the assets exceed the external liabilities. The firm’s solvency position is not favourable it the external liabilities exceed the assets.

Question 6.
What is meant by grouping and Marshalling of assets and liabilities?
Answer:

1. The assets and liabilities shown in the balance sheet are grouped and presented in a particular order.
2. The term ‘grouping’ means showing the items of similar nature under a common heading.
3. For example, the amount due from various customers will be shown under the head ‘Sundry debtors/ Similarly, under the head ‘Current assets’, the balance of cash, bank, debtors, stock and other current assets will be shown.
4. ‘Marshalling’ is the arrangement of various assets and liabilities in a proper order.
5. Marshalling can be made in one of the following two ways:

a) In the order of liquidity:

• According to this method, an asset which is most easily convertible into cash, i.e., cash in hand is shown first and then will follow those assets which are comparatively less easily convertible, So that the least liquid asset i.e., goodwill is shown last.
• In the same way, the liabilities which are to be paid at the earliest will be shown first. In other words, current liabilities are shown first, then fixed or long-term liabilities and finally the proprietor’s capital.

b) In the order of permanence:

• This method is exactly the reverse of the first method.
• Asset which is more permanent, i.e., goodwill is shown first followed by assets which are less permanent. Similarly, those liabilities which are to be paid last will be shown first.
• In other words, the proprietor’s capital is shown first, then fixed or long-term liabilities and lastly the current liabilities. Joint stock companies are required under the Companies Act to prepare their balance sheet in the order of permanence.

IV. Exercises

Question 1.
Prepare trading account in the books of Sivashankar from the following figures

Solution:
Trading account

Question 2.
Prepare trading account in the books of Mr. Sanjay for the year ended 31^st December 2017

Solution:
Trading account as on 31^st Dec 2017

Question 3.
From the following balances taken from the books of Saravanan, calculate gross profit for the year ended December 31, 2017

Solution:
Trading account as on 31^st Dec 2017

Question 4.
From the following details for the year ended 31^st March, 2018, prepare trading account.

Solution:
Trading account as on 31^st Mar 2018

Question 5.
Ascertain gross profit or gross loss from the following:

Solution:
Trading account

Question 6.
From the following balances taken from the books of Victor, prepare trading account for the t year ended December 31, 2017:

Solution:
Trading account as on 31^st Dec 2017

Question 7.
Compute cost of goods sold from the following information

Solution:
Compute cost of goods sold from the following information.
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 10,000 + 80,000 + 7,000 -15,000 .
= ₹ 82,000

Question 8.
Find out the amount of sales from the following information:

Solution:
Find out the amount of sales from the following information:
Cost of goods sold = Opening stock + Net purchases – Closing stock
= 30,000 + 2,00,000 – 20,000
= ₹ 2,10,000
Let the sales be = 100
Less: Gross profit (30% on sales i.e,100) = 30
Cost of goods sold = ₹ 70
Therefore percentage of gross profit on cost of goods sold is \(\frac { 30 }{ 70 }\) x 100
= 42.86%
Gross profit = 42.86% on ₹ 2,10,000
i.e = \(\frac { 42.86 }{ 100 }\) x 2,10,000
= 90,000
Sales = Cost of goods sold + Gross profit
= 2,10,000 + 90,000
Sales = ₹ 3,00,000

Question 9.
Prepare profit and loss account in the books of Kirubavathi for the year ended 31^st December, 2016 from the following information:

Solution:
Trading account as on 31^st Dec 2016

Question 10.
Ascertain net profit or net loss from the following:

Solution:
Trading account

Question 11.
From the following details, prepare profit and loss account.

Solution:
Trading account

Question 12.
From the following information, prepare profit and loss account for the year ending 31^st December, 2016.

Solution:
Trading account as on 31^st Dec 2016

Question 13.
From the following balances obtained from the books of Mr. Ganesh, prepare trading and profit and loss account.

Solution:
Trading and Profit & loss account

Question 14.
From the following balances extracted from the books of a trader, ascertain gross profit and net profit for the year ended March 31^st, 2017.

