Bhagya

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TN State Board 12th English Model Question Paper 4

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1= 20]

Choose the correct synonyms for the underlined words from the options given:

Question 1.
Of the rest, many will end up as mental or physical cripples.
(a) rare gifts (b) able-bodied (c) disabilities (d) perfect health
Answer:
(c) disabilities

Question 2.
Little dangling baskets under the spout to catch the stray leaves.
(a) loose (b) gathered (c) harmful (d) fresh
Answer:
(a) loose

Question 3.
They were doing brisk business in the public square.
(a) dull (b) indolent (c) lethargic (d) active
Answer:
(d) active

Choose the correct antonyms for the underlined words from the options given:

Question 4.
The trolley was commandeered by the two boys.
(a) snatched (b) abandoned (c) usurped (d) hijacked
Answer:
(b) abandoned

Question 5.
The advent of brain-machine interfaces is certain to blur the boundary between humans and machines.
(a) departure (b) danger (c) drawback (d) dispute
Answer:
(d) dispute

Question 6.
Makalu was unexplored.
(a) expelled (b) explained (c) explored (d) expedited
Answer:
(c) explored

Question 7.
Choose the correct combination for the compound word light show.
(a) Noun + Verb (b) Verb + Noun (c) Noun- + Gerund (d) Preposition + Noun
Answer:
(a) Noun + Verb

Question 8.
Choose the correct expansion of NSC.
(a) National Savings Certificate (b) National Service Certificate (c) National Savings Career (d) National Service Certificate
Answer:
(a) National Savings Certificate

Question 9.
Choose the meaning of the foreign word in the sentence: My cousin Charlotte is a real klutz.
(a) stranger (b) thief (c) clumsy fool (d) coward
Answer:
(c) clumsy fool

Question 10.
Choose the right combination for the blended word Bionic.
(a) Bio + Electric (b) Biology + Electronic (c) Bio + Nic (d) Biology + Nic
Answer:
(b) Biology + Electronic

Question 11.
Choose the clipped word for dormitory.
(a) dormi (b) doritory (c) dory (d) dorm
Answer:
(d) dorm

Question 12.
A craze for establishing banks is known as …………………… .
(a) monotonist (b) islomania (c) bancomania (d) barbarian
Answer:
(c) bancomania

Question 13.
Form a derivative by adding the right prefix to the word ‘open’.
(a) re- (b) en- (c) un (d) dis-
Answer:
(a) re-

Question 14.
Fill in the blanks with a suitable relative pronoun.
I have a friend …………………….. cat is very cute.
(a) which (b) whom (c) whose (d) when
Answer:
(c) whose

Question 15.
Fill in the blanks with a suitable preposition.
I heard that news …………………….. the radio.
(a) in (b) on (c) at (d) through
Answer:
(b) on

Question 16.
Choose the correct question tag for the following statement.
He reads a lot of comics, ……………………..?
(a) should he (b) won’t he (c) will he (d) doesn’t he
Answer:
(d) doesn’t he

Question 17.
Choose the suitable meaning or idiom found in the following sentence.
Sumathi was sick of bending over backwards to entertain her nieces.
(a) doing gymnastics (b) dancing (c) trying very hard (d) doing yoga
Answer:
(c) trying very hard

Question 18.
Substitute the underlined word with the appropriate polite alternative.
Till the next trial, the suspects will be kept in the prison camp.
(a) relocation center (b) jail (c) dungeon (d) dormitory
Answer:
(a) relocation center

Question 19.
Choose the correct sentence pattern for the following sentence.
We aren’t complaining.
(a) SVIODO (b) SVC (c) SVO (d) SVCA
Answer:
(b) SVC

Question 20.
Fill in the blank with a suitable phrasal verb.
The following afternoon we …………………….. the tiny village set high upon the hillside.
(a) set aside (b) called on (c) got away (d) drove to
Answer:
(d) drove to

Part II
Section -1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“For what, we thought, had we to fear
With our arms and provender, load on load,”
(a) What was the mood of the soldiers?
(b) What made the soldiers confident?
Answer:
(a) The soldiers were confident of winning the war.
(b) They had enough grains to survive a siege and plenty of arms to fight the war. These things made them confident.

Question 22.
“…………….. Free imaginations
Bringing changes into a world resenting change.”
(a) How does free imagination help the world?
(b) Identify the figure of speech.
Answer:
(a) Free imagination brings changes in the world.
(b) Personification

Question 23.
“Death closes all: but something ere the end,
Some work of noble note, may yet be done,
Not unbecoming men that strove with Gods.”
(a) The above lines convey the undying spirit of Ulysses. Explain.
(b) Pick out the words in alliteration in the above lines.
Answer:
(a) Ulysses is aware of ageing and substantial decrease in his physical strength. He knows that will close in on him sooner or later. But before that happens, he wants to sail beyond the sunset/horizon and if possible meet warriors like Achilles. He wants to achieve something worthy of those who challenged and fought with Gods. Thus these lines show the undying spirit of Ulysses.
(b) ere, end noble, note are the words that alliterate.

Question 24.
“And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow.”
(a) What does the lover do for his mistress?
(b) Explain, ‘sighing like furnace’.
Answer:
(a) The lover is always sighing and longing for his beloved. He writes a sad ballad describing the eyebrow of his mistress.
(b) It means moaning, breathing deeply and sadly like a fire place.

Question 25.
“But not because of its magnificence Dear is the Casuarina to my soul:
Beneath it we have played; though years may roll,”
(a) What is not the cause for Toru Dutt’s love for the Casuarina tree?
(b) What makes the tree dear to the poet?
Answer:
(a) As children, the poet and her friends had played under the tree. This experience has made the tree dear to the poet.
(b) The poet is unable to forget the wonderful time she had under the tree with her friends.

Question 26.
“Just as perhaps he mused, ‘My plans’
That soar, to earth may fall,”
(а) What may hamper the soaring plans of Napoleon?
(b) What is the figure of speech employed in these lines?
Answer:
(a) The negative outcome of battle at Ratisbon may hamper his soaring plans.
(b) ‘plans that soar’ – Metaphor.

Section – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the following dialogue:
Answer:
Lady to the Judge : This man is responsible for the accident.
Man : No, my Lord. This lady does not know how to drive a car.

The lady told the Judge that, that man was responsible for the accident. The man denied the lady’s statement respectfully and said that that lady did not know how to drive a car.

Question 28.
Obey the traffic rules. Otherwise, you will be prosecuted, (combine using if)
Answer:
If you do not obey the traffic rules, you will be prosecuted.

Question 29.
Rewrite the sentence making an inversion in the conditional clause.
If you don’t wish to sign the contract, you must let them know by the end of this month.
Answer:
Should you not wish to sign the contract, you must let them know by the end of this month.

Question 30.
When he walked through the wood, he saw a fox that was following him.(Change the following into a simple sentence)
Answer:
Walking through the wood, he saw a fox following him.

Part-III
Section -1

Explain any two of the following with reference to the context. [2 x 3 = 6]

Question 31.
Unto thy honor, Tree, beloved of those Who now in blessed sleep for aye repose,
Answer:
Reference: These lines are from the poem, ‘Our Casuarina Tree’ written by Torn Dutt.

Context: The poet says this while consecrating the memories of the tree to the dear departed.

Explanation: The poet wants to freeze the love for the Casuarina tree in her poem. She remembers with pain three younger Dutts who had succumbed to Tuberculosis. She remembers with poignance the numerous days they had spent under the Casuarina tree.

Question 32.
We are not now that strength which in old days Moved earth and heaven;
Answer:
Reference: These lines are from the poem, ‘Ulysses’ written by Alfred Tennyson.

Context: The poet says these words through Ulysses when he wants to justify the reasons for resuming the daring voyage.

Explanation: Ulysses admits the decline in the compatriots’ physical strength with which they were able to move heaven and earth in their youth. He asks his compatriots to ignore the infinity of age and draw on their inner spiritual strength to resume their voyage beyond sunset.

Question 33.
Then off there flung in smiling joy,
And held himself erect
Answer:
Reference: These words are from the poem, ‘Incident of the French Camp” written by Robert Browning.

Context: The narrator says these words while describing the arrival of a boy soldier at the mound where Napoleon was anxiously awaiting news about the battle.

Explanation: Even though the boy-soldier was split into two, he sped fast in a horse amidst the smoke of cannon fire. On seeing Napoleon he jumped off the horse with a beaming face. Being at the verge of embracing death, he declared the good news that they had won the battle.

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
What were the boys doing at night in a deserted square?
Answer:
The boys were resting on the stone pavement in the windy and deserted square beneath the street light. They were waiting for the last bus from Padua to sell the unsold newspapers.

Question 35.
What are the subsidiary uses of tea leaves?
Answer:
Telling fortunes, predicting the arrival of visitors, feeding rabbits, healing bums and sweeping the carpet are some of the subsidiary’s uses of tea leaves.

Question 36.
When did Hillary feel a sense of freedom and well being?
Answer:
Firsdtly, their partly-filled bottle of oxygen got exhausted. They had only one oxygen bottle to carry. With reduced load of 20 litre bottle, Hillary cut steps down off the South Summit. So, he felt a sense of freedom and well-being.

Section – 3

Answer any three of the following questions in about 30 words. [3 x 3 = 9]

Question 37.
Study the pie chart given and answer the questions that follow.

Popular Brands of Bluetooth Speakers
(a) Which is the most favourite brand of bluetooth speaker?
(b) Name the brand preferred by the least number of customers?
(c) Which two brands have equal number of customers?
Answer:
(a) UE Boom 3 is the most favourite brand preferred by the customers.
(b) Marshall Kilburn is the least preferred brand.
(c) UE Wonderbbom and Bose SoundLink Mini II.

Question 38.
Write a dialogue of minimum 3 exchanges between two friends discussing about a Book Fair.
Answer:
Rajini : Did you visit the Book Fair that started last week?
Ravi : Yes, I did. I was in fact waiting for the Book Fair to start.
Rajini : I know that you are a Voracious reader. How many did you buy?
Ravi : I bought a couple of mystery and detective novels by Tamizhvanan, Agatha Christie and Sujatha. What are you planning to buy?
Rajini : I need to buy a good set of Encyclopedias
Ravi : Do you want me to accompany you? I too want to buy some books published by Scholastic India.

Question 39.
Describe the process of making ice cream.
Answer:
Boil one litre of milk till it becomes thick and creamy and is reduced to one fourth its quantity. Add sugar to taste.
Keep aside five tablespoons of cold milk and mix two heaped spoons of ice cream powder. Pour the ice cream powder mixed milk to the boiling milk and stir continuously till it thickens. Add two drops of vanilla, strawberry or chocolate essence to taste and stir well.
After it cools, freeze for four hours.

Question 40.
Complete the proverbs using the word given below.
(a) He who hesitates is (gone, lost, dead)
(b) If God had meant us to fly, he would have given us (wings, cars, tails)
Answer:
(c) Little pitchers have big (ears, mouths, bottom)
(a) lost (b) wings (c) bottom

Part – IV

Answer the following questions: [7 x 5 = 35]
Answer in a paragraph in about 150 words.

Question 41.
What are the aspects that contribute to humour in the essay?
Answer:
The author claims the best manner of making tea is the subject matter of violent disputes. He compliments china tea for being economical but also says one does not feel rejuvenated after drinking it. One doesn’t feel braver, wiser, more optimistic or stimulated after drinking the Chinese tea. He says army tea tastes of grease and white wash. In Britain there are two schools of thought (i.e.) milk first school and tea first school (i.e.) one school claiming that milk must be poured first and tea added later and another school claiming that tea must be poured first and milk to be added next.

He says those who take tea with sugar for its sweetness alone as misguided people. He says that they could very well add salt and pepper to tea and drink it. He claims that there are some mysterious social etiquette surrounding teapot. It is vulgar to drink tea out of one’s saucer. There are some subsidiary uses of tea leaves such as telling fortunes, predicting the arrival of visitors, feeding rabbits, healing burns and sweeping the carpet.

[OR]

“My right to swing my fist ends, where your nose begins.” Elucidate with reference to, ‘On the Rule of the Road’.
Answer:
Rights are not completely individual affairs. In order to enjoy one’s rights one has to respect the rights of others too. The lady in Petrograd had the right to walk on the pavement. The right to move stops when the other person’s right to drive starts. A person may have a walking stick and roll it too. But his right just stops where the other person’s nose begins. No one has the right to violate the rights of others. The right, one exercises, must not affect or erode the rights of others.

One should not think of one’s own rights but also the rights of others. A.G.Gardiner beautifully illustrates this idea by emphasising the metaphor of traffic rules. Rules of road are in fact rules of politeness and unselfishness. One may have absolute freedom in the choice of food, religion, fashionable dress, up keep of hair, funny hairstyle, etc. But one must be conscious of the rights of others. So, the statement “my right to swing my fist ends, where your nose begins” fits well with the central theme of the essay ‘ On the Rule of the Road’.

Question 42.
Human greed led to the mighty fall of the citadel. Explain.
Answer:
The loyal soldiers and their brave captain expected enemies from outside the castle. Their arms and army was ready to fight them. But they could not identify the enemy within. The soldiers were proud that no might would tear their castle down. But they were unaware of the invisible soul-dead enemy within. The ingredients of personal downfall went unnoticed by them.

If a person never looks within, the faults that can be their doom go overlooked. Their reality could crumble while they gaze outward and pride themselves on their sureness. This is what happened precisely with the soldiers of the castle. They only focused on the strength of their physical surroundings and what was beyond the castle. Human greed-propelled betrayal from within caused the castle’s downfall.

[OR]

Explain how the poet guides his son who is at the threshold of manhood, to face the challenges of life.
Answer:
The poet shares his wisdom with his son who is at the threshold of manhood. He persuades his son to be hard like steel or rock to withstand challenges and unforeseen betrayals in life. A person with soft heart will crumble before a breach of trust. Similarly he wants his son to be discerning enough to be soft when needed to grow like a frail flower plant splitting a rock.

