Search Results for “” – Samacheer Kalvi

Samacheer Kalvi 10th English Model Question Paper 5

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th English Model Question Paper 5 Pdf, Samacheer Kalvi 10th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamil Nadu Samacheer Kalvi 10th English Model Question Paper 5

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 I to 14 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 four sections. The questions are of two marks each. Question numbers 15 to 18 in Section I and Question numbers 19 to 22 in Section II are to be answered in about one or two sentences each. Question numbers 23 to 28 in Section III and IV are to be answered as directed.
Question numbers 29 to 45 in Part III are of five marks each and have been divided in five sections. These are to be answered as directed.
Question numbers 46 and 47 in Part IV are of eight marks each. Question number 47 has four questions of two marks each. These are to be answered as directed.

Time: 2.30 Hours
Maximum Marks: 100

Part – I

Answer all the questions. [14 x 1= 14]
Choose the most suitable answer and write the code with the corresponding answer.
Choose the appropriate synonyms for the italicised words.

Question 1.
The mother seagull swooped upwards.
(a) leap
(b) rush
(c) move very quickly
(d) ascend
Answer:
(a) leap

Question 2.
The attic has always been favourite with children.
(a) loft
(b) terrace
(c) apartment
(d) Strong Room
Answer:
(a) loft

Question 3.
It is a 55-foot sailing vessel built indigenously in India.
(a) fully
(b) collectively
(c) innately
(d) specially
Answer:
(c) innately

Choose the appropriate antonym for the italicised words.

Question 4.
She screamed back mockingly.
(a) disrespectfully
(b) ridiculously
(c) jeeringly
(d) respectfully
Answer:
(d) respectfully

Question 5.
We don’t have to use any means of repulsion.
(a) attraction
(b) distaste
(c) hate
(d) horror
Answer:
(a) attraction

Question 6.
I indulged in banking.
(a) took part
(b) participated
(c) abstained
(d) yielded
Answer:
(c) abstained

Question 7.
Choose the correct plural form of alga from the following:
(a) algum
(b) algi
(c) algae
(d) algas
Answer:
(c) algae

Question 8.
Form a derivative by adding the right suffix to the word – document.
(a) -ory
(b) -ise
(c) -ation
(d) -ly
Answer:
(c) -ation

Question 9.
Choose the correct expansion of the abbreviation SIM.
(a) Subscriber Information Module
(b) Subscriber Identification Module
(c) Student Identification Module
(d) School Identification Module
Answer:
(b) Subscriber Identification Module

Question 10.
Complete the following sentence with the most appropriate phrasal verb given below:
The crew …………………….. of water and food before they could complete their expedition.
(a) ran on
(b) ran about
(c) ran in
(d) ran out
Answer:
(d) ran out

Question 11.
Choose the suitable option to pair it with the word ‘ watch ’ to form a compound word.
(a) hall
(b) house
(c) man
(d) clock
Answer:
(c) man

Question 12.
Fill in the blank with the most appropriate preposition given below:
Mulan heard this ………….. her tent.
(a) by
(b) from
(c) at
(d) for
Answer:
(b) from

Question 13.
Complete the following sentence using the most appropriate tense form of the verb given below:
After he ………………. his lunch, he went across to the window.
(a) will finish
(b) finish
(c) was finishing
(d) had finished
Answer:
(d) had finished

Question 14.
Choose the most appropriate linker from the given four alternatives.he is ninety years old, he is in the pink of health.
(a) When
(b) Since
(c) Even though
(d) yet
Answer:
(c) Even though

PART – II [10 x 2 = 20]
Section -1

Answer any THREE of the following questions in a sentence or two, [3 x 2 = 6]

Question 15.
Mention the special features of INSV Tarini.
Answer:
Indian Naval Ship Vessel Tarini commissioned to the Indian Navy service on 18th February, 2017, is the second sailboat of the Indian Navy. It is a 55-foot sailing vessel built indigenously in India by M/s Aquarius Shipyard Pvt. Ltd, located in Goa. After undergoing extensive sea trials, she was commissioned to the Indian Navy service.

Question 16.
What prompted the seagull to fly finally?
Answer:
The seagull was quite hungry and yearned for food. When he saw a piece of fish in the beak of his mother, the sight was quite tempting for him. He was maddened at the sight of the food and suddenly dived at the fish forgetting that he didn’t know how to fly. Thus the young seagull was prompted to fly finally into space.

Question 17.
What was the daily routine of Mr. Sanyal?
Answer:
Sanyal would come every day to Nagen’s Tea Shop, sit at a comer table, have tea and biscuits, pay for the same and leave before it got dark. This was his regular duty after his wife and only son passed away a year ago.

Question 18.
What were the various things that tempted Franz to spend his day outdoors?
Answer
Franz found the warm and bright day to be a temptation to spend his day outdoors. The birds were chirping at the edge of woods. The Pmssian soldiers were drilling in the open field at the back of sawmill. He felt he could gladl> spend life outdoors. Above all, Franz was not prepared for the test on participles.

Section – II

Read the following sets of poetic lines and answer any THREE of the following.
[3 x 2 = 6]

Question 19.
“Let us learn to walk with a smile and a song,
No matter if things do sometimes go wrong;”
(a) What does the poet want everyone to learn?
(b) What should we do when things go wrong?
Answer:
(a) The poetess expects everyone to learn to walk with a smile and a song even when things go wrong.
(b) Even when things go wrong, we need to feel happy and be cheerful.

Question 20.
“She’s a lioness; don’t mess with her.
She ’ll not spare you if you ’re a prankster. ”
(a) How is a woman described here?
(b) Who is a prankster?
Answer:
(a) A woman is described as lionesses whom others should be afraid to play with.
(b) A prankster is a mischievous or malicious person who plays tricks on others.

Question 21.
“Not a flower could he see,
Not a leaf on a tree.”
(a) Who does ‘he’ refer to?
(b) Mention the season when he could not see a flower or a leaf on a tree.
Answer:
(a) ‘He’ refers to the Cricket.
(b) The season was winter. So he could not see a flower or a leaf on the tree.

Question 22.
“Beside the house sits a tree It never grows leaves.”
(а) What is found near the house?
(b) Why does it never grow leaves?
Answer:
(a) Near the house is found a tree with no leaves.
(b) It never grows leaves because it is eerie.

Section – III

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

Question 23.
Rewrite the following sentence to the other voice.
Answer:
Please assemble in the ground.
You are requested to assemble in the ground.

Question 24.
Rewrite using indirect speech.
Answer:
“Where are we going, sir?” asked the aero-coachman.
The aero-coachman asked the passenger where they were going.

Question 25.
Punctuate the following sentence.
Answer:
Wherefore said miranda did they not that hour destroy us
“Wherefore,” said Miranda, “did they not that hour destroy us?”

Question 26.
Transform the following sentence into a simple sentence.
Answer:
As Catherin is a voracious reader, she buys a lot of books.
Being a voracious reader. Catherin buys a lot of books.

Question 27.
Rearrange the words in the correct order to make meaningful sentences:
(a) he saw / When / in the / platform/ the train / he rushed.
(b) as /I/ healthy / are you / am / as.
(a) He rushed, when he saw the train in the platform.
Or
(a) When he saw the train in the platform he rushed.
(b) I am as healthy as you are.

Section – IV

Answer the following question:

Question 28.
A stranger wants to visit the library. Write the steps to guide him to reach him destination.
Answer

Go straight and take the first right on Anna Road.
Proceed further and take the second left after the bank on Big street.
You will find the Library on the left after you take the right turn.

Section -I
Part – III [10 x 5 = 50]

Answer any TWO of the following in utmost 10 lines. [2 x 5 = 10]

Question 29.
Describe the struggles undergone by the young seagull to overcome its fear of flying.
Answer:
The Young Seagull was afraid to fly and was alone on his ledge. He was more frightened than his siblings. His attempts failed. He had taken a little run forward and tried to flap his wings. But that was all he could do. He felt that his wings would not support him. He failed to muster up the courage and fly. His parents taunted, scolded and threatened him to leave him starving at the ledge unless he flew away. But nothing could make him fly.

The seagull helplessly watched his parents flying with his brothers and sisters. The whole family went on taunting him for his cowardice. Only his mother was looking at him. She had picked a piece of fish and was flying across to him with it. He leaned out eagerly. The mother was very near to him with the fish in her beak. Maddened by hunger, he dived at the fish. With a loud scream, he fell outwards and downwards into space. A terror seized him.

His heart stood still. His mother swooped past him. He answered her with another scream. He saw his two brothers and sister flying around him. The seagull completely forgot that he was not able to fly. He let himself free to dive, soar and curve at will. He was shrieking shrilly. He saw a green sea beneath him. He was tired and weak with hunger. His feet sank into the green sea and his belly touched it. He sank no farther. Now, his family was praising him and their beaks were offering him scraps of fish. He had made his first flight.

“Flying is learning how to throw yourself to the ground and miss.”

Question 30.
‘Technology is a boon to the disabled’. Justify.
Answer:
The differently abled, Alisha and David’s life has been transformed because of Technology. It is a boon to the disabled. Assistive technology is designed to help people with disabilities. Typing was impossible, but now Dragon Dictate helps the disabled to speak for words to be printed on screen. Technology can control a computer screen even with Eye Gaze.

Technology is vital to be free and independent. For verbal communication, Liberator Communication Device, with eye movements to communicate verbally is used. It has a Bluetooth adaptor and took a couple of weeks to learn using it. Communicating with people was very difficult before. An ACTIV controller also in the headrest of a mobility chair is used to control TV, Blu-Ray and music players.

Augmentative and Alternative Communication is also the product of technological advancements and a boon to the disabled. EC02 linked to an interactive whiteboard is used to teach PE lessons. Technology also helps control the Play Station, MP3, electric wheelchair and ECO point Eye Gaze to communicate, access the computer. Thus/Technology is a boon to the disabled’.
“Technology’ is not just a tool. It can give learners a voice
that they may not have had before.”

Question 31.
How did Watson help his friend to arrest the criminal?
Answer:
Dr. Watson is called by Mrs. Hudson to tend Holmes, who is apparently dying of a rare tropical disease, Tarpaunli fever, contracted while he was on a case at Rotherhithe. Holmes forbids Watson to go near him because the illness is highly infectious. Although Watson wishes to examine Holmes himself or send for a specialist, Holmes demands that Watson wait several hours before seeking help.

While Watson waits, he examines several objects in Holmes’s room. Holmes is angered when Watson touches things on his table stating his dislike for people fidgeting his things. At six o’clock, Holmes asks Watson to turn the gaslight on, but only half-full. He then instructs Watson to bring Mr. Culverton Smith from 13 Lower Burke Street to see Holmes but to make sure that Watson returns to Baker Street before Smith arrives. Watson goes to Smith’s address. Although Smith refuses to see anyone, Watson forces his way in. Once Watson explains his duty on behalf of Sherlock Holmes, Smith’s assertiveness changes significantly. Smith agrees to come to Baker Street within half an hour.

Watson excuses himself, saying that he has another appointment, and returns to Baker Street before Smith’s arrival. Believing that they are alone, Smith is frank with Holmes. It soon appears, to the hiding Watson’s horror, that Holmes has been sickened by the same illness that killed Smith’s nephew Victor. At the fend, Holmes calls for Watson to come out from behind the screen, to present himself as another witness to the conversation. Holmes explains his illness was feigned as a trick to induce Smith to confess to his nephew’s murder. Starving himself for three days and claiming to be suffering from a deadly infectious disease was to keep Watson from examining him and discovering the ruse, since, as he clarifies, he has every respect for his friend’s medical skills.
‘A detectives perspective traps the criminal.’

Question 32.
‘Man does change with time’-What were the various changes that came about in Aditya?
Answer:
Man does change with time. According to Sanyal, Aditya was a boy who could never compete him in scholastic or non-scholastic activities. He was a liar and an envious personality. But, years later Aditya no longer is the same personality. He is more matured and responsible. His childish nature makes him feel guilty for stealing the medal that belonged to Sanyal. Aditya now wanted to restore Sanyal, his lost happiness. Knowing about his financial constraints, he decides to give him the cost of the silver medal. He drives with a firm determination to his ancestral house. When he reaches, he rushes to the attic. Aditya gets on top of a packing case

and pushes his hand inside the ventilator, upsetting a sparrow’s nest. However, he heaves a sigh of relief when he gets what he had been looking for. He goes straight to the Jeweller to determine the current rate of the medal in the market. All these surely show a remarkable change in him. Even when Sanyal behaves rudely, Aditya is courteous and friendly. His only intention is penitence. He is so mature that he even understands and justifies the grievances Sanyal had in his heart. When Sanyal prefers the medal, without second thoughts he restores it to Sanyal whole-heartedly.
“Change is universal.”

Section – II

Answer any TWO of the following in utmost 10 lines. [2 x 5 = 10]

Question 33.
How is mystery depicted in the poem ‘The House on Elm Street’.
Answer:
Nadia Bush the poetess talks about a house that stood alone in an isolated place on Elm Street. No one knows what happens inside that house and it is very mysterious. Even by looking from outside, one can easily say that it has a vast space but at the same time it is bare to the bone meaning there is no one living inside the mysterious house or no basic necessities inside the house on Elm Street.

Generally at night, the house looks like it is alive with people in if Lights are switched on and off. It gives a feel that someone is inside the house. When Nadia sees such a situation, she is tempted to go inside the house to just take a look and see what is actually happening inside the house. However, Nadia the poetess is frightened and never dares to do so. Every day the poetess, drives past the house. The house seems to look a bit brighter.

The very thought of this mysterious house plays with your mind since it is just one house of this kind in the area around. Next to the house, is also a tree with no leaves the whole year round. It doesn’t grow tall nor does it shrink in size. The tree too is mysterious like the house since it has no leaves in any of the seasons to make wonder how a tree could survive without any leaves or without any growth. Every day, the house also begins to fade making it all the more mysterious.
“The best secrets are the most twisted.”

Question 34.
Compare and contrast the attitude of the ant and the cricket.
Answer:
The poem, ‘The Ant and the Cricket’ is about a careless cricket and a hardworking ant. We all know that ants are hardworking creatures but on the other hand, the Cricket is portrayed as a lazy being. He made no efforts to plan for the future. Thus, the cricket is shown as a young and silly creature because he sang all through summer and spring with no worries in the world for winter that was to come. But when winter arrived, he began to complain that he would die of starvation and hunger.

He found that his cupboard was empty with no piece of bread to eat. He neither found a leaf, nor a flower. Everything was covered with snow. Therefore, the cricket cried and moaned as he perceived his bad future. Finally, deprived of hunger and starvation, being all wet and cold, the cricket journeyed to the house of the stingy ant. Becoming bold by nature, he begged for food and shelter. But the Ant is wise, foresees its needs for winter and prudently turned down the foolish Cricket. The Ant clearly stated that Ants neither borrow nor lend.

So the Ant is shown as a straightforward and outspoken personality. Perhaps the cricket here is portrayed as a person who makes false promises and the Ant is portrayed to be a strong personality not carried away by the false promises of one. Therefore the Ant could never be exploited though it was humble enough to admit it to be a servant and friend of the Cricket. The Cricket was surely careless and reckless against the Ant who was judicious, discrete and level-headed.
“Accept responsibility for your life. Know that it is
you who will get you where you want to go, no one else”

Question 35.
Read the following stanza and answer the questions given below.
“The weather is always too hot or cold;
Summer and winter alike they scold.
Nothing goes right with the folks you meet Down on the gloomy Complaining street.’’
(i) Pick out the rhyming words from the above lines.
(ii) Write the rhyme scheme of the given stanza.
(iii) Identify the figure of speech employed in the fourth line of the given stanza.
(iv) Pick out the alliterating words.
Answer:
(i) The rhyming words are cold-scold and meet-street.
(ii) The rhyme scheme of the poem is aabb.
(iii) The figure of speech employed in the fourth line of the given stanza is Personification.
(iv) There aren’t any consonant sounds repeated in the words in a line. However the vowel ‘ sounds in and alike are repeated in the second line.

Question 36.
Paraphrase the following stanza.
We can pull and haul and push and lift and drive,
We can print and plough and weave and heat and light,
We can run and race and swim and fly and dive,
We can see and hear and count and read and write!
Answer:
In this stanza, the machines elucidate how they can serve human race by doing all the possible human activities mastered by them. They say they can do various tasks such as pulling, carrying, pushing, lifting, driving, printing, ploughing, weaving, heating, lighting, running, racing, swimming, flying, diving, seeing, hearing, counting, reading and writing.

Section – III

Answer any ONE of the following: [1 x 5 = 5]

Question 37.
Rearrange the following sentences in coherent order.
(i) Using his powers, Prospero released the good spirits from large bodies of trees.
(ii) Prospero and Miranda came to an island and lived in a cave.
(iii) He raised a violent storm in the sea to wreck the ship of his enemies.
(iv) The king of Naples and Antonio the false brother, repented the injustice they had done to Prospero.
(v) He ordered Ariel to torment the inmates of the ship.
Answer:
Rearranged number sequence: (ii), (i), (iii), (v), (iv)
(ii) Prospero and Miranda came to an island and lived in a cave.
(i) Using his powers, Prospero released the good spirits from large bodies of trees.
(iii) He raised a violent storm in the sea to wreck the ship of his enemies.
(v) He ordered Ariel to torment the inmates of the ship.
(iv) The king of Naples and Antonio the false brother, repented the injustice they had done to Prospero.

Question 38.
Read the following passage and answer the questions that follow.
Answer
The country Shining was governed by a despotic leader who though a warrior, had a great and cowardly shrinking from anything suggestive of failing health and strength. This caused him to send out a cruel proclamation. The entire province was given strict orders to immediately put to death all aged people. Those were barbarous days, and the custom of abandoning old people to die was not uncommon.

The poor farmer loved his aged mother with tender reverence, and the order filled his heart with sorrow. But no one ever thought twice about obeying the mandate of the governor, so with many deep and hopeless sighs, the youth prepared for what at that time was considered the kindest mode of death.

(i) Who governed Shining?
(ii) What was the cowardly act of the governor?
(iii) What proclamation did the governor send out?
(iv) How did the poor farmer treat his mother?
(v) Did the people obey the governor’s order?
Answer:
(i) Shining was governed by a despotic leader.
(ii) The governor had a great and cowardly shrinking from anything suggestive of failing health and strength.
(iii) The governor sent out a cruel proclamation in the entire province giving strict orders to immediately put to death all aged people.
(iv) The poor farmer loved his aged mother with tender reverence.
(v) Yes, the people obeyed the governor’s order.

Section – IV

Answer any FOUR of the following. [4 x 5 = 20]

Question 39.
Prepare an attractive advertisement using the hints given below.
Home appliances – Aadi Sale – 20-50% – Special Combo Offers – Aadhi & Co., Raja Street, Chennai, Chennai.
Answer:

Question 40.
Write a letter to the manager of a famous daily, ordering subscription for your school library.
Answer:
From
The Librarian
Indira Gandhi Hr. Sec. School,
Rajajipuram Chennai – 600 087
12.11.2020
To
The Higgin Bothams,
Anna Salai,
Chennai-600 002 Respected Sir,
Sub: Annual subscription for daily.
I would like to place an order for the annual subscription of three English dailies and one Tamil daily to our School from the 1st of January 2020.

The Hindu	3 copies	@ 435/- each p.a.	1305.00
The Times of India	3 copies	@ 365/- each p.a.	1095.00
The New Indian Express	3 copies	@ 425/- each p.a.	1275.00
Thina Thanthi	3 copies	@ 175/- each p.a.	525.00
			4200.00

Please find enclosed the Annual subscription amount for all the four dailies which amount to Rs. 3780/- (Rupees Three Thousand Seven Hundred and Eighty Only) after deducting the usual 10% discount that you give us.
Thank you very much for your continued support to our institution.
Yours sincerely,
Mrs. Anand – Librarian

Address on the envelope:
The Higgin Bothams,
Anna Salai,
Chennai-600 002

Question 41.
You are Adhira / Athiran, school pupil leader of GHSS, Trichy. Prepare a notice on behalf of your school inviting the grandparents of the students to celebrate World Elders’ Day in your school auditorium on the 20^th of next month.
Answer:

Government Higher Secondary School, Trichy
Notice
World Elder’s Day
23^rd January, 2020

On behalf of the school, I invite every’ grandparent of the students in this school to be a part of the World Elder’s Day. We will be celebrating World Elder’s Day in our school auditorium on the 20^th of February, 2020. Please bring in your grandparents to school at 10 a.m. Lots of fun-filled activities await your grandparents. The programme will end with a buffet arranged by the management of the school to honour them. Adhira/Athiran School Pupil Leader

Question 42.
Look at the following picture and express your views an it in about five sentences
Answer:

Though we pray for rain, it’s such a sad state to see people stranded when an area gets flooded. There is about three feet water and people are moving out to safer places. Fathers carry their sons and daughters over the shoulders in order to keep them safe. If there is proper drainage system and proper maintenance by the Corporation, we can surely avoid such disasters. In certain states, natural disaster such as flooding is a frequently witnessed problem.

Question 43.
Make notes or write a summary of the following passage.
Answer:
There are many different kinds of books that are published each year. These are the new’ titles available for us to read. Besides these, there are books that have been published through the years. Together, there are millions of books available throughout the world in as many languages as are spoken by people. There are different genres in which books are published. There are fiction and non-fiction categories in books, and each of these categories has many different genres of books.

The academic books we study at school belong to the text book category. We smdy them to complete our syllabus and pass the examinations at the end of each academic session. There are other books that we read for our pleasure and enrichment. We read story books of different types. There are comedy, horror, detective and thriller stories in prose, plays and poetry forms. Books are our best friends.
Notes
Title: Books Galore
Dif. Kinds A Books
(i) New Titles;
(ii) Publd Every Yr;
(iii) Publd Thru’ The Yrs;
(iv) Publd Thru’out the World;
(v) Publd In Many Langs
Dif. Genres Of Books
(i) Fiction;
(ii) Non-Fiction Categs

Academic Books – Text Book Categy.
For Study – Pres. Syl.
Pass Exams – End A Academic Session.
Books For Pleasure &Enrichment
Books Of Different Types
Comedy, Horror, Detective & Thriller Stories
Various Forms
(i) Prose
(ii) Plays; &
(iii) Poetry Forms Conclusion: Books are our best friends.
Abbreviations used: Dif. – different; A -of; publd.- published; yr=year; thru’- through; langs.- languages; pres. – prscribed; syl-syllabus; & – and

Summary

Title: Books Galore
Rough Draft
There are different kinds of books published every year for us to read. Together, there are millions of books-avaiable throughout the world in many languages. There are different genres such as fiction and non friction categories. The academic books we study at school belong to the text book category. We study them to be tested at the end of each academic session. We read story books for pleasure and enrichment/Here-areaymiedy, horror, detective and thriller stories in prose, plays and poetry forms. Books are our best friends .

Fair Draft:
Title: Books Galore

Millions of books are published every year in different languages. There are different genres such as fiction and non-fiction. The academic books, belong to the text book category. We study them to be tested at the end of the academic year. We read various types of story books for pleasure. Books are our best friends.
No. of words written in the summary: 55

Question 44.
Identify and correct the errors in the following sentences.
(а) You may speak politely to the elders.
(b) This is the boy whom won the race
(c) He come late to school every day.
(d) Though he was hungry hut he did not eat.
(e) Is this a book that you wanted to buy.
Answer
(a) You must speak politely to the elders.
(b) This is the boy who won the race
(c) He comes late to school every day.
(d) Though he was hungry he did not eat.
(e) Is this the book that you wanted to buy.

Section – V

Quote from memory. [1 x 5 = 5]

Question 45.
Let me but live ……………. back in fear.
Answer:
Let me but live my life from year to year,
With forward face and unreluctant soul;
Not hurrying to, nor turning from the goal; .
Not mourning for the things that disappear
In the dim past, nor holding back in fear

Part – IV

Write a paragraph of about 150 words by developing the following hints. [2 x 8 = 16]

Question 46.
(а) Many years ago – China – the emperor ordered – one man from -family – join army
– Mulan heard -told father – she join army – father objected – she is a girl – Mulan – fathers robes cuts her hair – convinced father – she has learnt – Kung Fu – no one will find – she is a girl. – Mulan left – village – fought bravely – war – given top position – very soon – fever swept – the army -Mulan – sick – doctor examines – finds the truth – spreads the news in the army – everyone objects – to follow a girl leader – Mulan stood tall – gave command – soldiers – followed her – attacked enemies – won the battle- emperor glad – offered Mulan positions – court – Mulan refused – went back – village – royal – gifts.
Answer:
The Story of Mulan portrays the legendary Chinese warrior Hua Mulan. This old Chinese folktale is about the story of the young Chinese maiden who leams that her wizened, old and frail father is to be called up into the army in order to fight the invading Huns by the Chinese Emperor. When the Huns invade China, one man from every family is called to arms. She hears of the order that every family must send one man to the army while washing clothes.

Mulan’s father, who is frail and aged decides to fight for his country though it is clear that he will not survive an enemy encounter. He decides to go to war but is prevented by her daughter with her outrageous decision. Knowing her father’s frail state, she decides to disguise herself and join in his place without second thoughts. In the army, Mulan proves to be a brave soldier who is later put in charge of other soldiers. Her battles goes so well that more soldiers are added. After a few years, Mulan becomes the General of the entire army. Suddenly, bad fever swept through the army. Many soldiers including Mulan become a prey. The arrival of the doctor brings to light the hidden truth.

Many soldiers disapprove such a thought, though some soldiers see the winning chances. Just then a soldier announces the surprise attack by the enemies. With no time to debate, the soldiers spring to action at the command of the General who hears this from inside her tent. She gets dressed and though not strong, she stands tall. She instructs the soldiers to attack knowing very well her strategic planning that all her soldiers acknowledge and win the battle. It was such a big victory that the enemy gave up, at last. The war was over, and China was saved! The Emperor forgives Mulan and was glad that Mulan had ended the long war. He wanted Mulan to stay with him in the palace and be an advisor but as she chose to go to her family, the emperor gave her six horses and six fine swords so that her people will know that he thinks of her.

[OR]

(b) Holland – land – below sea level – dikes protected the country – everyone did best to -protect – Holland – Years ago – boy Peter – lived Holland – His father – attended – dykes gates – opened- closed dikes – one day – Peter Mother – gave cakes to Peter – to be given – old blind friend of Peter – across the dike Peter happily left home – Peter visited – old man – returned near by the dike – heard water trickling – stopped to see – small hole – dike – called for help – in vain -he put his little finger – throughout the
night – slept near the dike – morning – found by passer by – alerted the people – Peter and Holland -saved.
Answer:
Peter is a little boy who lived in Holland. His father took care of the dikes called sluices so that ships could pass out of Holland’s canals into the sea. On a beautiful day in Autumn, Peter was asked to go and give cakes to his blind friend who lived on the other side of the dike by his mother. After about an hour he returned home, but the climate had changed. It was raining and the water in the channel was rising. All of a sudden he heard the sound of dribbling water and he wondered from where the sound came. He then saw a small hole in the dike.

He knew what that meant and due to the pressure of the water the hole would not stay the same. He feared danger. He climbed onto the dike and put his finger in the hole and hoped for someone to come to his help and cried out aloud. His mother mistook him to have stayed back with his blind friend and retired to bed. The boy was sure that he had to stay awake the entire night and keep his finger in the hole to arrest the water from flooding Holland.

The water in the canal was rising and if he would remove his finger from the hole in the sluice, the water would gush through and make the hole bigger and bigger. The town would obviously flood. When dawn broke, a man going to work heard the sound of Peter groaning and wondered what the little boy was up to. He was shocked at his reply but understood the danger and called for help. People came in with shovel and mended the hole. Peter was carried home and they all hailed him as the brave boy who saved Holland from drowning.

Question 47.
Read the following passage and answer the questions given below:
Kung Fu – ‘kung’ meaning ‘energy’ and ‘fu’ meaning ‘time’ – is a Chinese martial art whose recorded history dates back to around 525 CE, during the Liang dynasty. The man credited with introducing martial arts to China is said to be an Indian monk known as Bodhidarma.

Many people have a misconception that Chinese Kung Fu is about fighting and killing. It is actually based on Chinese philosophy and is about improving wisdom and intelligence. Taoist philosophy is deeply rooted in and had a profound influence on the culture of Chinese martial arts.

The five traditional animal styles of Shaolin Kung Fu are the dragon, the snake, the tiger, the leopard and the crane. The union of the five animal forms clearly displayed the efficacy of both hard and soft movements, of both internal and external energy – this form of Chinese martial arts was known as Shaolin Kung Fu, named after the temple in which it was developed.

Questions.
(a) Which country does the martial art Kung Fu belong to?
(b) What is the meaning of the term “Kung Fu”?
(c) Write any two martial arts of India?
(d) What are the five animal styles followed in Shaolin Kung Fu?
Answer:
(a) The martial art Kung Fu belongs to the country China.
(b) The term ‘kung’ means ‘energy’ and ‘fu’ means ‘time’.
(c) Kalari and Silambam are two martial arts in India.
(d) The five traditional animal styles of Shaolin Kung Fu are the dragon, the snake, the tiger, the leopard and the crane.

[OR]

Read the following poem and answer the questions given below:

If you can’t be a pine on the top of the hill,
Be a scrub in the valley – but be
The best little scrub by the side of the rill;
Be a bush, if you can’t be a tree.
If you can’t be a bush, be a bit of the grass,
And some highway happier make;
If you can’t be a muskie, then just be a bass- But the liveliest bass in the lake!
We can’t all be captains, we’ve got to be crew,
There’s something for all of us here.
There’s big work to do and there’s lesser to do
And the task we must do is the near.
If you can’t be a highway, then just be a trail,
If you can’t be the sun, be a star;
It isn’t by size that you win or you fail- Be the best of whatever you are!

Questions.
(a) Where does the best scrub grow?
(b) What makes a highway traveller happy?
(c) Does size matter? Give reason.
(d) What is the underlying theme of the poem?
Answer:
(a) The best scrub grows by the side of a rill.
(b) The green grass on the sides of a highway makes a highway traveller happy.
(c) No, size doesn’t matter because its winning and being the best in whatever you do that matters.
(d) Be your best in all that you try to do and accomplish in life.

Samacheer Kalvi 12th Computer Applications Solutions Chapter 3 Introduction to Database Management System

Leave a Comment / Class 12 / By Prasanna

Students can Download Computer Applications Chapter 3 Introduction to Database Management System Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 3 Introduction to Database Management System

Samacheer Kalvi 12th Computer Applications Introduction to Database Management System Text Book Back Questions and Answers

PART – I
I. Choose The Correct Answer

Question 1.
Which language is used to request information from a Database?
(a) Relational
(b) Structural
(c) Query
(d) Compiler
Answer:
(c) Query

Question 2.
The ………………………. diagram gives a logical structure of the database graphically?
(a) Entity-Relationship
(b) Entity
(c) Architectural Representation
(d) Database
Answer:
(a) Entity-Relationship

Question 3.
An entity set that does not have enough attributes to form primary key is known as
(a) Strong entity set
(b) Weak entity set
(c) Identity set
(d) Owner set
Answer:
(b) Weak entity set

Question 4.
…………………………. Command is used to delete a database.
(a) Delete database databasename
(b) Delete database_name
(c) drop database database name
(d) drop database name
Answer:
(c) drop database database_name

Question 5.
Which type of below DBMS is MySQL?
(a) Object Oriented
(b) Hierarchical
(c) Relational
(d) Network
Answer:
(c) Relational

Question 6.
MySQL is freely available and is open source.
(a) True
(b) False
Answer:
(a) True

Question 7.
……………………….. represents a “tuple” in a relational database?
(a) Table
(b) Row
(c) Column
(d) Object
Answer:
(b) Row

Question 8.
Communication is established with MySQL using
(a) SQL
(b) Network calls
(c) Java
(d) API’s
Answer:
(a) SQL

Question 9.
Which is the MySQL instance responsible for data processing?
(a) MySQL Client
(b) MySQL Server
(c) SQL
(d) Server Daemon Program
Answer:
(b) MySQL Server

Question 10.
The structure representing the organizational view of entire database is known as in MySQL database.
(a) Schema
(b) View
(c) Instance
(d) Table
Answer:
(a) Schema

PART – II
II. Short Answer:

Question 1.
Define Data Model and list the types of data model used?
Answer:
Data models define how the logical structure of a database is modeled.
Data models define how data is connected to each other and how they are processed and stored inside the system. The various data models are;

Hierarchical Database Model,
Network Model,
Relational Model and
Object-oriented Database Model.

Question 2.
List few disadvantages of file processing system?
Answer:
Data Duplication – Same data is used by multiple resources for processing, thus created multiple copies of same data wasting the spaces.
High Maintenance – Access control and verifying data consistency needs high maintenance cost.
Security – less security provided to the data.

Question 3.
Define Single and multi valued attributes?
Answer:
A single valued attribute contains only one value for the attribute and they don’t have multiple numbers of values. For Example:Age-
Amulti valued attribute has more than one value for that particular attribute. F or Example :Degree

Question 4.
List any two DDL and DML commands with its Syntax?
Answer:
Commands:

CREATE
DROP

Syntax:

CREATE database databasename
DROP database databasename

DML COMMANDS List:
Commands:

INSERT
DELETE

Syntax:

Syntax 1: INSERT INTO tablename (column1, column2, column3) VALUES (value 1, value2, value3);
Syntax 2: INSERT INTO tablename VALUES (value1, value2, value3);
DELETE from tablename WHERE columnname=”value”;

Question 5.
What are the ACID properties?
Answer:
ACID Properties – The acronym stands for Atomicity, Consistency, Isolation and Durability. Atomicity follows the thumb rule “All or Nothing”, while updating the data in database for the user performing the update operation. Consistency ensures that the changes in data value to be constant at any given instance. Isolation property is needed during concurrent action. Durability is defied as the system’s ability to recover all committed actions during the failure of storage or the system.