Closing stock on December 31.12.2017 was ₹ 4,500
Solution:
Trading and Profit & loss A/c for the year ended 31^st Mar 2017

Question 15.
From the following particulars, prepare balance sheet in the books of Bragathish as on 31^st December, 2017

Solution:
Balance Sheet as on 31^st Dec 2017

Question 16.
Prepare trading and profit and loss account in the books of Ramasundari for the year ended 31^st December, 2017 and balance sheet as on that date from the following information:

Solution:
Trading and Profit & loss A/c for the year ended 31^st Dec 2017

Balance Sheet as on Ramasundari on 31^st  December 2017

Question 17.
From the Trial balance, given by Saif, prepare final accounts for the year ended 31^st March, 2018 in his books.

Closing stock (31-12-2017) ₹ 14,500
Solution:
Trading and Profit & loss A/c for the year ended 31^st Mar 2018

Balance sheet of saif on 31^st Mar 2018

Question 18.
Prepare trading and profit and loss account and balance sheet in the books of Deri, a trader, from the following balances as on March 31, 2018.

Closing stock (31^st March, 2018) ₹ 8,000
Solution:
Trading and Profit & loss A/c for the year ended 31^st Mar 2018

Balance sheet of Deri as on 31^st March 2018.

11th Accountancy Guide Final Accounts of Sole Proprietors – I Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
Carriage outwards will be shown ________.
a) In the trading account
b) In the profit and loss account
c) On the liabilities side
d) On the assets side
Answer:
b) In the profit and loss account

Question 2.
Opening stock is ________.
a) Debited in trading account
b) Credited in trading account
c) Credit ¡n profit and loss account
d) Debited in profit and loss account
Answer:
a) Debited in trading account

Question 3.
___________ account enables the trader to find out gross profit or loss.
a) Trading Account
b) Profit and loss Account
c) Balance sheet
d) Trial balance
Answer:
a) Trading Account

Question 4.
__________ account enables the trader to find out Net profit or loss.
a) Trading Account
b) Profit and loss Account
c) Balance sheet
d) Trial balance
Answer:
b) Profit and loss Account

Question 5.
Fixed assets does not include ________.
a) Plant
b) Stock
c) Furniture
d) Computer
Answer:
c) Furniture

Question 6.
Current Liabilities does not include ________.
a) Sundry Creditors
b) Bills Payable
c) Debentures
d) Outstanding Expenses
Answer:
c) Debentures

Question 7.
All incomes are ________ in the profit and loss account.
a) Debited
b) Credited
c) Assets
d) Liabilities
Answer:
b) Credited

Question 8.
Bad debt is a ________ expense.
a) Office expenses
b) Administrative expenses
c) Selling expenses
d) Distribution expenses
Answer:
c) Selling expenses

Question 9.
Wages is an example of ________.
a) Capital expenses
b) Indirect expenses
c) Direct expenses
d) Revenue expenses
Answer:
c) Direct expenses

Question 10.
Fixed assets have ________.
a) Short life
b) long life
c) no life
d) All of these
Answer:
b) long life

Question 11.
________ refers to buying and selhng of goods with the intention of making profit.
a) Trading
b) Trial balance
c) Profit and loss account
d) Balance sheet
Answer:
a) Trading

Question 12.
The goods remaining unsold at the end of the accounting period are known as _________
a) Opening stock
b) Closing stock
c) Average stock
d) None of these
Answer:
b) Closing stock

Question 13.
________ is the arrangement of various assets and liabilities in a proper order.
a) Marshalling
b) Grouping
c) Recording
d) Packing
Answer:
a) Marshalling

Question 14.
Net profit or Net loss ¡s traflsferred to the ________ account.
a) Trading
b) Profit and loss
c) Capital
d) None of these
Answer:
c) Capital

Question 15.
Gross profit or Grosš loss is transferred to the _______ account.
a) Trading
b) Profit and loss
c) Capital
d) None of these
Answer:
b) Profit and loss

II. Very Short Answer Type Questions

Question 1.
Definition of trading accounting?
Answer:
According to J. R. Batliboi, “The trading account shows the results of buying and selling of goods. In preparing this account, the general establishment charges are ignored and only the transactions in goods are included.”

Question 2.
What is opening stock?
Answer:
The stock of goods remaining unsold at the end of the previous year is the opening stock of the current year. This item will not be there in a newly started business. It will not appear if it is adjusted with pur-chases. As opening stock would have been sold during the year, the cost of opening stock is included in trading account.

Question 3.
What do you mean by direct expenses?
Answer:
All the expenses incurred on the purchase of goods and for bringing the goods to the go down or place of business and to make them to saleable condition are known as direct expenses.