Occasionally one has to go with the current because life is at times fertile with a lot of opportunities to grow even among the harshest circumstances. ‘Rich soft wanting’ can help a person to win against all odds. He reiterates this idea by explaining how gentleness can reform a hardened criminal when lashes would, in contrast, harden them further.

Question 43.
Write a paragraph of about 150 words by developing the following hints:
Pi convinced – water on board – divining rod – Pi’s mind – genuine regulation lifeboat – outfitted with supplies – captain – ensure safety and survival – ship chandler – extra money-saving lives – water on board.
Answer:
Pi had never before experienced physical hell than that putrid taste and pasty feeling in the mouth. It was an unbearable pressure at the back of his throat. The divining rod in Pi’s mind dipped sharply and spring gushed forth. He remembered that he was on a genuine regulation lifeboat and such a lifeboat was surely outfitted with supplies. A captain would never fail in so elementary way of preserving water to ensure safety and survival of his crew in the event of a disaster. Besides, it is natural that a ship chandler would think of making a little extra money under the noble pretext of saving lives. It was settled beyond doubt that there was water on board.

[OR]

Bond of friendship – Baldwin and Gresham – boyhood chums – school together – lifetime bondage – thirty five years – baptism – 60 dollars a week – Governor of the private bank – Gresham siphoning money – loyalty and honesty – part ways – upright character – close the bank – return the deposit – reorganization of the bank – arrest – court could not nail Gresham – one hundred thousand dollars – three words – slapped him – body language – not compromise with conscience.
Answer:
Baldwin and Gresham were boyhood chums. They went to school together. Their friendship flowered into a lifetime bondage as it was sustained for thirty five years. When Baldwin’s son was being baptised in the church, Gresham was present. Baldwin expressed his wish that his son John Gresham Baldwin grew up to Gresham’s standard in life. Baldwin was paid only 60 dollars a week for working as a Governor of the private bank founded by Gresham.

Accidentally, Baldwin found out how Gresham was siphoning money very cleverly from the deposits technically without really causing any loss to the depositors. But Baldwin was loyal to Gresham only as long as he was honest. Once he parted ways with honesty, Baldwin would also part ways with Gresham. By the force of his upright character, Baldwin advised him to close the bank and return the deposit to the customers.

Third National Bank was ready to help the reorganization of the bank. Meanwhile, Gresham was arrested. The court could not nail Gresham as there was not a shred of evidence to prove his guilt. If Baldwin testified, the court would indict him. Just before being arrested Gresham offered him one hundred thousand dollars to just say “I don’t remember”. The three words to let him off the hook. He claimed it was the difference between the salary he had been paid and what he ought to have received. Had it been someone else, Baldwin would have slapped him.

But it was Gresham who understood his body language and did not press the matter further. He was proud of the fact that Gresham understood how he could not compromise with his conscience. His family members Martha, Evie and John felt that it was a compliment that cost him one hundred thousand dollars. Baldwin said that Gresham’s compliment about his uprightness was worth a hundred thousand dollars.

Question 44.
Write a summary or Make notes of the following passage.
Answer:
Occasional self-medication has always been part of normal living. The making and selling of drugs has a long history and is closely linked, like medical practice itself, with belief in magic. Only during the last hundred years or so, as the development of scientific techniques made it possible, diagnosis has become possible. The doctor is now able to follow up the correct diagnosis of many illnesses – with specific treatment of their causes. In many other illnesses of which the causes remain unknown, he is still limited, like the unqualified prescriber, to the treatment of symptoms.

The doctor is trained to decide when to treat symptoms only and when to attack the cause. This is the essential difference between medical prescribing and self-medication. The advance of technology has brought about much progress in some fields of medicine, including the development of scientific drug therapy. Parallel with such beneficial trends are two which have an adverse effect. One is the use of high pressure advertising by the pharmaceutical industry which has tended to influence both patients and doctors and has led to the overuse of drugs generally.

The other is emergence of eating, insufficient sleep, excessive smoking and drinking. People with disorders arising from faulty habits such as these, as well as from unhappy human relationships, often resort to self-medication. Advertisers go to great lengths to catch this market. Clever advertising, aimed at chronic sufferers who will try anything because doctors have not been able to cure them, can induce such faith in a preparation, particularly if steeply priced, that it will produce -by suggestion- a very real effect in some people.

It is doubtful whether taking these things ever improves a person’s health, it may even make it worse. Worse, because the preparation may contain unsuitable ingredients; worse because the taker may become dependent on them; worse because they might be taken excess; worse because they may cause poisoning, and worst of all because symptoms of some serious underlying cause may be asked and therefore medical help may not be sought. Self-diagnosis is a greater danger than self-medication.

Summary

No. of words given in the original passage: 347
No. of words to be written in the summary: 347/3 = 115 ± 5
Rough Draft
Self-medication is part of normal living. Medicinal experts are required for diagnosis and treatment of diseases according to symptoms and causes. The development of drug therapy and improvement in public health organizations and nutritional standards have helped progress in medicinal science. Excessive advertising by pharmaceutical companies and emergence of the sedentary society are two counter trends. Self-medication is dangerous as the preparation may be toxic or contain unsuitable ingredients; the user becomes dependent and consumes medicine in excess. Self-diagnosis is worse than self-medication.

No. of words in the summary: 83
Notes:
Title: Self-Medication

1. Self-medication
(a) Part of normal living—last 100 yrs
(b) Advance in diagnostic technology
(c) Doctors required – diagnosis & treatment of disease
(d) Self-medication differs from medical prescription

2. Technological Advmnt in medicine
(a) Drug therapy
(b) Improvement in public health organizations
(c) Increase in nutritional standards

3. Clever advertising by pharmaceutical companies
(a) Take advantage of people’s need
(b) Chronic sufferers
(c) Faulty lifestyle
(i) Lack of exercise, overeating, insufficient sleep, etc.
(ii) Stress, unhappy rela’ps, etc.

4. Dangers of self-medication
(a) Preparation of some drugs contains unsuitable ingredients
(b) Taker becomes dependent
(c) Taker consumes medicine in excess
(d) Preparations may cause poisoning .
(e) Real cause of illness gets suppressed or untreated

Abbreviations used: yrs – years; Advmnt – Advertisement; rela’ps – relapses

Question 45.
Write a letter to the Port Trust, Chennai, asking permission for a group of 25 students of your school to visit a ship that has collection of books of all languages of the world. The ship would remain in Chennai only for a fortnight. Get permission mentioning the date, time and the number of visitors the teachers in – charge of the bona fide students of your school.
Answer:
14th January, 2020
From
The Principal
TM Hr. Sec. School
Chennai – 45

To
The Naval Officer
Port Trust of India
Chennai
Respected Officer,

Sub: Reg. Permission for visiting the Ship – Queen Mary

Given to understand that the Ship named Queen Mary with a collection of books of all languages of the world has arrived at the Chennai Port for public viewing, I hereby request permission for our school children to visit the ship on one of the working days.
We would be grateful if you can permit us to view the books in the ship on Friday, 6th February, 2020 at 10 a.m. There will be 15 students from Std IX accompanied by two teachers with one male office staff.

Please find enclosed the bona fide certificates and Photocopy of the staff identity card accompanying them.
Staff in-charge: Mrs. Shalini Varma and Mrs. Sharmila Das.

Students of Std IX:

• Anitha.M
• Asha.L
• Balaji.G
• Bharath.U
• Chandhini.G
• Christopher. Y
• rthy.T
• Firoz.P
• Janaki.R
• Kalai.A
• Lalitha.G
• Leena.B
• Mohan.C
• Niranjan.F
• Rohit.R

Looking forward to an enriching experience and safe visit. Kindly send us a reply acknowledging our request.

Yours sincerely,
Meena
(Principal)

Address on the envelope:
The Naval Officer
Port Trust of India
Chennai

[OR]

Write a paragraph of 150 words on “Punctuality”.
Answer:
It has been said, “Never put off for tomorrow what you can do today”. Yet there are many people who enjoy postponing things. Such people do not realize the dangers of delaying. Work does not disappear if we postpone it. The difficulty of the work does not get reduced by postponing it. The more we postpone, the more the work piles up. Finally, the load of work seems too much. We then have to work for many days at a stretch under great strain and tension. Finally we do it hurriedly and in a careless manner.

It is as bad as not doing the work at all. Again, when some work is delayed, the time which would have been profitably used is wasted. Time wasted is time lost forever. Hence procrastination, that is the habit of postponing things, is rightly called the thief of time. Precious time wasted means opportunities lost, and lost forever. By the time we realize how much time we have wasted and how many opportunities we have missed by delaying word, it may be too late.

And then nothing remains for us except to regret and repent. Sometimes a very heavy price has to be paid for delaying things. The person who delays insuring his factory will regret it when the factory is gutted by a sudden fire. Delay in the treatment of a disease may make it worse, and may even result in death. There are other proverbs conveying similar meaning. Thus we say: “Time and tide wait for no man” and “A stitch in time saves nine”. All these proverbs warn us against the dangers of delay in actions, and stress the importance of timely action and punctuality.

Question 46.
Spot the errors and rewrite the sentences correctly
(a) Many a man have been arrested at the meeting.
(b) Most of the work have been completed.
(c) Neither of the books are illustrated.
(d) One of the thieves, were arrested.
(e) Rani sat besides the new student.
Answer:
(a) Many a man had been arrested at the meeting.
(b) Most of the work has been completed.
(c) Neither of the books is illustrated.
(d) One of the thieves, was arrested.
(e) Rani sat beside the new student.

[OR]

Fill in the blanks correctly.
(а) Someone who decides to …………….. a car has committed a crime, but auto parts are made of aluminium and …………….. (steel/steal)
(b) They look just the same! I am sure that boy …………….. be his son. (Use a modal in the given blank.)
(c) All my clothes are dirty! I …………….. do some laundry, (use a semi-modal)
(d) He …………….. never …………….. (study) art before he came to college. (use a proper tense)
Answer:
(a) steal/steel (b) must (c) need to (d) had/studied

Question 47.
Identify each of the following sentences with the fields given below:
(a) Shakespeare’s plays are read by many people.
(b) The yield of wheat has increased.
(c) Fast food is a growing health hazard.
(d) My brother is planning to go to the US.
(e) Dhoni was declared the Man of the Series.
[Sports, Nutrition and Dietetics, Travel, Literature, Agriculture]
Answer:
(a) Literature
(b) Agriculture
(c) Nutrition and Dietetics
(d) Travel
(e) Sports

[OR]

Read the following passage and answer the questions in your own words.
Answer:
Housed in an 18th century style Heritage building with wooden paneling running through the thirty seat dining area, “Lean and Lovely” is the latest attraction in Siddhapuram Nagar. Says the owner Chef Virina “People come to us for the unusual fare that we serve. Fairly ordinary Indian recipes are ignited with a dash of sauce and spiced with colour. There is harmony and balance between taste, lightness and tradition. Wholly organically grown vegetables and flour are used. We use the freshest of ingredients that are cut and cooked so as to display their colour and individual texture.

Questions:
a. What is special about the vegetables and flour used by “Lean and Lovely”?
b. In which town is “Lean and Lovely” located?
c. Who is a ‘Chef’?
d. How do the people in “Lean and Lovely” make the ordinary Indian recipes more attractive?
e. How do they keep the colour and texture of the food items?
Answers:
(a) Wholly organically grown vegetables and flour is the speciality of “Lean and Lovely’.
(b) “Lean and Lovely” is located in Siddhapuram Nagar.
(c) A chef is a skilled cook, who prepares a variety of dishes in a restaurant and manages them.
(d) Fairly ordinary Indian recipes are ignited with a dash of sauce and spiced with colour in “Lean and Lovely”. There is also harmony and balance between taste, lightness and tradition attracting the people.
(e) Freshest of ingredients to retain the colour and texture of the food items.