Question 6.
Which command is used to make permanent changes done by a trAnswer:action?
Answer:
COMMIT

Question 7.
What is view in SQL?
Answer:
Views – A set of stored queries.

Question 8.
Write the difference between SQL and MySQL?
Answer:
SQL:

SQL is a query language
SQL is used to query and operate database system.

MySQL:

MySQL is DBMS software.
MySQL allows data handling, storing, modifying, deleting, etc.

Question 9.
What is Relationship and List its types?
Answer:
In ER Model, relationship exists between two entities.

One-to-One relationship,
One-to-Many relationship and
Many-to-Many relationship.

Question 10.
State few advantages of Relational databases?
Answer:

High Availability
High Performance
Robust TrAnswer:actions and support
Ease of management
Less cost

PART – III
III. Explain in Brief Answer

Question 1.
Explain on Evolution of DBMS?
Answer:

The concept of storing the data started before 40 years in various formats.
In earlier days they have used punched card technology to store the data.
Then files were used. The file systems were known as predecessor of database system.
Various access methods in file system were indexed, random and sequential access.
The file systems has limitations like duplication, less security. To overcome this, DBMS was introduced.

Question 2.
What is relationship in databases? List its types?
Answer:
In ER Model, relationship exists between two entities. The various types of relationships are;

One-to-One relationship.
One-to-Many relationship.
Many-to-Many relationship.

Question 3.
Discuss on Cardinality in DBMS?
Answer:
Cardinality is defined as the number of items that must be included in a relationship, i.e. number of entities in one set mapped with the number of entities of another set via the relationship. Three classifications in Cardinality are one-to-one, one-to-many and Many-to-Many.

In the above example we have two entities Person and Vehicle. If we consider the current vehicle, the driver is operating, then we have one-to-one relationship between Person and Vehicle.
In the above example, Customer places the Order is a one-to-many relationship. Here the customer can place multiple orders and the order is related to only one customer. The example of many-to-many relationship is Students registering the Courses. A student can register more than one courses and A course can be registered by many students. Hence it is many-to-many.

Question 4.
List .any 5 privileges available in MySQL for the User?
List of privileges available in MySQL?
Answer:
Privileges:

Select_priv
Insert_priv
Update_priv
Delete_priv
Create_priv
Alter_priv

Action Performed (If Granted):

User can select rows from database tables.
User can insert rows into database tables.
User can update rows of database tables.
User can delete rows of database tables.
User can create new tables in database.
User can make changes to the database structure.

Question 5.
Write few commands used by DBA to control the entire database.
USE Database – This command is used to select the database in MySQL for working.
Syntax:
mysql > use test;
Database changed
mysql>
SHOW Databases – Lists all the databases available in the database server.
Syntax:
mysql > show databases;

SHOW Tables – Lists all the tables available in the current database we are working in.
mysql > show tables;

SHOW COLUMNS FROM tablename – Lists all the attributes, attribute type, Is Null value
permitted, key information, default value and other information for the given table.
Syntax:
mysql > show columns from sports;

SHOW INDEX FROM tablename – The query shows all the indexes for the given table.
Syntax:
mysql > show indexes from sports;

SHOW TABLE STATUS LIKE tablename\G – This command provides with detailed report on the performance of the table.

PART – IV
IV. Explain in detail

Question 1.
Discuss on various database models available in DBMS?
Answer:
DBMS Database Models:
The database technology came into existence in terms of models with relational and object- relational behavior. The major database models are listed below:

(i) Hierarchical Database Model The famous Hierarchical database model was IMS (Information Management System), IBM’s first DBMS. In this model each record has information in parent/child relationship like a tree structure. The collection of records was called as record types, which are equivalent to tables in relational model. The individual records are equal to rows.

In the above model we have many advantages like less redundant data, efficient search, data integrity and security. This model also has few limitations like complex to implement and difficulty in handling many to many relationships.

(ii) Network model:
The first developed network data model was IDS (Integrated Data Store) at Honeywell. Network model is similar to Hierarchical model except that in this model each member can have more than one owner. The many to many relationships are handled in a better way. This model identified the three database components Network schema, Sub schema and Language for data management. Network schema – schema defines all about the structure of the database.
Sub schema – controls on views of the database for the user.
Language – basic procedural for accessing the database.
The major advantage of this model is the ability to handle more relationship types, easy data access, data integrity and independence. The limitation of network model is difficulty in design and maintenance.

(iii) Relational model:
Oracle and DB2 are few commercial relational models in use. Relational model is defined with two terminologies Instance and Schema.
Instance – A table consisting of rows and columns
Schema – Specifies the structure including name and type of each column.
A relation (table) consists of unique attributes (columns) and tuples (rows).

(iv) Object-oriented database model:
This model incorporates the combination of Object Oriented Programming (OOP’s) concepts and database technologies.
Practically, this model serves as the base of Relational model. Object oriented model uses small, reusable software known as Objects.
These are stored in object oriented database.
This model efficiently manages large number of different data types. Moreover complex behaviors are handled efficiently using OOP’s concepts.

Question 2.
List the basic concepts of ER Model with suitable example?
Answer:
ER Modeling basic concepts
The basic concepts of ER model consists of

Entity or Entity type
Attributes
Relationship

These are the general concepts which help to create an ER diagram and produce an ER model. With the help of these any database design can be created and viewed to know the concept in that database design.

(i) Entity or Entity type
An Entity can be anything a real-world object or animation which is easily identifiable by anyone even by a common man. Eg: In a company’s database Employee, HR, Manager are considered as entities, where each of these entity will be having their own attributes . An entity is represented by a rectangular box.

Types of Entity:

Strong Entity
Weak Entity
Entity Instance

1. Strong Entity:
A Strong entity is the one which doesn’t depend on any other entity on the schema or database and a strong entity will have a primary key with it (i.e. a unique id which other entities will not have in their attributes).It is represented by one rectangle. In the above example it is a strong entity because it has a primary key(a unique id) as the roll no because for every one roll no varies and it will not be same.

2. Weak Entity:
A weak entity is dependent on other entities and it doesn’t have any primary key like the Strong entity. It is represented by double rectangle. For Example: Here the marks is the weak entity and there are no unique id or primary – key for that entity. So they are dependent on the existence of the other entity.

(ii) Attributes
An attribute is the information about that entity and it will describe, quantify, qualify, classify, and specify an entity. An attribute will always have a single value, that value can be a number or character or string.
Types of attributes:

Key Attribute
Simple Attributes
Composite Attributes
Single Valued Attribute
Multi Valued Attribute

1. Key Attribute:
Generally a key attribute describes a unique characteristic of an entity.

2. Simple Attribute:
The simple attributes cannot be separated it will be having a single value for their entity. For Example: Let us consider the name as the attribute for the entity employee and here the value for that attribute is a single value.

3. Composite Attributes:
The composite attributes can be sub-divided into simple attributes without change in the meaning of that attribute. For Example: In the above diagram the employee is the entity with the composite attribute Name which are sub-divided into two simple attributes first and last name.

4. Single Valued Attributes:
A single valued attribute contains only one value for the attribute and they don’t have multiple numbers of values. For Example: Age- It is a single value for a person as we cannot given number of ages for a single person,therefore it is a single valued attribute.

Arrtibute:

Age
Roll no

Values:

In the above table are the some examples for single valued attributes.

5. Multi Valued Attributes:
A multi valued attribute has more than one value for that particular attribute. For Example:
Degree – A person can hold n number of degrees so it is a multi-valued attribute.

Attributes and Values:

Attributes:

Degree
BankAccount

Values:

B. Tech, MBA
SBI, HDFC

(iii) Relationship Type:
In ER Model, relationship exists between two entities. Three types of relationships are – available and the Entity-Relationship(ER) diagram is based on the three types listed below.

(iv) One-to-One relationship:
Consider two entities A and B. one-to-one (1:1) relationship is said to exist in a relational database design, if 0 or 1 instance of entity A is associated with 0 or 1 instance of entity B, and 0 or 1 instance of entity B is associated with 0 or 1 instance of entity A.

(v) One-to-Many relationship:
Consider two entities A and B. one-to-many (1:N) relationship is said to exist in a relational database design, for 1 instance of entity A there exists 0 or 1 or many instances of entity B, but for 1 instance of entity B there exists 0 or 1 instance of entity A.

(vi) Many-to-Many relationship:
Consider two entities A and B. many-to-many (M:N) relationship is said to exist in a relational database design, for 1 instance of entity A there exists 0 or 1 or many instances of entity B, and for 1 instance of entity B there exists 0 or 1 or many instance of entity A.
In reality one-to-one are in less usage, where as one-to-many and many-to-many are commonly used. However in relational databases, many-to-many are converted into one-to- many relationships.

Question 3.
Discuss in detail on various types of attributes in DBMS?
Answer:
Types of attributes:

Key Attribute
Simple Attributes
Composite Attributes
Single Valued Attribute
Multi Valued Attribute

1. Key Attribute:
Generally a key attribute describes a unique characteristic of an entity.

Attribute:

Degree
Bank_Account

Values:

5. Multi Valued Attributes:
A multi valued attribute has more than one value for that particular attribute. For example:
Degree – A person can hold n number of degrees so it is a multi-valued attribute.
In the below table are some examples for multi valued attributes.

Attribute:

Degree
Bank_Account

Values:

B.TEch, MBA
SBI, HDFC

Question 4.
Write a note on open source software tools available in MySQL Administration?
MYSQL Administration open source software tools:
Answer:
Types of software tools:
Many open source tools are available in the market to design the database in a better and efficient manner. PhpMyAdmin is most popular for Web Administration. The popular Desktop Application tools are MySQL Workbench and HeidiSQL.

PHPMYADMIN (Web Admin):

This administrative tool of MySQL is a web application written in PHP.
They are used predominantly in web hosting.
The main feature is providing web interface, importing data from CSV and exporting data to various formats.
It generates live charts for monitoring MySQL server activities like connections, processes and memory usage.
It also helps in making the complex queries easier.

MySQL Workbench (Desktop Application):

It is a database tool used by developers and DBA’s mainly for visualization.
This tool helps in data modeling, development of SQL, server configuration and backup for MySQL in a better way.
Its basic release version is 5.0 and is now in 8.0 supporting all Operating Systems.
The SQL editor of this tool is very flexible and comfortable in dealing multiple results set.

HeidiSQL (Desktop Application):

This open source tools helps in the administration of better database systems.
It supports GUI (Graphical User Interface) features for monitoring server host, server connection, Databases, Tables, Views, Triggers and Events.

Question 5.
Explain in detail on Sub Queries with suitable examples?
Answer:
Sub queries:
The SQL query is written within a main Query. This is called as Nested Inner/SubQuery.The sub query is executed first and the results of sub query are used as the condition for main query.
The sub query must follow the below rules:

Subqueries are always written within the parentheses.
Always place the Subquery on the right side of the comparison operator.
ORDER BY clause is not used in sub query, since Subqueries cannot manipulate the results internally.

Consider the Employee table with the fields EmpID, Name, Age and Salary.

In the below Query, we use sub query in an SELECT statement.
SELECT * from Employee
where EmpID IN (SELECT EmpID from Employee WHERE Salary < 20000);
First, the inner query is executed. As a result EmpID 101 and 103 are retrieved. Now the external or outer query is executed. Internally the query is SELECT * from Employee where EmpID IN(101,103) and the output is drawn below.
Select Record List

Similarly the subqueries are used with INSERT, UPDATE and DELETE.

Samacheer Kalvi 12th Computer Applications Introduction to Database Management System Additional Question and Answer:wers

I. Choose the Best Answer

Question 1.
Expand DBMS?
(a) DataBase Management System
(b) Data Manipulation Schema
(c) Data Base Management Schema
(d) Data Base Manipulation Schema
Answer:
(a) DataBase Management System

Question 2.
Expand RDBMS.
(a) Relational Data Base Manipulation System
(b) Relational Data Base Management Schema
(c) Relational Data Base Management System
(d) Record Data Base Managing Schema
Answer:
(c) Relational Data Base Management System

Question 3.
Expand ODBMS.
(a) Object Data Base Management System
(b) Objective Data Base Management System
(c) Object Oriented Data Base Management System
(d) Objective Data Manipulation System
Answer:
(a) Object Data Base Management System

Question 4.
Update operation is otherwise called as …………………………
Answer:
transaction

Question 5.
Two types of trAnswer:action operations are ……………………… and ………………………
Answer:
commits and aborts

Question 6.
To prevent the conflict in database update, the trAnswer:actions are isolated from other user and serialized. This is called as ……………………….
(a) Atomicity
(b) Consistency
(c) Isolation
(d) Degree of Consistency
Answer:
(d) Degree of Consistency

Question 7.
ensures that the changes in data value to be constant at any given instance.
(a) Atomicity
(b) Consistency
(c) Isolation
(d) Durability
Answer:
(b) Consistency

Question 8.
Match the following:
1. Atomicity – (i) commit / aborts
2. TrAnswer:action – (ii) constant data
3. Consistency – (iii) recover all committed trAnswer:actions
4. Durability – (iv) All or Nothing
(a) 1-(iv), 2-(i), 3-(ii), 4-(iii)
(b) 1-(i), 2-(ii), 3-(iii), 4-(iv)
(c) 1 -(iv), 2-(iii), 3-(ii),4-(i)
(d) 1-(i), 2-(ii), 3-(iv), 4-(iii)
Answer:
(a) 1-(iv), 2-(i), 3-(ii), 4-(iii)

Question 9.
Concurrency control and locking are needed
(a) Consistency
(b) data sharing
(c) data hiding
(d) TrAnswer:action
Answer:
(b) data sharing

Question 10.
How many Database models are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 11.
The famous Hierarchical database model was ……………………
Answer:
IMS

Question 12.
IMS stands for ……………………..
Answer:
Information Data Store

Question 13.
IBM’s first DBMS is ……………………..
(a) IMS
(b) IVS
(c) BMS
(d) VMS
Answer:
(a) IMS

Question 14.
Hierarchical database model uses …………………….. structure.
(a) star
(b) tree
(c) HoneyComb
(d) Hair like
Answer:
(b) tree

Question 15.
Parent child relationship is established in …………………….. database Models.
(a) Hierarchical
(b) Network
(c) Relational
(d) Object
Answer:
(a) Hierarchical

Question 16.
The collection of records in Hierarchical models are called as ……………………..
(a) field types
(b) tuples
(c) record types
(d) ACID
Answer:
(c) record types

Question 17.
In ACID properties I indicates
(a) Inverter
(b) Interface
(c) Isolation
(d) Identification
Answer:
(c) Isolation

Question 18.
The first Network Model was developed at ……………………..
Answer:
Honey well

Question 19.
…………………….. is the first developed Network Data Model.
(a) IMS
(b) IDS
(c) DB2
(d) OOP’s
Answer:
(b) IDS

Question 20.
IDS stands for ……………………..
Answer:
Integrated Data Store

Question 21.
Which of the two databases models are similar?
(a) Hierarchical, Network
(b) Network, relational
(c) Relational, Object oreinted
(d) Network, Object oreinted
Answer:
(a) Hierarchical, Network

Question 22.
Network database model is identified by ………………………..
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 23.
Find the Statement which is wrong?
(a) Network Schema deals with locking
(b) Subschema controls on view of the database
Answer:
(a) Network Schema deals with locking

Question 24.
Which one of the following is the basic procedural for accessing the database in Network model?
Answer:
(a) Network schema
(b) Sub schema
(c) Language
(d) Super schema
Answer:
(c) Language

Question 25.
……………….. and …………………… are the commercial relational database models.
(a) Oracle, DB2
(b) DB2, IDS
(c) IDS, IMS
(d) Oracle, IBME
Answer:
(a) Oracle, DB2

Question 26.
Find which of the following statements are true with respect to Relational model
(i) Instance – A table with rows and columns
(ii) Schema – Specifies the sjtructure
(a) (i) is true
(b) (ii) is true
(c) both are true
(d) both are false
Answer:
(c) both are true

Question 27.
Columns are otherwise called as …………………….
Answer:
attributes

Question 28.
OOP’s stands for …………………….
Answer:
Object Oriented Language.

Question 29.
Object oriented model uses small, reusable software known as ………………………
(a) objects
(b) software
(c) class
(d) private
Answer:
(a) objects

Question 30.
The vertical entity of a table is known as ………………………. or ……………………….
Answer:
Attribute, Column

Question 31.
A primary key which is a combination of more than one attribute is called a ………………………..
Answer:
composite primary key

Question 32.
Each super key is called as ……………………… key.
(a) Sub
(b) Candidate
(c) Schema
(d) Constraint
Answer:
(b) Candidate

Question 33.
What is the another name for Candidate Key?
(a) Super key
(b) Sub key
(c) Minimal super key
(d) Maximal super key
Answer:
(c) Minimal super key

Question 34.
Composite key is otherwise called as ………………………
(a) compound key
(b) super key
(c) sub key
(d) foreign key
Answer:
(a) compound key

Question 35.
How many basic concepts are there in ER model?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 36.
How many types of entities are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 37.
What is the symbol for weak Entity?
(a) square
(b) double square
(c) rectangle
(d) double rectangle
Answer:
(d) double rectangle

Question 38.
………………………. denotes the category values for the given entity.
(a) Schema
(b) Structure
(c) Entity Instance
(d) Existence
Answer:
(c) Entity Instance

Question 39.
……………………. describes a unique characteristic of an entity.
Answer:
Key Attribute.

Question 40.
How many types of attributes are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 41.
How many relationship types are there in ER model?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 42.
The number of entity types involved is known as …………………….
Answer:
degree of relationship

Question 43.
Customer places the order is an example of ……………………. relationship.
(a) one to one
(b) one to many
(c) many to many
(d) two to many
Answer:
(b) one to many

Question 44.
MySQL is founded by ……………………….
(a) My Widenius
(b) Monty Widenius
(c) Marley Minto
(d) My Minto
Answer:
(b) Monty Widenius

Question 45.
SQL meAnswer: ……………………….
Answer:
Structured Query Language.

Question 46.
The structured collection of data is ……………………..
(a) data
(b) table
(c) row
(d) database
Answer:
(d) database

Question 47.
The databases are broadly classified into …………………….. and …………………….. databases.
Answer:
Heavy, light

Question 48.
The light databases that supports the web applications are also known as ……………………. databases.
(a) web
(b) Heavy
(c) Application
(d) Super
Answer:
(a) web

Question 49.
DBA stands for ……………………..
Answer:
DataBase Administrators.

Question 50.
All the query will terminate with ……………………
(a) colon
(b) semicolon
(c) bracket
(d) dot
Answer:
(b) semicolon

Question 51.
…………………. reboots the server.
Answer:
Flush privileges

Question 52.
…………………… is the most popular web administration tool.
Answer:
Phpmyadmin

Question 53.
The popular Desktop Application tools are
(a) MySQL workbench
(b) HeidiSQL
(c) PlSQL
(d) both a and b
Answer:
(d) both a and b

Question 54.
The process of creating, implementing and maintaining the enterprise data in the system is known as ……………………… of databases.
Answer:
desingning

Question 55.
Expand Answer:
(a) American National Standards Institute
(b) African National Standard Institute
(c) American North Sound Institute
(d) All-National Standard Institute
Answer:
(a) American National Standards Institute

Question 56.
…………………… is used to temporarily save a trAnswer:action.
(a) Commit
(b) Rollback
(c) TrAnswer:action
(d) SavePoint
Answer:
(d) SavePoint

Question 57.
When we want to select data from more than 2 tables ……………………. clause is used.
(a) group
(b) merge
(c) combine
(d) sql join
Answer:
(d) sql join

Question 58.
Multiple numbers of values are not allowed in ………………………. attributes.
(a) single valued
(b) composite
(c) multi valued
(d) key
Answer:
(a) single valued

Question 59.
Match the following attributes with their values.
1. simple – (i) Degree
2. composite – (ii) Name
3. Single valued – (iii) First Name, Last Name
4. Multi valued – (iv) Age
(a) 1-(ii), 2-(iii), 3-(iv), 4-(i)
(b) 1-(i), 2-(ii), 3-(iii), 4-(iv)
(c) 1-(iv), 2-(ii), 3-(i), 4-(iii)
(d) 1-(iv), 2-(i), 3-(ii), 4-(iii)
Answer:
(a) 1-(ii), 2-(iii), 3-(iv), 4-(i)

Question 60.
…………………….. is the symbol for Relationship in ER Model.
Answer:
Rhombus

Question 61.
M : N relationship is said to be ………………………. relationship.
Answer:
many to many

Question 62.
The most popular DBMS is ……………………….
Answer:
MySQL

Question 63.
The least popular DBMS as per the statistical data is ……………………….
Answer:
cassandra

Question 64.
………………… lists all the databases available in the databases.
Answer:
Show databases

Question 65.
shows all the indexes for the given table.
Answer:
Show index from tablename

Question 66.
Match the following
1. DDL – (a) commit
2. DML – (b) alter
3. DQL – (c) insert
4. TCL – (d) select
(a) 1-(b) 2 (c) 3-(d) 4-(a)
(b) 1-(a) 2-(d) 3-(b) 4-(c)
(c) 1-(c)2-(b)3-(a)4-(d)
(d) 1-(d) 2-(b) 5-(a) 4-(c)
Answer:
(a) 1-(b) 2 (c) 3-(d) 4-(a)

Question 67.
Find the wrong one.
(a) DDL – create command
(b) DML – update
(c) TCL – solve point
(d) DCL – select
Answer:
(d) DCL – select

Question 68.
DDL stands for ……………………….
Answer:
Data Definition Language

Question 69.
DML stands for ……………………….
Answer:
Data Manipulation Language

Question 70.
DQL stands for ……………………….
Answer:
Data Query Language

Question 71.
TCL stands for ……………………….
Answer:
Transaction Control Structure

Question 72.
DCL stands for ……………………….
Answer:
Data Control Language

Question 73.
………………………. clause is used in select and update query statement for the condition.
Answer:
where

Question 74.
The existing record in a table is removed from the table using ………………………. command.
Answer:
DELETE

Question 75.
………………………. is not a logical operator.
(a) And
(b) Between
(c) Plus
(d) Unique
Answer:
(c) Plus

Question 76.
IN is ………………………. operator.
Answer:
logical

Question 77.
<>is a ………………………. operator.
Answer:
comparision

Question 78.
………………………. is used to take out permission from the specific users.
Answer:
Revoke

Question 79.
………………………. is used to delete all table records.
Answer:
Truncate

Question 80.
………………………. deletes database or table.
Answer:
drop

II. Short Answer

Question 1.
Define database?
Answer:
“A database management system (DBMS) is system software for creating and managing databases. The DBMS provides users and programmers with a systematic way to create, retrieve, update and manage data”.

Question 2.
Name the various access methods in file system?
Answer:
Various access methods in file system were indexed, random and sequential access.

Question 3.
Define degree of consistency?
Answer:
To prevent the conflict in database update, the trAnswer:actions are isolated from other user and serialized. This is also known as Degree of Consistency.

Question 4.
Give advantages of Hierarchical database models?
Answer:

Less redundant data
data integrity
efficient search
security

Question 5.
Give limitations of Hierarchical db model?
Answer:

complex to implement
difficult to handle many to many relationships.

Question 6.
Define relational database?
Answer:
Any database whose logical organization is based on relational data model is known as Relational Database.

Question 7.
Define Table?
Answer:
In relational database model, table is defined as the collection of data organized in terms of rows and columns. Table is the simple representation of relations.

Question 8.
Define row in a table?
Answer:
A single entry in a table is called as Row or Record or Tuple. Set of related data’s are represented in a row or tuple. The horizontal entity in a table is known as Record or row.

Question 9.
Define Primary key?
Answer:

The candidate key that is chosen to perform the identification task is called the primary key and any others are Alternate keys.
Every tuple must have, a unique value for its primary key.

Question 10.
Define Foreign key?
Answer:

A foreign key is a “copy” of a primary key that has been exported from one relation into another to represent the existence of a relationship between them.
Foreign keys can also be null.

Question 11.
Define Composite key?
Answer:
A key with more than one attribute to identify rows uniquely in a table is called Composite key. This is also known as Compound Key.

Question 12.
Name the different types of entity?
Answer:
Types of Entity:

Strong Entity
Weak Entity
Entity Instance

Question 13.
Name the different types of attributes?
Answer:
Types of attributes:

Key Attribute
Simple Attributes
Composite Attributes
Single Valued Attribute
Multi Valued Attribute

Question 14.
Define degree of relationship?
Answer:
The number of entity types involved is known as Degree of relationship. .

One-Unary,
Two-Binary,
Three -Ternary.

Question 15.
Classify databases (or) Compare heavy database with light database?
Answer:
The databases are broadly divided into Heavy and Light databases. Heavy databases support all the desktop applications whereas the web applications are supported by Light databases.

Question 16.
List some commonly used databases?
Answer:
The lists of commonly used databases.

DB2
MySQL
Oracle
PostgreSQL
SQLite
SQL Server
Sybase

Question 17.
Give the syntax for inserting record?
Answer:
The Syntax for inserting record is
INSERT INTO table name (Parameterl,Parameter2, Parameter3..) VALUES (Value!, Value2, Value3..).

Question 18.
Give the syntax for USE database?
Answer:
Syntax:
mysql > use database name test;
Database changed
mysql>

Question 19.
Define designing of databases?
Answer:
The process of creating, implementing and maintaining the enterprise data in a system is known as Designing of databases.

III. Explain in Brief

Question 1.
Explain three database components of the network model?
Answer:
The three database components of Network models are: Network schema, Sub schema and Language.
Network schema – schema defines all about the structure of the database.
Sub schema – controls on views of the database for the user.
Language – basic procedural for accessing the database.

Question 2.
What are the advantages and limitations of the network model?
Answer:
Advantages:

The ability to handle more relationship types,
Easy data access,
Data integrity and independence.

Disadvantages:
The limitation of network model is difficulty in design and maintenance.

Question 3.
What are the two terminologies related With relational model?
Answer:
Relational model is defined with two terminologies Instance and Schema.

Instance – A table consisting of rows and columns
Schema – Specifies the structure including name arid type of each column.

Question 4.
Define Super key?
Answer:

An attribute or group of attributes, which is sufficient to distinguish every tuple in the relation from every other one is known as Super Key.
Each super key is called a candidate key.
A candidate key is selected from the set of Super Key.

Question 5.
Define Strong Entity?
Answer:
A Strong entity is the one which doesn’t depend on any other entity on the schema or database and a strong entity will have a primary key with it (i;e. a unique id which other entities will not have in their attributes). It is represented by one rectangle.

Question 6.
Define relationship instance with example?
Answer:
Each instance of the relationship between members of these entity types is called a relationship instance.
E.g if ‘B’ is the relationship between the Employee entity and the department entity, then Ram works for Comp. Sc department, Shyam works for Electrical department etc. are relationship instances of the relationship, works for.

Question 7.
Give the roles and responsibilities of DBA?
Answer:
Database Administrators (DBA’s) takes care of configuration, installation, performance, security and data backup.
DBA’s posses the skills on database design, database queries, RDMS, SQL and networking.
The primary task is the creation of new user and providing them with access rights.

Question 8.
List the components that make up a database (or) parts of the database?
Answer:
The three major parts that forms a database are Tables, Queries and Views.

Tables – similar to an excel sheet, containing multiple rows and columns. Where each row is a record and each column is an attribute.
Queries – It is a question with multiple conditions posted to the database. The records in the database that satisfies the passed conditions are retrieved.
Views -A set of stored queries.

Question 9.
List the functions of SQL?
Answer:
Functions performed using SQL are listed below:

Executes queries against a database.
Retrieves data from database.
Inserts and updates records in a database
Delete records from database.
Creates new databases and new tables in a database.

Question 10.
List the types of SQL commands?
Answer:
Data Definition Language (DDL),
Data Manipulation Language (DML),
Data Query Language (DQL),
Action Control Language (TCL),
Data Control Language (DCL).

Question 11.
Give the SQL DCL commands?
Answer:
SQL DCL COMMANDS List

Commands:

Grant
Revoke

Description:

Used to give permission to specific users on specific database objects like table, view etc
Used to take out permission from specific users on specific database objects like table, view etc.

Question 12.
How will you modify the records in the table?
Answer:
Modifying Record:
SQL provides us with modifying and updating the existing records in a table using UPDATE command. The age of Krishna in Biodata table is changed using the below Syntax.
Syntax:
UPDATE tablename
SET column1= “new value”
Where column2=“value2”;
Example: mysql>UPDATE Biodata SET age=13 WHERE firstname=“Krishna”;

Question 13.
Write note on operators in SQL?
MySQL Operators
Answer:

Question 14.
How will you arrange the records in ascending or descending order in a table?
Answer:
The Query results are listed in Ascending or Descending order using the command ORDER ‘ BY. In some databases the results are sorted by default in Ascending order.

Question 15.
How will you delete the record in the table?
Answer:
Deleting Record:
The existing record in a table is removed from the table using DELETE , command. Entire record or specified columns in the table can be deleted. If we want to perform delete operation on specific columns, then that condition is given in the WHERE condition. If the condition is not specified, then the entire data will be deleted.
Syntax: DELETE from tablename WHERE columnname= “value”;

Question 16.
What are the rules to be followed by sub query?
Answer:
The sub query must follow the below rules:

Subqueries are always written within the parentheses.
Always place the Subquery on the right side of the comparison operator.
ORDER BY clause is not used in sub query, since Subqueries cannot manipulate the results internally.

IV. Answer in detail

Question 1.
Explain RDBMS Jargons in detail?
Answer:
RDBMS Jargons
Database:
The most popular Relational Database is MySQL. It is an open source SQL database supporting different platforms like Windows, Linux and MAC Operating Systems. The other relational databases available are Oracle, MS SQL Server and MS Access. The features of RDBMS are

High Availability
High Performance
Robust TrAnswer:actions and support
Ease of management
Less cost

Table:
In relational database model, table is defined as the collection of data organized in terms of rows and columns. Table is the simple representation of relations. The true relations cannot have duplicate rows where as the table can have. The example of Employee table is shown below in Table.

Table Structure :

Column:
The table consists of several rows and columns. Table can be divided into smaller parts, in ‘ terms of columns. Each column is known as attributes. In the Employee table four attributes are available namely Id, Name, Age and Salary. The attribute is defined in a table to hold values of same type. This is known as Attribute Domain. In the Employee table, the Name field will hold only characters not the numbers in it. The vertical entity in a table is known as Attribute or Column.

Row:
A single entry in a table is called as Row or Record or Tuple. Set of related data’s are represented in a row or tuple. The horizontal entity in a table is known as Record or row.
Row Structure

Primary key:
The candidate key that is chosen to perform the identification task is called the primary key and any others are Alternate keys. Every tuple must have, by definition, a unique value for its primary key. A primary key which is a combination of more than one attribute is called a composite primary key.

Foreign Key:
A foreign key is a “copy” of a primary key that has been exported from one relation into another to represent the existence of a relationship between them. A foreign key is a copy of the whole of its parent primary key he if the primary key is composite, then so is the foreign key. Foreign key values do not (usually) have to be unique. Foreign keys can also
be null. A composite foreign key cannot have some attribute(s) null and others non-null.

Super Key:
An attribute or group of attributes, which is sufficient to distinguish every tuple in the relation from every other oiie is known as Super Key. Each super key is called a candidate key. A candidate key is selected from the set of Super Key. While selecting candidate key, redundant attributes should not be taken. The candidate key is also known as minimal super keys.

Composite Key:
A key with more than one attribute to identify rows uniquely in a table is called Composite key. This is also known as Compound Key.

Question 2.
Explain ER – Modeling diagram Notations?
Answer:
Entities, Attributes and Relationship forms the components of ER Diagram and the defined symbols and shapes are summarized below in Table.
ER diagram Notations.

Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

Instructions

The question paper comprises of four parts.
You are to attempt all the parts. 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 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART – 1

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
If there are 10 relations from a set A = {1, 2, 3,4, 5} to a set B, then the number of elements in B is ………….. .
(1) 3
(2) 2
(3) 4
(4) 8
Answer:
(2) 2

Question 2.
If g = {(1,1),(2, 3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are ………….. .
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2,-1)

Question 3.
If 7^4k = ………….. (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1

Question 4.
The next term ot the sequence $\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}$ is ………….. .
(1) $\frac{1}{24}$
(2) $\frac{1}{27}$
(3) $\frac{2}{3}$
(4) $\frac{1}{81}$
Answer:
(2) $\frac{1}{27}$

Question 5.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of K is ………….. .
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5

Question 6.
Find the matrix X if 2X + $\left[ \begin{matrix} 1 & 3 \\ 5 & 7 \end{matrix} \right] =\left[ \begin{matrix} 5 & 7 \\ 9 & 5 \end{matrix} \right]$ ………….. .
(1) $\left[ \begin{matrix} -2 & -2 \\ 2 & -1 \end{matrix} \right] $
(2) $\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] $
(3) $\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} \right] $
(4) $\left[ \begin{matrix} 2 & 1 \\ 2 & 2 \end{matrix} \right] $
Answer:
(2) $\left[ \begin{matrix} 2 & 2 \\ 2 & -1 \end{matrix} \right] $

Question 7.
The two tangents from an external points P to a circle with centre at O are PA and PB. If ∠APB = 70° then the value of ∠AOB is ………….. .
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°

Question 8.
The equation of a line passing through the origin and perpendicular to the line 7x – 3y + 4 = 0 is ………….. .
(1)7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0

Question 9.
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) $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$

Question 10.
The ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height is ………….. .
(1) 1:2:3
(2) 2:1:3
(3) 1:3:2
(4) 3:1:2
Answer:
(4) 3:1:2

Question 11.
The probability of getting a job for a person is $\frac { x }{ 3 }$. If the probability of not getting the job is $\frac { 2 }{ 3 }$ then the value of x is ………….. .
(1) 2
(2) 1
(3) 3
(4) 1.5
Answer:
(2) 1

Question 12.
Variance of the first 11 natural numbers is ………….. .
(1) √5
(2) √10
(3) 5√2
(4) 10
Answer:
(4) 10

Question 13.
If α and β are the roots of the equation ax² + bx + c = 0 then (α + β)² is ………….. .
(1) $-\frac{b^{2}}{a^{2}}$
(2) $\frac{c^{2}}{a^{2}}$
(3) $\frac{b^{2}}{a^{2}}$
(4) $\frac{b c}{a}$
Answer:
(3) $\frac{b^{2}}{a^{2}}$

Question 14.
If K(x) = 3x – 9 and L(x) = 7x – 10 then Lok is ………….. .
(1) 21x + 73
(2) -21x + 73
(3) 21x – 73
(4) 22x – 73
Answer:
(3) 21x – 73

PART – II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
Let A = {1,2,3,4,…, 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Answer:
A = {1,2,3,4. . . .45} .
The relation is defined as “is square of’
R = {(1, 1) (2, 4) (3, 9) (4, 16) (5, 25) (6, 36)} .
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1,4,9,16,25,36}

Question 16.
If f(x) = 3x – 2, g(x) = 2x + k and if fog = gof, then find the value of k.
Answer:
f(x) = 3x – 2, g(x) = 2x + k
fog(x) = f(g(x)) = f(2x + k) = 3(2x + k) – 2 = 6x + 3k – 2
Thus, fog(x) = 6x + 3k – 2.
gof(x) = g(3x – 2) = 2(3x – 2) + k
Thus, gof(x) = 6x – 4 + k.
Given that fog = gof
Therefore, 6x + 3k – 2 = 6x – 4 + k
6x – 6x + 3k – k = – 4 + 2 ⇒ k = – 1

Question 17.
Find the rational form of the number $0 . \overline{123}$.
Answer:
Letx = $0 . \overline{123}$
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P
Here a = 0.123, r = $\frac{0.000123}{0.123}$ = 0.001
S_n = $\frac{a}{1-r}=\frac{0.123}{1-0.001}=\frac{0.123}{0.999}=. \frac{41}{333}$

Question 18.
How many consecutive odd integers beginning with 5 will sum to 480?
Answer:
First term (a) = 5
Common difference (d) = 2 (consecutive odd integer)
S_n = 480

$\frac { n }{ 2 }$ [2a + (n – 1)d] = 480
$\frac { n }{ 2 }$ [10 + (n – 1)2] = 480
$\frac { n }{ 2 }$ [10 + 2n – 2] = 480
$\frac { n }{ 2 }$ (8 + 2 n) = 480
n( 4 + n) = 480
4n + n² – 480 = 0
n² + 4n – 480 = 0
(n + 24) (n – 20) = 0
n + 24 = 0 or n – 20 = 0
n = -24 or n = 20 [number of terms cannot be negative]
∴ Number of consecutive odd integers is 20

Question 19.
Simplify $\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}$
Answer:

Question 20.
Solve the following quadratic equations by factorization method √2x² + 7x + 5 √2 = 0
Answer:
√2x² + 7x + 5√2 = 0
√2x² + 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2) + 5(x + √2) = 0

(x + √2) or 5(x + √2) = 0 (equate the product of factors to zero)
x = -√2 or √2x = -5 x ⇒ x = $\frac{-5}{\sqrt{2}}$
The roots are -√2, $\frac{-5}{\sqrt{2}}$

Question 21.
Find the value of a, b, c, d from the equation $\begin{pmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{pmatrix}=\begin{pmatrix} 1 & 5 \\ 0 & 2 \end{pmatrix}$
Answer:
The given matrices are equal. Thus all corresponding elements are equal.
Therefore, a – b = 1 …(1)
2a + c = 5 …(2)
2a – b = 0 ….(3)
3c + d = 2 …(4)
(3) gives 2a – b = 0 …(4)
2 a = b …(5)
Put 2a = b in equation (1), a – 2a = 1 gives a = -1
Put a = -1 in equation (5), 2(-1) = b gives b = -2 .
Put a = -1 in equation (2), 2(-1) + c = 5 gives c = 7
Put c = 7 in equation (4), 3(7) + d = 2 gives = -19
Therefore, a = -1, b = -2, c = 7, d = -19

Question 22.
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC If AD = 8x -7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Answer:

Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
$\frac{A D}{D B}=\frac{A E}{E C}$
$\frac{8 x-7}{5 x-3}=\frac{4 x-3}{3 x-1}$
(8x – 7) (3x -1) = (4x -3) (5x -3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 20x² – 29x + 27x + 7 – 9 = 0
4x² – 2x – 2 = 0 .
2x² – x – 1 = 0 (Divided by 2)
2x² – 2x + x – 1 = 0

2x(x – 1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = $\frac { 1 }{ 2 }$ (Negative value will be omitted)
The value of x = 1

Question 23.
The hill in the form of a right triangle has its foot at (19, 3) . The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Answer:

Slope of AB (m) = tan 45° = 1
Equation of the hill joining the foot and the top is 45°
y – y₁ = m(x – x₁
y – 3 = 1(x – 19)
y – 3 = x – 19
– x + y – 3 + 19 = 0
– x + y + 16 = 0
x – y – 16=0
The required equation is x – y – 16 = 0

Question 24.
Prove that $\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}$ = sec θ + tan θ
Answer:

Question 25.
If the total surface area of a cone of radius 7cm is 704 cm², then find its slant height.
Answer:
Given that, radius r = 7 cm
Now, total surface area of the cone = πr(l + r)sq. units
T.S.A = 704 cm²
704 = $\frac { 22 }{ 7 }$ × 7(l + 7)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Question 26.
The first term of an A.P is 6 and the common difference is 5. Find the A.P and its general term.
Answer:
Given a = 6, d = 5
General term t_n = a + (n – 1) d
= 6 + (n – 1)5
= 6 + 5n – 5
= 5n + 1
The general form of the A.P is a, a + d, a + 2d ………
The A.P. is 6, 11, 16, 21 …. 5n + 1

Question 27.
If θ is an acute angle and tan θ + cot θ = 2 find the value of tan⁷θ + cot⁷θ
Answer:
Given tan θ + cot θ = 2
tan θ + $\frac{1}{\tan \theta}$ = 2
$\frac{\tan ^{2} \theta+1}{\tan \theta}$ = 2
tan²θ + 1 = 2 tan θ
tan²θ – 2 tan θ + 1 = 0
(tan θ – 1)² = 0
∴ tanθ – 1 = 0
tanθ = 1
tanθ = tan45 ⇒ θ = 45°
tan⁷θ + cot⁷θ = tan⁷45° + cot ⁷45°
= (1)⁷ + (1)^{7

= 2}

Question 28.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly one card is drawn from this box. Find the probability that the number on the card is a number which is a perfect square.
Answer:
Sample space = {2, 3, 4,… 101}
n(s) = 100
Let A be the event of getting perfect square numbers
A= {4, 9, 16, 25, 36, 49, 64, 81, 100}
n(A) = 9
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{100}$
Probability of getting a card marked with a number which is a perfect square is $\frac{9}{100}$

PART – III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Consider the functions f(x) = x², g(x) = 2x and h(x) = x + 4 Show that (fog)oh = fo(goh)
Answer:
f(x) = x² ; g(x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f( 2X)
= (2x)²
= 4x²
(fog) oh (x) = fog [h(x)]
= fog(x + 4)
= 4(x + 4)²
= 4[x² + 8x + 16]
= 4x² + 32x + 64 ….(1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh)x = fo[goh(x)}
= f[2x + 8}
= (2x + 8)²
= 4x² + 32x + 64 ……. (2)
From (1) and (2) we get (fog) oh = fo(goh)

Question 30.

(i) f(4)
(ii) f(-2)
(iii) f(4) + 2f(1)
(iv) $\frac{f(1)-3 f(4)}{f(-3)}$
Answer:
The function f is defined by three values in intervals I, II, III as shown by the side

For a given value of x = a, find out the interval at which the point a is located, there after find f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x – 2 ; f(4) = 3(4) – 2 = 10

(ii) x = -2 lies in the second interval.
Therefore, f(x) = x² – 2 ; f(-2) = (-2)² – 2 = 2

(iii) From (i), f(4) = 10.
To find f(1), first we see that x = 1 lies in the second interval.
Therefore,f(x) = x² – 2 => f(1) = 1² – 2 = -1
So, f(4) + 2f(1) = 10 + 2 (-1) = 8

(iv) We know that f(1) = -1 and f(4) = 10.
For finding f(-3), we see that x = -3 , lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, $\frac{f(1)-3 f(4)}{f(-3)}=\frac{-1-3(10)}{1}=-31$

Question 31.
If (m + 1)^th term of an A.P. is twice the (n + 1)^th> term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term.
Answer:
t_n = a + (n – 1)d
Given t_m+1 = 2 t_n+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md = 2a + 2nd
md – 2nd = a
d(m – 2n) = a ……. (1)
To prove that t_{3m + 1} = 2(t_{3m + n + 1})
L.H.S. = t_{3m + 1}
= a + (3m + 1 – 1 )d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)

R.H.S. = 2 (t_{m + n + 1})
= 2[a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2md – nd] = 2d (2m – n)
R.H.S = L.H.S
∴ t_{(3m + 1)}= 2t_{(m + n + 1)}

Question 32.
Find the sum to n terms of the series 5 + 55 + 555 + ….
Answer:
The series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.
5 + 55 + 555 + …. + n terms = 5[1 + 11 + 111 + …. + n terms]
= $\frac { 5 }{ 9 }$[9 + 99 + 999 + …. + « terms]
= $\frac { 5 }{ 9 }$[(10 – 1) + (100 – 1) + (1000 – 1) + …. + n terms)]
= $\frac { 5 }{ 9 }$[(10 + 100 + 1000 + …. +n terms) -n]
= $\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{(10-1)}-n\right]=\frac{50\left(10^{n}-1\right)}{81}-\frac{5 n}{9}$

Question 33.
Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. If 4 years ago Vani’s grandfather was four times as old as Vani then how old are they all now?
Answer:
Let the age of Vani be”x” years
Vani father age = “y” years
Vani grand father = “z” years
By the given first condition.
$\frac{x+y+z}{3}=53$
x + y + z = 159….(1)
By the given 2nd condition.
$\frac{1}{2} z+\frac{1}{3} y+\frac{1}{4} x=65$
Multiply by 12
6z + 4y +3x = 780
3x + 4y + 6z = 780 ….(2)
By the given 3^rd condition
z – 4 = 4 (x – 4) ⇒ z – 4 = 4x – 16
-4x + z = -16 + 4
4x – z = 12 ….(3)

Substitute the value of x = 24 in (3)
4 (24) – z = 12
96 – z = 12
-Z = 12 – 96
z = 84
Substitute the value of
x = 24 and z = 84 in (1)
24 + y + 84 = 159
y + 108 = 159
y = 159 – 108
= 51
Vani age = 24 years
Vani’s father age = 51 years
Vani – grand father age = 84 years

Question 34.
If A = $\left( \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \end{matrix} \right)$ and B = $\left( \begin{matrix} 2 & -1 \\ -1 & 4 \\ 0 & 2 \end{matrix} \right)$ show that (AB)^T = B^TA^T
Answer:

Question 35.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.
Answer:

In the right ∆ OTP,
PT² = OT² – OP²
= 13² – 5²
= 169 – 25 = 144
PT = √144 = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x .
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB .
∠OEA= 90°
∠AET = 90°
In ∆AET, AT² = AE² + ET²
In the right triangle AET,
AT² = AE² + ET²
(12 – x)² = x² + (13 – 5)²
144 – 24x + x² = x² + 64
24x = 80 ⇒ x = $\frac{80}{24}=\frac{20}{6}=\frac{10}{3}$
BE = $\frac{10}{3}$ cm
AB = AE + BE
= $\frac{10}{3}+\frac{10}{3}=\frac{20}{3}$
∴ Length of AB = $\frac{20}{3}$ cm

Question 36.
Find the area of the quadrilateral whose vertices are at (-9, 0), (-8, 6), (-1, -2) and (-6,-3)
Answer:

Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB
= $\frac { 1 }{ 2 }$ [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄ )]

= $\frac { 1 }{ 2 }$ [27 + 12- 6 + 0 -(0 + 3 + 16 – 54)]
= $\frac { 1 }{ 2 }$ [33 -(-35)]
= $\frac { 1 }{ 2 }$ [33 + 35] = $\frac { 1 }{ 2 }$ × 68 = 34 sq. units.
Area of the Quadrilateral = 34 sq. units

Question 37.
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point ‘A’ on the ground is 60° and the angle of depression to the point ‘A’ from the top of the tower is 45°. Find the height of the tower. (√3 = 1.732)
Answer:
Let BC be the height of the tower and CD be the height of the pole.
Let‘A’be the point of observation.
Let BC = x and AB = y.
From the diagram,

∠BAD = 60° and ∠XCA = 45° = ∠BAC ,
In right triangle ABC, tan 45° = $\frac{B C}{A B}$
gives 1 = $\frac { x }{ y }$ so, x = y …… (1)
In right triangle ABD, tan60° = $\frac{B D}{A B}=\frac{B C+C D}{A B}$
gives √3 =  so, √3y = x + 5
we get √3x = x + 5 [From (1)]

Hence, height of the tower is 6.83 m.

Question 38.
A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.
Answer:

Radius of the lower end of the frustum (r) = 1 cm
Radius of the upper end of the frustum (R) = 2.5 cm
Height of the frustum (h) = 6 cm
Let “l” be the slant height of the frustum
l = $\sqrt{h^{2}+(\mathrm{R}-r)^{2}}$
= $\sqrt{6^{2}+(2.5-1)^{2}}$
= $\sqrt{36+2.25}$ = $\sqrt{38.25}$
= 6.18 cm
External surface area of shuttle cock = C.S.Aof the frustum + C.S.Aof a hemisphere
= πl(R + r) + 2 πr²
= π [6.18 (2.5 + 1) + 2 × 1²] cm²
= $\frac { 22 }{ 7 }$[6.18 × 3.5 + 2]
= $\frac { 22 }{ 7 }$ × (21.63 + 2)
= $\frac { 22 }{ 7 }$ × 23.63 cm² = 74.26 cm²
External surface area = 74.26 cm²

Question 39.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4,10,12 and 14, then find the remaining two observations.
Answer:
Let the missing two observation be ‘a’ and ‘b’
Arithmetic mean = 8

$\frac{2+4+10+12+14+a+b}{7}$ = 8 ⇒ $\frac{42+a+b}{7}$ = 8
a + b + 42 = 56
a + b = 56 – 42
a + b = 14 ……… (1)
Variance = 16
Variance = $\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$

560 – 460 = a² + b²
a² + b²= 100 ⇒ (a + b)² – 2ab = 100 [a² + b² = (a+b)² – 2 ab]
14² – 2 ab = 100 ⇒ 196 – 2 ab = 100 [a + b = 14(from (1)]
196 – 100 = 2ab
96 = 2ab ⇒ ab = $\frac{96}{2}$ = 48
∴ b = $\frac{48}{a}$ …… (2)
Substitute thr value of b = $\frac{48}{a}$ in (1)
a + $\frac{48}{a}$ = 14 ⇒ a² + 48 = 14a
a² – 14a + 48 = 0 ⇒ (a – 6) (a – 8) = 0
a = 6 or 8
when a = 6
b = $\frac{48}{a}=\frac{48}{6}=8$ = 8

when a = 8
b = $\frac{48}{a}=\frac{48}{6}=8$ = 6

∴ Missing observation is 8 and 6 (or) 6 and 8

Question 40.
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find (i) the volume of the remaining solid also, find (ii) the whole surface area.
Answer:
Circular cylinder
Radius of the base (r) = 6 cm
Height of the cylinder (h) = 10 cm

Circular Cone
Radius of the base (R) = 6 cm
Height of the cone (H) = 10 cm

(i) Volume of the remaining solid = Volume of the cylinder – Volume of the cone

= πr²h – $\frac { 1 }{ 3 }$πR²H
= π(r²h – $\frac { 1 }{ 3 }$R²H)
= $\frac { 22 }{ 7 }$ [6 × 6 × 10 – $\frac { 1 }{ 3 }$ × 6 × 6 × 10] cm²
= $\frac { 22 }{ 7 }$ [360 – 120] cm³
= $\frac { 22 }{ 7 }$ × 240 cm³
= $\frac { 5280 }{ 7 }$ cm³ = 754.29 cm³

(ii) Slant height of a cone = $\sqrt{h^{2}+r^{2}}$
= $\sqrt{10^{2}+6^{2}}$
= $\sqrt{100+36}$
= $\sqrt{136}$ = 11.66 cm
Whole surface area of the solid = curved surface area of the cylinder + curved surface area of the cone + base area
= 2 πrh + πRI + πr²
= π[2 × 6 × 10 + 6 × 11.66 + 6 × 6]
= $\frac { 22 }{ 7 }$[120 + 69.96 + 36] cm²
= $\frac { 22 }{ 7 }$ × 225.96 cm²
= $\frac { 4971.12 }{ 7 }$ cm²
= 710.16 cm²
(i) Volume of the remaining solid = 754.29 cm³
(ii) whole surface area = 710.16 cm²

Question 41.
If a and p are the roots of the equation 3x² – 6x + 1=0 form the equation whose roots are 2α + β and 2β + α
Answer:
α and β are the roots of 3x² – 6x + 1 = 0
α + β = $\frac { 6 }{ 3 }$ = 2
αβ = $\frac { 1 }{ 3 }$
Given the roots are 2α + β; 2β + α
Sum of the roots = 2α + β + 2β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6
Product of roots = (2α + β) (2β + α)
= 4αβ + 2α² + 2β² + αβ
= 5αβ + 2(α² + β²)
5αβ + 2[(α + β)² – 2αβ]
= 5 × $\frac { 1 }{ 3 }$ + 2 (4 – 2 × $\frac { 1 }{ 3 }$)
= $\frac { 5 }{ 3 }$ + $2\left(\frac{12-2}{3}\right)$
= $\frac { 5 }{ 3 }$ + 2 × $\frac { 10 }{ 3 }$
= $\frac{5}{3}+\frac{20}{3}$
= $\frac { 25 }{ 3 }$
The quadratic polynomial is x² – (sum of the roots)x – product of the roots = 0
x² – 6x + $\frac { 25 }{ 3 }$ = 0
3x² – 18x + 25 = 0

Question 42.
Tw o dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:
Sample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3.1) , (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5.1) , (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
Let A be the event of getting the sum is divisible by 3 ‘
A = { (1,2) (2,1) (1,5) (5,1) (2,4) (4,2) (3,3) (3,6) (6,3) (4,5) (5,4) (6,6)}
n (A) = 12
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{12}{36}$
Let B be the event of getting a sum is divisible by 4.
B = {(1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6,6)} n (B) = 9
n(B) = 9
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{9}{36}$
A ∩ B = {(6,6)}
n(A ∩ B) = 1
P(A ∩ B) = $\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{1}{36}$
P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
$=\frac{12}{36}+\frac{9}{36}-\frac{1}{36}$
$=\frac{12+9-1}{36}=\frac{20}{36}$
Neither divisible by 3 nor by 4
P(A’ ∩ B’) = P(A ∪ B)’
= 1 – p(A ∪ B) = 1 – $\frac{20}{36}=\frac{36-20}{36}$
= $\frac{16}{36}=\frac{4}{9}$

PART-IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:

Steps of construction:

With. O as centre, draw a circle of radius 5 cm.
Draw a line OP = 10 cm.
Draw a perpendicular bisector of OP, which cuts OP at M.
With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
Join AP and BP. AP and BP are the required tangents.

Verification: In the right ∆ OAP
PA² = OP² – OA²
= 10² – 5² = $\sqrt{100-25}$ = √75 = 8.7 cm.
Length of the tangent is 8.7 cm

[OR]

(b) Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:

Steps of construction

Draw a line segment PQ = 4.5 cm
At P, draw PE such that ∠QPE = 60°
At P, draw PF such that ∠EPF = 90°
Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
With O as centre and OP as radius draw a circle.
From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
Join PR and RQ. PQR is the required triangle.

Question 44.
(a) Draw the graph of y = x² and hence solve x² – 4x – 5 = 0.
Answer:
Given equations are y = x² and x² – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.

(ii) Plot the points (-4, 16), (- 3, 9), (-2,4), (-1, 1), (0, 0), (1, 1), (2,4), (3, 9), (4, 16), (5, 25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations

(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.

[OR]

(b) Draw the graph of y – x² -5x – 6 and hence solve x² – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5, -6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Samacheer Kalvi 11th Accountancy Solutions Chapter 8 Bank Reconciliation Statement

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

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

I. Multiple Choice Questions
Choose the Correct Answer

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

Question 2.
A bank reconciliation statement is prepared with the help of ………………
(a) Bank statement
(b) Cash book
(c) Bank statement and bank column of the cash book
(d) Petty cash book
Answer:
(c) Bank statement and bank column of the cash book

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

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

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

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

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

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

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

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

II. Very Short Answer Questions

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

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

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

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

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

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

III. Short Answer Questions

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

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

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

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

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

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

IV. Exercises

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:
Bank reconciliation statement

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

Bank Statement

Answer:

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

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

Textbook Case Study Solved

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

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

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

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

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

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

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

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

Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Leave a Comment / Class 12 / By Prasanna

Students can Download Computer Applications Chapter 5 PHP Function and Array Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Samacheer Kalvi 12th Computer Applications PHP Function and Array Text Book Back Questions and Answers

I. Choose The Correct Answer

Question 1.
Which one of the following is the right way of defining a function in PHP?
(a) function {function body }
(b) data type functionName(parameters) {function body}
(c) functionName(parameters) {function body }
(d) function functionName(parameters) { function body }
Answer:
(d) function functionName(parameters) { function body }

Question 2.
A function in PHP which starts with ………. (double underscore) is known as …………………………
(a) Magic Function
(b) Inbuilt Function
(c) Default Function
(d) User Defined Function
Answer:
(a) Magic Function

Question 3.
PHP’s numerically indexed array begin with position ………………………..
(a) 1
(b) 2
(c) 0
(d) -1
Answer:
(c) 0

Question 4.
Which of the following are correct ways of creating an array?
(i) state[0] = “Tamilnadu”;
(ii) $state[ ] = arfay(“Tamilnadu”);
(iii) $state[0] = “Tamilnadu”;
(iv) $state = array(“Tamilnadu”);
(a) (iii) and (iv)
(b) (ii) and (iii)
(c) Only (i)
(d) (ii), (iii) and (iv)
Answer:
(a) (iii) and (iv)

Question 5.
What will be the output of the following PHP code? <?php
$a=array(“A”,“Cat”,“Dog”,“A”,“Dog”);
$b=array(“A”,“A”,“Cat”,“A”,“Tiger”);
$c=array_combine($a,$b);
print_r(array_count_values($c));
?>
(а) Array ([A] => 5 [Cat] => 2 [Dog] => 2 [Tiger] => 1)
(b) Array ([A] => 2 [Cat] => 2 [Dog] => 1 [Tiger] => 1)
(c) Array ([A] => 6 [Cat] => 1 [Dog] => 2 [Tiger] => 1)
(d) Array ([A] => 2 [Cat] => 1 [Dog] => 4 [Tiger] => 1)
Answer:
(а) Array ([A] => 5 [Cat] => 2 [Dog] => 2 [Tiger] => 1)

Question 6.
For finding nonempty elements in array we use
(a) is_array ( ) function)
(b) sizeof( ) function
(c) array_count ( ) function
(d) count ( ) function
Answer:
(d) count ( ) function

Question 7.
Indices of arrays can be either strings or numbers and they are denoted as
(a) $my_array {4}
(b) $my array [4]
(c) $my_array |4|
(d) None of these
Answer:
(b) $my array [4]

Question 8.
PHP arrays are also called as
(a) Vector arrays
(b) Perl arrays
(c) Hashes
Answer:
(c) Hashes

Question 9.
As compared to associative arrays, vector arrays are much
(a) Faster
(b) Slower
(c) Stable
(d) None of them
Answer:
(b) Slower

Question 10.
What functions count elements in an array?
(a) count
(b) both a and b
(c) Array Count
(d) Count_array
Answer:
(b) both a and b

PART – II
I. Short Answer

Question 1.
Define Function in PHP?
Answer:
In most of the programming language, a block of segment in a program that performs a specific operation tasks (Insert, Execute, Delete, Calculate, etc.). This segment is also known as Function. A Function is a type of sub routine or procedure in a program.

Question 2.
Define User defined Function?
Answer:
User Defined Function:
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module.

Function Declaration:
A user-defined Function declaration begins with the keyword “function”.

Question 3.
What is parameterized Function?
Answer:
Parameterized Function:

PHP Parameterized functions are the functions with parameters or arguments.
Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables.
The arguments are mentioned after the function name and inside of the parenthesis.
There is no limit for sending arguments, just separate them with a comma notation.

Question 4.
List out System defined Functions?
Answer:

round
cos
tan
is_number
rand etc
sqrt
sin
pi
number_format

Question 5.
Define Array in PHP?
Answer:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array in PHP.

Indexed Arrays
Associative Array and
Multi-Dimensional Array

‘

Question 7.
Usage of Array in PHP?
Answer:
One of the most useful aspects of using arrays in PHP is when combined with the foreach statement. This allows you to quickly loop though an array with very little code.

Question 8.
List out the types of array in PHP?
Answer:

Question 9.
Define associative array?
Answer:
Associative arrays are a key-value pair data structure. Instead of having storing data in a linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Question 10.
Write array Syntax in PHP?
Answer:
Array defines with the keyword array( )

PART – III
III. Explain in Brief Answer

Question 1.
Write the features of System defined Functions?
Answer:
System Defined Functions: A function is already created by system it is a reusable piece or block of code that performs a specific action. Functions can either return values when called or can simply perform an operation without returning any value.
Features of System defined functions:

System defined functions are otherwise called as predefined or built-in functions.
PHP has over 700 built in functions that performs different tasks.
Built in functions are functions that exists in PHP installation package.

Question 2.
Write the purpose of parameterized Function?
Answer:

Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables. The arguments are mentioned after the function name and inside of the parenthesis.
There is no limit for sending arguments, just separate them with a comma notation.

Question 3.
Differentiate user define and system defined Functions?
Answer:
System defined:

These functions are already present in the system (comes along with installation package).
Cannot be edited.
Name is provided by the developer Eg. isnum( ), isalpha( ).

System defined:

These functions are already present in the system (comes along with installation package).
Cannot be edited.
Name is provided by the developer Eg. isnum( ), isalpha( ).

Question 4.
Write Short notes on Array?
Array in PHP
Answer:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array concepts in PHP.

Indexed Arrays,
Associative Array and
Multi-Dimensional Array.

Question 5.
Differentiate Associate array and Multidimensional array?
Answer:
Associate array:
Associative arrays are a key-value pair data structure. Instead of having storing data in a* linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Multidimensional array:
A multidimensional array is an array containing one or more arrays. PHP understands multidimensional arrays that are two, three, four, five, or more levels deep. However, arrays more than three levels deep are hard to manage for most people.

PART – IV
IV. Explain in detail

Question 1.
Explain Function concepts in PHP?
Answer:
Functions in PHP
In most of the programming language, a block of segment in a program that performs a specific operation tasks (Insert, Execute, Delete, Calculate, etc.). This segment is also known as Function. A Function is a type of sub routine or procedure in a program.

Function will be executed by a call to the Function and the Function returns any data type values or NULL value to called Function in the part of respective program.
The Function can be divided into three types as follows:

User defined Function,
Pre-defined or System or built-in Function, and
Parameterized Function.

1. User Defined Function:
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module. Two important steps the Programmer has to create for users define Functions are: Function declaration and Function calling.

2. System Define Functions:
A function is already created by system it is a reusable piece or block of code that performs a specific action. Functions can either return values when called or can simply perform an operation without returning any value.

3. Parameterized Function:
PHP Parameterized functions are the functions with parameters or arguments.

Question 2.
Discuss in detail about User defined Functions?
Answer:
User Defined Function:

User can create their own functions.
User Defined Function (UDF) in PHP gives a privilege to user to write own specific operation inside of existing program module.
Two important steps the Programmer has to create for users define Functions are:

Function Declaration:
A user-defined Function declaration begins with the keyword “function”. User can write any custom logic inside the function block.
Syntax:
function functionName( )
{
Custom Logic code to be executed;
}
Function Calling:
A function declaration part will be executed by a call to the function. Programmer has to create Function Calling part inside the respective program.
Syntax:
functionName( );
Example:
<?php
function insertMsg( )
{
echo “Student Details Inserted Successfully!”;
}
insertMsg( ); // call the function
?>
Parameterized Function:

Required information can be shared between function declaration and function calling part inside the program.
The parameter is also called as arguments, it is like variables.
The arguments are mentioned after the function name and inside of the parenthesis. There is no limit for sending arguments, just separate them with a comma notation.

Question 3.
Explain the Multidimensional Array?
Answer:
Multidimensional Arrays:

A multidimensional array is an array containing one or more arrays.
PHP understands multidimensional arrays that are two, three, four, five, or more levels deep.
However, arrays more than three levels deep are hard to manage for most people.

Example:
<?php
// A two-dimensional array Sstudent-array
(
array(“Iniyan”, 100,96),
array(“Kavin”,60,59),
array(“Nilani”,1313,139)
);
echo $$student[0][0].“: Tamil Mark: “.$student [0][1]English mark: “.$student [0] [2].” ”;

echo $$student[1][0].“: Tamil Mark: “.$student [1][1].”. English mark: “.$student [1] [2].” ”;

echo $$student[2][0].“: Tamil Mark: “.$student [2][1]English mark: “.$student [2] [2].” ”;
?>

Question 4.
Explain Array concepts and their types?
Answer:
Arrays in PHP:
Array is a concept that stores more than one value of same data type (homogeneous) in single array variable. They are 3 types of array concepts in PHP.

Indexed Arrays,
Associative Array and
Multi-Dimensional Array.

SYNTAX:
Array Syntax:
Array defines with the keyword array( )

1. Indexed Arrays:
Arrays with numeric index for the available values in array variable which contains key value pair as user / developer can take the values using keys.

2. Associative Arrays:

Associative arrays are a key-value pair data structure.
Instead of having storing data in a linear array, with associative arrays you can store your data in a collection and assign it a unique key which you may use for referencing your data.

Question 3.
Multidimensional Arrays?
Answer:

A multidimensional array is an array containing one or more arrays. .
PHP understands multidimensional arrays that are two, three, four, five, or more levels deep.
However, arrays more than three levels deep are hard to manage for most people.

Question 5.
Explain Indexed array and Associate array in PHP?
Answer:
Indexed Arrays:
Arrays with numeric index for the available values in array variable which contains key value pair as user / developer can take the values using keys.
Example:
<?php
$teacher_name=array(“Iniyan”, “Kavin”, “Nilani”);
echo “The students name are “ . $teacher_name[0]. “, “ . $$teacher_name[l]. “ and” . $teacher_name[2].
?>

Associative Arrays:

Associative arrays are a key-value pair data structure.
Instead of having storing data in a, linear array, with associative arrays you can store your data.

Example:
<?php
$Marks=array(“Studentl”=>“35”,“Student2”==>“17”,“Student3”=>“43”);
echo “Studentl mark is” . $Marks[‘Studentl’]. “ is eligible for qualification”;
echo “Student2 mark is” . $Marks[‘Student2’]. “ is not eligible for qualification”;

Samacheer Kalvi 12th Computer Applications Solutions PHP Function and Array Additional Questions and Answers

I. Choose the Best Answer

Question 1.
………………………. are functions that exist in PHP installation package.
Answer:
Built-in functions

Question 2.
PHP has over ………………………. built in functions.
(a) 200
(b) 500
(c) 700
(d) 900
Answer:
(c) 700

Question 3.
A function is a type of …………………………… in a program.
(a) sub routine
(b) procedure
(c) both a & b
(d) array
Answer:
(c) both a & b

Question 4.
How many classification of functions are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 5.
Find the wrong statement from the following?
(a) pre-defined functions are called as built-in functions
(b) pre-defined functions are called as system functions
(c) parameterized functions are called system functions
Answer:
(c) parameterized functions are called system functions

Question 6.
UDF stands for …………………………..
Answer:
User defined functions

Question 7.
Pick the odd one out related to functions.
(a) userdefined
(b) pre-defined
(c) system
(d) builtin
Answer:
(a) userdefined

Question 8.
Find the statement which is correct.
I: A user-defined function declaration begins with the keyword “user”
II: user can write any custom logic inside the function block
(a) I-True, II-False
(b) I-Flase, II-True
(c) I, II-True
(d) I, II-False
Answer:
(b) I-Flase, II-True

Question 9.
The parameter is also called as
(a) Arguments
(b) Value
(c) Calling function
(d) return
Answer:
(a) Arguments

Question 10.
THe arguments are separated by
(a) .
(b) :
(c) ,
(d) ;
Answer:
(c) ,

Question 11.
The arguments have to be enclosed within
(a) [ ]
(b) <>
(c) { }
(d) ( )
Answer:
(d) ( )

Question 12.
The function blocks are given by
(a) [ ]
(b) <>
(c) { }
(d) ( )
Answer:
(c) { }

Question 13.
How many arguments can be send to return a value in a function?
(a) 2
(b) 3
(c) 4
(d) No limit
Answer:
(d) No limit

Question 14.
Which statement can be used to return a value in a function?
(a) pass
(b) call
(c) return
(d) give
Answer:
(c) return

Question 15.
Which is used to store more than one value in a single variable of same data type?
(a) Array
(b) Functions
(c) Class
(d) Object
Answer:
(a) Array

Question 16.
How many types of arrays are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 17.
Pick the odd one out.
Indexed, Associative, Multi functional, Multi dimensional
Answer:
Multi functional

Question 18.
What is the keyword for creating an array?
(a) Arr( )
(b) array( )
(c) a( )
(d) a[ ]
Answer:
(b) array( )

Question 19.
One of the most useful aspects of using array is when it is combined with the …………………………. statement.
(a) if
(b) while
(c) for
(d) foreach
Answer:
(d) foreach

Question 20.
…………………………. specifies the value in the syntax of Association Array.
(a) value
(b) val
(c) value at
(d) value of
Answer:
(a) value

Question 21.
The key in the Associative Array may be of …………………………… or …………………………..
Answer:
numeric or string

Question 22.
………………………… arrays are a key value pair data structure.
(a) Indexed
(b) Numeric
(c) Associative
(d) Multidimensional
Answer:
(c) Associative

Question 23.
In PHP ………………………….. are Information hiding.
(a) Functions
(b) Variables
(c) Keywords
(d) datatypes
Answer:
(a) Functions

II. Short Answer

Question 1.
Classify functions?
Answer:
The Function can be divided in to three types as follow

User defined Function,
Pre-defined or System or built-in Function, and
Parameterized Function.