Question 4.
What is Carriage inwards or Freight inwards?
Answer:
Amount paid for transporting the goods purchased to the go down or business premises is called carriage inwards or carriage on purchases or freight inwards.

Question 5.
What is Wages?
Answer:
Amount paid to workers who are directly engaged in loading, unloading and handling of goods purchased is known as wages.

Question 6.
What is Dock Charges?
Answer:
These are the charges levied for shipping the cargo while entering or leaving docks. When they are paid on import of goods, they are treated as direct expenses.

Question 7.
What do you mean by direct expenses?
Answer:
The goods remaining unsold at the end of the accounting period are known as closing stock. They are valued at cost price or net realisable value (market price) whichever is lower.

Question 8.
Definition of Profit and Loss?
Answer:
According to Prof. Carter, “A Profit and Loss Account is an account into which all gains and losses are collected, in order to ascertain the excess of gains over the losses or vice-versa”.

Question 9.
Definition of Balance Sheet?
Answer:
According to J.R. Batliboi, “A Balance Sheet is a statement prepared with a view to measure the exact financial position of a business on a certain fixed date.”

Question 10.
State Methods of drafting a balance sheet.
Answer:
The balance sheet of business concern can be presented in the following two forms.

1. Horizontal form
2. Vertical form

Question 11.
Explain the Tangible fixed assets?
Answer:
Tangible fixed assets are those which have physical existence or which can be seen and felt. Examples: plant and machinery, building and furniture.

III. Short Answer Questions

Question 1.
What do you mean by current assets?
Answer:

1. Current assets are those assets which are either in the form of cash or can be easily converted into cash in the normal course of business or within one year.
2. In the words of Howard and Upton, “The current assets are usually defined as those assets which are convertible into cash through the normal course of business within a short time, ordinarily in a year.”
3. Current assets include cash in hand, cash at bank, short-term investments, bills receivable, debtors, prepaid expenses, accrued income, closing stock, etc.
4. Among these, closing stock is valued at cost or realisable value whichever is lower and debtors are shown after deducting a reasonable provision for bad and doubtful debts.

Question 2.
Explain the Intangible fixed assets?
Answer:

1. Intangible fixed assets are those which do not have any physical existence or which cannot be seen or touched.
2. Examples: goodwill, trade-marks, copy rights and patents. Intangible assets are as much valuable as tangible assets because they also help the firm in earning profits.
3. For example, goodwill helps in attracting customers and patents represent the know-how which helps in producing the goods.

Question 3.
What is the Need for preparation of trading account?
Answer:
i) Provides information about gross profit or gross loss:

• It shows the gross profit or gross loss of the business for an accounting year.
• This helps the business persons to find out gross profit ratio by expressing the gross profit as a percentage of sales.
• It helps to compare and analyse with the ratios of the previous years.
• Thus, it provides data for com-parison, analysis and planning for a future period.

ii) Provides an opportunity to safeguard against possible losses:

• If the ratio of gross profit has decreased in comparison to the preceding years, effective measures can be taken to safeguard against future losses.
• For example, the sale price of goods may be increased or steps may be taken to analyse and control the direct expenses.

iii) Provides information about direct expenses and direct incomes:

• All the expenses incurred on the purchase of goods are direct expenses. They are recorded in the trading account.
• Trading account also shows sales revenue, which is a direct income. With the help of trading account, percentage of such expenses on sales revenue can be calculated and compared with similar ratios of the previous years.
• Thus, it enables the management to have control over the direct expenses.

Question 4.
What is the Need for preparation of profit and loss account?
Answer:
i) Ascertainment of net profit or net loss:

• The profit and loss account discloses the net profit available to the proprietor or net loss to be borne by him.
• Ascertainment of profitability helps in planning for the growth and efficiency of a business enterprise.
• Inter-firm comparison and intra-firm comparison of profit and loss account items help in assessing efficiency in comparison with other enterprises and other departments of the same enterprise respectively.

ii) Comparison of profit – The net profit of the current year can be compared with the profit of the previous years. It helps to know whether the business is conducted efficiently or not.

iii) Control on expenses – Profit and loss account helps in comparing various expenses with the expenses of the previous years. The percentage of individual expenses to net sales can be calculated and compared with the similar ratios of previous years. Such a comparison will be helpful in taking effective steps for controlling unnecessary expenses.

iv) Helpful in the preparation of balance sheet – A balance sheet can be prepared only after ascertaining the net profit or loss through profit and loss account. Net profit or loss is shown in the balance sheet. Thus, it facilitates preparation of balance sheet.