Students can Download Tamil Nadu 12th English Model Question Paper 5 Pdf, Tamil Nadu 12th 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 12th English Model Question Paper 5

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1= 20]

Choose the correct synonyms for the underlined words from the options given:

Question 1.
The policeman is not a symbol of tyranny.
(a) autocracy (b) plutocracy (c) aristocracy (d) democracy
Answer:
(d) democracy

Question 2.
I dragged our oxygen sets into the tent.
(a) pushed (b) pulled (c) lauded (d) loaded
Answer:
(b) pulled

Question 3.
His shoulder and arm were amputated.
(a) frustrated (b) attached (c) transplanted (d) removed
Answer:
(d) removed

Choose the correct antonyms for the underlined words from the options given:

Question 4.
She was eager to do so.
(a) anxious (b) apathetic (c) impatient (d) repulsive
Answer:
(b) apathetic

Question 5.
Not the flat, shallow type…
(a) wide (b) wide (c) direct (d) deep
Answer:
(d) deep

Question 6.
Why people should suffer.
(a) undergo pain (b) enjoy (c) heal (d) struggle
Answer:
(b) enjoy

Question 7.
Choose the correct combination for the compound word ‘milk white’.
(a) Noun+ Gerund (b) Noun + Adjective (c) Gerund + Noun (d) Preposition + Noun
Answer:
(b) Noun + Adjective

Question 8.
Choose the correct expansion of TPS.
(a) Thermal Power Supply (b) Thermal Private Sector (c) Thermal Power Station (d). Thermal Power Sector
Answer:
(c) Thermal Power Station

Question 9.
Choose the meaning of the foreign word in the sentence:
I planned to go downtown to meet with Betty, but I ran into a glitch.
(a) big problem (b) small problem (c) gutter (d) accident
Answer:
(b) small problem

Question 10.
Choose the correct combination for the blended word Globish.
(a) Globe+ Fish (b) Globe + frisk (c) Globe + ish (d) Global + English
Answer:
(d) Global + English

Question 11.
Choose the clipped form for graduate.
(a) gradud (b) gad (c) grad (d) duate
Answer:
(c) grad

Question 12.
A desire to read and enjoy eulogy is known as ……………………. .
(a) eulogomania (b) misogamist (c) oleograph (d) monogram
Answer:
(a) eulogomania

Question 13.
Form a derivative by adding the right prefix to the word ‘closed’.
(a) in- (b) re- (c) un- (d) en-
Answer:
(d) en-

Question 14.
Fill in the blanks with a suitable relative pronoun. This is the house ………………… my father built.
(a) where (b) by (c) that (d) who
Answer:
(c) that

Question 15.
Fill in the blanks with a suitable preposition. You should explain this ……………… them.
(a) for (b) with (c) from (d) to
Answer:
(d) to

Question 16.
Choose the correct question tag for the following statement.
Heera can read many languages ……………………….?
(a) won’t she (b) can’t she (c) isn’t she (d) wasn’t she
Answer:
(b) can’t she

Question 17.
Choose the suitable meaning or idiom found in the following sentence.
When the management wanted to cut down on the man power, they offered a golden hand shake to many of their newly recruited employees.
(a) extension of job
(b) renewal of contract
(c) large amount of money given to a person when they leave
(d) give part-time work
Answer:
(c) large amount of money given to a person when they leave

Question 18.
Substitute the underlined word with the appropriate polite alternative.
I think the garbage man didn’t come on his usual rounds.
(a) cleaner (b) sanitizer (c) garbage disposer (d) sanitation engineer
Answer:
(d) sanitation engineer

Question 19.
Choose the correct sentence pattern for the following sentence.
There was a grand finale of scattered plates and silverware.
(a) SVIODO (b) SVAA (c) SVCA (d) AVSC
Answer:
(c) SVCA

Question 20.
Fill in the blank with a suitable phrasal verb.
I ………………………… to the fountain to have my shoes shined.
(a) went across (b) went over (c) went into (d) put aside
Answer:
(b) went over

Part II
Section – 1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“And the enemy half a mile away They seemed no threat to us at all”
(a) What were the soldiers expecting?
(b) Where were the enemies?
Answer:
(a) The soldiers were expecting a war.
(b) The enemies were half a mile away.

Question 22.
“And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances
(a) Whom does justice refer to?
(b) Describe his appearance.
Answer:
(a) Justice refers to a man in his fifth stage when he becomes critical of everyone else’s opinion in life.
(b) He has a pot belly and is fond of eating delicacies.

Question 23.
“A creeper climbs, in whose embraces bound No other tree could live.”
(a) How does the tree survive the tight hold of the creeper?
(b) Why does Toru Dutt use the expression ‘a creeper climbs’?
Answer:
(a) The tree takes the tight hold of the creeper like the embraces of a lady love. So, it doesn’t hurt the tree. It grows stronger bearing the bite marks of love.
(b) A creeper cannot grow without the support of another tree or a pole. While climbing, it tries to sap the energy from the living tree. It twines its body around the tree and keeps climbing.

Question 24.
“That ever with a frolic welcome took
The thunder and the sunshine, and opposed”
(a) What do ‘thunder’ and ‘sunshine’ refer to?
(b) What do we infer about the attitude of the sailors?
Answer:
(a) Thunder and sunshine refers to misfortunes and happy days. Ulysses and his comrades had undergone both kinds of experiences.
(b) The sailors shared the undying quest for exploration, adventure and for seeking newer knowledge in the untravelled world. They even welcomed dangers in fighting with Gods. They enjoyed the thrill of action and never worried about the outcome of battles or quests. They have equal temper of heroic hearts.

Question 25.
“Just as perhaps he mused, ‘My plans That soar, to earth may fall,”
(а) Who does ‘he’ refer to?
(b) What may hamper the soaring plans of Napoleon?
Answer:
(a) ‘He’ refers to Napoleon Bonaparte.
(b) The negative outcome of battle at Ratisbon may hamper his soaring plans.

Question 26.
“Tell him to be a fool ever so often and to have no shame over having been a fool yet learning something out of every folly hoping to repeat none of the cheap follies”
(a) Is it a shame to be a fool at times?
(b) What does one learn from every folly?
Answer:
(a) No, everyone does commit funny mistakes in life. One might just laugh at them.
(b) Every folly teaches a person his limitations and vulnerabilities. By making conscious efforts to avoid them in future, one will become stronger and wiser.

Section – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the following dialogue:
Answer:
James : Can you come to my apartment tomorrow?
Simon : Thank you, I will come tomorrow after lunch.
James asked Simon if he could come to his apartment the next day. Simon thanked him and said that he would come to his apartment the following day after lunch.

Question 28.
Sundar wants to pass the test. He has to work hard, (combine using ‘if’)
Answer:
If Sundar wants to pass the test, he has to work hard.

Question 29.
Rewrite the sentence making an inversion in the conditional clause.
If we were to have children, we would certainly need to move to a bigger house.
Answer:
Were we to have children, we would certainly need to move to a bigger house.

Question 30.
The house was destroyed in the fire, but the whole family was saved. (Change the following into a complex sentence)
Though the house was destroyed in the fire, the whole family was saved.

Part-III
Section -1

Explain any two of the following with reference to the context. [2 x 3 = 6]

Question 31.
I will maintain until my death
Answer:
Reference: This line is from Edwin Muir’s poem “The Castle”.

Context: The poet says this while recounting the shameful act of betrayal by the aged warder of the wicket gate.

Explanation: The narrator was overconfident of the invincibility of their castle, their stock piled arms and well-stocked granary. They had a brave captain and loyal soldiers. In an open war, they need not fear defeat as their friendly neighbours were also willing to join them during a war against any invader. But their enemy was within their fortified castle. They were sold for a bag of gold. Without a groan the citadel was captured. It was a shameful way to lose one’s side. So, the narrator prefers not to recount this shameful secret to anyone so long as he is alive.

Question 32.
Last scene of all,
That ends this strange eventful history,
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.

Context: The poet says these words while describing the preparedness of the old man in the last stage of life to exit from this lonely planet.

Explanation: The poet beautifully says the “eventful history” (i.e.) life which was spiced up with many interesting things is now coming to a dramatic close. The eternal jewel of life, ‘the soul’, is going to depart the body which had kept it imprisoned for long. The soul celebrates the joy of freedom in death.

Question 33.
He will be lonely enough to have time for the work
Answer:
Reference: These lines are from the poem ‘A Father to his Son’ written by Carl August Sandburg.

Context: The poet says these words to explain how creative thinkers and those who strive to bring about changes are left alone to fend for themselves.

Explanation: The poet asks his son to take advantage of the loneliness. He must enjoy the advantages of solitude. Solitude would help him to be creative. He would invariably learn that final decisions are always taken in silent rooms. In solitude he can pursue his creative imagination and succeed like Shakespeare, Wright brothers, Pasteur, Pavlov and Faraday.

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
How was Dr. Barnard’s attitude to suffering different from that of his father’s?
Answer:
Dr. Barnard’s father accepted suffering as God’s will. He also believed that suffering ennobles humans. But Dr. Barnard found no meaning in the agony and suffering of patients and especially of the young children.

Question 35.
What was thrilling to watch 8000 feet below them?
Answer:
In a number of places, the overhanging ice cornices were very large. In order to escape them, Hillary cut a line of steps down to where the snow met the rocks on the west. It was a great thrill to look straight down that enormous rock face to see 8000 ft below them the tiny tents of Camp 4 in the Western Cwm.

Question 36.
‘Curtailment of private liberty is done to establish social order’ – Do you agree?
Answer:
Yes, curtailment of private liberty is done to establish social order. Liberty is not a personal affair only, it is a social contract. There are a lot of people in the world who need to accommodate their liberty to the liberties of others. A reasonable consideration of the rights or feelings of others in the foundation of social conduct. When one has submitted to the curtailment of private liberty, one may enjoy a social order which makes his liberty, a reality.

Section – 3

Answer any three of the following questions in about 30 words. [3 x 3 = 9]

Question 37.
Study the pie-chart given below and answer the questions that follow:

(a) How much percentage in this group have AB blood group?
(b) What is the percentage of people in this group who do not have the donor group?
(c) According to the given percentage, how many people have either A or B blood groups?
Answer:
(a) 19% of the people have AB Blood group.
(b) 60% of the people do not have the donor group that is Type O.
(c) Approximately Question Question 41.1 % of people do not have either A or B blood groups.

Question 38.
Write a dialogue of minimum 3 exchanges between a Mr. Karnan and a Policeman.
Answer:
Mr. Karnan : Thank you sir for responding to us at this time. There has been a theft in my house.
Policeman : It’s our duty, sir. I hope you haven’t touched anything inside the house.
Mr. Karnan : No sir, I made a call from my mobile as soon as we returned from the night show and found the front door open.
Policeman : Did you enter the house?
Mr. Karnan : Yes sir, just now. The burglars have stolen many of my household articles including money and jewellery.
Policeman: Don’t worry. We will find the culprits at the earliest.

Question 39.
Describe the process of making an omelette.
Answer:
Take an egg and beat the egg well with a fork or egg beater.
Finely chop a small onion, one green chilly, one pod of garlic, coriander leaves Mix the chopped ingredients to the beaten egg.
Add salt to taste and beat.
Heat a tava and add a spoon of oil.
Pour the mixture into the hot tava like a dosa.
Flip sides till it gets cooked on both sides.

Question 40.
Complete the proverbs using the word given below.
(а) Familiarity breeds (friendship, contempt, malice)
(b) Fools rush in where angels fear to (tread, exit, enter)
(c) Give the devil his (share, profit, due)
Answer:
(a) contempt (b) tread (c) due

Part – IV

Answer the following questions: [7 x 5 = 35]
Answer in a paragraph in about 150 words.

Question 41.
How did a casual incident in a hospital help Dr. Barnard perceive a new dimension of life?
Answer:
Initially Dr. Barnard was grumbling. He wondered why on earth he and his wife should have been subjected to agony and inconvenience. He couldn’t take his dad’s view that suffering ennobles human beings. But the little boys taught him a profound lesson of life. One should get on with the business of living irrespective of whatever misfortune strikes one.

You don’t become a better person because you suffered, your suffering does not ennoble you. But you become a better person because you have experienced suffering. It is not what you have lost is important. What is important is what you have left. We can appreciate light better once we have experienced darkness. Similarly, we can appreciate warmth only after experiencing cold.

[OR]

The ridge had Hillary and Tenzing two and half hours, but it seemed like lifetime. Why?
Answer:
Both Hillary and Tenzing reached a wide ledge. Hillary deeply felt the fierce determination that nothing could stop them from scaling the Himalayas. He took a stronghold on the ledge and signalled Tenzing to come up. With great difficulty Tenzing reached the way up the crack and collapsed like a giant fish hauled up from the ocean. There were giant cornices on the right and steep rock sloped on the left. The ridge curved away to the right. They had no idea where the peak was.

As Hillary cut around the back of one hump, another would swing anew to his view. Time seemed never-ending. Their original zest started diminishing. The climb was becoming a grim struggle. To their surprise, the ridge ahead now dropped sharply away. A few more whacks of the ice-axe in the firm snow, they stood on top. It was Question 11.30, the ridge had taken two and a half hours. But it seemed like a lifetime as they had to negotiate numerous ridges on the way.

Question 42.
Give the summary of the poem, “Our Casuarina Tree’.
Answer:
The poet Torn Dutt describes the perpetually young Casuarina tree in this poem. The tree is tall and mighty. A creeper winds around its huge and wide trunk like a python. Birds chirp happily. The poet is ecstatic seeing the tree from her casement / window. In fact, she greets the tree everyday soon after getting up from bed. She finds a baboon sitting like a statue on top of the tree. The cows are grazing in the lush green meadows around the tree. Water lilies add to the charm of the pond. She recalls that the tree is dear not because of its majestic looks but because of the nostalgic memories attached to it.

She remembers the happy times she had spent with her siblings, who are now no more, under the tree. They died of consumption. Their absence makes the tree much more endearing to the poet. Mysteriously she hears the grief- filled lament of the tree when she is on the shores of France and Italy. She consecrates the tree’s memory to her dear departed siblings. Taking inspiration from William Wordsworth’s poem ‘Yew-trees’, she also wishes to immortalize the Cauarina tree. So she sings about it in her poem.

[OR]

What is the role of the young soldier in the victory of the French at Ratisbon?
Answer:
The young soldier was one of the soldiers in the infantry division leading the battle. On storming Ratisbon, unmindful of the cannon fire, he climbed the flag post with French flag and hoisted it. He received the bullets in turn for his service to the emperor and French army. He did not succumb to the bullets immediately, he galloped on horse back to convey the news to emperor Napoleon Bonaparte himself.

He held on to life till he reached Napoleon and conveyed the happy news. He waited with abated breath to know the reaction of his great leader. When he expressed his sadness, his pride was hurt. He denied the emperor’s sympathy and said emphatically that he was killed. He fell down beside emperor Napoleon with a smiling face and died.

Question 43.
Write a paragraph of about 150 words by developing the following hints:
Ivan – tongue-tied – knife – moral shock – innocent victim – faith in justice – wife’s suspicion – loses hopes – belief in God – divine justice. Tsar’s judgement- mills of justice grind slow – Semyonich confesses – cruel joke
Answer:
Ivan is tongue-tied when the police finds knife in his bag. The moral shock he undergoes as an innocent victim of circumstances allows him to have faith injustice. But the moment his wife expresses her suspicion over his involvement in the murder, he loses all hopes. It is then that he starts believing in God and divine justice.

Tsar can’t give him freedom. His judgement is based on evidences given by crooked people. God need not be given any evidences of innocence. He knows the truth but his mills of justice grind very slow. Ivan loses interest in escaping or leaving the prison. He had no one to go back to. At this juncture, Semyonich confesses his guilt of murdering the merchant. When the pardon arrives as a cruel joke, Ivan is dead. So, ‘God knows the truth but waits’ is the most appropriate title for this story.