Question 2.
Define Magic functions?
Answer:

A function in PHP which starts with underscore ( ) is called as magic function.
Magic function is present inside the class.

III. Explain in Brief Answer

Question 1.
Write a PHP program for function with one arguments?
Answer:
<?php
function School_Name($sname) {
echo $sname.“in Tamilnadu. ”;
}
SchoolName (“Government Higher Secondary School Madurai”);
SchoolName (“Government Higher Secondary School Trichy”);
School Name (“Government Higher Secondary School Chennai”);
School Name (“Government Higher Secondary School Kanchipuram”);
School Name (“Government Higher Secondary School Tirunelveli”);
?>

Question 2.
Give an example for function with two arguments?
Answer:
<?php
function School_Name($sname,$Stfength) {
echo $sname.“in Tamilnadu and Student Strength is”.$Strength;
}
School Name (“Government Higher Secondary School Madurai”,200);
School Name (“Government Higher Secondary School Trichy”,300);
School Name (“Government Higher Secondary School Chennai”,250);
School Name (“Government Higher Secondary School Kanchipuram”, 100);
School Name (“Government Higher Secondary School Tirunelveli”,200);
?>

Question 3.
Write a PHP program to explain function with return values?
Answer:
For a function to return a value, use the return statement
<?php
function sum($x, $y) {
$z = $x + $y;
return $z;
}
echo “5 + 10 = “ . sum(5, 10). “ ”;
echo “7 + 13 = “ . sum(7, 13). “ ”;
echo “2 + 4 = “ . sum(2, 4);
?>

Question 4.
Give the Syntax for Associative Arrays?
Answer:
Associative Arrays Syntax:
array(key ⇒ value,key ⇒ value,key ⇒ value,etc.);
key = Specifies the key (numeric or string)
value = Specifies the value.

Question 5.
What is the use of array combine( ) in PHP?
Answer:
The array_combine( ) is an inbuilt function in PHP.
It is used to combine two arrays and create a new array by using one array for keys and another for values.
Eg. $c = array_combine($a, $b);

IV. Explain in detail

Question 1.
Give important characteristics of PHP functions?
Answer:

PHP Functions are Reducing duplication of code.
PHP Functions are Decomposing complex problems into simpler pieces.
PHP Functions are Improving clarity of the code.
PHP Functions are Reuse of code.
PHP Functions are Information hiding.

Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Question 1.
Haemophilia is more common in males because it is a ____________
(a) Recessive character carried by Y-chromosome
(b) Dominant character carried by Y-chromosome
(c) Dominant trait carried by X-chromosome
(d) Recessive trait carried by X-chromosome
Answer:
(d) Recessive trait carried by X-chromosome

Question 2.
ABO blood group in man is controlled by ____________
(a) Multiple alleles
(b) Lethal genes
(c) Sex linked genes
(d) Y-linked genes
Answer:
(a) Multiple alleles

Question 3.
Three children of a family have blood groups A, AB and B. What could be the geno types of their parents?
(a) I^AI^B and ii
(b) I^A1^O and I^BI^O
(c) I^B I^B and I^AI^A
(d) I^AI^A and ii
Answer:
(b) I^AI^O and I^BI^O

Question 4.
Which of the following is not correct?
(a) Three or more alleles of a trait in the population are called multiple alleles.
(b) A normal gene undergoes mutations to form many alleles
(c) Multiple alleles map at different loci of a chromosome
(d) A diploid organism has only two alleles out of many in the population
Answer:
(c) Multiple alleles map at different loci of a chromosome

Question 5.
Which of the following phenotypes in the progeny are ____________
(a) A and B only
(b) A,B only and AB
(c) AB only
(d) A,B,AB and O
Answer:
(d) A,B,AB and O

Question 6.
Which of the following phenotypes is not possible in the progeny of the parental genotypic combination I^AI^O × l^Al^B ?
(a) AB
(b) O
(c) A
(d) B
Answer:
(b) O

Question 7.
Which of the following is true about Rh factor in the offspring of a parental combination DdXDd (both Rh positive)?
(a) All will be Rh positive
(b) Half will be Rh positive
(c) About 3/4 will be Rh negative
(d) About one fourth will be Rh negative
Answer:
(d) About one fourth will be Rh negative

Question 8.
What can be the blood group of offspring when both parents have AB blood group?
(a) AB only
(b) A, B and AB
(c) A, B, AB and O
(d) A and B only
Answer:
(b) A, B and AB

Question 9.
If the childs blood group is ‘O’ and fathers blood group is ‘A’ and mother’s blood group is ‘B’ the genotype of the parents will be _________
(a) I^AI^A and I^AI^O
(b) I^AI^O and I^BI^O
(c) I^AI^O and I^OI^O
(d) I^OI^O and I^BI^B
Answer:
(b) I^AI^O and I^BI^O

Question 10.
XO type of sex determination and XY type of sex determination are examples of _________
(a) Male heterogamety
(b) Female heterogamety
(c) Male homogamety
(d) Both (b) and (c)
Answer:
(a) Male heterogamety

Question 11.
In an accident there is great loss of blood and there is no time to analyse the blood group which blood can be safely transferred?
(a) ‘O’ and Rh negative
(b) ‘O’ and Rh positive
(c) ‘B’ and Rh negative
(d) ‘AB’ and Rh positive
Answer:
(a) ‘O’ and Rh negative

Question 12.
Father of a child is colour blind and mother is carrier for colour blindness, the probability of the child being colour blind is _________
(a) 25%
(b) 50%
(c) 100%
(d) 75%
Answer:
(b) 50%

Question 13.
A marriage between a colour blind man and a normal woman produces _________
(a) All carrier daughters and normal sons
(b) 50% carrier daughters, 50% normal daughters
(c) 50% colour blind sons, 50% normal sons
(d) All carrier off springs
Answer:
(a) All carrier daughters and normal sons

Question 14.
Mangolism is a genetic disorder which is caused by the presence of an extra chromosome number.
(a) 20
(b) 21
(C) 4
(d) 23
Answer:
(b) 21

Question 15.
Klinefelters’ syndrome is characterized by a karyotype of _________
(a) XYY
(b) XO
(c) XXX
(d) XXY
Answer:
(d) XXY

Question 16.
Females with Turners’syndrome have _________
(a) Small uterus
(b) Rudimentary ovaries
(c) Underdeveloped breasts
(d) All of these
Answer:
(d) All of these

Question 17.
Pataus’ syndrome is also referred to as _________
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 18.
Who is the founder of Modem Eugenics movement?
(a) Mendel
(b) Darwin
(c) Fransis Galton
(d) Karl pearson
Answer:
(c) Fransis Galton

Question 19.
Improvement of human race by encouraging the healthy persons to marry early and produce large number of children is called _________
(a) Positive eugenics
(b) Negative eugenics
(c) Positive euthenics
(d) Positive euphenics
Answer:
(a) Positive eugenics

Question 20.
The _________ deals with the control of several inherited human diseases especially inborn errors of metabolism.
(a) Euphenics
(b) Eugenics
(c) Euthenics
(d) All of these
Answer:
(a) Euphenics

Question 21.
“Universal Donor” and “Universal Recipients” blood group are _________ and _________ respectively.
(a) AB, O
(b) O, AB
(c) A, B
(d) B, A
Answer:
(b) O, AB

Question 22.
ZW-ZZ system of sex determination occurs in _________
(a) Fishes
(b) Reptiles
(c) Birds
(d) All of these
Answer:
(d) All of these

Question 23.
Co-dominant blood group is _________
(a) A
(b) AB
(c) B
(d) O
Answer:
(b) AB

Question 24.
Which of the following is incorrect regarding ZW-ZZ type of sex determination?
(a) It occurs in birds and some reptiles
(b) Females are homogametic and males are heterogametic
(c) Male produce two types of gametes
(d) It occurs in gypsy moth
Answer:
(b) Females are homogametic and males are heterogametic

Question 25.
What is haplodiploidy?
Answer:
In haplodiploidy, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid).

Question 26.
Distinguish between heterogametic and homogametic sex determination systems
Answer:
Heterogametic Sex:

Organisms producing two different types of gametes.
Example: Human male.
Sperm with X chromosome Sperm with a Y chromosome

Heterogametic Sex:

Organisms producing two different types of gametes.
Example: Human female.
Every egg produced contain X chromosomes.

Question 27.
What is Lyonisation?
Answer:
Lyonisation is a process of inactivation of one of the X chromosomes in some females.

Question 28.
What is criss-cross inheritance?
Answer:

Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild.
E.g., X-linked gene inheritance.

Question 29.
Why are sex-linked recessive characters more common in male human beings?
Answer:
Sex-linked inherited traits are more common in males than females because males are hemizygous and therefore express the trait when they inherit one mutant allele.

Question 30.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 31.
Mention the symptoms of Phenylketonuria.
Answer:
Severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 32.
Mention the symptoms of Down’s syndrome.
Answer:
Severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 33.
Differentiate Intersexes from Supersexes.
Answer:
Intersexes :
Intersexes refers to the individuals having the characteristics of both female and male sexes and their sexual anatomy does not seem to fit the typical definition of male or female.
Example: Super males in humans human beings have 44+XYY chromosomes.

Supersexes:
Supersexes are formed as a result of an abnormal combination of sex chromosomes.
Example: Super males in humans human beings have 44+XYY chromosomes.

Question 34.
Explain the genetic basis of ABO blood grouping in man.
Multiple allele inheritance of ABO blood groups
Answer:
Blood differs chemically from person to person. When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells. Karl Landsteiner discovered two kinds of antigens called antigen ‘A’ and antigen ‘B’ on the surface of RBC’s of human blood. Based on the presence or absence of these antigens three kinds of blood groups, type ‘A’, type ‘B’, and type ‘O’ (universal donor) were recognized. The fourth and the rarest blood group ‘AB’ (universal recipient) was discovered in 1902 by two of Landsteiner’s students Von De Castelle and Sturli.

Bernstein in 1925 discovered that the inheritance of different blood groups in human beings is determined by a number of multiple allelic series. The three autosomal alleles located on chromosome 9 are concerned with the determination of blood group in any person. The gene controlling blood type has been labeled as ‘L’ (after the name of the discoverer, Landsteiner) or I (from isoagglutination). The I gene exists in three allelic forms, IA, IB and 1°. IA specifies A antigen. IB allele determines B antigen and 1° allele specifies no antigen. Individuals who possess these antigens in their fluids such as the saliva are called secretors.

Each allele (IA and IB) produces a transferase enzyme. IA allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and IB allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

From the phenotypic combinations it is evident that the alleles I^A and I^B are dominant to 1°, but co-dominant to each other (I^A = I^B). Their dominance hierarchy can be given as (I^A=I^B> 1^O). A child receives one of three alleles from each parent, giving rise to six possible genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A, I^AI^O, I^BI^B, I^BI^O, I^AI^B and I^OI^O

Question 35.
How is sex determined in human
Answer:

Genes determining sex in human beings are located on two sex chromosomes, called allosomes.
In mammals, sex determination is associated with chromosomal differences between the two sexes, typically XX females and XY males. 23 pairs of human chromosomes include 22 pairs of autosomes (44A) and one pair of sex chromosomes (XX or XY). Females are homogametic producing only one type of gametes (egg), each containing one X chromosome while the males are heterogametic producing two types of sperms with X and Y chromosomes. An independently evolved XX: XY system of sex chromosomes also exist in Drosophila.

Question 36.
Explain male heterogamety.
Answer:
Male heterogamety (XY males) is a type of sex determination in which males produce two different types of gametes. For example, human males produce two kinds of sperms that is sperm with X-chromosome and sperms with Y-chromosome.

Question 37.
Brief about female heterogamety.
Answer:
Female heterogamety (ZO females) refers to the condition, where the female produces two types of egg cells. Some with Z chromosome and some without the Z chromosome.

Question 38.
Give an account of genetic control of Rh factor?
Genetic control of Rh factor
Answer:

Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature. In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1.

The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Wiener Hypothesis:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^Z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ ^phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 39.
Explain the mode of sex determination in honeybees.
Haplodiploidv in Honeybees:
Answer:
Haplodiploidv in Honeybees:
In hymenopteran insects such as honeybees, ants and wasps, a mechanism of sex determination called haplodiploidy mechanism of sex determination is common. In this system, the sex of the offspring is determined by the number of sets of chromosomes it receives. Fertilized eggs develop into females (Queen or Worker) and unfertilized eggs develop into males (drones) by parthenogenesis. It means that the males have half the number of chromosomes (haploid) and the females have double the number (diploid), hence the name haplodiplody for this system of sex determination.

This mode of sex determination facilitates the evolution of sociality in which only one diploid female becomes a queen and lays the eggs for the colony. All other females which are diploid having developed from fertilized eggs help to raise the queen’s eggs and so contribute to the queen’s reproductive success and indirectly to their own, a phenomenon known as Kin Selection. The queen constructs their social environment by releasing a hormone that suppresses fertility of the workers.

Question 40.
Discuss the genic balance mechanism of sex determination with reference to Drosophila?
Answer:
XX-XY type (Lygaeus Type) sex determination is seen in Drosophila. The females are homogametic with XX chromosome, while the males are heterogametic with X and Y

chromosome. Homogametic females produce only one kind of egg, each with one X chromosome, while the heterogametic males produce two kinds of sperms some with X chromosome and some with Y chromosome.

The sex of the embryo depends on the fertilizing sperm. An egg fertilized by an ‘X’ bearing sperm produces a female, if fertilized by a ‘Y’ bearing sperm, a male is produced.

Question 41.
What are the applications of Karyotyping?
Answer:

Karyotyping helps in gender identification.
It is used to detect chromosomal aberrations like deletion, duplication, translocation, non-disjunction of chromosomes.
It helps to identify the abnormalities of chromosomes like aneuploidy.
It is also used in predicting the evolutionary relationships between species.
Genetic diseases in human beings can be detected by this technique.

Question 42.
Explain the inheritance of sex-linked characters in human beings.
Answer:
Haemophilia is commonly known as bleeder’s disease, which is more common in men than women. This hereditary disease was first reported by John Cotto in 1803. Haemophilia is caused by a recessive X-linked gene. A person with a recessive gene for haemophilia lacks a normal clotting substance (thromboplastin) in blood, hence minor injuries cause continuous ’bleeding, leading to death. The females are carriers of the disease and would transmit the disease to 50% of their sons even if the male parent is normal. Haemophilia follows the characteristic criss-cross pattern of inheritaitce.

Question 43.
What is extrachromosomal inheritance? Explain with an example.
Answer:
The cytoplasmic extranuclear genes have a characteristic pattern of inheritance which does not resemble genes of nuclear chromosomes and are known as Extrachromosomal/ Cytoplasmic inheritance.

Question 44.
Comment on the methods of Eugenics.
Answer:
Eugenics refers to the study of the possibility of improving the qualities of human population. Methods of Eugenics:

Sex-education in school and public forums.
Promoting the uses of contraception.
Compulsory sterilization for mentally retarded and criminals.
Egg donation.
Artificial insemination by donors.
Prenatal diagnosis of genetic disorders and performing MTP
Gene therapy
Cloning
Egg/sperm donation of healthy individuals.

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

1 – Mark Questions

Question 1.
If a colorblind female marries a normal male, their sons will be _________
(a) All normal visioned
(b) All color blinded
(c) One-half normal visioned another half colorblind
(d) Three fourth colorblind one fourth normal
Answer:
(c) One-half normal visioned another half colorblind

Question 2.
Excess hair growth on pinna is a feature noticed only in males because _________
(a) Males produce more testosterone
(b) a gene responsible for the character is located in Y-chromosome
(c) Estrogen suppresses the character in females
(d) females act only as a carrier for this character
Answer:
(b) gene responsible for the character is located in Y-chromosome

Question 3.
ABO blood group is a classical example for _________
(a) Multiple allelism
(b) Pleotropism
(c) Incomplete dominance
(d) Polygenic mechanism
Answer:
(a) Multiple allelism

Question 4.
Unit of heredity is _________
(a) allele
(b) allelomorph
(d) gene
Answer:
(d) gene

Question 5.
Identify the proper dominance hierarchy.
(a) I^B = 1° > I^B
(b) I^A = I^B> 0
(c) I^O = I^B >
(d) I^B = I^A > O
Answer:
(b) I^A = I^B> 0

Question 6.
Haemophilia is more common in human males than human females. The reason is due to ______
(а) X-linked dominant gene
(b) X-linked recessive gene
(c) Y-linked recessive gene
(d) Allosomal abnormality
Answer:
(b) X-linked recessive gene

Question 7.
Identify the correct statement.
(a) Homozygous sex chromosome (XX) produce males in Drosophila
(b) Homozygous sex chromosome (ZZ) determine female sex in birds
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper
(d) Heterozygous sex chromosome (ZW) determine male sex in gypsy moth
Answer:
(c) Heterozygous sex chromosome (XO) determine male sex in grasshopper

Question 8.
Which blood group does not possess antibodies?
(a) I^AI^B
(b) I^OI^O
(c) I^AO
(d) I^BI^B
Answer:
(a) I^AI^B

Question 9.
Assertion (A): On diagnosis, Ramu is reported to have underdeveloped testis and gynecomastia.
Reason (R): His karyotype reveals XXY condition.
(а) A is right but R is wrong
(b) R explains A
(c) Both A and R are wrong
(d) Both and R are right but R is not the correct explanation of A
Answer:
(b) R explains A

Question 10.
Pick out the odd man.
(a) Klinefelter’s syndrome
(b) Turner’s syndrome
(c) Huntington’s chorea
(d) 13-Trisomy
Answer:
(c) Huntington’s chorea

Question 11.
Pick the odd one out regarding Mendelian disorder.
(a) Thalassemia
(b) phenylketonuria
(c) Albinism
(d) Huntington’s chorea
Answer:
(d) Huntington’s chorea

Question 12.

Answer:
(a) A – iv, B – ii, D – i, D – iii

Question 13.
Identify the proper ratio of normal visioned individuals against colorblind individuals, if colorblind carrier female marries a normal male.
(a) 1 : 1
(b) 3 : 1
(c) 1 : 3
(d) All four are normal visioned
Answer:
(c) 1 : 3

Question 14.
Pick out the correct statement.
(i) Karyotyping helps in gender identification
(ii) Holandric genes are located on X-chromosome
(iii) Trisomy-21 is an allosomal abnormality
(iv) Cooley’s anaemia is an autosomal recessive disorder
(a) i, iii, iv are correct
(c) i and iv are correct
Answer:
(c) i and iv are correct

Question 15.
DOPA stands for _________
(a) 3,4 – dihydroxy phenylacetate
(b) 3,4 – dihydroxy phenyle alanine
(c) 3,4 – dihydroxy phenyl asparate
(d) 3,4 – dihydroxy phenyle aldehyde
Answer:
(b) 3,4 – dihydroxy phenyle alanine

Question 16.
The type of antibody generated against Rh antigen is _________
(a) IgE
(b) IgG
(c) IgA
Answer:
(b) IgG

Question 17.
Which of the following symbol is used in pedigree analysis to represent unspecified sex?

Answer:

Question 18.
A colorblind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson being colorblind?
(a) 1/4
(b) 3/4
(c) 2/4
(d) 4/4
Answer:
(a) 1/4

Question 19.
Multiple alleles are located _________
(a) at different loci on homologous chromosome
(b) at same locus on homologous chromosome
(c) at different loci on non-homologous chromosome
(d) at different chromosomes
Answer:
(b) at same locus on homologous chromosome

Question 20.
Identify the incorrect statement regarding haplodiploidy.
(g) Haplodiploidy is noticed in honeybees and drosophila
(b) Unfertilized eggs develop into drones
(c) Fertilized eggs develop into queen and worker bees
(d) Males have half the total chromosomal number
Answer:
(a) Haplodiploidy is noticed in honeybees and drosophila

Question 21.
IA and IB genes of ABO blood group are _________
(a) Co-dominant
(b) Pleotropic
(c) Dominant and recessive
(d) Epistatic
Answer:
(a) Co-dominant

Question 22.
Which one of the following crosses show 3 : 1 ratio of normal visioned versus carrier blind?
(a) X^C X^C × X⁺Y
(b) X⁺ X^C × X^C Y
(c) X⁺X^C × X⁺Y
(d) X⁺X⁺× X^CY
Answer:
(c) X⁺X^C × X⁺Y

2 – Mark Questions

Question 1.
Define multiple allelism.
Answer:
When three or more alleles of a gene that control a particular trait occupy the same locus on the homologous chromosome of an organism, they are called multiple alleles and their inheritance is called multiple allelism.

Question 2.
Name the discoverers of antigen A, B and AB.
Answer:
Antigens A and Antigen B was discovered by Karl Landsteiner. Antigen AB was discovered by Von De Castelle and Sturli.

Question 3.
What happens if type A blood is injected to a person having B blood group? Explain the reason.
Answer:
When two different incompatible blood types are mixed, agglutination (clumping together) of erythrocytes (RBC) occurs. The basis of these chemical differences is due to the presence of antigens (surface antigens) on the membrane of RBC and epithelial cells.

Question 4.
State the allelic forms of I gene and mention its chromosomal location.
Answer:
The I gene exists in three forms: IA, IB and I°. The alleles are located on chromosome 9.

Question 5.
Write the possible genotypes for a person having B-blood group.
Answer:
The possible genotypes of a B-blood group person are IBIB or IBI°.

Question 6.
State Wiener Hypothesis on Rh-factor.
Answer:
Wiener proposed the existence of eight alleles (R¹, R², R⁰, R^z, r, r¹, r¹¹, r^y) at a single Rh locus. All genotypes carrying a dominant ‘R allele’ (R¹, R² ,R⁰ ,R^z) will produce ‘Rh-positive’ phenotype and double recessive genotypes (rr, rr¹, rr¹¹, rr^y) will give rise to Rh-negative phenotype.

Question 7.
Distinguish between homogametic and heterogametic condition with example.
Answer:
Homogametic organism:

Organism producing only one type of gametes.
e.g. Human female (Only X)

Heterogametic organism:

Organism producing two different types of gametes.
e.g. Human Male (X and Y)

Question 8.
Name any four organism expressing ZW-ZZ type of sex determination.
Answer:
Gypsy moth, fishes, reptiles and birds.

Question 9.
Expand

Answer:

SRY – Sex Determining region
TDF – Testes Determining Factor

Question 10.
Define Barr body.
Answer:
In 1949, Barr and Bertram first observed a condensed body in the nerve cells of female cat which was absent in the male. This condensed body was called sex chromatin by them and was later referred as Barr body.

Question 11.
Based on Lyon’s hypothesis, mention the number of Barr bodies in XXY males, XO females.
Answer:

XXY males – One Barr body.
XO females – No Barr body.

Question 12.
State Lyon’s hypothesis.
Answer:
Lyon’s hypothesis states that in mammals the necessary dosage compensation is accomplished by the inactivation of one of the X chromosome in females so that both males and females have only one functional X chromosome per cell.

Mary Lyon suggested that Barr bodies represented an inactive chromosome, which in females becomes tightly coiled into a heterochromatin, a condensed and visible form of chromatin (Lyon’s hypothesis). The number of Barr bodies observed in cell was one less than the number of X-Chromosome. XO females have no Barr body, whereas XXY males have one Barr body.

Question 13.
Mention few X-linked inherited diseases.
Answer:
Red-green colour blindness or daltonism, haemophilia and Duchenne’s muscular dystrophy.

Question 14.
Define Karyotyping.
Answer:
Karyotyping is a technique through which a complete set of chromosomes is separated from a cell and the chromosomes are arranged in pairs. An idiogram refers to a diagrammatic representation of chromosomes.

Question 15.
Explain the inheritance pattern of Y-linked genes with example.
Answer:
Genes in the non-homologous region of the Y-chromosome are inherited directly from male to male. In humans, the Y-linked or holandric genes for hypertrichosis (excessive development of hairs on pinna of the ear) are transmitted directly from father to son, because males inherit the Y chromosome from the father. Female inherits only X chromosome from the father and are not affected.

Question 16.
Observe the symbol used in pedigree analysis and give the proper terms they represent.

Answer:

Question 17.
Write a brief note on pedigree analysis.
Answer:
Pedigree is a “family tree”, drawn with standard genetic symbols, showing the inheritance pathway for specific phenotypic characters. Pedigree analysis is the study of traits as they have appeared in a given family line for several past generations.

Question 18.
What do you mean by ‘Mendelian disorder’.
Answer:
Alteration or mutation in a single gene causes Mendelian disorders. These disorders are transmitted to the off springs on the same line as the Mendelian pattern of inheritance.
E.g., Thalassemia.

Question 19.
Name any four Mendelian disorders.
Answer:

Thalassemia
Albinism
sickle cell anaemia
Huntington’s chorea

Question 20.
What is the phenotype of

I^AI^O
I^OI^O

Answer:

I^AI^O – A blood group person
I^OI^O – O blood group person

Question 21.
On which chromosomes does HBA1 gene and HBB genes are located?
Answer:

HBA1 gene is located on chromosome 16.
HBB gene is located on chromosome 11.

Question 22.
Complete the equation.

Answer:
(a) A = Phenylalanine hydroxylase
(b) B = Tyrosinase

Question 23.
Write a note on Huntington’s chorea.
Answer:
Huntington’s chorea is inherited as an autosomal dominant lethal gene in man. It is characterized by involuntary jerking of the body and progressive degeneration of the nervous system, accompanied by gradual mental and physical deterioration. The patients with this disease usually die between the age of 35 and 40.

Question 24.
Comment on Trisomy-21.
Answer:
Trisomic condition of chromosome – 21 results in Down’s syndrome. It is characterized by severe mental retardation, defective development of the central nervous system, increased separation between the eyes, flattened nose, ears are malformed, mouth is constantly open and the tongue protrudes.

Question 25.
Mention the genetic makeup of Turner’s syndrome person and Klinefelter’s syndrome , person.
Answer:
Klinefelter’s syndrome – 44AA+XXY Turner’s syndrome – 44AA+XO

Question 26.
List out any four clinical symptoms of Klinefelter’s syndrome.
Answer:
Gynaecomastia, high pitched voice, under developed genetalia and tall with long limbs.

3 – Mark Questions

Question 27.
Write the types of sex-determination mechanisms does the following crosses as shown. Give an example for each.

Female XX with Male XO
Female ZW with Male ZZ

Answer:

Male heterogamety. e.g., Human beings.
Female heterogamety. e.g., Birds.

Question 28.
What are the enzymes encoded by the alleles I^A, I^B and I°?
Answer:
I^A allele produces N-acetyl galactose transferase and can add N-acetyl galactosamine (NAG) and I^B allele encodes for the enzyme galactose transferase that adds galactose to the precursor (i.e. H substances). In the case of I^O/I^O allele no terminal transferase enzyme is produced and therefore called “null” allele and hence cannot add NAG or galactose to the precursor.

Question 29.
Draw a tabular column representing various types of blood group in human beings, their phenotypes, genotypes, antigens and respective antibodies.
Answer:
Genetic basis of the human ABO blood groups:

Question 30.
Give an account on Rhesus factor.
Answer:
Rhesus or Rh – Factor: The Rh factor or Rh antigen is found on the surface of erythrocytes. It was discovered in 1940 by Karl Landsteiner and Alexander Wiener in the blood of rhesus monkey, Macaca rhesus and later in human beings. The term ‘Rh factor’ refers to “immunogenic D antigen of the Rh blood group system. An individual having D antigen are Rh D positive (Rh+) and those without D antigen are Rh D negative (Rh”)”. Rhesus factor in the blood is inherited as a dominant trait. Naturally occurring Anti D antibodies are absent in the plasma of any normal individual. However if an Rh” (Rh negative) person is exposed to Rh+ (Rh positive) blood cells (erythrocytes) for the first time, anti D antibodies are formed in the blood of that individual. On the other hand, when an Rh positive person receives Rh negative blood no effect is seen.

Question 31.
How Erythroblastosis foetalis can be prevented?
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 32.
Explain XX-XO type of sex determination.
Answer:

XX-XO method of sex determination is seen in bugs, some insects such as cockroaches and grasshoppers. Pi The female with two X chromosomes are homogametic Gametes (XX) while the males with only one X chromosome are heterogametic (XO). The presence of an unpaired X chromosomes determines the male sex. The males PI Generation with unpaired ‘X’ chromosome produce two types of sperms, one half with X chromosome and other half without X chromosome. The sex of the offspring depends upon the sperm that fertilizes the egg.

Question 33.
Name the type of sex-determination mechanism of the following organisms.

Gypsy moth
Human beings
Butterflies

Answer:

Gypsy moth -ZW – ZZ type (ZW-females, ZZ – males)
Human beings – XX – XY type (XX-females, XY – males)
Butterflies – ZO – ZZ type (ZO-females, ZZ – males)

Question 34.
Complete the following cross.
AAZW × AAZZ (female) (male)
Answer:

Question 35.
Role of Y- chromosome is crucial for maleness – Justify.
Answer:
Current analysis of Y chromosomes has revealed numerous genes and regions with potential genetic function; some genes with or without homologous counterparts are seen on the X. Present at both ends of the Y chromosome are the pseudoautosomal regions (PARs) that are similar with regions on the X chromosome which synapse and recombine during meiosis. The remaining 95% of the Y chromosome is referred as the Non-combining Region of the Y (NRY).

The NRY is divided equally into functional genes (euchromatic) and non-functional genes (heterochromatic). Within the euchromatin regions, is a gene called Sex determining region Y (SRY). In humans, absence of Y chromosome inevitably leads to female development and this SRY gene is absent in X chromosome. The gene product of SRY is the testes determining factor (TDF) present in the adult male testis.

Question 36.
Color blindness is a perfect example for criss-cross of inheritance – Justify the statement.
Answer:
A marriage between a colour blind man and a normal visioned woman will produce normal visioned male and female individuals in F₁ generation but the females are carriers. The marriage between a F₁ normal visioned carrier woman and a normal visioned male will produce one normal visioned female, one carrier female, one normal visioned male and one colour blind male. Colour blind trait is inherited from the male parent to his grandson through carrier daughter, which is an example of criss-cross pattern of inheritance.
image 12

Question 37.
How the Karyotype of lymphocytes was prepared by Tjio and Levan?
Answer:
Preparation of Karyotype Tjio and Levan (1960) described a simple method of culturing lymphocytes from the human blood. Mitosis is induced followed by addition of colchicine to arrest cell division at metaphase stage and the suitable spread of metaphase chromosomes is photographed. The individual chromosomes are cut from the photograph and are arranged in an orderly fashion in homologous pairs. This arrangement is called a karyotype. Chromosome banding permits structural definitions and differentiation of chromosomes.

Question 38.
What is a genetic disorder? Mention its types?
Answer:
A genetic disorder is a disease or syndrome that is caused by an abnormality in an individual -DNA. Abnormalities can range from a small mutation in a single gene to the addition or subtraction of an entire chromosome or even a set of chromosomes. Genetic disorders are of two types namely, Mendelian disorders and chromosomal disorders.

Question 39.
Explain the genetic basis of Phenylketonuria.
Answer:
Phenylketonuria is an inborn error of Phenylalanine metabolism caused due to a pair of autosomal recessive genes. It is caused due to mutation in the gene PAH (phenylalanine hydroxylase gene) located on chromosome 12 for the hepatic enzyme “phenylalanine hydroxylase”.

This enzyme is essential for the conversion of phenylalanine to tyrosine. Affected individual lacks this enzyme, so phenylalanine accumulates and gets converted to phenylpyruvic acid and other derivatives. It is characterized by severe mental retardation, light pigmentation of skin and hair. Phenylpyruvic acid is excreted in the urine.

Question 40.
Give an account on Patau’s syndrome.
Answer:
Trisomic condition of chromosome 13 results in Patau’s syndrome. Meiotic non-disjunction is thought to be the cause for this chromosomal abnormality. It is characterized by multiple and severe body malformations as well as profound mental deficiency. Small head with small eyes, cleft palate, malformation of the brain and internal organs are some of the symptoms of this syndrome.