Question 5.
What is the Need for preparation of balance sheet?
Answer:
a) The main purpose of preparing a balance sheet is to ascertain the true financial position of the business at a particular point of time.

b) It helps in comparing the cost of various assets of the business such as the amount of closing stock, amount due from debtors, amount of fictitious assets, etc.

Moreover as assets and liabilities of similar nature are grouped and presented in balance sheet, a comparative study of these assets and liabilities is facilitated. It helps in comparing the various liabilities of the business.

c) It helps in finding out the solvency position of the firm. The firm’s solvency position is favourable if the assets exceed the external liabilities. The firm’s solvency position is not favourable it the external liabilities exceed the assets.

Question 6.
What are the Characteristics of balance sheet?
Answer:
a) A balance sheet is a part of the final accounts. However, the balance sheet is a statement and not an account. It has no debit or credit sides and as such the words ‘To’ and ‘By’ are not used before the names of the accounts shown therein.

b) A balance sheet is a summary of the personal and real accounts, which have balances. Personal and real accounts having debit balances are shown on the right hand side known as assets side, whereas personal and real accounts having credit balances are shown on the left hand side known as liabilities side.

c) The totals of the two sides of the balance sheet must be equal. If the totals are not equal, it indicates existence of error. It must satisfy the accounting equation, ie., Assets = Capital + Liabilities, following the dual aspect concept.

d) Balance sheet is prepared on a particular date and not for a fixed period. It discloses the financial position of a business on a particular date. It gives the balances only for the date on which it is prepared.

e) It shows the financial position of the business according to the going concern concept.

Question 7.
What is the Classification of assets?
Answer:
a) Fixed assets – Fixed assets are those assets which are acquired or constructed for continued use in the business and last for many years such as land and building, plant and machinery, motor vehicles, furniture, etc. According to Finley and Miller, “Fixed assets are assets of a relatively permanent nature used in the operations of business and not intended for sale”.

b) Current assets – Current assets are those assets which are either in the form of cash or can be easily converted into cash in the normal course of business or within one year. In the words of Howard and Upton, “The current assets are usually defined as those assets which are convertible into cash through the normal course of business within a short time, ordinarily in a year.”

Current assets include cash in hand, cash at bank, short-term investments, bills receivable, debtors, prepaid expenses, accrued income, closing stock, etc.

c) Liquid assets – Liquid assets are the assets which are either in the form of cash or which can be immediately converted into cash within a very short period of time, such as cash at bank, bills receivable, short-term investments, debtors and accrued incomes.

In other words, if prepaid expenses and closing stock are excluded from current assets, the balance is known as liquid assets.

d) Investments – Amount invested outside the business in shares, debentures, bonds and other securities is called investments.

If it is invested for a period more than a year they are called long-term investments. If they are invested for a period less than a year they are short term investments and shown under current assets.

e) Wasting assets – These are the assets which get exhausted gradually in the process of excavation. Examples: mines and quarries.

Question 8.
Explain the type of liabilities.
Answer:
a) Fixed or long-term liabilities – The liabilities which are to be repaid after one year or more are termed as long-term liabilities. Example: Long-term loans.

b) Current or short-term liabilities – The liabilities which are expected to be paid within the normal operating cycle or one year are termed as current or short-term liabilities. These include bank overdraft, creditors, bills payable, outstanding expenses, etc.

c) Contingent liabilities – These are the liabilities which are not certain at the time of preparation of balance sheet. These liabilities may or may not occur.

These are the liabilities which will become payable only on the happening of some specific event which itself is not certain, otherwise these need not be paid. Such liabilities are as follows:

IV. Problems and solutions

Question 1.
From the following particulars prepare the trading account and calculate the gross profit.

Solution:
Trading Account

Question 2.
From the following figures, ascertain the gross profit

Solution:
Trading Account

Question 3.
Front the information given below prepare trading account.

Solution:
Trading Account

Question 4.
From the following particulars calculate gross profit.

Solution:
Trading Account

Question 5.
Calulate the Grose profit from the fllowing figures

Solution:
Trading Account

Question 6.
Prepare profit and loss account for the year ending 31.3.2017

Solution:
Trading Account

Question 7.
From the following information, prepare the Profit and Loss Account of a Trader for the year ending 31st March, 2017.