[OR]

Sermon on diligence – Roger demands half-day leave – value of hard work – an impartial judge – switches off attention – take notes – self-indulgent absent-mindedness – escapes – rewinds memory – hoodwinks the judge – interview the award committee – judge was quite impressed – Mr. Clay’s office – orderly mind – excel in any learned profession
Answer:
In the beginning of the play Roger demands half-day leave and that was enough for the judge to start his sermon on the value of hard work and his standing in the society as an impartial judge. Roger conveniently switches off his attention and pretends to take notes of the instructions of the judge. The judge almost catches in his self-indulgent absent-mindedness. Roger escapes by the skin of his teeth by just rewinding from his memory the oft-repeated story of his success.

He hoodwinks the judge as to the purpose of his demand for half-a-day leave by saying that he wanted to personally interview the award committee as to why they had not forwarded the documents. The judge was quite impressed. The judge asked him to walk to Mr. Clay’s office in the afternoon. He reiterated that diligence, courage and attention to details are required to cultivate an orderly mind. Without an orderly mind no man can hope to excel in any learned profession.

Question 44.
Write a summary or Make notes of the following passage.
Answer:
Teaching is the noblest of vocations. A teacher has a sacred duty to perform. It is he on whom rests the responsibility of shaping the character of young children. Apart from developing their intellect, he can inculcate in them qualities of good citizenship, remaining neat and clean, talking decently and sitting properly. These virtues are not easy to be absorbed.

Only he who himself leads a life of simplicity, purity and rigid discipline can successfully cultivate these habits in his pupils. Besides a teacher always remains young. He may grow old in age, but not in spite. Perpetual contact with budding youth keeps him happy and cheerful. There are moments when domestic worries weigh heavily on his mind but the delightful company of innocent children makes him overcome his temporary moods of despair.

Summary

No. of words given in the original passage: 133
No. of words to be written in the summary: 133/3 = 41 ± 5

Rough Draft
Teaching is the noblest vocation. A teacher has a sacred duty to perform the responsibility of shaping the character of young children. Apart from developing their intellect, he can inculcate good citizenship. Besides a teacher always remains young perpetual contact with the youth keeps him overcome his temporary moods of despair.

Fair draft Noble Profession
Teaching is the noblest vocation. A teacher himself leading a simple, pure and disciplined life can shape the character of the young children and make them neat and good mannered citizens. Besides he remains every young forgetting his own domestic worries in the constant company of the young.

No. of words in the summary: 48

Or

Notes
Title: Noble Profession
Teachg – Noble Profession
sacred duty – mouldg charac
qlts A good citizenship
neat, clean,

Teachr – Forever Young
grow old in age but not in spite.
contact c budg youth .
domestic worries weigh heavily-delightful co. A innocent children overcomes

Abbreviations used: Mouldg – moulding; charac – character; A – of; qlts – qualities; Teachg – teaching; Teachr – teacher; c – with; budg-budding; co. – company

Question 45.
You are Roshini of 27-Railway Quarters, ICF, Chennai. Last week you bought a bluetooth speaker from ‘Universal Mobile’, 20L, Luz Corner, Chennai. The bluetooth speaker developed a problem within a few days of its purchase. Write a complaint letter to the dealer giving details of the nature of the problem and asking him/her to rectify the defect or replace the phone.
Answer:
26th March, 2019
From
Roshini
27-Railway Quarters
ICF
Chennai

To
M/s Universal Mobile
20L, Luz Corner
Chennai
Dear Sirs,

Sub: Defective Mobile Phone
I am a resident of ICF, Chennai. I purchased a JBL Clip 3 bluetooth speaker from Universal Mobile on 19th March, 20Question 20.1 am sorry to say that the speaker developed a problem within a few days of its purchase. The sound is quite unclear and irritating. I feel cheated to have such a defective mobile phone after spending more than twelve thousand mpees.

It is quite unfortunate that even after sending two reminders, you have shown no urgency to rectify the defects or replace the defective mobile set at the earliest. I hope you will do the needful within a week. I am sure you will not compel me to knock the doors of the Consumer Court for this unpleasant lapse of time on your part.

Yours sincerely
Roshini

Address on the envelope:
To
Universal Mobile, Luz Comer
Chennai

[OR]

Write a paragraph of 150 words on “My ambition in life.”.
Answer:

My ambition in life

Taking a photo or posing for a photo is something that I have always enjoyed even as a small little kid. There’s one thing that gets me to do and excites me. It is photography and that really stands out from all the rest. I always think of going all over the world and take shots of everything worth-seeing. I want to stick to photography all the time. I think I will never get tired of it. Even as a kindergartener, I used to admire pictures taken and collect pictures.

I always asked my parents to get me a camera. For my 14th birthday, I was gifted an SLR because of my passion for photography. I know there’s so much I have to leam about photography and I am trying to leam more and more about it. I joined a photography class after my tenth exams. I always used to take part in competitions and win prizes. I think there are many people in my community who are really good at photography and present outstanding photographs regarding nature’s beauty and others.

Many of them held photography contests to motivate and encourage new photographers. I now have a craze to do Candid Photography. I know for sure that I can pursue this as a hobby as well as my profession. If you love your job, you will never feel the pressure of working and hence I have decided to become a photographer for my life.

Question 46.
Spot the errors and rewrite the sentences correctly
(a) If I had known you were ill, I would visit you.
(b) He spoke the English perfectly.
(c) I am writing the essay when the bell rang.
(d) I and Sheela have booked tickets for the latest Harry Potter movie.
(e) It is an great honour.
Answer:
(a) If I had known you were ill, I would have visited you.
(b) He spoke English perfectly.
(c) I was writing the essay when the bell rang.
(d) Sheela and I have booked tickets for the latest Harry Potter movie.
(e) It is a great honour.

[OR]

Fill in the blanks correctly.
(а) My daughter who is just four …………………… (ears/years) old loves to play in the (sun/son)
(b) If it were not cold outside, I …………………… be much happier! (Use a modal in the given blank.)
(c) You …………………… finish reading this book by this weekend, (use a semi-modal)
(d) They …………………… (buy) sponges this week, (use a proper tense)
Answer:
(a) years/sun (b) would (c) need to (d) will be buying

Question 47.
Identify each of the following sentences with the fields given below:
(a) At the Sarangkheda horse fair in Maharashtra, local breeds attract buyers from as far as Saudi Arabia.
(b) My brother was awestruck at the western classical concert.
(c) Dhoni was declared the Man of the Series.
(d) French actress Catherine’s clothing was designed by her friend Yas for her next project.
(e) The investigations included megascopic and microscopic studies of rocks.
[Geology; Tourism; Music; Sports; Costume]
Answer:
(a) Tourism (b) Music (c) Sports (d) Costume (e) Geology

[OR]

Read the following passage and answer the questions in your own words.
Answer:
Black holes are one of the most mysterious and powerful forces in the universe. But what exactly are they? Are they simply holes that are black? Well, there’s more to it than that. A black hole is not a regular object with a surface area, like a planet. Instead, it’s an area in space where matter has literally collapsed onto itself. Black holes are formed when giant stars run out of energy. They end their life cycle and explode. This massive explosion is called a supernova. If the star has enough mass, it will collapse onto itself into a super small size.

Imagine an enormous mass squeezed into a tiny space. That space would become compact and dense. This causes the gravitational pull to be extremely strong. It will gobble up and absorb anything in its way, even gigantic stars. Nothing can move fast enough to escape its gravity. Not even light, the fastest thing in the universe! Black holes will gobble up anything and everything in their path, including light.

What happens to an object that gets sucked into a black hole? It will be literally stretched to its breaking point. Luckily, we don’t have to worry about black holes swallowing up our solar system. That’s because the nearest black hole is about 27,000 light-years away! Here’s a strange fact: black holes cannot actually be seen. That’s because they don’t deflect light. So how can we be certain they exist? Scientists discovered them by observing light and objects around them.

They noticed that black holes affected their surroundings like nearby dust, stars, and galaxies. Two different scientists from the 18th century named John Michell and Pierre-Simon Laplace first observed them. Then in 1967, a physicist named John Archibald Wheeler came up with the term “black hole”.

Black holes come in different sizes. Some are about the mass of one star. These are called “stellar” black holes. Others can grow and become huge. They may continue to absorb light, mass, and even stars around them. These are called “super-massive black holes.” They can be one million times more massive than our sun. Many scientists believe these super-massive black holes often exist right at the center of galaxies. In fact, we have one right in the middle of our very own Milky Way.

Questions:
a. Based on the information in the article, what is a black hole?
b. Who first discovered the existence of black holes?
c. Why are black holes dangerous?
d. Are all black holes the same size? Explain.
e. Why can’t black holes be seen?
Answers:
(a) A black hole is one of the most mysterious and powerful forces in the universe, an area in space where matter has collapsed in on itself.
(b) Two different scientists from the 18th century named John Michell and Pierre-Simon Laplace first discovered the existence of black holes.
(c) Black holes pull in anything that gets too close and stretch things to their breaking point and hence they are dangerous.
(d) No. Some black holes are the size of a single star. These are called stellar black holes. Others are the size of a million stars the size of our sun, or larger. These are called super- massive black holes.
(e) Black holes cannot actually be seen because they don’t deflect light. They can only be discovered by observing light and objects around them.

Students can download 10th Science Chapter 2 Optics Questions and Answers, Notes, Samacheer Kalvi 10th Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 2 Optics

Samacheer Kalvi 10th Science Optics Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
The refractive index of four substances A, B, C and D are 1.31,1.43,1.33, 2.4 respectively. The speed of light is maximum in:
(a) A
(b) B
(c) C
(d) D
Answer:
(a) A

Question 2.
Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens:
(a) f
(b) 2f
(c) infinity
(d) between f and 2f
Answer:
(b) 2f

Question 3.
Where should an object be placed so that a real and inverted image of the same size is obtained by a convex lens ______.
(a) f
(b) 2f
(c) infinity
(d) between f and 2f.
Answer:
(b) 2f

Question 4.
Magnification of a convex lens is _____.
(a) positive
(b) negative
(c) either positive or negative
(d) zero.
Answer:
(b) negative

Question 5.
A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at:
(a) focus
(b) infinity
(c) at 2f
(d) between f and 2f
Answer:
(b) infinity

Question 6.
Power of a lens is -4D, then its focal length is:
(a) 4 m
(b) -40 m
(c) -0.25 m
(d) -2.5 m
Answer:
(d) -2.5 m

Question 7.
In a myopic eye, the image of the object is formed _____.
(a) behind the retina
(b) on the retina
(c) in front of the retina
(d) on the blind spot.
Answer:
(c) in front of the retina

Question 8.
The eye defect ‘presbyopia’ can be corrected by:
(a) convex lens
(b) concave lens
(c) convex mirror
(d) Bi focal lenses
Answer:
(d) Bi focal lenses

Question 9.
Which of the following lens would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 5 cm
(b) A concave lens of focal length 5 cm
(c) A convex lens of focal length 10 cm
(d) A concave lens of focal length 10 cm
Answer:
(d) A concave lens of focal length 10 cm

Question 10.
If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?
(a) V_B = V_G = V_R
(b) V_B > V_G > V_R
(c) V_B < V_G < V_R
(d) V_B < V_G > V_R
Answer:
(c) V_B < V_G < V_R

II. Fill in the blanks.

1. The path of the light is called as ………
2. The refractive index of a transparent medium is always greater than ……….
3. If the energy of incident beam and the scattered beam are same, then the ………. scattering of light is called as scattering ……….
4. According to Rayleigh’s scattering law, the amount of scattering of light is inversely proportional to the fourth power of its ……….
5. Amount of light entering into the eye is controlled by ……….

Answer:

1. ray
2. unity
3. elastic
4. wavelength
5. iris

III. True or False. If false correct it.

1. Velocity of light is greater in denser medium than in rarer medium.
2. The power of lens depends on the focal length of the lens.
3. Increase in the converging power of eye lens cause ‘hypermetropia’
4. The convex lens always gives small virtual image.

Answer:

1. False – Velocity of light is greater in rarer medium than in denser medium.
2. True
3. True
4. False – The convex lens does not give small virtual image always.

IV. Match the following.

Answer:
1. d
2. a
3. e
4. b
5. c

V. Assertion and reasoning type.

Mark the correct choice as-
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small
Reason: Refractive index of the medium is inversely proportional to the velocity of the light.

2. Assertion: Myopia is due to the increase in the converging power of eye lens.
Reason: Myopia can be corrected with the help of concave lens.
Answer:
1. (a)
2. (a)

VI. Answer Briefly.

Question 1.
What is refractive index?
Answer:
Refractive index gives us an idea of how fast or how slow light travels in a medium.

Question 2.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer:

Question 4.
Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

Question 5.
State Rayleigh’s law of scattering.
Answer:
Rayleigh’s scattering law states that, “The amount of scattering of light is inversely proportional to the fourth power of its wavelength”.
Amount of scattering ‘S’ ∝\(\frac{1}{λ^4}\)

Question 6.
Differentiate convex lens and concave lens.
Answer:

Question 7.
What is the power of accommodation of the eye?
Answer:

• The ability of the eye lens to focus nearby as well as the distant objects is called the power of accommodation of the eye.
• This is achieved by changing the focal length of the eye lens with the help of ciliary muscles.

Question 8.
What are the causes of ‘Myopia’?
Answer:

1. The lengthening of eye ball.
2. The focal length of eye lens is reduced.
3. The distance between eye lens and retina increases.
4. The far point will not be at infinity.
5. The far point comes closer.

Question 9.
Why does the sky appear blue in colour?
Answer:
When sunlight passes through the atmosphere, the blue colour (shorter wavelength) is scattered to a greater extent than the red colour (longer wavelength). This scattering causes the sky to appear blue in colour.