Question 41.
Define aneuploidy.
Answer:
Failure of chromatids to segregate during cell division resulting in the gain or loss of one or more chromosomes is called aneuploidy. It is caused by non-disjunction of chromosomes.

Question 42.
What do you mean by “syndrome”? Give two examples.
Answer:
Group of signs and symptoms that occur together and characterize a particular abnormality is called a syndrome, e.g., Down’s syndrome and Turner’s syndrome.

5 – Mark Questions

Question 42.
Explain in detail about Erythroblastosis foetalis.
Answer:
Rh incompatability has great significance in child birth. If a woman is Rh negative and the man is Rh positive, the foetus may be Rh positive having inherited the factor from its father. The Rh negative mother becomes sensitized by carrying Rh positive foetus within her body. Due to damage of blood vessels, during child birth, the mother’s immune system recognizes the Rh antigens and gets sensitized. The sensitized mother produces Rh antibodies. The antibodies are IgG type which are small and can cross placenta and enter the foetal circulation. By the time the mother gets sensitized and produce anti ‘D’ antibodies, the child is delivered.

Usually no effects are associated with exposure of the mother to Rh positive antigen during the first child birth, subsequent Rh positive children carried by the same mother, may be exposed to antibodies produced by the mother against Rh antigen, which are carried across the placenta into the foetal blood circulation. This causes haemolysis of foetal RBCs resulting in haemolytic jaundice and anaemia. This condition is known as Erythroblastosis foetalis or Haemolytic disease of the new bom (HDN).

Question 43.
Decribe female heterogamy and its types.
Answer:
Heterogametic Females:
In this method of sex determination, the homogametic male possesses two ‘X’ chromosomes as in certain insects and certain vertebrates like fishes, reptiles and birds producing a single type of gamete; while females produce dissimilar gametes. The female sex consists of a single ‘X’ chromosome or one ‘X’ and one ‘Y’ chromosome. Thus the females are heterogametic and produce two types of eggs. Heterogametic females are of two types, ZO-ZZ type and ZW-ZZ type.

ZO-ZZ Type:
This method of sex determination is seen in certain moths, butterflies and domestic chickens. In this type, the female possesses
single ‘Z’ chromosome in its body cells and is heterogametic (ZO) producing two kinds of eggs some with ‘Z’ chromosome and some without ‘Z’ chromosome, while the male possesses two ‘Z’ chromosomes and is homogametic (ZZ).

ZW-ZZ type:
This method of sex determination occurs in certain insects (gypsy moth) and in vertebrates such as fishes, reptiles and birds. In this method the female has one ‘Z’ and one ‘ W’ chromosome (ZW) producing two types of eggs, some carrying the Z chromosomes and some carry the W chromosome. The male sex has two ‘Z’ chromosomes and is homogametic (ZZ) producing a single type of sperm.

Question 44.
Write elaborately about the following Mendelian disorders.
Answer:
(a) Thalassemia
(b) Albinism
Answer:
(a) Thalassemia: Thalassemia is an autosomal recessive disorder. It is caused by gene mutation resulting in excessive destruction of RBC’s due to the formation of abnormal haemoglobin molecules. Normally haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains. Thalassemia patients have defects in either the alpha or beta globin chain causing the production of abnormal haemoglobin molecules resulting in anaemia.

Thalassemia is classified into alpha and beta based on which chain of haemoglobin molecule _ is affected. It is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16. Mutation or deletion of one or more of the four alpha gene alleles causes Alpha Thalassemia. In Beta Thalassemia, production of beta globin chain is affected. It is controlled by a single gene (HBB) on chromosome 11. It is the most common type of Thalassemia and is also known as Cooley’s anaemia. In this disorder, the alpha chain production is increased and damages the membranes of RBC.

(b) Albinism: Albinism is an inborn error of metabolism, caused due to an autosomal recessive gene. Melanin pigment is responsible for skin colour. Absence of melanin results in a condition called albinism. A person with the recessive allele lacks the tyrosinase enzyme system, which is required for the conversion of dihydroxyphenyl alanine (DOPA) into melanin pigment inside the melanocytes. In an albino, melanocytes are present in normal numbers in their skin, hair, iris, etc., but lack melanin pigment.

Question 45.
Discuss any two Allosomal anomalies in human.
Allosomal abnormalities in human beings
Answer:
Mitotic or meiotic non-disjunction of sex chromosomes causes allosomal abnormalities. Several sex chromosomal abnormalities have been detected.
E.g. Klinefelter’s syndrome and Turner’s syndrome.

1. Klinefelter’s Syndrome (XXY Males): This genetic disorder is due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47,XXY. Persons with this syndrome have 47 chromosomes (44AA+XXY). They are usually sterile males, tall, obese, with long limbs, high pitched voice, under developed genetalia and have feeble breast (gynaecomastia) development.

2. Turner’s Syndrome (XO Females): This genetic disorder is due to the loss of a X chromosome resulting in a karyotype of 45,X. Persons with this syndrome have 45 chromosomes (44 autosomes and one X chromosome) (44AA+XO) and are sterile females. Low stature, webbed neck, under developed breast, rudimentary gonads lack of menstrual cycle during puberty, are the main symptoms of this syndrome.

Higher Order Thinking Skills (HOTs) Questions

Question 1.
On analysis, a person’s karyotype reveals an extra one chromosome of twenty first pair. What does this condition represents? which type of symptoms can be noticed in the person?
Answer:

Trisomy-21 or Down’s syndrome.
Symptoms – Mental retardation, malformed ears, protruded tongue, mouth is constantly open etc.

Question 2.
A female whose blood group is AB- got conceived and later it is diagnoised that her – foetus possess B+. What measures would be taken to prevent the foetus from Haemolytic
disease of Newborn (HDN)
Answer:
If the mother is Rh negative and foetus is Rh positive, anti D antibodies should be administered to the mother at 28th and 34th week of gestation as a prophylactic measure. If the Rh negative mother delivers Rh positive child then anti D antibodies should be administered to the mother soon after delivery. This develops passive immunity and prevents the formation of anti D antibodies in the mothers blood by destroying the Rh foetal RBC before the mother’s immune system is sensitized. This has to be done whenever the woman attains pregnancy.

Question 3.
The following table shows the genotypes for ABO blood grouping and this phenotypes. Complete the table by filling the gaps.
Answer:

Question 4.
Give one example for each of the following group of drugs,

Stimulants
Analgesic
Hallucinogens

Answer:

Stimulants – Eg : Nicotine
Analgesic – Eg : Opium
Hallucinogens – Phencyclidine

Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Leave a Comment / Class 11 / By Prasanna

Students can Download Bio Botany Chapter 3 Vegetative Morphology Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Which of the following k polycarpic plant?
(a) Mangifera
(b) Bambusa
(c) Musa
(d) Agave
Answer:
(a) Mangifera

Question 2.
Roots are …………… .
(a) descending. negatively geotrophic, positively phototrophic
(b) descending, positively geotrophic, negatively phototrophic
(c) ascending, positively geotrophic, negatively phototrophic
(d) ascending, negatively geotrophic, positively phototrophic
Answer:
(b) descending, positively geotrophic, negatively pholotrophic

Question 3.
Bryophyllum and Dioscorea are example for …………… .
(a) foliar bud, apical bud
(b) foliar bud, cauline bud
(c) cauline bud, apical bud
(d) cauline bud, foliar bud
Answer:
(b) foliar bud, cauline bud

Question 4.
Which of the following is correct statement?
(a) In Pisum sativum leaflets modified into tendrils
(b) In Atalantia terminal bud is modified into thorns
(c) In Nepenthes midrib is modified into lid
(d) In Smilax inflorescence axis is modified into tendrils
Answer:
(b) In Pisum sativum leaflets modified into tendrils

Question 5.
Select the mismatch pair.

Answer:
(d) Allamanda – (iv) Ternate phyllotaxy

Question 6.
Draw and label the parts of regions of root.
Answer:
The parts of regions of root:

Question 7.
Write the similarities and differences between
1. Avicennia and Trapa
2. Radical buds and foliar buds
3. Phylloclade and cladode
Answer:
1. Avicennia and Trapa
2. Radical buds and foliar buds
3. Phylloclade and cladode

Question 8.
How root climbers differ from stem climbers?
Answer:
Root climbers:

Plants climbing with the help of adventitious roots (arise from nodes)
Example: Piper betel

Stem climbers (twiners):

These climbers lack specialised structure for climbing and the stem itself coils around the support.
Example: Ipomoea

Question 9.
Compare sympodial branching with monopodial branching.
Answer:
Sympodial branching:
The terminal bud cease to grow after a period of growth and the further growth is taken care by successive or several lateral meristem or buds. This type of growth is also known as sympodial branching. Example: Cycas.

Monopodial branching:
The terminal bud grows uninterrupted and produce several lateral branches. This type of growth is also known as monopodial branching. Example: Polyalthia.

Question 10.
Compare pinnate unicostate venation and palmate multicostate venation.
Answer:
Pinnate Unicostate Venation and Palmate Multicostate Venation:

Pinnate unicostate venation: In pinnate unicostate there is only one prominent midrib.
Palmate multicostate venation: In palmate multicostate there are many midribs running parallel to each other.

Text Book Activities Solved

Question 1.
Collection of medicines prepared from root, stem, leaf of organic plants.
Answer:
Medicines Prepared From Root, Stem, Leaf Of Organic Plants:

Question 2.
Prepare a report of traditional medicines.
Answer:
Traditional medicines: Ayurvcda, Siddha, Unani, Acupuncture, Homeopathy, Naturopathy, Chinese or oriental medicine.

Question 3.
Growing micro greens In class room – project work. (Green seed sprouts)
Answer:
Definition: Micro green are miniature plants of greens, herbs or vegetables. They arc rich nutrient source and also added as flavouring agent in food. Growing micro greens in classroom:

Step 1. Take a shallow tray.
Step 2. Place a inch of organic potting soil in the bottom of tray.
Step 3. Scatter the seeds (celery, lettuce, etc.) over the soil surface.
Step 4. Sprinkle some water. Cover the seeds with a thin layer of soil.
Step 5. Place the whole setup near the lighted window.
Step 6. Within a day or two, seeds with germinate.
Step 7. Maintain the moisture of soil. Greens can be harvested within 2 – 3 weeks.

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The study about external features of an organism is known as …………… .
(a) morphology
(b) anatomy
(c) physiology
(d) taxonomy
Answer:
(a) morphology

Question 2.
…………… deals with the study of shape. site and structure of plant parts.
(a) External morphology
(b) Internal morphology
(c) External anatomy
(d) Internal anatomy
Answer:
(a) External morphology

Question 3.
The branch of science that deals with the classification of organisms is called as …………… .
(a) taxonomy
(b) morphology
(c) physiology
(d) anatomy
Answer:
(a) taxonomy

Question 4.
…………… deals with the study about the root and shoot system.
(a) Vegetative morphology
(b) Reproductive morphology
(c) External morphology
(d) Internal morphology
Answer:
(a) Vegetative morphology

Question 5.
…………… deals with the study about flowers, fruits and seeds of a plant.
(a) Reproductive morphology
(b) Vegetative morphology
(c) External morphology
(d) Internal morphology
Answer:
(a) Reproductive morphology

Question 6.
The general form of a plant is referred to as …………… .
(a) habitat
(b) structure
(c) habit
(d) shape and size
Answer:
(c) habit

Question 7.
…………… are soft stemmed plants with less wood or no wood.
(a) Shrubs
(b) Trees
(c) Herbs
(d) Climbers
Answer:
(c) Herbs

Question 8.
Perennial herbs having a bulb, corm. rhizome or tuber as the underground stem are termed as …………… .
(a) thallophytes
(b) rhodophytes
(c) geophytes
(d) cyanophytes
Answer:
(c) geophytes

Question 9.
Climbers are also called as …………… .
(a) herbs
(b) trees
(c) vines
(d) shrubs
Answer:
(c) vines

Question 10.
…………… is a perennial, woody plant with several main stems arising from the ground level.
(a) Herb
(b) Runner
(c) Climber
(d) Shrub
Answer:
(d) Shrub

Question 11.
…………… is an example for vines.
(a) Hibiscus
(b) Phyllanthus
(c) Palmyra
(d) Ventilago
Answer:
(d) Ventilago

Question 12.
…………… grows on rocks.
(a) Mesophytes
(b) Lithophytes
(c) Xerophytes
(d) Psammophytcs
Answer:
(b) Lithophytes

Question 13.
The plants that grows on dry habitats are called …………… .
(a) meophytes
(b) xerophytes
(c) lithophytes
(d) psammophytes
Answer:
(b) xerophytes

Question 14.
Mesophytes grows on …………… .
(a) sand
(b) soils with sufficient water
(c) rocks
(d) dry habitats
Answer:
(b) soils with sufficient water

Question 15.
…………… grows on sand.
(a) Mesophytes
(b) Psammophytes
(c) Lithophytes
(d) Xerophytes
Answer:
(b) Psammophytes

Question 16.
Azadirachta indica is an example for …………… .
(a) psammophytes
(b) mesophytes
(c) lithophy es
(d) xerophytes
Answer:
(b) mesophytes

Question 17.
Ipomoea pes – caprae is an example for …………… .
(a) mesophyles
(b) psammophytes
(c) lithophytes
(d) serophytes
Answer:
(b) psammophytes

Question 18.
…………… is an example for xerophytes.
(a) Lichens
(b) Euphorbia
(c) Ficus
(d) Ipomoea
Answer:
(b) Euphorbia

Question 19.
Plants growing emergent in marshy saline habitat are called as …………… plants.
(a) free floating
(b) submerged
(c) emergent
(d) mangroves
Answer:
(d) mangroves

Question 20.
…………… type of aquatic plants with roots or stems are anchored to the substrate under water with aerial shoots growing above water.
(a) Submerged
(b) Free floating
(c) Emergent
(d) Mangroves
Answer:
(c) Emergent

Question 21.
Hydrilla and Vallisneria are the examples of …………… type of aquatic plants.
(a) emergent
(b) free floating
(c) submerged
(d) mangroves
Answer:
(c) submerged

Question 22.
Plants that grows completely under water are called as ……………. type of aquatic plants.
(a) emergent
(b) free floating
(c) submerged
(d) mangroves
Answer:
(c) submerged

Question 23.
Plants growing on the water surface are called as …………… type of aquatic plants.
(a) emergent
(b) submerged
(c) free floating
(d) mangroves
Answer:
(c) free floating

Question 24.
Therophyte is a plant that completes its life cycle in …………… growing season.
(a) one
(b) two
(c) three
(d) several
Answer:
(a) one

Question 25.
A plant that grows vegetatively during the first season and undergoes flowering and fruiting during the second season is called as …………… .
(a) biennial
(b) therophyte
(c) perennial
(d) geophyte
Answer:
(a) biennial

Question 26.
Spinach, carrot and lettuce are the examples of …………… .
(a) biennial
(b) annual
(c) geophytes
(d) ephemerals
Answer:
(a) biennial

Question 27.
A plant that grows for many years that flowers and set fruits for several season during the life span is called as …………… .
(a) geophytes
(b) annual
(c) biennial
(d) ephemerals
Answer:
(a) geophytes

Question 28.
When perennial plants bear fruits every year they are called as …………… .
(a) polycarpic
(b) ephemerals
(c) annual
(d) therophyte
Answer:
(a) polycarpic

Question 29.
Talipot palm, Bamboo and Agave are examples of …………… .
(a) polycarpic geophytes
(b) therophytes
(c) monocarpic geophytes
(d) biennial
Answer:
(c) monocarpic geophytes

Question 30.
Watermelon is a / an …………… plant.
(a) biennial
(b) perennial
(c) geophytic
(d) ephemeral
Answer:
(d) ephemeral

Question 31.
…………… is an example for polycarpic perennial.
(a) Peas
(b) Fennel
(c) Agave
(d) Sapota
Answer:
(d) Sapota

Question 32.
Flowering plants are also called as …………… .
(a) sporophytes
(b) thallophytes
(c) magnoliophytes
(d) phaeophytes
Answer:
(c) magnoliophytes

Question 33.
The part of a plant that arises from radicle is …………… .
(a) stem
(b) root
(c) bud
(d) flower
Answer:
(b) root

Question 34.
Root cap is made up of …………… cells.
(a) parenchyma
(b) collenchyma
(c) sclerenchyma
(d) chlorenchyma
Answer:
(a) parenchyma

Question 35.
In …………… plant multiple root cap is seen.
(a) Pandanus
(b) Pistia
(c) Bougainvillea
(d) Pea
Answer:
(a) Pandanus

Question 36.
Root pockets are seen in …………… .
(a) Pandanus
(b) Pistia
(c) Bougainvillea
(d) Pea
Answer:
(b) Pistia

Question 37.
In …………… zone of the root, the cells get differentiated.
(a) Root hair zone
(b) Elongation zone
(c) Meristematic zone
(d) Maturation zone
Answer:
(d) Maturation zone

Question 38.
Roots developing from any part of the plant other than the radicle is called …………… root.
(a) fibrous
(b) adventitious
(c) tap
Answer:
(b) adventitious

Question 39.
Match the following topics with correct examples.

Answer:
(d) a – ii, b – iv, c – i, d – iii.

Question 40.
The cells of this region undergo active cell division …………… .
(a) root hair zone
(b) maturation zone
(c) elongation zone
(d) meristematic zone
Answer:
(d) meristematic zone

Question 41.
Tap root system develops from this part of embryo …………… .
(a) plumule
(b) coleoptile
(c) node
(d) radicle
Answer:
(d) radicle

Question 42.
Negatively geotropic roots are seen in plant like …………… .
(a) Beta vulgaris
(b) Hibiscus
(c) Rhizophora
(d) Euphorbia
Answer:
(c) Rhizophora

Question 43.
Respiratory roots are found in …………… .
(a) Sweet potato
(b) Bruguiera
(c) Mango
(d) Dahlia
Answer:
(b) Bruguiera

Question 44.
Beaded root are also called as …………… root.
(a) annulated
(b) moniliform
(c) tuberous
(d) fasciculated
Answer:
(b) moniliform

Question 45.
Match the following.

Answer:
(c) a – iv, b – iii, c – ii, d – i.

Question 46.
Match the following.

Answer:
(d) a – v, b – i, c – ii, d – iii, e – iv.

Question 47.
The roots that grows vertically downwards from the lateral branches into the soil is called as …………… roots
(a) brace
(b) climbing
(c) buttress
(d) prop
Answer:
(d) prop

Question 48.
…………… roots are thick roots growing obliquely from the basal nodes of their main stem.
(a) pillar
(b) stilt
(c) buttress
(d) prop
Answer:
(b) stilt

Question 49.
Clasping roots are also called as …………… .
(a) Pillar
(b) Stilt
(c) Clinging
(d) Buttress
Answer:
(c) Clinging

Question 50.
Match the following.

Answer:
(c) a – iii, b – iv, c – i, d – ii.

Question 51.
Which part of embryo develops into stem?
(a) Radicle
(b) Micropyle
(c) Ostia
(d) Plumule
Answer:
(d) Plumule

Question 52.
The point from which leaf arises is called as …………… .
(a) internode
(b) intranode
(c) node
(d) code
Answer:
(c) node

Question 53.
Which is not a character of stem?
(a) Exogenous branches
(b) Descending portion
(c) Nodes
(d) buds
Answer:
(b) Descending portion

Question 54.
Which is not a function of stem?
(a) Support
(b) transport of food
(c) Transport of water
(d) Absorption of water
Answer:
(d) Absorption of water

Question 55.
Which is the primary function of stem?
(a) Storage
(b) Perennation
(c) Photosynthesis
(d) Water transport
Answer:
(d) Water transport

Question 56.
Collateral bud is a …………… bud.
(a) terminal
(b) lateral
(c) extra axillary
(d) accessory
Answer:
(d) accessory

Question 57.
Cauline buds arise from …………… .
(a) root
(b) stem
(c) leaf
(d) nodes
Answer:
(b) stem

Question 58.
Which of the following plant produces bulbils?
(a) Bryophyllum
(b) Begonia
(c) Allium proliferum
(d) Solanum americanum
Answer:
(c) Allium proliferum

Question 59.
…………… plants produce foliar buds.
(a) Allium
(b) Solanum
(c) Citrus
(d) Begonia
Answer:
(d) Begonia

Question 60.
Radical buds develop from …………… .
(a) root
(b) stem
(c) leaf
(d) plumule
Answer:
(a) root

Question 61.
In Polyalthia longifolia, the stem is …………… .
(a) decurrent
(b) caudex
(c) excurrent
(d) culm
Answer:
(c) excurrent

Question 62.
Which of the following is not a creeper?
(a) Cynodon
(b) Oxalis
(c) Indigofera
(d) Centella
Answer:
(c) Indigofera

Question 63.
Clockwise coiling climbers are called …………… .
(a) hexose
(b) dextrose
(c) ministrose
(d) sinistrose
Answer:
(b) dextrose

Question 64.
In Artabotrys, …………… is modified into hook.
(a) leaflets
(b) inflorescence axis
(c) petiole
(d) axillary buds
Answer:
(b) inflorescence axis

Question 65.
…………… are woody stem climbers.
(a) Lianas
(b) Tendrils
(c) Phylloclade
(d) Phyllode
Answer:
(a) Lianas

Question 66.
……………. is a characteristic adaptation of xerophytes.
(a) Hook
(b) Thorn
(c) Cladode
(d) Phylloclade
Answer:
(d) Phylloclade

Question 67.
Flattened cladode is present in …………… .
(a) Asparagus
(b) Atalantia
(c) Carissa
(d) Ruscus
Answer:
(d) Rusus

Question 68.
Musa is an example for …………… .
(a) climber
(b) runner
(c) stolon
(d) sucker
Answer:
(d) sucker

Question 69.
Eichhornia exhibit …………… type of stem modification.
(a) stolon
(b) offset
(c) runner
(d) sucker
Answer:
(b) offset

Question 70.
Underground stems are generally called as …………… .
(a) root caps
(b) root stocks
(c) root pockets
(d) root modification
Answer:
(b) root stocks

Question 71.
Which is not a character of root stock?
(a) Nodes
(b) Internodes
(c) Terminal bud
(d) Root cap
Answer:
(d) Root cap

Question 72.
Which is an example for compound tunicated bulb?
(a) Allium cepa
(b) Allium sativum
(c) Tulipa sps.
(d) Bulbophyllum
Answer:
(b) Allium sativum

Question 73.
…………… is a pseudobulb.
(a) Allium cepa
(b) Allium sativum
(c) Tulipa sps.
(d) Bulbophyllum
Answer:
(d) Bulbophyllum

Question 74.
…………… is a horizontally growing underground stem.
(a) Corm
(b) Rhizome
(c) Bulb
(d) Tuber
Answer:
(b) Rhizome

Question 75.
All the leaves of a plant together called as …………… .
(a) phyllome
(b) phyllode
(c) phylloclade
(d) Phyllanthus
Answer:
(a) phyllome

Question 76.
Which Is not a primary function of leaf?
(a) Photosynthesis
(b) Transpiration
(c) Storage
(d) Gas exchange
Answer:
(c) Storage

Question 77.
Lamina of leaf is called as …………… .
(a) hypopodium
(b) mesopodium
(c) endopodiurn
(d) epipodium
Answer:
(d) epipodium

Question 78.
Swollen, broad leaf base Is called as …………… .
(a) phyttome
(b) pulvinus
(c) stipule
(d) bract
Answer:
(b) pulvinus

Question 79.
Pulvinus is a characteristic feature of …………… .
(a) Malvaceae
(b) Fabaceae
(c) Musaceae
(d) Solanaceae
Answer:
(b) Fabaceae

Question 80.
Stalk of leaf is called as …………… .
(a) pedicel
(b) pctiole
(c) phyllum
(d) perianth
Answer:
(b) petiote

Question 81.
Poaceac is seen in …………… family.
(a) Malvaceae
(b) Fabaceae
(c) Musaccac
(d) Sotanaccac
Answer:
(c) Musacese

Question 82.
Which of the following plant possess sessile leases?
(a) Hibiscus
(b) mangifera
(c) Psidiun
(d) Gloriosa
Answer:
(d) Gloriosa

Question 83.
Arrangement of veins on lamina is called …………… .
(a) venalion
(b) aesivation
(c) placentation
(d) phyllotaxy
Answer:
(a) venation

Question 84.
Parallel venatlon is the characteristic feature of …………… .
(a) angiospertns
(b) gymnosperms
(c) dicots
(d) monocots
Ans.
(d) monocots

Question 85.
In greek, ‘taxis’ means …………… .
(a) crowding
(b) spreading
(c) arrangement
(d) attachment
Answer:
(c) arrangement

Question 86.
Palmately parallel divergent venation is seen in …………… .
(a) Carica papaya
(b) Borassus flabellifer
(c) Zizyphus
(d) Cinnamomum
Answer:
(b) Borassus flabellifer

Question 87.
Spiral arrangement of leaves show vertical rows called …………… .
(a) decussate
(b) bifarious
(c) orthostichies
(d) distichous
Answer:
(c) orthostichies

Question 88.
Nerium exhibits …………… phyllotaxy.
(a) ternate
(b) whorled
(c) decussate
(d) alternate
Answer:
(a) ternate

Question 89.
In Allamanda, …………… phyllotaxy is seen.
(a) ternate
(b) verticillate
(c) alternate
(d) opposite
Answer:
(b) verticillate

Question 90.
The secondary rachii are called as …………… .
(a) stipel
(b) ligule
(c) petiole
(d) pinnae
Answer:
(d) pinnae

Question 91.
Tripinnate compound leaves are seen in …………… .
(a) Citrus
(b) Oxalis
(c) Oroxylum
(d) Allamanda
Answer:
(c) Oroxylum

Question 92.
In Gloriosa, …………… is modified into tendril.
(a) apical leaflet
(b) entire leaf
(c) leaf tip
(d) Petiole
Answer:
(c) leaf tip

Question 93.
In Euphorbia, …………… are modified into spines.
(a) stipels
(b) ligules
(c) stipules
(d) Petiole
Answer:
(c) stipules

Question 94.
Storage leaves are observed in …………… family.
(a) Solanaceae
(b) Cucurbitaceae
(c) Crassulaceae
(d) Musaceae
Answer:
(c) Crassulaceae

Question 95.
Phyllodes are the modification of …………… .
(a) pedicel
(b) pinnae
(c) petiole
(d) stipule
Answer:
(c) petiole

Question 96.
…………… is an example for pitcher.
(a) Sarracenia
(b) Acacia
(c) Parkinsonia
(d) Sedum
Answer:
(a) Sarracenia

Question 97.
Stamens are equivalent to …………… .
(a) megasporophyll
(b) microsporophyll
(c) microsporangium
(d) megasporangium
Answer:
(a) megasporophyll

Question 98.
Rolling or folding of individual leaves is called …………… .
(a) venation
(b) phyllotaxy
(c) ptyxis
(d) branching
Answer:
(c) ptyxis

Question 99.
In Mimusops, the leaves are …………… .
(a) fagacious
(b) evergreen
(c) deciduous
(d) marcescent
Answer:
(b) evergreen

Question 100.
Heterophylly is mainly seen in …………… .
(a) xerophytes
(b) mesophytes
(c) lithophytes
(d) hydrophytes
Answer:
(d) hydrophytes

Question 101.
Isobilateral leaf is seen in …………… .
(a) onion
(b) pine
(c) tridax
(d) grass
Answer:
(d) grass

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define morphology.
Answer:
The study of various external features of the organism is known as morphology.

Question 2.
What is plant morphology?
Answer:
Plant morphology also known as external morphology deals with the study of shape, size and structure of plants and their parts like (roots, stems, leaves, flowers, fruits and seeds).

Question 3.
Differentiate between vegetative morphology & reproductive morphology.
Answer:
between vegetative morphology & reproductive morphology:

Vegetative morphology: Vegetative morphology deals with the shoot system and root system.
Reproductive morphology: Reproductive morphology deals with flowers, inflorescence, fruits & seals.

Question 4.
Define shrub. Give an example.
Answer:
A shrub is a perennial, woody plant with several main stems arising from the ground level. e.g.. Hibiscus

Question 5.
Classify plants based on habitat.
Answer:
Depending upon where plants grow habitats may be classified into major categories:

Terrestrial and
Aquatic.

Question 6.
Distinguish between terrestrial and aquatic plants.
Answer:
Between terrestrial and aquatic plants:

Terrestrial plants: Plants growing on land are called terrestrial plants.
Aquatic plants: Plants living in water environment are called aquatic plants.

Question 7.
Classify the terrestrial plants based on their adaptation.
Answer:
Mcsophytes, serophytes, psammophytes and lithophytes.

Question 8.
What do you mean by caudex?
Answer:
If the trunk of a plant remains unbranched it is said to be caudex. e.g.. Palmyra and coconut.

Question 9.
What is a liana? Mention its importance.
Answer:
Liana is a vine that is perennial and woody. Lianas are major components in the tree canopy layer of some tropical forests. e.g. Ventilago.

Question 10.
What are therophytes?
Answer:
Therophyte is a plant that completes its life cycle in one growing season. e.g. Peas.

Question 11.
Name the two primary functions of roots.
Answer:
The two primary functions of roots:

Absorption of water and nutrients from soil.
Anchorage of plants in soil.

Question 12.
Where does the roots develop from?
Answer:
Root develops from the radicle which is the first structure that comes out when a seed is placed in the soil.

Question 13.
Give an example for
(a) Free floating plants
(b) Submerged plants
(c) Mangroves and
(d) Emergent plants?
Answer:
(a) Free floating plant – e.g Pistia.
(b) Submerged plant – e.g: Hydrilla
(c) Mangroves – e.g: Avicennia
(d) Emergent – e.g: Typha

Question 14.
What is root cap? Mention its role.
Answer:
Root tip is covered by a dome shaped parenchymatous cells called root Cap. It protects the meristematic cells in the apex.

Question 15.
Explain the nature of root based on tropism.
Answer:
Based on tropism. roots are positively geotropic and negatively phototropic.

Question 16.
What are adventitious roots?
Answer:
Roots developing from any part of the plant other than radicle is called adventitious roots.

Question 17.
Why do roots modify their structure? Name the types of root modification.
Answer:
Roots modify their structure to perform secondary functions. The two types of root modification are tap root modification and fibrous root modification.

Question 18.
Name the places from which adventitious roots arise.
Answer:
Adventitious roots may rise from base of stem, nodes or internodes.

Question 19.
What are prop roots? Give an example.
Answer:
Prop (Pillar) roots grow vertically downward from the lateral branches into the soil, e.g., Ficus benghalensis (banyan tree) and Indian rubber.

Question 20.
How velamen helps the Vanda plant?
Answer:
Velamen is a spongy tissue developed by epiphytic roots of Vanda. It helps in absorption of moisture from atmosphere.

Question 21.
Which type of adventitious roots are seen in Bryophyllum plants?
Answer:
Bryophyllum produces foliar roots which arise from the veins or lamina of leaf for the formation of new plants.

Question 22.
How Cuscuta survives?
Answer:
Cuscuta is a parasitic plant which produces haustorial roots from stem that penetrates into the tissue of host plant and suck nutrients. Thus Cuscuta survives.

Question 23.
“Roots perform photosynthesis” – Justify with example.
Answer:
Roots of some climbing or epiphytic plants develop chlorophyll and turn green which help in photosynthesis. These roots are called as photosynthetic roots. e.g.,Tinospora.

Question 24.
Which part of embryo gives rise to root and shoot?
Answer:
Radicle gives rise to root. Plumule gives rise to shoot.

Question 25.
Define node & internode.
Answer:
Node & Internode:

Node: The point from which the leaf arises is called node.
Internode: The region between two adjacent nodes is called intemode.

Question 26.
Explain stem based on tropism.
Answer:
Stem is positively phototropic & negatively geotropic.

Question 27.
Classify buds based on their origin & function.
Answer:
Buds based on their origin & functio:

Based on origin, buds are classified into (a) Terminal or Apical bud and (b) Lateral or Axillary or Axil bud.
Based on function, buds are classified into (a) Vegetative bud and (b) Floral or Reproductive bud.

Question 28.
What are adventitious buds? Give an example.
Answer:
Buds arising at any part other than stem are known as adventitious bud. e.g., Begonia.

Question 29.
Mention various types of stem seen in angiosperms.
Answer:
Majority of angiosperm possess upright, vertically growing erect stem. They are:

Excurrent
Decurrent
Caudex and
Culm.

Question 30.
Why do certain plants climb?
Answer:
Climbing helps to display the leaves towards sunlight and to position the flower for effective pollination.

Question 31.
What are creepers? Give example.
Answer:
Creepers are plants growing horizontally closer to the ground and produces roots at each node. e.g., Oxalis.

Question 32.
Distinguish between dextrose & sinistrose climbers.
Answer:
Between dextrose & sinistrose climbers:

Dextrose: Clockwise coiling climbers are called dextrose, e.g. Dioscorea alata.
Sinistrose: Anti – clockwise coiling climbers are called sinistrose. e.g. Dioscorea bulbifera.

Question 33.
What are root stocks? What are its function?
Answer:
Perennial and some biennial herbs have underground stems, which are generally known as root stocks. Root stock functions as a storage and protective organ.

Question 34.
What is a bulb? Mention its types.
Answer:
Bulb is a condensed, conical or convex stem surrounded by fleshy scale leaves. There are two types:

Tunicated (coated) bulb and
Scaly bulb.

Question 35.
Define branching. Mention its types.
Answer:
The mode of arrangement of branches on a stem is known as branching. There are two main types of branching:

Lateral branching and
Dichotomous branching.