Solution:

Question 8.
Prepare Trading and Profit Si Loss account from the following information:

Solution:
Trading and Profit & loss A/c Cr

Question 9.
From the following information, prepare a Balance Sheet of Mr.A as at 31st March 2016.

Solution:

Question 10.
From the following information prepare balance sheet

Solution:
Balance Sheet

Question 11.
From the following information prepare trading account for the year ended 31.12.2016.

Solution:
Trading A/c for the year ended 31st Dec 2016

Question 12.
From the following balance extracted from the books of M/S Lavanya and sons, prepare trading account for the year ended 31st March 2017.

Solution:
Trading A/c for the year ended 31st Mar 2017

Question 13.
Prepare trading account for the year ended 31st December 2017 from the following.

Closing stock is valued at ₹ 6,00,000
Solution:
Trading A/c for the year ended 31st Dec 2017

Note: Selling expenses, carriage on sales advertisement and office rent will not appear in trading account as they are indirect expenses.

Question 14.
Following in then extract of a trial balance as on 31st December 2017 prepare trading account.

Solution:
Trading A/c for the year ended 31st Dec 2017

Note:
Closing stock will not appear

Question 15.
From the following information prepare trading account for the year ending 31st December, 2017.

Solution:
Trading A/c for the year ended 31st Dec 2017

Question 16.
Compute cost of goods sold from the following.

Solution:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 8,000 + 60,000 + 5,000 – 9,000
= ₹ 64,000

Note : Indirect expenses do not form part of cost of goods sold.

Question 17.
Find the amount of sales from the following.

Solution:
Cost of goods sold = Opening stock + Net purchases + Direct expenses – Closing stock
= 20,000 + 70,000 + 10,000 – 30,000
= ₹ 70,000
Let the sales be less Gross profit (20% on sales i.e,100) (100 – 20 = 80)
cost of goods sold
Therefore percentage of Gross profit on cost of goods sold is \(\frac { 20 }{ 80 }\) x 100 = 25%
Gross profit = 25% on 70,000 (Ex) \(\frac { 25 }{ 100 }\) x 70,000 = 17,500
Sales = Cost of goods sold + Gross profit
= 70,000 + 17,500
= ₹ 87,500

Question 18.
Following the information prepare profit and loss account for 31st March 2018.

Solution:
Profit and loss account for the year ended 31st march 2018

Question 19.
Prepare the profit and loss account for the year ended 31st December 2017.

Solution:
Profit and loss account for the year ended 31st December 2017

Note: Carriage inwards will not appear in profit and loss account as is a direct expense.

V. Long Answers

Question 1.
The following trial balance of Mr.A is extracted on 31.12.2017. Prepare Trading and Profit and Loss account. The closing stock is valued at ₹ 35,000,

Solution:
Trading and Profit & loss A/c for the year ended 31 Dec 2017

Balance Sheet of Mr. A as on 31.12.2017

Question 2.
From the following balances extracted from the accounts of Shri & Co for year ending 31.03.2018, prepare Trading and Profit & loss account and also Balance sheet as on that date.

Solution:
Trading Account of Shri & Co for the year ended 31 Mar 2018

Balance Sheet as on 31 Mar 2018

Question 3.
From the trial balance f Thiru.Vetri for the year ending 31.12.2017 prepare trading & profit & Loss account for that period and also Balance sheet as on that date.

Closing stock Rs. 1,970; outstanding rent ₹ 60
Solution:
Trading and Profi & loss A/c of Mr. Vetri for the year ended 31 Dec 2017

Balance Sheets as on 31 Dec 2017

Question 4.
The following particulars prepare profits and loss account year ended 31st December 2017.

Solution:
Profit and loss account for the year ended 31st Dec 2017

Question 5.
Following balance of Niruban, prepare balance sheet as on 31st December 2017.