Question 10.
Why are traffic signals red in colour?
Answer:

• Red light has the highest wavelength.
• It is scattered by atmospheric particles.
• So red light is able to travel the longest distance through a fog, rain etc.

VII. Give the answer in detail.

Question 1.
List any five properties of light?
Answer:

• Light is a form of energy.
• Light always travels along a straight line.
• Light does not need any medium for its propagation. It can even travel through a vacuum.
• The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1
• Since light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = vλ (c = velocity of light).
• Different coloured light has a different wavelength and frequency.

Question 2.
Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
Answer:
Rule-1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation.

Rule-2: When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus.

Rule-3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis.

Question 3.
Differentiate the eye defects: Myopia and Hypermetropia.
Answer:

Question 4.
Explain the construction and working of a ‘Compound Microscope’.
Answer:
Construction : A compound microscope consists of two convex lenses. The lens with the shorter focal length is placed near the object, and is called as ‘objective lens’ or ‘objective piece’. The lens with larger focal length and larger aperture placed near the observer’s eye is called as ‘eye lens’ or ‘eye piece’. Both the lenses are fixed in a narrow tube with adjustable provision.

Working : The object (AB) is placed at a distance slightly greater than the focal length of objective lens (u > F₀). A real, inverted and magnified image (A’B’) is formed at the other side of the objective lens. This image behaves as the object for the eye lens. The position of the eye lens is adjusted in such a way, that the image (B’B’) falls within the principal focus of the eye piece. This eye piece forms a virtual, enlarged and erect image (A”B”) on the same side of the object.

Compound microscope has 50 to 200 times more magnification power than simple microscope.

VIII. Numerical Problems.

Question 1.
An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.
Answer:
Distance of an object u = 20 cm
Focal length of a convex lens f = 10 cm
Let the image distance be v
We know

v = 20 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{20}{20}\) = 1
Hence a real image of same size is formed at 20 cm.
Image distance = 20 cm

Question 2.
An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.
Answer:
Object distance u = 10 cm
Focal length of a concave lens f= -15 cm
Let v be the image distance,

Distance of image v = 6 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{6}{10}\) = 0.6
And Magnification m = \(\frac{h’}{h}\)
Where h’ – height of image
h – height of object
0.6 = \(\frac{h’}{3}\)
∴ h’ = 3 × 0.6 = 1.8 cm
∴ Height of image = 1.8 cm

IX. Higher order thinking (HOT) questions.

Question 1.
While doing an experiment for the determination of focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens,
(a) can he get the image?
(b) Is there any change in the focal length?
Answer:
(a) He can get the image.
(b) The focal length of the lens will be doubled.

Question 2.
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
Answer:

• The large pupil opens wider and allows the maximum amount of light to enter the eye in the dark.
• Their lens is large and situated near the retina. This also allows a lot of light to register on the retina. The retina contains 2 types of light-sensing cells rods and cones.
• Cones are responsible for the coloured vision and require bright, focused light.
• Rods are extremely sensitive to light and have a photosensitive pigment called rhodopsin which plays a vital role in night vision.

Samacheer Kalvi 10th Science Optics Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The scattering of sun light by the atoms or molecules of the gases in the Earth’s atmosphere is known as:
(a) Mie scattering
(b) Tyndall scattering
(c) Rayleigh scattering
(d) Raman scattering
Answer:
(c) Rayleigh scattering

Question 2.
Mie scattering is responsible for the _____ appearance of the clouds.
(a) red
(b) blue
(c) colourless
(d) white.
Answer:
(d) white

Question 3.
In an inelastic scattering the energy of the incident beam of light is ……….. that of scattering beam.
(a) greater than
(b) less than
(c) equal to
(d) different from
Answer:
(d) different from

Question 4.
As per Rayleigh’s scattering law, amount of scattering is:
(a) directly proportioanl to fourth power of wavelength
(b) inversely proportioanl to fourth power of wavelength
(c) inversely proportioanl to square of wavelength
(d) directly proportional to square of wavelength
Answer:
(b) inversely proportioanl to fourth power of wavelength

Question 5.
The refractive index of a medium is dependent on the _____ of the light.
(a) wavelength
(b) strength
(c) density
(d) refraction.
Answer:
(a) wavelength

Question 6.
The scattering of light by colloidal particles in the colloidal solution is called:
(a) Raman scattering
(b) Tyndall scattering
(c) Mie scattering
(d) Elastic scattering
Answer:
(b) Tyndall scattering

Question 7.
A piece of transparent material bounded by curved surfaces is called:
(a) mirror
(b) prism
(c) slab
(d) lens
Answer:
(d) lens

Question 8.
If the energy of the incident and the scattered beam of light are not the same, then it is called as _____.
(a) Elastic
(b) Raman
(c) Inelastic
(d) Mie.
Answer:
(c) Inelastic

Question 9.
A convex lens does not produce:
(a) real magnified image
(b) virtual magnified image
(c) virtual diminished image
(d) real diminished image
Answer:
(c) virtual diminished image

Question 10.
A lens which is thicker in the middle than at the edges is known as:
(a) concave lens
(b) convex lens
(c) bifocal lens
(d) cylindrical lens
Answer:
(b) convex lens

Question 11.
The object is always placed on the _____ side of the lens.
(a) left
(b) right
(c) top
(d) bottom.
Answer:
(a) left

Question 12.
The parallel rays from the outer edge are deviated towards the middle in a:
(a) convex mirror
(b) concave lens
(c) concave mirror
(d) convex lens
Answer:
(d) convex lens

Question 13.
The light rays passing through the optic centre will:
(a) diverged
(b) scattered
(c) converged
(d) emerge undeviated
Answer:
(d) emerge undeviated

Question 14.
All the distances are measured from the ______ of the lense.
(a) centre of curvature
(b) optical centre
(c) principal focus
(d) infinity.
Answer:
(b) optical centre

Question 15.
A ray passing through the principal focus and incident on the lens will:
(a) converge
(b) diverge
(c) emerge parallel to the principal axis
(d) not emerge out
Answer:
(c) emerge parallel to the principal axis

Question 16.
When the object is placed at infinity from the convex lens, the image is formed at:
(a) F
(b) C
(c) infinity
(d) between F and 2F
Answer:
(a) F

Question 17.
The human eye is ____ in nature.
(a) convex
(b) concave
(c) transparent glass
(d) Plano – concave.
Answer:
(a) convex

Question 18.
The image formed by a concave lens is:
(a) virtual
(b) diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(c) virtual and diminished

Question 19.
To get a real image using convex lens, the object must be placed at:
(a) infinity
(b) principal focus
(c) beyond principal focus and infinity
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 20.
_____ is the centre part of the iris.
(a) cornea
(b) retina
(c) pupil
(d) eye lens.
Answer:
(c) pupil

Question 21.
For a convex lens the point at which the parallel rays converge is called of the lens.
(a) pole
(b) centre of curvature
(c) principal focus
(d) none
Answer:
(c) principal focus

Question 22.
A real image formed by a convex lens is always:
(a) erect
(b) magnified
(c) inverted
(d) diminished
Answer:
(c) inverted

Question 23.
The law of distances is given by:

Answer:
(b) \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

Question 24.
The unit of focal length is:
(a) dioptre
(b) metre
(c) ohm
(d) ampere
Answer:
(b) metre

Question 25.
The sign of focal length of a convex lens is ………. sign.
(a) negative
(b) positive
(c) negative or positive
(d) none of the above
Answer:
(b) positive

Question 26.
The focal length of concave lens has ……… sign.
(a) positive
(b) negative
(c) positive or negative
(d) none of the above
Answer:
(b) negative

Question 27.
The magnification in terms of object distance u and image distance v is m :
(a) \(\frac{u}{v}\)
(b) u + v
(c) \(\frac{v}{u}\)
(d) uv
Answer:
(c) \(\frac{v}{u}\)

Question 28.
In terms of object distance u and focal length/, magnification is given by m =

Answer:
(b) \(\frac{f}{u-f}\)

Question 29.
The magnification in terms of v and/:
(a) f
(b) v – f
(c) \(\frac{f}{v-f}\)
(d) \(\frac{v-f}{f}\)
Answer:
(d) \(\frac{v-f}{f}\)

Question 30.
The unit of power is:
(a) m
(b) ohm
(C) dioptre
(d) ampere
Answer:
(C) dioptre

Question 31.
If the focal length of a convex lens is 1 m then its power is:
(a) 1 dioptre
(b) 0.1 dioptre
(c) 10 dioptre
(d) 0.01 dioptre
Answer:
(a) 1 dioptre

Question 32.
In a simple microscope, the magnification can be increased by:
(a) lens of long focal length
(b) lens
(c) lens of short focal length
(d) lens of infinite focal length
Answer:
(c) lens of short focal length

Question 33.
Convex lenses are used:
(a) as camera lenses
(b) as magnifying lenses
(c) to correct hypermetropia
(d) all the above
Answer:
(d) all the above

Question 34.
Which lens is used in wide angle spyhole in doors?
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) parabolic lens
Answer:
(b) concave lens

Question 35.
The mathematical form of lens maker’s formula is:

Answer:
(a)

Question 36.
If f is the focal length of the lens then its power is given by:
(a) P = \(\frac{2}{f}\)
(b) p = \(\frac{1}{f}\)
(c) p = f
(d) p = f₁
Answer:
(b) p = \(\frac{1}{f}\)

Question 37.
Which part of the human eye changes the focal length of the eye lens?
(a) pupil
(b) retina
(c) ciliary muscles
(d) cornea
Answer:
(c) ciliary muscles

Question 38.
On which part of human eye, image is formed?
(a) cornea
(b) iris
(c) retina
(d) pupil
Answer:
(c) retina

Question 39.
For normal human eye the value of near point is:
(a) 25 cm
(b) 25 m
(c) 2.5 m
(d) 25 mm
Answer:
(a) 25 cm

Question 40.
In hypermeteropia, the focal length of the eye lens is:
(a) decreased
(b) remains the same
(c) increased
(d) none of the above
Answer:
(c) increased

Question 41.
Presbyopia can be corrected by using:
(a) convex lens
(b) bifocal lens
(c) concave lens
(d) cylindrical lens
Answer:
(b) bifocal lens

Question 42.
Astigmatism can be corrected by using:
(a) bifocal lens
(b) cylindrical lens
(c) convex lens
(d) concave lens
Answer:
(b) cylindrical lens

Question 43.
The magnifying power of compound microscope is:
(a) 10
(b) 20
(c) 50
(d) 50 to 200
Answer:
(d) 50 to 200

Question 44.
The accuracy of travelling microscope is of the order of:
(a) 0.01 cm
(b) 0.01 mm
(c) 0.1 mm
(d) 0.1 cm
Answer:
(b) 0.01 mm

II. Fill in the blanks.

1. The velocity of light in vacuum is ……….
2. If v is the frequency and λ is the wavelength then velocity of the wave is c = ……….
3. Among colours of visible light ……… colour has the highest wavelength.
4. According to Snell’s law refractive index, µ₂ = ……….
5. In a medium having high value of refractive index then speed of light in that medium is ……….
6. Angle of refraction is the smallest for ……… and the highest for ……….
7. The refractive index depends on ………. of light.
8. Colours having shorter wavelength scattered more than longer wavelength colours according to ……….. law.
9. After passing through a convex lens ……….. rays ………. at the principal focus.
10. For a convex lens, as the object distance increases, the image distance ……….
11. A ray passing through the optic centre of a lens emerges …………
12. ……… is due to irregular curvature of the surface of the eye lens.
13. When a parallel beam of light passes through a convex lens, the rays from the outer edges are …………
14. A ray parallel to the principal axis of a convex lens after refraction passes through …………
15. When the object is placed between ………….. and ………… of a convex lens a virtual image will be formed.
16. For a convex lens, as the object approaches the lens the image becomes …………
17. In a phographic camera ………… lens is used.
18. The shorter the focal length, the ………. is the magnification.
19. The nature of the image formed by a simple microscope is ……….., ………… and …………
20. Real images are formed by a ……….. lens.
21. Concave lens produces ………… images.
22. The value of power of a lens having focal length one metre is ………..
23. For a normal eye the value of far point is …………
24. …………. is known as short sightedness.
25. Hyper metropia is known as ………..
26. The mathematical form of focal length of a concave lens used to correct myopia is f = ……….
27. ……….. lenses are used to correct astigmatism.
28. For a normal eye, the value of least distance of distinct vision is ………..
29. The objective of the compound microscope has …………. focal length.
30. The focal length of ………. is greater in a compound microscope.
31. ……… is an optical instrument to see the distant objects.
32. A terrestrial telescope produces ……… image.
33. Elaborate view of galaxies and planets is obtained by ………
Answer:
1. 3 × 10⁸ m/s
2. vλ
3. red
4. \(\frac{sin i}{sin r}\)
5. low
6. red, violet
7. wavelength
8. Rayleigh scattering
9. parallel, converge
10. will decrease
11. undeviated
12. Astigmatism
13. deviated towards the centre of the lens
14. the principal focus
15. principal focus, optical centre
16. bigger
17. biconvex
18. greater
19. virtual, erect, magnified
20. convex
21. virtual
22. One dioptre
23. infinity
24. Myopia
25. long sightedness
26. xy/x – y
27. Cylindrical
28. 25 cm
29. shorter
30. eye piece
31. Telescope
32. an erect
33. Telescope

III. True or False. If false correct it.

1. Light does not travel along a straight line.
2. All coloured light has same wavelength.
3. In refraction incident ray, refracted ray and normal lie in the same plane.
4. Velocity of light is greater in rarer medium is greater than that in denser medium.
5. For red colour angle of refraction is the least.
6. The refractive index of a medium ¡s independent of wavelength.
7. Tyndall scattering, is the scattering of light by colloids.
8. According to Rayleigh’s scattering law, red colour is scattered to a greater extent than blue colour.
9. Mie scattering takes place when the diameter is larger than the wavelength of the incident light.
10. The lines in Raman scattering having frequencies lower than the incident frequency are called Antistoke’s lines.
11. In front of a convex lens when the object is placed at infinity the formed image is smaller than that of the object.
12. When an object is placed at finite distance from the concave lens a virtual image is formed between optical centre and focus of the concave lens.
13. Pupil of human eye bends the incident light on to the lens.
14. For a normal human eye, the value of far point is 25 cm.
15. Astigmatism is corrected by cylindrical lenses.
Answer:
1. False – Light always travels along a straight lines.
2. False – Different coloured light has different wavelength.
3. True
4. True
5. True
6. False – The refractive index of a medium depends on wavelength.
7. True
8. False – According to Rayleigh’s scattering law, blue colour is scattered to a greater extent than red colour.
9. True
10. False – The lines in Raman scattering having frequencies higher than the incident frequency are called Antistoke’s lines.
11. True
12. True
13. False – Cornea of human eye bends the incident light on to the lens.
14. False – For a normal human eye, the value of near point is 25 cm.
15. True

IV. Match the following.

Question 1.
Match the column I with column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 2.
Position of the object placed infront of a convex lens are given in Column
I. Match them with the natures of the images formed by the convex lens given in column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 3.
Match the following:

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 4.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

Question 5.
Match the Column I with Column II.