Question 36.
Define phyllome.
Answer:
All the leaves of a plant together are referred as phyllome.

Question 37.
List out any four primary functions of leaves.
Answer:
Photosynthesis, transpiration, gaseous exchange and protection of buds.

Question 38.
What is a leaf base?
Answer:
The part of the leaf attached to the node of the stem is called leaf base. It protects growing buds at its axil.

Question 39.
What is pulvinus?
Answer:
In legumes leaf base become broad, thick and swollen which is known as pulvinus. e.g., Clitoria.

Question 40.
Which leaf part acts a bridge between leaf & stem? Define.
Answer:
Petiole is the bridge between leaf and stem. Petiole or leaf stalk is a cylindrical or subcylindrical or flattened structure of a leaf which joins the lamina with the stem.

Question 41.
Mention the types of leaves based on petiole.
Answer:
A leaf with petiole is said to be petiolate. e.g. Hibiscus. Leaves that do not possess petiole is said to be sessile, e.g. Calotropis.

Question 42.
What are stipules? State its functions.
Answer:
Stipules are the two lateral appendages develop at the leaf base of dicot plants. Stipules protects the leaf in bud condition.

Question 43.
Define venation. Mention its types.
Answer:
The arrangement of veins and veinlets on the leaf blade or lamina is called venation. Venation is classified into two types:

Reticulate venation and
Parallel venation.

Question 44.
Define ligule.
Answer:
In some grasses (Monocots) an additional out growth is present between leaf base and lamina. It is called ligule.

Question 45.
What are stipulate & exstipulate leaves?
Answer:
Stipulate & Exstipulate Leaves:

Leaves with stipules are called stipulate leaves.
Leaves without stipules are called exstipulate or estipulate leaves.

Question 46.
Define phyllotaxy. Mention its types.
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy. There are 4 types of phyllotaxy:

Alternate
Opposite
Temate and
Whorled.

Question 47.
How phyllotaxy helps the plants?
Answer:
Phyllotaxy is to avoid over crowding of leaves and expose the leaves maximum to the sunlight for photosynthesis.

Question 48.
You are given a mango leaf. Which type does it belongs to? Define it.
Answer:
Mango leaf is a simple leaf. A simple leaf is the one, where the petiole bears a single lamina.

Question 49.
Define heterophylly. Which type of plants show this adaptation?
Answer:
Occurence of two different kinds of leaves in the same plants is called heterophylly. Heterophylly is an adaptation of aquatic plants.

Question 50.
When a leaf is said to be centric?
Answer:
When the leaf is more or less cylindrical and directed upwards or downwards, as in pine and onion, etc., the leaf is said to be centric.

Question 51.
Which type of leaf is common among monocots? Define it.
Answer:
When the leaf is directed vertically upwards, as in many monocotyledons, it is said to be ‘ isobilateral leaf. Example: Grass.

Question 52.
Define ptyxis.
Answer:
Rolling or folding of individual leaves is called as ptyxis.

Question 53.
Classify plants based on habit.
Answer:
Based on habit plants are classified into herbs, shrubs, climbers (vines) and trees.

Question 54.
Musa is a monocarpic perennial. Give possible reason.
Answer:
Musa is a monocarpic perennial, since it produces flowers and fruits only once and die after a vegetative growth of several years.

Question 55.
What are the parts that constitute the typical leaf?
Answer:
There are three main parts in a typical leaf:

Leaf base (Hypopodium)
Petiole (Mesopodium) and
Lamina (Epipodium).

Question 56.
What is a pseudobulb?
Answer:
Pseudobulb is a short erect aerial storage or propagating stem of certain epiphytic and terrestrial sympodial orchids, e.g. Bulbophyllum.

Question 57.
Which factor determines the branching patterns?
Answer:
Branching pattern is determined by the relative activity of apical meristem.

Question 58.
Mention the secondary functions of leaf with an example for each.
Answer:
Functions:

Storage
Protection
Support
Reproduction

Examples:

Aloe
Opuntia
Nepenthes
Bryophyllum

Question 59.
List out the families that possess sheathing leafbase.
Answer:
Arecaceae, Musaceae, Zingiberaceae and Poaceae.

Question 60.
What are stipels?
Answer:
Sometimes, small stipule like outgrowths are found at the base of leaflets of a compound leaf. They are called stipels.

Question 61.
Compare the stem nature of Corm and Rhizome
Answer:
Corm:

Stem is succulent underground
Presence of erect growing tips.

Rhizome:

Stem is horizontal underground
Presence of lateral growing tips.

III. Short Answer Type Questions (3 Marks)

Question 1.
Morphological study is important in Taxonomy. Why?
Answer:
Morphological features are important in determining productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.

Question 2.
Differentiate between polycarpic and monocarpic perennial.
Answer:
Polycarpic perennial:

When they bear fruits every year, they are called polycarpic.
e.g. mango, sapota, etc.

Monocarpic perennial:

Some plants produce flowers and fruits only once and die after a vegetative growth of several years. These plants are called monocarpic.
e.g. Bamboo, Agave, Musa, Talipot palm.

Question 3.
List down the key difference between roots and shoots.
Answer:
Roots:

Positively geotropic
Negatively phototropic
Non – green in colour
Nodes, intemodes and buds are absent

Shoots:

Negatively geotropic
Positively phototropic
Green in colour
Nodes, intemodes and buds are present

Question 4.
Name the three distinct zones of root.
Answer:
The three distinct zones of root:

Meristematic Zone
Zone of Elongation
Zone of Maturation

Question 5.
Draw a simplified diagram showing the various regions of root.
Answer:
The various regions of root:

Question 6.
Give a brief account on tap root system.
Answer:
Primary root is the direct prolongation of the radicle. When the primary root persists and continues to grow as in dicotyledons, it forms the main root of the plant and is called the tap root. Tap root produces lateral roots that further branches into finer roots. Lateral roots along with its branches together called as secondary roots.

Question 7.
How does the fibrous root system develops?
Answer:
Most of the monocots the primary root of the seedling is short lived and lateral roots arise from various regions of the plant body. These are bunch of thread – like roots equal in size which are collectively called fibrous root system generally found in grasses. Example: Oryza sativa.

Question 8.
Briefly explain the development process of leaf primordium.
Answer:
The plumule of the embryo of a germinating seed grows into stem. The epicotyl elongates after embryo growth into the axis (the stem) that bears leaves from its tip, which contain the actively dividing cells of the shoot called apical meristem. Further cell divisions and growth result in the formation of mass of tissue called a leaf primordium.

Question 9.
Name the three types of Adventitious buds.
Answer:
The three types of Adventitious buds:

Radical buds
Foliar buds
Cauline buds

Question 10.
Write a brief note on Bulbils.
Answer:
Bulbils (or specialized buds): Bulbils are modified and enlarged bud, meant for propagation. When bulbils detach from parent plant and fall on the ground, they germinate into new plants and serve as a means of vegetative propagation, e.g., Agave.

Question 11.
Distinguish between root climbers and stem climbers.
Answer:
Root climbers:

Plants climbing with the help of adventitious roots (arise from nodes).
e.g. Piper betel, Piper nigrum, Hedera helix, Pothos, Hoya.

Stem climbers:

These climbers lack specialised structure for climbing and the stem itself coils around the support.
e.g. Ipomoea, Convolvulus, Dolichos, Clitoria, Quisqualis.

Question 12.
Explain the three different types of trailers with an example.
Answer:
Types of Trailers:

Prostrate (Procumbent): A stem that grows flat on the ground, e.g. Evolvulus alsinoides, Indigofera prostrata.
Decumbent: A stem that grows flat but becomes erect during reproductive stage. e.g., Portulaca, Tridax, Lindenbergia.
Diffuse: A trailing stem with spreading branches, e.g. Boerhaavia diffusa, Merremia tridentata.

Question 13.
Cladode is a stem modification. Comment on it.
Answer:
Cladode is a flattened or cylindrical stem similar to Phylloclade but with one or two inteENodes only. Their stem nature is evident by the fact that they bear buds, scales and flowers. e.g. Asparagus (cylindrical cladode), Ruscus (flattened Cladode).

Question 14.
Comment on Corm with an example.
Answer:
Corm is a succulent underground stem with an erect growing tip. The corm is surrounded by scale leaves and exhibit nodes and intemodes. e.g., Amorphophallus, Gladiolus, Colacasia, Crocus, Colchicum.

Question 15.
Differentiate between monopodial and sympodial branching.
Answer:
Monopodial branching:

The terminal bud grows uninterrupted and produce several lateral branches. This type of growth is also known as monopodial branching.
e.g. Polyalthia, Swietenia, Antiaris.

Sympodial branching:

The terminal bud caese to grow after a period of growth and the further growth is taken care by successive or several lateral meristem or buds. This type of growth is also known as sympodial branching.
e.g. Cycas.

Question 16.
Draw and label the parts of a leaf.
Answer:
The parts of a leaf:

Question 17.
Describe the pattern of leaf arrangement in mosaic leaf.
Answer:
In mosaic leaf, leaves tend to fit in with one another and adjust themselves in such a way that they may secure the maximum amount of sunlight with minimum amount of overlapping. The lower leaves have longer petioles and successive upper leaves possess decreasing length petioles, e.g., Acalypha, Begonia.

Question 18.
How the leaf hooks helps the Bignonia plant?
Answer:
In cat’s nail (Bignonia unguiscati) an elegant climber, the terminal leaflets become modified into three, very sharp, stiff and curved hooks, very much like the nails of a cat. These hooks cling to the bark of a tree and act as organs of support for climbing.

Question 19.
Which types of plants develop tendril? How does it helps the plant?
Answer:
In some plants Stem is very weak and hence they have some special organs for attachment to the support. So some leaves are partially or wholly modified into tendril. Tendril is a slender wiry coiled structure which helps in climbing the support.

Question 20.
Rosa species plants are not eaten by herbivores. Why?
Answer:
Rosa species plants develop Prickles. Prickles are small, sharp structure which are the outgrowths from epidermal cells of stem or leaf. It helps the plant in scrambling over other plants. It is also protective against herbivory.

Question 21.
Certain plants like Aloe and Agave can survive in extreme dry conditions. How ?
Answer:
Aloe and Agave are Xerophytes. They develop fleshy and swollen leaves. These succulent leaves store water, mucilage and food. They also resist desiccation.

Question 22.
Write a brief note on Phyllode.
Answer:
Phyllodes are flat, green – coloured leaf – like modifications of petioles or rachis. The leaflets or lamina of the leaf are highly reduced or caducous. The phyllodes perform photosynthesis and other functions of leaf.
Example: Acacia auriculiformis (Australian acacia), Parkinsonia.

Question 23.
Briefly describe the leaf modification in Nepenthes.
Answer:
The leaf becomes modified into a pitcher in Nepenthes. In Nepenthes the basal part of the leaf is laminar and the midrib continues as a coiled tendrillar structure. The apical part of the leaf as modified into a pitcher the mouth of the pitcher is closed by a lid which is the modification of leaf apex.

Question 24.
How the leaves of Utricularia helps in its nourishment?
Answer:
In bladderwort (Utricularia), a rootless free – floating, aquatic plant the leaf is very much segmented. Some of these segments are modified to form bladder – like structures, with a trap – door entrance that traps aquatic animalcules.

Question 25.
What do you understand by the term developmental heterophylly.
Answer:
In plants like Ficus heterophylla leaves vary from entire to variously lobed structures during different developmental stages. Young leaves are usually lobed or dissected and the mature leaves are entire. Such type is known as developmental heterophylly.

Question 26.
List out few secondary functions of stem.
Answer:
Foods storage, perennation, water storage, photosynthesis, buoyancy, protection and support.

Question 27.
Make a tabular column showing types of terrestrial plants and their environmental adaptation with examples.
Answer:
Types of terrestrial plants and their environmental adaptation with examples:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Make a list of aquatic plants and their environmental adaptation with an example.
Answer:
Aquatic plants and their environmental adaptation with an example:

Question 2.
Compare the location, cellular types and the functions of different zones of root.
Answer:
The location, cellular types and the functions of different zones of root:

Question 3.
Draw a flow chart depicting the various types of root modification.
Answer:
The various types of root modification:

Question 4.
Describe the tap root modification for storage purpose with diagram.
Answer:
Tap root modification – Storage roots

Conical Root – These are cone like, broad at the base and gradually tapering towards the apex. e.g. Daucus carota.
Fusiform root – These roots are swollen in the middle and tapering towards both ends. e.g. Raphanus sativus
Napiform root – It is very broad and suddenly tapers like a tail at the apex. e.g. Beta vulgaris

Question 5.
Define bud. Explain the types of buds based on location.
Answer:
Buds are the growing points surrounded by protective scale leaves:

Terminal bud or Apical bud: These buds are present at the apex of the main stem and at the tips of the branches.
Lateral bud or Axillary bud: These buds occur in the axil of the leaves and develop into a branch or flower.
Extra axillary bud: These buds are formed at nodes but outside the axil of the leaf as in Solanum americanum.
Accessory bud: An extra bud on either side (collateral bud) or above (superposed bud or serial bud) the axillary bud. e.g: Citrus and Duranta.

Question 6.
Draw a flow chart illustrating stem modifications.
Answer:
A flow chart illustrating stem modifications:

Question 7.
List out the characteristics of leaf.
Answer:
The characteristics of leaf:

Leaf is a lateral appendage of the stem.
It is borne at the node of the stem.
It is exogenous in origin.
It has limited growth.
It does not possess apical bud.
It has three main parts namely, leaf base, petiole and lamina.
Lamina of the leaf is traversed by vascular strands, called veins.

Question 8.
Define phyllotaxy. Explain its types
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy (Greek. Phyllon = leaf; taxis = arrangement). Phyllotaxy is to avoid over crowding of leaves and expose the leaves maximum to the sunlight for photosynthesis. The four main types of phyllotaxy are:

Alternate
Opposite
Temate and
Whorled.

1. Alternate phyllotaxy: In this type there is only one leaf per node and the leaves on the successive nodes are arranged alternate to each other. Spiral arrangement of leaves show vertical rows are called orthostichies. They are two types:

Alternate spiral: In which the leaves are arranged alternatively in a spiral manner. e.g. Hibiscus and Ficus.
Alternate distichous or Bifarious: In which the leaves are organized alternatively in two rows on either side of the stem, e.g. Monoon longifolium (Polyalthia longifolia).

2. Opposite phyllotaxy: In this type each node possess two leaves opposite to each other. They are organized in two different types:

Opposite superposed: The pair of leaves arranged in succession are in the same direction, that is two opposite leaves at a node lie exactly above those at the lower node. e.g. Psidium (Guava), Eugenia jambolana (Jamun) and Quisqualis (Rangoon creeper).
Opposite decussate: In this type of phyllotaxy one pair of leaves is placed at right angles to the next upper or lower pair of leaves, e.g., Calotropis, Zinnia and Ocimum

3. Ternate phyllotaxy: In this type there are three leaves attached at each node. e.g. Nerium.

4. Whorled (verticillate) type of phyllotaxy: In this type more than three leaves are present in a whorl at each node forming a circle or whorl, e.g. Allamanda and Alstonia scholaris.

Question 9.
Define ptyxis & explain its types.
Answer:
Rolling or folding of individual leaves is called ptyxis. There are seven types of ptyxis as follows:

Reclinate – When the upper half of the leaf blade is bent upon the lower half as in loquat (Eriobotrya japonica).
Conduplicate – When the leaf is folded lengthwise along the mid – rib, as in guava, sweet potato and camel’s foot tree (Bauhinia).
Plicate or Plaited – When the leaf is repeatedly folded longitudinally along ribs in a zig – zag manner, as in Borassus flabellifer.
Circinate – When the leaf is rolled from the apex towards the base like the tail of a dog, as in ferns.
Convolute – When the leaf is rolled from one margin to the other, as in banana, aroids and Indian pennywort. Musa and members of Araceae.
Involute – When the two margins are rolled on the upper surface of the leaf towards the mid – rib or the centre of the leaf, as in water lily, lotus, Sandwich Island Climber (Antigonon) and Plumbago.
Crumpled – When the leaf is irregularly folded as in cabbage.

Question 10.
How the duration of leaf is determined? Classify leaves according to duration.
Answer:
Leaves may stay and function for few days to many years, largely determined by the adaptations to climatic conditions.

Cauducuous (Fagacious): Falling off soon after formation, e.g., Opuntia and Cissus quadrangularis.
Deciduous: Falling at the end of growing season so that the plant (tree or shrub) is leafless in winter / summer season, e.g., Maple, Plumeria, Launea and Erythrina.
Evergreen: Leaves persist throughout the year, falling regularly so that tree is never leafless. e.g., Mimusops and Calophyllum.
Marcescent: Leaves not falling but withering on the plant as in several members of Fagaceae.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Roots are non – green coloured. Is there is any green coloured root? Explain.
Answer:
Yes, roots of certain epiphytic & climbing plants develop chlorophyll and turn green to perform the function of photosynthesis. Such a root is called photosynthetic root or assimilatory root. E.g., Tinospora.

Question 2.
Which part of ginger and onion are edible?
Answer:
The edible part of onion is stem covered by fleshy leaves (bulb). The edible part of ginger is underground stem (rhizome).

Question 3.
Name the body parts of the following plants which is modified for food storage.
Answer:
Plant:

Carrot
Colocasia
Aloe

Modified part for food storage:

Root
Stem
Leaves

Question 4.
Give two examples for angiospermic plants producing adventitious roots.
Answer:
Two examples for angiospermic plants producing adventitious roots:

Buttress root of Bombax.
Prop (Pillar) root of Ficus benghalensis.

Question 5.
Rhizome of ginger is like roots of other plants grown underground. Despite this fact ginger is a stem not a root – Justify.
Answer:
Rhizome of ginger is a underground stem not a root because, it possess nodes, internodes, scale leaves & buds, which are the characteristics of stem.

Question 6.
Vanda is an epiphyte. Epiphytes are the plants growing on branches of trees. They do not have direct contact with soil. How they obtain water for its photosynthetic activity?
Answer:
Epiphytic plants like Vanda develop special type of roots containing sponge – like tissue called velamen. These spongy tissue helps in absorbing the atmospheric moisture and utilize it for their photosynthetic activity.

Question 7.
How does a pneumatopore work?
Answer:
Pneumatopores are the special above-ground roots growing above the surface of water seen in plants growing in water logged soils. These pneumatopores has small pores that facilitate the intake of oxygen by roots.

Question 8.
Carnivorous plants like Nepenthes have nutritional adaptations. Which part of Nepenthes plant is modified to solve this problem?
Answer:
The apical part of the leaf is modified into pitcher and the leaf tip is modified into lid of pitcher.

Question 9.
Why do we use the term ‘monocarpic perennial’ for Musa?
Answer:
Musa is a monocarpic perennial because it grows for several years but produces flowers and fruits once in its life time.

Question 10.
Mention any two morphological characters to differentiate monocots from dicots.
Answer:
Two morphological characters to differentiate monocots from dicots:

Monocots have fibrous roots and dicots have tap root system.
Monocot leaves show parallel venation, whereas dicot leaves show reticulate venation.

Question 11.
Why potato tuber is considered as a stem? Although it is an underground plant part.
Answer:
Though potato tuber is found underground, it is a stem since it possess axillary buds & scale leaves.

Question 12.
Fibrous roots are adventitious in origin – Explain.
Answer:
Adventitious roots are those arising from plant parts other than radicle. Fibrous roots are seen in monocots. In monocots, the primary root arising from radicle is short – lived & soon replaced by the lateral roots arising in bunches from the base of stem. Thus fibrous roots are adventitious in origin.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Leave a Comment / Class 11 / By Prasanna

Students can Download Bio Botany Chapter 2 Plant Kingdom Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Samacheer Kalvi 11th Bio Botany Plant Kingdom Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Which of the plant group has gametophyte as a dominant phase?
(a) Pteridophytes
(b) Bryophytes
(c) Gymnosperm
(d) Angiosperm
Answer:
(b) Bryophytes

Question 2.
Which of following represent gametophytic generation in pteridophytes?
(a) Prothallus
(b) Thallus
(c) Cone
(d) Rhizophore
Answer:
(a) Prothallus

Question 3.
The haploid number of chromosome for an Angiosperm is 14, the number of chromosome in its endosperm would be …………… .
(a) 7
(b) 14
(c) 42
(d) 28
Answer:
(c) 42

Question 4.
Endosperm in Gymnosperm is formed …………… .
(a) at the time of fertilization
(b) before fertilization
(c) after fertilization
(d) along with the development of embryo
Answer:
(b) before fertilization

Question 5.
Differentiate halpontic and diplontic life cycle.
Answer:
Halpontic cycle:

Gametophyte is dominant
Sporophyte is represented by zygote

Diplontic cycle:

Sporophyte is dominant
Gametophyte is represented by few cell gametophyte

Question 6.
What is Plectostele? Give example.
Answer:
Plectostele: Xylem plates alternates with phloem plates. Example: Lycopodiurn clavatum.

Question 7.
What do you infer from the term Pycnoxylic?
Answer:
Pycnoxylic wood is compact with narrow medullary ray. Example: Pinus

Question 8.
Mention two characters shared by Gymnosperms and Angiosperms.
Answer:
Gymnosperms:

Vessels are absent (except Gnetales)
Phloem lacks companion cells

Angiosperms:

Vessels are present
Companion cells are present

Question 9.
Do you think shape of chloroplast is unique for algae? Justify your answer.
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup shaped (chlamydomonas). Discoid ((Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 10.
Do you agree with the statement ‘Bryophytes need water for fertilization’? Justify your answer.
Answer:
Yes, in Bryophytes. water plays a vital role in fertilisation, since water film is needed for the transfer of spermatium (male sex cell) to the egg cell.

Entrance Examination Questions Solved

Question 1.
Which of the following are found in extreme saline conditions? (NEET – 2017)
(a) Archaebacteria
(b) Eubacteria
(c) Cyanobacteria
(d) Mycobacteria
Answer:
(a) Archaebacteria

Question 2.
Select the mismatch …………… . (NEET – 2017)

Answer:
(b) Rhodospirillum – Mycorrhiza

Question 3.
Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen? (NEET – 2017)
(a) Bacillus
(b) Pseudomonas
(c) Mycoplasma
(d) Nostoc
Answer:
(c) Mycoplasma

Question 4.
Read the following statements (A to E) and select the option with all correct statements: (AIPMT – 2015).
A. Mosses and Lichens are the first organisms to colonise a bare rock.
B. Selaginella is a homosporous pteridophyte,
C. Coralloid roots in Cycas have VAM.
D. Main plant body in bryophytes is gametophytic, whereas in pteridophytes it is sporophytic.
E. In gymnosperms, male and female gametophytes arc present within sporangia located on sporophyte.

(a) B, C and E
(b) A, C and D
(c) B, C and D
(d) A, D and E
Answer:
(d) A, D and E

Question 5.
An example of colonial algae is …………… . (NEET – 2017)
(a) Chlorella
(b) Volvox
(c) Ulothrix
(d) Spirogyra
Answer:
(b) Volvox

Question 6.
Five kingdom system of classification suggested by RH. Whittaker is not based on …………… . (AIPMT- 2014)
(a) Presence or absence of a well defined nucleus
(b) Mode of reproduction
(c) Mode of nutrition
(d) Complexity of body organisation
Answer:
(a) Presence or absence of a well defined nucleus

Question 7.
Mycorrhizae are the example of …………… . (NEET – 2017)
(a) Fungitasis
(b) Amensalism
(c) Antibiosis
(d) Mutualism
Answer:
(d) Mutualism

Question 8.
Which of the following shows coiled RNA strand and capsomeres? (AIPMT – 2014)
(a) Polio virus
(b) Tobacco mosaic virus
(c) Measles virus
(d) Retrovirus
Answer:
(b) Tobacco mosaic virus

Question 9.
Viroids differ from viruses in having: (NEET – 2017)
(a) DNA molecules with protein coat
(b) DNA molecules without protein coat
(c) RNA molecules with protein coat
(d) RNA molecules without protein coat
Answer:
(d) RNA molecules without protein coat

Question 10.
Select the mismatch: (NEET – 2017)
(a) Pinus – Dioecious
(b) Cycas – Dioecious
(c) Salvinia – Heterosporous
(d) Equisetum – Homosporous
Answer:
(a) Pinus – Dioecious

Question 11.
Life cycle of Ectocarpus and Fucus respectively are …………… . (NEET – 2017)
(a) Haplontic, Diplontic
(b) Diplontic, Haplodiplontic
(c) Haplodiplontic, Diplontic
(d) Haplodiplontic, Halplontic
Answer:
(c) Haplodiplontic, Diplontic

Question 12.
Zygote meiosis Is characterisitic of …………… . (NEET – 2017)
(a) Marchantia
(b) Fucus
(c) Funaria
(d) Chlamydomonas
Answer:
(d) Chlamydomonas

Question 13.
Which of the following is correctly matched for the product produced by them? (NEET – 2017)
(a) Acetobacter aceti : Antibiotics
(b) Merthanobacterium : Lactic acid
(c) Penicillium natatum : Acetic acid
(d) Saccharomyces cerevisiae : Ethanol
Answer:
(d) Saccharomyces cerevisiae : Ethanol

Question 14.
Which of the following components provides sticky character to the bacterial cell? (NEET – 2017)
(a) Cell wall
(b) Nuclear membrane
(e) Plasma membrane
(d) Glycocalyx
Answer:
(d) Glycocalyx

Question 15.
Which of the following statements is wrong for viroids? (NEET – 2016)
(a) They lack a protein coat
(b) They are smaller than viruses
(c) They causes infections
(d) Their RNA is a high molecular weight
Answer:
(d) Their RNA Is a high molecular weight

Question 16.
In bryophytes and pteridophytes, transport of male gametes require …………… . (NEET – 20L6)
(a) Wind
(b) Insects
(c) Birds
(d) Water
Answer:
(d) Water

Question 17.
How many organisms in the list below are autotrophs? (AIPMT Mains 2012)
Answer:
Lactobacillus, Nostoc, Chora, Nitrosomonas, Nitrobacter, Streptomyces, Saccharomyces, Trypanosoma, Porphyra, Wolffia.
(a) Four
(b) Five
(c) Six
(d) Three
Answer:
(c) Six

Question 18.
Which of the following would appear as the pioneer organisms on bare rocks? (NEET – 2016)
(a) Lichens
(b) Liverworts
(c) Mosses
(d) Green algae
Answer:
(a) Lichens

Question 19.
Monoecious plant of Chara shows occurrence of …………… .(NEET – 2013)
(a) Stamen and carpel on the same plant
(b) Upper antheridium and lower oogonium on the same plant
(c) Upper oogonium and lower antheridium on the same plant
(d) Antheridiophore and archegoniophore on the same plant
Answer:
(c) Upper oogonium and lower antheridium on the same plant

Question 20.
Read the following five statement (A – E) and answer as asked next to them …………… . (AIPMT Prelims – 2012)
(a) In Equisetum, the female gametophyte is retained on the parent sporophyte
(b) In Ginkgo, male gametophyte is not independent
(c) The sporophyte in Riccia is more developed than that in Polytrichum
(d) Sexual reproduction in Volvox is isogamous
(e) The spores of slime moulds lack cell walls
How many of the above statement are correct? (AIPMT Prelims – 2012)
(a) Two
(b) Three
(c) Four
(d) One
Answer:
(d) One

Question 21.
One of the major components of cell wall of most fungi is …………… . (NEET – 2016)
(a) Chitin
(b) Peptidoglycan
(c) Cellulose
(d) Hemicellulose
Answer:
(a) Chitin

Question 22.
Which one of the following statements is wrong? (NEET – 2016)
(a) Cyanobacteria are also called blue – green algae
(b) Golden algae are also called desmids
(c) Eubacteria are also called false bacteria
(d) Phycomycetes are also called algal fungi
Answer:
(c) Eubacteria are also called false bacteria

Question 23.
Flagellated male gametes are present in all the three of which one of the following sets? (AIPMT Prelims – 2007)
(a) Riccia, Dryopteris and Cycas
(b) Anthoceros, Funaria and Spirogyra
(c) Zygnema, Saprolegnia and Hydrilla
(d) Fucus, Marsilea and Calotropis
Answer:
(a) Riccia, Dryopteris and Cycas

Question 24.
Ectophloic siphonostele is found in …………… . (AIPMT Prelims – 2005)
(a) Adiantum and Cucurbitaceae
(b) Osmunda and Equisetum
(c) Marsilea and Botrychium
(d) Dicksonia and maiden hair fern
Answer:
(b) Osmunda and Equisetum

Question 25.
Which part of the tobacco plant is infected by Meloidogyne incognita? (NEET – 2016)
(a) Flower
(b) Leaf
(c) Stem
(d) Root
Answer:
(d) Root

Question 26.
Select the correct statement: (NEET – 2016)
(a) Gymnosperms are both homosporous and heterosporous
(b) Salvinia, Ginkgo and Pinus all are gymnosperms
(c) Sequoia is one of the tallest trees
(d) The leaves of gymnosperms are not well adapted to extremes of climate
Answer:
(c) Sequoia is one of the tallest trees

Question 27.
Seed formation without fertilization in flowering plants involves the process of …………… . (NEET – 2016)
(a) Sporulation
(b) Budding
(c) Somatic hybridization
(d) Apomixis
Answer:
(d) Apomixis

Question 28.
Chrysophytes, Euglenoids, Dinoflagellates and Slime moulds are included in the kingdom …………… . (NEET – 2016)
(a) Animalia
(b) Monera
(c) Protista
(d) Fungi
Answer:
(b) Monera

Question 29.
The primitive prokaryotes responsible for the production of biogas from the dung of ruminant animals, include the …………… . (NEET – 2016)
(a) Halophiles
(b) Thermoacidophiles
(c) Methanogens
(d) Eubacteria
Answer:
(c) Methanogens

Samacheer Kalvi 11th Bio Botany Plant Kingdom Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
Gametophytic phase is …………… .
(a) triploid
(b) tetraploid
(c) haploid
(d) diploid
Answer:
(c) haploid

Question 2.
Haplodiplontic life cycle is seen in …………… .
(a) algae
(b) gymnosperm
(c) bryophytes
(d) angiosperm
Answer:
(c) bryophytes

Question 3.
Which algae leads an endozoic life in Hydra?
(a) Chlorella
(b) Gracilaria
(c) Ulothrix
(d) Chlamydomonas
Answer:
(a) Chlorella

Question 4.
Study of algae is called …………… .
(a) biology
(b) mycology
(c) bryology
(d) phycology
Answer:
(d) phycology

Question 5.
Siliceous walls are present in …………… .
(a) Chara
(b) Chlamydomonas
(c) Dunaliella
(d) Diatoms
Answer:
(d) Diatoms

Question 6.
In Chara, thallus is encrusted with …………… .
(a) calcium carbonate
(b) hydrogen sulphate
(c) silica
(d) ammonium carbonate
Answer:
(a) calcium carbonate

Question 7.
Pyrenoids are present in …………… .
(a) mitochondrion
(b) chloroplast
(c) ribosomes
(d) lysosomes
Answer:
(b) chloroplast

Question 8.
Type of vegetative reproduction seen in ulothrix is …………… .
(a) bulbils
(b) fission
(c) fragmentation
(d) tubers
Answer:
(c) fragmentation

Question 9.
…………… are thin walled non – motile spores.
(a) Zoospores
(b) Akinetes
(c) Aplanospores
(d) Genunae
Answer:
(c) Aplanospores

Question 10.
Fusion of either morphologically or physiologically dissimilar gametes is called as …………… .
(a) isogamy
(b) anisogamy
(c) syngamy
(d) oogamy
Answer:
(b) anisogamy

Question 11.
According to Fritsch, the algae are classified into …………… classes.
(a) 10
(b) 12
(c) 11
(d) 9
Answer:
(c) 11

Question 12.
Photosynthetic part of the phaeophyceae thallus Is called as
(a) holdfast
(b) stipes
(c) lamina
(d) fronds
Answer:
(d) fronds

Question 13.
A characteristic pigment of phaeophyceae is …………… .
(a) xanthophyle
(b) carotenoid
(c) fucoxanthin
(d) chlorophyll
Answer:
(c) fucoxanthin

Question 14.
…………… is used as single cell protein.
(a) Chlorella
(b) Kelps
(c) Chlamydomonas
(d) Spirogyra
Answer:
(a) Chlorella

Question 15.
Gelidium belongs to …………… members.
(a) Rhodophyccae
(b) Phaeophyceae
(c) Cyanophyccae
(d) Dinophyceae
Answer:
(a) Rhodophyceae

Question 16.
Carrageenan is obtained from …………… .
(a) Chlorella
(b) Chara
(c) Chondrus
(d) Chlamydomonas
Answer:
(c) Chondrus

Question 17.
…………… are the amphibians of the plant kingdom.
(a) Pteridophytes
(b) Algae
(c) Gymnosperms
(d) Bryophytes
Answer:
(d) Bryophytes

Question 18.
Marchantia vegetatively propagates by …………… .
(a) tubers
(b) gemmae
(c) buds
(d) brood bodies
Answer:
(b) gemmae

Question 19.
Peat is obtained from …………… .
(a) Anthoceros
(b) Dendroceros
(c) Sphagnum
(d) Funaria
Answer:
(c) Sphagnum

Question 20.
…………… is a bryophyte used to cure pulmonary tuberculosis.
(a) Marchantia polymorpha
(b) Polytrichum
(c) Sphagnum
(d) Bryum
Answer:
(a) Marchantia polymorpha

Question 21.
Type of stele seen in Marsilea is …………… .
(a) Protostele
(b) Siphonostele
(c) Adiantum
Answer:
(b) Siphonostele

Question 22.
Which of the following pteridophyte is used as a biofertiliser?
(a) Marsilea
(b) Pteridium
(c) Pteris
(d) Azolla
Answer:
(d) Azolla

Question 23.
Which of the following is naked seed producing plant?
(a) Angiosperm
(b) Gymnosperm
(c) Pteridophytes
(d) Bryophytes
Answer:
(b) Gymnosperm

Question 24.
Amber is obtained from …………… .
(a) Angiosperm
(b) Gymnosperm
(c) Pteridophytes
(d) Bryophytes
Answer:
(b) Gymnosperm

Question 25.
Coralloid roots of cycas have symbiotic association with …………… .
(a) Blue green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(a) Blue green algae

Question 26.
Pinus roots are in symbiotic relationship with …………… .
(a) Blue green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(b) Mycorrhiza

Question 27.
Which is not a class of Gymnosperm?
(a) Lycopodia
(b) Cycadopsida
(c) Coniferopsida
(d) gnetopsida
Answer:
(a) Lycopodia

Question 28.
The endosperm of gymnosperm is …………… .
(a) haploid
(b) triploid
(c) diploid
(d) Poliploidy
Answer:
(a) haploid

Question 29.
Shiwalik fossil park is located at …………… .
(a) Madhya Pradesh
(b) Himachal Pradesh
(c) Rajmahal hills
(d) Jharkhand
Answer:
(b) Himachal Pradesh

Question 30.
When does the angiosperm appeared on Earth?
(a) Devonian
(b) Cambrian
(c) Early cretaceous
Answer:
(c) Early cretaceous

Question 31.
Which is also called as vascular cryptogam?
(a) Gymnosperms
(b) Pteridophytes
(c) Bryophytes
(d) Algae
Answer:
(b) Pteridophytes

Question 32.
Which is not a cryptogam?
(a) Algae
(b) Bryophytes
(c) Pteridophyta
(d) Angiospermae
Answer:
(d) Angiospermae

Question 33.
…………… is a haiophytic alga.
(a) Chlamydomonas nivalis
(b) Dunaliella salina
(c) Coleochaete
(d) Volvox
Answer:
(b) Dunaliella salina

Question 34.
Who is called as the Father of Indian Phycology?
(a) M.O. Parthasarathy
(b) Y. Bharadwaja
(c) V.S. Sundaralingam
(d) V. Desikachary
Answer:
(a) M.O. Parthasarathy

Question 35.
Wedge shaped modified branches developed by Sphacelaria are called as …………… .
(a) Buds
(b) Akinetes
(c) Tubers
(d) Bulbils
Answer:
(d) Bulbils

Question 36.
Pteridophytes were abundant in the …………… period.
(a) Cambrian
(b) Precambrian
(c) Devonian
(d) Cretaceous
Answer:
(c) Devonian

Question 37.
Heterospory is originated in …………… .
(a) Gymnosperms
(b) Pteridophytes
(c) Bryophytes
(d) Algae
Answer:
(b) Pteridophytes

Question 38.
Sago is obtained from …………… .
(a) Cycas revoluta
(b) Pinus roxburghii
(c) Finus insularis
(d) Cedrus deodara
Answer:
(a) Cycas revoluta

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define alternation of generation.
Answer:
Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 2.
Name any two marine algae.
Answer:
Two marine alga:

Gracilaria and
Sargassum

Question 3.
Name any two fresh water algae.
Answer:
Two fresh water algae:

Oedogonium and
Ulothrix

Question 4.
Mention any two endozoic algae.
Answer:
Two endozoic algae:

Chlorella and
Cladophora crispata.

Question 5.
Define Phycology.
Answer:
The study of algae is called as phycology or algology.

Question 6.
Define epiphytic algae with an example.
Answer:
Algae growing on the surface of aquatic plants are called as epiphytic algae.
Example: Coleochaete.

Question 7.
Name few eminent algologist.
Answer:
F.E. Fritsch, F.E. Round, Y. Bharadwaja and T.V. Desikachary

Question 8.
Write the chemical composition of algae cell wall.
Answer:
Cellulose and hemicellulose.

Question 9.
List out the criteria involved in algal classification.
Answer:
Pigmentation, reserve food materials and flagellation pattern.

Question 10.
What are pyrenoids? Mention its role.
Answer:
Pyrenoids are proteinaceous bodies found in chromatophores of algae and assist in the synthesis and storage of starch.

Question 11.
Distinguish between Isogamy and Oogamy with example.
Answer:
Between Isogamy Oogamy with example:

Isogamy	Oogamy
Fusion of morphologically and physiologically similar gametes. e.g. Ulothrix	Fusion of both morphologically and physiologically dissimilar gametes. e.g. Sargassum

Question 12.
Which is the reserve food material of phaeophyceae members.
Answer:
Mannitol and Laminarin starch.

Question 13.
Name the male & female sex organ of Rhodophyceae members.
Answer:
Male sex organ is called spermatangium. Female sex organ is called carpogonium.

Question 14.
Which is responsible for pigmentation of Brown algae?
Answer:
A golden brown pigment called fucoxanthin is present and it gives shades of colour from olive green to brown to the algal members of Phaeophyceae.

Question 15.
Mention any two algae members used in Agar – Agar production.
Answer:
Two algae member used in Agar – Agar production:

Gracilaria and
Gigartina.

Question 16.
Bryophytes are amphibians of plant kingdom – Justify.
Answer:
Bryophytes are called as ‘amphibians of plant kingdom’ because they need water for completing their life cycle.

Question 17.
Why bryophytes are called as Non – vascular cryptogam?
Answer:
Vascular tissue like xylem and phloem are completely absent in bryophytes, hence called as ‘Non – vascular cryptogams’.

Question 18.
Which type of sexual reproduction occurs in Bryophytes. Name the male & female parts.
Answer:
Sexual reproduction is Oogamous. Male sex organ is called as Antheridia. Female sex organ is called as Archegonia.

Question 19.
What are sporophylls?
Answer:
Sporophylls are the special leaves on which spore bearing sporangia are borne. Sporophylls organize to form strobilus or cone.

Question 20.
Compare Eusporangiate and Leptosporangiate.
Answer:
Eusporangiate and Leptosporangiate:

Eusporangiate	Leptosporangiate
Development of sporangium from group of initials	Development of sporangium from single initial

Question 21.
Differentiate homospory and heterospory with example.
Answer:
Homospory and Heterospory With Example:

Homospory	Heterospory
Production of one type of spores. e.g. Lycopodium	Production of two types of spores. e.g. Selaginella

Question 22.
Which period, does the pteridophytes dominate the surface?
Answer:
Devonian period of Paleozoic era.

Question 23.
List out the ways of vegetative propagation by Pteridophytes.
Answer:
Fragmentation, resting buds, root tubers and adventitious buds.

Question 24.
Define Stele & mention its types.
Answer:
Stele refers to the central cylinder of vascular tissues consisting of xylem, phloem. pericycle and sometimes medullary rays with pith. There are two types of steles:

Protostele
Siphonostele

Question 25.
Distinguish between Protostele & Siphonostele.
Answer:

Protostele	Siphonostele
In protostele, xylem surrounds phloem.	In siphonostele, phloem surrounds xylem.

Question 26.
DefIne Eustele?
Answer:
The stele is split into distinct collateral vascular bundles around the pith. Example: Dicot stem.

Question 27.
What Is amber? Which group of plants produce amber?
Answer:
Amber is a plant secretion that is a efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Question 28.
Which period, does the gymnosperm dominate the Earth?
Answer:
Jurassic and Cretaceous periods of Mesozoic era.

Question 29.
Distinguish between Manoxylic & Pycnoxylic.
Answer:

Manoxylic	Pycnoxylic
Porous, soft wood	Compact hard wood
More parenchyma with wide medullary rays.	Compact with narrow medullary rays.

Question 30.
Define Siphonogamous condition.
Answer:
Siphonogainy refers to the development of pollen tubes for the transfer of male nuclei to egg cell.

Question 31.
Mention any two common features for both gymnosperm & angiosperm.
Answer:
Two common features for both gymnosperm & angiosperm.:

Production of seeds
Presence of Eustele

Question 32.
What is Canada balsam. Add a note on it.
Answer:
Canada balsam is a resin obtained from Abies balsamea. It is used as mounting medium in permanent slide preparation.

Question 33.
Why do we use the term ‘form genera’ for fossil plants?
Answer:
The term ‘form genera’ is used to name the fossil plants because the whole plant is not recovered as fossils instead organs or parts of the extinct plants are obtained in fragments.

Question 34.
Name few fossil sites of india.
Answer:
few fossil sites of india:

1. Shiwalik fossil park	1. Himachal Pradesh
2. Mandla fossil park	2. Madhya Pradesh
3. Rajmahal hills	3. Jharkhand
4. Ariyalur	4. Tamil Nadu

Question 35.
Mention the names of any two fossil gymnosperm.
Answer:
Medullosa, Lepidocarpon and Lepidodendron.

Question 36.
Which group of plants dominate the Earth today? Define it.
Answer:
Angiosperms are the group of plants producing ovules enclosed by ovary.

Question 37.
What is an open vascular bundle?
Answer:
A vascular bundle is open when it has Cambium.

Question 38.
What is a closed vascular bundle?
Answer:
A vascular bundle is closed when it does not have Cambium.

Question 39.
Mention any two morphological differences between Dicot & Monocot.
Answer:
Two morphological differences between Dicot:

Leaves show reticulate venation
Flowers are tetramerous or pentamerous

Two morphological differences between Monocot:

Leaves show parallel venation
Flowers are trimerous

Question 40.
Name the two divisions of spermatophyta?
Answer:
The two divisions are:

Gymnospermae and
Angiospermae

Question 41.
What are brood bodies?
Answer:
Brood bodies are the small detachable branches which help in vegetative propagation.
e.g., Bryopteris fruticulosa.

Question 42.
What are gemmae?
Answer:
Gemmae are small propagative structures which help in asexual reproduction.
e.g., Marchantia.

III. Short Answer Type Questions (3 Marks)

Question 1.
What are cryptogam? Mention its division.
Answer:
Cryptogams are non – flowering or non – seed producing plants. It has been divided into Algae, Bryophytes and Pteridophytes.

Question 2.
In which group of plants we can observe Haplodiplontic life cycle? Draw a diagram of Haplodiplontic life cycle.
Answer:
Bryophytes and Pteridophytes:

Question 3.
Name the 3 types of life cycles seen in plants?
Answer:
The 3 types of life cycles seen in plant:

Haplontic life cycle
Diplontic life cycle
Haplodiplontic life cycle

Question 4.
Where can we see cryophytic & halophytic algae? Give example.
Answer:
Cryophytic & Halophytic Algae:

Cryophytic algae grow on snow. e.g., Chlamydomonas nivalis.
Halophytic algae grow in salt pans, e.g., Dunaliella salina.

Question 5.
List out the various types of vegetative reproduction seen in algae.
Answer:
Fission, fragmentation, budding, bulbils, tubers.

Question 6.
List out the various asexual spores produced by algae.
Answer:
Zoospores, aplanospores, autospores, hypnospores, tetraspores and akinetes.

Question 7.
Write any three differences between chlorophyceae and phaeophyceae members?
Answer:

Question 8.
Define exoscopic embryogeny.
Answer:
In exoscopic embryogeny, the first division of the zygote is transverse & form inner and outer cell. The apex of the embryo develops from outer cell.

Question 9.
Name the three classes of Bryophytes, according to Proskauer.
Answer:
Three Classes of Bryophytes, According to Proskauer:

Hepaticopsida
Anthocerotopsida and
Bryopsida.

Question 10.
How peat is obtained? Write its economic value.
Answer:
A large amount of dead thallus of Sphagnum gets accumulated and compressed, hardened to form peat. It is used as fuel in commercial scale (Netherlands). Nitrates, brown dye and tanning materials are derived from peat. Sphagnum and peat are also used in horticulture as packing material because of their water holding capacity.

Question 11.
Mention any three Pteridophytes and their economic value.
Answer:
Economic Importance of Pteridophyte:
Pteridophyte:

Marsilea
Azolla
Pteris vittata

Uses:

Food
Biofertilizer
Removal of heavy metals from soils – Bioremediation

Question 12.
How the vascular plants dominate the Earth?
Answer:
The success and dominance of vascular plants is due to the development of,

Extensive root system.
Efficient conducting tissues.
Cuticle to prevent desiccation.
Stomata for effective gaseous exchange.

Question 13.
Name the three classes of gymnosperms.
Answer:
Three Classes of Gymnosperms:

Cycadospsida
Coniferopsida and
Gnetopsida.

Question 14.
Name any three economically important products & uses of the gymnosperm plants.
Answer:
Three economically important products & uses of the gymnosperm plants:

Question 15.
Compare the anatomical features between Dicots & Monocots.
Answer:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain in detail about the various life cycle patterns in plants.
Answer:
Life cycle patterns in plants: Alternation of Generation: Alternation of generation is common in all plants. Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation. Following type of life cycles are found in plants:

(a) Haplontic life cycle: Gametophytic phase is dominant. photosynthetic and independent, whereas sporophytic phase is represented by the zygote. Zygote undergoes meiosis to restore haploid condition. Example: Volvox and Spimgyra.

(b) Diplontic life cycle: Sporophytic phase (2n) is dominant, photosynthetic and independent. The gametophytic phase is represented by the single to few celled gametophyte. The gametes fuse to form zygote which develops into sporophyte. e.g., Fucus, gymnosperms and angiosperms.

(c) Haplodiplontic life cycle: This type of life cycle is found in Bryophytes and pteridophytes which is intermediate between haplontic and diplontic type. Both the phases are multicellular, but they differ in their dominant phase.

In Bryophytes dominant independent phase is gametophyte and it alternates with short – lived multicellular sporophyte totally or partially dependent on the gametophyte. In Pteridophytes sporophyte is the independent phase. It alternates with multicellular saprophytic or autotrophic, independent, short lived gametophyte (n).

Question 2.
Write a note on diversified thallus organisation seen in algae with examples.
Answer:
A wide range ofthallus organisation is found in algae. Unicellular motile (Chlamydomonas), unicellular non-motile (Chlorella), Colonial motile (Volvox), Colonial non-motile (Hydrodictyon), siphonous (Vaucheria), unbranched filamentous (Spirogyra), branched filamentous (Cladophora), discoid (Coleochaete) heterotrichous (Fritschiella), Foliaceous (Ulva) to Giant Kelps (Laminaria and Macrocystis).

Question 3.
Describe the various types of sexual reproduction observed in algae.
Answer:
Sexual reproduction in algae are of three type:

Isogamy: Fusion of morphologically and Physiologically similar gametes E.g. Ulothrix.
Anisogamy: Fusion of either morphologically or physiologically dissimilar gametes E.g. Pandorina
Oogamy: Fusion of both morphologically and physiologically dissimilar gametes. E.g. Sargassum.

The life cycle shows distinct alternation of generation.

Question 4.
Describe the salient features of Chlorophyceae members.
Answer:
The salient features of Chlorophyceae members:

Chlorophyceae commonly called as green algae.
Mostly aquatic (fresh water or marine), few terrestrial.
Shape of chloroplast differs. It may be cup shaped (Chlamydomonas) or girdle – shaped or reticulate, or stellate etc.
Chlorophyll ‘a’ and ‘b’ are photosynthetic pigments.
Pyrenoids store starch & also proteins.
Outer cell wall is made of pectin and inner is cellulose.
Vegetative reproduction is by fragmentation.
Asexual reproduction by zoospores, aplanospores and akinetes.
Sexual reproduction may be isogamous, anisogamous or oogamous. E.g. Chlamydomonas, Volvox and Spirogyra.

Question 5.
Describe the salient features of Phaeophyceae members.
Answer:
The salient features of Phaeophyceae members:

Phaeophyceae commonly called as Brown algae.
Majority are marine habitats. Pleurocladia is a fresh water form.
Thallus may be filamentous, frond – like or giant kelps.
Thallus is differentiated into photosynthetic part-frond, stalk – like structure – stipe and a holdfast for attachment.
Chlorophyll ‘a’ and ‘c’, carotenoids and Xanthophylls are photosynthetic pigments.
A golden brown fucoxanthin pigment gives olive green to brown colour.
Mannitol and Laminarin starch is the storage material.
Motile spores with unequal flagella (one whiplash and one tinsel) are present.
Oogamous is the major type of sexual reproduction. Isogamy is also seen.
Alternation of generation is seen. Example: Sargassum, Fucus, Laminaria and Dictyota.

Question 6.
Describe the salient features of Rhodophyceae.
Answer:
The salient features of Rhodophyceae:

Rhodophyceae commonly called as red algae.
Mostly marine habitats.
The thallus is multicellular, macroscopic, and may be filamentous, ribbon – like etc.
Chlorophyll ‘a’, r-phycoerythrin and r-phycocyanin are photosynthetic pigments.
Asexual reproduction is by means of monospores, neutral spores and tetraspores.
Floridean starch is the storage material
Sexual reproduction in oogamous.
Male sex organ is spermatangium producing spermatium.
Female sex organ is carpogonium.
Spermatium is carried by water and fuses with egg forming zygote.
Zygote undergoes meiosis forming carpospores.
Alternation of generation is seen. Example: Ceramium, Gelidium and Gigartina.

Question 7.
Tabulate the economic importance of algae.
Answer:
Economic importance of Algae:

Question 8.
Enumerate the general character of Bryophytes.
Answer:
The general character of Bryophytes:

Bryophytes are non – vascular cryptogams due to absence of xylem & phloem.
The plant body is a gametophyte and it is conspicuous, long – lived.
Plant body is undifferentiated into root, stem & leaves. Thalloid forms with rhizoids are seen in liverworts & hornworts. Leaf – like and stem – like structures are seen in mosses.
Vegetative reproduction is by adventitious buds, tubers, brood bodies or by gemmae.
Sexual reproduction is oogamous producing Antheridia & Archegonia in multicellular protective coverings.
Antheridia produces biflagellate antherozoids which swims in water & fuse with egg forming diploid zygote.
Water is essential for fertilization.
Zygote is the first cell of sporophyte. Zygote undergoes mitotics forming undifferentiated embryo, forming sporophyte. The embryogeny is exoscopic.
Sporophyte is dependent on gametophyte.
Sporophyte is differentiated into foot, seta & capsule.
Capsule of Sporophyte produces haploid spores by meiosis.
Bryophtyes are homosporous which are dispersed by elaters.
Spores germinate producing haploid gametophyte.
Heterologous alternation of generation.
Proskauer classified bryophytes into 3 classes, Hepaticopsida(Riccia),Anthocerotopsida (Anthoceros) and Bryopsida (Funaria).

Question 9.
List out the general characters of Pteridophytes.
Answer:
General characteristic features of Pteridophytes:

Plant body is sporophyte (2n) and it is the dominant phase. It is differentiated into root, stem and leaves.
Roots are adventitious.
Stem shows monopodial or dichotomous branching.
Leaves may be microphyllous or megaphyllous.
Stele is protostele but in some forms siphonostele is present (Marsilea)
Tracheids are the major water conducting elements but in Selaginella vessels are found.
Sporangia, spore bearing bag like structures are borne on special leaves called sporophyll. The sporophylls gets organized to form cone or strobilus. e.g., Selaginella and Equisetum.
They may be homosporous (produce one type of spores – Lycopodium) or Heterosporous (produce two types of shorts – Selaginella). Heterospory is the origin for seed habit.
Development of sporangia may be eusporangiate (development of sporangium from group of initials) or leptosporangiate (development of sporangium from single initial).
Spore mother cells undergo meiosis and produce spores (n).
Spore germinates to produce haploid, multicellular green, cordate shaped independent gametophytes called prothallus.
Fragmentation, resting buds, root tubers and adventitious buds help in vegetative reproduction.
Sexual reproduction is Oogamous. Sex organs, namely antheridium and archegonium are produced on the prothallus.
Antheridium produces spirally coiled and multiflagellate antherozoids.
Archegonium is flask shaped with broad venter and elongated narrow neck. The venter possesses egg or ovum and neck contain neck canal cells.
Water is essential for fertilization. After fertilization a diploid zygote is formed and undergoes mitotic division to form embryo.
Pteridophytes show apogamy and apospory.

Question 10.
Write a note on economic importance of Pteridophytes.
Answer:
A note on economic importance of Pteridophytes:
Pteridophyte:

Rumohra adiantiformis (leather leaf fem)
Marsilea
Azolla
Dryopteris filix – mas
Pteris vittata
Pteridium sp.
Equisetum sp.
Psilotum, Lycopodium, Selaginella, Angiopteris, Marattia.

Uses:

Cut flower arrangements
Food
Biofertilizer
Treatment for tapeworm
Removal of heavy metals from soils – Bioremediation
Leaves yield green dye
Stems for scouring
Ornamental plants

Question 11.
What is protostele? Explain its types.
Answer:
In protostele xylem surrounds phloem. The type includes Haplostele, Actinostele, Plectostele and mixed protostele.

Haplostele: Xylem surrounded by phloem is known as haplostele. E.g. Selaginella.
Actinostele: Star shaped xylem core is surrounded by phloem is known as actinostele. E.g. Lycopodium serratum.
Plectostele: Xylem plates alternates with phloem plates. E.g. Lycopodium clavatum.
Mixed prototostele: Xylem groups uniformly scattered in the phloem. E.g. Lycopodium cernuum.

Question 12.
Define Siphonostele. Explain its types.
Answer:
In siphonostele xylem is surrounded by phloem with pith at the centre. It includes Ectophloic siphonostele, Amphiphloic siphonostele, Solenostele, Eustele, Atactostele and Polycylic stele.

Ectophloic siphonostele: The phloem is restricted only on the external side of the xylem. Pith is in centre. E.g. Osmunda.
Amphiphloic siphonostele: The phloem is present on both the sides of xylem. The pith is in the centre. E.g. Marsilea.
Solenostele: The stele is perforated at a place or places corresponding the origin of the leaf trace.
- Ectophloic solenostele – Pith is in the centre and the xylem is surrounded by phloem. E.g. Osmunda.
- Amphiphloic solenostele – Pith is in the centre and the phloem is present on both sides of the xylem. E.g. Adiantum pedatum.
- Dictyostele – The stele is separated into several vascular strands and each one is called meristele. E.g. Adiantum capillus – veneris.
Eustele: The stele is split into distinct collateral vascular bundles around the pith. E.g. Dicot stem.
Atactostele: The stele is split into distinct collateral vascular bundles and are scattered in the ground tissue. E.g. Monocot stem.
Polycyclic stele: The vascular tissues are present in the form of two or more concentric cylinders. E.g. Pteridium.

Question 13.
Point out the general characters of Gymnosperms.
Answer:
General characteristic features:

Most of the gymnosperms are evergreen woody trees or shrubs. Some are lianas (Gnetum)
The plant body is sporophyte and is differentiated into root, stem and leaves.
A well developed Tap root system is present. Coralloid Roots of Cycas have symbiotic association with blue green algae. In Pinus the roots have mycorrhizae.
The stem is aerial, erect and branched or unbranched (Cycas) with leaf scars.
In conifers two types of branches namely branches of limited growth (Dwarf shoot) and Branches of unlimited growth (Long shoot) is present.
Leaves are dimorphic, foliage and scale leaves are present. Foliage leaves are green, photosynthetic and borne on branches of limited growth. They show xerophytic features.
The xylem consists of tracheids but in Gnetum and Ephedra vessels are present.
Secondary growth is present. The wood may be Manoxylic (Porous, soft, more parenchyma with wide medullary ray – Cycas) or Pycnoxylic (compact with narrow medullary ray – Pinus).
They are Heterosporous. The plant may be monoecious (Pinus) or dioecious (Cycas).
Microsporangia and Megasporangia are produced on Microsporophyll and Megasporophyll respectively.
Male and female cones are produced.
Anemophilous pollination is present.
Fertilization is siphonogamous and pollen tube helps in the transfer of male nuclei.
Sporne (1965) classified gymnosperms into 3 classes, 9 orders and 31 families. The classes include
- Cycadospsida
- Coniferopsida
- Gnetopsida.

Question 14.
List out the features common for both Gymnosperms & Angiosperms.
Answer:
Gymnosperms resemble with angiosperms in the following features:

Presence of well organised plant body which is differentiated into roots, stem and leaves
Polyembryony (presence of many embryo). The naked ovule develops into seed. The endosperm is haploid and develop before fertilization.
The life cycle shows alternation of generation. The sporophytic phase is dominant and gametophytic phase is highly reduced.
Presence of cambium in gymnosperms as in dicotyledons.
Flowers in Gnetum resemble to the angiosperm male flower. The Zygote represent the first cell of sporophyte.
Presence of integument around the ovule.
Both plant groups produce seeds
Pollen tube helps in the transfer of male nucleus in both.
Presence of Eustele.

Question 15.
Differentiate the characters of Gymnosperm & Angiosperm.
Answer:
Difference between Gymnosperms and Angiosperms:
Gymnosperms:

Vessels are absent [except Gnetales]
Phloem lacks companion cells
Ovules are naked
Wind pollination only
Double fertilization is absent
Endosperm is haploid
Fruit formation is absent
Flowers absent

Angiosperms:

Vessels are present
Companion cells are present
Ovules are enclosed within the ovary
Insects, wind, water, animals etc., act as pollinating agents
Double fertilization is present
Endosperm is triploid
Fruit formation is present
Flowers present

Question 16.
List out the economic importance of Gymnosperms.
Answer:
Economic importance of Gymnosperms:

Question 17.
List Out the salient’ features of Anglosperms.
Answer:
Salient features of Anglosperms:

Vascular tissue (Xylem and Phloem) is well developed.
Flowers are produced instead of cone.
The embryosac (Ovule) remains enclosed in the ovary.
Pollen tube helps in fertilization, so water is not essential for fertilization.
Double fertilization is present. The endosperm is triploid.
Angiosperms are broadly classified into two classes namely Dicotyledons and Monocotyledons.

Question 18.
Distinguish between Dicotyledons and Monocotyledons.
Answer:
Angiosperms:

V. Higher Order Thinking Skills (HOTs)

Question 1.
State which were the first true land plants? Mention two Characteristics features of these plants.
Answer:
Pteridophytes are the first true land plants. Pteridophytes are the first plants to acquire vascular tissue. Heterosporous condition was developed from Pteridophytes.

Question 2.
Give a comparative account of the following:
(a) Marchantia and Marsilea
(b) Cycas and rose
Answer:
(a) Marchantia is a Bryophyte whereas Marsilea is a Pteridophyte.
(b) Cycas is a gymnospermic plant and rose is a angiospermic plant.

Question 3.
Why are angiosperms so called? In which structures do the seeds develop?
Answer:
Angiosperms are so called because these plants have covered seeds. Seed of angiosperms develop within the ovary which later modify into fruit.

Question 4.
Name the gymnosperms that are exception with regard – to vascular tissue.
Answer:
All the gymnosperms possess tracheids as conducting tissues whereas gymnosperms like Gnetum & Ephedra possess vessels as their conducting tissues.

Question 5.
Both gymnosperms & angiosperms are seed bearers. Yet they are classified separately. Why?
Answer:
Gymnosperms and angiosperms are classified separately because the seeds of the angiosperms are enclosed by ovary (fruit wall) whereas the gymnospermic seeds are naked (not covered by ovary).

Question 6.
Bryophytes maintain soil texture – comment.
Answer:
Bryophytes play a major role in soil formation through succession and help in soil conservation.

Question 7.
Why heterosporous condition is advanced?
Answer:
Heterospory refers to the development of two different types of spores. Heterospory is the origin for seed habit.

Question 8.
Associate the following features with groups in which they first appeared.
(a) Vascular tissues
(b) Seeds inside fruits
(c) Heterospore production
Answer:
(a) Vascular tissues – Pteridophytes
(b) Seeds inside fruits – Angiosperms
(c) Heterospory – Pteridophytes.

Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Science Model Question Paper 4 English Medium Pdf, Samacheer Kalvi 10th Science Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamil Nadu Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

General Instructions:

The question paper comprises of four parts
You are to attempt all the questions 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 12 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.
Question numbers 13 to 22 in Part II are of two marks each. Any one question should be answered compulsorily.
Question numbers 23 to 32 in Part III are of four marks each. Any one question should be answered compulsorily.
Question numbers 33 to 35 in Part IV are of seven marks each. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 75

Part – I

(i) Answer all the questions. [12 × 1 = 12]
(ii) Choose the most suitable answer and write the code with the corresponding answer.

Question 1.
Impulse is equal to _________.
(a) rate of change of momentum
(b) rate of force and time
(c) change of momentum
(d) rate of change of mass
Answer:
(c) change of momentum

Question 2.
The value of universal gas constant ________.
(a) 3.81 mol^-1 KJ^-1
(b) 8.03 J mol^-1 K^-1
(c) 1.38 mol^-1 KJ^-1
(d) 8.31 J mol^-1 K^-1
Answer:
(d) 8.31 J mol^-1 K^-1

Question 3.
The safe limit of receiving the radiation is about _______.
(a) 1R
(b) 0.1 R
(c)100 R
(d) 10 R
Answer:
(b) 0.1 R

Question 4.
The gram molecular mass of oxygen molecule is ________.
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g
Answer:
(c) 32 g

Question 5.
The basis of modem periodic law is _________.
(a) atomic number
(b) atomic mass
(c) Isotopic mass
(d) number of neutrons
Answer:
(a) atomic number

Question 6.
A solution is a ______ mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non-homogeneous
Answer:
(a) homogeneous

Question 7.
Which is the sequence of correct blood flow _______.
(a) ventricle – atrium – vein – arteries
(b) atrium – ventricle – veins – arteries
(c) atrium – ventricle – arteries – vein
(d) ventricles – vein – atrium – arteries
Answer:
(c) atrium – ventricle – arteries – vein

Question 8.
Polyphagia is a condition seen in ______.
(a) Diabetes insipidus
(b) Diabetes mellitus
(c) Obesity
(d) AIDS
Answer:
(b) Diabetes mellitus

Question 9.
The xylem and phloem arranged side by side on same radius is called _______.
(a) radial
(b) amphivasal
(c) conjoint
(d) None of these
Answer:
(c) conjoint

Question 10.
Which is formed during anaerobic respiration?
(a) Carbohydrate
(b) Ethyl alcohol
(c) Acetyl CoA
(d) Pyruvate
Answer:
(b) Ethyl alcohol

Question 11.
Which is used to build scripts?
(a) Script area
(b) Block palette
(c) stage
(d) Sprite
Answer:
(a) Script area

Question 12.
Rice normally grows well in alluvial soil, but _______ is a rice variety produced by mutation breeding that grows well in saline soil.
(a) Atlas 66
(b) Triticale
(c) Protina
(d) Atomita-2-rice
Answer:
(d) Atomita-2-rice

Part – II

Answer any seven questions. (Q.No: 22 is compulsory) [7 × 2 = 14]

Question 13.
Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
Answer:
This is because turning effect to tighten the screws depends upon the perpendicular distance of the applied force from the axis of rotation in power arm. Larger the power arm less is the force required to turn the screws. So spanner is provided with a long handle.

Question 14.
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.
$\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}$

Question 15.
What is co-efficient of cubical expansion?
Answer:
The ratio of increase in volume of the body per degree rise in temperature to its unit volume is called as coefficient of cubical expansion.

Question 16.
Define Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to
l/12th part of the mass of a carbon-12 atom. It is denoted as Ar.

Question 17.
Differentiate aqueous and non-aqueous solution:
Answer:
Aqueous solution:

Water acts as a solvent
E.g. Common salt in water

Non -Aqueous solution:

Any liquid, other than water, acts as a solvent.
E.g. Sulphur dissolved in carbon disulphide

Question 18.
How is diastema formed in rabbit?
Answer:
Diastema is a gap between the incisors and premolar. It is formed due to the absence of canine. It helps in mastication and chewing of food in herbivorous animals.

Question 19.
What is the role of parathormone?
Answer:
The parathormone regulate calcium and phosphorus metabolism in the body. They act on bone, kidney and intestine to maintain blood calcium levels.

Question 20.
Identify the parts A, B, C and D?
Answer:

A – Head
B – Acrosome
C – Nucleus
D – Mitochondria

Question 21.
Write any two importance of Ethnobotany.
Answer:

It provides traditional uses of plant.
It gives information about certain unknown and known useful plants.

Question 22.
The focal length of the concave lens is 7 m. Calculate the power of lens.
Answer:
Given f = -7 m
Power of lens = $\frac{1}{f}=\frac{1}{-7}$ =
= -0.14 dioptre or -0.14 D

Part – III

Answer any seven questions (Q.No: 32 is compulsory) [7 × 4 = 28]

Question 23.
Write the applications of universal law of gravitation.
Answer:

Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, radius of the Earth, acceleration due to gravity, etc. can be calculated with a higher accuracy. .
Helps in discovering new stars and planets.
One of the irregularities in the motion of stars is called ‘Wobble’ that lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
Helps to explain germination of roots is due to the property of geotropism, which is the property of a root responding to the gravity.
Helps to predict the path of the astronomical bodies.

Question 24.
(i) Differentiate convex lens and concave lens.
Answer:

Convex lens	Concave lens
The lens which is thicker at the centre than at the edges..	The lens which is thinner at the centre than at the edges
A beam of light passing through it is converged to a point.	A beam of light passing through it, is diverged or spreadout.
It is called as converging lens.	It is called as diverging lens.

(ii) 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 25.
(i) Three resistors of resistance 5 ohm, 3 ohm and 2 ohm are connected in series with 10 V battery. Calculate their effective resistance and the current flowing through the circuit.
Answer:
R₁ = 5 Ω, R₂ = 3 Ω, R₃ = 2 Ω, V = 10 V
R_s = R₁ + R₂ + R₃
= 5 + 3 + 2= 10 Ω
Hence Rs = 10 Ω
The current, I = $\frac{V}{R_{s}}=\frac{10}{10}$ = IA

(ii) An alloy of nickel and chromium is used as the heating element. Why?
Answer:
An alloy of nickel and chromium is used as the heating element. Because:

It has high resistivity
It has a high melting point
It is not easily oxidized.

Question 26.
State and explain the applications of Avogadro’s law.
Answer:
Avogadro’s law : Equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Applications of Avogadro’s law:

It is used to determine the atomicity of gases.
It is helpful in determining the molecular formula of gaseous compounds.
It establishes the relationship between the vapour density and molecular mass of a gas.
It gives the value of molar volume of gases at STP.
Molar volume of a gas at STP = 22.4 litres.
It explains Gay lussac’s law effectively.

Question 27.
Explain the types of double displacement reactions with examples.
Answer:
There are two major classes of double displacement reactions. They are,
(i) Precipitation Reactions:
When aqueous solutions of two compounds are mixed, if they react to form an insoluble compound and a soluble compound, then it is called precipitation reaction.
pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s)↓+ 2KNO₃_(aq)

(ii) Neutralisation Reactions:
Another type of displacement reaction in which the acid reacts with the base to form a salt and water. It is called ‘neutralisation reaction’ as both acid and base neutralize each other.
NaOH_(aq) + HCl_1(aq) → NaCl_(aq) + H₂O_(l)

Question 28.
(i) Explain the structure of an atom bomb.
Answer:
Principle:

Uncontrolled chain reaction
Release large amount of energy in short time

Structure:

Fissile material whose mass is sub – critical.
Cylindrical fissile material is fit into void.
Cylinder is injected into the void, using conventional explosive
Two pieces together to form super – critical mass, leads to explosion.
Tremendous amount of energy is released in the form of heat, light and radiation.
γ radiation, affects the living creatures
This type of atom bombs were exploded at Hiroshima and Nagasaki in Japan during Second World War.

(ii) Calculate the pH of 1.0 × 10^-4 molar solution of HNO3.
Answer:
[H⁺] = 1.0 × 10^-4 m, pH = ?
pH = – log₁₀ [H⁺]
= – log₁₀ 1 × 10^-4
[M⁺] = – log₁₀ ¹– log₁₀ 10^-4
= 0 – (-4) log₁₀ ¹⁰ ∴ log₁₀ ¹⁰= 1
= 4 × 1
pH =4

Question 29.
With a neat labelled diagram explain the structure of a neuron.
Answer:

A neuron or nerve cell is the structural and functional unit of the nervous system. It consists of Cyton, Dendrites and Axon.

(a) Cyton: Cyton has a central nucleus with the abundant – cytoplasm called neuroplasm. The cytoplasm has large granular body called Nissel’s granules and the other cell organelles like mitochondria, ribosomes, lysosomes, and endoplasmic reticulum.

(b) Dendrites: These are the numerous branched cytoplasmic processes, that project from the surface of the cell body. They conduct nerve impulses, towards the cyton.

(c) Axon: The axon is a single, elongated, slender projection. The end of axon terminates as fine branches, which terminate into knob like swellings called synaptic knob. The plasma membrane of axon is called axolemma, while the cytoplasm is called axoplasm. It carries impulses away from the cyton.

The axon may be covered by a protein sheath called myelin sheath, which is further covered by a layer of Schwann cells called Neurilemma. Myelin sheath breaks at intervals, by depressions called Nodes of Ranvier. The region between the nodes is called as internode.

Question 30.
(i) What are the various routes by which transmission of human immuno deficiency virus takes place?
Answer:
AIDS is not transmitted by touch or any physical contact. HIV is generally transmitted by

Sexual contact with infected person.
Use of contaminated or infected blood or blood products, needles or syringes.
By the use of contaminated needles or syringes.
From infected mother to her child through placenta.

(ii) Name three improved characteristics of wheat that helped India to achieve high productivity.
Answer:

High yielding
Semi-dwarf
Fertilizer responsive

Question 31.
(i) Kavitha gave birth to a female baby. Her family members say that she can give birth to only female babies because of her family history. Is the statement given by her family members true. Justify your answer.
Answer:
No, the statement is not true. Sex determination is a chance of probability as to which category of sperm fuses with egg.
If the egg [X] is fused by the X-bearing sperm, an XX individual (female) is produced.

If the egg [X] is fused by the Y-bearing sperm an XY individual (male) is produced.

The sperm, produced by the father, determines the sex of the child. The mother is not responsible in determining the sex of the child.

(ii) Who discovered Rh factor? Why was it named so?
Answer:
The Rh factor was discovered by Landsteiner and Wiener in 1940 in Rhesus monkey. Hence the name Rh factor.

Question 32.
Describe and name three stages of cellular respiration that aerobic organisms use to obtain energy?
Answer:
Stages of cellular respiration
(i) Glycolysis: It takes place in cytoplasm of the cell. It is the break down of one molecule of Glucose (6 Carbon) into two molecules of Pyruvic acid (3 Carbon).

(ii) Kreb’s cycle: This occurs in mitochondria matrix. During this cycle, oxidation of pyruvic acid result into CO₂ and water.

(iii) Electron transport chain(ETC):
ETC is located on the inner membrane of the mitochondria.

(a) NADH₂ and FADH₂ molecules formed during Glycolysis and Kreb’s cycle are oxidised to NAD⁺ and FAD⁺ to release the energy via electrons.

(b) The electrons, as they move through the system, release energy which is trapped by ADP to synthesize ATP. This is called Oxidative Phosphorylation. During this process, oxygen which is the acceptor of electrons get reduced to water.

Part – IV

(1) Answer all the questions. [3 × 7 = 21]
(2) Each question carries seven marks.
(3) Draw diagram wherever necessary.

Question 33.
(a) (i) State Newton’s third law of motion.
Answer:
‘For every action, there is an equal and opposite reaction. They always act on two different bodies’.

(ii) Deduce the equation of a force using Newton’s second law of motion.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed ‘u’. After a time interval of ‘t’ the velocity of the body changes to ‘v’ due to the impact of an unbalanced external force F.

Initial momentum of the body P_i = mu
Final momentum of the body P_f = mv
Change in momentum Δp = P_f – P_i
= mv – mu

By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
F ∝ $\frac{m v-m u}{t}$
F = $\frac{k m(v-u)}{t}$

Here, k is the proportionality constant, k = 1 in all systems of units. Hence,
F = $\frac{m(v-u)}{t}$
Since, acceleration = change in velocity / time, a = (v – u)/t. Hence, we have
F = m × a
Force = mass × acceleration

No external force is required to maintain the motion of a body moving with uniform velocity.
When the net force acting on a body is not equal to zero, then definitely the velocity of the body will change.
Thus, change in momentum takes place in the direction of the force. The change may take place either in magnitude or in direction or in both.

[OR]

(b) A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
Answer:
Specific resistance (ρ) = $\frac{\mathrm{RA}}{l}$
R = $\frac{\rho l}{\mathrm{A}}$
When the length is increased by three and then the area of cross section is reduced by three
Resistance of wire R = 10 Ω
New resistance R’ = $\frac{\rho(3 l)}{(A / 3)}$
= ρ (3l) x $\frac{3}{\mathrm{A}}$
R’ = $\frac{9 \rho l}{\mathrm{A}}$
R’ = 9 x R = 9 x 10
R’ = 90 Ω

Question 34.
(a) (i) Calculate the mass of the 0.023 x 10²⁰ molecules of H₂O.
Answer:
Molecular mass of H₂O = H × 2 + 0 × 1
= 1 × 2 + 16 × 1
= 2 + 16 = 18

= 10 × 18
Mass of an atom = 0.018 g

(ii) Calculate the moles of 12 g of magnesium.
Answer:
Number of moles = $\frac{\text { Given mass }}{\text { molecular mass }}$
= $\frac{12}{24}=\frac{1}{2}$ = 0.5
∴ 12 g magnesium makes 0.5 moles.

(iii) Calculate the number of molecules present in 50 g of Fe.
Answer:

= 5.387 × 1023
∴ 50 g Fe contain 5.387 × 10²³ atoms/molecules

[OR]

(b) Explain smelting process..
Answer:
Smelting: The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.
There are three important regions in the furnace.

(a) The Lower Region (Combustion Zone): The temperature is at 1500°C. In this region, coke bums
with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

(b) The middle region (Fusion Zone): The temparature prevails at 1000° C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

(c) The Upper Region (Reduction Zone): The temperature prevails at 400°C . In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.

The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

Question 35.
(a) (i) Write any two physiological effects of ethylene.
Answer:
Physiological effects of ethylene:

It promotes the ripening of fruits. E.g., Tomato, Apple, Mango, Banana, etc.
It breaks the dormancy of buds,seeds and storage organs.

(ii) What is Autogamy?
Answer:
Autogamy or self pollination is the transfer of pollen grains from the anther to the stigma of same flower or another flower borne on the same plant. Eg. Hibiscus.

(iii) How will you prevent soil erosion?
Answer:
Prevention of soil erosion:

Retain vegetation cover, so that soil is not exposed.
Cattle grazing should be controlled.
Crop rotation and soil management improve soil organic matter.
Run off water should be stored in the catchment.
Reforestation, terracing and contour ploughing.
Wind speed can be controlled by planting trees in the form of a shelter belt.

[OR]

(b) (i) How do you differentiate Homologous organs from Analogous organs.
Answer:

Homologous organs	Analogous organs
1. Homologous organs are those inherited from common ancestors, with similar developmental pattern in embryos.	1. The analogous organs look similar and perform similar functions but they have different origin and development pattern.
2. A human hand, a front leg of cat, flipper of a whale and a bat’s, wing look dissimilar and adapted for different functions.	2. The function of the wings of a bat, the wings of a bird and wings of an insect are similar, but their basic structures are different.

(ii) Why are the rings of cartilage found in trachea of rabbit?
Answer:
The tracheal walls of rabbit are supported by rings of cartilage which help in the free passage of air.

Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium

Leave a Comment / Class 10 / By Prasanna

Students can Download Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium Pdf, Samacheer Kalvi 10th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium

Instructions

The question paper comprises of four parts.
You are to attempt all the parts. 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 14 in Part I are Multiple Choice Quèstions of one-mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and.writing the option code and the corresponding answer.
Question numbers 15 to 28 in Part II àre two-marks questions. These are to be answered in about one or two sentences.
Question numbers 29 to 42 in Part III are five-marks questions. These are to be answered in about three to five short sentences.
Question numbers 43 to 44 in Part IV are eight-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 100

PART – I

I. Choose the correct answer. Answer all the questions. [14 × 1 = 14]

Question 1.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, 5} then n[(A ∪ C) × B] is ……………. .
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12

Question 2.
If a, b, c are in G.P., then $\frac{a-b}{b-c}$ is equal to ……………. .
(1) $\frac{a}{b}$
(2) $\frac{b}{a}$
(3) $\frac{a}{c}$
(4) $\frac{c}{b}$
Answer:
(1) $\frac{a}{b}$

Question 3.
If k + 2, 4k – 6, 3k – 2 are the 3 consecutive terms of an A.P, then the value of k is ……………. .
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3

Question 4.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of k is ……………. .
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5

Question 5.
If A is a 2 × 3 matrix and B is a 3 × 4 matrix, how many columns does AB have ……………. .
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4

Question 6.
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is ……………. .
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Answer:
(2) 4 cm

Question 7.
(2, 1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 1 = 0
Answer:
(2) x + y = 3; 3x + y = 7

Question 8.
If the ratio of the height of a tower and the length of its shadow is √3 : 1 , then the angle of elevation of the sun has measure ……………. .
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°

Question 9.
A spherical ball of radius r₁ units is melted to make 8 new identical balls each of radius r₂ units. Then r₁ : r₂ is ……………. .
(1) 2:1
(2) 1:2
(3) 4:1
(4) 1:4
Answer:
(1) 2:1

Question 10.
The standard deviation of a data is 3. If each value is multiplied by 5 then the new variance is ……………. .
(1) 3
(2) 15
(3) 5
(4) 225
Answer:
(4) 225

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is ……………. .
(1) $\frac { 3 }{ 10 }$
(2) $\frac { 7 }{ 10 }$
(3) $\frac { 3 }{ 9 }$
(4) $\frac { 7 }{ 9 }$
Answer:
(2) $\frac { 7 }{ 10 }$

Question 12.
The range of the relation R = {(x, x³)/x} is a prime number less than 13} is ……………. .
(1) {2,3,5,7,11}
(2) {4,9,25,49, 121}
(3) {8, 27,125, 343,1331}
(4) {1, 8, 27, 125, 343, 1331}
Answer:
(3) {8, 27,125, 343,1331}

Question 13.
If 1 + 2 + 3 + …. + n = k then 1³ + 2³ + 3³ + ……… n³ is equal to ……………. .
(1) K²
(2) K³
(3) $\frac{k(k+1)}{2}$
(4) (k + 1)K³
Answer:
(1) K²

Question 14.
Two dice are thrown simultaneously. The probability of getting a doublet is ……………. .
(1) $\frac { 1 }{ 36 }$
(2) $\frac { 1 }{ 3 }$
(3) $\frac { 1 }{ 6 }$
(4) $\frac { 2 }{ 3 }$
Answer:
(3) $\frac { 1 }{ 6 }$

PART-II

II. Answer any ten questions. Question No. 28 is compulsory. [10 × 2 = 20]

Question 15.
ARelation R is given by the set {{x,y)/y = x + 3, x ∈ {0,1,2,3,4,5}}. Determine its domain and range.
Answer:
x = {0, 1, 2, 3, 4, 5} and y = x + 3
when x = 0 ⇒ y = 0 + 3 = 3
when x = 1 ⇒ y = 1 + 3 = 4 .
when x = 2 ⇒ y = 2 + 3 = 5
when x = 3 ⇒ y = 3 + 3 = 6
whenx = 4 ⇒ y = 4 + 3 = 7
when x = 5 ⇒ y = 5 + 3 = 8
R = {(0, 3) (1, 4) (2, 5) (3, 6) (4, 7) (5, 8)}
Domain = {0, 1, 2, 3, 4, 5}
Range = {3, 4, 5, 6, 7, 8}

Question 16.
Let f = {(-1,3), (0, -1), (2, – 9)} be a linear function from Z to Z. Find f(x).
Answer:
The linear equation is f(x) = ax + b
f(-1) = 3
a(-1) + b = 3
-a + b = 3 ….(1)
f(0) = -1
a(0) + b = -1
0 + b = -1
b = -1
Substitute the value of b = -1 in (1)
– a – 1 = 3
– a = 3 + 1 ⇒ – a = 4
a = -4
∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)

Question 17.
Solve 8x = 1 (mod 11)
Answer:
8x ≡ 1 (mod 11) can be written as 8x – 1 = 11k, for some integer k.
x = $\frac{11 k+1}{8}$
When we put k = 5, 13, 21, 29,… . then 11 k + 1 is divisible by 8.
x = $\frac{11 \times 5+1}{8}$ = 7
x = $\frac{11 \times 13+1}{8}$ = 18
Therefore, the solutions are 7,18,29,40, …….

Question 18.
How many terms of the series 1³ + 2³ + 3³ +……….. should be taken to get the sum 14400?
Answer:
1³ + 2³ + 3³ +. . . + n³ = 14400
$\left[\frac{n(n+1)}{2}\right]^{2}$ = 14400
$\frac{n(n+1)}{2}$ = √14400
$\frac{n(n+1)}{2}$ = 120 ⇒ n² + n = 240
n² + n – 240 =0

(n + 16) (n – 15) =0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

Question 19.
Given the L.C.M and G.C.D of the two polynomials are a³ – 10a² + 11a + 70 and a – 7, the polynomial p(x) is a² – 12a + 35 then find q(x).
Answer:

L.C.M. = a³ – 10a² + 11a + 70
= (a – 7) (a² – 3a – 10)

= (a – 7) (a – 5) (a + 2)
G.C.D. = (a – 7)

p(x) = a² – 12a + 35
= (a – 5) (a – 7)
q(x) = $\frac{\mathrm{LCM} \times \mathrm{GCD}}{p(x)}$
= $\frac{(a-7)(a-5)(a+2) \times(a-7)}{(a-5)(a-7)}$
q(x) = (a + 2) (a – 7)

Question 20.
Find the square root of 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:
$\sqrt{9 x^{2}-24 x y+30 x z-40 y z+25 z^{2}+16 y^{2}}$
= $\sqrt{(3 x)^{2}+(4 y)^{2}+(5 z)^{2}-2(3 x)(4 y)-2(4 y)(5 z)+2(3 x)(5 z)}$
= $\sqrt{(3 x-4 y+5 z)^{2}}$ [using (a – b + c)² = a² + b² + c² – 2ab – 2bc + 2ca]
= |3x – 4y + 5z|

Question 21.
Construct a 3 × 3 matrix whose elements are a_ij = i² j²
Answer:
The general 3 × 3 matrix is given by A = $\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$ a_ij = i² j²

a₁₁ = 1² × 1² = 1 × 1 = 1;
a₁₂ = 1² × 2² = 1 × 4 = 4;
a₁₃ = 1² × 3² = 1 × 9 = 9;

a₂₁ = 2² × 1² = 4 × 1 = 4;
a₂₂ = 2² × 2² = 4 × 4 = 16;
a₂₃ = 2² × 3² = 4 × 9 = 36;

a₃₁ = 3² × 1² = 9 × 1 = 1;
a₃₂ = 3² × 2² = 9 × 4 = 36;
a₃₃ = 3² × 3² = 9 × 9 = 81;
Hence the required matrix is A = $\left[ \begin{matrix} 1 & 4 & 9 \\ 4 & 16 & 36 \\ 9 & 36 & 81 \end{matrix} \right] $

Question 22.
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.
Answer:
Given that AE = 3 cm, EC = 4 cm, CD = 10 cm, DB = 3 cm, AF = 5 cm.
Let FB be x
Using Ceva’s theorem we have

$\frac{A E}{E C} \times \frac{C D}{D B} \times \frac{B F}{A F}=1$
$\frac{3}{4} \times \frac{10}{3} \times \frac{x}{5}=1$ $\Rightarrow \frac{2 x}{4}=1$
2x = 4 ⇒ x = $\frac{4}{2}$ = 2
The value of BF = 2

Question 23.
Find the area of the triangle whose vertices are (-3, 5), (5, 6) and (5, -2)
Answer:
Plot the points in a rough diagram and take them in counter-clockwise order. Let the vertices be A(-3,5), B(5,-2), C(5,6)
The area of ∆ABC is

= $\frac { 1 }{ 2 }$ {(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁ y₃)}
= $\frac { 1 }{ 2 }$ {(6 + 30 + 25) – (25 – 10 – 18)}
= $\frac { 1 }{ 2 }$ {(61 + 3)}
= $\frac { 1 }{ 2 }$ (64) = 32 sq. units

Question 24.
From the top of a rock 50 √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° = $\frac{A B}{B C}$
$\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}$
X = 50 √3 × √3 = 50 × 3
= 150 m
∴ Distance of the car from the rock =150 m

Question 25.
If n = 5, x̄ = 6, Σx² = 765, then calculate the coefficient of variation.
Answer:
Standard deviation (σ) = $\sqrt{\frac{\Sigma x^{2}}{n}-\left(\frac{\Sigma x}{n}\right)^{2}}$
$=\sqrt{\frac{765}{5}-(6)^{2}}=\sqrt{153-36}=\sqrt{117}$
σ = 10.816
Coefficient of variation = $\frac{\sigma}{\bar{x}} \times 100 \%$
= $\frac{10.816}{6} \times 100 \%$ = 180.266%
Coefficient of variation = 180. 27%

Question 26.
The roots of the equation x² + 6x – 4 = 0 are α, β. Find the quadratic equation whose roots are $\frac{2}{\alpha}$ and $\frac{2}{\beta}$.
Answer:
α and β are the roots of x² + 6x – 4 = 0
α + β = -6 ; αβ = -4
Sum of the roots = $\frac{2}{\alpha}+\frac{2}{\beta}$
$\frac{2 \beta+2 \alpha}{\alpha \beta}=\frac{2(\alpha+\beta)}{\alpha \beta}=\frac{2(-6)}{-4}$
= 3
Product of the roots = $\frac{2}{\alpha} \times \frac{2}{\beta}=\frac{4}{\alpha \beta}$
= $\frac{4}{-4}=-1$
The quadratic equation is x² – (sum of the roots) x + product of the roots = 0
∴ x² – 3x – 1 = 0 .

Question 27.
If the points A(2, 5), B(4, 6) and C(8, a) are collinear, find the value of “a” using slope concept.
Answer:
Since the three points are collinear
Slope of line = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Slope of AB = Slope of BC
$\frac{6-5}{4-2}=\frac{a-6}{8-4}$
$\frac{1}{2}=\frac{a-6}{4}$
2a – 12 = 4 ⇒ 2a = 16
a = $\frac{16}{2}$ = 8
The value of a = 8

Question 28.
The king, Queen and Jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards, find the probability of getting (i) a card of clubs (ii) a queen of diamond
Answer:
Sample space (S) = (52 – 3) = 49
n (S) = 49
(i) Let A be the event of getting a card of clubs.
n(A) = (13 – 3) = 10
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{49}$
(ii) Let B be the event of getting a queen of diamond
n (B) = 1
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{49}$

PART-III

III. Answer any ten questions. Question No. 42 is compulsory. [10 × 5 = 50]

Question 29.
Find x if gff (x) = fgg (x), given f(x) = 3x + 1 and g(x) = x + 3.
Answer:
gff(x) = g [f {f{x)}} (This means “g of f of f of x”)
= g[f(3x + 1)] = g[3(3x + 1) + 1] = g(9x + 4)
g (9x + 4) = [ (9x + 4) + 3] = 9x + 7
fgg (x) = f[g{g (x)}] (This means “f of g of g of x”)
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

Question 30.
Let A = {-l,l}and B = {0,2}. If the function f : A → B defined by f(x) = ax + b is an onto function? Find a and b.
Answer:
A = {-1, 1}; B = {0,2}
f(x) = ax + b
0 = -a + b
a – b = 0 …….. (1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ………. (2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
1 – b = 0 ⇒ -b = -1 ⇒ 6 = 1
The value of a = 1 and b = 1

Question 31.
Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.
Answer:

Monthly savings form an A.P.
2000, 2600, 3200 ….
a = 2000 ; d = 2600 – 2000 = 600
Given t_n = 20,000
t_n = a + (n – 1) d
20000 = 2000 + (n – 1) 600
20000 = 2000 + 600n – 600
= 1400 + 600n
20000 – 1400 = 600n
1860o = 600n
n = $\frac{18600}{600}$ = 31
He will take 31 years to save ₹20,000 per month

Question 32.
Find the sum to n terms of the series 0.4 + 0.44 + 0.444 + ……… to n terms
Answer:

Question 33.
Find the GCD of 6x³ – 30x² + 60x – 48 and 3x³ – 12x² + 21x – 18.
Answer:
Let, f(x) = 6x³ – 30x² + 60x – 48 = 6(x³ – 5x² + 10x – 8) and
g(x) = 3x³ – 12x² + 21x – 18 = 3(x³ – 4x² + 7x – 6)
Now, we shall find the GCD of x³ – 5x² + 10x – 8 and x³ – 4x² + 7x – 6

Here, we get zero as remainder.
GCD of leading coeffients 3 and 6 is 3
Thus GCD
[(6x³ – 30x² + 60x – 48, 3x³ – 12x² + 21x – 18)] = 3(x – 2)

Question 34.
If the roots of the equation (c² – ab)x² – 2(a² – bc)x + b² – ac = 0 are real and equal prove that either a = 0 (or) a³ + b³ + c³ = 3abc
Answer:
(c² – ab)x² – 2(a² – bc)x + b² – ac = 0
Here a = c² – ab ; b = -2 (a² – bc) ; c = b² – ac
Since the roots are real and equal
Δ = b² – 4ac
[-2 (a² – bc)]² – 4(c² – ab) (b² – ac) = 0
4(a² – bc)² – 4[c ²b² – ac³ – ab³+ a²bc] = 0
Divided by 4 we get
(a² – bc)² [c²b² – ac³ – ab² + a²bc] = 0
a⁴ + b²c² – 2a²bc – c²b² + ac³ + ab³ – a²bc = 0
a⁴ + ab³ + ac³ – 3a²bc = 0
a(a³ +b³ + c³) = 3a²bc
a³ + b³ + c³ = $\frac{3 a^{2} b c}{a}$
a³ + b³ + c³ = 3abc
Hence it is proved

Question 35.
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Answer:

ABCD is a quadrilateral. AB = AD.
AE and AF are the internal bisector of ∠BAC and ∠DAC.
To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By Angle bisector theorem, we have,
∴ $\frac{A B}{A C}=\frac{B E}{E C}$ …….. (1)
In ∆ ADC, AF is the internal bisector of ∠DAC
By Angle bisector theorem, we have,
$\frac{A D}{A C}=\frac{D F}{F C}$
∴ $\frac{A B}{A C}=\frac{D F}{F C}$ (AB = AD given) …….. (2)
From (1) and (2), we get
$\frac{B E}{E C}=\frac{D F}{F C}$
Hence In ∆ BCD,
BD || EF (By converse of BPT)

Question 36.
Find the equation of a straight line passing through (1,-4) and has intercepts which are in the ratio 2:5
Answer:
Let the x-intercept be 2a and the y intercept 5a .
The equation of a line is $\frac{x}{a}+\frac{y}{a}=1 \Rightarrow \frac{x}{2 a}+\frac{y}{5 a}=1$
The line passes through the point (1, -A)
$\frac{1}{2 a}+\frac{(-4)}{5 a}=1 \Rightarrow \frac{1}{2 a}-\frac{4}{5 a}=1$
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a .
a = $\frac{-3}{10}$
The equation of the line is $\frac{x}{2(-3 / 10)}+\frac{y}{5(-3 / 10)}=1$
$\frac{x}{-3 / 5}+\frac{y}{-3 / 2}=1 \Rightarrow \frac{5 x}{-3}+\frac{2 y}{-3}=1$
$\frac{-5 x}{3}-\frac{2 y}{3}=1$
Multiply by 3
-5x – 2y = 3 ⇒ -5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

Question 37.
If $\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a}$, then prove that $\frac{a^{2}-1}{a^{2}+1}$ = sin θ
Answer:

Question 38.
A toy is in the shape of a cylinder surmounted by a hemisphere. The height of the toy is 25 cm. Find the total surface area of the toy if its common diameter is 12 cm.
Answer:

Let r and h be the radius and height of the cylinder respectively.
Given that, diameter d = 12 cm, radius r = 6 cm
Total height of the toy is 25 cm
Therefore, height of the cylindrical portion = 25 – 6 = 19 cm
T.S.A. of the toy = C.S.A. of the cylinder + C.S.A. of the hemisphere + Base Area of the cylinder
= 2πrh + 2πr² + πr²
= πr(2h + 3r) sq.unis
= $\frac { 22 }{ 7 }$ × 6(38 + 18)
= $\frac { 22 }{ 7 }$ × 6 × 56 = 1056
Therefore, T.S.A. of the toy is 1056 cm²

Question 39.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n (S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
P(A) = $\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}$
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
P(B) = $\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{7}{8}$
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3
P(C) = $\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{3}{8}$
A ∩ B = { HHT, HTH, THH}
n(A ∩ B) = 3
p(A ∩ B) = $\frac{n(\mathrm{A} \cap \mathrm{B})}{n(\mathrm{S})}=\frac{3}{8}$
B ∩ C = {HHT, THH}
n(B ∩ C) = 2
p(B ∩ C) = $\frac{n(\mathrm{B} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
A ∩ C = {HHT, THH}
n(A ∩ C) = 2
P(A ∩ C) = $\frac{n(\mathrm{A} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
(A ∩ B ∩ C) = 2
P(A ∩ B ∩ C) = $\frac{n(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})}{n(\mathrm{S})}=\frac{2}{8}$
P(A∪B∪C) = P(A) + P(B) + P(C) -P(A ∩ B) – p(B ∩ C) – P(A ∩ C) + P(A∩B∩C)
$=\frac{3}{8}+\frac{7}{8}+\frac{3}{8}-\frac{3}{8}-\frac{2}{8}-\frac{2}{8}+\frac{2}{8}$
$=\frac{3}{8}+\frac{7}{8}-\frac{2}{8}$
$=\frac{10-2}{8}=\frac{8}{8}$ = 1
The probability is 1

Question 40.
The following table show the marks obtained by 48 students in a quiz competition in mathematics calculate the standard deviation.
Answer:

Ans.
Let us form the following table using given data

Question 41.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4m, and slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of ₹500 per m².
Answer:

For conical portion
Radius (r) = 2m
Slant height (l) = 2.8 m
For cylindrical portion
Radius (R) = 2m
Height (H) = 2.1 m
Area of canvas used for the tent = Curved surface area of a cone + curved area of the cylinder
= πrl + 2πRH
= π(2 × 2.8 + 2 × 2 × 2.1)
= $\frac { 22 }{ 7 }$ (5.6+ 8.4)
= $\frac { 22 }{ 7 }$ × 14 sq.m
= 44 sq. m
Cost of the canvas at the rate of ₹500 per m² = ₹ 500 × 44
= ₹ 22000

Question 42.
Seven years ago Ramkumar’s age was five time the square of Daniel’s age. Three years hence Daniel age will be two fifth of Ramkumar’s age. Find their present ages.
Answer:
Seven years ago, let Daniel’s age be x years
Seven years ago, Ramkumar age was 5x² years
Daniel’s present age = (x + 7) year
Ramkumar’s present age = 5x² + 7 years
Three years hence
Daniel’s age = (x + 7 + 3) year = (x + 10) years
Ramkumar’s age = (5x² + 7 + 3) year = (5x² + 10) years
By the given condition
x + 10 = $\frac { 2 }{ 5 }$ (5x² +10)
5x + 50 = 10x² + 20
10x² – 5x – 30 = 0
2x² – x – 6 = 0
2x² – 4x + 3x – 6 = 0
2x (x – 2) + 3(x – 2) = 0
(x – 2) (2x + 3) = 0
x – 2 = 0 or 2x + 3 = 0
x = 2 or x = $\frac { -3 }{ 2 }$
∴ x = 2 (age will not be negative)
Hence Daniel age = 9 years (2 + 7)
Ramkumar’s age = 27 years [5(2)² + 7]

PART – IV

IV. Answer all the questions. [2 × 8 = 16]

Question 43.
(a) Construct a ∆PQR in which PQ = 8 cm, ∠R = 60° and the median RG from R to PQ is 5.8 cm. Find the length of the altitude from R to PQ.
Answer:

Step 1 : Draw a line segment PQ = 8 cm.
Step 2 : At P, draw PE such that ∠QPE = 60°.
Step 3 : At P, draw PF such that ∠EPF = 90°.
Step 4 : Draw the perpendicular bisector to PQ, which intersects PF at O and PQ at G.
Step 5 : With O as centre and OP as radius draw a circle.
Step 6 : From G mark arcs of radius 5.8 cm on the circle. Mark them as R and S.
Step 7 : Join PR and RQ. Then ∆PQR is the required triangle.
Step 8 : From R draw a line RN perpendicular to LQ.
LQ meets RN at M
Step 9: The length of the altitude is RM = 3.5 cm.

[OR]

(b) Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 3cm, Distance = 5 cm.

Steps of construction:

With O as centre, draw a circle of radius 3 cm.
Draw a line OP = 5 cm.
Draw a perpendicular bisector of OP, which cuts OP at M.
With M as centre and MO as radius draw a circle which cuts previous circles at A and B.
Join AP and BP, AP and BP are the required tangents.
The length of the tangent PA = PB = 4 cm

Verification: In the right angle triangle OAP
PA² = OP² – OA²
= 5² – 3² .
= 25 – 9 = 16
PA = √l6 = 4 cm
Length of the tangent = 4 cm

Question 44.
(a) Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below.

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x² + 2x – 3

(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)

[OR]

(b) Draw the graph of y = x² +x and hence solve x² + 1=0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x This x² + 1 has no real roots.

Search Results for:

Tamil Nadu Samacheer Kalvi 10th English Model Question Paper 5

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 3 Introduction to Database Management System

Samacheer Kalvi 12th Computer Applications Introduction to Database Management System Text Book Back Questions and Answers

Samacheer Kalvi 12th Computer Applications Introduction to Database Management System Additional Question and Answer:wers

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 4 English Medium

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

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

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 5 PHP Function and Array

Samacheer Kalvi 12th Computer Applications PHP Function and Array Text Book Back Questions and Answers

Samacheer Kalvi 12th Computer Applications Solutions PHP Function and Array Additional Questions and Answers

Tamilnadu Samacheer Kalvi 12th Bio Zoology Solutions Chapter 4 Principles of Inheritance and Variation

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Text Book Back Questions and Answers

Samacheer Kalvi 12th Bio Zoology Principles of Inheritance and Variation Additional Questions and Answers

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Text Book Back Questions and Answers

Samacheer Kalvi 11th Bio Botany Vegetative Morphology Additional Questions and Answers

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

Samacheer Kalvi 11th Bio Botany Plant Kingdom Text Book Back Questions and Answers

Samacheer Kalvi 11th Bio Botany Plant Kingdom Additional Questions and Answers

Tamil Nadu Samacheer Kalvi 10th Science Model Question Paper 4 English Medium

Tamilnadu Samacheer Kalvi 10th Maths Model Question Paper 3 English Medium