Solution:
Balance Sheet as on 31^st Dec 2017

Question 6.
Prepare trading and profit and loss A/c of about Rahuman for the year ending 31st December, 2016 and balance sheet as on that date. The closing stock on 31st December 2016 was valued at ₹ 2,000

Solution:
Balance Sheet as on 31st Dec 2016

Balance sheet as on 31st December, 2016

Question 7.
Trial balance of sharn, prepare trading and profit and loss account for the year ending 31st December 2013 was valued at 25,00,000

Solution:
Profit and loss account for the year ended 31st Dec 2013

Balance sheet as on sharan 31st December 2017

Question 8.
The trial balance of Ms. Kalpana shows the following balance on March 31.2017

Adjustment:
The closing stock was valued at ₹ 60,000

Balance sheet of Ms. Kalpana as on 31st March 2017

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Accountancy Guide Pdf Chapter 11 Capital and Revenue Transactions Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 11 Capital and Revenue Transactions

11th Accountancy Guide Capital and Revenue Transactions Text Book Back Questions and Answers

I. Multiple Choice Questions

Choose the correct answer.

Question 1.
Amount spent on increasing the seating capacity in a cinema hall is _______.
a) Capital expenditure
b) Revenue expenditure
c) Deferred revenue expenditure
d) None of the above.
Answer:
a) Capital expenditure

Question 2.
Expenditure incurred ₹ 20,000 for trial run of a newly installed machinery will be _______.
a) Preliminary expense
b) Revenue expenditure
c) Capital expenditure
d) Deferred revenue expenditure
Answer:
c) Capital expenditure

Question 3.
Interest on bank deposits is _______.
a) Capital receipt
b) Revenue receipt
c) Capital expenditures
d) Revenue expenditures
Answer:
b) Revenue receipt

Question 4.
Amount received from IDBI as a medium term loan for augmenting working capital _______.
a) Capital expenditures
b) Revenue expenditures
c) Revenue receipts
d) Capital receipt
Answer:
d) Capital receipt

Question 5.
Revenue expenditure is intended to benefit _______.
a) Past period
b) Future period
c) Current period
d) Any period
Answer:
c) Current period

Question 6.
Pre – operative expenses are _______.
a) Revenue expenditure
b) Prepaid revenue expenditure
c) Deferred revenue expenditure
d) Capital expenditure
Answer:
d) Capital expenditure

II. Very Short Answer Type Question

Question 1.
What is meant by revenue Expenditure?
Answer:

1. The expenditure incurred for day to day running of the business or for maintaining the earning capacity of the business is known as revenue expenditure.
2. It is recurring in nature. It is incurred to generate revenue for a particular accounting period. The revenue expenditure may be incurred in relation with revenue or in relation with a particular accounting period.
3. For example, cost of purchases is a revenue expenditure related to sales revenue. Rent and salaries are related to a particular accounting period.

Question 2.
What is capital expenditure?
Answer:

1. It is an expenditure incurred during an accounting period, the benefits of which will be available for more than one accounting period.
2. It includes any expenditure resulting in the acquisition of any fixed asset or contributes to the revenue earning capacity of the business. It is non- recurring in nature.

Question 3.
What is capital profit?
Answer:
Capital profit is the profit which arises not from the normal course of the business. Profit on sale of fixed asset is an example for capital profit.

Question 4.
Write a short note on revenue receipt.
Answer:
Receipts which are obtained in the normal course of business are called revenue receipts. It is recurring in nature. The amount received is generally small.

Examples:

• Proceeds from sale of goods
• Interest on investments received
• Respet Received
• Dividend from investment in shares.

Question 5.
What is meant by deferred revenue expenditure?
Answer:

1. An expenditure, which is revenue expenditure in nature, the benefit of which is to be derived over a subsequent period or periods is known as deferred revenue expenditure.
2. The benefit usually accrues for a period of two or more years. It is for the time being, deferred from being charged against income. It is charged against income over a period of certain years.

III. Short Answer Questions

Question 1.
Distinguish between capital expenditure and revenue expenditure.
Answer:

Question 2.
Distinguish between capital receipt and revenue receipt.
Answer:

Question 3.
What is deferred revenue expenditure? Give two examples.
Answer:
1. An expenditure, which is revenue expenditure in nature, the benefit of which is to be derived over a subsequent period or periods is known as deferred revenue expenditure.

2. The benefit usually accrues for a period of two or more years. It is for the time being, deferred from being charged against income. It is charged against income over a period of certain years.

Examples:

• Considerable amount spent on advertising
• Major repairs to plant and machinery

IV. Exercises

Question 1.
State whether the following expenditures are capital, revenue or deferred revenue.