Answer:
A. (iv)
B. (i)
C. (ii)
D. (iii)

Question 6.
Match the column I with column II.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)

Question 7.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

V. Assertion and Reasoning type.

Question 1.
Assertion : The sun looks bigger in size at sunrise and sunset than during day.
Reason : In detraction light rays bend around the edges of the obstacle.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 2.
Assertion: Colours can be scan in thin layers of oil on the surface water. Reason: White light is composed of several colours.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion: Raman spectrum of a liquids contains lines whose frequencies are not equal to that of incident radiation.
Reason: If a photon strikes an atom in a liquid that is in existed state photon losses energy.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The refractive index of a prism depends only on the material of the prism.
Reason: The refractive index of a prism depends upon the refracting angle and angel of minimum deviation.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 5.
Assertion: A single lens produces a coloured image of an object illuminated by white light.
Reason: The refractive index of material of lens is different for different wavelength of light.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 6.
Assertion: If a convex lens is placed in water, its convergence power decrease.
Reason: Focal length of lens is independent of refractive index of the medium.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 7.
Assertion: Light waves travel in straight lines.
Reason: Rectilinear propagation of light confirm the above mentioned properly.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 8.
Assertion: Raman scattering the scattering of monochromatic light by atoms and molecule of a liquid.
Reason: The wavelength of Raman lines is same.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 9.
Assertion: Power of a lens is the reciprocal of its focal length.
Reason: The unit of power is one dioptre when the unit of focal length is one metre.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 10.
Assertion: Presbyopia is due to ageing of human beings.
Reason: For those persons, ciliary muscles of the eye become weak.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

VI. Answer Briefly.

Question 1.
What is meant by refraction?
Answer:
When a ray of light travels from one transparent medium into another obliquely, the path of the light undergoes deviation. This deviation of ray of light is called refraction.

Question 2.
State laws of refraction.
Answer:
First law of refraction: The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.

Second law of refraction: The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Define refractive index of a medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in a medium is defined as refractive index ‘p’ of that medium.

Question 4.
What is meant by monochromatic source?
Answer:
If a source of light produces a light of single colour, it is known as a monochromatic source.

Question 5.
When white light is refracted by a transparent medium what will you get? Why?
Answer:

1. When white light is refracted by a transparent medium, a spectrum is obtained.
2. This is because, different coloured lights are bent through different angles.

Question 6.
What is scattering of light?
Answer:
When sunlight enters the Earth’s atmosphere, the atoms and molecules of different gases present in the atmosphere refract the light in all possible directions. This is called as ‘Scattering of light’.

Question 7.
State the types of scattering.
Answer:

1. Elastic scattering
2. Inelastic scattering

Question 8.
What is elastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are same, then it is called as ‘elastic scattering’.

Question 9.
What is inelastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as ‘inelastic scattering’.

Question 10.
How are different types of scattering formed? Mention the types of scattering.
Answer:
The nature and size of the scatterer results in different types of scattering. They are

1. Rayleigh scattering
2. Mie scattering
3. Tyndall scattering
4. Raman scattering

Question 11.
What is Rayleigh scattering?
Answer:
The scattering of sunlight by the atoms or molecules of the gases in the earth’s atmosphere is known as Rayleigh scattering.

Question 12.
Why the colour of the Sun is red at sunrise and sunset?
Answer:
At sunrise and sunset, the light rays from the Sun have to travel a larger distance in the atmosphere than at noon. Hence, most of the blue lights are scattered away and only the red light which gets least scattered reaches us. Therefore, the colour of the Sun is red at sunrise and sunset.

Question 13.
When does Mie scattering take place?
Answer:
Mie scattering takes place when the diameter of the scatterer is similar to or larger than the wavelength of the incident light.

Question 14.
What are the causes of Mie scattering?
Answer:
Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.

Question 15.
Why the clouds have white appearance?
Answer:
Mie scattering is responsible for the white appearance of the clouds. When white light falls on the water drop, all the colours are equally scattered which together form the white light.

Question 16.
What is Tyndall Scattering?
Answer:
The scattering of light rays by the colloidal particles in the colloidal solution is called Tyndall Scattering or Tyndall Effect.

Question 17.
What is meant by colloid? State few examples.
Answer:
Colloid is a microscopically small substance that is equally dispersed throughout another material. Eg: Milk, Ice cream, muddy water, smoke.

Question 18.
What is meant by Raman Scattering?
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

Question 19.
Define Raman Scattering.
Answer:
Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

Question 20.
What is Rayleigh line?
Answer:
The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’.

Question 21.
What are Raman lines?
Answer:
The spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’.

Question 22.
What are stokes lines and Antistokes lines?
Answer:
The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 23.
What is a lens?
Answer:
A lens is an optically transparent medium bounded by two spherical refracting surfaces or one plane and one spherical surface.

Question 24.
How is lens classified?
Answer:
Lens is basically classified into two types. They are:

1. Convex Lens
2. Concave Lens.

Question 25.
What is biconvex lens?
Answer:
Convex or bi-convex lens: It is a lens bounded by two spherical surfaces such that it is thicker at the centre than at the edges. A beam of light passing through it, is converged to a point. So, a convex lens is also called as converging lens.

Question 26.
What is meant by biconcave lens?
Answer:
Concave or bi-concave Lens: It is a lens bounded by two spherical surfaces such that it is thinner at the centre than at the edges. A parallel beam of light passing through it, is diverged or spread out. So, a concave lens is also called as diverging lens.

Question 27.
What are
(i) Plano-convex lens?
(ii) Plano-concave lens?
Answer:
(i) If one of the faces of a bi-convex lens is plane, it is known as a plano-convex lens.
(ii) If one of the faces of a bi-concave lens is plane, it is known as a plano-concave lens.

Question 28.
State the applications of convex lenses.
Answer:

1. Convex lenses are used as camera lenses.
2. They are used as magnifying lenses.
3. They are used in making microscope, telescope and slide projectors.
4. They are used to correct the defect of vision called hypermetropia.

Question 29.
Draw diagrams of different types converging lenses.
Answer:

Question 30.
Represent different types of lenses by diagram.
Answer:

Question 31.
What is the nature of the image formed by an object is placed behind the centre of curvature of a convex leas. Draw a ray diagram.
Answer:
When an object is placed behind the center of curvature (beyond C), a real and inverted image is formed between the center of curvature and the principal focus. The size of the image is the same as that of the object.

Question 32.
Draw a ray diagram to indicate the nature of the image formed when an object is placed in between the centre of curvature and principal focus of a convex lens.
Answer:
When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

Question 33.
Draw a ray diagram for the formation of image (by the concave lens) when object is at infinity.
Answer:
When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Question 34.
What are the applications of concave lens?
Answer:

1. Concave lenses are used as eye lens of‘Galilean Telescope’.
2. They are used in wide angle spy hole in doors.
3. They are used to correct the defect of vision called ‘myopia’.

Question 35.
What do you know about lens formula?
Answer:
The lens formula gives the relationship among distance of the object (u), distance of the image (v) and the focal length (f) of the lens. It is expressed as
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)

Question 36.
Define magnification.
Answer:
It is defined as the ratio of the height of the image to the height of an object. Magnification is denoted by the letter ‘m’. If height of the object is h and height of the image is h’, the magnification produced by lens is,

Question 37.
What is lens formula?
Answer:
The lens formula relates the focal length of a lens with the distance of object and image.

where µ is the refractive index of the material of the lens; R₁ and R₂ are the radii of curvature of the two faces of the lens; f is the focal length of the lens.

Question 38.
What is meant by power of lens?
Answer:
The ability of a lens to converge (convex lens) or diverge (concave lens) light rays is called as its power.

Question 39.
Define power of a lens. State its unit.
Answer:
Power of a lens is numerically defined as the reciprocal of its focal length.
P = \(\frac{1}{f}\)
The SI unit of power of a lens is dioptre.

Question 40.
What is meant by dioptre.
Answer:
Dioptre is the power of a lens, whose focal length is 1 metre.
1 Dioptre = 1 m^-1.

Question 41.
Differentiate convex lens from concave lens.
Answer:

Question 42.
What are
(i) Pupil &
(ii) Retina?
Answer:
(i) Pupil: It is the centre part of the Iris. It is the pathway for the light to retina.
(ii) Retina: This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Question 43.
What is persistence of vision?
Answer:
If the time interval between two consecutive light pulses is less than 0.1 second, human eye cannot distinguish them separately. It is called persistence of vision.

Question 44.
What is least distance of distinct vision?
Answer:
The minimum distance required to see the objects distinctly without strain is called least distance of distinct vision. It is called as near point of eye. It is 25 cm for normal human eye.

Question 45.
What is far point?
Answer:
The maximum distance up to which the eye can see objects clearly is called as far point of the eye.

Question 46.
What is Presbyopia?
Answer:
Due to ageing, ciliary muscles become weak and the eye-lens become rigid (inflexible) and so the eye loses its power of accommodation. Because of this, an aged person cannot see the nearby objects clearly. So, it is also called as ‘old age hypermetropia’.

Question 47.
What is meant by astigmatism?
Answer:
In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the . refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses (Torrid lenses).

Question 48.
State the principle of microscope. How is it classified?
Answer:
It works under the principle of angular magnification of lenses. It is classified as

1. Simple microscope
2. Compound microscope

Question 49.
What are the uses of simple microscope?
Answer:
Simple microscopes are used

1. By watch repairers and jewellers.
2. To read small letters clearly.
3. To observe parts of flower, insects etc.
4. To observe finger prints in the field of forensic science.

Question 50.
How is telescope classified?
Answer:
According to optical property, it is classified into two groups:

1. refracting telescope
2. reflecting telescope

Question 51.
Mention the advantages of telescope.
Answer:

1. Elaborate view of the Galaxies, Planets, stars and other heavenly bodies is possible.
2. Camera can be attached for taking photograph for the celestial objects.
3. Telescope can be viewed even with the low intensity of light.

Question 52.
What are the disadvantages of telescope?
Answer:

1. Frequent maintenance is needed.
2. It is not easily portable.

VII. Give the answer in Detail.

Question 1.
State the Laws of Refraction.
Answer:
The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.
Second law of Refraction:

1. The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
   \(\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}\)
2. Refractive index gives us an idea of how fast or how slow light travels in a medium. The ratio of the speed of light in a vacuum to the speed of light in a medium is defined as the refractive index ‘µ’ of that medium.
3. The speed of light in a medium is low if the refractive index of the medium is high and vice versa.
4. When light travels from a denser medium into a rarer medium, the refracted ray is bent away from the normal drawn to the interface.
5. When light travels from a rarer medium into a denser medium, the refracted ray is bent towards the normal drawn to the interface.

Question 2.
Describe Raman Scattering.
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

The scattered light contains some additional frequencies (or wavelengths) other than that of incident frequency (or wavelength). This is known as Raman scattering or Raman Effect.

Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’ and the spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’. The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 3.
With the help of ray diagram, explain the nature, size and position of the image formed by a convex lens. When object is placed at
(i) infinity
(ii) beyond C
(iii) placed at C
(iv) Placed between F and C,
(v) placed at F
(vi) placed between F and optical centre O.
Answer:
(i) Object at infinity: When an object is placed at infinity, a real image is formed at the principal focus. The size of the image is much smaller than that of the object.

(ii) Object placed beyond C (>2F): When an object is placed behind the center of curvature(beyond C), a real and inverted image is formed between the center of curvature and the principal focus. Th e size of the image is the same as that of the object.

(iii) Object placed at C: When an object is placed at the center of curvature, a real and inverted image is formed at the other center of curvature. The size of the image is the same as that of the object.

(iv) Object placed between F and C: When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

(v) Object placed at the principal focus F: When an object is placed at the focus, a real image is formed at infinity. The size of the image is much larger than that of the object.

(vi) Object placed between the principal focus F and optical centre O: When an object is placed in between principal focus and optical centre, a virtual image is formed. The size of the image is larger than that of the object.

Question 4.
Explain the formation of images formed by a concave lens.
Answer:
Object at Infinity: When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Object anywhere on the principal axis at a finite distance: When an object is placed at a finite distance from the lens, a virtual image is formed between optical center and focus of the concave lens. The size of the image is smaller than that of the object.

But, as the distance between the object and the lens is decreased, the distance between the image and the lens also keeps decreasing. Further, the size of the image formed increases as the distance between the object and the lens is decreased.

Question 5.
Explain Mie Scattering.
Answer:
Mie scattering:

1. Mie scattering takes place when the diameter of the Scatterer is similar to or larger than the wavelength of the incident light. It is also an elastic scattering.
2. The amount of scattering is independent of wavelength.
3. Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.
4. Mie scattering is responsible for the white appearance of the clouds.
5. When white light falls on the water drop, all the colours are equally scattered which, together form the white light.

Question 6.
With the help of a diagram, explain the structure and working of human eye.
Answer:
Structure of the eye:
The eye ball is approximately spherical in shape with a diameter of about 2.3 cm. It consists of a tough membrane called sclera, which protects the internal parts of the eye.

Cornea : This is the thin and transparent layer on the front surface of the eyeball as shown in figure. It is the main refracting surface. When light enters through the cornea, it refracts or bends the light on to the lens.

Iris : It is the coloured part of the eye. It may be blue, brown or green in colour. Every person has a unique colour, pattern and texture. Iris controls amount of light entering into the pupil like camera aperture.

Pupil : It is the centre part of the Iris. It is the pathway for the light to retina.

Retina : This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Ciliary muscles : Eye lens is fixed between the ciliary muscles. It helps to change the focal length of the eye lens according to the position of the object.

Eye Lens : It is the important part of human eye. It is convex in nature.

Working of the eye : The transparent layer cornea bends the light rays through pupil located at the centre part of the Iris. The adjusted light passes through the eye lens. Eye lens is convex in nature. So, the light rays from the objects are converged and a real and inverted image is formed on retina. Then, retina passes the received real and inverted image to the brain through optical nerves. Finally, the brain senses it as erect image.

Question 7.
Describe simple microscope.
Answer:
Simple microscope: It has a convex lens of short focal length. It is held near the eye to get enlarged image of small objects.
Let an object (AB) is placed at a point within the principal focus (u < f) of the convex lens and the observer’s eye is placed just behind the lens. As per this position the convex lens produces an erect, virtual and enlarged image (A’B’), The image formed is in the same side of the object and the distance equal to the least distance of distinct vision (D) (For normal human eye D = 25 cm).

Question 8.
Write short notes on
(i) Astronomical telescope
(ii) Terrestrial telescope.
Answer:
(i) Astronomical Telescope: An astronomical telescope is used to view heavenly bodies like stars, planets, galaxies and satellites.

(ii) Terrestrial Telescope: The image in an astronomical telescope is inverted. So, it is not suitable for viewing objects on the surface of the Earth. Therefore, a terrestrial telescope is used. It provides an erect image. The major difference between astronomical and terrestrial telescope is erecting the final image with respect to the object.

VIII. Numerical Problems.

Question 1.
A needle of size 5 cm is placed 45 cm from a lens produced an image on a screen placed 90 cm away from the lens.
Answer:
(i) Identify the types of lens.
Calculate focal length of the lens.
Height of the object h₁ = 5 cm
Distance of the object u = -45 cm
Distance of the image v = 90 cm
We know that

Focal length of the lens = 30 cm
Since focal length is positive the lens is convex lens.

(ii) Identify the size of the image

∴ h₂ = -10 cm
The negative sign indicates that the image is real and inverted.

Question 2.
A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image 10 cm from the lens?
Answer:
v = -10 cm; f =-15 cm; u = ?
Lens formula:
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)
Type of lens: Concave lens

u =-30 cm
Thus, the object distance is 30 cm.

Question 3.
An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. An erect and virtual image is formed at a distance of 10 cm from the lens. Calculate the magnification.
Answer:
Type of lens is Cancave lens.
Formula:
Magnification m = \(\frac{v}{u}\)
Object distance u = -30 cm
Image distance v = -10 cm
m = \(\frac{-10}{-30}\) = \(\frac{1}{3}\) = +0.33

Question 4.
The focal length of a concave lens is 2 cm. Calculate the power of the lens.
Answer:
Formula:
P = \(\frac{1}{f}\)
Type of lens is concave lens.
Focal length of concave lens,
f = -2 m power of the lens.
P = \(\frac{1}{-2 m}\)
P = -0.5 dioptre

Question 5.
A needle placed at 30 cm from the lens forms an image on a screen placed 60 cm on the other side of the lens. Identify the type of lens and determine the focal length.
Answer:
u = -30 cm
v = 60 cm
u is negative because image is formed on the on the other side of the lens.

It is a convex lens.

Question 6.
A 3 cm tall bulb is placed at a distance of 20 cm from a diverging lens having a focal length of 10.5 cm. Determine the distance of the image.
Answer:
u = -20 cm
f = -10.5 cm

The distance of the image is -6.88 cm

Question 7.
A ray from medium 1 is refracted below while passing through medium 2. Find the refractive index of the second medium with respect to medium 1.
Answer:
Refractive index µ

Refractive index = 0.707

Question 8.
The optical prescription of a pair of spectacle is
Right eye: -3.5 D, Left eye: -4.00 D.
(i) Name the defect of the eye.
Answer:
Shortsighted (Myopia)

(ii) Are these lenses thinner at the middle or at the edges?
Answer:
These lenses are thinner in the middle.

(iii) Which lens has a greater focal length?
Answer:
power = \(\frac{1}{focal length}\)
Right eye: power P = -3.5 D

Left eye: Power P = -4 D

Hence the lens having power of -3.5 D has greater focal length.

Question 9.
The radii of curvature of two surfaces of a double convex lens are 10 cm each. Calculate its focal length and power of the lens in air and liquid. Refractive indices of glass and liquid are 1.5 and 1.8 respectively.
Answer:
Radius of curvature of first surface R₁ = 10 cm
Radius of curvature of second surface R₂ = 10 cm
In air

p_l = -3.33 d

Question 10.
An object 2 cm tall is placed 10 cm in front of a convex lens of focal length 15 cm. Find the position, size and nature of the image formed.
Answer:
Focal length of a convex lens f = 15 × 10^-2 m
Weight of the object ho = 2 × 10^-2 m
Let weight of the image be h_v
Distance of the object u = 10 × 10^-2 m
Distance of the image v = 15 × 10^-2 m
We know

∴ v = -30 × 10^-2 m
Distance of the image = 30 × 10^-2 m

Hence a virtual image 6 × 10^-2 m height is formed at a distance of 30 × 10^-2 m from the lens on the same side of the lens.

IX. Higher order thinking (HOT) questions.

Question 1.
Ramu passes white light through a quartz prism. For which colour refractive index is greater?
Answer:
Refractive index is maximum for violet light when white light passes through a quartz prism.

Question 2.
Sita has kept a stud consists of diamond. What will she observe? Give reason.
Answer:
The diamond stud appears bright because of total internal reflection.

Question 3.
Guna passes a ray light through a glass slab. Which optical phenomenon will take place? What can he observe with reference to wavelength?
Answer:
When a ray of light enters a glass slab he can observe refraction of light. He observed that wavelength of light decreases.

Question 4.
A prism is placed in the minimum deviation position. Chari has passed a ray of light at an angle of 45°, then what is the value of angle of emergence? Why?
Answer:
The angle of emergence = 45°.
Since, in the minimum deviation positron, angle of incidence is equal to angle of emergence.

Question 5.
Mani is using a lens of power 2 dioptre. What is the focal length of the lens?
Answer:
Focal length = \(\frac{1}{power}\)
F= \(\frac{1}{2}\) = 0.5 m

Question 6.
Surya has placed a lens of power 1 D in side water. What will happen to power of the lens?
Answer:
The power of the lens will be more than original power.

Question 7.
Sonu has observed some lines in solar spectrum are absorbed by the elements present in the atmosphere. What are the lines?
Answer:
The lines are Fraunhofer lines.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.7

Question 1.
Solve the following system of homogeneous equations.
(i) 3x + 2y + 7z = 0; 4x – 3y – 2z = 0; 5x + 9y + 23z = 0
Solution:

ρ(A) = 2 ρ[A | B] = 2
ρ(A) ρ[A | B] = 2 < n
The system is consistent. It has non trivial solution.
Writing the equivalent equations from echelon form
3x + 2y + 7z = 0 ………… (1)
-17y – 34z = 0 ……….. (2)
Put z = t
(2) ⇒ -17y = 34t
y = \(\frac {34t}{-17}\) = -2t
(1) ⇒ 3x + 2(-2t) + 7t = 0
3x – 4t + 7t = 0
3x + 3t = 0
3x = -3t
x = -t
(x, y, z) (-t, -2t, t) ∀ t ∈ R

(ii) 2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:

ρ(A) = 3 ρ[A | B] = 3
ρ(A) = ρ[A | B] = 3
The system is consistent. It has trivial solution.
x = 0, y = 0, z = 0

Question 2.
Determine the values of λ for which the following system of equations.
x + y + 3z = 0; 4x + 3y + λz = 0, 2x + y + 2z = 0 has
(i) a unique solution
(ii) a non-trivial solution.
Solution:

Case (i):
if λ ≠ 8
ρ(A) = 3 ρ(A | B) = 3
ρ(A) = ρ(A | B) = 3 = n
The system is consistent. It has unique (trivial) solution.
∴ Solution x = 0, y = 0, z = 0

Case (ii):
if λ = 8
ρ(A)= ρ(A | B) = 2
ρ(A) = ρ(A | B) = 2 < n
The system is consistent. It has non trivial solution.

Question 3.
By using Gaussian elimination method, balance the chemical -reaction equation:
C₂H₆ + O₂ → H₂O + CO₂.
Solution:
We are searching for positive integers x₁, x₂, x₃ and x₄
x₁ C₂H₆ + x₂ O₂ → x₃ H₂O + x₄ CO₂ ……….(1)
The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1) so we get a linear
homogeneous equation.
2x₁ x₄ = 2x₁ – x₄ = 0 ……..(2)
6x₁ = 2x₃ = 6x₁ – 2x₃ = 0
÷ 2 ⇒ 3x₁ – x₃ = 0 ………(3)
2x₂ = x₃ + 2x₄ ⇒ 2x₂ – x₃ – 2x₄ = 0 ……… (4)
Equation (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns.
Augmented matrix

The system is consistent and has an infinite number of solutions.
Writing the equations using the echelon form we get
-2x₃ + 3x₄ = 0 ……….. (1)
2x₂ – x₃ – 2x₄ = 0 ………… (2)
2x₁ – x₄ = 0 …………. (3)
Put x₄ = t
(3) ⇒ 2x₁ – t = 0
x₁ = \(\frac {t}{2}\)
(1) ⇒ -2x₃ + 3x₄ = 0
-2x₃ = -3t
x₃ = \(\frac {3}{2}\) t
(2) ⇒ 2x₂ – x₃ – 2x₄ = 0
2x₂ – \(\frac {3}{2}\) t – 2t = 0
2x₂ = \(\frac {3}{2}\) t + 2t = \(\frac {7t}{2}\)
x₂ = \(\frac {7t}{4}\)
(x₁, x₂, x₃, x₄) = (\(\frac {t}{2}\), \(\frac {7t}{4}\), \(\frac {3}{2}\)t, t) ∀ t ∈ R
since x₁, x₂, x₃ and x₄ are positive integers.
Let us choose t = 4

So the balanced equation is
2C₂H₆ + 7O₂ → 6H₂O + 4CO₂.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.6

Question 1.
Test for consistency and if possible, solve the following systems of equations by rank method.
(i) x – y + 2z = 2, 2x + y + 4z = 7, 4x – y + z = 4
Solution:
Matrix form

The system is consistent.
ρ(A) ρ[A|B] = 3 = n
it has unique solution.
Writing the equivalent equations from echelon form
x – y + 2z = 2 ………… (1)
3y = 3 ⇒ y = 1
-7z = -7
z = 1
(1)⇒ x – y + 2z = 2
x – 1 + 2 = 2
x = 1
∴ Solution is x = 1, y = 1, z = 1

(ii) 3x + y + z = 2, x – 3y + 2z = 1, 7x – y + 4z = 5
Solution:

ρ(A) = 2 ρ[A | B] = 2
ρ(A) = ρ[A | B] = 2 < n
The system is consistent. It has infinitely many solution.
Writing the equivalent equations from echelon form.
x – 3y + 2z = 1 ………. (1)
10y – 5z = -1 ………. (2)
Put z = t.
(2) ⇒ 10y – 5z = -1
10y = -1 + 5z = 5t – 1

(iii) 2x + 2y + z = 5, x – y + z = 1, 3x + y + 2z = 4
Solution:

ρ(A) = 2 ρ[A | B] = 3
ρ(A) ≠ ρ[A | B] = 2 < n
∴ The system is inconsistent. It has no solution.

(iv) 2x – y + z = 2, 6x – 3y + 3z = 6, 4x – 2y + 2z = 4
Solution:

ρ(A) = 1 ρ[A | B] = 1
ρ(A) = ρ[A | B] = 1 < n.
∴ The system reduces into a single equation.
∴ It is consistent and has infinitely many solutions.
Writing the equivalent equations from echelon form
2x – y + z = 2
Put y = s, z = t
2x – s + t = 2
2x = 2 + s – t
x = \(\frac {2+s-t}{2}\)
(x, y, z) = (\(\frac {2+s-t}{2}\), s, t) ∀ s, t ∈ R

Question 2.
Find the value of k for which the equations kx – 2y + z = 1, x – 2ky + z = -2, x – 2y + kz = 1 have
(i) no solution
(ii) unique solution
(iii) infinitely many solution.
Solution:

[∵ 2 – k – k² = -(k² + k – 2)
= -(k + 2)(k – 1)
= (k + 2)(1 – k)]
case (i)
If k = 1
ρ(A) = 2, ρ(A | B) = 3.
ρ(A) ≠ ρ(A | B)
The system is inconsistent and it has no solution.

Case (ii)
If k ≠ 1, k ≠ -2
ρ(A) = 3, ρ(A | B) = 3 = n
The system is consistent and it has unique solution.

Case (iii)
If k = -2
ρ(A) = 2, ρ(A | B) = 2
The system is consistent and it has infinitely many solution.