1. Advertising expenditure, the benefits of which will last for three years.
2. Registration fees paid at the time of registration of a building.
3. Expenditure incurred on repairs and whitewashing at the time of purchase of an old building in order to make it usable.

Solution:

1. Deferred revenue expenditure
2. Capital Expenditure
3. Capital Expenditure

Question 2.
Classify the following items into capital and revenue.

1. Registration expenses incurred for the purchase of land.
2. Repairing charges paid for remodeling the old building purchased.
3. Carriage paid on goods purchased.
4. Legal expenses paid for raising of loans

Solution:

1. Capital
2. Capital
3. Revenue
4. Capital

Question 3.
State whether they are capital and revenue.
Answer:

1. Construction of building ₹ 10,00,000.
2. Repairs to furniture ₹ 50,000.
3. White-washing the building ₹ 80,000
4. Pulling down the old building and rebuilding ₹ 4,00,000

Solution:

1. Capital
2. Revenue
3. Revenue
4. Capital

Question 4.
Classify the following items into capital and revenue.

1. ₹ 50,000 spent for railway siding.
2. Loss on sale of old furniture
3. Carriage paid on goods sold.

Solution:

1. Capital
2. Revenue
3. Revenue

Question 5.
State whether the following are capital, revenue and deferred revenue.

1. Legal fees paid to the lawyer for acquiring a land ₹ 20,000.
2. Heavy advertising cost of ₹ 12,00,000 spent on introducing a new product.
3. Renewal of factory licence ₹ 12,000.
4. A sum of ₹ 4,000 was spent on painting the factory.

Solution:

1. Capital
2. Deferred Revenue
3. Revenue
4. Revenue

Question 6.
Classify the following receipts into capital and revenue.

1. Sale proceeds of goods ₹ 75,000.
2. Loan borrowed from bank ₹ 2,50,000
3. Sale of investment ₹ 1,20,000.
4. Commission received ₹ 30,000.
5. ₹ 1,400 wages paid in connection with the erection of new machinery.

Solution:

1. Revenue
2. Capital
3. Capital
4. Revenue
5. Capital

Question 7.
Identify the following items into capital or revenue.

1. Audit fees paid ₹ 10,000.
2. Labour welfare expenses ₹ 5,000.
3. ₹ 2,000 paid for servicing the company vehicle.
4. Repair to furniture purchased second hand ₹ 3,000.
5. Rent paid for the factory ₹ 12,000

Solution:

1. Revenue
2. Revenue
3. Revenue
4. Capital
5. Revenue

11th Accountancy Guide Capital and Revenue Transactions Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
Expenses on research and development will be classified under _______.
a) Preliminary expense
b) Revenue expenditure
c) Capital expenditure
d) Deferred revenue expenditure
Answer:
d) Deferred revenue expenditure

Question 2.
Depreciation on fixed asset is a _______ expenditure.
a) Capital expenditure
b) Revenue expenditure
c) Deferred revenue expenditure
d) None of the above.
Answer:
b) Revenue expenditure

Question 3.
Revenue receipts are _______ in the business.
a) non-recurring
b) recurring
c) neither of the above
d) A AND B
Answer:
b) recurring

Question 4.
An plant worth ₹ 8,000 is sold for ₹ 8,500 the capital receipt amounts to _______.
a) ₹ 8,000
b) ₹ 8,500
c) ₹ 500
d) ₹ 165
Answer:
c) ₹ 500

Question 5.
An asset worth ₹ 1,00,000 is sold for ₹ 85,000 the capital loss amounts to _______.
a) ₹ 85,000
b) ₹ 1,00,000
c) ₹ 15,000
d) ₹ 70000
Answer:
c) ₹ 15,000

Question 6.
An asset worth ₹ 1,00,000 is sold for ₹ 75,000 the capital loss amounts to
a) ₹ 1,75,000
b) ₹ 1,00,000
c) ₹ 75,000
d) ₹ 25,000
Answer:
c) ₹ 75,000

Question 7.
Transaction which provide benefit to*the business for more than one year is called as _______.
a) Capital expenditure
b) Revenue expenditure
c) Deferred revenue expenditure
d) None of the above
Answer:
c) Deferred revenue expenditure

Question 8.
Revenue expenditure is intended to benefit.
a) Subsequent year
b) previous’ year
c) current year
d) None of the above
Answer:
c) current year

II. Very Short Answer Type Questions

Question 1.
What is revenue loss?
Answer:
Revenue losses are the losses that arise from the normal course of the business. In other words, ‘net loss’ – i.e., excess of revenue expenditures over revenue receipts.

Question 2.
Write a short note on Capital receipt.
Answer:
Receipt which is not revenue in nature is called capital receipt. It is non-recurring in nature. The amount received is normally substantial. It is shown on the liabilities side of the balance sheet.

Question 3.
Write the Features of capital expenditure?
Answer:

1. It gives benefit for more than one accounting period.
2. It includes acquisition of fixed assets and all expenditure incurred upto the point an asset is ready for use.
3. It contributes to the revenue earning capacity of the business.
4. It is non-recurring in nature.
5. It is shown on the assets side of the balance sheet.

Question 4.
Write the Features of revenue expenditure?
Answer:

1. It is recurring in nature.
2. It is incurred for maintaining the earning capacity of the business.
3. Its benefit expires in the same accounting period.
4. It is shown on the debit side of the trading and profit and loss account.

Question 5.
Write the Features of deferred revenue expenditure?
Answer:

1. It is a revenue expenditure, the benefit of which is to be derived over a subsequent period or periods.
2. It is not fully written off in the year of actual expenditure. It is written off over a period of certain years.
3. The balance available after writing off (i.e., Actual expenditure – Amount written off) is shown on the assets side balance sheet.

Question 6.
Distinguish Capital, Revenue 8i Deferred revenue expenditure.
Answer:

III. Short Answer Questions

Question 1.
Classify the following expenditures and receipts as capital or revenue

1. ₹ 10,000 spent as travelling expenses of the directors on trips abroad for the purchase of fixed assets.
2. Amount received from trade receivables during the year.
3. Amount spent on demolition of building to construct a large building on the same site.
4. Insurance claim received on account of machinery damaged by fire.

Solution:

1. Capital expenditure
2. Revenue receipt
3. Capital expenditure
4. Capital receipt.

Question 2.
Classify the following expenses as capital or revenue.
(i) The sum of ₹ 3,200 has been spent on a machine as follows:

• ₹ 2,000 for additions to double the output.
• ₹ 1,200 for repairs necessitated by negligence.

(ii) Overhauling expenses of ₹ 25,000 for the engine of a motor car to get better fuel efficiency.
Solution:
(i) a. capital expenditure
b. revenue expenditure

(ii) capital expenditure.

Question 3.
State whether the following are capital or revenue items.

1. ₹ 5,000 spent towards additions to buildings.
2. Second-hand motor car purchased for ₹ 30,000 and paid ₹ 2,000 as repairs immediately.
3. ₹ 10,000 was spent on painting the new factory.
4. Freight and cartage on the new machine ₹ 150, erection charges ₹ 200.
5. ₹ 150 spent on repairs before using a second hand car purchased recently.

Solution:

1. Capital expenditure.
2. Capital expenditure.
3. Capital expenditure.
4. Capital expenditures.
5. Capital expenditure.

Question 4.
State whether the following are capital, revenue or deferred revenue expenditure.

1. Carriage of ₹ 1,000 spent on machinery purchased and installed.
2. Office rent paid ₹ 2,000.
3. Wages of ₹ 5,000 paid to machine operators.
4. Hire charges for the use of motor vehicle, hired for five years, but paid yearly.

Solution:

1. Capital expenditure.
2. Revenue expenditure.
3. Revenue expenditure.
4. Revenue expenditure.

Question 5.
State with reasons whether the following are capital or revenue expenditure

1. Expenses incurred in connection with obtaining a licence for starting the factory for ₹ 25,000.
2. A factory shed was constructed at a cost of ₹ 2,00,000. A sum of ₹ 10,000 had been incurred in the construction of temporary huts for storing building material.
3. Overhaul expenses of second-hand machinery purchased amounted to ₹ 5,000.

Solution:

1. Capital expenditure.
2. Capital expenditure.
3. Capital expenditure.

Question 6.
State with reasons whether the following are capital or revenue or deferred revenue expenditure

1. Advertisement expenses amounted to ₹ 10 crores to introduce a new product.
2. Expenses on freight for purchasing new machinery.
3. Freight and insurance on the new machinery and cartage paid to bring the new machinery to the factory.

Solution:

1. Deferred revenue expenditure.
2. Capital expenditure.
3. Capital expenditure.