Question 3.
Investigate the values of λ and µ the system of linear equations 2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
Solution:

Case (i)
If λ = 5, µ ≠ 9
ρ(A) = 2, ρ(A | B) = 3
ρ(A) ≠ ρ(A | B)
The system is inconsistent. It has no solution.

Case (ii)
If λ = 5, µ ≠ 9
ρ(A) = 3, ρ(A | B) = 3
ρ(A) = ρ(A | B) = 3 = n
The system is consistent. It has unique solution.

Case (iii)
If λ = 5, µ = 9
ρ(A) = 2, ρ(A | B) =2
ρ(A) = ρ(A | B) = 2 < n
The system is consistent. It has infinitely many solution.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
Solution:
Augmented matrix

Writing the equivalent equations from echelon from.
x – y + 2z = 3 …………. (1)
5y – 6z = -4 ………….. (2)
-z = -4
z = 4
(2) ⇒ 5y – 6z = -4
5y – 24 = -4
5y = -4 + 24
5y = 20
y = 4
(1) ⇒ x – y + 2z = 3
x – 4 + 8 = 3
x = 3 + 4 – 8
x = -1
∴ x = -1, y = 4, z = 4

(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2.
Solution:
Augmented matrix

Writing the equivalent equations from echelon from.
x + 2y + 3z = 11 …………. (1)
y – 2z = -3 ………….. (2)
11z = 22
z = 2
(2) ⇒ y – 2z = -3
y – 4 = -3
y = -3 + 4
y = 1
(1) ⇒ x + 2y + 3z = 11
x + 2(1) + 3(2) = 11
x + 2 + 6 = 11
x = 11 – 8 = 3
∴ x = 3, y = 1, z = 2

Question 2.
If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Solution:
Given: f(x) = ax² + bx + c
In Remainder Theorem
f(-3) = 21
a(-3)² + b(-3) + c = 21
9a – 3b + c = 21 ……….. (1)
f(5) = 61
25a + 5b + c = 61 …………. (2)
f(1) = 9
a + b + c = 9 ………… (3)
Augmented matrix

Writing the equivalent equations from echelon from.
a + b + c = 9 …………. (1)
b + 2c = 5 ………….. (2)
-4c = -8
c = 2
(2) ⇒ b + 2c = 5
b + 4 = 5
b = 5 – 4
b = 1
(1) ⇒ a + b + c = 9
a + 1 + 2 = 9
a = 9 – 3
a = 6
a = 6, b = 1, c = 2

Question 3.
An amount of Rs 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is Rs 5,000. The income from the third bond is Rs 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amounts of 3 bounds be x, y, z
x + y + z = 65,000

Writing the equivalent equations from echelon from.
x + y + z = 65000 …………. (1)
2y + 3z = 90000 ………….. (2)
21z = 42000
z = 20000
(2) ⇒ 2y = 90000 – 3 × 20000
2y = 30000
y = 15000
(1) ⇒ x + 15000 + 20000 = 65000
x = 30000
∴ x = 30000, y = 15000, z = 20000

Question 4.
A boy is walking along the path y = ax² + bx + c through the points (-6, 8),(-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Solution:
y = ax² + bx + c
At(-6, 8) ⇒ 8 = 36a – 6b + c ………… (1)
At(-2, -12) ⇒ -12 = 4a – 2b + c ………… (2)
At(3, 8) ⇒ 8 = 9a + 3b + c ………… (3)
Augmented matrix

Writing the equivalent equations from the echelon.
36a – 6b + c = 8 …………. (1)
3b – 2c = 29 ………….. (2)
5c = -50
c = -10
(2) ⇒ 3b – 2c = 29
3b – 20 = 29
3b = 9
b = 3
(1) ⇒ 36a – 18 – 10 = 8
36a = 8 + 18 + 10
36a = 36
a = 1
At P (7, 60), y = ax² + bx + c
60 = 1(7²) + 3(7) – 10
60 = 49 – 21 – 10
60 = 60
He will meet his friend at P (7, 60)

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Multiple Choice Questions

Question 1.
The value of sin2 θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to ………………
(1) tan² θ
(2) 1
(3) cot² θ
(4) 0
Answer:
(2) 1
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1

Question 2.
tan θ cosec² θ – tan θ is equal to ………………
(1) sec θ
(2) cot² θ
(3) sin θ
(4) cot θ
Answer:
(4) cot θ
Hint:
tan θ cosec² θ – tan θ = tan θ (cosec² θ – 1)
= tan θ × cot² θ = \(\frac{1}{\cot \theta}\) × cot² θ = cot θ

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
(sin α + cos α)² + (cos α + sec α)²
= sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α
= 1 + cosec² α + 2 + sec² α + 2
= 1 + cot² α + 1 + 2 + tan² α + 1 + 2
= 7 + tan² α + cot² α
k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b (a2 – 1) is equal to ……………
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a
Hint:
b (a² – 1) = (sec θ + cosec θ) [(sin θ + cos θ)² – 1]
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\) [sin² θ + cos² θ + 2 sin θ cos θ – 1]

Question 5.
If 5x = sec θ and \(\frac { 5 }{ x } \) = tan θ, then x2 – \(\frac{1}{x^{2}}\) is equal to …………….
(1) 25
(2) \(\frac { 1 }{ 25 } \)
(3) 5
(4) 1
Answer:
(2) \(\frac { 1 }{ 25 } \)
Hint:



Question 6.
If sin θ = cos θ , then 2 tan² θ + sin² θ – 1 is equal to ………………
(1) \(\frac { -3 }{ 2 } \)
(2) \(\frac { 3 }{ 2 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { -2 }{ 3 } \)
Answer:
(2) \(\frac { 3 }{ 2 } \)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then …………..
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:
x = a tan θ
\(\frac { x }{ a } \) = tan θ
\(\frac{x^{2}}{a^{2}}\) = tan² θ
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = sec² θ – tan² θ = 1
y = b sec θ
\(\frac{y}{b}\) = sec θ
\(\frac{y^{2}}{b^{2}}\) = sec² θ

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to ……………
(1) 0
(2) 1
(3) 2
(4) -1
Answer:
(3) 2
Hint:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b-a
Solution:
(2) b² – a²
(a cot θ + b cosec θ)² = p²
(b cot θ + a cosec θ )² = q²
p² – q² = a² cost²θ + a² cot²θ + 2ab cot θ cosec θ – (b²cot²θ + a² cosec²θ + 2ab cot θ cosec θ) = (a² – b²) cot²θ + (b² – a²)cosec²θ = (a² – b²) (cosec²θ – 1) + (b² – a²) (cosec²θ)
= (a² – b²)cosec²θ – (a² – b²) – (a² – b²) cosec²θ
= b² – a²

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt { 3 }\) : 1, then the angle of elevation of the sun has a measure
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°
Hint:
Ratio of length of the tower : length of the shadow = \(\sqrt { 3 }\) : 1

Let the tower be \(\sqrt { 3 }\) x and the shadow be x
tan C = \(\frac { AB }{ BC } \) ⇒ tan C = \(\frac{\sqrt{3} x}{x}\) = \(\sqrt { 3 }\)
tan C = tan 60° ⇒ ∴ ∠C = 60°

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘6’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to ……………
(1) \(\sqrt { 3 }\) b
(2) \(\frac { b }{ 3 } \)
(3) \(\frac { b }{ 2 } \)
(4) \(\frac{b}{\sqrt{3}}\)
Answer:
(3) \(\frac { b }{ 2 } \)
Hint:
Let the height of the pole BC be h
AC = b + h
Let CD be x
In the right ∆ BCD, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ x } \)
x = \(\sqrt { 3 }\) h ………. (1)

In the right ∆ ACD, tan 60° = \(\frac { AC }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { b+h }{ x } \)
x = \(\frac{b+h}{\sqrt{3}}\) ………(2)
From (1) and (2) we get
\(\sqrt { 3 }\) h = \(\frac{b+h}{\sqrt{3}}\) ⇒ 3 h = b + h
2 h = b ⇒ h = \(\frac { b }{ 2 } \)

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to
(1) 41. 92 m
(2) 43. 92 m
(3) 43 m
(4) 45. 6 m
Answer:
(2) 43. 92 m
Hint:
In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \) = \(\frac { 60 }{ x+y } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 60 }{ x+y } \) ⇒ x + y = 60 \(\sqrt { 3 }\)

y = 60 \(\sqrt { 3 }\) – x …….(1)
In the right ∆ ABD, tan 45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 60 }{ y } \) ⇒ y = 60 ………..(2)
From (1) and (2) we get
60 = 60 \(\sqrt { 3 }\) – x
x = 60 \(\sqrt { 3 }\) – 60 = 60 (\(\sqrt { 3 }\) – 1) = 60 (1.732 – 1)
= 60 × 0.732
x = 43.92 m

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is …………….
(1) 20,10\(\sqrt { 3 }\)
(2) 30, 5 \(\sqrt { 3 }\)
(3) 20, 10
(4) 30, 10\(\sqrt { 3 }\)
Answer:
(4) 30, 10\(\sqrt { 3 }\)
Hint:
Let the height of the multistoried building AB be “h”
AE = h – 20
Let BC be x
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \) ⇒ \(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ………..(1)

In the right ∆ ABC, tan 30° = \(\frac { AE }{ ED } \) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
x = (h – 20) \(\sqrt { 3 }\) ………(2)
From (1) and (2) we get,
\(\frac{h}{\sqrt{3}}\) = (h – 20) \(\sqrt { 3 }\)
h = 3h – 60 ⇒ 60 = 2 h
h = \(\frac { 60 }{ 2 } \) = 30
Distance between the building (x) = \(\frac{h}{\sqrt{3}}=\frac{30}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}=10 \sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is ……………….
(1) \(\sqrt { 2 }\)x
(2) \(\frac{x}{2 \sqrt{2}}\)
(3) \(\frac{x}{\sqrt{2}}\)
(4) 2 x
Answer:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:
Consider the height of the 2^nd person ED be “h”
Height of the second person is 2 h
C is the mid point of BD
In the right ∆ ABC, tan θ = \(\frac { AB }{ BC } \)

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β . The angle of depression of its reflection in the lake is 45° . The height of the location of the cloud from the lake is ………….

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 1.
Solve the following systems of linear equations by Cramer’s rule:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
Solution:

(ii) \(\frac{3}{x}\) + 2y =12, \(\frac{2}{x}\) + 3y = 13
Solution:
Let \(\frac{1}{x}\) = a
3a + 2b = 12
2a + 3b = 13

(iii) 3x + 3y – z = 1 1, 2x – y + 2z = 9, 4x + 3y + 2z = 25
Solution:
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= 3(-2 – 6) -3 (4 – 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 – 10 = -22 ≠ 0
Δ_x = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= 11 (-2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 – 52 = -44
Δ_y = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) -11(4) -1(14)
= -96 + 44 – 14 = -66
Δ_x = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
= 3(-25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 – 42 + 110
= -88

Solution:

3a – 4b – 2c = 1 ……….. (1)
a + 2b + c = 2 …………… (2)
2a – 5b – 4c = -1 ………….. (3)
Δ = \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| \)
= 3(-8 + 5) + 4 (-4 – 2) – 2(-5 – 4)
= 3(-3) +4(-6) -2(-9)
= -9 – 24 + 18
= -15 ≠ 0
Δ_a = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= 1(-8 + 5) + 4(-8 + 1) – 2(-10 + 2)
= 1(-3) + 4(-7) – 2(-8)
= -3 – 28 + 16
= -15
Δ_b = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
= 3(-8 + 1) – 1(-4 – 2) – 2(-1 – 4)
= 3(-7) -1(-6) – 2(-5)
= – 21 + 6 + 10 = -5
Δ_c = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4)
= 24 – 20 – 9 = -5

Question 2.
In a competitive examination, one mark is awarded for every correct answer while \(\frac{1}{4}\) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer’s rule to solve the problem).
solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x
and the No. of questions answered wrongly be y
Here, x + y = 100 and x – \(\frac { 1 }{ 4 }\) y = 80
(i.e) x + y = 100 and 4x – y = 320

correct questions = 84
wrong questions = 16.

Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramer’s rule to solve the problem).
solution:
Let two solutions x and y
x + y = 10 …….. (1)
0.25 x + (0.50)y = (0.40) ……….. (2)
(2) × 100 ⇒ 25x + 50y = 400
(2) ÷ 5 ⇒ 5x + 10y = 80 …………. (3)

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer’s rule to solve the problems).
solution:
pump A fills \((\frac {1}{x})^{th}\) of the tank in 1 hour.
pump B fills \((\frac {1}{y})^{th}\) of the tank in 1 hour.
Both can filled \((\frac {1}{10})^{th}\) of the tank in 1 hour.

using Cramer’s rule

Pump A takes 15 minutes
Pump B takes 30 minutes

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
solution:
Let the Cost of dosai, Idlies and vadais be x, y, z
2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= 2(4 – 16) – 3(4 – 20) + 2(8 – 10)
= 2(-12) – 3(-16) + 2(-2)
= -24 + 48 – 4
= 20 ≠ 0
Δ_x = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
= 150(4 – 16) – 3(400 – 1000) + 2(800 – 500)
= 150(-12) – 3(-600) + 2(300)
= -1800 + 1800 + 600
= 600
Δ_y = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| \)
= 2(400 – 1000) – 150(4 – 20) + 2(500 – 1000)
= 2(-600) – 150(-16) + 2(-500)
= -1200 + 2400 – 1000
= 200
Δ_z = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| \)
= 2(500 – 800) – 3(500 – 1000) + 150(8 – 10)
= 2(-300) – 3(-500) + 150(-2)
= -600 + 1500 – 300
= 600

x = Rs 30, y = Rs 10, z = Rs 30
There are 3 dosai, 6 idlies and 6 vadais
= 3x + 6y + 6z
= 3(30) + 6(10) + 6 (30)
= 90 + 60 + 180
= Rs. 330
They can eat within the amount.

Students can download Maths Chapter 6 Trigonometry Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 6 Trigonometry Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile