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Odisha State Board CHSE Odisha Class 12 Approaches to English Book 1 Solutions Unit 4 Text A: The Year 2050-Reflections of a Futurist Textbook Activity Questions and Answers.

Class 12th Alternative English Chapter 10 The Year 2050-Reflections of a Futurist Question Answers CHSE Odisha

The Year 2050-Reflections of a Futurist Class 12 Questions and Answers

Activity-1

Vocabulary:
Choose the word from the passage, which more or less mean the following. The paragraph numbers have been given in brackets.
(i) One who studies changes in population in an area (5).
(ii) long existence (5)
(iii) to be flooded with something (6)
(iv) the things that develop from a particular thing (9)
(v) natural potency to behave in a particular way (17)
(vi) the act of controlling or influencing somebody or something by clever or unfair mean (20)

Answer:
(i) One who studies changes in population in an area – demographer.
(ii) long existence – longevity
(iii) to be flooded with something – inundating
(iv) the things that develop from a particular thing-evolve
(v) natural potency to behave in a particular way-genetics
(vi) the act of controlling or influencing somebody or something by clever or unfair mean – manipulation

Activity-2

Facts And Opinions:
Some facts as well as some opinions to the writer have been presented in the essay. Put the facts and opinions in different columns below:

Facts	Opinion
New 18% of Americans reach the age of 90.	More than 50% of people who were born in 1960 will be alive by 2050

Answer:

Facts	Opinion
New 18% of Americans reach the age of 90	(i) more than 50% of people who were born
Stroke deaths and rheumatic heart disease will reduce by 20% and 50%	(ii) Deaths from cardiovascular diseases, hypertension heart diseases will reduce by hypertension, and heart stroke will reduce considerably.

Activity – 3

Remedial Grammar:
Fill in each with the appropriate verb phrase from the following list.
must have          let might have been      wouldn’t be
won’t be             would happen               mustn’t have rung

(a) Sunita: Do you know a girl of Standard V was knocked down by a town bus in front of our school gate this afternoon?
Binita: Oh no! I always said this _________ sooner or later.
Sunita: She is badly injured but she _________ they say. But she _________ out of hospital of a few weeks.
(b) Gopi: There is a letter on the floor outside the door. The postman _________ it.
Moti: Well, he _________ it outside. Someone _________ it. Why didn’t he ring the bell?
Gopi: he always rings the bell. You _________ out when he came.
Moti: I haven’t been out. So he _________ the bell.

Answer:
(a) Sunita: Do you know a girl of Standard V was knocked down by a town bus in front of our school gate this afternoon?
Binita: Oh no! I always said this would happen sooner or later.
Sunita: She is badly injured but she will live they say. But she won’t be out of the hospital of a few weeks.
(b) Gopi: There is a letter on the floor outside the door. The postman must have left it.
Moti: Well, he shouldn’t have left it outside. Someone might have taken it. Why didn’t he ring the bell?
Gopi: he always rings the bell. You might have been out when he came.
Moti: I haven’t been out. So he mustn’t have run the bell.

Extra Activity – 3(A)
Make sentences with the following expression from Text A (in sentences of your own). Don’t copy out exact sentences from the text.

Remarkable	Inexhaustible
Survive	Unending
Spectacular	Solar
Availability	Progeny
Antibiotics	Argument
Span	Measure
Longevity	Orbit
Varieties	Contamination
Non-depletable	Cosmos, Spawn

Answer:
Remarkable — Invention of the computer is a remarkable achievement of modern science.
Survive — We cannot survive without oxygen.
Spectacular — Your performance is really spectacular.
Availability — You will be given arrears only in the availability of funds.
Antibiotics — Antibiotics are administered in the treatment of many kinds of diseases.
Span — He lived a long span of 120 years
Longevity — Man doesn’t live only by longevity of years.
Varieties — This dish is made from varieties of ingredients.
Non-depletable — The ozone layer is not non-depletable
Inexhaustible — He continues working still as if his energy were inexhaustible.
Unending — Money is not an unending flow.
Solar — Today many types of work are conducted by using solar power.
Progeny — On need not give birth to numerous progeny in the days of population explosion.
Argument — He argues his income by earning from myriads of sources.
Measure — Can you measure his temperature?
Orbit — Every planet has its own orbit.
Contamination — Contamination of water is a great offense.
Cosmos — One should keep the cosmos pure at any cost.
Spawn — Reptiles usually spawn eggs.

Extra Activity – 3(B)
1. (i) Derive Adjectives from the following nouns.

Bride	charity
electricity	episode
bureaucrat	friend
Minister	authenticity
inclusion	legend

Answer:

Nouns	Adjectives
Bride	brides
episode	episodic
bureaucrat	bureaucratic
inclusion	inclusive
authenticity	authentic
charity	chaste
electricity	electric
Minister	ministerial
friend	friendly
legend	legendary

(ii) Give antonyms of the following:

right	short
seriously	lull
auspicious	above
special	down
senior	strong
alive	transparent
nearness	resistible
include	logical
ascend	legal
persuade	personal

Answer:

Words	Antonyms
right	Wrong
seriously	lightly
auspicious	inauspicious
special	ordinary
senior	Junior
alive	dead
nearness	remoteness /distance
include	exclude
ascend	descend
persuade	dissuade
short	long
full	empty
above	below
down	up
strong	weak
transparent	opaque
resistible	irresistible
logical	illogical
legal	illegal
personal	impersonal

(iii) Substitute the following expressions with one word each:
(a) strong dislike
(b) sympathy for someone who has experienced great sorrow.
(c) to say that something is very bad.
(d) to show pity.
(e) likely to bring good luck.

Answer:
(a) strong dislike – disgust
(b) sympathy for someone who has experienced great sorrow – condolence
(c) to say that something is very bad – rubbish
(d) to show pity-relent
(e) likely to bring good luck – auspicious

(iv) The following words are wrongly spelled. Rewrite them correctly.

burocrat	protected
condolence	sholder
cooperative	dinastic
Goodby	vegetarian
gimiks	colloqual

Answer:

Word	Correct Form
burocrat condolence cooperative good by gimiks protected sholder dynastic vegetarian colloquial	Bureaucracts condolence co-operative good bye gimmicks protracted shoulders dynastic vegetarian cotloquial

Section-A

How will you look after 50 years?
What will be the major changes in the world by that time?
Think of the possible changes in the fields of agriculture medicine and transport. List out three of the possible changes.
(i) _____________________________
(ii) ____________________________
(iii) ___________________________

The Year 2050-Reflection Of A Futurist Summary in English

Summary:

A remarkable feature of 2050 will be that most of the 1960 babies will still be alive because of a biomedical revolution that is underway. Death rules of different life-killing diseases have dropped considerably. This has become possible due to the availability of antibiotics, better health care, attention to diet, jogging, and exercise the effects in the United States are clear-cut and lasting. But increased longevity and improve health are like to have several drawbacks:

• world population will be larger than it might have been
• low birth rate and increased longevity combine to raise the average age of the population
• A period of difficult social adjustment will be likely.

However, there is a really good chance that a huge increase in food production can come from such development as:

• new plant varieties, obtained through genetic engineering which are photosynthetically efficient use less water and tend to be self-fertilizing,
• improved uses of the ocean, including the domestication of seeing animals aquaculture and
• tropical agriculture which will open to the world many billions of acres of land currently unusable.

Before nondepletable alternatives and commercially developed, new synthetic fuel industries for the conversion of coal into gaseous and liquid fuels and the extraction of petroleum of liquids from oil shale are likely to arise. Solar, geothermal, wind power, and fusion are electric-producing items. By the year 2050, we should be well along toward utilizing two virtually inexhaustible energy resources; solar electric power and nuclear power resources between now and 2050. They are (i) electronics, genetics, and psychology.

By the early in century machines will be available that perform better than human beings. In fact, genetics is a science about to become a technology. This technology will lead to the ability to ‘design’ plants and animals to perform human functions. In agriculture, scientists will be able to produce plants that have improved photosynthetic efficiency, minimum water requirements, self-fertilizing characteristics, and a desired spectrum of nutrient qualities. In mining, organisms will metabolize desired ones tells and thus concertable them for later ‘harvesting’.

In the production of pharmaceuticals, microorganisms will be used as factory workers to produce chemicals normally found only in natural body and plant processes. Finally, in medicine, scientists will intervene in the process by which genetic diseases such as sickle cell anemia. Tay Sachs disease and mongolism are passed from parents to their progeny to cure these diseases before conception. They will also address other diseases such as cancer or heart disease and even aging itself.  Of course, Psychology by 2050 will be ready to take off. The ‘trigger’ discovery will help us know how memory is recorded and retrieved.

It is not clear till now whether memory is chemical, electrical, or physical knowledge of sharing and retrieving of memory will improve education, persuasion, rehabilitation, personality development, and knowledge itself and open the huge and exciting possibility of expanding mental capacity closer to the limits of human potential. Perhaps by 2050, observers in the orbital city cloud follow the world food supply and predict harvest size and crop diseases. Many things can be controlled from the orbit. The boom – babies will face significant challenges in the years ahead.

Analytical Outlines:

• Most of the 1960 babies will still be alive in 2050.
• 2050 will be remarkable for a biomedical revolution.
• Death rates of different life-killing diseases have dropped considerably.
• This has become possible due to the availability of antibiotics.
• It will be possible to better health care.
• It will be possible to taking attention to diet.
• It will be possible due to jogging.
• It will be possible due to exercise.
• It will entirely affect the United States.
• But increased longevity has several dements.
• Improved health has also some demerits.
• The world population will be much longer.

• Low – birth rate will raise the average age ofthe population.
• Increased longevity will also help to raise it.
• A period of difficult social adjustment will be likely.
• This increase in population will develop something.
• It will develop a huge increase in food production.
• It will increase new plant varieties.
• These varieties will be obtained through genetic engineering.
• They are photosynthetically efficient.
• They will use less water.
• They will tend to be self-fertilizing.
• Improved uses of the ocean will take place.
• It will include the domestication of seeing animals in aquaculture.
• It will be possible for tropical agriculture.
• It will utilize billions of acres of unused land in agriculture.
• It will commercially develop non-depletable alternatives.
• New synthetic fuel industries will be there.
• It will convert coal to gases.
• It will also convert coal to liquid fuel.
• The extraction of petroleum of liquids from oil shale is likely to rise.
• There will be various electric-producing items.
• The use of solar power will be there.
• Geothermal use will be there.
• The use of wind power will be there.
• The use of fusion will also be there.
• By 2050 are will be using two virtually exhaustible energy resources.
• One is solar electric power.
• Another is nuclear fusion.
• These are, actually, highly expensive.
• The author also predicts three more power resources.
• One is electronics.
• The other is genetics.
• The other one is psychology.
• There will be the use of machines.
• They will perform better than human beings.
• In feet, genetics is a science about technology.
• This technology will lead to the ability to ‘design’ plants.
• It will also design animals to perform human functions.
• In agriculture, scientists will be able to produce plants.
• It has improved photosynthetic efficiency.
• It has improved minimum water requirements.
• It has improved self-fertilizing characteristics.
• It has developed a desired spectrum of nutrient qualities.
• In mining, organisms will metabolize desired metals.
• It will convertible them for later ‘harvesting’

• We can find changes in the production of pharmaceuticals.
• Here, microorganisms will be used as natural bodies.
• It will be also used in plant processes.
• Finally, scientists will intervene in medicine.
• They will try to cure some genetic diseases.
• One such disease is sickle cell anemia.
• Another is Tay Sachs disease.
• Other such one is mongolism.
• These diseases are passed from parents to their progeny.
• They will try to cure these diseases before conception.
• They will also address other diseases.
• One such disease is cancer.
• Other one is heart disease.
• Of course, Psychology by 2050 will be ready to take off.
• We will have the‘trigger’discovery.
• It will help us to know how memory is recorded and retrieved.
• It is not clear still now whether memory is chemical.
• Or it is electrical.
• The physical knowledge of storing and retrieving of memory will improve education.
• It will improve persuasion.
• It will improve rehabilitation.
• It will improve personality development.
• It will improve knowledge itself

• It will open the huge exciting possibility of expanding mental capacity.
• It will be closer to the limits of human potential.
• Perhaps by 2050, observers in the orbital city cloud follow the world food supply.
• It will predict harvest size.
• It will predict crop disease.
• Many things can be controlled from the orbit.
• The boom-babies will free significant changes in the years ahead.

Meanings Of Difficult Words:

nascent – just beginning and expected to become stronger and bigger.
arable – land suitable for growing crops.
augment – to grow longer, to increase the value or effectiveness of something.
Luddites – those who are strongly opposed to using modem machines and methods.
spawned – laid eggs (fish, frog, salmon, etc).
spectacular – very important, showy, eye-catching.
cardiovascular diseases – diseases of the heart.
hypertensive rheumatic – disease relating to tension and blood pressure.
heart disease – heart disease giving too much pain.
antibodies – medicine administered against micro-bacteria and other living organisms causing disease in human bodies.
optimistic – a hopeful inclination.
drawbacks – demerits, weaknesses, faults, etc.
longevity – living a very long span of life.
computed – calculated, reckoned, estimated.

aquaculture – water culture, treatment of water.
non-depletabIe – that which cannot be depleted or exhausted
solar – of the sun, the power coming from the sun.
geothermal – ‘geo’ means earth and thermal means heat. Hence, the energy emanates from the heat emitted from the earth.
progeny – the successor of a kind of parentage.
human potential – energy of human beings.
contamination – defiling or pocketing something.
accomplish – to attain, to gain, to have
perspective – a bright and hopeful future
cosmos – universe
decade – a period of ten years
exploration – discovery, finding something from a search.
illustrate – to explain, exemplify
utility – vainness, something without results.
infuse – mix, bind, amalgamate
stagger – move unsteadily due to heavy load overhead.

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• Text A: The Year 2050-Reflections of a Futurist Question Answer
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Odisha State Board BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 9 Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c)

Question 1.
ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଠିକ୍ ଉତ୍ତରଟିକୁ ବାଛି ଲେଖ ।

(i) x² – 3x + 2 ର ଉତ୍ପାଦକ ଦ୍ଵୟ
(a) (x – 2) ଓ (x + 1)
(b) (x + 2) ଓ (x – 1)
(c) (x – 2) ଓ (x – 1)
(d) (x + 2) ଓ (x + 1)
ସମାଧାନ:
(x – 2) ଓ (x + 1)
x² – 3 + 2 = x² – (2 + 1) x + 2 . 1 = (x – 2) (x – 1)

(ii) ଏକ ଦ୍ୱିଘାତୀ ପଲିନୋମିଆଲ୍‌ର ଉତ୍ପାଦକ ଦ୍ବୟ (x – 1) ଓ (x – 3) ହେଲେ ପଲିନୋମିଆଲ୍‌ଟି
(a) x² – 4x – 3
(b) x² – 4x + 3
(c) x² + 4x – 3
(d) x² + 4x + 3
ସମାଧାନ:
x² – 4x + 3
(x – 1) (x – 3) = x² – (1 + 3) x + (1) (3) = x² – 4x + 3

(iii) x – y ର ଉତ୍ପାଦକ ମାନ
(a) (x² + y²) (x + y) (x – y)
(b) (x² – y²) (x – y) (x + y)
(c) (x² + y²) (x + y)²
(d) (x² + y²) (x – y)
ସମାଧାନ:
(x² + y²) (x + y) (x – y)
x⁴ – y⁴ = (x²)² – (y²)² = (x² + y²) (x² – y²) = (x² + y²) (x + y) (x – y)

(iv) 8a³ – b³ – 12a²b + 6ab² ର ଉତ୍ପାଦକଗୁଡ଼ିକ
(a) (2a – b), (2a + b), (2a + b)
(b) (2a + b) (2a + b) (2a + b)
(c) (2a – b), (2a – b), (2a + b)
(d) (2a – b), (2a – b), (2a – b)
ସମାଧାନ:
(2a – b) (2a – b) (2a – b)
8a³ – b³ – 12a²b + 6ab² = (2a)³ – b³ – 3.2a.b(2a – b)
= (2a – b)³ = (2a – b) (2a – b) (2a – b)

(v) 625 + 25x⁴ + x⁸ ର ଉତ୍ପାଦକଗୁଡ଼ିକ
(a) (25 + 5x² + x⁴) (25 – 5x² + x⁴)
(b) (25 + 5x² + x⁴) (25 + 5x² – x⁴)
(c) (25 + 5x⁴ + x⁴) (25 – 5x⁴ + x⁴)
(d) (25 – 5x⁴ + x⁴)(25 + 5x⁴ – x⁴)
ସମାଧାନ:
(25 + 5x² + x⁴) (25 – 5x² + x⁴)
625 + 25x⁴ + x⁸ = 5⁴ + 5²(x²)² + (x²)⁴
= {5² + 5x² + (x²)²} {5² – 5x² + (x²)²} = (25 + 5x² + x⁴) (25 – 5x² + x⁴)

(vi) 1 – a³ + b³ + 3ab ର ଗୋଟିଏ ଉତ୍ପାଦକ
(a) (1 – a + b)
(b) (1 – a – b)
(c) (1 + a + b)
(d) (1 + a – b)
ସମାଧାନ:
(1 – a + b)
1 – a³ + b³ + 3ab
= (1 – a + b) {1² + a² + b² – (1) (-a) – (-a) (b) – (b) (1)}
= (1 – a + b) (1 + a² + b² + a + ab – b)

(vii) (2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³ ର ଉତ୍ପାଦକଗୁଡ଼ିକ ହେଲେ
(a) 6(2x – 3y)(3y – 4z) (2z – x)
(b) 3(2x – 3y) (3y – 4z) (2z – x)
(c) 60xyz
(d) ଏଥୁମଧ୍ୟରୁ କୌଣସିଟି ନୁହେଁ
ସମାଧାନ:
6 (2x – 3y) (3y – 4z) (2z – x)
2x – 3y + 3y – 4z + 4z – 2x = 0
(2x – 3y)³ + (3y – 4z)³ + (4z – 2x)³ = 3 (2x – 3y) (3y – 4z) (4z – 2x)
= 6 (2x – 3y) (3y – 4z) (2z – x)(∵ a + b + c = 0 ହେଲେ a³ + b³ + c³ = 3abc)

(viii) (28)³ + (-15)³ + (-13)³ ର ସରଳୀକୃତ ମାନ
(a) 8190
(b) 16380
(c) 24570
(d) 4095
ସମାଧାନ:
16380
28 – 15 -13 = 0
∴ (28)³ + (-15)³ + (-13)³ = 3 (28) (-15) (-13) = 16380
(∵ a + b + c = 0 ହେଲେ a³ + b³ + c³ = 3abc)

(ix) (a – b)³ + (b – c)³ + (c – a)³ ର ମାନ
(a) 3abc
(b) 3a³b³c³
(c) 3(a – b) (b – c)(c – a)
(d) {a – (b + c)}³
ସମାଧାନ:
3(a – b) (b – c) (c – a)
(a – b)³ + (b – c)³ + (c – a)³ = 3(a – b) (b – c) (c – a)
(∵ (a – b) + (b – c) + (c – a) = 0)

(x) 2x² – x – 1 ର ଗୋଟିଏ ଉତ୍ପାଦକ
(a) 2x – 1
(b) x + 1
(c) x – 1
(d) x + 2
ସମାଧାନ:
(x – 1)
2x² – x – 1 = 2x² – 2x + x – 1
= 2x (x – 1) + 1 (x – 1) = (x – 1) (2x + 1)
∴ 2x² – x – 1 ର ଏକ ଉତ୍ପାଦକ (x – 1)

Question 2.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।

(i) 2x² – x – 1
ସମାଧାନ:
2x² – x – 1 = 2x² – 2x + x – 1
= 2x(x – 1) + 1(x – 1) = (x – 1) (2x + 1)

(ii) 2x² – 3x + 1
ସମାଧାନ:
2x² – 3x + 1 = 2x² – 2x – x + 1
= 2x (x – 1) – 1 (x – 1) = (x – 1) (2x – 1)

(iii) 5x² – x – 4
ସମାଧାନ:
5x² – x – 4 = 5x² – 5x + 4x – 4
= 5x (x – 1) + 4 (x – 1) = (x – 1) (5x + 4)

(iv) 4x² – 5x – 6
ସମାଧାନ:
4x² – 5x – 6 = 4x² – 8x + 3x – 6
= 4x(x – 2) + 3 (x – 2) = (4x + 3) (x – 2)

(v) 3x² + 11x + 6
ସମାଧାନ:
3x² + 11x + 6 = 3x² + 9x + 2x + 6
= 3x (x + 3) + 2 (x + 3) = (3x + 2) (x + 3)

(vi) 7x² + x – 6
ସମାଧାନ:
7x² + x – 6 = 7x² + 7x – 6x – 6
= 7x (x + 1) – 6 (x + 1) = (7x – 6) (x + 1)

(vii) 2x² + 5x – 7
ସମାଧାନ:
2x² + 5x – 7 = 2x² + 7x – 2x – 7
= x (2x + 7) – 1 (2x + 7) = (2x + 7) (x – 1)

(viii) 4x² – 5x + 1
ସମାଧାନ:
4x² – 5x + 1 = 4x² – 4x – x + 1
= 4x (x – 1) – 1 (x – 1) = (4x – 1) (x – 1)

(ix) 4x² – 3x – 7
ସମାଧାନ:
4x² – 3x – 7 = 4x² – 7x + 4x – 7
= x (4x – 7) + 1 (4x – 7) = (4x – 7) (x + 1)

Question 3.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
[a³ + b³ = (a + b) (a² – ab + b²), a³ – b³ = (a – b) (a² + ab + b²)]

(i) 25a⁴ – 16b²
ସମାଧାନ:
25a⁴ – 16b² = (5a²)² – (4b)² = (5a² + 4b)(5a² – 4b)

(ii) 9 – 64p²q²
ସମାଧାନ:
9 – 64p²q² = (3)² – (8pq)² = (3 + 8pq)(3 – 8pq)

(iii) 8x³ + 27y³
ସମାଧାନ:
8x³ + 27y³ = (2x)³ + (3y)³ = (2x + 3y){(2x)² – 2x.3y + (3y)²}
= (2x + 3y)(4x² – 6xy + 9y²)

(iv) 8x³ – 27y³
ସମାଧାନ:
8x³ – 27y³ = (2x)³ – (3y)³ = (2x – 3y){(2x)² + 2x.3y + (3y)²}
= (2x – 3y)(4x² + 6xy + 9y²)

(v) (a + b)² – 9
ସମାଧାନ:
(a + b)² – 9 = (a + b)² – 32 = (a + b + 3) (a + b – 3)

(vi) (2a + 5)² – 16
ସମାଧାନ:
(2a + 5)² – 16 = (2a + 5)² – 42 = (2a + 5 + 4) (2a + 5 – 4) = (2a + 9) (2a + 1)

(vii) (x + 2y)² – (x – y)²
ସମାଧାନ:
(x + 2y)² – (x – y)² = {(x + 2y) + (x – y)} {x + 2y) – (x – y)}
= (x + 2y + x – y) (x + 2y – x + y) = (2x + y) (3y) = 3y (2x + y)

(viii) 4(a + 2p)² – 9 (2a – p)²
ସମାଧାନ:
4 (a + 2p)² – 9 (2a – p)² = {2 (a + 2p)}² – {3 (2a – p)}²
= (2a + 4p)² – (6a – 3p)² = (2a + 4p + 6a- 3p) (2a + 4p – 6a + 3p) = (8a + p) (7p – 4a)

(ix) 75 (2a – b + 1)² – 12 (a + b)²
ସମାଧାନ:
75 (2a – b + 1)² – 12 (a + b)² = 3 {25 (2a – b + 1)² – 4 (a + b)²}
= 3 [{5 (2a – b + 1)}² – {2 (a + b)²}] = 3 {(10a – 5b + 5)² – (2a + 2b)²]
= 3 (10a – 5b + 5 + 2a + 2b) (10a – 5b + 5 – 2a – 2b) = 3 (12a – 3b + 5) (8a – 7b + 5)

(x) (a + b)³ – 8c³
ସମାଧାନ:
(a + b)³ – 8c³ = (a + b)³ – (2c)³
= (a + b – 2c) {(a + b)² + (a + b)2c + (2c)²}
= (a + b – 2c) (a² + 2ab + b² + 2ca + 2bc + 4c²)
= (a + b – 2c) (a² + b² + 4c² + 2ab + 2bc + 2ca)

(xi) p⁴ – 27pq⁶
ସମାଧାନ:
p⁴ – 27pq⁶ = p(p³ – 27q⁶)
= p{p³ – (3q²)³} = p(p – 3q²) {p² + 3pq² + (3q²)²} = p(p – 3q²) (p² + 3pq² + 9q⁴)

(xii) 1 – (a + 2)³
ସମାଧାନ:
1 – (a + 2)³ = 1³ – (a + 2)³ = (1 – a – 2) {1² + a + 2 + (a + 2)²}
= (- a – 1) (1 + a + 2 + a² + 4a + 4) = -(a + 1) (a² + 5a + 7)

(xiii) 8 – (2x – 3)³
ସମାଧାନ:
8 – (2x – 3)³ = 2³ – (2x – 3)³ = (2 – 2x + 3) {2² + 2 (2x – 3) + (2x – 3)²}
= (-2x + 5) (4 + 4x – 6 + 4x² – 12x + 9) = (5 – 2x) (4x² – 8x + 7)

(xiv) 320p⁶q – 5p²q⁷
ସମାଧାନ:
320 p⁶q – 5p²q⁷ = 5p²q (64p⁴ – q⁶)
= 5p²q {(8p²)² – (q³)²} = 5p²q (8p² + q³) (8p² – q³)

(xv) 1 + (a + 2)³
ସମାଧାନ:
1 + (a + 2)³ = 1³ + (a + 2)³ = (1 + a + 2) {1² – 1(a + 2) + (a + 2)²}
= (a + 3) (1 – a – 2 + a² + 4a + 4) = (a + 3) (a² + 3a + 3)

(xvi) 8 + (2x – 3)³
ସମାଧାନ:
8 + (2x – 3)³ = 2³ + (2x – 3)³ = (2 + 2x – 3) {2² – 2 (2x – 3) + (2x – 3)²}
= (2x – 1) (4 – 4x + 6 + 4x² – 12x + 9) = (2x – 1) (4x² – 16x + 19)

(xvii) a³ + 6a²b + 12ab² + 8b³
ସମାଧାନ:
a³ + 6a²b + 12ab² + 8b³ = a³ + 3a² (2b) + 3 . a . (2b)² + (2b)³
= (a + 2b)³ = (a + 2b) (a + 2b) (a + 2b)

(xviii) a³ + 9a² + 27a + 27
ସମାଧାନ:
a³ + 9a² + 27a + 27 = a³ + 3a² . 3 + 3 . a . 3² + 3³ = (a + 3)³
= (a + 3) (a + 3) (a + 3)

(xix) 8 – 36p + 54p² – 27p³
ସମାଧାନ:
8 – 36p + 54p² – 27p³ = 2³ – 3. 2² 3p + 3 . 2 (3p)² – (3p)³
= (2 – 3p)³ = (2 – 3p) (2 – 3p) (2 – 3p)

(xx) (b – q)³ – (c – q)³ – 3 (b – c) (b – q) (c – q)
ସମାଧାନ:
(b – q)³ – (c – q)³ – 3 (b – c) (b – q) (c – q)
ମନେକର b – q = x ଓ c – q = y
∴ x – y = (b – q) – (c – q) = b – q – c + q = b – c
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍‌ଟି x³ – y³ – 3 (x – y) (xy) = x³ – y³ – 3xy (x – y) = (x – y)³
= {(b – q) – (c – q)}³ (x ଓ yର ମାନ ବସାକଳେ)
= (b – c)³ = (b – c) (b – c) (b – c)

Question 4.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² – ab + b²)

(i) a⁴ + a² + 1
ସମାଧାନ:
a⁴ + a² + 1
= a⁴ + 1 + a² = (a²)² + (1)² + a² = (a² + 1)² – 2 . a². 1 + a²
= (a² + 1)² – 2a² + a² = (a² + 1)² – a² = (a² + 1 + a) (a² + 1 – a)
= (a² + a + 1)(a² – a + 1)

(ii) a⁴b⁴ + a²b² + 1
ସମାଧାନ:
a⁴b⁴ + a²b² + 1 = (ab)4 + (ab)² 1² + 1⁴
= (a²b² + ab + 1) (a²b² – ab + 1)

(iii) 16a⁴ + 36a²b² + 81b⁴
ସମାଧାନ:
16a⁴ + 36a²b² + 81b⁴ = (4a²)² + 36a²b² + (9b²)²
= (4a²)² + (9b²)² + 72a²b² – 72a²b² + 36a²b² [72a²b² ଯେ।ଗ ଓ ବିୟେ।ଗ କରି]
= (4a² + 9b²)² – 36a²b² = (4a² + 9b²)² – (6ab)²
= (4a² + 9b² + 6ab) (4a² + 9b² – 6ab) = (4a² + 6ab + 9b²) (4a² – 6ab + 9b²)
ବିକଳ୍ପ ସମାଧାନ –
16a⁴ + 36a²b² + 81b⁴ = (4a²)² + (9b²)² + 36a²b²
= (4a² + 9b²)² – 2 . 4a² . 9b² + 36a²b² = (4a² + 9b²)² – 36a²b²
= (4a2 + 9b²)² – (6ab)² = (4a² + 9b² + 6ab) (4a² + 9b² – 6ab)
= (4a² + 6ab + 9b²) (4a² – 6ab + 9b²)

(iv) a⁸ + a⁴ + 1
ସମାଧାନ:
a⁸ + a⁴ + 1 = (a²)⁴ + (a²)² . 1² + 1⁴
= {(a²)² + a² . 1 + 1²} {(a²)² – a² . 1 + 1²} = (a⁴ + a² + 1) (a⁴ – a² + 1)
= (a⁴ + a² . 1² + 1⁴) (a⁴ – a² + 1) = (a² + a + 1) (a² – a + 1) (a⁴ – a² + 1)

(v) x⁴ + 4
ସମାଧାନ:
x⁴ + 4 = (x²⁾² + (2)² = (x²)² + (2)² + 4x² – 4x²
= (x² + 2)² – (2x)² = (x² + 2 + 2x) (x² + 2 – 2x) = (x² + 2x + 2) (x² – 2x + 2)

(vi) 2a⁴ + 8b⁴
ସମାଧାନ:
2a⁴ + 8b⁴ = 2(a⁴ + 4b⁴) = 2{(a²)² + (2b²)² + 4a²b² – 4a²b²}
= 2{(a² + 2b²)² – (2ab)²} = 2 {a² + 2b² + 2ab) (a² + 2b² – 2ab)}
= 2(a² + 2ab + 2b²) (a² – 2ab + 2b²)

(vii) 36a⁴ + 9b⁴
ସମାଧାନ:
36a⁴ + 9b⁴ = 9 (4a⁴ + b⁴) = 9 {(2a²)² + (b²)²}
= 9 {(2a² + b²)² – 2 (2a²) b²} = 9 {(2a² + b²)² – 4a²b²}
= 9 {(2a² + b²)² – (2ab)²} = 9(2a² + b² + 2ab) (2a² + b² – 2ab)
= 9 (2a² + 2ab + b²) (2a² – 2ab + b²)

(viii) 4a⁴ + 7a² + 16
ସମାଧାନ:
4a⁴ + 7a² + 16 = (2a²)² + (4)² + 16a² – 16a² + 7a²
= (2a² + 4)² – 9a² = (2a² + 4)² – (3a)² = (2a² + 4 + 3a) (2a² + 4 – 3a)
= (2a² + 3a + 4) (2a² – 3a + 4)

(ix) a⁴ + 2a²b² + 9b⁴
ସମାଧାନ:
a⁴ + 2a²b² + 9b⁴ = a⁴ + 9b⁴ + 2a²b²
= (a²)² + (3b²)² + 6a²b² – 6a²b² + 2a²b² =  (a² + 3b²)² – 4a²b²
= (a² + 3b²)² – (2ab)² = (a² + 3b² + 2ab) (a² + 3b² – 2ab)
= (a² + 2ab + 3b²) (a² – 2ab + 3b²)

(x) a⁴ – 3a² + 1
ସମାଧାନ:
a⁴ – 3a² + 1 = (a²)² + (1)² – 3a²
= (a²)² + (1)² – 2a² + 2a² – 3a² (2a² ଯେ।ଗ ଓ ବିୟେ।ଗ କରି)
= (a² – 1)² – (a)² = (a² – 1 + a) (a² – 1 – a) = (a² + a – 1) (a² – a – 1)

(xi) 25a⁴ – 19a²b² + 9b²
ସମାଧାନ:
25a⁴ – 19a²b² + 9b⁴ = (5a²)² + (3b²)² – 19a²b²
= (5a² + 3b²)² + 2 . 5 a² . 3b² – 19 a²b²
= (5a² + 3b²)² – 49 a²b² = (5a² + 3b²) – (7ab)²
= (5a² + 3b² + 7ab) (5a² + 3b² – 7ab) = (5a² + 7ab + 3b²) (5a² – 7ab + 3b²)

(xii) 9x² + y² + 6xy – 4z²
ସମାଧାନ:
9x² + y² + 6xy – 4z² = (3x)² + y² + 2 . 3x . y – (2z)²
= (3x + y)² – (2z)² = (3x + y + 2z) (3x + y – 2z)

(xiii) 16 – x² – 24y + 9y²
ସମାଧାନ:
16 – x² – 24y + 9y² = 16 – 24y + 9y² – x2
= 4² – 2 . 4 . 3y + (3y)² – x² = (4 – 3y)² – x² = (4 – 3y + x) (4 – 3y – x)

(xiv) (a² – b²) (x² – y²) – 4abxy
ସମାଧାନ:
(a² – b²) (x² – y²) – 4abxy = a²x² – a²y² – b²x² + b²y² – 4abxy
= a²x² + b²y² – 2abxy – a²y² – b²x² – 2abxy
= (a²x² + b²y² – 2abxy) – (a²y² + b²x² + 2abxy)
= {(ax)² + (by)² – 2(ax) (by)} – {(ay)² + (bx)² + 2(ay) (bx)}
= (ax – by)² – (ay + bx)² = (ax – by + ay + bx) (ax – by – ay – bx)

(xv) (a² + b²)(x² – y²) – 2ab (x² + y²)
ସମାଧାନ:
(a² + b²)(x² – y²) – 2ab (x² + y²)
= a²x² – ay² + b²x² – b²y² – 2abx² – 2aby²
= a²x² + b²x² – 2abx² – a²y² – b²y² – 2aby²
= (a²x² + b²x² – 2abx²) – (a²y² + b²y² + 2aby²)
= {(ax)² + (bx)² – 2ax . bx} – {(ay)² + (by)² + 2ay . by}
= (ax – bx)² – (ay + by)² = (ax – bx + ay + by) (ax – bx – ay – by)
= {x (a – b) + y (a + b)} {x (a – b) – y(a + b)}

Question 5.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
(a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

(i) a³ + b³ + x³ – 3abx
ସମାଧାନ:
a³ + b³ + x³ – 3abx
= (a + b + x) (a² + b² + x² – ab – bx – ax)

(ii) 8a³ + b³ + c³ – 6abc
ସମାଧାନ:
8a³ + b³ + c³ – 6abc = (2a)³ + b³ + c³ – 3(2a) (b) (c)
= (2a + b + c) { (2a)² + b² + c² – 2a . b. bc – c . 2a}
= (2a + b + c) (4a² + b² + c² – 2ab – bc – 2ca)

(iii) a³ + b³ – 8 + 6ab
ସମାଧାନ:
a³ + b³ – 8 + 6ab = a³ + b³ + (-2)³ – 3a . b(-2)
= (a + b – 2) { a² + b² + (-2)² – ab – b (-2) – (-2) a}
= (a + b – 2) (a² + b² + 4 – ab + 2b + 2a)

(iv) l – 27m³ – n³ – 9 lmn
ସମାଧାନ:
l³ – 27m³ – n³ – 9lmn = l³ + (- 3m)³ + (- n)³ – 3l(-3m) (-n)
= {l + (-3m) + (-n)} {l² + (-3m)² + (- n)² – l(-3m) – (-3m) (-n) – (-n) l}
= (l – 3m – n) (l² + 9m² + n² + 3lm – 3mn + nl)

(v) (a – b)³ + (c – b)³ + (a – c)³ – 3 (a – b) (b – c) (c – a)
ସମାଧାନ:
(a – b)³ + (c – b)³ + (a- c)³ – 3 (a – b) (b – c) (c – a)
ମନେକର a – b = x, b – c = y, c – a = z ⇒ c – b = -y ⇒ a – c = -z
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍‌ଟି x³ + (-y)³ + (-z)³ – 3(x) (-y) (-z)
= {x + (-y) + (-z)} {x² + (-y)² + (-z)² – x (-y) – (-y) (-z) – (-z) x}
= (x – y – z) (x² + y² + z² + xy – yz + zx)
= {(a – b) – (b – c) – (c – a)} {(a – b)² + (b – c)² + (c – a)² +(a – b) (b – c) – (b – c) (c – a) + (c – a) (a – b) x, y ଓ zର ମାନ ବସାକଳେ,
= (a – b – b + c – c + a) (a² – 2ab + b² + b² – 2bc + c² + c² – 2ca + a² + ab – ca – b² + bc – bc + ab + c² – ca + ca – bc – a² + ab)
= (2a – 2b) (a² + b² + c² + ab – 3bc – 3ca)
= 2 (a – b) (a² + b² + c² + ab – 3bc – 3ca)

(vi) a⁶ + 4a³ – 1
ସମାଧାନ:
a⁶ + 4a³ – 1 = a⁶ + a³ – 1 + 3a³ = (a2)³ + a³ + (-1)³ – 3. a² .a(-1)
= (a² + a – 1) {(a²)² + (a)² + (-1)² – a².a – a(-1) – (-1)a²}
= (a² + a – 1) (a⁴ + a² + 1 – a³ + a + a²) = (a² + a – 1)(a⁴ – a³ + 2a² + a + 1)

(vii) x³ + 72 – 24x
ସମାଧାନ:
x³ + 72 – 24x = x³ + 64 + 8 – 24x
= x³ + 4³ + 2³ – 3. x. 4. 2 = (x + 4 + 2)(x² +  4² + 2² – x. 4 – 4. 2 – 2. X)
= (x + 4 + 2)(x² + 16 + 4 – 4x – 8 – 2x)
= (x + 6)(x² + 16 + 4 – 8 – 6x)
= (x + 6)(x² – 16 + 12)

(viii) m⁶ + 7m³ – 8
ସମାଧାନ:
m⁶ + 7m³ – 8 = m⁶ + m³ – 8 + 6m³
= (m²)³ + m³ + (-2)³ – 3m² .m(-2)
= (m² + m – 2) {(m²)² + m² + (2)² – m² . m – m (-2) – (-2) m²}
= (m² + m – 2)(m⁴ + m² + 4 – m³ + 2m + 2m²)
= (m² + 2m – m – 2)(m⁴ – m³ + 3m² + 2m + 4)
= {m(m + 2) – 1(m + 2)}(m⁴ – m³ +3m² + 2m + 4)
= (m + 2)(m – 1)(m⁴ – m³ + 3m² + 2m + 4)

(ix) a⁶ + \(\frac{1}{a^6}\) + 2 (a ≠ 0)
ସମାଧାନ:

(x)  r⁶ + 45r³ – 8
ସମାଧାନ:
r⁶ + 45r³ – 8 = r⁶  + 27r³ – 8 + 18r³
= (r²)³ + (3r)³ + (-2) – 3 . r² . 3r (-2)
= (r² + 3r – 2) {(r²)² + (3r)² + (2)² – r² 3r – 3r (-2) – (-2) r²)
= (r² + 3r – 2)(r⁴ + 9r² + 4 – 3r³ + 6r + 2r²)
= (r² + 3r – 2)(r⁴ – 3r³ – t – 11r² + 6r + 4)

(xi) 16x³ – 54y⁶ – 2z³ – 36xy²z
ସମାଧାନ:
16x³ – 54y⁶ – 2z³ – 36xy²z = 2 { 8x³ – 27y⁶ –  z³ – 18xy²z)}
= 2 {(2x)³ + (-3y²)³ + (-z)³ – 3 (2x) (-3y²) (-z))
= 2{2x + (-3y²) + (-z)) {(2x)² + (-3y²)² + (-z)² – (2x) (-3y²) – (-3y²) (-z) – (-z) (2x)}
= 2(2x – 3y² – z) (4x² + 9y⁴ + z² + 6xy² – 3y²z + 2zx)

(xii) a³ + b³ – \(\frac{1}{27}\) c³ + abc
ସମାଧାନ:

(xiii) 27a³ – 8b⁶ + 125 c³ + 90ab²c
ସମାଧାନ:
27a³ – 8b⁶ + 125 c³ + 90 ab²c = (3a)³ + (- 2b²)³ + (5c)³ – 3(3a) (- 2b²) (5c)
= {3a + (-2b²) + 5c} {(3a)² + (-2b²)² + (5c)² – (3a) (-2b²) – (-2b²) (5c) – (5c) (3a)}
= (3a – 2b² + 5c) (9a² + 4b⁴ + 25c² + 6ab² + 10b²c – 15 ca)

(xiv) (2x + 3)³ + (3x – 2)³ – (5x + 1)³
ସମାଧାନ:
(2x + 3)³ + (3x – 2)³ – (5x + 1)³
ମନେକର 2x + 3 = a, 3x- 2 = b, 5x + 1 = c
∴ a + b – c = 2x + 3 + 3x – 2 – 5x – 1=0
a + b – c = 0 ହେଲେ a³ + b³ – c³ = -3abc
∴(2x + 3)³ + (3x – 2)³ – (5x + 1)³ = -3 (2x + 3) (3x – 2) (5x + 1)
(a, b ଓ cର ମାନ ବସାକଳେ)

Question 6.
a + b + c = 0 ହେଲେ ଦର୍ଶାଅ ଯେ, a³ + b³ + c³ = 3 abc
ସମାଧାନ:
L.H.S. = a³ + b³ + c³ = a³ + b³ + c³ – 3abc + 3abc
= (a + b + c) (a² + b² + c² –  ab – bc – ca) + 3abc
= (0) (a² + b² + c² – ab – bc – ca) + 3abc
= 0 + 3abc = 3abc = R.H.S. (ପ୍ରମାଣିତ)

Question 7.
(x – y)³ + (y – z)³ + (z – x)³ ର ଉତ୍ପାଦକଗୁଡ଼ିକୁ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ:
ମନେକର x – y = a, y – z = b, z – x = c
∴ a + b + c = x – y + y – z + z – x ⇒ a + b + c = 0
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍‌ଟି = a³ + b³+ c³ = 3abc (∵ a + b + c = 0)
∴ (x – y)³ + (y – z)³ + (z – x)³ = 3(x – y) (y – z) (z – x)

Question 8.
ଦର୍ଶାଅ ଯେ, x³ + y³ + z³ – 3xyz = \(\frac{1}{2}\) {(x – y)² + (y – z)² + (z – x)²}
ସମାଧାନ:
x³ + y³ + z³ – 3xyz
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= \(\frac{1}{2}\) (x + y + z) (2x² + 2y² + 2z² – 2xy – 2yz – 2zx)
= \(\frac{1}{2}\) (x + y + z) (x² – 2xy + y² + y² – 2yz + z² + z² – 2zx + x²}
= \(\frac{1}{2}\) (x + y + z) {(x – y)² + (y – z)² + (z – x)²}
∴ x³ + y³ + z³ = \(\frac{1}{2}\) {(x – y)² + (y – z)² + (z – x)²}

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 5 ପରିମିତି Ex 5(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 5 ପରିମିତି Ex 5(a)

Question 1.
(a) ହଉଇ ବ୍ୟାସାଦ
(i) 10 ସେ.ମି.
(ii) 2.8 ସେ.ମି.
(iii) 14 ସେ.ମି.
(iv) 4.2 ସେ.ମି. ହେଲେ ପରିଧ୍ କେତେ ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ଏକକ ହେଲେ, ବୃତ୍ତର ପରିଧ୍ = 2πr ଏକକ

(i) ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 10 ସେ.ମି.
∴ ବୃତ୍ତର ପରିଧୂ = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 10 = \(\frac { 440 }{ 7 }\) ସେ.ମି. = 62 \(\frac { 6 }{ 7 }\) ସେ.ମି |

(ii) ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 2.8 ସେ.ମି.
∴ ବୃତ୍ତର ପରିଧୂ = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 2.8 = 17.6 ସେ.ମି

(iii) ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 14 ସେ.ମି.
∴ ବୃତ୍ତର ପରିଧୂ = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 14 = 88 ସେ.ମି

(iv) ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 4.2 ସେ.ମି.
∴ ବୃତ୍ତର ପରିଧୂ = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 4.2 = 26.4 ସେ.ମି

(b) ବୃତ୍ତର ପରିଧ୍ (i) 34.9 ସେ.ମି
(ii) 1047 ସେ.ମି
(iii) 25.128 ସେ.ମି
(iv) 15.705 ସେ.ମି ହେଲେ ପରିଧ୍ କେତେ ? (π ≃ 3.141)
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = ପରିଧ୍ / 2π

(i) ମନେକର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି., ତେବେ ପରିଧୂ = 2πr ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 2πr = 34.9 ⇒ 2 × 3.141 × r = 34.9
⇒ r = \(\frac{34.9}{2 \times 3.141}\) = \(\frac { 349 }{ 10 }\) × \(\frac { 500 }{ 3141 }\) = \(\frac { 17450 }{ 3141 }\) = \(\frac { 50 }{ 9 }\) ବା 5\(\frac { 5 }{ 9 }\) ସେ.ମି
∴ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 166\(\frac { 2 }{ 3 }\) ସେ.ମି |

(ii) ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି. ∴ ହେଲେ ପରିଧ୍ = 2πr ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 2πr = 1047 ⇒ 2 × 3.141 × r = 1047 ⇒ r = \(\frac{1047}{2 \times 3.141}\)
⇒ r = \(\frac{1047 \times 500}{3141}\) = \(\frac { 500 }{ 3 }\) = 166\(\frac { 2 }{ 3 }\) ସେ.ମି
∴ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 166\(\frac { 2 }{ 3 }\) ସେ.ମି |

(iii) ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି. .. ବୃତ୍ତର ପରିସ୍ = 2πr ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 2лr = 25.128 ⇒ 2 × 3.141 × r = 25.128 ⇒ r = \(\frac{25.128}{2 \times 3.141}\) = 4 ସେ.ମି.
∴ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 4 ସେ.ମି. ।

(iv) ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 2лr = 15.705
∴ ହେଲେ ପରିଧ୍ = 2πr ସେ.ମି.
⇒ 2 × 3.141 × r = 15.705 ⇒ r = \(\frac{15.705}{2 \times 3.141}\) = 2.5 ସେ.ମି.

Question 2.
ଏକ ବୃତ୍ତର ଚାପର ଦୈର୍ଘ୍ୟ L, ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r, ଚାପର ଡିଗ୍ରୀ ପରିମାପ θ ଦ୍ଵାରା ସୂଚିତ ହେଲେ ନିମ୍ନଲିଖ ପ୍ରଶ୍ନମାନଙ୍କର ସମାଧାନ କର । (л ≃ \(\frac { 22 }{ 7 }\))
(a) r = 56 ସେ.ମି. , θ = 45° ହେଲେ L କେତେ ?
(b) L = 110 ମି. θ = 75° ହେଲେ r କେତେ ?
(c) 2r=9 ସେ.ମି. L = 22 ସେ.ମି. ହେଲେ θ କେତେ ?
Solution:
(a) r=56 ସେ.ମି. θ = 45°
ଚାପର ଦୈର୍ଘ୍ୟ (L) = \(\frac { θ }{ 180 }\) × лr = \(\frac { 45 }{ 180 }\) × \(\frac { 22 }{ 7 }\) × 56 = 44 ସେ.ମି.

(b) L = 110, θ = 75°
ଆମେ ଜାଣିଛେ, L = \(\frac { θ }{ 180 }\) × лr
⇒ 110 = \(\frac { 75 }{ 180 }\) × \(\frac { 22 }{ 7 }\) × r ⇒ r = \(\frac{110 \times 180 \times 7}{75 \times 22}\) ⇒ r = 84 ମି.

(c) 2r = 9 ସେ.ମି. L = 22 ସେ.ମି. ⇒ r = \(\frac { 9 }{ 2 }\) ଡେ.ମି.
ଆମେ ଜାଣିଛେ, L = \(\frac { θ }{ 180 }\) × лr ⇒ 22 = \(\frac { θ }{ 180 }\) × \(\frac { 27 }{ 7 }\) × \(\frac { 9 }{ 2 }\)
⇒ θ = \(\frac{22 \times 180 \times 7 \times 2}{22 \times 9}\) = 280°

Question 3.
ନିମ୍ନଲିଖତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ । (л ≃ \(\frac { 22 }{ 7 }\))
(a) କୌଣସି ଏକ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 10.5 ସେ.ମି. ହେଲେ ସେହି ବୃତ୍ତର 11 ସେ.ମି. ପରିମିତ ଚାପର ଡିଗ୍ରୀ
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 10.5 ସେ.ମି.
ଚାପର ଦୈର୍ଘ୍ୟ (L) = 11 ସେ.ମି.
ମନେକର ଚାପର ଡିଗ୍ରୀ ପରିମାପ = θ°
ଆମେ ଜାଣିଛେ, L = \(\frac { θ }{ 180° }\) × лr
⇒ 11 = \(\frac { θ }{ 180 }\) × \(\frac { 22 }{ 7 }\) × 10.5 ⇒ θ = \(\frac{11 \times 180 \times 7}{22 \times 10.5}\) = 60°

(b) 21 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 72° ହେଲେ ଚାପଟିର ଦୈର୍ଘ୍ୟ କେତେ ହେବ ?
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଷ (r) = 21 ସେ.ମି., ଚାପର ଡିଗ୍ରୀ ପରିମାପ (θ) = 72°
∴ ଚାପର ଦୈର୍ଘ୍ୟ (L) = \(\frac { θ }{ 180 }\) × лr = \(\frac { 72 }{ 180 }\) × \(\frac { 22 }{ 7 }\) × 21 = 26.4 ସେ.ମି.

(C) ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ କେତେ ହେଲେ ସେହି ବୃତ୍ତର 11 ସେ.ମି. ପରିମିତ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 10° ହେବ ?
Solution:
ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି., ବୃତ୍ତର ଚାପର ଡିଗ୍ରୀ ପରିମାପ = (θ) = 10°
ଚାପର ଦୈର୍ଘ୍ୟ (L) = 11 ସେ.ମି.
ଆମେ ଜାଣିଛେ, L = \(\frac { θ }{ 180° }\) × лr
⇒ 11 = \(\frac { 10 }{ 180 }\) × \(\frac { 22 }{ 7 }\) × r ⇒ r = \(\) = 63 ସେ.ମି.
∴ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ 63 ସେ.ମି. ।

(d) ଏକ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ x ଏକକ, ଚାପର ଦୈର୍ଘ୍ୟ y ଏକକ, ଚାପର ଡିଗ୍ରୀ ପରିମାପ Z ଡିଗ୍ରୀ ହେଲେ ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ л ମାଧ୍ୟମରେ ନିର୍ଣ୍ଣୟ କର ।
Solution:
L = \(\frac { θ }{ 180° }\) × лr
(ଏଠାରେ L = y ଏକକ, θ = z°, r = x ଏକକ)
⇒ y = \(\frac { z }{ 180 }\) × лr ⇒ x = \(\frac { 180y }{ лz }\)
∴ ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ \(\frac { 180y }{ лz }\) |

(e) r ଏକକ ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତରେ a ଏକକ ଦୀର୍ଘ ବାହୁ ବିଶିଷ୍ଟ ଏକ ବର୍ଗଚିତ୍ର ଅନ୍ତର୍ଲିଖ୍ ହେଲେ a ଏବଂ r ମଧ୍ଯରେ ସମ୍ପର୍କ ନିଶ୍ଚୟ କର ।
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ଏକକ ଓ ବର୍ଗଚିତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ = a ଏକକ ।
ଏକ ବୃତ୍ତ ମଧ୍ଯରେ ଗୋଟିଏ ବର୍ଗାକାର କ୍ଷେତ୍ର ABCD ଅନ୍ତର୍ଲିଖ୍ ହେଲେ, ବର୍ଗକ୍ଷେତ୍ରର କର୍ପୂର ଦୈର୍ଘ୍ୟ AC = ବୃତ୍ତର ବ୍ୟାସ ହେବ ।

ବୃତ୍ତର ବ୍ୟାସ = 2r ଏକକ
ବର୍ଗକ୍ଷେତ୍ରର କଣ୍ଠ = √2 × ବାହୁର ଦୈର୍ଘ୍ୟ = √2a ଏକକ
∴ √2a = 2r ⇒ a = √2.r

Question 4.
ବିଷୁବରେଖାଠାରେ ପୃଥ‌ିବୀର ବ୍ୟାସ 12530 କି.ମି. ହେଲେ ବିଷୁବ ବୃତ୍ତର ପରିସ୍ କେତେ ? (л ≃ \(\frac { 22 }{ 7 }\))
Solution:
ପୃଥ‌ିବୀର ବ୍ୟାସ (d) = 12530 ସେ.ମି.
∴ ବିଷୁବ ବୃତ୍ତର ପରିଧ୍ = x × d = \(\frac { 22 }{ 7 }\) × 12530 = 22 × 1790 = 39380 ସେ.ମି.
∴ ବିଷୁବ ବୃତ୍ତର ପରିଧ୍ 39,380 ସେ.ମି. |

Question 5.
44 ମି. ଦୀର୍ଘ ତାରରୁ 5 ସେ.ମି. ବ୍ୟାସାର୍କ ବିଶିଷ୍ଟ କେତୋଟି ବୃତ୍ତ ତିଆରି କରାଯାଇ ପାରିବ ?(л ≃ \(\frac { 22 }{ 7 }\))
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 5 ସେ.ମି., ତାରର ଦୈର୍ଘ୍ୟ = 44 ମି. = 4400 ସେ.ମି.
ଦ୍ରଭର ପରିଧ୍ = 2лr = 2 × \(\frac { 22 }{ 7 }\) × 5 = \(\frac { 220 }{ 7 }\) ସେ.ମି.
∴ ଦୃଭ ସଖ୍ୟା = ପାଇର ଦୈଶ୍ୟ / ଦରକ ପରିସ୍ = \(\frac{\frac{4400}{220}}{7}\) = 4400 × \(\frac {7 }{ 220 }\) = 140ଟି

Question 6.
ଗୋଟିଏ ବୃତ୍ତାକାର ରାସ୍ତାର ବାହାର ଓ ଭିତର ପରିଧୂ ଯଥାକ୍ରମେ 396 ଓ 352 ମିଟର ହେଲେ ରାସ୍ତାର ପ୍ରସ୍ଥ ନିର୍ଣ୍ଣୟ କର । (л ≃ \(\frac { 22 }{ 7 }\))
Solution:
ମନେକର ବହିଃବୃତ୍ତ ଏବଂ ଅନ୍ତଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଯଥାକ୍ରମେ R ମି. ଏବଂ r ମି. ।

Question 7.
ଦୁଇଟି ବୃତ୍ତର ପରିତ୍ରର ଅନ୍ତର 44 ମିଟର ଏବଂ ସେମାନଙ୍କର ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ଵୟର ସମଷ୍ଟି 77 ମିଟର ହେଲେ ପରିଧ୍ 22 ଦ୍ଵୟ ନିର୍ଣ୍ଣୟ କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ମନେକର ବହିଃବୃତ୍ତ ଏବଂ ଅନ୍ତଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଯଥାକ୍ରମେ R ମି. ଓ r ମି. ।
ବହିଃବୃତ୍ତର ପରିସ୍ = 2πR ମି. ଓ ଅନ୍ତଃବୃତ୍ତର ପରିଧ୍ = 2πr ମି.
ପ୍ରଣାଳୁମାରେ, 2лR – 2лг = 44 ମି.
⇒ 2 × \(\frac { 22 }{ 7 }\) (R – r) = 44 ⇒ R – r = 44 × \(\frac { 7 }{ 44 }\) = 7 ମି.
⇒ R – r = 1 ….(i)
ପୁନଶ୍ଚ R + r = 77 ….(2)
ସମୀକରଣ (1) ଓ (2) କୁ ଯୋଗକଲେ , 2R = 84 ⇒ R = 42 ମି.
ସମୀକରଣ (2) R = 42 ମି. ବସାଇଲେ , r = 77 – 42 = 35 ମି.
∴ ବହିଃବୃତ୍ତର ପରିସ୍ = 2πR = 2 × \(\frac { 22 }{ 7 }\) × 42 = 264 ମି.
ଓ ଅନ୍ତଃବୃତ୍ତର ପରିସ୍ = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 35 = 220 ମି.

Question 8.
ଦୁଇଟି ଏକକେନ୍ଦ୍ରିକ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ବୟର ଅନୁପାତ 3 : 4 । ସେମାନଙ୍କର ପରିଧ୍ ଦ୍ଵୟର ସମଷ୍ଟି 308 ସେ.ମି. ହେଲେ ବଳୟର ପ୍ରସ୍ଥ କେତେ ହେବ ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ମନେକର ଅନ୍ତଃବୃତ୍ତ ଏବଂ ବହିଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଯଥାକ୍ରମେ 3x ସେ.ମି. ଓ 4x ସେ.ମି. ।
ଅନ୍ତଃବୃତ୍ତର ପରିଧୂ = 2π × 3x = 6πx ସେ.ମି. ଓ ବହିଃବୃତ୍ତର ପରିଧୂ = 2π × 4x = 8πx ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 6πx + 8πx = 308
⇒ 14лx = 308 ⇒ 14 × \(\frac { 22 }{ 7 }\) × x = 308 ⇒ x = \(\frac { 308 }{ 44 }\) = 7 ସେ.ମି.
∴ ଦଳାଯାଇ ପ୍ରସ୍ଥ = (R – r) = (4x – 3x) = x = 7 ସେ.ମି.

Question 9.
ଗୋଟିଏ ବଳୟ ଆକାରର ରାସ୍ତାର ବାହାର ଓ ଭିତର ବୃତ୍ତର ପରିଧୂ ଯଥାକ୍ରମେ 300 ମିଟର ଓ 200 ମିଟର ହେଲେ, ରାସ୍ତାର ପ୍ରସ୍ଥ କେତେ ? (π ≃ √10)
Solution:
ମନେକର ବହିଃବୃତ୍ତ ଓ ଅନ୍ତଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଯଥାକ୍ରମେ R ମି. ଓ r ମି. |
ବହିଃବୃତ୍ତର ପରିଧ୍ = 2лR ମି. ଓ ଅନ୍ତଃବୃତ୍ତର ପରିଧ୍ = 2лr ମି. ।
ପ୍ରଶ୍ନନୁସାରେ, 2лR = 300 ମି.
⇒ 2 × √10 × R = 300 ⇒ R = \(\frac{300}{2 \times \sqrt{10}}\) = 15√10 ମି.
ପୁନଶ୍ଚ, 2лR = 200 ମି.
⇒ 2 × √10 × r = 200 ⇒ r = \(\frac{200}{2 \times \sqrt{10}}\) = 10√10 ମି.
∴ ଦଳାଯାଇ ପ୍ରସ୍ଥ = R – r = (15√10 – 10√10) ମି. = 5√10 ମି.

Question 10.
7 ମି. ବ୍ୟାସାର୍କବିଶିଷ୍ଟ ବୃତ୍ତ ଉପରେ କେତେଥର ଘୂରିଲେ 11 କି.ମି. ଦୂରତା ଅତିକ୍ରମ କରିହେବ ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 7 ମି.
ବୃତ୍ତର ପରିସ୍ = 2лr = 2 × \(\frac { 22 }{ 7 }\) × 7 = 44 ମି.
ବୃତ୍ତ ଉପରେ ଥରେ ଘୂରିଲେ 44 ମି. ବାଟ ଅତିକ୍ରମ କରିବ ।
11 କି.ମି. ଅର୍ଥାତ୍ (11 × 1000) ମି. ବାଟ ଅତିକ୍ରମ କଲେ ଘୂର୍ଣ୍ଣନ ସଂଖ୍ୟା = \(\frac{11 \times 1000}{44}\) = 250 ଥର
∴ 7ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବୃତ୍ତ ଉପରେ 250 ଥର ଘୂରିଲେ 11 କି.ମି. ଦୂରତା ଅତିକ୍ରମ କରିହେବ ।

Question 11.
ଗୋଟିଏ ସାଇକେଲ୍‌ର ପ୍ରତ୍ୟେକ ଚକ ମିନିଟ୍‌ରେ 80 ଥର ଘୂରନ୍ତି । ଚକର ବହିଃ ବ୍ୟାସ 42 ସେ.ମି. ହେଲେ ସାଇକେଲ୍‌ର ଘଣ୍ଟାପ୍ରତି ବେଗ ନିର୍ଣ୍ଣୟ କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ସାଇକେଲ ଚକର ବହିଃ ବ୍ୟାସ = 42 ସେ.ମି.
ସାଇକେଲ ଚକର ପରିସ୍ = л × ବ୍ୟାସ = \(\frac { 22 }{ 7 }\) × 42 = 132 ସେ.ମି.
ସାଇକେଲ ଚକ ଥରେ ଘୂରିଲେ 132 ସେ.ମି. ଯାଏ ।
ସାଇକେଲ ଚକ.80 ଥର ଘୂରିଲେ ଯାଏ = 132 × 80 ସେ.ମି. = 10560 ସେ.ମି. ବାଟ ଅତିକ୍ରମ କରିବ ।
ସାଇକେଲ୍‌ର ମିନିଟ୍ ପ୍ରତି ବେଗ = 10560 ସେ.ମି.
∴ ସାଇକେଲ୍‌ର ମିନିଟ୍ ପ୍ରତି ବେଗ = 10560 × 60 ସେ.ମି. = 633600 ସେ.ମି. = 6336 ମି. 6.366 ମି.

Question 12.
ଗୋଟିଏ ଗାଡ଼ିର ବଡ଼ ଚକ ଓ ସାନ ଚକର ପରିଧର ଅନୁପାତ 4 : 1 | 440 ମିଟର ରାସ୍ତା ଅତିକ୍ରମ କରିବାରେ ସାନ ଚକ ବଡ଼ ଚକ ଅପେକ୍ଷା 15 ଥର ଅବ୍ଲକ ଘୂରେ । ପ୍ରତ୍ୟେକ ଚକର ପରିସ୍ ନିର୍ଣ୍ଣୟ କର | (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ମନେକର ଗାଡ଼ିର ବଡ଼ ଚକର ପରିଧୂ = 4x ମି. ଓ ସାନ ଚକର ପରିଧ୍ = x ମି. |
ଥରେ ଘୂରିଲେ ବଡ଼ ଚକ ଓ ସାନ ଚକ ଦ୍ଵୟ ଯଥାକ୍ରମେ 4x ମି. ଓ x ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବ ।
440 ମି. ରାସ୍ତା ଅତିକ୍ରମ କରିବାରେ ବଡ଼ ଚକ ଏବଂ ସାନ ଚକ ଯଥାକ୍ରମେ \(\frac { 440 }{ 4x }\) ଏବଂ \(\frac { 440 }{ x }\)
ସାନ ଚକ ଯଥାକ୍ରମେ , \(\frac { 440 }{ x }\) ଏବଂ \(\frac { 440 }{ 4x }\) = 15
⇒ \(\frac { 440 }{ x }\) – \(\frac { 440 }{ 4x }\) = 15 ⇒ \(\frac { 440 }{ x }\) (1 – \(\frac { 1 }{ 4 }\)) = 15
⇒ \(\frac { 440 }{ x }\) – \(\frac { 3 }{ 4 }\) = 15 ⇒ x = \(\frac{440 \times 3}{4 \times 15}\) = 22 ମି.
∴ ଦତ ଚଳଉ ପରିଧ = 4x ମି. = 4 × 22 ମି. = 88 ମି. ଓ ଦତ ଚଳଉ ପରିଧ = x ମି. = 22 ମି. |

Question 13.
ଗୋଟିଏ ଅଦ୍ଧବୃତ୍ତାକାର ଜମିର ଚାରିପାଖରେ ବାଡ଼ ଦେବା ଖର୍ଚ୍ଚ ମିଟରକୁ 75 ପଇସା ହିସାବରେ 216 ଟଙ୍କା ଖର୍ଚ୍ଚ ହେଲେ ଅର୍ଥବୃତ୍ତାକାର ଜମିର ବ୍ୟାସ ନିର୍ଣ୍ଣୟ କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ଅର୍ଥବୃତ୍ତାକାର ଜମିର ପରିସୀମା = \(\frac { 21600 }{ 75 }\) = 288 ମି.
ଯଦି ଅର୍ଥବୃତ୍ତାକାର ଜମିର ବ୍ୟାସାର୍ଦ୍ଧ r ମି. ହୁଏ ତେବେ ପରିସୀମା = (лr + 2r) ମି. ।
ପ୍ରଶ୍ନନୁସାରେ, лr + 2r = 288
⇒ r(\(\frac { 22 }{ 7 }\) + 2) = 288 ⇒ r(\(\frac{22+14}{7}\)) = 288 ⇒ r × \(\frac { 36 }{ 7 }\) = 288 ⇒ r = \(\frac{288 × 7}{36}\) = 56
∴ ଅର୍ଥବୃତ୍ତାକାର ଜମିର ବ୍ୟାପ = 56 × 2 = 112 ମି.

Question 14.
ଗୋଟିଏ ଘୋଡ଼ା ବୃତ୍ତ ଉପରେ ଥରେ ଘୂରିଆସି ସିଧା ଯାଇ କେନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ ତାକୁ 10 ମିନିଟ୍ 12 ସେକେଣ୍ଡ ସମୟ ଲାଗିଲା । ସେ କେବଳ ବୃତ୍ତ ଉପରେ ଘୂରିଥଲେ ତାକୁ କେତେ ସମୟ ଲାଗିଥାନ୍ତା ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ମନେକର ଘୋଡ଼ାଟିକୁ ବ୍ୟାସାର୍ଷ ଉପରେ ଯିବାକୁ r ମିନିଟ୍ ସମୟ ଲାଗିଥା’ନ୍ତା ।
ତେବେ ପରିଧ୍ ଉପରେ ଯିବାକୁ 2лr ମିନିଟ୍ ସମୟ ଲାଗିବ ।
ବୃତ୍ତାକାର ପଥରେ ଥରେ ଘୂରି ସିଧା କେନ୍ଦ୍ରରେ ପହଞ୍ଚିବା ପାଇଁ (2πr + r) ମିନିଟ୍ ସମୟ ଲାଗିବ ।
ପ୍ରଶ୍ନନୁସାରେ, (2πr + r) = 10\(\frac { 12 }{ 60 }\) ⇒ r(2 × \(\frac { 22 }{ 7 }\) + 1) = \(\frac { 51 }{ 5 }\)
⇒ r(\(\frac{44+7}{7}\)) = \(\frac { 51 }{ 5 }\) ⇒ r = \(\frac { 51 }{ 5 }\) × \(\frac { 7 }{ 51 }\) ⇒ r = \(\frac { 7 }{ 5 }\) ମିନିଟ୍
ବୃତ୍ତ ଉପରେ ଘୂରିବାପାଇଁ ଅର୍ଥାତ୍ 2πr = 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 7 }{ 5 }\) = \(\frac { 44 }{ 5 }\) = ମିନିଟ୍ = 8\(\frac { 4 }{ 5 }\) ମିନିଟ୍
= 8 ମିନିଟ୍ 48 ସେକେଣ୍ଡ ସମୟ ଲାଗିବ ।
∴ ଘୋଡ଼ାଟି କେବଳ ବୃତ୍ତ ଉପରେ ଘୂରିଥଲେ ତାକୁ 8 ମିନିଟ୍ 48 ସେକେଣ୍ଡ ସମୟ ଲାଗିଥା’ନ୍ତା ।

Question 15.
ଗୋଟିଏ ବୃତ୍ତ ଉପରେ ଥରେ ଭ୍ରମଣ କରିବାକୁ ଯେତେ ସମୟ ଲାଗେ ବୃତ୍ତଟିର ବ୍ୟାସ ପରିମିତ ପଥ ଅତିକ୍ରମ କରିବାକୁ 45 ସେକେଣ୍ଡ କମ୍ ଲାଗେ । ଯଦି ଲୋକଟିର ବେଗ ଏକ ମିନିଟ୍‌ରେ 80 ମିଟର ହୁଏ ତେବେ ବୃତ୍ତର ବ୍ୟାସ କେତେ ହେବ ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ଲୋକଟି 1 ମିନିଟ୍‌ରେ 80 ମିଟର ପଥ ଅତିକ୍ରମ କଲେ 45 ସେକେଣ୍ଡରେ ଯିବ
=(80 × \(\frac { 45 }{ 60 }\)) ମିଟର = 60 ମି. ।
ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r ମି., ବୃତ୍ତର ପରିଧୂ = 2πr ମି. ଓ ବ୍ୟାସ = 2r ମି. |
ପ୍ରଶ୍ନନୁସାରେ, 2πr – 2r = 60
⇒ 2r (\(\frac { 22 }{ 7 }\) – 1) = 60 ⇒ 2r (\(\frac{22-7}{7}\)) = 60
⇒ 2r = \(\frac{60 × 7}{15}\) ⇒ r = \(\frac { 28 }{ 2 }\) = 14 ମି.
∴ ଦୁଇର ବ୍ୟାସ = 2r = 2 × 14 ମି. = 28 ମିଟର |

Question 16.
ଖଣ୍ଡେ ତାରକୁ ସମବାହୁ ତ୍ରିଭୁଜାକୃତି କଲେ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ 1936√3 ବର୍ଗମିଟର ହୁଏ । ଉକ୍ତ ତ୍ରିଭୁଜର ପରିସୀମା ସହ ସମାନ ପରିଧ୍ ଥ‌ିବା ବୃତ୍ତଟିର ବ୍ୟାସ କେତେ ହେବ ? (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ସମବାହୁ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = 1936√3 ବର୍ଗ ମି.
⇒ \(\frac{\sqrt{3}}{4}\) × (ବାହୁର ଦୈର୍ଘ୍ୟ)² = 1936√3
⇒ (ବାହୁର ଦୈର୍ଘ୍ୟ)² = \(\frac{1936 \sqrt{3} \times 4}{\sqrt{3}}\)
⇒ (ବାହୁର ଦୈର୍ଘ୍ୟ) = \(\sqrt{1936 \times 4}\) = 88 ମି.
∴ ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁର ଦୈର୍ଘ୍ୟ 88 ମି. ।
ସମବାହୁ ତ୍ରିଭୁଜର ପରିସୀମା = 3 × 88 ମି. = 264 ମି. = ବୃତ୍ତର ପରିସ୍
ଯଦି ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r ମି. ହୁଏ ତେବେ ପରିଧ୍ 2πr ମି. ହେବ ।
ପ୍ରଶ୍ନନୁସାରେ, 2πr = 264
⇒ 2 × \(\frac { 22 }{ 7 }\) × r = 264 ⇒ r = 264 × \(\frac{7}{2 \times 22}\) = 42 ମି.
∴ ଦୁଇର ବ୍ୟାସ = 2r = 2 × 42 = 84 ମି. |

Question 17.
20 ସେ.ମି. ଦୀର୍ଘ ବାହୁ ବିଶିଷ୍ଟ ଏକ ବର୍ଗଚିତ୍ର ମଧ୍ୟରେ ଏକ ବୃତ୍ତ ଅନ୍ତର୍ଲିଖ୍ତ ହେଲେ ବୃତ୍ତର ପରିସ୍ କେତେ ହେବ ? (n ≃ 3.14)
Solution:

ABCD ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ = 20 ସେ.ମି. ।
ଗୋଟିଏ ବର୍ଗକ୍ଷେତ୍ର ମଧ୍ୟରେ ବୃତ୍ତଟିଏ ଅନ୍ତର୍ଲିଖ୍ତ ହେଲେ ବର୍ଗକ୍ଷେତ୍ରର ବାହୁର ଦୈର୍ଘ୍ୟ = ବୃତ୍ତର ବ୍ୟାସ ।
ବୃତ୍ତର ବ୍ୟାସ = MN = 20 ସେ.ମି.
∴ ଚଉର ପରିଧି = π × ବ୍ୟାସ = 3.14 × 20 ସେ.ମି. = 62.8 ସେ.ମି. |

Question 18.
42 ସେ.ମି. ଦୀର୍ଘ ବାହୁ ବିଶିଷ୍ଟ ଏକ ସମବାହୁ ତ୍ରିଭୁଜର ପରିଲିଖ ଓ ଅନ୍ତର୍ଲିଖ୍ତ ବୃତ୍ତର ପରିସ୍ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ : ସମବାହୁ ତ୍ରିଭୁଜର ସମାନ ବାହୁର ଦୈର୍ଘ୍ୟ = 42 ସେ.ମି. |
ବୃତ୍ତରେ ABC ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ଅନ୍ତର୍ଲିଖ୍ ।
ପରିବୃତ୍ତ ଓ ଅନ୍ତଃବୃତ୍ତର କେନ୍ଦ୍ର ୦ ଏବଂ AD ⊥ BC|
ABC ଇଉତା (AD) = \(\frac{\sqrt{3}}{2}\) × ସମାନ ବାହୁର ଦୈର୍ଘ୍ୟ = \(\frac{\sqrt{3}}{2}\) × 42 ସେ.ମି.= 21, √3 ସେ.ମି.
O, △ABC ର ଭରକେନ୍ଦ୍ର ହେତୁ, AD = 3 OD
⇒ 3 OD = 21 √3
⇒ OD = \(\frac{21 \sqrt{3}}{3}\) = 7√3 ସେ.ମି. |

∴ ଅନ୍ତର୍ଲିଖ ବୃତ୍ତର ପରିସ୍ = 2 × \(\frac { 22 }{ 7 }\) × 7√3 ସେ.ମି. = 44√3 ସେ.ମି.
ପରିହରର ବ୍ୟାସାଦ = OA = 2 × OD = 2 × 7√3 = 14√3 ସେ.ମି.
∴ ଅନ୍ତର୍ଲିଖ ଚଉର ପରିଧି = 2 × \(\frac { 22 }{ 7 }\) × 14√3 ସେ.ମି. = 88√3 ସେ.ମି.

Question 19.
(a) 21 ସେ.ମି. ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ବୃତ୍ତକଳାର ପରିସୀମା 64 ସେ.ମି. ହେଲେ, ବୃତ୍ତକଳାର ଚାପର ଡିଗ୍ରୀ ପରିମାପ ସ୍ଥିର କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ବୃତ୍ତକଳାର ବ୍ୟାସାର୍ଦ୍ଧ (r) = 21 ସେ.ମି. ।
ମନେକର ବୃତ୍ତକଳାର ଚାପର ଦୈର୍ଘ୍ୟ = L ସେ.ମି. ଓ ଚାପର ଡିଗ୍ରୀ ପରିମାପ = m\(\widehat{\mathrm{ACB}}\) θ° |
ବୃତ୍ତକଳାର ପରିସୀମା = (L + 2r) ସେ.ମି.
= (L+ 2 × 21) ସେ.ମି.. = (L + 42) ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, L + 42 = 66 ସେ.ମି.

⇒ L = 64 – 42 = 22 ସେ.ମି.
ଆମେ ଜାଣିଛେ, L= \(\frac{\theta}{180^{\circ}}\) × πr
⇒ 22 = \(\frac { θ }{ 180 }\) × \(\frac { 22 }{ 7 }\) × 21
⇒ θ = \(\frac{22 \times 180 \times 7}{22 \times 21}\) = 60° ⇒ m\(\widehat{\mathrm{ACB}}\) 60°
∴ ବୃତ୍ତକଳାର ଚାପର ଡିଗ୍ରୀ ପରିମାପ 60° |

(b) ଏକ ବୃତ୍ତରେ ଯେଉଁ ବୃତ୍ତକଳାର ଚାପର ଡିଗ୍ରୀ ପରିମାପ 40°, ସେହି ବୃତ୍ତକଳାର ପରିସୀମା 26.98 ସେ.ମି. ହେଲେ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ କେତେ ? (π ≃ 3.14)
Solution:
ମନେକର ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ = r ସେ.ମି. ଓ ବୃତ୍ତକଳାର ଚାପର ଦୈର୍ଘ୍ୟ = L ସେ.ମି.
ବୃତ୍ତକଳାର ଚାପର ଡିଗ୍ରୀ ପରିମାପ (θ) = 40°
∴ ବୃତ୍ତକଳାର ପରିସୀମା = (L + 2r) ସେ.ମି.
କିନ୍ତୁ L = \(\frac { θ }{ 180 }\) × πr = \(\frac { 40 }{ 180 }\) × 3.14 × r = \(\frac { 2 }{ 9 }\) × 3.14r ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, L + 2r = 26.98 ⇒ r(\(\frac{6.28+18}{9}\)) = 26.98
⇒ r = 26.98 × \(\frac { 9 }{ 24.28 }\) = \(\frac { 24282 }{ 2428 }\) = 10 ସେ.ମି.
∴ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ 10 ସେ.ମି. ।

Question 20.
କୌଣସି ଏକ ବୃତ୍ତକଳାର କେନ୍ଦ୍ରସ୍ଥ କୋଣର ପରିମାଣ 90° । ଏହାର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ 5 ସେ.ମି. ହେଲେ ବୃତ୍ତକଳାର ପରିସୀମା ନିର୍ଣ୍ଣୟ କର । (π ≃ 3.1416)
Solution:
ବୃତ୍ତକଳାର କେନ୍ଦ୍ରସ୍ଥ କୋଣର ପରିମାଣ (θ) = 90°
ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ (r) = 5 ସେ.ମି.
ବୃତ୍ତକଳାର ଚାପର ଦୈର୍ଘ୍ୟ (L) = \(\frac { θ }{ 180 }\) × πr
= \(\frac { 90 }{ 180 }\) × 3.1416 × 5 = 7.854 ସେ.ମି.
∴ ବୃତ୍ତକଳାର ପରିବାପା = L + 2r = (7.854 + 2 × 5) ସେ.ମି. = 17.854 ସେ.ମି.

Question 21.
କୌଣସି ଏକ ବୃତ୍ତର ଏକ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 40 ଏବଂ ଅନ୍ୟ ଏକ ବୃତ୍ତର ସମଦୈର୍ଘ୍ୟ ବିଶିଷ୍ଟ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 60° ହେଲେ ଉଭୟ ବୃତ୍ତର ବ୍ୟାସାର୍ଷର ଅନୁପାତ ନିର୍ଣ୍ଣୟ କର ।
Solution:
ମନେକର ପ୍ରଥମ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ r₁ ସେ.ମି. ଓ ଦ୍ୱିତୀୟ
ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ r₂ ସେ.ମି.
ପ୍ରଥମ ବୃତ୍ତର ଚାପର ଡିଗ୍ରୀ
ପରିମାପ (θ₁) = 40°
ଦିତାଯ ହଇର ଚାପର କିସ୍ଥା
ପରିମାପ (θ₂) = 60°

ପ୍ରଥମ ବୃତ୍ତର ଚାପର ଦୈର୍ଘ୍ୟ (L₁) = \(\frac{\theta_1}{180}\) × πr₁ = (\(\frac { 40 }{ 180 }\) × πr₁) ସେ.ମି.
ପ୍ରଥମ ବୃତ୍ତର ଚାପର ଦୈର୍ଘ୍ୟ (L₂) = \(\frac{\theta_2}{180}\) × πr₂ = (\(\frac { 60 }{ 180 }\) × πr₂) ସେ.ମି.
ମନେକର ପ୍ରଥମ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧର ଦୈର୍ଘ୍ୟ L₁ = L₂
⇒ \(\frac { 40 }{ 180 }\) × πr₁ = \(\frac { 60 }{ 180 }\) × πr₂ ⇒ \(\frac{r_1}{r_2}\) = \(\frac { 60 }{ 180 }\) × \(\frac { 180 }{ 40 }\) = \(\frac { 3 }{ 2 }\)
∴ r₁ : r₂ = 3:2

Question 22.
ଗୋଟିଏ ଘଣ୍ଟାର ମିନିଟ୍ କଣ୍ଟାର ଅଗ୍ରଭାଗ 5 ମିନିଟ୍‌ରେ 7\(\frac { 1 }{ 3 }\) ସେ.ମି. ଦୈର୍ଘ୍ୟ ବିଶିଷ୍ଟ ଏକ ଚାପ ଅଙ୍କନ କରେ । ମିନିଟ୍ କଣ୍ଟାର ଦୈର୍ଘ୍ୟ ନିର୍ଣ୍ଣୟ କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
60 ମିନିଟ୍‌ରେ ମିନିଟ୍ କଣ୍ଟାଟି 360° କୋଣ ସୃଷ୍ଟି କରେ ।
5 ମିନିଟ୍‌ରେ ମିନିଟ୍‌ରେ କଣ୍ଟାଟି କେନ୍ଦ୍ରରେ ଉତ୍ପନ୍ନ କରୁଥିବା କୋଣର ପରିମାଣ = \(\frac { 360 }{ 60 }\) × 5 = 30°
ମିନିଟ୍ କଣ୍ଟାର ଅଗ୍ରଭାଗ ଦ୍ବାରା ଅତିକ୍ରାନ୍ତ ପଥ (L) = 7\(\frac { 1 }{ 3 }\) ସେ.ମି. = \(\frac { 22 }{ 3 }\) ସେ.ମି. |
ମନେକର ମିନିଟ୍ କଣ୍ଟାର ଦୈର୍ଘ୍ୟ = r ସେ.ମି.
ଆମେ ଜାଣିଛେ, L = \(\frac { θ }{ 180 }\) × πr ⇒ \(\frac { 22 }{ 3 }\) = \(\frac { 30 }{ 180 }\) × \(\frac { 22 }{ 7 }\) × r = \(\frac{22 \times 180 \times 7}{3 \times 30 \times 22}\) = 14 ସେ.ମି.
∴ ମିନିଟ୍ କଣ୍ଟାର ଦୈର୍ଘ୍ୟ 14 ସେ.ମି. |
ମିନିଟ୍ କଣ୍ଟାର ଅଗ୍ରଭାଗ 5 ମିନିଟ୍‌ରେ 7\(\frac { 1 }{ 3 }\) ସେ.ମି. = \(\frac { 22 }{ 3 }\) ସେ.ମି. ଦୈର୍ଘ୍ୟବିଶିଷ୍ଟ ଚାପ ଅଙ୍କନ କରେ ।
60 ମିନିଟ୍‌ରେ କଣ୍ଟାଟି ଦୈର୍ଘ୍ୟ ଅତିକ୍ରମ କରିବ = \(\frac { 22 }{ 3 }\) × 12 = 88 ସେ.ମି.
ଏହି ଦୈର୍ଘ୍ୟଟି ସମ୍ପୂର୍ଣ ଚାପ ଅର୍ଥାତ୍ ବୃତ୍ତର ପରିସ୍ ଅଟେ ।
∴ 2лr = 88
⇒ 2 × \(\frac { 22 }{ 7 }\)r = 88 ⇒ r = 88 × \(\frac { 7 }{ 44 }\) = 14 ସେ.ମି. |
∴ ମିନିଟ୍ କଣ୍ଟାର ଦୈର୍ଘ୍ୟ 14 ସେ.ମି. ।

Question 23.
ଗୋଟିଏ ବୃତ୍ତର ପରିସ୍ ଅନ୍ୟ ଏକ ବୃତ୍ତର ପରିଧୂର ତିନିଗୁଣ । ପ୍ରଥମ ବୃତ୍ତର 10 ସେ.ମି. ପରିମିତ ଚାପର ଡିଗ୍ରୀ ପରିମାପ 30° ହେଲେ ଦ୍ବିତୀୟ ବୃତ୍ତର ପରିସ୍ କେତେ ହେବ ନିର୍ଣ୍ଣୟ କର । (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
ପ୍ରଥମ ବୃତ୍ତର ଚାପର ଦୈର୍ଘ୍ୟ (L) = 10 ସେ.ମି. ଓ ଚାପର ଡିଗ୍ରୀ ପରିମାପ (θ) = 30°
L = \(\frac { θ }{ 180 }\) × πr ⇒ 10 = \(\frac { 30 }{ 180 }\) × πr ⇒ r = \(\frac{180 \times 10}{30 \times \pi}\) = \(\frac { 60 }{ π }\) ସେ.ମି.
ପ୍ରଥମ ବୃତ୍ତର ପରିସ୍ = 2πr = 2 π × \(\frac { 60 }{ π }\) = 120 ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, ପ୍ରଥମ ବୃତ୍ତର ପରିଧ ଦ୍ବିତୀୟ ବୃତ୍ତର ପରିଧର 3 ଗୁଣ ।
∴ ଦ୍ୱିତୀୟ ବୃତ୍ତର ପରିସ୍ = \(\frac { 120 }{ 3 }\) = 40 ସେ.ମି. |

Question 24.
ଗୋଟିଏ ବୃତ୍ତର ପରିଧ୍ 6.282 ସେ.ମି. ହେଲେ ଓ ଏହା ଏକ ସମବାହୁ ତ୍ରିଭୁଜ ମଧ୍ୟରେ ଅନ୍ତର୍ଲିଖ୍ତ ହେଲେ, ତ୍ରିଭୁଜର ବାହୁର ଦୈର୍ଘ୍ୟ କେତେ ? (π ≃ 3.141)
Solution:
ମନେକର ଅନ୍ତର୍ଲିଖ୍ତ ବୃତ୍ତର ବ୍ୟାସାର୍କର ଦୈର୍ଘ୍ୟ = r ସେ.ମି.
ତେବେ ପରିସ୍ଥି = 2πr ସେ.ମି. = 2 × 3.141 × r = 6.282 r ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, 6.282 r = 6.282 ⇒ r = 1 ସେ.ମି.
ସମବାହୁ ତ୍ରିଭୁଜର ଅନ୍ତଃକେନ୍ଦ୍ର, ଭରକେନ୍ଦ୍ର ଓ ଲମ୍ବବିନ୍ଦୁ O ଅଭିନ୍ନ ଅଟେ ।

\(\overline{\mathrm{AD}}\) ଲମ୍ବ ଉପରିସ୍ଥ ‘O’ ଭରକେନ୍ଦ୍ର ହେତୁ OD = \(\frac { 1 }{ 3 }\) AD
ଆଥାତ୍ (AD) = 3 × OD = 3 × r = 3 × 1 = 3 ସେ.ମି.
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁର ଦୈର୍ଘ୍ୟ = ଉଚ୍ଚତା × \(\frac{2}{\sqrt{3}}\) = 3 × \(\frac{2}{\sqrt{3}}\) ସେ.ମି. = 2√3 ସେ.ମି.
∴ ତ୍ରିଭୁଜର ବାହୁର ଦୈର୍ଘ୍ୟ 2√3 ସେ.ମି. |

Question 25.
ଗୋଟିଏ ବୃତ୍ତକଳାର ଚାପର ଡିଗ୍ରୀ ପରିମାପ 60° । ଏହାର ଦୁଇ ବ୍ୟାସାର୍ଦ୍ଧ ଓ ଚାପକୁ ସ୍ପର୍ଶକରି ଏକ ବୃତ୍ତ ଅନ୍ତର୍ଲିଖ୍ । ପ୍ରମାଣ କର ଯେ, ଏହି ବୃତ୍ତର ପରିସ୍ ଓ ବୃତ୍ତକଳାର ପରିସୀମାର ଅନୁପାତ 11 : 16 | (π ≃ \(\frac { 22 }{ 7 }\))
Solution:
OACB ବୃତ୍ତକଳାର ଡିଗ୍ରୀ ପରିମାପ 60°
ଅର୍ଥାତ୍ m∠AOB = 60°
ବୃତ୍ତକଳାର ଅନ୍ତର୍ଲିଖ୍ତ ବୃତ୍ତର କେନ୍ଦ୍ର M ଓ D ସ୍ପର୍ଶବିନ୍ଦୁ ।
\(\overline{\mathrm{OC}}\) ବୃତ୍ତକୁ E ବିନ୍ଦୁରେ ଛେଦ କରୁଛି ।
\(\overline{\mathrm{DE}}\) ଅଙ୍କନ କରାଯାଉ ।
ବର୍ତ୍ତମାନ △OMDରେ ∠MOD = \(\frac{60^{\circ}}{2}\) = 30°
ମନେକର ବୃତ୍ତକଳାର ବ୍ୟାସାର୍ଦ୍ଧ R ଏକକ ଏବଂ ଅନ୍ତର୍ଲିଖ ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ r ଏକକ ।
ବୃତ୍ତକଳାର ବ୍ୟାସାର୍ଦ୍ଧ (R) = OC = OM + MC = 2r + r = 3r

[∵ △OMDରେ sin 30° = \(\frac { MD }{ OM }\) ⇒ \(\frac { 1 }{ 2 }\) = \(\frac { r }{ OM }\) ⇒ OM = 2r]
\(\overparen{A C B}\) ର ଦୈର୍ଘ୍ୟ = \(\frac { 60 }{ 180 }\) × π × 3r = πr ଏକକ
ବୃତ୍ତକଳାର ପରିସୀମା = OA + OB + l \(\overparen{A C B}\)
= 2 × 3r + лr = 6r + πr = r (6 + \(\frac { 22 }{ 7 }\)) = r (\(\frac{42+22}{7}\)) = \(\frac { 64 }{ 7 }\) r ଏକକ
ଅନ୍ତର୍ଲିଖ ବୃତ୍ତର ପରିଧୂ = 2лr = 2 × \(\frac { 22 }{ 7 }\) × r = \(\frac { 44r }{ 7 }\) ଏକକ ।
∴ ବୃତ୍ତର ପରିଧୂ / ଦରଜକାର ପରିମାମା = \(\frac{\frac{44 \mathrm{r}}{7}}{\frac{64 \mathrm{r}}{7}}\) = \(\frac { 44 }{ 64 }\) = \(\frac { 11 }{ 16 }\) = 11 : 16
∴ ବୃତ୍ତର ପରିଧୂ : ଦରଜକାର ପରିମାମା = 11 : 16 |

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି Ex 6(c)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ସଠିକ୍ ଉତ୍ତରଟି ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) ଗୋଟିଏ ତ୍ରିଭୁଜର ଶୀର୍ଷବିଦୁ୍ୟତ୍ରୟ (2, 5), (-3, 5) ଓ (0, 5) ହେଲେ, ତ୍ରିଭୁଜଟିର କ୍ଷେତ୍ରଫଳ _____ ହେବ ।
[-5, 3, 0, 10]
(ii) ଯଦି a = _____ ହୁଏ, ତେବେ (a, -2), (2, 5) ଓ (2, 10) ବିନ୍ଦୁଦ୍ଵୟ ଏକ ସରଳରେଖାରେ ରହିବେ ।
[0, 3, 2, -2]
(iii) yର ମାନ _____ ପାଇଁ (-2, -2), (0, y) ଓ (3, 3) ବିନ୍ଦୁଦ୍ଵୟ ଏକ ସରଳରେଖାରେ ରହିବେ ।
[0, 2, 2, 3]
(iv) kର ମାନ _____ ପାଇଁ (k, -2), (1, 4) ଏବଂ (-2, 7) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ହେବେ ।
[3, -3, 2, -2]
(v) ଥର ମାନ _____ ପାଇଁ (4, -5), (1, a) ଏବଂ (-2, 7) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିଦୁ୍ୟତ୍ରୟ ଗୋଟିଏ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁ ହେବେ ନାହିଁ ।
[1, 2, 3, 4]
ଉତ୍ତର:
(i) 0
(ii) 2
(iii) 0
(iv) 3
(v) 1

Question 2.
ନିମ୍ନରେ କେତେକ ତ୍ରଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର ସ୍ଥାନଙ୍କ ଦିଆଯାଇଛି । ପ୍ରତ୍ୟେକ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ ନିର୍ଣ୍ଣୟ କର ।
(i) (3, 0), (4, 5) ଓ (2, 0)
(ii) (0, 0), (1, 0) ଓ (1, 1)
(iii) (-2, 1), (2, -3) ଓ (4, – 4)
(iv) (5,7),(6, 4) ଓ (2, -5)
(v) (5, 2), (1, 3) ଓ (1,-2)
ସମାଧାନ :
(i) ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର ତ୍ରୟ (3, 0), (4, 5), (2, 0) 1.
ଏଠାରେ x₁ = 3, x₂ = 4, x₃ = 2, y₁ = 0, y₂ = 5, y₃ = 0
∆ର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\)|{3(5 – 0) + 4(0 – 0) + 2(0 – 5)}| = \(\frac{1}{2}\)|{15 + 0 – 10}|
= \(\frac{1}{2}\) × 5 = \(\frac{5}{2}\)
∴ ∆ର କ୍ଷେତ୍ରଫଳ \(\frac{5}{2}\) ବର୍ଗ ଏକକ ।

(ii) ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର ତ୍ରୟ (0, 0), (1, 0), (1, 1) |
ଏଠାରେ x₁ = 1, x₂ = 1, y₁ = 0, y₂ = 1
∴ ∆ ର ଏକ ଶୀର୍ଷବିନ୍ଦୁ ମୂଳବିନ୍ଦୁ ହୋଇଥିବାରୁ ଏହାର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|(x₁y₂ – x₂y₁)|
= \(\frac{1}{2}\)|(1 × 1 – 0 × 1)| = \(\frac{1}{2}\)|(1 – 0)| = \(\frac{1}{2}\) × 1 = \(\frac{1}{2}\) ବର୍ଗ ଏକକ ।

ବିକତ୍ଵ ସମାଧାନ :
ଏଠାରେ x₁ = 0, x₂ = 1, x₃ = 1, y₁ = 0, y₂ = 0, y₃ = 1
∴ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\)|{0(0 – 1) + 1(1 – 0) + 1(0 – 0)}| = \(\frac{1}{2}\)(0 + 1 + 0) = \(\frac{1}{2}\)

(iii) ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର ତ୍ରୟ (-2, 1), (2, -3), (4, -4) ।
ଏଠାରେ x₁ = -2, x₂ = 2, x₃ = 4, y₁ = 1, y₂ = -3, y₃ = -4
∴ ∆ର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\)|{-2(-3 + 4) + 2(-4 – 1) + 4(1 + 3)}|
= \(\frac{1}{2}\)|(-2) + (-10) + 16| = \(\frac{1}{2}\) × 4 = 2 ବର୍ଗ ଏକକ ।

(iv) ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର (5,7), (6, 4), (2,-5) ।
ଏଠାରେ x₁ = -5, x₂ = 6, x₃ = 2, y₁ = 7, y₂ = 4, y₃ = -5
∴ ∆ର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\)|{5(4 + 5) + 6(-5 – 7) + 2(7 – 4)}|
= \(\frac{1}{2}\)(45 – 76 + 6) = \(\frac{1}{2}\)|(51 – 72)| = \(\frac{1}{2}\)|-21|
= \(\frac{21}{2}\) = 10.5 ବର୍ଗ ଏକକ ।

(v) ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁତ୍ରୟର (5, 2), (1, 3) ଓ (1,-2) ।
ଏଠାରେ x₁ = 5, x₂ = -1, x₃ = 1, y₁ = 2, y₂ = 3, y₃ = -2
∴ ∆ର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\)|{5 (3 + 2) + (-1) (-2 – 2) + 1(2 – 3)}|
= \(\frac{1}{2}\)|{25 + 4 – 1}| = \(\frac{1}{2}\) × 28 = 14 ବର୍ଗ ଏକକ ।

Question 3.
ଦର୍ଶାଅ ଯେ, ଦତ୍ତ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।
(i) (1, 1), (4, 3) ଓ (-2, -1)
(ii) (-1,-5), (0, -3) ଓ (4, 5)
(iii) (1,4), (3, -2) ଓ (-3, 16)
(iv) (-4a, – 6a), (-a, -2a) ଓ (5a, 6a)
(v) (-a, 2b), (0, b) ଓ (\(\frac{a}{2}, \frac{b}{2}\))
ସମାଧାନ :
(i) ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନୀଙ୍କ A (1, 1), B (4, 3) ଓ C (-2, -1) ।
ଏଠାରେ x₁ = 1, x₂ = 4, x₃ = -2, y₁ = 1, y₂ = 3, y₃ = -1
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | { 1 (3 + 1) + 4 (-1 – 1 ) + ( -2) (1 – 3)}| = \(\frac{1}{2}\) |4 – 8 + 4| = \(\frac{1}{2}\) (0) = 0
∴ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।

(ii) ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନୀଙ୍କ (-1, -5), B (0, -3) ଓ C (4, 5) ।
ଏଠାରେ x₁ = -1, x₂ = 0, x₃ = 4, y₁ = -5, y₂ = -3, y₃ = 5
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | { (-1) (-3 – 5) + 0 (5 + 5) + 4 (-5 + 3)}|
= \(\frac{1}{2}\) |(-1)(-8) + 0 + 4(-2)| = \(\frac{1}{2}\) (8 – 8) = \(\frac{1}{2}\) (0) = 0
∴ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।

(iii) ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନୀଙ୍କ A(1, 4), B (3, -2) ଓ C (-3, 16) ।
ଏଠାରେ x₁ = 1, x₂ = 3, x₃ = -3, y₁ = 4, y₂ = -2, y₃ = 16
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) |{ 1 (-2 – 16) + 3 (16 – 4) + (- 3) (4 + 2)}|
= \(\frac{1}{2}\) |(- 18 + 36 – 18)| = \(\frac{1}{2}\) (0) = 0
∴ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।

(iv) ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନୀଙ୍କ A(-4a, -6a), B (-a, -2a) ଓ C (5a, 6a) ।
ଏଠାରେ x₁ = -4a, x₂ = -a, x₃ = 5a, y₁ = – 6a, y₂ = – 2a, y₃ = 6a
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ
= \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) |{(- 4a) (-2a – 6a) + (- a) (6a + 6a) + 5a (- 6a + 2a)}|
= \(\frac{1}{2}\) | (32a² – 12a² – 20a²) | = \(\frac{1}{2}\) (0) = 0
∴ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।

(v) ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନୀଙ୍କ A (-a, 2b), B (0, b) ଓ C(\(\frac{a}{2}, \frac{b}{2}\))
ଏଠାରେ x₁ = -a, x₂ = 0, x₃ = \(\frac{a}{2}\), y₁ = 2b, y₂ = b, y₃ = \(\frac{b}{2}\)

∴ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ଅବସ୍ଥିତ ।

Question 4.
ଗୋଟିଏ ତ୍ରିଭୁଜର ଶୀର୍ଷନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ (1, -3), (2, -5) ଓ (x, 1) ଏବଂ କ୍ଷେତ୍ରଫଳ 4 ବର୍ଗ ଏକକ ହେଲେ, xର ମାନ ନିରୂପଣ କର ।
ସମାଧାନ :
ଗୋଟିଏ ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ A(1, – 3), B(2, -5) ଓ C (x, 1)।
ଏବଂ କ୍ଷେତ୍ରଫଳ 4 ବର୍ଗ ଏକକ ।
ଏଠାରେ x₁ = 1, x₂ = 2, x₃ = x, y₁ = – 3, y₂ = – 5, y₃ = 1
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | 1 (-5 – 1) + 2 (1 + 3) + x (- 3 + 5) |
= \(\frac{1}{2}\) | 1 -6 + 8 + 2x | = \(\frac{1}{2}\) | 2x + 21 | = \(\frac{1}{2}\) × 2 (x + 1) = x + 1
ପ୍ରଶ୍ନନୁସାରେ, x + 1=4 ⇒ x = 4 – 1 = 3 ∴ x = 3

Question 5.
k ର କେଉଁ ମୂଲ୍ୟ ପାଇଁ (3, – 5), (k, 0) ଓ (-4, 7) ସ୍ଥାନାଙ୍କ ବିନ୍ଦୁମାନଙ୍କ ଦ୍ୱାରା ଗଠିତ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ \(\frac{95}{2}\)ବର୍ଗ ଏକକ ହେବ ?
ସମାଧାନ :
ତ୍ରିଭୁଜର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟ A (3, -5), B (k, 0) ଓ C (-4, 7) ।
ଏଠାରେ x₁ = 3, x₂ = k, x₃ = -4, y₁ = – 5, y₂ = 0, y₃ = 7
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) |{3 (0 – 7) + k (7 + 5) + (-4) (-5 – 0)}|
= \(\frac{1}{2}\) |{- 21 + 12 k + 20}| = \(\frac{1}{2}\) (12 k – 1)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{2}\) (12 k – 1) = \(\frac{95}{2}\) ⇒ 12k – 1 = 95 ⇒ 12 k = 96 ⇒ k = \(\frac{96}{12}\) = 8
∴ kର ମାନ 8 ହେଲେ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ \(\frac{95}{2}\) ବର୍ଗ ଏକକ ହେବ ।

Question 6.
(2, 3), (0, 5) ଓ (1, y) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ରହିଲେ, Yର ମାନ ନିରୂପଣ କର ।
ସମାଧାନ :
ବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନାଙ୍କ A (2, 3), B (0, 5) & C (1, y) |
ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | 2 (5 – y) + 0 ( y – 3) + 1 (3 – 5) | = \(\frac{1}{2}\) | 10 – 2y + 0 – 21 = \(\frac{1}{2}\) | -2y + 8 |
= \(\frac{1}{2}\) (2y – 8) = \(\frac{1}{2}\) × 2 (y – 4) = y – 4
ପ୍ରଶ୍ନନୁସାରେ, y – 4 = 0 = y = 4
∴ yର ମୂଲ୍ୟ 4 ହେଲେ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ 0 ହେବ ।
ଅର୍ଥାତ୍ (2, 3), (0, 5) ଓ (1, y) ଏକ ସରଳରେଖାରେ ରହିବେ ।

Question 7.
k ର କେଉଁ ମୂଲ୍ୟ ପାଇଁ (2, 3), (3, k) ଏବଂ (5, 9) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖା ଉପରେ ଅବସ୍ଥିତ ହେବେ ?
ସମାଧାନ :
ଦଉ ବିନ୍ଦୁତ୍ରୟ A (2, 3), B (3, k) ଏବଂ C (5, 9) ।
ଏଠାରେ x₁ = 2, x₂ = 3, x₃ = 5, y₁ =3, y₂ = k, y₃ = 9
∆ ABC ର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\)|{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | {2 (k – 9) + 3(9 – 3) + 5 (3 – k)} |
= \(\frac{1}{2}\) | (2k – 18 + 18 + 15 – 5k) | = \(\frac{1}{2}\) (-3k + 15) | = \(\frac{1}{2}\) (3k – 15)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{2}\) (3k – 15) = 0 ⇒ 3k – 15 = 0 ⇒ k = \(\frac{15}{3}\) = 5
∴ kର ମାନ 5 ପାଇଁ (2, 3), (3, k) ଏବଂ (5, 9) ଏକ ସରଳରେଖାରେ ରହିବେ ।

Question 8.
କେଉଁ ସର୍ଭରେ (1, 1), (3, 5) ଓ (x, y) ସ୍ଥାନାଙ୍କ ବିଶିଷ୍ଟ ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ରହିବେ, ସ୍ଥିର କର ।
ସମାଧାନ :
ଦତ୍ତ ବିନ୍ଦୁତ୍ରୟ A (1, 1), B (3, 5) ଓ C (x, y) ।
ଏଠାରେ x₁ = 1, x₂ = 3, x₃ = x, y₁ = 1, y₂ = 5, y₃ = y
ABC ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) |{x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)}|
= \(\frac{1}{2}\) | 1 (5 – y) + 3 ( y – 1 ) + x (1 – 5) | = \(\frac{1}{2}\) | (5 – y + 3y – 3 + x – 5x) |
= \(\frac{1}{2}\) | (2 y – 4x + 2) | = \(\frac{1}{2}\) × 2 (2x – y – 1) = (2x – y – 1)
ପ୍ରଶ୍ନନୁସାରେ, 2x – y – 1=0 ⇒ 2x – y = 1
∴ 2x – y = 1 ହେଲେ (1, 1), (3, 5) ଓ (x, y) ବିନ୍ଦୁତ୍ରୟ ଏକ ସରଳରେଖାରେ ରହିବେ ।

Question 9.
ଏକ ଚତୁର୍ଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ (1, 0), (2, 4), (0, 5) ଓ (-2, 1) ହେଲେ, ତାହାର କ୍ଷେତ୍ରଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ଚତୁର୍ଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ (1, 0), (2, 4), (0, 5) 8 (-2, 1) ।
∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନାଙ୍କ A (1, 0), B (2, 4), C (0, 5) ।
ଏଠାରେ x₁ = 1, x₂ = 2, x₃ = 0, y₁ = 0, y₂ = 4, y₃ = 5

Question 10.
ଏକ ଚତୁର୍ଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ (- 2, 3), (3, -2), (7, 4) ଓ (1, 5) ହେଲେ, ତାହାର କ୍ଷେତ୍ରଫଳ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
ଏକ ଚତୁର୍ଭୁଜର ଶୀର୍ଷବିନ୍ଦୁମାନଙ୍କର ସ୍ଥାନାଙ୍କ A (-2, 3), B (3, -2), C (7, 4) ଓ D (1, 5) ।
AC କଣ୍ଠକୁ ଯୋଗକଲେ ଚତୁର୍ଭୁଜଟି ∆ ABC ଓ ∆ ACD ଦୁଇଟି ତ୍ରିଭୁଜରେ ପରିଣତ ହେବ ।
∆ ABCର ଶୀର୍ଷବିନ୍ଦୁ ତ୍ରୟର ସ୍ଥାନାଙ୍କ A(-2, 3), B (3, – 2) ଓ C (7, 4) ।
ଏଠାରେ x₁ = -2, x₂ = 3, x₂ = 7, y₁ = 3, y₂ = -2, y₃ = 4

∴ ABCD ଚତୁର୍ଭୁଜଟିର କ୍ଷେତ୍ରଫଳ = ∆ ABCର କ୍ଷେତ୍ରଫଳ + ∆ ACDର କ୍ଷେତ୍ରଫଳ
= 25 ବର୍ଗ ଏକକ + 7.5 ବର୍ଗ ଏକକ = 32.5 ବର୍ଗ ଏକକ ।

Question 11.
∆ ABCରେ A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (1, 1) ଓ AB, AC ବାହୁ ଦ୍ବୟର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ D (-1, – 2) ଓ E (3, 2) ହେଲେ B ଓ Cର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କରି ∆ ABCର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
∆ ABCର A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (1, 1) ।
AB ଓ ACର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ D (-1, -2) ଓ E (3, 2) ।
ମନେକର B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x₂, y₂) ଏବଂ C ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x₃, y₃) ।

Question 12.
(3, 0), (5, – 1) ଓ (p, p) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ହେଲେ pର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
A (3, 0), B (5, -1), C (p, p) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ।
∆ ABCର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) | 3 (- 1 – p) + 5 (p – 0) + p (0 + 1) | = 0 ହେବ।
⇒ \(\frac{1}{2}\) | – 3 – 3p + 5p + p | ⇒ \(\frac{1}{2}\) | 3p – 3 | = 0
⇒ \(\frac{3p-3}{2}\) = 0 ⇒ 3p = 3 ⇒ p = \(\frac{3}{3}\) =1
∴ pର ମୂଲ୍ୟ 1 ହେଲେ (3, 0), (5, -1), (p, p) ଏକରେଖ୍ୟ ହେବେ ।

Question 13.
(p, 2p), (3p, 3p) ଓ (3, 1) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ହେଲେ pର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
A (p, 2p), B (3p, 3p), C (3, 1) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ।
∆ ABCର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) | p(3p – 1) + 3p (1 – 2p) + 3 (2p – 3p) | = 0 ହେବ ।
⇒ \(\frac{1}{2}\) | 3p² – p + 3p – 6p² – 3p | ⇒ \(\frac{1}{2}\) |- 3p² -p | = 0
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{2}\) (3p² + p) = 0 ⇒ p (3p + 1) = 0 ⇒ p =0 ବା p = \(– \frac{1}{3}\)

Question 14.
(x, -1), (2, -1), (2, 1) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ହେଲେ xର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
A (x,-1), B (2, 1), C (2, 1) ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖ୍ୟ ।
A ABCର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) | x(- 1 – 1 ) + 2 (1 + 1) + 2 (- 1 + 1) | = 0 ହେବ ।
= \(\frac{1}{2}\) | -2x + 4 + 0 | = 0 ⇒ \(\frac{1}{2}\) | – (2x – 4)|= 0
⇒ \(\frac{1}{2}\) (2x – 4) = 0 ⇒ x – 2 =0 ∴ xର ମୂଲ୍ୟ 2 ।

Question 15.
(x, y) ବିନ୍ଦୁଟି (a, 0) ଓ (0, b) ବିନ୍ଦୁ ଦୁଇଗୋଟିର ସଂଯୋଗକାରୀ ସରଳରେଖା ଉପରେ ଅବସ୍ଥିତ ହେଲେ ପ୍ରମାଣ କର ଯେ, \(\frac{x}{a}+\frac{y}{b}=1\) ।
ସମାଧାନ :

ଦତ୍ତ ଚିତ୍ରରେ O ମୂଳବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, 0) ।
A ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (a, 0), B ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (0, b) ।
ଏଠାରେ OA = a ଏକକ, OB = b ଏକକ
P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ।
P ବିନ୍ଦୁରୁ y-ଅକ୍ଷ ପ୍ରତି ଲମ୍ବ PQ ।
∴ OQ = y ଏବଂ PQ = x
⇒ BQ = OB – OQ = b – y
∆ BOP ଓ A BOA ମଧ୍ୟରେ m∠BQP = m∠BOA (ସମକୋଣ)
∠A ଉଭୟ ତ୍ରିଭୁଜର ସାଧାରଣ କୋଣ ।
∴ ଅବଶିଷ୍ଟ m∠BPQ = m∠BAO ∴ ∆ BOP ~ ∆ BOA
⇒ \(\frac{PQ}{OA}=\frac{BQ}{OB}\) ⇒ \(\frac{x}{a}=\frac{b-y}{b}\) ⇒ \(\frac{x}{a}=\frac{ b}{b}-\frac{y}{b}\) ⇒ \(\frac{x}{a}+\frac{y}{b}=1\)

Question 16.
ପ୍ରମାଣ କର ଯେ, (a, b), (a, b) ଓ (a – a, b – b) ବିଦୁ୍ ତ୍ରୟ ଏକରେଖୀୟ ନୁହେଁ ।
ସମାଧାନ :
A (a, b), B (a’, b’) ଓ (a – a’, b – b’) ବିଦୁ୍ତ୍ରୟ ଏକ ରେଖୀୟ ହେବେ
ଯଦି A, B ଓ C ଶୀର୍ଷବିନ୍ଦୁ ବିଶିଷ୍ଟ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ () ହେବ ।
∴ ∆ ABCର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) | a (b’ – b + b’) + a’ (b – b’ – b) + (a – a’) (b – b’) |
= \(\frac{1}{2}\) | (ab’ – ab + ab’ – a’b’ + ab – ab’ – a′b + a’b’) |
= \(\frac{1}{2}\) | ab’ + ab’ – ab’ – ab + ab – ab’ + a’b’ – ab ] = \(\frac{1}{2}\) (ab’ – a’b)
∴ ଏଠାରେ ବିଦୁ୍ୟତ୍ରୟ ଏକରେଖୀୟ ନୁହଁନ୍ତି । କାରଣ, ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ ନୁହେଁ ।

Question 17.
A (p + 1, 1), B (2p + 1, 3) ଓ C (2p + 2, 2p) ବିନ୍ଦୁ ତ୍ରୟ ଏକ ରେଖୀୟ ହେଲେ pର ମାନ ସ୍ଥିର କର ।
ସମାଧାନ :
A (p + 1, 1), B (2p + 1, 3) ଓ C (2p + 2, 2p) ବିଦୁ୍ତ୍ରୟ ଏକରେଖୀୟ ।
∴ ∆ ABCର କ୍ଷେତ୍ରଫଳ = 0
⇒ \(\frac{1}{2}\) | (p + 1) (3 – 2p) + (2p + 1) (2p – 1) + (2p + 2) (1 – 3) | = 0
⇒ \(\frac{1}{2}\) | 13p + 3 – 2p² – 2p + 4p² – 1 – 4p – 4 | = 0
⇒ \(\frac{1}{2}\) | 2p² – 3p – 2 | = \(\frac{1}{2}\) (2p² – 3p – 2) = 0
⇒ \(\frac{1}{2}\) (2p² – 3p – 2) = 0
⇒ 2p² – 3p – 2 = 0 ⇒ 2p² – 4p + p – 2 = 0
⇒ 2p (p – 2) + 1 (p – 2) = 0 ⇒ (p – 2) (2p + 1) = 0
⇒ p – 2 = 0 କିମ୍ବା 2p + 1 = 0 ⇒ p = 2 କିମ୍ବା P = \(– \frac{1}{2}\)

Question 18.
(x, y), (3, 4) 3 (-5, -6) ବିନ୍ଦୁ ତ୍ରୟ ଏକ ରେଖୀୟ ହେଲେ ପ୍ରମାଣ କର ଯେ 5x – 4y+ 1 = 0 ।
ସମାଧାନ :
A (x, y), B(3, 4) 3 C (-5, -6) ବିଦୁ୍ତ୍ରୟ ଏକରେଖୀୟ ।
∆ ABCର କ୍ଷେତ୍ରଫଳ = \(\frac{1}{2}\) | x (4 + 6) + 3 (-6 – y) + (5) (y – 4) | = 0 ହେବ ।
⇒ \(\frac{1}{2}\) | 10 x – 18 – 3y – 5y + 20 | = 0 ⇒ \(\frac{1}{2}\) | 10x – 8y + 2 | = 0
⇒ \(\frac{1}{2}\) (10x – 8y + 2) = 0 ⇒ 5x – 4y + 1 = 0 (ପ୍ରମାଣିତ)

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text C: Psychobabble School Textbook Activity Questions and Answers.

Class 11th Alternative English Chapter 15 Psychobabble School Question Answers CHSE Odisha

Psychobabble School Class 11 Questions and Answers

Activity-10

a) Psychoanalysis is no longer used for curing mental diseases. (✓)
b) There is no end to an analysis. (✓)
c) Change in behavior ¡s only produced by self-knowledge. (✓)
d) Psychoanalysis is a waste of time. (✓)
e) Brief counseling is an honest form of talking cure. (✓)
f) Only doctors can become analysts. (✓)
g) Freud used psychoanalysis to cure a wide variety of psychological problems. (✓)

Activity-11

a)“ __________ it goes without saying that his research contributed enormously to our understanding of the subconscious.” (approval).
b) “But the analysis was then adopted for all sorts of psychological problems to which it was entirely insulted” (disapproval)
c) “ __________ if your problem is morbid introspection then the worst thing you can do is to spend hours talking about yourself. (disapproval)
d) “You create new problems for yourself as fast as you solve them, and the phony sense of progress is one of the things that makes it so addictive.” (disapproved)
e) “And all you get rid of ¡s the fee for another two years of treatment.” (approval)
f) In America it was finally the health insurance companies who called a halt to this madness. (approval)
g) “ __________ it involves a maximum of 25 sessions and sometimes just one.” (disapproval).

Activity-12

a) Which of the following in your opinion is the main purpose of the article?
(i) to describe a typical analysis session.
(ii) to amuse the reader.
(iii) to shock the reader.
(iv) to criticize psychoanalysis.
(v) to convince the reader that psychoanalysis is a waste of memory.
Answer:
(v) to convince the reader that psychoanalysis is a waste of memory.

b) How would you describe the writer’s attitude towards
(i) psychoanalysis,
(ii) Brief counseling.
Disapproving       Admiring
Approving            Indifferent
Contemptuous     Prejudiced
Uncompromising.
Answer:
Critical.

Activity-13
Cohesive Devices: Link Words

In Unit III you have looked at reference aa a device that binds the sentences of a text together. There are hints at the use of discourse markers as the author’s important device of text cohesion. Discourse markers (also called indicators in discourse) are easily recognized. “Signposts indicate how the writer has organized the text and what’s” he intends to say. They include link words such as ‘however, although furthermore but, newly’. They also include expressions such as “the second fact is ‘that’, which shows that the writer is introducing a second point in his discourse. In the following text, some link words are missing. Put in the link words from the

Instead of      When         But         Then         Yet        However      That’s how.

Television was invented by John Logie Baird. When he was young he built an airplane. He tried to fly in it, But it crashed down below. Baird was fortunate not to be killed. __________ he was older, he became a businessman. __________ his business failed, __________ he thought of working at television. His family advised him not to do it. He did not listen to them. __________ he rented an attic and brought the apparatus he needed. He started working. One day, he saw a picture on his screen. He rushed out to get someone he could ‘televise’. He found an office boy and took him back to the office. __________ no image of the boy appeared on the screen. The boy terrified, had put his head down. He put it up again. His picture appeared on the screen. __________ television had been invented.
Answer:
Television was invented by John Logic Baird. When he was young he built an airplane. He tried to fly in it. But it crashed down below. Baird was fortunate not to be killed. When he was older, he became a businessman. But his business failed, Then he thought of working at television. His family advised him not to do it instead. He did not listen to them. However, he rented an attic and bought the apparatus he needed. He started working. One day, he saw a picture on his screen. He rushed out to get someone he could ‘televise’. He found an office boy and took him back to the office.  Yet no image of the boy appeared on the screen. The boy terrified, had put his head down. He put it up again. His picture appeared on the screen. That’s how television had been invented.

Section – C
Introduction:
In this section, you will have the pleasure of reading an interesting article, ‘Psychobabble’. For the present, however, read-only its opening paragraph and guess what the article is about.

Psychobabble Summary in English

An unhappy man lies on a sofa. He is allowed to ramble on for an hour about thinking about something. He thinks about how an amount of $35 will be $30,000 after four or five years. It is actually, very odd. But people have been falling for it for a century. Freud invented Psychoanalysis in 1895. His research contributed enormously to the understanding of the subconscious. But it is obvious whether this analysis has any place in modern medical treatment.

Fraud and his co-worker’s ‘The Talking Cure’ was designed specifically to uncover the cause of hysterical symptoms and had a few successes. George Gershwin who was psychoanalyzed by doctors died, at the age of 39. Psychoanalysis was also administered as a cure for Schizophrenia and mental deficiency on which there was no effect at all. Woody Allen, a Western intellectual who is himself living proof that you can be analyzed until you are semicomatose and still end up with your personal life.

They believed that understanding will produce change which is highly doubtful. Any drunk driver who gets pulled over may well understand that he has behaved irresponsibly. But this understanding does not reduce the pleasure of drinking. It is considered a bad form to talk about what .you will achieve. The other thing that hooks people in the analysis is the phenomenon of transference. Psychoanalysts who expect and even encourage this will tell you it’s how the patient ultimately gets rid of those feelings.

In America, it was finally the Health Insurance Companies who called a halt to all this madness. The analysts were forced to admit that treatment was open-ended and the benefits uncertain. The dominant psychological problems are identified right from the start and a time limit is set for sorting them out. Learning from experiences is encouraged and strategies are worked out that will stop one from repeating self-destructive behavior. Most of our problems arise from making the same stupid mistake again and again.

Analytical outlines of the text:

• An unhappy man lies on a sofa.
• He is allowed to think for one hour.
• He thinks about how an amount of $35 will be $30,000 after four or five years.
• It is, actually, very odd.
• But people have been falling for it for a century.
• Freud invented psychoanalysis in 1895.
• His research helps to understand the subconscious.
• But it is obvious whether this analysis has any place in modern medical treatment.
• Freud and his coworkers produced “The talking cure”.
• It was specially designed to uncover the cause of hysterical symptoms.
• It had, however, a few successes.
• George Gershwin was psychoanalyzed by doctors.
• But he died at the age of 39.
• Psychoanalysis was administered as a cure for Schizophrenia.
• It is also applicable to mental deficiency.

• Actually, there was no effect at all.
• Woody Allen, a Western intellectual is himself living proof.
• One can be analyzed until one is semi-comatose.
• One can analyze it till ends up with his personal life.
• They believe that understanding will produce change.
• Actually, it is highly doubtful.
• Any drunk driver can understand his irresponsible behavior.
• But this understanding does not reduce the pleasure of drinking.
• It is considered a bad form.
• The other thing that hooks people in the analysis is the phenomenon of transference.
• Psychoanalysts encourage the patients to get rid of these feelings ultimately.
• The health insurance companies in America have stopped all this madness.
• To analysts, treatment is open-ended.
• But benefits are uncertain to them.
• The dominant psychological problems are identified.
• Learning from experiences is encouraged.
• Strategies are worked out not to repeat self-destructive behavior.
• Most of our problems arise from making the same stupid mistake again and again.

Meaning of difficult words:

ramble – to travel, to wander, to trail, here talk about ceaselessly.
confused – perplexed, disordered, disturbed.
enormously – immensely, atrociously, greatly.
hysterical – excitement, morbidity, terrible mental excitement.
symptoms – signs, characteristics, and traits of a desire.
adopted – taken up, received, used, employed.

Schizophrenia – violent mental problem.
depression – pressing down, saddening, mental frustration.
diminish – lessen, reduce, decrease.
compensate – make amounts for, replace the loss with something.
diagnosis – finding out the cause of an ailment, and identification of disease by symptoms.
therapists – persons treating diseases in a certain way.

Read More:

• Text A: That Lean, Hungry Look Question Answer
• Text B: What is Art? Question Answer
• Text C: Psychobabble Question Answer
• Text D: The Case against Man Question Answer

Odisha State Board BSE Odisha 10th Class Maths Notes Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି will enable students to study smartly.

BSE Odisha Class 10 Maths Notes Algebra Chapter 6 ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତି

→ କାର୍ଟେଜୀୟ ସମତଳ ଓ କାର୍ଟେଜୀୟ ସ୍ଥାନାଙ୍କ (Cartesian Plane and Cartesian Co-ordinates) :

→ବିଷୟବସ୍ତୁର ରୂପରେଖ

• କାର୍ଟେଜୀୟ ସମତଳ ଓ କାର୍ଟେଜୀୟ ସ୍ଥାନାଙ୍କ
• ଦୁଇଟି ଦତ୍ତ ବିନ୍ଦୁ ମଧ୍ୟରେ ଦୂରତ ।
• ବିଭାଜନ ସୂତ୍ର
• ତ୍ରିଭୁକର କ୍ଷେତ୍ରଫଳ

→ ଯେକୌଣସି ବାସ୍ତବ ସଂଖ୍ୟା ଏକ ସରଳରେଖା ଉପରେ ଗୋଟିଏ ବିନ୍ଦୁଦ୍ୱାରା ସୂଚିତ ହୋଇପାରିବ ଏବଂ ବିପରୀତ କ୍ରମେ ସରଳରେଖାର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁ ଏକ ବାସ୍ତବ ସଂଖ୍ୟାଦ୍ବାରା ସୂଚିତ ହୋଇପାରିବ । କିନ୍ତୁ ଉକ୍ତ ସରଳରେଖାର ବାହାରେ, ସମତଳ ଉପରେ ଅବସ୍ଥିତ କୌଣସି ବିନ୍ଦୁକୁ ଗୋଟିଏ ସଂଖ୍ୟା ସାହାଯ୍ୟରେ ସୂଚିତ କରାନଯାଇ ଦୁଇଟି ବାସ୍ତବ ସଂଖ୍ୟାଦ୍ଵାରା ସୂଚିତ କରାଯାଏ ।

→ ଏଠାରେ ଆମେ ଦୁଇଗୋଟି ସଂଖ୍ୟାରେଖା \(\overleftrightarrow{\mathbf{X’OX}}\) ଓ \(\overleftrightarrow{\mathbf{Y’OY}}\) ନେବା ଯେପରିକି ସେମାନେ ସମକୋଣରେ ପରସ୍ପରକୁ ଠ ବିନ୍ଦୁରେ ଛେଦ କରିବେ । \(\overleftrightarrow{\mathbf{X’OX}}\) ଓ \(\overleftrightarrow{\mathbf{Y’OY}}\) ସଂଖ୍ୟାରେଖା ଦ୍ଵୟକୁ ଯଥାକ୍ରମେ x-ଅକ୍ଷ (x- axis) ଓ y-ଅକ୍ଷ (y-axis) କୁହାଯାଏ ଏବଂ O ବିନ୍ଦୁକୁ ମୂଳବିନ୍ଦୁ (Origin) କୁହାଯାଏ । ଯେହେତୁ ଏହି ସମତଳଟି ଦୁଇଟି ବାସ୍ତବ ସରଳରେଖା ଦ୍ୱାରା ସୂଚିତ ହୁଏ, ତେଣୁ ଏହାକୁ R × R ବା R² -ସମତଳ (R² – Plane) ମଧ୍ୟ କୁହାଯାଏ । ଏହି ଅକ୍ଷଦ୍ବୟ R’-ସମତଳକୁ ଚାରିଭାଗରେ ବିଭକ୍ତ କରେ । ପ୍ରତ୍ୟେକ ଭାଗକୁ ପାଦ (Quadrant) କୁହାଯାଏ । ପ୍ରଥା ଅନୁସାରେ XOY ପାଦକୁ ପ୍ରଥମ ପାଦ (First quadrant, Q₁), YOX’ କୁ ଦ୍ଵିତୀୟ ପାଦ (Second quadrant, Q₂), X′OY’ କୁ ଦ୍ଵିତୀୟ ପାଦ (Third quadrant, Q₃) ଓ Y’OXକୁ ଚତୁର୍ଥ ପାଦ (Fourth quadrant, (Q₄) କୁହାଯାଏ ।

→ ମନେକର କାଗଜର ଉପର ପୃଷ୍ଠତଳ ଆମର ଆଲୋଚ୍ୟ ସମତଳ ଓ ଏହି ସମତଳ ଉପରିସ୍ଥ ଏକ ବିନ୍ଦୁ P ର ଅବସ୍ଥିତି ନିରୂପଣ କରିବା । P ବିନ୍ଦୁରୁ x-ଅକ୍ଷ ପ୍ରତି PM ଲମ୍ବ ଓ y-ଅକ୍ଷ ପ୍ରତି PN ଲମ୍ବ ଅଙ୍କନ କର । ଯଦି x-ଅକ୍ଷର ଅବସ୍ଥିତି M ବିନ୍ଦୁ ବାସ୍ତବ ସଂଖ୍ୟା xକୁ ସୂଚାଏ ଏବଂ y-ଅକ୍ଷରେ ଅବସ୍ଥିତି N ବିନ୍ଦୁ ବାସ୍ତବ ସଂଖ୍ୟା yକୁ ସୂଚାଏ, ଅର୍ଥାତ୍ OM = NP = x ଏବଂ ON =MP = y, ତେବେ ଆମେ P ବିନ୍ଦୁକୁ ଦୁଇଟି କ୍ରମିତ ଯୋଡ଼ି (ordered pair) (x, y) ଦ୍ଵାରା ସୂଚିତ କରିପାରିବା ଏବଂ ଲେଖୁଲାବେଳେ ଆମେ ଏହାକୁ P(x, y) ଦ୍ଵାରାପ୍ରକାଶ କରିବା ।

→ ବାସ୍ତବ ସଂଖ୍ୟା xକୁ P ବିନ୍ଦୁର x-ସ୍ଥାନାଙ୍କ (x-co-ordinate) ବା ଭୁଜ (abscissa) ଏବଂ ବାସ୍ତବ ସଂଖ୍ୟା yକୁ P ବିନ୍ଦୁର y ସ୍ଥାନାଙ୍କ (y-coordinate) ବା କୋଟି (ordinate) ବୋଲି କୁହାଯାଏ ।

→ P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଦ୍ଵୟ ଗୋଟିଏ ନିର୍ଦ୍ଦିଷ୍ଟ କ୍ରମରେ (ପ୍ରଥମ x ଓ ପରେ y) ଆବଦ୍ଧ ହେଉଥ‌ିବାରୁ ଏହାକୁ ଏକ କ୍ର ମିତ ସଂଖ୍ୟାଯୋଡ଼ି (ordered pair) ବୋଲି କୁହାଯାଏ । ତେଣୁ କୌଣସି ବିନ୍ଦୁ ସ୍ଥାନାଙ୍କର ପ୍ରଥମ ସଂଖ୍ୟାଟି x ଓ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି y-ସ୍ଥାନାଙ୍କକୁ ବୁଝାଏ ।

ସର୍ବପ୍ରଥମେ ସ୍ଥାନାଙ୍କଦ୍ୱାରା ସୂଚିତ କରାଯାଇଥିବା ସ୍ଥାନାଙ୍କ ଜ୍ୟାମିତିର ଜନକ ଡେକାର୍ଟେଙ୍କ ନାମାନୁସାରେ P ବିନ୍ଦୁର ଏହି ସ୍ଥାନାଙ୍କକୁ କାର୍ଟେଜୀୟ ସ୍ଥାନାଙ୍କ (Cartesian co-ordinates) କୁହାଯାଏ । ବସ୍ତୁଟି ଯେଉଁ ସମତଳରେ ଅବସ୍ଥିତ, ସେ ସମତଳକୁ କାର୍ଟେଜୀୟ ସମତଳ (Cartesian Plane) କୁହାଯାଏ !

→ ସମତଳ :

• ସମତଳଟି ଅସଂଖ୍ୟ ବିନ୍ଦୁମାନଙ୍କର ସେଟ୍ । ସ୍ଥାନାଙ୍କ ସମତଳରେ ପ୍ରତ୍ୟକ ବିନ୍ଦୁପାଇଁ ଏକ କ୍ରମିତ ସଂଖ୍ୟାଯୋଡ଼ି ରହିଛି । ତେଣୁ ସେଟ୍ ଭାଷାରେ ସମତଳକୁ ଲେଖିଲେ-
  ସମତଳ = {(x, y) | x, y ∈ R}
• (x, y) ∈ R × R ବା R² । ତେଣୁ ଏହି ସ୍ଥାନାଙ୍କ ସମତଳକୁ R-ସମତଳ (R-Plane) ବା କାର୍ଟେଜୀୟ ସମତଳ (Cartesian Plane) ମଧ୍ୟ କୁହାଯାଏ ।
  A × B = {(a, b) | a ∈ A, ∈ B}
  B × A = {(b, a) | a ∈ A, ∈ B}
  ଯଦି A = B = R (ବାସ୍ତବ ସଂଖ୍ୟା ସେଟ୍) ତେବେ କାର୍ଟେଜୀୟ ଗୁଣଫଳ ସେଟ୍ R × R = {(x, y) | x, y = R} ଓ ଏହାକୁ R? ରୂପେ ମଧ୍ୟ ଲେଖାଯାଏ ।

→ \(\overleftrightarrow{\mathbf{X’OX}}\) ଅକ୍ଷର \(\overrightarrow{\mathrm{OX}}\) କୁ x-ଅକ୍ଷର ଧନଦିଗ, \(\overrightarrow{ OX’ } \) କୁ ଋଣଦିଗ କୁହାଯାଏ । ସେହିପରି \(\overrightarrow { OY } \) ଏବଂ \(\overrightarrow{ OY’ } \) କୁ \(\overleftrightarrow{\mathbf{Y’OY}}\) ଅକ୍ଷର ଯଥାକ୍ରମେ ଧନଦିଗ ଓ ଋଣଦିଗ ଭାବେ ନିଆଯାଇଥାଏ ।

→ପ୍ରଥମେ x-ଅକ୍ଷରେ ଅବସ୍ଥିତ ଯେକୌଣସି ଏକ ବିନ୍ଦୁ Mର ସ୍ଥାନାଙ୍କ ନିର୍ଣ୍ଣୟ କରାଯାଉ । ଏହାର x-ସ୍ଥାନାଙ୍କ x ଏବଂ y-ସ୍ଥାନାଙ୍କ ଶୂନ । କାରଣ x-ଅକ୍ଷଠାରୁ \(\overrightarrow{ OY } \) ବା \(\overrightarrow{ OY’ } \) ଦିଗରେ M ବିନ୍ଦୁର ଦୂରତା ଶୂନ ହୋଇଥିବାରୁ ଏହାର y-ସ୍ଥାନାଙ୍କ ଶୂନ ।

• ତେଣୁ x- ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ଯେକୌଣସି ବିଦୁର ସ୍ଥାନାଙ୍କ (x, 0) ଅର୍ଥାତ୍ x-ଅକ୍ଷର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର y-ସ୍ଥାନାଙ୍କ = 0 ।
• ସେହିପରି y- ଅକ୍ଷର ଯେକୌଣସି ବିଦୁର ସ୍ଥାନାଙ୍କ (0, y) ଅର୍ଥାତ୍ y-ଅକ୍ଷର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର x ସ୍ଥାନାଙ୍କ 0 ।
• ମୂଳବିନ୍ଦୁ ( ଉଭୟ ଅକ୍ଷର ପରସ୍ପର ଛେଦବିନ୍ଦୁରେ ଥିବାରୁ ଏହାର ସ୍ଥାନାଙ୍କ (0, 0) ଅଟେ ।
  • ପ୍ରଥମ ପାଦରେ ଅବସ୍ଥିତ ଯେକୌଣସି ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ପାଇଁ x > 0, y > 0 । ଅର୍ଥାତ୍‌x ଓ y ଉଭୟେ ଧନାତ୍ମକ ।
  • ଦ୍ଵିତୀୟ ପାଦରେ ଅବସ୍ଥିତ ଯେକୌଣସି ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ପାଇଁ x < 0, y > 0 ଅର୍ଥାତ୍ x ଋଣାତ୍ମକ ଓ y ଧନାତ୍ମକ ।
  • ତୃତୀୟ ପାଦରେ ଅବସ୍ଥିତ ଯେକୌଣସି ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ପାଇଁ x < 0, y < 0 । ଅର୍ଥାତ୍‌ x ଓ y ଉଭୟେ ଋଣାତ୍ମକ ।
  • ଚତୁର୍ଥ ପାଦରେ ଅବସ୍ଥିତ ଯେକୌଣସି ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) ପାଇଁ x > 0, y < 0 । ଅର୍ଥାତ୍‌ x ଧନାତ୍ମକ ଓ y ଋଣାତ୍ମକ ।
• ଅକ୍ଷଦ୍ବୟ ଉପରିସ୍ଥ ଯେକୌଣସି ବିନ୍ଦୁ କୌଣସି ପାଦରେ ଅନ୍ତର୍ଭୁକ୍ତ ନୁହଁନ୍ତି ।
• x- ଅକ୍ଷର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର y-ସ୍ଥାନାଙ୍କ (0) ହେତୁ x-ଅକ୍ଷର ସମୀକରଣ ହେଉଛି y = 0 ।
  ସେହିପରି y- ଅକ୍ଷର ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର x-ସ୍ଥାନାଙ୍କ ( ହେତୁ y-ଅକ୍ଷର ସମୀକରଣ ହେଉଛି x = 0 ।

→ ଦୁଇଟି ଦତ୍ତ ବିନ୍ଦୁ ମଧ୍ୟରେ ଦୂରତ | (Distance between two given points) :

• ଉପପାଦ୍ୟ 1 :
  
• ସ୍ଥାନାଙ୍କ ସମତଳରେ P₁(x₁, y₁) ଓ P₂(x₂, y₂) ଦୁଇଟି ଦତ୍ତ ବିନ୍ଦୁ ହେଲେ, ସେମାନଙ୍କ ଦୂରତା
  P₁P₂ = \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) ବା \(\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\)
• ଦତ୍ତ : ସ୍ଥାନାଙ୍କ ସମତଳରେ P₁(x₁, y₁) ଏବଂ P₂(x₂, y₂) ଦୁଇଟି ବିନ୍ଦୁ ।
• ପ୍ରାମାଣ୍ୟ : P₁P₂ = \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
• ଅଙ୍କନ : ଏକ ସମତଳରେ P₁ ଓ P₂ ଦୁଇଟି ବିନ୍ଦୁ । ସେମାନଙ୍କର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (x₁, y₁) ଓ (x₂, y₂) । P₁P₂ ରେଖାଖଣ୍ଡ ଅଙ୍କନ କର ।
  P₁ ଓ P₂ ବିଦୁ୍ୟଦ୍ବୟରୁ x-ଅକ୍ଷ ପ୍ରତି ଯଥାକ୍ରମେ \(\overline{{P}_1 {M}_1}\) ଓ \(\overline{{P}_2 {M}_2}\) ଲମ୍ବ ଅଙ୍କନ କର । ପୁନଶ୍ଚ, P₁ ବିନ୍ଦୁରୁ P₂M₂ ପ୍ରତି x-ଅକ୍ଷ ସହ ସମାନ୍ତର କରି PAR ରେଖାଖଣ୍ଡ ଅଙ୍କନ କର ।
• ପ୍ରମାଣ : OM₁ = x₁, OM₂ = x₂, M₁P₁ = y₁ ଓ M₂P₂ = y₂
  ତେଣୁ P₁R = M₁M₂ = OM₂ – OM₁ = x₂ – x₁
  ଏବଂ RP₂ = M₂P₂ – M₂R = M₂P₂ – M₁P₁ = y₂ – y₁
  ଯେହେତୁ ∆ P₁RP₂ ରେ m∠P₁RP₂ = 90°
  ତେଣୁ ପିଥାଗୋରାସ୍ ଉପପାଦ୍ୟ ଅନୁସାରେ (P₁P₂)² = (P₁R)² + (RP₂)²= (x₂ – x₁)² + (y₂ – y₁)²
  ଯେହେତୁ ଦୂରତା ଏକ ଧନାତ୍ମକ ସଂଖ୍ୟା ।
  ତେଣୁ P₁P₂ = \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) ଅଥବା \(\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\)
  

→ ଅନୁସିଦ୍ଧାନ୍ତ 1 : ମୂଳବିନ୍ଦୁ O(0, 0)ରୁ ଯେକୌଣସି ବିନ୍ଦୁ P(x, y) ର ଦୂରତା OP = \(\sqrt{x^2 + y^2}\) ହେବ ।
→ ଅନୁସିଦ୍ଧାନ୍ତ 2 : P₁, P₂ ବିନ୍ଦୁଦ୍ଵୟ x-ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ହେଲେ P₁P₂ = | x₂ – x₁ | ଓ y-ଅକ୍ଷ ଉପରେ ଅବସ୍ଥିତ ହେଲେ, P₁P₂ = | y₂ – y₁ | ହେବ ।

→ ବିଭାଜନ ସୂତ୍ର (Division Formula) :
ସଂଜ୍ଞା : ଅନ୍ତର୍ବିଭାଜନ :
ଯଦି A-P-B ହୁଏ, ଅର୍ଥାତ୍ AB ଉପରେ A ଓ B ବିନ୍ଦୁଦ୍ଵୟର ମଧ୍ୟବର୍ତୀ P ବିନ୍ଦୁ ହୁଏ, ତେବେ AB ରେଖାଖଣ୍ଡ P ବିନ୍ଦୁରେ \(\overline{\mathrm{AP}}\) ଓ \(\overline{\mathrm{PB}}\) ରେଖାଖଣ୍ଡରେ ଅନ୍ତର୍ବିଭକ୍ତ ହୁଏ ।
ଏ କ୍ଷେତ୍ରରେ AP + PB = AB ହୁଏ ଓ ଅନ୍ତର୍ବିଭକ୍ତ ହୋଇଥିବା ଦୁଇ ରେଖାଖଣ୍ଡର ଦୈର୍ଘ୍ୟର ଅନୁପାତ AP : PB ।
ଯଦି P ବିନ୍ଦୁ AB ରେଖାଖଣ୍ଡକୁ m : n ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରେ, ଆମେ ଲେଖୁବା ଯେ, \(\frac{PA}{PB}=\frac{m}{n}\) ।
କିନ୍ତୁ P ବିନ୍ଦୁ BA ରେଖାଖଣ୍ଡକୁ r : s ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କଲେ, ଆମେ ଲେଖୁବା ଯେ, \(\frac{PB}{PA}=\frac{r}{s}\) ।

ଉପପାଦ୍ୟ 2 :
A (x,, y) ଓ B(x,, y,) ବିନ୍ଦୁଦ୍ଵୟକୁ ଯୋଗ କରୁଥିବା ରେଖାଖଣ୍ଡ AB, ଯଦି P(x, y) ବିନ୍ଦୁଦ୍ୱାରା m : n ଅନୁପାତରେ ଅନ୍ତର୍ବିଭାଜିତ ହୁଏ, ତେବେ P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ \(\frac{{mx}_2+{nx_1}}{m+n}, \frac{{my}_2+{ny_1}}{m+n}\) ହେବ ।
ଦତ୍ତ : ସ୍ଥାନାଙ୍କ ସମତଳରେ AB ରେଖାଖଣ୍ଡ ଉପରିସ୍ଥ P ଏପରି ଏକ ବିନ୍ଦୁ ଯେପରି \(\frac{AP}{BP}=\frac{m}{n}\) ।

ମନେକର A, B ଓ P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ ଯଥାକ୍ରମେ (x₁, y₁), (x₂, y₂) ଓ (x, y) ।
ପ୍ରାମାଣ୍ୟ : P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) = (\(\frac{{mx}_2+{nx_1}}{m+n}, \frac{{my}_2+{ny_1}}{m+n}\))

→ ଅଙ୍କନ : A, P ଓ Bରୁ x-ଅକ୍ଷପ୍ରତି ଯଥାକ୍ରମେ AC, PM ଏବଂ BD ଲମ୍ବ ଏବଂ AS ⊥ PM, PT ⊥ BD ଅଙ୍କନ କର ।
ପ୍ରମାଣ : A ASP ଏବଂ A PTB ତ୍ରିଭୁଜଦ୍ଵୟ ମଧ୍ଯରେ m∠PAS = m∠BPT = 90°
m∠PAS = m∠BPT (ଅନୁରୂପ)
∴ ∆ ASP ଓ ∆ PTB ଦ୍ଵୟ ସଦୃଶ, ଅର୍ଥାତ୍ ∆ ASP – ∆ PTB
ତେଣୁ \(\frac{\mathrm{AS}}{\mathrm{PT}}=\frac{\mathrm{PS}}{\mathrm{BT}}=\frac{\mathrm{PA}}{\mathrm{PB}}=\frac{\mathrm{m}}{\mathrm{n}}\) ଅର୍ଥାତ୍ \(\frac{\mathrm{AS}}{\mathrm{PT}}=\frac{\mathrm{m}}{\mathrm{n}}\) ଏବଂ \(\frac{\mathrm{PS}}{\mathrm{BT}}=\frac{\mathrm{m}}{\mathrm{n}}\)
ମାତ୍ର AS = CM = x – x₁, PT = MD = x₂ – x
ଏବଂ PS = PM – SM = PM – AC = y – y₁
BT = BD – TD = TD – PM = y₂ – y
\(\frac{\mathrm{AS}}{\mathrm{PT}}=\frac{x-x_1}{x_2-x}=\frac{m}{n}\) ⇒ mx₂ – mx = nx – nx₁ ⇒ mx₂ + nx₁ = mx + nx
⇒ x (m + n) = mx₂ + nx₁, PX = \(\frac{mx_2+nx_1}{m+n}\)
ସେହିପରି, \(\frac{\mathrm{AS}}{\mathrm{PT}}=\frac{y-y_1}{y_2-y}=\frac{m}{n}\) ⇒ my₂ – my = ny – ny₁
⇒ my₂ + ny₁ = my + ny ⇒ y (m + n) = my₂ + ny₁ ⇒ y = \(\frac{my_2+ny_1}{m+n}\)

→ ତେଣୁ A(x₁, y₁) ଓ B(x₂, y₂) ବିନ୍ଦୁଦ୍ଵୟର ସଂଯୋଗକାରୀ ରେଖାଖଣ୍ଡ ABକୁ m : n ଅନୁପାତରେ ଅନ୍ତର୍ବିଭକ୍ତ କରୁଥ‌ିବା ବିନ୍ଦୁ P ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ (x, y) = \(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\) ଅଟେ ।

• ଯଦି A – B – P ହୁଏ, ଅର୍ଥାତ୍‌ \(\overrightarrow{AB}\) ଉପରିସ୍ଥ P ଏକ ବିନ୍ଦୁ ହୁଏ, ତେବେ \(\overline{\mathrm{AB}}\), P ବିନ୍ଦୁଦ୍ଵାରା AP ଓ BP ରେଖାଖଣ୍ଡରେ ବହିର୍ବିଭକ୍ତ ହୋଇଛି ବୋଲି କୁହାଯାଏ ।
  
• ଏଠାରେ ବହିର୍ବିଭାଜନର ଅନୁପାତ AP : BP ହେବ ଓ AP – PB = AB ହେବ ।
• \(\frac{AB}{BP}\) < 1 ହେଲେ P – A – B ଏବଂ \(\frac{AB}{BP}\) > 1 ହେଲେ A – B – P ହେବ ।
• A(x₁, y₁) ଓ B(x₂, y₂) ବିଦୁ୍ୟଦ୍ୱୟକୁ ଯୋଗକରୁଥିବା ରେଖାଖଣ୍ଡ AB, ଯଦି P(x, y) ଦ୍ଵାରା m : n ଅନୁପାତରେ ବହିର୍ବିଭାଜିତ ହୁଏ, ତେବେ P(x, y) ବିନ୍ଦୁର ସ୍ଥାନାଙ୍କ \(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}\) ହେବ ।
• ଯଦି P ବିଦୁଟି \(\overline{\mathrm{AB}}\) ରେଖାଖଣ୍ଡର ମଧ୍ୟବିନ୍ଦୁ ହୁଏ, ସେ କ୍ଷେତ୍ରରେ m = n ହୁଏ ଏବଂ ମଧ୍ୟବିନ୍ଦୁ Pର ସ୍ଥାନାଙ୍କ (x, y) = \(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\) ହୁଏ ।

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text A: That Lean, Hungry Look School Textbook Activity Questions and Answers.

Class 11th Alternative English Chapter 13 That Lean, Hungry Look School Question Answers CHSE Odisha

That Lean, Hungry Look School Class 11 Questions and Answers

Activity – 1
Purpose And Attitude And The Text Type

Question 1.
Which of the following described the writer’s attitude to thin people?
(i) impressed
(ii) complementary
(iii) disapproving
(iv) condemning
(v) approving
(vi) noncommittal
Answer:
(iv) condemning.

Question 2.
Which of the following phrases best expresses the writer’s purpose?
(i) to present objective information
(ii) to present both sides of a controversial issue
(iii) to shock the reader with an unconventional point of view
(iv) to persuade the reader that fat people are better than thin people.
(v) to express his dislike of thin people.
Answer:
(iv) to persuade the reader that fat people are better than people.

Question 3.
Which of the following categories of text type would you say the article belongs to?
(i) informative
(ii) imaginative
(iii) expressing an opinion
(iv) descriptive
(v) narrative
Answer:
(iii) expressing an opinion.

Question 4.
What is the general tone of the article?
(i) ironic
(ii) humorous
(iii) matter of fact
(iv) Passionate
(v) serious
Answer:
(iii) matter of fact.

Activity-3
Getting The Main Ideas Of Paragraphs

Match the paragraph in column A with the titles in column B and then say whether a title refers to thin people or fat people.

Answer:

A	B
1.	(vi) dangerous people.
2.	(v) no absolute truth.
3.	(ix) speedy metabolism
4	(vii) relaxed and fun-loving
5.	(xi) seeing all sides
6.	(xii) life is illogical and unfair.
7.	(i) a long list of logical things
8.	(viii) happiness is elusive.
9.	(iv) muddling through rather than saving time.
10.	(x) not enough time for work.
11.	(iii) love of math and morality.
12.	(ii) loving and accepting

Activity 4
Understanding Patterns Of Comparison And Contrast

There are two important ways of developing a comparison and contrast text, namely (i) the block method and the point-by-point method. In the Block method, you single out one basic way in which the two objects are alike or different. For example, if you are comparing two people at work, the introductory paragraph would tell the reader what your article would be about. The first body paragraph of the article would show something about one person’s approach to work, the following body paragraph would focus on the other person’s approach.

And in the concluding paragraph, you would briefly summarise the topic and give a dominant impression about the similarities and/or differences in the two worker’s approaches to their jobs. However, instead of deciding to compare and contrast the two objects one after another, you may decide not to separate the two objects you are discussing. You may then adopt the point-by-point method and treat’ both objects together as you present each point of comparison or contrast. You may have discovered that both of these methods have been employed in Text A.

a) Which patterns of comparison and Contrast does the writer use in paragraphs 2-5 and paragraphs 8-14?
Answer:
Point-by-point method.

b) Which pattern does she use in paragraphs 6,7,11 and 12?
Answer:
Block method.

c) Which of these two patterns do you find more effective and why?
Answer:
Both these patterns are effective in dealing with a problem. However, the point-by-point pattern is more effective because the comparison and contrast will be clear in the treatment and approach in this pattern.

d) Does the writer state the thesis explicitly? If so, where does she state it?
Answer:
The writer states the thesis of this work of art explicitly. It appears in paragraph -1.

e) How does the conclusion support the thesis? Write a few words on the appropriateness or otherwise of the conclusion.
Answer:
The conclusion almost sums up the nature and pursuit of the thin and fat people described in the previous paragraphs. The concluding paragraph is eloquent of the strong contrasts between fat and thin people.

Section – A
The paragraph below is the beginning paragraph of Text-A. Read it quickly and try to guess what Text-A is about.

Ceasar was right. Thin people need watching. I have been watching them for most of my adult life and 1 do not like what I see. When these narrow fellows spring at me, quiver to my toes. Thin people come in all personalities, most of them menacing. You have got your ‘together’ thin person, your mechanical thin person. Your condescending thin person, your efficiency expert thin person. All of them are dangerous.
Now read Text-A, which is adopt?  from an article in news week in the year 197. in order to check whether your prediction made above is right.

That Lean, Hungry Look Summary in English

According to Ceasar, thin people need watching. The writer has been watching such people for most of his adult life and never likes what he sees. He says when these thin fellows spring at him he trembles to his toes. Thin people come in all personalities and most if they are dangerous. Thin people in the first place are not fun. They have always got to be going something. They make others tired. They get speedily little metabolisms that cause them to burtle briskly. Sluggish, inert, easy-going fat people are preferable to thin ones.

Fat people don’t chattel all day long. Thin people turn mean and hard at a young age because they never learn the value of a hot fudge Sunday for easing tension. They are firm and fresh and dull like carrots. Thin people believe in logic, fat people see all sides. Fat people realize that life is illogical and unfair. They know well that God is not in heaven and all is not right with the world. If God was up there, fat people could leave two doughnuts and a big orange drink the time they wanted it.

Thin people have a long list of logical things they are always spouting off to me. They hold up one finger at a time as they reel off these things. They speak slowly as if to a young child. The list is long and full of holes. They also think these 2,000-point plans lead to happiness. Fat people know happiness is elusive at best and even if ey could get the kind of thin people to talk about, they wouldn’t want it. Fat people see that such programs are too dull, too hard, and too off the mark. They are never better than a whole cheesecake.

However, fat people know all about the mystery of life. They get acquainted with the night, luck, and fate, and play them by ears. The main problem with people is that they oppress. Thin people are downers. They like math and morality and reasoned evaluation of the limitations of human beings. They expound prognosis, probe, and prick. Fat people are friendly and cheerful. Fat people will talk continuously, trade quickly, laugh loudly, gyrate, and gossip. They are generous, giving, and gallant. They are gluttonous, goodly, and great.

Analytical outlines:

• According to Ceasar, thin people need to be watched minutely.
• He has been watching such people for most of his adult life.
• He calls them as narrow fellows.
• When they spring at him, he trembles to his toes.
• They appear in all personalities.
• Most of them are dangerous.
• Thin people in the first place are not having fun.
• They have always got to be doing something.
• Give them a coffee break.
• They will job around the block.
• They make others tired.
• They have got a speedily little metabolism.
• It makes them to burtle briskly.

• They have forever been rubbing their bony hands together.
• They have also been eying new problems to tackle.
• But the fat people are sluggish, inert, easy going.
• So, they are preferable to the thin ones.
• Fat people don’t chartle all day long.
• Thin people turn mean and hard at a young age.
• Because they never learn the value of a hot fudge Sunday for casing tension.
• They are firm and fresh and dull like carrots.
• They go straight to the heart of the matter.
• But fat people let things stay all blurry, hazy, and vague.
• They want to race the truth.
• Fat people know there is no truth.
• Thin people believe in logic.
• Fat people see all sides.
• Fat people realize that life is illogical and unfair.
• They know very well that God is not in heaven.
• They consider all is not right with the world.
• If God was up there, fat people could have two doughnuts and a big orange drink any time they wanted it.
• Thin people have a long list of logical things.
• They are always spouting off to him.
• They hold up one finger. at a time as they reel off these things.
• They speak slowly as if to a young child.
• Their list is long and full of holes.
• They think about 2000-point plans.
• They think it must lead them to happiness.
• Fat people know happiness is elusive at best.

• They don’t want as the thin people talk about it.
• To fat people, such programs are too dull, too hard, and too off the mark.
• They are never better than a whole cheesecake.
• Fat people know all about the mystery of life.
• They get acquainted with might, luck, and fate, and playing them by ears.
• But the main problem with the thin men is that they oppress.
• Their good intentions, bony torsos, tight, ships, neat corners, cerebral machinations, and pet solutions loom like dark clouds over the
• loose, comfortable, spread out, soft world of the fat.
• Thin people are downers.
• They like math and morality.
• They also like reasoned evaluation of the limitations of human beings.
• They have their skinny little acts together.
• They expound prognoses, probes, and prick.
• Fat people are convivial that is jovial.
• They even like irregular people.
• They will come up with a good reason.
• Fat people are generous, giving, and gallant.
• They are also gluttonous, goodly, and great.
• They are friendly and cheerful.
• Fat people will gab, giggle, guffa, gyrate, and gossip.
• They have plenty of room to be free and frank.

Meaning of difficult words:

to goof off – to make a trivial mistake.
to burtle – to move around quickly.
sluggish – moving or reacting more slowly than normal.
chartling – bulging out of amusement.
wizened – small and thin and wrinkled.
shrivel led – dried up and bent, became small.
gooey – sticky, soft, and sweet.
not-fudge sonde- a hot and soft creamy light
brown sweet dish made from the ice-cream, fruits, and nuts.
crunchy – firm and fresh.
nebulous – not clear or exact, fainted.
doughnuts – small and cakes.
elusive – difficult to achieve.
muscled – covered the ground with

decaying leaves to improve its quality.
double-fudged – with two layers of chocolate or cream dressing.
cerebral machination – secret and clever plans made by the brain.
rutabagas – a king of roots.
punch line – the last few words of a joke or story.
dovners – a person who stops your
feeling cheerful or happy.
convivial – friendly and cheerful.
gab – talk continuously.
guffaw – laugh loudly.
gyrate – turn around fast in circles.
giggle – moving from side to side with quick short movements.

Read More:

• Text A: That Lean, Hungry Look Question Answer
• Text B: What is Art? Question Answer
• Text C: Psychobabble Question Answer
• Text D: The Case against Man Question Answer

Odisha State Board CHSE Odisha Class 12 Invitation to English 1 Solutions Chapter 5 Development of Polio Vaccines Textbook Exercise Questions and Answers.

Class 12th Invitation English Chapter 5 Development of Polio Vaccines Question Answers CHSE Odisha

Development of Polio Vaccines Class 12 Questions and Answers

Unitwise Gist and Glossary

UNIT -I
Gist:
The writer takes us back to 1921 when America became a victim of polio. Franklin D. Roosevelt, a young politician, was not an exception. A tiring day made him take a cold The writer takes us back to 1921 when America became a victim of polio. Franklin D. Roosevelt, a young politician, was not an exception. A tiring day made him take a cold swim and lie leisurely in his wet swimsuit at home. He went to bed feeling as if he had caught a cold.

In a few days, he came to know that he was suffering from polio. The early symptoms of poor attack are headache, nausea, vomiting, and fever. Air, food, and water are the carriers of poliovirus. One percentage of polio-infected people actually suffer from severe polio. There are two forms of polio: spinal and bulbar; the former affects the limbs and the latter lung. Muscle pain, and stiff neck and back characterize both forms. The symptoms of both forms of polio are similar.
ସାରମର୍ମ :
ଲେଖକ ଆମକୁ ୧୯୨୧ ମସିହାର ପୃଷ୍ଠଭୂମିକୁ ଫେରାଇ ନେଇଛନ୍ତି ଯେତେବେଳେ ଆମେରିକା ପୋଲି ଓ ବ୍ୟାଧର ଶିକାର ହୋଇଥିଲା । ଯୁବ ରାଜନୀତିଜ୍ଞ Franklin D. Roosevelt ଏହାର ବ୍ୟତିକ୍ରମ ନ ଥିଲେ । ଏକ କ୍ଲାନ୍ତ ଦିବସରେ ସେ ଥଣ୍ଡା ପାଣିରେ ପହଁରିଥିଲେ ଓ ଓଦା ସନ୍ତରଣ-ପୋଷାକ ପରିଧାନ କରି ଘରେ ଶୋଇ ରହିଥିଲେ । ସେ ଯେତେବେଳେ ଶୋଇବାକୁ ଗଲେ, ସେ ଅନୁଭବ କଲେ ଯେପରିକି ତାଙ୍କୁ ଥଣ୍ଡା ହୋଇଛି । କିଛି ଦିନ ଭିତରେ ସେ ପୋଲିଓ ରୋଗରେ ପୀଡ଼ିତ ହୋଇଥିବାର ଜାଣିବାକୁ ପାଇଲେ । ପୋଲିଓ ସଂକ୍ରମଣର ପ୍ରାକ୍-ଲକ୍ଷଣଗୁଡ଼ିକ ହେଲା – ମୁଣ୍ଡବ୍ୟଥା, ଅସ୍ୱସ୍ଥି ଲାଗିବା, ବାନ୍ତି ଏବଂ ଜ୍ଵର । ବାୟୁ, ଖାଦ୍ୟ ଏବଂ ଜଳ ହେଉଛି ପୋଲିଓ ଭୂତାଣୁର ବାହକ । ପୋଲିଓ ଆକ୍ରାନ୍ତ ବ୍ୟକ୍ତିମାନଙ୍କ ମଧ୍ୟରୁ ବାସ୍ତବରେ ୧ ପ୍ରତିଶତ ମାରାତ୍ମକ ପୋଲିଓରେ ପୀଡ଼ିତ ହୋଇଥା’ନ୍ତି । ପୋଲିଓ ଦୁଇପ୍ରକାର; ଯଥା – spinal (ମେରୁଦଣ୍ଡ ସମ୍ବନ୍ଧୀୟ) ଓ bulbar (ଫୁସ୍‌ଫୁସ୍ ସମ୍ବନ୍ଧୀୟ) । ପ୍ରଥମଟି ଅଙ୍ଗପ୍ରତ୍ୟଙ୍ଗକୁ ଏବଂ ଦ୍ବିତୀୟଟି ଫୁସ୍‌ଫୁସ୍‌ ଆକ୍ରାନ୍ତ କରିଥାଏ । ମାଂସପେଶୀ ଯନ୍ତ୍ରଣା, ବେକ ଓ ପିଠି ଲାଠି ହୋଇଯିବା ଉଭୟ ପ୍ରକାର ପୋଲିଓର ବୈଶିଷ୍ଟ୍ୟ । ଉଭୟ ପ୍ରକାର ପୋଲିଓର ଲକ୍ଷଣ ସମାନ ।

Glossary:
outbreak : sudden occurrence (ବ୍ୟାପିବା)
plagued : caused trouble for a period of time (କିଛି ସମୟ ପାଇଁ ଅସୁବିଧା ସୃଷ୍ଟି କରିଥିଲା )
poliomyelitis : polio virus (ପୋଲି ଓ ଭୂତାଣୁ )
taking a cold swim : ଥଣ୍ଡା ପାଣିରେ ପହଁରିବା
relief : ଆରାମ
swimsuit : ସନ୍ତରଣ ପୋଷାକ
strike : deliver a blow (ଆଘାତ ଦେବା )
vinis : ଭୂତାଣୁ
incubate : hatch (ଅଣ୍ଡା ଫୁଟାଇବା )
a symptomatic: In medicine, a disease is considered asymptomatic if a patient is a carrier for a disease or infection but experiences no symptoms.
symptoms :
stage : condition (ଅବସ୍ଥା )
infected: ସଂକ୍ରମିତ
droplets: a very small drop of a liquid (ବୁନ୍ଦା)
antibodies : antibodies (1g) are found in blood or other bodily fluids of vertebrates, and are used by the immune system to identify and neutralize foreign objects, such as
bacteria and viruses.
unlucky : ଭାଗ୍ୟହୀନ
permanently : ସବୁଦିନ ପାଇଁ
paralyzed : ଅଚଳ ହୋଇଗଲା
spinal cord : ମେରୁଦଣ୍ଡସ୍ଥ ସ୍ନାୟୁ ସଂସ୍ଥାନ
multiplies : increases (ବଢ଼ିଯାଏ)
nerves : ସ୍ନାୟୁ
affected : ପ୍ରଭାବିତ ହେଲା
pain : ଯନ୍ତ୍ରଣା
stiff : unable to move easily (ଅଚଳ)
lungs : ଶ୍ଵାସନଳୀ
breathe : ନିଃଶ୍ଵାସ ନେବା
physical therapy: ଶାରୀରିକ ଚିକିତ୍ସା
recover : cure (ଭଲ ହୋଇଯିବା)
varies : changes (ପରିବର୍ତ୍ତନ ହୁଏ)

Think it:
Question 1.
What is poliomyelitis ?
Answer:
Poliomyelitis means poliovirus. It causes an acute, viral, infectious disease that spreads from person to person, basically via the fecal-oral route.

Question 2.
When did Roosevelt find out that he had polio?
Answer:
On a tiring day, Roosevelt took a cold swim and lie leisurely in his wet swimsuit at home. When he went to bed, he felt as if he had caught a cold. After a few days, he found out that he had polio.

Question 3.
What are the early symptoms of a polio attack?
Answer:
The early symptoms of polio attack are flue – like symptoms like headache, nausea, vomiting, and fever.

Question 4.
Which things are the carriers of poliovirus?
Answer:
The carriers of poliovirus are air, water, and food.

Question 5.
What percentage of people attacked by poliovirus actually suffer from severe polio?
Answer:
One percentage of people attacked by poliovirus actually suffer from severe polio.

Question 6.
What are the two forms of polio?
Answer:
The two forms of polio are spinal and bulbar.

Question 7.
Are the symptoms of both forms of polio similar or different? Which form of the disease is more dangerous?
Answer:
The symptoms of both forms of polio are similar. The bulbar form of the disease is more dangerous.

UNIT – II

Gist:
Roosevelt was keen on not allowing dangerous polio to dominate him. It left his long-term career as the President of the USA untouched. He led the fight against polio by increasing public awareness of the deadly disease and promoting research. Polio never destroyed the large bulk of the population like plague or influenza. Instead, it was a fearful, highly contagious disease that affected all, irrespective of rich or poor, when it occurred in a deadly manner. It seemed as if advances in science paled into insignificance before this disease. The first half of the nineteenth century witnessed tremendous advancement in basic hygiene methods and knowledge. A great change had come in our civilization for the first time. The people’s mere hope for good health gave way to expectations. Children were the worst sufferers of polio. The panic-stricken parents were afraid of sending their children to schools in the early 1950s because there was no vaccine for the treatment of polio.
ସାରମର୍ମ :
ବିପଜ୍ଜନକ ପୋଲିଓକୁ ତାଙ୍କ ଉପରେ ପ୍ରଭାବ ବିସ୍ତାର କରାଇ ନ ଦେବାକୁ Roosevelt ଦୃଢ଼ସଂକଳ୍ପ ଥିଲେ । ଯୁକ୍ତରାଷ୍ଟ୍ର ଆମେରିକାର ରାଷ୍ଟ୍ରପତି ଭାବେ ତାଙ୍କର ଦୀର୍ଘ ରାଜନୈତିକ ଜୀବନ ନିରାପଦ ରହିଥିଲା । ବିପଜ୍ଜନକ ବ୍ୟାଧ୍ ସମ୍ବନ୍ଧରେ ଜନସଚେତନତା ସୃଷ୍ଟି କରି ଏବଂ ଗବେଷଣାକୁ ପ୍ରୋତ୍ସାହନ ଦେଇ ସେ ପୋଲିଓ ବିରୁଦ୍ଧରେ ସଂଗ୍ରାମ କଲେ । ପ୍ଲେଗ୍ ବା ଇନ୍‌ଫ୍ଲୁ ଏଞ୍ଜା ଭଳି ଏହା ଅଧିକାଂଶ ଲୋକଙ୍କୁ ମୃତ୍ୟୁମୁଖକୁ ଠେଲି ଦେଉନଥିଲା । ଅପରପକ୍ଷେ ଏହା ଏକ ଭୟ ଉଦ୍ରେକକାରୀ ପ୍ରବଳ ସଂକ୍ରାମକ ରୋଗ ଥିଲା ଯାହାକି ଭୟଙ୍କର ଭାବେ ବ୍ୟାପୁଥ‌ିବା ସମୟରେ ଧନୀ, ଦରିଦ୍ର, ନିର୍ବିଶେଷରେ ସମସ୍ତଙ୍କୁ ଆକ୍ରାନ୍ତ କରୁଥିଲା । ବିଜ୍ଞାନର ଅଗ୍ରଗତି ଏହି ରୋଗ ନିକଟରେ ପରାଭୂତ ହେଲା ଭଳି ପ୍ରତୀୟମାନ ହେଉଥିଲା । ଊନବିଂଶ ଶତାବ୍ଦୀର ଆଦ୍ୟ ଭାଗରେ ମୌଳିକ ପରିମଳ ବ୍ୟବସ୍ତା ଓ ଜ୍ଞାନ କ୍ଷେତ୍ରରେ ପ୍ରଭୃତ ଅଗ୍ରଗତି ଘଟିଥିଲା । ପ୍ରଥମ ଥର ପାଇଁ ଆମ ସଭ୍ୟତାରେ ଏକ ବିରାଟ ପରିବର୍ତ୍ତନ ଆସିଥିଲା । ଉତ୍ତମ-ସ୍ୱାସ୍ଥ୍ୟ ପାଇଁ ଲୋକମାନଙ୍କ ଆଶା ଅନେକ ସମ୍ଭାବନା ସୃଷ୍ଟି କରିଥିଲା । ପୋଲିଓଦ୍ୱାରା ପିଲାମାନେ ଦୟନୀୟଭାବେ ପୀଡ଼ିତ ହେଉଥିଲେ । ୧୯୫୦ଦଶକ ପ୍ରାରମ୍ଭରେ ଭୀତତ୍ରସ୍ତ ପିତାମାତାମାନେ ସେମାନଙ୍କ ପିଲାମାନଙ୍କୁ ବିଦ୍ୟାଳୟକୁ ପଠାଇବାକୁ ଭୟ କରୁଥିଲେ । କାରଣ ପୋଲିଓର ଚିକିତ୍ସା ନିମନ୍ତେ ସେତେବେଳେ କୌଣସି ପ୍ରତିଷେଧକ ନ ଥିଲା ।

Glossary:
determined : keen (ଦୃଢ଼ସଂକଳ୍ପ କରିଥିଲେ)
let : allow (ଅନୁମତି ଦେବା)
major : ବଡ଼
get the best of : conquer (ଜୟ କର )
illustrious : famous and much admired (ବିଖ୍ୟାତ ଓ ପ୍ରଶଂସିତ)
political career : ରାଜନୈତିକ ଜୀବନ
long term : ଦୀର୍ଘଦିନ
spearhead (h) : lead an attack (ନେତୃତ୍ୱ ନେବା)
fight : ଯୁଦ୍ଧ କର
public awareness : ଜନସଚେତନତା
deadly : ପ୍ରାଣଘାତୀ
promoting : ପ୍ରୋତ୍ସାହିତ କରିବା
research : ପ୍ରୋତ୍ସାହିତ କରିବା
devastated : destroyed completely (ସମ୍ପୂର୍ଣ୍ଣ ଭାବରେ ନଷ୍ଟ
population : ଜନସଂଖ୍ୟା
plague : ମହାମାରୀ
influenza : ବ୍ୟାପକ ସର୍ଦ୍ଦି
contagious disease : disease that spreads by people touching each other (ସଂକ୍ରାମକ ବ୍ୟାଧ୍)
terrifying : ଭୟଙ୍କର
in spite of : ସତ୍ତ୍ବେ
advance : ପ୍ରଗତି
century : hundred years (ଶତାବ୍ଦୀ)
basic : ପ୍ରାରମ୍ଭିକ
tremendously : ବହୁଳ ପରିମାଣରେ
civilization : ସଭ୍ୟତା
instead of : ପରିବର୍ତ୍ତେ
vulnerable : easily hurt physically (ସହଜରେ ସଂକ୍ରମିତ ହେବା / ସହଜରେ ପ୍ରଭାବିତ ହେବା)
heartbreaking : ହୃଦୟବିଦାରକ
crutches : ଆଶାବାଡ଼ି
lacking : want (ଅଭାବ)
panicked : ଭୟଭୀତ ହୋଇଗଲେ
cumbersome : large and heavy (ବଡ଼ ଓ ଭାରୀ )

Think it out:
Question 1.
Did polio affect Roosevelt’s political career?
Answer:
Polio did not affect Roosevelt’s political career. Instead, he enjoyed a distinguished long-term Presidency of the United States.

Question 2.
What was the highest position did Roosevelt achieve in his political career?
Answer:
Roosevelt became one of the most distinguished Presidents of the United States. His position was not short-lived. He enjoyed a long-term political career at the highest level.

Question 3.
How did he spearhead the fight against polio?
Answer:
He spearheaded the fight against polio by increasing public awareness of the dangerous disease and promoting research.

Question 4.
Who are the most vulnerable to polio?
Answer:
Children are the most vulnerable to polio.

Question 5.
Why were parents in the early 1950s afraid of sending their children to schools?
Answer:
Parents in the early 1950s were afraid of sending their children to school because there were no vaccines for the treatment of polio.

UNIT – III

Gist:
The story of vaccine developments for polio dates back to the early 1900s. Early attempts fell through because of the lack of researchers’ knowledge that there existed more than one virus. It is now known that polio is attributed to three strains of completely stable viruses. Ironically, some children were immune to polio before the 1900s, because of deplorable sanitary conditions and primitive efforts in connection with sewage and water treatments. With the improvement in sanitary methods, children were not exposed to poliovirus any longer. As a result, they did not develop antibodies to the virus. Paradoxically, their exposure to the virus during later childhood and adulthood made them run the risk of contracting polio. In the meantime, March of Dimes, a foundation in the United States, emerged on the scene. Thanks to President Roosevelt, this organization worked for the elimination of polio by enlisting the services of a competent researcher who was sure of finding a danger-free vaccination.
ସାରମର୍ମ :
ପୋଲିଓ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତିର କାହାଣୀ ଊନବିଂଶ ଶତାବ୍ଦୀର ଆଦ୍ୟଭାଗରୁ ଆରମ୍ଭ ହୋଇଥିଲା । ପ୍ରାରମ୍ଭିକ ପ୍ରୟାସ ବ୍ୟର୍ଥ ହୋଇଥିଲା କାରଣ ପୋଲିଓର ଏକାଧ୍ଵକ ପ୍ରକାର ଭୂତାଣୁ ଥିବା କଥା ଗବେଷକମାନେ ଜାଣିନଥିଲେ । ଏବେ ଜଣାପଡ଼ିଲା ଯେ ପୋଲିଓ ତିନିପ୍ରକାର ସମ୍ପୂର୍ଣ୍ଣ ସ୍ଥିର ଭୂତାଣୁଦ୍ବାରା ସୃଷ୍ଟି ହେଉଛି । ବିଡ଼ମ୍ବନାର କଥା ଯେ ଊନବିଂଶ ଶତାବ୍ଦୀ ପୂର୍ବରୁ ଦୁଃସ୍ଥ ପରିମଳ ବ୍ୟବସ୍ଥା, ନାଳନର୍ଦ୍ଦମା ପରିଷ୍କରଣର ପୁରୁଣାକାଳିଆ ବ୍ୟବସ୍ଥା, ଜଳବିଶୋଧନ ଅବ୍ୟବସ୍ଥା କାରଣରୁ କେତେକ ପିଲା ପୋଲିସ ସଂକ୍ରମଣରୁ ମୁକ୍ତ ଥିଲେ । ପରିମଳ ବ୍ୟବସ୍ଥାର ଉନ୍ନତି ପରେ ପିଲାମାନେ ପୋଲିଓ ଭୂତାଣୁ ସଂସର୍ଗରୁ ରକ୍ଷା ପାଇପାରିଲେ । ଫଳରେ ସେମାନଙ୍କ ଶରୀରରେ ଭୂତାଣୁ ପ୍ରତିରୋଧକ ଶକ୍ତି ସୃଷ୍ଟି ହେଲା ନାହିଁ । ଆଶ୍ଚର୍ଯ୍ୟର କଥା ଯେ ପରବର୍ତ୍ତୀ ବାଲ୍ୟାବସ୍ଥା ଓ ବୟସ୍କାବସ୍ଥାରେ ସେମାନଙ୍କର ଭୂତାଣୁ ସଂସର୍ଗରେ ଆସିବାଦ୍ଵାରା ପୋଲିଓ ରୋଗାକ୍ରାନ୍ତ ହେବାର ବିପଦ ବଢ଼ିଗଲା । ଇତ୍ୟବସରରେ March of Dimes ନାମକ ଆମେରିକାର ଏକ ଅନୁଷ୍ଠାନର-ଦୃଶ୍ୟପଟ୍ଟରେ ଆବିର୍ଭାବ ହେଲା । ରାଷ୍ଟ୍ରପତି Rooseveltଙ୍କଦ୍ଵାରା ଅନୁପ୍ରାଣିତ ହୋଇ ଏହି ସଂସ୍ଥା ପୋଲିଓର ଦୂରୀକରଣ ନିମନ୍ତେ ଜଣେ ଯୋଗ୍ୟ ଗବେଷକଙ୍କ ସହାୟତାରେ କାର୍ଯ୍ୟରତ ଥିଲା ଯିଏକି ଏକ ବ୍ୟବସ୍ଥା ବାହାର କରିପାରିବେ ବୋଲି ନିଶ୍ଚିତ ଥିଲେ ।

Glossary:
quite : ସମ୍ପୂର୍ଣ୍ଣ
strain : a distinct breed (ଭିନ୍ନ ପ୍ରକାରର )
stable : ସ୍ଥିର
enterovirus : viruses that cause gastro-intestinal illness (ପରିପାକ ଓ ଅନ୍ତନଳୀ ଜନିତ ଅସୁସ୍ଥତା ସୃଷ୍ଟିକାରୀ ଭୂତାଣୁ)
RNA : Ribonucleic acid (RNA) is one of the three major macromolecules (along with DNA and proteins) essential for all known forms of life. (ରାଇବୋନ୍ୟୁକ୍ଲିକ୍ ଏସିଡ୍)
effective : ଫଳପ୍ରଦ
ironically : ବିଡମ୍ବନାର ବିଷୟ
sanitation condition : ପରିମଳ ଅବସ୍ଥା
effort : ପ୍ରଚେଷ୍ଟା
sewage : ନର୍ଦ୍ଦମାର ମଇଳା
primitive : ପୁରୁଣାକାଳିଆ
infant : ଶିଶୁ
consequently : ଫଳସ୍ୱରୂପ
later : ପରବର୍ତ୍ତୀ
childhood : ବାଲ୍ୟକାଳ
adulthood : ବାଲ୍ୟାବସ୍ଥା
painstaking : ଆୟସସାଧ୍ୟ
funded : ଆର୍ଥିକ ସାହାଯ୍ୟ ଯୋଗାଇ
March of Dimes: a Foundation in the United States that works to improve the health of mothers and babies. ft was originally founded by Franklin D. Roosevelt in 1938 to
combat polio. (ମା’ ଓ ଶିଶୁମାନଙ୍ଗ ସ୍ୱାସ୍ଥ୍ୟବସ୍ଥାର ଉନ୍ନତି କଳ୍ପେ କାର୍ଯ୍ୟ କରୁଥିବା ଏକ ଆମେରିକୀୟ ସଂସ୍ଥା)
grassroot : basic level (ତୃଣମୂଳ / ପ୍ରାଥମିକ ସ୍ତର )
enlist : (here) utilise
respected : ସମ୍ମାନୀୟ
safe : ନିରାପଦ

Think it out:
Question 1.
Why did early attempts at the development of the polio vaccine fail?
Answer:
Early attempts at the development of the polio vaccine failed because of the researcher’s lack of knowledge concerning the existence of more than one virus.

Question 2.
Why were some children immune to polio before the 1900s?
Answer:
Some children were immune to polio before the 1900s because sanitary conditions and primitive efforts in connection with sewage and water treatments were quite deplorable. Besides, through breastfeeding their mothers’ antibodies passed onto them played a great role.

Question 3.
Did improved sanitation help to avoid polio attacks before the 1900s?
Answer:
Improved sanitation did not help to avoid polio attacks before the l900s, because of not being exposed to poliovirus in their infancy, they did not develop antibodies to the viruses. As a result, when they were exposed to the virus in later childhood and adulthood they ran the risk of suffering from polio.

Question 4.
How did the March of Dimes work for the elimination of polio?
Answer:
The March of Dimes worked for the elimination of polio with its decision to utilize the services of a venerable researcher who was sure of finding a safe vaccination.

UNIT – IV

Gist:
This part begins with the editors’ eloquent reference to Dr. Jonas Salk who started his medical career studying immunology. In 1947, while at the University of Pittsburgh, he began his research on poliovirus. He used a method of developing poliovirus in cell culture. In 1952, Salk developed a successful vaccine resorting to a blend of three types of virus, found in monkey kidney cultures. He was instrumental in developing a process with the use of formalin, a chemical that made the whole virus inactive.

Dr. Salk’s research gave rise to the scope of clinical trials which created an unprecedented history in medical science. This resulted in the spectacular reduction of polio. However, there was a problem with the original Salk vaccine. The vaccine in reality caused 260 cases of polio and consequently 10 deaths. Some virus particles were not completely inactivated. This did not last long. The problem was solved. Ever since the result has been quite effective. Salk’s vaccine is given in two intramuscular injections administered in a span of one month and in need of boosters every five years. Then came Albert Bruce Sabin who developed or oral form of the vaccine.

The advantages of an oral vaccine are its long-lasting immunity, the prevention of reinfection of the digestive tract, and the administering of the vaccine at a cheap cost. But it is not an unmixed blessing. A live, oral virus is not applicable to patients with a compromised immune system. The Sabin oral vaccine has another disadvantage. Patients suffering from enterovirus infection of the gastrointestinal tract may not develop the immune response at the time of taking the oral vaccine. Last but not least, both vaccines have their merits and demerits.
ସାରମର୍ମ :
ଏହି ଅଂଶର ଆରମ୍ଭରେ ଲେଖକ ଡକ୍ଟର Jonas Salkଙ୍କ ବିଷୟରେ ସ୍ପଷ୍ଟ ମନ୍ତବ୍ୟ ଦେଇଛନ୍ତି ଯିଏକି ସଂକ୍ରମଣ ପ୍ରତିରୋଧ ବିଷୟରେ ଅଧ୍ୟୟନ କରି ଚିକିତ୍ସକର ଜୀବନ ଆରମ୍ଭ କରିଥିଲେ । ୧୯୪୭ ମସିହା ପିଟସ୍ଵର୍ଗ ବିଶ୍ବବିଦ୍ୟାଳୟରେ ଥ‌ିବାବେଳେ ସେ ପୋଲିଓ ଭୂତାଣୁ ଉପରେ ଗବେଷଣା ଆରମ୍ଭ କରିଥିଲେ । ସେ କୋଷୀୟ ପୋଷଣ ମାଧ୍ୟମରେ ପୋଲିଓ ଭୂତାଣୁକୁ ବଢ଼ାଇବାର ପଦ୍ଧତି ବ୍ୟବହାର କରିଥିଲେ । ୧୯୫୨ ମସିହାରେ Salk ମାଙ୍କଡ଼ ବୃକ୍‌କରେ ବଢ଼ୁଥ‌ିବା ତିନି ପ୍ରକାର ଭୂତାଣୁ ସମ୍ମିଶ୍ରଣରେ ଏକ ଫଳପ୍ରଦ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତ କରିଥିଲେ । ଫରମାଲିନ୍ ନାମକ ରାସାୟନିକ ପଦାର୍ଥ ବ୍ୟବହାର କରି ସେ ଏକ ପଦ୍ଧତି ଆବିଷ୍କାର କରିବାରେ ମୁଖ୍ୟ ଥିଲେ ଯାହାକି ସମସ୍ତ ଭୂତାଣୁଙ୍କୁ ନିଷ୍କ୍ରିୟ କରିଦେଉଥ୍ଲା ।

ଡକ୍ତର Salkଙ୍କ ଗବେଷଣା ଚିକିତ୍ସା କ୍ଷେତ୍ରରେ ପରୀକ୍ଷଣ ପାଇଁ ସୁଯୋଗ ସୃଷ୍ଟି କରିଥଳା, ଯାହାକି ଭେଷଜ ବିଜ୍ଞାନ କ୍ଷେତ୍ରରେ ଅଭୂତପୂର୍ବ ଇତିହାସ ସୃଷ୍ଟି କରିଥିଲା । ମାତ୍ର ମୂଳ Salk ପ୍ରତିଷେଧକରେ ସମସ୍ୟା ଥିଲା । ବାସ୍ତବରେ ଏହା ୨୬୦ ଜଣଙ୍କୁ ପୋଲିଓ ବ୍ୟାଧୁଗ୍ରସ୍ତ କରାଇଥିଲା ଏବଂ ୧୦ ଜଣ ମୃତ୍ୟୁବରଣ କରିଥିଲେ । କେତେକ ଭୂତାଣୁ ସମ୍ପୂର୍ଣ୍ଣରୂପେ ନିଷ୍କ୍ରିୟ ହୋଇନଥିଲେ । କିନ୍ତୁ ଏହି ସମସ୍ୟା ବେଶିଦିନ ରହିନଥିଲା । ଏହାର ସମାଧାନ ହୋଇଥିଲା । ସେହିଦିନଠାରୁ ଫଳାଫଳ ଉତ୍ତମ ହୋଇଆସୁଛି । Salkଙ୍କ ପ୍ରତିଷେଧକ ଏକ ମାସ ବ୍ୟବଧାନରେ ଦୁଇଟି ଟୀକା ଜରିଆରେ ମାଂସପେଶୀ ମଧ୍ଯରେ ଦିଆଯାଉଥିଲା ଏବଂ ପ୍ରତି ପାଞ୍ଚ ବର୍ଷରେ ବୁଷ୍ଟର ଟୀକା ଦେବାର ଆବଶ୍ୟକତା ଥିଲା । ୧୯୫୭ ମସିହାରେ ଦୃଶ୍ୟପଟ୍ଟରେ Albert Bruce Sabinଙ୍କର ଆବିର୍ଭାବ ହେଲା

ଯିଏକି ପାଟିରେ ଖିଆଯାଉଥ‌ିବା ଏକପ୍ରକାର ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତ କରିଥିଲେ । ପାଟିରେ ଖିଆଯାଉଥ‌ିବା ପ୍ରତିଷେଧକର ଉପକାରଗୁଡ଼ିକ ଦୀର୍ଘ ପ୍ରତିଷେଧକ ଶକ୍ତି, ପରିପାକ ତନ୍ତ୍ରର ପୁନଃସଂକ୍ରମଣରୁ ରକ୍ଷା ଏବଂ ପ୍ରତିଷେଧକ ପ୍ରଦାନରେ କମ୍ – ଖର୍ଚ୍ଚାନ୍ତ ହେବା । ଏହା ଏକ ଅପକାରିତାଶୂନ୍ୟ ଆଶୀର୍ବାଦ ନ ଥିଲା । ଦୁର୍ବଳ ପ୍ରତିରୋଧକ ଶକ୍ତିସମ୍ପନ୍ନ ରୋଗୀମାନଙ୍କ କ୍ଷେତ୍ରରେ ଏକ ଜୀବନ୍ତ ପାଟିରେ ଖିଆଯାଉଥ‌ିବା ଭୂତାଣୁ ପ୍ରୟୋଗ କରାଯାଇପାରିବ ନାହିଁ । ଏହାର ଅନ୍ୟ ଏକ ଅପକାରିତା ଥିଲା । ଅନ୍ତନଳୀରେ ସଂକ୍ରମଣ ଭୋଗୁଥିବା ରୋଗୀମାନଙ୍କୁ ପାଟିରେ ଖିଆଯାଉଥ‌ିବା ପ୍ରତିଷେଧକ ଦେଲେ ସେମାନଙ୍କଠାରେ ପ୍ରତିରୋଧକ ଶକ୍ତି ବୃଦ୍ଧି ପାଉନଥିଲା । କିନ୍ତୁ ନିଃସନ୍ଦେହରେ ଏହା କୁହାଯାଇପାରେ ଯେ ଉଭୟ ପ୍ରତିଷେଧକର ଉଭୟ ଭଲ ଏବଂ ମନ୍ଦ ଗୁଣ ଅଛି ।

Glossary:
Pr. Jonas Salk: Jonas Edward Salk (1914-1995) was an American medical researcher and virologist, best known for his discovery and development of the first sale and most effective polio vaccine.
immunology: the scientific study of protection against disease (ରୋଗ ପ୍ରତିରୋଧ
cell culture: Cell culture is the complex process by which cells are grown under controlled conditions, generally outside of their natural environment. ପୋଷଣ )
massive: ବହୁଳ ଭାବରେ
unprecedented : ଅଭୂତପୂର୍ବ
spectacularly : ଆଶ୍ଚର୍ଯ୍ୟଜନକ ଭାବରେ
oral form of vaccine: ପାଟିରେ ଖିଆଯାଉଥ‌ିବା
Albert Bruce Sabin : Albert Bruce Sabin (1906-1993) was an American medical researcher best known for having developed an oral polio vaccine.
intramuscular injections : medicinal doses injected into the muscle
booster : a second dose of medicine given w strengthen the earlier dose (ଶକ୍ତିବର୍ଦ୍ଧକ ଔଷଧ)
endemic region : intected area (ସଂକ୍ରମିତ ଅଞ୍ଚଳ)
long-lasting : ଦୀର୍ଘସ୍ଥାୟୀ
reinfection : ପୁନଃସଂକ୍ରମଣ
digestive tract : ପରିପାକ ତନ୍ତ୍ର
administering : ଦେବା
sterile syringes : disinfccted injection syringes
disadvantage : demerit (ଅପକାର)
contact : ସଂସର୍ଗ
transniincd : ସ୍ଥାନାନ୍ତରିତ ହୋଇଯିବା
immunocompromised: a body that does not have good immunity, the ability to defend against illness (ଦୁର୍ବଳ ପ୍ରତିରୋଧକ କ୍ଷମତାସମ୍ପନ୍ନ )
relative : ଆପେକ୍ଷିକ
safety : ନିରାପଦ
cost : ମୂଲ୍ୟ

Think about it:
Question 1.
What method did Jonas Salk use to develop the polio vaccine?
Answer:
Jonas Salk used a method of growing poliovirus in cell culture to develop the polio vaccine.

Question 2.
How did Salk develop a successful vaccine?
Answer:
Salk developed a successful vaccine using a combination of the three types of virus, grown in monkey kidney cultures.

Question 3.
How was the first polio vaccine accepted?
Answer:
The first polio vaccine was subjected to clinical tests in the United States and parts of Canada on a massive scale. As it was successful, the Government promptly granted permission for the vaccine to be given away among the children.

Question 4.
What was the problem with the original Salk vaccine?
Answer:
The problem with the original Salk vaccine was that it brought 260 cases of polio including 10 deaths due to incomplete inactivation of some virus particles.

Question 5.
How was the Salk vaccine given?
Answer:
Salk vaccine was injected into the muscle in two medical doses, each at an interval of one month, and requires boosters every five years.

Question 6.
How is the Sabin vaccine given?
Answer:
Sabin vaccine is given in three doses in the first two years of life and a booster is given when the child starts school.

Question 7.
What are the advantages of an oral vaccine?
Answer:
The advantages of an oral vaccine are long-lasting immunity, the prevention of reinfection of the digestive system, and oral administration at a cheaper price.

Question 8.
What is its major disadvantage?
Answer:
Its major disadvantage is that it is not applicable to patients with weak immune systems, since it is a live virus that is likely to cause disease in these patients.

UNIT – V

Gist:
Despite the never-ending debate between safety and cost, it is fortunate for us to choose from two good alternatives. Both vaccines invented by Salk and Sabin, are accepted all over the world. The United States prefers the Sabin vaccine to the Salk one, but other countries like the former. Relentless research is carried out for the development of these vaccines. The development of more fruitful culturing and purification methods induces a higher level of antibody formation. The latest research in the development of the polio vaccine is to combine E. Coli’s genes with genes of poliovirus by which E. Coli can synthesize viral capsid proteins to be used in making a vaccine. Thanks to the invention and use of the polio vaccine, America was declared to be free from polio in 1994. The World Health Organization is also concerned about it. This unit ends on a positive note. Like smallpox complete eradication of poliomyelitis, the editors’ hope, is possible.
ସାରମର୍ମ :
ନିରାପତ୍ତା ଏବଂ ମୂଲ୍ୟ ମଧ୍ୟରେ ଚାଲିଥିବା ଅସରନ୍ତି ବିତର୍କ ସତ୍ତ୍ବେ, ଆମ୍ଭମାନଙ୍କର ସୌଭାଗ୍ୟ ଯେ ଆମେ ଦୁଇଟି ବିକଳ୍ପ ମଧ୍ୟରୁ ଗୋଟିଏକୁ ବାଛିପାରୁଛୁ । Salk ଏବଂ Sabinଙ୍କଦ୍ବାରା ଉଦ୍ଭାବନ ହୋଇଥ‌ିବା ଉଭୟ ପ୍ରତିଷେଧକ ପୃଥ‌ିବୀର ଚାରିଆଡ଼େ ବ୍ୟବହାର କରାଯାଉଛି । ଆମେରିକାରେ Salkଙ୍କ ପ୍ରତିଷେଧକ ଅପେକ୍ଷା Sabinଙ୍କ ପ୍ରତିଷେଧକର ଆଦର ବେଶୀ । କିନ୍ତୁ ଅନ୍ୟାନ୍ୟ ଦେଶରେ ପ୍ରଥମଟିକୁ ପସନ୍ଦ କରାଯାଉଛି । ଏଇ ଦୁଇଟି ପ୍ରତିଷେଧକର ଉନ୍ନତିକରଣ ପାଇଁ ଅବିଶ୍ରାନ୍ତ ଗବେଷଣା ଜାରି ରହିଛି । ଫଳପ୍ରଦ ପୋଷଣ ଓ ବିଶୋଧ ପଦ୍ଧତିର ଉନ୍ନତିକରଣ ଶରୀରରେ ଅଧ‌ିକ ମାତ୍ରାରେ ରୋଗ ପ୍ରତିରୋଧକ ଶକ୍ତି ସୃଷ୍ଟି କରିଥାଏ। ପୋଲିଓ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତିର ସର୍ବାଧୁନିକ ଗବେଷଣା ହେଉଛି E. Coliର ଜିକୁ ପୋଲିଓ ଭୂତାଣୁ ଜିନ୍ ସହିତ ମିଶ୍ରିତ କରିବା, ଯଦ୍ବାରା E. Coli ଭୂତାଣୁର ବାହ୍ୟ ପ୍ରୋଟିନ୍ ସ୍ତରକୁ ସଂଶ୍ଳେଷଣ କରିପାରିବ ଯାହାକି ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତି ପାଇଁ ବ୍ୟବହୃତ ହୋଇପାରିବ । ପୋଲିଓ ପ୍ରତିଷେଧକ ଉଦ୍ଭାବନ ଓ ବ୍ୟବହାରଦ୍ୱାରା ୧୯୯୪ ମସିହାରେ ଆମେରିକା ପୋଲିଓମୁକ୍ତ ଭାବେ ଘୋଷିତ ହେଲା । ବିଶ୍ଵ ସ୍ବାସ୍ଥ୍ୟ ସଙ୍ଗଠନ ମଧ୍ଯ ଏ ଦିଗରେ ଉଦ୍ୟମରତ ରହିଛି । ଲେଖକ ଆଶା କରନ୍ତି ଯେ ବସନ୍ତ ଭଳି ପୋଲିଓର ସମ୍ପୂର୍ଣ ଦୂରୀକରଣ ସମ୍ଭବ ହେବ ।

Glossary:
debate : ବିତର୍କ
alternative : ବିକଳ୍ପ
currently : at present (ବର୍ତ୍ତମାନ୍)
preferred : ପସନ୍ଦ କରାଯାଇଛି
induce : produce (ସୃଷ୍ଟି କରେ)
exciting. : causing eagerness (ଆଗ୍ରହ ସୃଷ୍ଟିକାରୀ)
recombinant biotechiogy : an advanced process of vaccine production (ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତିର ଉନ୍ନତ ପ୍ରକ୍ରିୟା)
genetic cloning: the process of producing similar populations of genetically identical individuals that occur in nature when organisms such as bacteria, insects, or plants reproduce asexually. (ଜିନୀୟ ପ୍ରତିରୂପ ସୃଷ୍ଟି ପ୍ରକ୍ରିୟା)
synthesis: a combination of components to form a connected whole (ସଂଶ୍ଳେଷଣ )
excluding: ବାଦ୍ ଦେଇ
content: ବିଷୟବସ୍ତୁ
rare: ବିରଳ
likely : possibly (ସମ୍ଭବତ)
its footsteps: the complete eradication of polio would follow the complete extinction of small-pox (ଏହାର ପଦଚିହ୍ନ)

Think it out:
Question 1.
What induces higher levels of antibody formation?
Answer:
The development of more fruitful culturing and purification techniques induces higher levels of antibody formation.

Question 2.
What is the latest research in the development of the polio vaccine?
Answer:
The latest research in the development of the polio vaccine is to combine E. Coli’s genes with genes of poliovirus by which E. Coli can synthesize viral capsid proteins to be used in making a vaccine.

Question 3.
Is complete eradication of poliomyelitis possible?
Answer:
Yes, it is possible. Complete eradication of smallpox is a case in point.

Post-Reading Activities:

Doing with words :
(A) Study the following sentence and note the verbs in it :
Paradoxically, when sanitation improved, infants were no longer exposed at an age when they were protected, so they did not develop antibodies to the viruses. There are four finite verbs – improved, were exposed, were protected, and did develop Note ‘improved’, and ‘did develop’ are in the active voice; ‘were exposed’ and ‘were protected’ are in the passive voice. In passive voice, the form of the verb is ‘be’ + pp verb: any form of the auxiliary verb ‘be’ + past participle form of the main verb – were + expose(d), were + protect(ed).

Identify the finite verbs in the following sentences and write which verbs are in active voice, and which are in passive voice.
(i) The virus enters the body by nose or mouth and travels to the intestines, where it incubates.
(ii) In most cases, this stops the progression of the virus; lifelong immunity against the disease is acquired.
(iii) Babies were frequently exposed to polioviruses.
(iv) These infants did not contract the disease because their mothers’ antibodies were passed on to them through breastfeeding.
(v) He developed a process using formalin, a chemical that inactivated the whole virus.
(vi) The Salk vaccine is given in two intramuscular injections spaced one month apart and requires boosters every 5 years.
(vii) The Sabin oral vaccine is given in three doses in the first two years of life, and a booster is given when the child starts school.
Answer:
(i) (a) enters, travels, incubates — finite verbs
(b) All verbs in active voice,
(ii) (a) stops – finite verb – active voice
(b) is acquired – finite verb – passive voice
(iii) (a) were exposed – finite verb – passive voice
(iv) (a) did not contract – finite verb – active voice,
(b) were passed – finite verb – passive voice
(v) (a) developed, inactivated – finite verbs – active voice
(vi) (a) is given – finite verb – passive voice
(b) requires – finite verb – active voice
(vii) is given – finite verb – passive voice
(b) starts – finite verb – active voice

(B) Fill in the blanks with the verbs given in brackets in active voice. Use simple past tense.
(i) Ramakrishna ______________(fascinate) him.
(ii) Wanderlust ______________ (seize) him.
(iii) He ______________(feel) the presence of an inward power.
(iv) He ______________ (decide) to take part in the Parliament of Religions.
(v) A friendly Maharaja ______________(give) him his passage.
(vi) He ______________ (address) the audience as ‘Sisters and Brothers of America’.
(vii) Hundreds ______________ (rise) and ______________(applaud).
(viii) He ______________(use) no written text, not even notes.
(ix) America’s outward glitter ______________ not (deceive) him.
(x) He ______________(preach) Hindu philosophy.
Answer:
(i) Ramakrishna fascinated him.
(ii) Wanderlust seized him.
(iii) He felt the presence of inward power.
(iv) He decided to take part in the Parliament of Religions.
(v) A friendly Maharaja gave him his passage.
(vi) He addressed the audience as ‘Sisters and Brothers of America’.
(vii) Hundreds rose and applauded.
(viii) He used no written text, not even notes.
(ix) America’s outward glitter did not deceive him.
(x) He preached Hindu philosophy.

(C) Fill in the blanks with the verbs given in brackets in passive voice. Use the simple present tense.
(i) Civilization ______________not (inherit).
(ii) It ______________(learn) and ; ______________(earn) by each generation anew.
(iii) If the transmission of Civilization ______________ (interrupt) for one century, civilization will die and we will be savage again.
(iv) Therefore, importance (give) to higher education in our country.
(v) Colleges ______________ (design) to meet the needs of higher education.
Answer:
(i) Civilization is not inherited.
(ii) It is learned and earned by each generation anew.
(iii) If the transmission of Civilization is interrupted for one century, civilization will die and we will be savage again.
(iv) Therefore, importance is given to higher education in our country.
(v) Colleges are designed to meet the needs of higher education.

CHSE Odisha Class 12 English Development of Polio Vaccines Important Questions and Answers

I. Multiple-Choice Questions (MCQs) with Answers

Question 1.
Polio broke out in America in —
(A) 1920
(B) 1921
(C) 1934
(D) 1923
Answer:
(B) 1921

Question 2.
Roosevelt fell prey to —
(A) influenza
(B) hunting
(C) polio
(D) none of these
Answer:
(C) polio

Question 3.
He was unfortunate, because —
(A) he suffered from polio for long
(B) his legs suffered lifelong paralysis
(C) his body ached for several days
(D) none of these
Answer:
(B) his legs suffered lifelong paralysis

Question 4.
The percentage of people affected by a paralytic form of polio is –
(A) 2
(B) 3
(C) 1
(D) none of these
Answer:
(C) 1

Question 5.
People showing symptoms of muscular paralysis can be —
(A) hopeful
(B) hopeless
(C) sadistic
(D) none of these
Answer:
(A) hopeful

Question 6.
Roosevelt’s political career was —
(A) ruined
(B) short-lived
(C) distinguished
(D) none of these
Answer:
(C) distinguished

Question 7.
Polio was not so deadly as –
(A) plague
(B) influenza
(C) both (A) and (B)
(D)none of these
Answer:
(C) both (A) and (B)

Question 8.
Through the first half of the century, the advancement of hygiene methods and knowledge has been –
(A) marginal
(B) fine
(C) phenomenal
(D) moderate
Answer:
(C) phenomenal

Question 9.
The word ‘cumbersome’ means –
(A) ordinary
(B) superficial
(C) systematic
(D) none of these
Answer:
(D) none of these

Question 10.
The lack of a vaccine for polio pushed parents throughout America to a state of –
(A) optimism
(B) fear
(C) great fear
(D) hopelessness
Answer:
(C) great fear

Question 11.
Vaccine developments for polio date back to –
(A) 1900s
(B) the early 1900s
(C) 1901
(D) none of these
Answer:
(B) the early 1900s

Question 12.
That immunity was primarily during infancy before the 1900s sounded –
(A) controversial
(B) tragic
(C) ironic
(D) all of these
Answer:
(C) ironic

Question 13
Which one of the following statements is false?
(A) Efforts at water treatments are modem.
(B) Polio is caused by three strains of stable viruses.
(C) In the early 1900s, researchers were alive of the existence of more than one polio vims.
(D) The March of Dimes was set up to conduct research on polio.
Answer:
(A) Efforts at water treatments are modem.

Question 14.
The March of Dimes came into existence, thanks to –
(A) a respected researcher
(B) a specialist
(C) Roosevelt
(D) none of these
Answer:
(C) Roosevelt

Question 15.
The word ‘founded’ means –
(A) set out
(B) foundation
(C) established
(D) structured
Answer:
(C) established

Question 16.
Which one of the following statements is true?
(A) Dr. Salk started his career as an immunologist.
(B) His research was of immense help in 1949, 1952, and 1954.
(C) Salk’s research on polio was not innovative,
(D) Salk vaccine was used in America without any deliberation.
Answer:
(A) Dr. Salk started his career as an immunologist.

Question 17.
Which one out of the following statements is false?
(A) A method of growing polio vims in cell culture was of immense help in 1949.
(B) Formalin was a chemical that made the whole virus work.
(C) Salk vaccine was put to clinical trials in the USA and parts of Canada in great measure.
(D) In 1957, Albert Brace emerged on the polio scene.
Answer:
(B) Formalin was a chemical that made the whole virus work.

Question 18.
1963 witnessed the availability of –
(A) Salk vaccine
(B) different vaccines for polio
(C) Brace vaccine
(D) none of these
Answer:
(C) Brace vaccine

Question 19.
The word ‘endemic’ means-
(A) contagious
(B) infected
(C) acute
(D) inimical
Answer:
(B) infected

Question 20.
The most appropriate title of this extract is –
(A) Research on Polio
(B) Cure for Polio
(C) Contributions of Salk and Bruce in the field of polio
(D) Polio, a deadly disease
Answer:
(C) Contributions of Salk and Bruce in the field of polio

Question 21.
Research continues to improve these vaccines. The underlined word means –
(A) three
(B) polio
(C) Sabin and Salk
(D)both (A) and (B)
Answer:
(C) Sabin and Salk

Question 22.
The word ‘Recombinant’ means –
(A) reorganized
(B) reconciliation
(C) remembrance
(D) none of these
Answer:
(C) remembrance

Question 23.
We are fortunate, because –
(A) Polio vaccination is plentily available.
(B) Polio is curable.
(C) We are left with a choice between the Sabin vaccine and the Salk vaccine.
(D) all of these
Answer:
(C) We are left with a choice between the Sabin vaccine and the Salk vaccine.

Question 24.
This is theoretically impossible. The underlined word means –
(A) treatment of polio
(B) poliovirus
(C) a goal of the complete elimination of polio from the whole world by 2000
(D) none of these
Answer:
(C) a goal of the complete elimination of polio from the whole world by 2000

Question 25.
The word ‘eradicated’ means –
(A) rooted out
(B) abdicated
(C) cleared out
(D) combat
Answer:
(A) rooted out

II. Short Type Questions with Answers

Question 1.
Why did early attempts at the polio vaccine fail and what do we now know?
Answer:
Early attempts of developing a polio vaccine failed as a result of the researchers’ total lack of knowledge about the existence of more than one virus. We now know that three strains of quite stable viruses that constitute a part of the enterovirus family, having RNA as their genetic material give rise to polio.

Question 2.
How can a polio vaccine be effective?
Answer:
A polio vaccine can be effective as a result of a vaccine’s power to confer immunity against all three strains of completely stable viruses.

Question 3.
What made it possible for infants to acquire immunity from polio?
Answer:
Infants acquired immunity from polio because of their mothers’ breastfeeding which created antibodies in them.

Question 4.
Did improved sanitation help to avoid polio attacks before the 1990s? Give a reasoned answer.
Answer:
It was an irony that immunity was primarily acquired during infancy before the 1990s as sanitation conditions were poor and efforts at sewage and water treatments were primitive.

Question 5.
Why was ‘polio’ a frightening, contagious disease?
Answer:
‘Polio’ was ‘a frightening, contagious disease’, because it affected all irrespective of poor and rich, assumed devastating proportions and it seemed impossible to hold it in check, despite strides in the field of medicine.

Question 6.
What did the first half of the 19th century witness?
Answer:
The first half of the 19th century witnessed great advancement in basic hygiene methods and knowledge.

Question 7.
What did people begin to expect for the first time in civilization?
Answer:
For the first time in civilization, people expected good health. They stopped merely hoping for it.

Question 8.
What happened in the early 1950s?
Answer:
In the early 1950s, there were shocking posters of children on crutches or iron lungs, the large and heavy mechanical aids to help those whose lungs would not function at all.

Question 9.
What light does the extract throw on Roosevelt?
Answer:
Roosevelt had a spectacular political career. He became the President of the United States of America. Above all, he was a relentless crusader against polio.

Question 10.
Throw light on one of the disadvantages of oral vaccines.
Answer:
One of the disadvantages of oral vaccine is that those in close touch with immunocompromised patients cannot use it because of the shedding of the live virus in the vaccine in the feces of those who inject it and the possible transmission to the immunocompromised patient.

Question 11.
How can polio spread?
Answer:
Polio can spread either through contact with infected feces or infected droplets traveling through the air, in food, or in water.

Question 12.
Why was it impossible at times to put polio in check despite advances in medicine?
Answer:
It was impossible at times to put polio in check despite advances in medicine because it was a deadly, highly contagious disease that ignored the rich and poor alike.

Question 13.
What made the people expect sound health instead of merely hoping for it?
Answer:
Advancements in basic hygienic methods and knowledge in the first half of the 19th century made people expect sound health instead of merely hoping for it.

Question 14.
“The writer takes us back to the early 1950s.” Why?
Answer:
The writer takes us back to the early 1950s because it was a witness to the stunning pictures of children on crutches or iron lungs, the large and heavy mechanical aids to help those whose lungs would never function.

Introducing the Author:
Bonnie A. Maybury Okonek is an American microbilogist. He is also a fiction writer. His writings are marked by their clarity and lucidity.

About the Topic:
The piece ‘Development of Polio Vaccines’ acquaints the readers with poliomyelitis or polio,- an acute, viral, and infectious disease. This deadly disease can give rise to paralysis or death. Its story goes back to 1921. Franklin D. Roosevelt, the 32nd President of the United States spearheaded the fight against it. The researchers in the early 1900s tried their best to find out a vaccine for polio. Dr. Jonas Salk and Albert Bruce Sabin immensely contributed to its development. The days are not far off when polio can be rooted out like smallpox.

Summary:
It was 1921. America fell prey to polio. A young politician Franklin Delano Roosevelt was unable to withstand this deadly disease. The virus spread very fast and ultimately his legs were paralyzed. The early symptoms of a polio attack are headache, nausea, vomiting, and fever. The carriers of poliovirus are air, food, or water. One percent of people attacked by it actually suffers from severe polio. The writer throws light on two forms of polio: spinal and bulbar. Muscle pain, stiff neck and back, and possible paralysis characterize both forms.

The former affects the patient’s limbs and the latter the lungs. Polio at its worst has no treatment. The symptoms of both forms of polio are similar and of the two, the bulbar form is more dangerous. The writer dwells upon the role of Roosevelt in spearheading the fight against polio. He was keen on not allowing this terrible disease to conquer him. His distinguished political career as the President of the United States continued for a long. He tried his best to increase public awareness of the dangerous disease and promote research.

Children are the most vulnerable to polio. Driven by panic parents, in the early 1950s, were afraid of sending their children to schools. They desperately waited for a vaccine. The early 1900s witnessed vaccine developments for polio. Some baffling observations were made by the researchers at that time in their attempt to find out a vaccine for polio. Ironically, some children were immune to polio before the 1900s because of poor sanitary conditions and primitive efforts to deal with sewage and water treatments.

Above all mothers, breastfeeding played a crucial role. In spite of improvements in sanitation, exposure to the virus in later childhood and adulthood made them run the risk of contracting polio. The March of Dimes set up by Franklin D. Roosevelt was a foundation that worked for the elimination of polio with the services of a venerable researcher. The writer acquaints us with Dr. Jonas Salk who started her medical research career embarking on the scientific study of protection against polio.

He used a method of growing poliovirus in cell culture to develop a polio vaccine. He developed a successful vaccine using a combination of the three types of virus grown in monkey kidney cultures. In 1952, he was the first to do so in this field. Then he developed a process using formalin, a chemical that deprived the whole virus of being active. Dr. Salk’s invention gave rise to the testing of the vaccine in clinical trials in the U.S.A. and parts of Canada in great measure. The scope of the trials was unprecedented in the history of medical science.

The results were spectacular. In 1955, the government immediately agreed to distribute vaccines to the children of the United States. But the problem with the Salk vaccine was that it brought 260 cases of polio including 10 deaths. It was sorted out at once. The Salk vaccine was given in intramuscular injections in a span of one month. In 1957, Albert Bruce Sabin emerged on the scene. He started testing a live, oral form of vaccine as a result of which the infectious part of the virus was rendered inactive.

The writer brings out the advantages of an oral vaccine. They are long-lasting immunity, the prevention of reinfection of the digestive system, and oral administration of the vaccine at a cheaper cost. Its major disadvantage is that it is not applicable to patients with weak immune systems, since it is a live virus that is likely to cause disease in these patients. It is not conducive to those in close contact with patients who are incapable of defending against illness. Sabin’s oral vaccine is not good for enterovirus-infected patients.

But there is no denying the fact that both vaccines have their merits and demerits concerning relative safety and cost. The writer states that both vaccines are now used all over the world. The United States prefers the Sabin vaccine to the Salk one. Research is still going on improving these vaccines. There has been development in effective culturing and purification techniques paving the way for vaccines to induce higher levels of antibody formation.

The latest research in the development of the polio vaccine is to combine E. Coli’s genes with genes of poliovirus by which E. Coli can synthesize viral capsid proteins to be used in making a vaccine. The invention and use of the polio vaccine have resulted in the elimination of polio in America. In 1988, the World Health Organization set a target of eradicating polio from the whole world by 2000. The essay ends on a positive note. In the writer’s view, polio could be rooted in smallpox.

୧୯୨୧ ମସିହାର କଥା। ଆମେରିକା ପୋଲିଓ ବ୍ୟାଧର ଶିକାର ହୋଇଗଲା । Franklin Delano Roosevelt ନାମକ ଜଣେ ଯୁବ ରାଜନୀତିଜ୍ଞ ଏହି ଭୟଙ୍କର ରୋଗକୁ ପ୍ରତିରୋଧ କରିବାକୁ ସମର୍ଥ ହୋଇ ନଥିଲେ । ଭୂତାଣୁ ବହୁ ଶୀଘ୍ର ବ୍ୟାପିଗଲା ଏବଂ ଶେଷରେ ତାଙ୍କର ଗୋଡ଼ଦ୍ବୟ ଅଞ୍ଚଳ ହୋଇଗଲା । ପୋଲିଓ ସଂକ୍ରମଣର ପୂର୍ବ ଲକ୍ଷଣଗୁଡ଼ିକ ହେଉଛି ମୁଣ୍ଡବ୍ୟଥା, ଅସ୍ଵସ୍ଥିବୋଧ, ବାନ୍ତି ଏବଂ ଜ୍ଵର । ବାୟୁ, ଖାଦ୍ୟ କିମ୍ବା ଜଳ ହେଉଛି ପୋଲିଓ ଭୂତାଣୁର ବାହକ । ପ୍ରକୃତରେ ପୋଲିଓଦ୍ଵାରା ଆକ୍ରାନ୍ତମାନଙ୍କ ମଧ୍ୟରୁ ଏକ ପ୍ରତିଶତ ରୋଗୀ ମାରାତ୍ମକ ପୋଲିଓ ବ୍ୟାଶ୍‌ରେ ପୀଡ଼ିତ ହୁଅନ୍ତି । ଲେଖକ ଦୁଇ ପ୍ରକାର ପୋଲିଓ ଉପରେ ଆଲୋକପାତ କରିଛନ୍ତି; ଯଥା – spinal (ମେରୁଦଣ୍ଡ ସମ୍ବନ୍ଧୀୟ) ଏବଂ bulbar (ଫୁସ୍‌ଫୁସ୍ ସଂକ୍ରାନ୍ତୀୟ) ।

ମାଂସପେଶୀ ଯନ୍ତ୍ରଣା, ବେକ ଓ ପିଠି ଲାଠି ହୋଇଯିବା ଏବଂ ସମ୍ଭାବ୍ୟ ପକ୍ଷାଘାତ ଦୁଇ ପ୍ରକାର ପୋଲିଓର ଲକ୍ଷଣ । ପ୍ରଥମଟି ରୋଗୀର ଅଙ୍ଗପ୍ରତ୍ୟଙ୍ଗକୁ ଆକ୍ରାନ୍ତ କରୁଥିବାବେଳେ ଦ୍ୱିତୀୟଟି ଫୁସ୍‌ଫୁସ୍‌କୁ ଆକ୍ରାନ୍ତ କରେ । ଉଭୟ ପ୍ରକାର ପୋଲିଓର ଲକ୍ଷଣ ସମାନ ଏବଂ ଦୁଇଟି ମଧ୍ୟରୁ ଫୁସ୍‌ଫୁସ୍ ସଂକ୍ରାନ୍ତୀୟ (bulbar) ପୋଲିଓ ବେଶୀ ଭୟଙ୍କର । ପୋଲିଓ ବିରୁଦ୍ଧରେ ରୁଜ୍ଭେଲ୍‌ଙ୍କ ଲଢ଼େଇକୁ ଆଗେଇ ନେବାର ଭୂମିକା ସମ୍ବନ୍ଧରେ ଲେଖକ କହିଛନ୍ତି । ଏହି ଭୟଙ୍କର ବ୍ୟାଡ୍ ତାଙ୍କୁ ପରାଜିତ କରିନପାରୁ ବୋଲି ସେ ବଦ୍ଧପରିକର ଥିଲେ । ଯୁକ୍ତରାଷ୍ଟ୍ର ଆମେରିକାର ରାଷ୍ଟ୍ରପତିରୂପେ ତାଙ୍କର ସ୍ୱନାମଧନ୍ୟ ରାଜନୈତିକ ଜୀବନ ବହୁଦିନ ପାଇଁ ତିଷ୍ଠି ରହିଲା । ଏହି ମାରାତ୍ମକ ରୋଗ ସମ୍ବନ୍ଧରେ ଜନସଚେତନତା ସୃଷ୍ଟି କରିବା ଓ ଗବେଷଣାକୁ ପ୍ରୋତ୍ସାହନ ଦେବା ନିମନ୍ତେ ସେ ଯତ୍‌ପରୋନାସ୍ତି ଉଦ୍ୟମ କରିଥିଲେ ।

ପିଲାମାନେ ପୋଲିଓର ଶିକାର ହେବାର ସମ୍ଭାବନା ସବୁଠୁ ଅଧିକ । ୧୯୫୦ ଦଶକର ପ୍ରାରମ୍ଭରେ ଆତଙ୍କଗ୍ରସ୍ତ ପିତାମାତାମାନେ ସେମାନଙ୍କ ପିଲାମାନଙ୍କୁ ବିଦ୍ୟାଳୟକୁ ପଠାଇବାକୁ ଭୟ କରୁଥିଲେ । ଊନବିଂଶ ଶତାବ୍ଦୀର ଆଦ୍ୟଭାଗରେ ପୋଲିଓ ପ୍ରତିଷେଧକ ଉଦ୍ଭାବନ ପାଇଁ ଉଦ୍ୟମ ଆରମ୍ଭ ହେଲା । ପୋଲିଓର ପ୍ରତିଷେଧକଟିଏ ପାଇବାପାଇଁ ଗବେଷକମାନେ କେତେକ ଦ୍ୱାତ୍ମକ ପର୍ଯ୍ୟବେକ୍ଷଣର ଆଶ୍ରୟ ନେଇଥିଲେ । ରହସ୍ୟଜନକଭାବେ ଉନବିଂଶ ଶତାବ୍ଦୀ ପୂର୍ବରୁ କେତେକ ପିଲା ପୋଲିଓ ସଂକ୍ରମଣରୁ ମୁକ୍ତ ଥିଲେ । ସର୍ବୋପରି ମାତୃ ସ୍ତନ୍ୟପାନର ଭୂମିକା ଏ ଦିଗରେ ଗୁରୁତ୍ଵ ଭୂମିକା ନିର୍ବାହ କରୁଥିଲା । ପରିମଳ ବ୍ୟବସ୍ଥାର ଉନ୍ନତି ସତ୍ତ୍ଵେ ପରବର୍ତ୍ତୀ ବାଲ୍ୟାବସ୍ଥା ଏବଂ ବୟସ୍କାବସ୍ଥାରେ ସେମାନେ ପୋଲିଓ ଭୂତାଣୁ ସଂସର୍ଗରେ ଆସିଲେ ପୋଲିଓ ବ୍ୟାଧଗ୍ରସ୍ତ ହେବାର ବିପଦ ରହୁଥିଲା ।

ଫ୍ରାଙ୍କଲିନ୍‌ ରୁଜ୍ଭେଲ୍ସଙ୍କଦ୍ଵାରା ସ୍ଥାପିତ The March of Dimes ଯାହାକି ଜଣେ ସମ୍ମାନନୀୟ ଗବେଷକଙ୍କଦ୍ବାରା ପୋଲିଓ ବ୍ୟାଧ୍ୟାକୁ ସମ୍ପୂର୍ଣ୍ଣରୂପେ ମୂଳୋତ୍ପାଟନ ନିମନ୍ତେ କାର୍ଯ୍ୟରତ ଥିବା ଏକ ଅନୁଷ୍ଠାନ ଥିଲା । ଲେଖକ Dr. Jonas Salkଙ୍କ ସହିତ ଆମକୁ ପରିଚିତ କରାଇଛନ୍ତି ଯିଏକି ପୋଲିଓରୁ ରକ୍ଷା ପାଇଁ ବିଜ୍ଞାନସମ୍ମତ ଗବେଷଣା କରି ଚିକିତ୍ସକର ଜୀବନ ଆରମ୍ଭ କରିଥିଲେ । ସେ ପୋଲିଓ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତି ପାଇଁ କୋଷୀୟ ପୋଷଣ ଜରିଆରେ ପୋଲିଓ ଭୂତାଣୁ ବଢ଼ାଇବାର ଏକ ପଦ୍ଧତି ବ୍ୟବହାର କରିଥିଲେ । ମାଙ୍କଡ଼ର ବୃକ୍‌କରେ ପୋଷିତ ହେଉଥ‌ିବା ତିନିପ୍ରକାର ଭୂତାଣୁର ମିଶ୍ରଣରେ ଏକ ଫଳପ୍ରଦ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତ କରିଥିଲେ ।

୧୯୫୨ ମସିହାରେ ସେ ଏ କ୍ଷେତ୍ରରେ ପ୍ରଥମ ଥିଲେ । ତା’ପରେ ସେ ଫରମାଲିନ୍ ନାମକ ଏକ ରାସାୟନିକ ପଦାର୍ଥ ବ୍ୟବହାର କରିଥିଲେ ଯାହାକି ସମସ୍ତ ଭୂତାଣୁଙ୍କୁ କାର୍ଯ୍ୟକ୍ଷମ ହେବାରୁ ବଞ୍ଚିତ କରୁଥିଲା । ଡକ୍ଟର Salkଙ୍କର ଉଦ୍ଭାବନ ଯୁକ୍ତରାଷ୍ଟ୍ର ଆମେରିକା ଓ କାନାଡ଼ାର କେତେକ ଅଞ୍ଚଳରେ ଚିକିତ୍ସା କ୍ଷେତ୍ରରେ ପରୀକ୍ଷଣ ନିମନ୍ତେ ବହୁଳ ଭାବେ ପ୍ରୟୋଗ ହୋଇଥିଲା । ଏହି ପ୍ରୟୋଗର ଫଳାଫଳ ଚିକିତ୍ସାବିଜ୍ଞାନର ଇତିହାସରେ ଅଭୂତପୂର୍ବ ଥିଲା ଫଳାଫଳ ଚମତ୍କାର ଥିଲା । ଆମେରିକାର ପିଲାମାନଙ୍କୁ ଏହି ପ୍ରତିଷେଧକ ବାଣ୍ଟିବାପାଇଁ ସରକାର ତତ୍‌କ୍ଷଣାତ୍ ରାଜି ହେଇଗଲେ । କିନ୍ତୁ Saikଙ୍କ ପ୍ରତିଷେଧକର ସମସ୍ୟା ଥିଲା ଯେ ଏହାଦ୍ଵାରା ୧୦ ଜଣଙ୍କର ମୃତ୍ୟୁ ସହିତ ୨୬୦ ଜଣ ରୋଗଗ୍ରସ୍ତ ହେବାର ଜଣାପଡ଼ିଲା । ଏହି ସମସ୍ୟାର ଆଶୁ ସମାଧାନ କରାଯାଇଥିଲା ।

Saikଙ୍କ ପ୍ରତିଷେଧକ ଟୀକା ଆକାରରେ ଏକ ମାସ ସମୟ ବ୍ୟବଧାନରେ ମାଂସପେଶୀ ମଧ୍ଯରେ ଦିଆଯାଉଥିଲା । ୧୯୫୭ ମସିହାରେ ଦୃଶ୍ୟପଟ୍ଟରେ Albert Bruce Sabinଙ୍କର ଆବିର୍ଭାବ ହେଲା । ସେ ଏକପ୍ରକାର ଜୀବନ୍ତ ଓ ପାଟିରେ ଖିଆଯାଇ ପାରୁଥିବା ପ୍ରତିଷେଧକ ଉପରେ ପରୀକ୍ଷଣ ଚଳାଇଲେ ଯାହାକି ଭୂତାଣୁର ସଂକ୍ରମିତ ଅଂଶକୁ ନିଷ୍କ୍ରିୟ କରିଦେଉଥ୍ଲା । ଲେଖକ ପାଟିରେ ଖିଆଯାଉଥିବା ବୁନ୍ଦାର ଉପକାରିତା ବିଷୟରେ ବର୍ଣ୍ଣନା କରିଛନ୍ତି । ସେଗୁଡ଼ିକ ହେଲା – ଦୀର୍ଘ ପ୍ରତିଷେଧକ ଶକ୍ତି, ପରିପାକ ତନ୍ତ୍ରରେ ପୁନଃସଂକ୍ରମଣର ନିରାକରଣ, ପାଟିରେ ଖୁଆଇବା କାରଣରୁ କମ୍ ଖର୍ଚ୍ଚାନ୍ତ ହେବା । ଏହାର ମୁଖ୍ୟ ଅପକାରିତା ହେଲା ଦୁର୍ବଳ ପ୍ରତିରୋଧ ଶକ୍ତିସମ୍ପନ୍ନ ପିଲାମାନଙ୍କ କ୍ଷେତ୍ରରେ ଏହାକ ପ୍ରୟୋଗ କରାଯାଇପାରିବ ନାହିଁ, କାରଣ ଏହା ଏକ ଜୀବନ୍ତ ଭୂତାଣୁ ହୋଇଥ‌ିବାରୁ ସେମାନଙ୍କଠାରେ ରୋଗ ସୃଷ୍ଟି କରିବାର ଯଥେଷ୍ଟ ସମ୍ଭାବନା ରହିଛି ।

ପୋଲିଓ ବ୍ୟାଧ୍ୟାକୁ ପ୍ରତିରୋଧ କରିପାରୁନଥ‌ିବା ରୋଗୀମାନଙ୍କ ସଂସ୍ପର୍ଶରେ ଆସୁଥ‌ିବା ବ୍ୟକ୍ତିବିଶେଷଙ୍କ ଏହା ଲାଭପ୍ରଦ ନୁହେଁ । ଅନ୍ତନଳୀରେ‍ ଭୂତାଣୁ ସଂକ୍ରମିତ ରୋଗୀମାନଙ୍କ ନିମନ୍ତେ Sabinଙ୍କ ପାଟିରେ ଖିଆଯାଉଥ‌ିବା ପ୍ରତିଷେଧକ ଲାଭଦାୟକ ନୁହେଁ । ଆପେକ୍ଷିକ ନିରାପତ୍ତା ଓ ଖର୍ଚ୍ଚ ଦୃଷ୍ଟିରୁ ଉଭୟ ପ୍ରକାର ପ୍ରତିଷେଧକର ଭଲ ଓ ମନ୍ଦ ଗୁଣ ଯେ ରହିଛି ଏହା ନିଃସନ୍ଦେହରେ କୁହାଯାଇପାରେ । ଲେଖକ କହିଛନ୍ତି ଯେ ବର୍ତ୍ତମାନ ଉଭୟ ପ୍ରକାର ପ୍ରତିଷେଧକ ପୃଥ‌ିବୀରେ ସର୍ବତ୍ର ବ୍ୟବହାର କରାଯାଉଛି । ଯୁକ୍ତରାଷ୍ଟ୍ର ଆମେରିକାରେ Salkଙ୍କ ପ୍ରତିଷେଧକ ପରିବର୍ତ୍ତେ Sabinଙ୍କ ପ୍ରତିଷେଧକ ବ୍ୟବହାରକୁ ପସନ୍ଦ କରାଯାଇଛି । ପ୍ରତିଷେଧକର ଉନ୍ନତିକରଣ ନିମନ୍ତେ ଅନୁସନ୍ଧାନ ଜାରି ରହିଛି । ଫଳପ୍ରଦ ପୋଷଣ ଓ ବିଶୋଧନ କୌଶଳର ଉନ୍ନତି କରାଯାଇପାରିଛି ଯାହାକି ପ୍ରତିଷେଧକକୁ ଶରୀରରେ ରୋଗ ପ୍ରତିରୋଧକ ଶକ୍ତି ସୃଷ୍ଟି କରିପାରୁଛି ।

ପୋଲିଓ ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତିର ସର୍ବାଧୁନିକ ଗବେଷଣା ହେଉଛି E. Coli ଜିକୁ ପୋଲିଓ ଭୂତାଣୁର ଜିନ୍ ସହିତ ମିଶ୍ରିତ କରିବା, ଯଦ୍ବାରା E. Coli ଭୂତାଣୁର ବାହ୍ୟ ପ୍ରୋଟିନ୍ ସ୍ତରକୁ ସଂଶ୍ଳେଷଣ କରିପାରିବ ଯାହାକି ପ୍ରତିଷେଧକ ପ୍ରସ୍ତୁତି ପାଇଁ ବ୍ୟବହୃତ ହୋଇପାରିବ । ପୋଲିଓ ପ୍ରତିଷେଧକର ଉଦ୍ଭାବନ ଏବଂ ବ୍ୟବହାର ଫଳରେ ଆମେରିକରୁ ପୋଲିଓର ମୂଳୋତ୍ପାଟନ ହୋଇପାରିଛି । ୧୯୮୮ ମସିହାରେ ବିଶ୍ଵ ସ୍ବାସ୍ଥ୍ୟ ସଙ୍ଗଠନ ପକ୍ଷରୁ ୨୦୦୦ ମସିହା ସୁଦ୍ଧା ସମଗ୍ର ବିଶ୍ଵରୁ ପୋଲିଓର ମୂଳୋତ୍ପାଟନ ପାଇଁ ଲକ୍ଷ୍ୟ ରଖୁଥିଲା । ସକାରାତ୍ମକ ଭାବଧାରାକୁ ନେଇ ପ୍ରବନ୍ଧର ପରିସମାପ୍ତି ଘଟିଛି । ଲେଖକଙ୍କ ମତରେ ବସନ୍ତ ଭଳି ପୋଲିଓର ମୂଳୋତ୍ପାଟନ କରାଯାଇପାରିବ ।

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Odisha State Board BSE Odisha 10th Class History Important Questions Chapter 2 ଭାରତରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ଓ ଓଡ଼ିଶାରେ ଏହାର ପ୍ରଭାବ Important Questions and Answers.

BSE Odisha Class 10 History Important Questions Chapter 2 ଭାରତରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ଓ ଓଡ଼ିଶାରେ ଏହାର ପ୍ରଭାବ

Subjective Type Questions With Answers

ଦୀର୍ଘ ଉତ୍ତରମୂଳକ ପ୍ରଶ୍ନୋତ୍ତର
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନର ଉତ୍ତର ପ୍ରାୟ ୬୦ ଗୋଟି ଶବ୍ଦରେ ଲେଖ ।

୧। ‘ମଣ୍ଟେଗୁ-ଚେମ୍‌ସ୍ଫୋର୍ଡ଼ ସଂସ୍କାର ଆଇନ’ କାହାକୁ କୁହାଯାଏ ? ଏଥ‌ିରେ ଥ‌ିବା ତ୍ରୁଟିଗୁଡ଼ିକୁ ଦର୍ଶାଅ ।
Answer:

• ଇଂଲଣ୍ଡର ବୈଦେଶିକ ସଚିବ ଏଡ଼ଉଇନ୍ ମଣ୍ଟେଗୁ ଓ ଭାରତର ଭାଇସ୍‌ରାୟ ଲର୍ଡ଼ ଚେମ୍‌ସ୍‌ଫୋର୍ଡ଼ଙ୍କଦ୍ୱାରା ୧୯୧୯ ମସିହା ଡିସେମ୍ବର ୨୩ ତାରିଖରେ ପ୍ରବର୍ତ୍ତିତ ହୋଇଥିବା ସମ୍ବିଧାନ ସଂସ୍କାର ଆଇନକୁ ‘ମଣ୍ଟେଗୁ-ଚେମ୍‌ସ୍‌ଫୋର୍ଡ଼ ସଂସ୍କାର ଆଇନ’ କୁହାଯାଏ । ଏଥ‌ିରେ ଅନେକ ତ୍ରୁଟି ରହିଥିଲା ।
• ପ୍ରଥମ ତ୍ରୁଟି – ଭାଇସ୍‌ୟ ଓ ତାଙ୍କ କାର୍ଯ୍ୟକାରୀ ପରିଷଦ ଭାରତୀୟ ସଂସଦ ପରିବର୍ତ୍ତେ ବ୍ରିଟିଶ୍ ମନ୍ତ୍ରୀମଣ୍ଡଳ ନିକଟରେ ଉତ୍ତରଦାୟୀ ରହିଲେ ।
• ଦ୍ଵିତୀୟ ତ୍ରୁଟି – ପ୍ରାଦେଶିକ ସ୍ତରରେ ଦ୍ଵୈତ ଶାସନ କ୍ଷମତା ବ୍ୟବସ୍ଥା ଅନୁଯାୟୀ ରାଜ୍ୟପାଳଙ୍କ ଅଧୀନରେ ରଖାଯାଇଥିବା ବିଭାଗଗୁଡ଼ିକଠାରୁ ଅପେକ୍ଷାକୃତ କମ୍ ଗୁରୁତ୍ବପୂର୍ଣ୍ଣ ବିଭାଗ ମନ୍ତ୍ରୀମଣ୍ଡଳ ଅଧୀନରେ ରହିଲା ।
• ତୃତୀୟ ତ୍ରୁଟି – କେନ୍ଦ୍ରୀୟ ଓ ପ୍ରାଦେଶିକ ବ୍ୟବସ୍ଥାପକ ସଭାର ସଭ୍ୟମାନଙ୍କ ପାଇଁ ପୂର୍ବରୁ ଥିବା ମୁସଲମାନ ସମ୍ପ୍ରଦାୟ ବ୍ୟତୀତ ଶିଖ୍ ଓ ଖ୍ରୀଷ୍ଟିଆନ୍ ସମ୍ପ୍ରଦାୟଭିତ୍ତିକ ନିର୍ବାଚନମଣ୍ଡଳୀ ଗଠିତ ହେଲା ।
• ଚତୁର୍ଥ ତ୍ରୁଟି – ଭୋଟଦାନ ଅଧିକାରକୁ ଅଧିକ ସଙ୍କୁଚିତ କରାଗଲା ।

୨। ଓଡ଼ିଶାର ଅସହଯୋଗ ଆନ୍ଦୋଳନରେ ଗୋପବନ୍ଧୁ ଦାସଙ୍କ ଭୂମିକା ଉଲ୍ଲେଖ କର ।
Answer:

• ଗୋପବନ୍ଧୁ ଦାସ ୧୯୨୦ ମସିହା ଡିସେମ୍ବର ମାସରେ ଅନୁଷ୍ଠିତ ନାଗପୁର କଂଗ୍ରେସ ଅଧିବେଶନରେ ଓଡ଼ିଶାରୁ ୩୫ ଜଣ ପ୍ରତିନିଧିଙ୍କୁ ନେଇ ଯୋଗ ଦେଇଥିଲେ ।
• ସେହି ଅଧ୍ଵବେଶନରୁ ଫେରି ଗୋପବନ୍ଧୁ ୧୯୨୧ ମସିହା ଜାନୁୟାରୀ ୨୪ ତାରିଖରେ କଟକଠାରେ ଛାତ୍ରମାନଙ୍କୁ ଉଦ୍‌ବୋଧନ ଦେଇଥିଲେ ଏବଂ ଓଡ଼ିଶାରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସଙ୍ଗଠିତ କରିଥିଲେ ।
• ୧୯୨୧ ମସିହା ମାର୍ଚ୍ଚ ମାସରେ ତାଙ୍କ ଉଦ୍ୟମରେ ‘ଓଡ଼ିଶା ପ୍ରାଦେଶିକ କଂଗ୍ରେସ କମିଟି’ ଗଠିତ ହୋଇଥିଲା ।
• ସେ ଥିଲେ ଏହାର ପ୍ରଥମ ସଭାପତି ଏବଂ ଭାଗୀରଥ ମହାପାତ୍ର ସମ୍ପାଦକ ଭାବେ କାର୍ଯ୍ୟ କରିଥିଲେ ।
• ଅସହଯୋଗ ଆନ୍ଦୋଳନକୁ ଓଡ଼ିଶାରେ ଲୋକାଭିମୁଖୀ କରାଇବା ନିମନ୍ତେ ସେ ମହାତ୍ମା ଗାନ୍ଧିଙ୍କୁ ଓଡ଼ିଶାକୁ ଡକାଇ ଆଣିଥିଲେ । ତାଙ୍କ ଉଦ୍ୟମରେ ଓଡ଼ିଶାରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ବ୍ୟାପକ ହୋଇଥିଲା ।

କ୍ଷୁଦ୍ର ଉତ୍ତରମୂଳକ ପ୍ରଶ୍ନୋତ୍ତର
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନର ଉତ୍ତର ପ୍ରାୟ ୩୦ ଗୋଟି ଶବ୍ଦରେ ଲେଖ ।

୧ | “‘ନିଷ୍ଫଳ ଭାରତ ଖୁଲାଫତ୍ ସମ୍ମିଳନୀ’’ କେବେ ଅନୁଷ୍ଠିତ ହୋଇଥିଲା ? ଏଥରେ କିଏ ସଭାପତି ଭାବେ ନିର୍ବାଚିତ ହୋଇଥିଲେ ଓ ଏଥିରେ କ’ଣ ନିଷ୍ପରି ନିଆଯାଇଥିଲା ?
Answer:

1. ୧୯୧୯ ମସିହା ନଭେମ୍ବର ମାସରେ ‘ନିଷ୍ଫଳ ଭାରତ ଖୁଲାଫତ୍ ସମ୍ମିଳନୀ’ ଅନୁଷ୍ଠିତ ହୋଇଥିଲା ।
2. ଏଥରେ ଗାନ୍ଧିଜୀ ସଭାପତି ଭାବେ ନିର୍ବାଚିତ ହୋଇଥିଲେ ।
3. ତୁର୍କୀର ସୁଲତାନଙ୍କ ଅଧୀନରେ ଯଥେଷ୍ଟ ଅଞ୍ଚଳ, ବିଶେଷ କରି ମୁସଲମାନମାନଙ୍କ ପବିତ୍ର ସ୍ଥାନଗୁଡ଼ିକୁ ନ ରଖିଲେ ବର୍ଜନ ଓ ଅସହଯୋଗ କରିବାପାଇଁ ଏହି ସମ୍ମିଳନୀରେ ନିଷ୍ପତ୍ତି ନିଆଯାଇଥିଲା ।

୨। ଅସହଯୋଗ ଆନ୍ଦୋଳନକୁ ପ୍ରଥମ ଗଣ ଆନ୍ଦୋଳନ କୁହାଯାଉଥିଲା କାହିଁକି ?
Answer:

• ଅସହଯୋଗ ଆନ୍ଦୋଳନ ୧୯୨୦ ମସିହା ଅଗଷ୍ଟ ୧ ତାରିଖରେ ଆନୁଷ୍ଠାନିକଭାବେ ଆରମ୍ଭ ହୋଇଥିଲା ଓ ସମଗ୍ର ଦେଶରେ ହରତାଳ ଓ ଶୋଭାଯାତ୍ରା ସଙ୍ଗଠିତ ହୋଇଥିଲା ।
• ଏହି ଆନ୍ଦୋଳନ ବ୍ରିଟିଶ୍ ସାମ୍ରାଜ୍ୟର ମୂଳଦୁଆକୁ ଦୋହଲାଇ ଦେଇଥିଲା ।
• ଏହି ଆନ୍ଦୋଳନ ଇଂରେଜ ସରକାରଙ୍କ ବିରୋଧରେ ମହାତ୍ମା ଗାନ୍ଧିଙ୍କ ପରିଚାଳିତ ପ୍ରଥମ ଗଣଆନ୍ଦୋଳନ ଥିଲା । କାରଣ ଏଥିରେ ସମସ୍ତ ସଂପ୍ରଦାୟର ଲୋକେ ଅଂଶଗ୍ରହଣ କରିଥିଲେ ।

୩ । ହଣ୍ଟର କମିଟିର ରିପୋର୍ଟରେ ଭାରତୀୟମାନେ କ୍ଷୁବ୍ଧ ହୋଇଥିଲେ କାହିଁକି ?
Answer:

• ଜାଲିଆନାୱାଲାବାଗ୍ ହତ୍ୟାକାଣ୍ଡର ଏକ ସରକାରୀ ତଦନ୍ତ ପାଇଁ ୧୯୧୯ ମସିହା ଅକ୍ଟୋବର ୧୪ ତାରିଖରେ ଲର୍ଡ଼ ହଣ୍ଟରଙ୍କ ନେତୃତ୍ୱରେ ଏକ କମିଟି ଗଠନ କରାଯାଇଥିଲା ।
• ଏହି କମିଟି ଗଠନ ଇଂରେଜମାନଙ୍କର କେବଳ ଏକ ଧୂଆଁବାଣ ଥିଲା ।
• ଲର୍ଡ଼ସ୍‌ ସଭାରେ ଜେନେରାଲ ଡାୟାରଙ୍କ କାର୍ଯ୍ୟକୁ ଅନୁମୋଦନ କରାଯାଇଥିଲା ଓ ତାଙ୍କ ପାଇଁ ବ୍ରିଟିଶ୍ ଜନସାଧାରଣ ତିରିଶ ହଜାର ପାଉଣ୍ଡ ସଂଗ୍ରହ କରିଥିଲେ । ଏହା ଭାରତୀୟମାନଙ୍କୁ ଅତ୍ୟନ୍ତ କ୍ଷୁବ୍ଧ କରିଥିଲା ।

୪। ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ କେଉଁସବୁ ଗଠନମୂଳକ କାର୍ଯ୍ୟକ୍ରମ ଗ୍ରହଣ କରିବାପାଇଁ ସ୍ଥିର କରାଯାଇଥିଲା ?
Answer:

1. ଖଦୀ ଓ ଗ୍ରାମୋଦ୍ୟୋଗର ପ୍ରସାର ।
2. ହିନ୍ଦୁ ଓ ମୁସଲମାନମାନଙ୍କ ମଧ୍ୟରେ ଏକତା ସୃଷ୍ଟି ।
3. ଅସ୍ପୃଶ୍ୟତା ନିରାକରଣ ଓ ଜାତୀୟ ଶିକ୍ଷାନୁଷ୍ଠାନ ସ୍ଥାପନ ।

୫ । ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ କନିକା ପ୍ରଜା ବିଦ୍ରୋହ କିପରି ସଂଘଟିତ ହେଲା ଓ କନିକା ଅତ୍ୟାଚାର ଘଟଣା କାହା ମାଧ୍ୟମରେ ଲୋକଲୋଚନକୁ ଆସିପାରିଥିଲା ? ଏହାର ପରିଣାମ କ’ଣ ହେଲା ?
Answer:

• ଅସହଯୋଗ ଆନ୍ଦୋଳନ ଜନସ୍ରୋତରେ ଭାସିଯାଇ କନିକାର ପ୍ରଜାମାନେ ବିଦ୍ରୋହ କରିଥିଲେ । ଫଳସ୍ବରୂପ ଗୁଳିଚାଳନା ଯୋଗୁଁ ଦୁଇଜଣ ମୃତ୍ୟୁବରଣ କରିଥିବାବେଳେ ଅନ୍ୟମାନେ ଆହତ ହୋଇଥିଲେ ।
• ‘ସମାଜ’ ଓ ‘ଉତ୍କଳ ଦୀପିକା’ ମାଧ୍ୟମରେ କନିକା ଅତ୍ୟାଚାର ଘଟଣା ଲୋକଲୋଚନକୁ ଆସିଥିଲା ।
• ଏହିଠାରେ ଆଇନ ଅମାନ୍ୟ କରିଥିବା ଅଭିଯୋଗରେ ଗୋପବନ୍ଧୁ ଦାସ ଓ ଭାଗୀରଥ୍ ମହାପାତ୍ରଙ୍କୁ ଗିରଫ କରି ହଜାରିବାଗ ଜେଲକୁ ପଠାଯାଇଥିଲା ।

୬। ଓଡ଼ିଶାରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନକୁ ଦମନ କରିବାପାଇଁ ଇଂରେଜ ସରକାରଙ୍କର ତିନିଗୋଟି କାର୍ଯ୍ୟ ଉଲ୍ଲେଖ କର ।
Answer:

• ଓଡ଼ିଶାରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନକୁ ଦମନ କରିବାପାଇଁ ଇଂରେଜ ସରକାର ଜନସାଧାରଣଙ୍କ ଉପରେ ଲାଠିଚାଳନା ଓ ବେତ୍ରାଘାତ କରିଥିଲେ ।
• ଇଂରେଜ ସରକାର ଶହ ଶହ ସତ୍ୟାଗ୍ରହୀଙ୍କୁ କାରାରୁଦ୍ଧ କରିଥିଲେ ।
• ଇଂରେଜ ସରକାର ସମାଜ ସମ୍ବାଦପତ୍ରରେ ‘ସତ୍ୟ ହେଲେ ସାଂଘାତିକ’ ସ୍ତମରେ ସରକାରୀ ବିରୋଧୀ ଲେଖା ନିମନ୍ତେ ଗୋପବନ୍ଧୁ ଦାସଙ୍କୁ ୧ ମାସ ଜେଲକୁ ପଠାଇ ଦେଇଥିଲେ । ‘ସ୍ବରାଜ ସଙ୍ଗୀତ’ ନାମକ ପ୍ରାଚୀରପତ୍ର ଛାପିଥ‌ିବାରୁ ସମ୍ବଲପୁରର ମିଶ୍ର ପ୍ରେସ୍‌କୁ ୨୫ ଟଙ୍କା ଜୋରିମାନା କରାଯାଇଥିଲା ।

ସଂଷିପ୍ତ ଉତ୍ତରମୂଳକ ପ୍ରଶ୍ନୋତ୍ତର
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନର ଉତ୍ତର ପ୍ରାୟ ୨୦ ଗୋଟି ଶବ୍ଦରେ ଲେଖ ।

୧। ମୌଲାନା ଆବୁଲ୍ କାଲାମ୍ ଆଜାଦଙ୍କ ମୂଳନାମ କ’ଣ ଥିଲା ? ସେ ଭାରତର କେଉଁ ବିଭାଗର ମନ୍ତ୍ରୀ ଥିଲେ ?
Answer:

1. ମୌଲାନା ଆବୁଲ୍ କାଲାମ୍ ଆଜାଦ୍ରଙ୍କ ମୂଳନାମ ଥିଲା ମୋହିଉଦ୍ଦିନ୍ ଅହମ୍ମଦ ।
2. ସେ ସ୍ବାଧୀନ ଭାରତର ପ୍ରଥମ ଶିକ୍ଷାମନ୍ତ୍ରୀ ଥିଲେ ।

୨। ୧୯୨୦ ଜୁନ୍ ୯ ତାରିଖରେ କେଉଁଠାରେ ଏକ ସର୍ବଦଳୀୟ ବୈଠକର ଆୟୋଜନ କରାଯାଇଥିଲା ଓ ଏଥ‌ିରେ କ’ଣ ନିଷ୍ପତ୍ତି ନିଆଯାଇଥିଲା ?
Answer:

• ୧୯୨୦ ମସିହା ଜୁନ୍ ୯ ତାରିଖରେ ଆଲ୍ଲାହାବାଦଠାରେ ଏକ ସର୍ବଦଳୀୟ ବୈଠକର ଆୟୋଜନ କରାଯାଇଥିଲା ।
• ଏଠାରେ ଗାନ୍ଧିଜୀଙ୍କ ନେତୃତ୍ୱରେ ଖୋଲାଫତ୍ କମିଟି ଇଂରେଜ ସରକାରଙ୍କ ସହିତ ସମ୍ପର୍କ ଛିନ୍ନ କରିବାପାଇଁ ନିଷ୍ପତ୍ତି ନେଇଥିଲା ।

୩ । କଂଗ୍ରେସର ଅହମ୍ମଦାବାଦ୍ ବାର୍ଷିକ ଅଶନ କେବେ ବସିଥିଲା ଓ ଏଥ‌ିରେ କ’ଣ ନିଷ୍ପତ୍ତି ନିଆଯାଇଥିଲା ?
Answer:

• ୧୯୨୧ ଖ୍ରୀଷ୍ଟାବ୍ଦ ଡିସେମ୍ବର ମାସରେ କଂଗ୍ରେସର ଅହମ୍ମଦାବାଦ୍ ବାର୍ଷିକ ଅଧୂବେଶନ ବସିଥିଲା ।
• ଏଥିରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନକୁ ଚାଲୁରଖୁବା ନିମନ୍ତେ ନିଷ୍ପତ୍ତି ନିଆଯାଇଥିଲା ।

୪ । ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ କେଉଁ ହିଂସାତ୍ମକ ଘଟଣା ଗାନ୍ଧିଜୀଙ୍କୁ ବ୍ୟଥ୍‌ତ କରିଥିଲା ?
Answer:

1. ୧୯୨୨ ମସିହା ଫେବୃୟାରୀ ୫ ତାରିଖରେ ଉତ୍ତରପ୍ରଦେଶର ଗୋରଖପୁର ଅଞ୍ଚଳସ୍ଥିତ ଚୌରିଚୌରା ନାମକ ସ୍ଥାନରେ ଉତ୍ୟକ୍ତ ଆନ୍ଦୋଳନକାରୀମାନେ ଏକ ପୋଲିସ୍ ଷ୍ଟେସନରେ ନିଆଁ ଲଗାଇ ଦେଇଥିଲେ ।
2. ଏହା ଫଳରେ ୨୨ ଜଣ ପୋଲିସ୍ କର୍ମଚାରୀ ଜୀବନ୍ତ ଦଗ୍‌ଧ ହୋଇଥିଲେ । ଏହି ଘଟଣା ଗାନ୍ଧିଜୀଙ୍କୁ ବ୍ୟର୍ଥାତ କରିଥିଲା ।

୫ । କେଉଁ ଆନ୍ଦୋଳନରେ ଗାନ୍ଧିଜୀଙ୍କୁ ଗିରଫ କରାଯାଇ କେବେ ଓ କେତେ ବର୍ଷ ପାଇଁ ଜେଲଦଣ୍ଡ ଦିଆଯାଇଥିଲା ?
Answer:

• ଅସହଯୋଗ ଆନ୍ଦୋଳନରେ ୧୯୨୨ ମସିହା ମାର୍ଚ୍ଚ ୧୦ ତାରିଖରେ ଗାନ୍ଧିଜୀଙ୍କୁ ଗିରଫ କରାଯାଇଥିଲା ।
• ତାଙ୍କୁ ୬ ବର୍ଷ ପାଇଁ ଜେଲଦଣ୍ଡ ଦିଆଯାଇଥିଲା ।

୬। ତୁର୍କୀରେ ସାଧାରଣତନ୍ତ୍ର କେବେ ପ୍ରତିଷ୍ଠିତ ହୋଇଥିଲା ? ଏହାଦ୍ଵାରା କେଉଁ ବ୍ୟବସ୍ଥାର ଉଚ୍ଛେଦ ହୋଇଥିଲା ?
Answer:

• ୧୯୨୨ ମସିହାରେ ମୁସ୍ତାଫା କମାଲ ପାଶାଙ୍କ ନେତୃତ୍ୱରେ ତୁର୍କୀରେ ସାଧାରଣତନ୍ତ୍ର ପ୍ରତିଷ୍ଠିତ ହୋଇଥିଲା ।
• ଏହାଦ୍ୱାରା ସେଠାରେ ଥିବା ଖଲିଫା ବ୍ୟବସ୍ଥାର ଉଚ୍ଛେଦ ହୋଇଥିଲା ।

୭ । ଓଡ଼ିଶା ପ୍ରାଦେଶିକ କଂଗ୍ରେସ କମିଟି କେବେ ଗଠିତ ହୋଇଥିଲା ? ଏସ୍‌ରେ କିଏ ସଭାପତି ଓ ସମ୍ପାଦକ ଭାବେ କାର୍ଯ୍ୟ କରିଥିଲେ ?
Answer:

1. ୧୯୨୧ ମସିହା ମାର୍ଚ୍ଚ ମାସରେ ଓଡ଼ିଶା ପ୍ରାଦେଶିକ କଂଗ୍ରେସ କମିଟି ଗଠିତ ହୋଇଥିଲା ।
2. ଏହାର ପ୍ରଥମ ସଭାପତି ଓ ସମ୍ପାଦକ ଭାବେ ଯଥାକ୍ରମେ ଗୋପବନ୍ଧୁ ଦାସ ଓ ଭାଗୀରଥ୍ ମହାପାତ୍ର କାର୍ଯ୍ୟ କରିଥିଲେ ।

୮ । ଗାନ୍ଧିଜୀ କେବେ ଓଡ଼ିଶା ଆସିଥିଲେ ? ସେ କାହା ଉପରେ ଗୁରୁତ୍ଵ ଦେଇଥିଲେ ?
Answer:

• ଗାନ୍ଧିଜୀ ୧୯୨୧ ମସିହାରେ ଓଡ଼ିଶା ଆସିଥ୍‌ଲେ ।
• ଖଦୀର ବ୍ୟବହାର, ଚରଖାର ପ୍ରଚଳନ ଏବଂ ସ୍ଵଦେଶୀ ଦ୍ରବ୍ୟ ପ୍ରତି ଆଦର ଉପରେ ସେ ଗୁରୁତ୍ଵ ଦେଇଥିଲେ ।

Objective Type Questions with Answers
A. ଗୋଟିଏ ବାକ୍ୟରେ ଉତ୍ତର ଲେଖ ।

1. ଖୁଲାଫତ୍ ଆନ୍ଦୋଳନର ନେତୃତ୍ୱ ନେଇଥ‌ିବା ଅଲ୍ଲୀ ଭ୍ରାତାଦ୍ୱୟ କିଏ ଥିଲେ ।
Answer:
ଖଲାଫତ୍ ଆନ୍ଦୋଳନର ନେତୃତ୍ୱ ନେଇଥିବା ଅଲ୍ଲୀ ଭ୍ରାତାଦ୍ବୟ ଯଥାକ୍ରମେ ସୌକତ ଅଲ୍ଲୀ ଓ ମହମ୍ମଦ ଅଲ୍ଲୀ ଥିଲେ ।

2. ୧୯୯୨ ମସିହାରେ କାହାକୁ ମରଣୋତ୍ତର ଭାବେ ‘ଭାରତରତ୍ନ’ ସମ୍ମାନରେ ଭୂଷିତ କରାଯାଇଥିଲା ?
Answer:
ମୌଲାନା ଆବୁଲ କାଲାମ ଆଜାଦ୍‌ଙ୍କୁ ୧୯୯୨ ମସିହାରେ ମରଣୋତ୍ତର ଭାବେ ‘ଭାରତରତ୍ନ’ ସମ୍ମାନରେ ଭୂଷିତ କରାଯାଇଥିଲା ।

3. ୧୯୧୯ ଭାରତ ଶାସନ ଆଇନଦ୍ୱାରା ଭାରତୀୟମାନଙ୍କର କେଉଁ ଅଧିକାରକୁ ସଙ୍କୁଚିତ କରାଯାଇଥିଲା ?
Answer:
୧୯୧୯ ଭାରତ ଶାସନ ଆଇନଦ୍ୱାରା ଭାରତୀୟମାନଙ୍କର ଭୋଟଦାନ ଅଧିକାରକୁ ସଙ୍କୁଚିତ କରାଯାଇଥିଲା ।

4. କାହା ନେତୃତ୍ୱରେ ଜାଲିଆନାଓ୍ବାଲାବାଗ୍ ହତ୍ୟାକାଣ୍ଡର ସରକାରୀ ତଦନ୍ତ କରାଯାଇଥିଲା ?
Answer:
ଲର୍ଡ଼ ହଣ୍ଟରଙ୍କ ନେତୃତ୍ବରେ ଜାଲିଆନାଓ୍ବାଲାବାଗ୍ ହତ୍ୟାକାଣ୍ଡର ସରକାରୀ ତଦନ୍ତ କରାଯାଇଥିଲା ।

5. ଅସହଯୋଗ ଆନ୍ଦୋଳନର ମୂଳନୀତି କ’ଣ ଥିଲା ?
Answer:
ଇଂରେଜ ଅନୁଷ୍ଠାନ ଓ ବିଦେଶୀ ଦ୍ରବ୍ୟ ବର୍ଜନ ଅସହଯୋଗ ଆନ୍ଦୋଳନର ମୂଳନୀତି ଥିଲା ।

6. ଅସହଯୋଗ ଆନ୍ଦୋଳନର ମୁଖ୍ୟ ଗଠନମୂଳକ କାର୍ଯ୍ୟକ୍ରମଟି କ’ଣ ଥିଲା ?
Answer:
ଖଦୀ ଓ ଗ୍ରାମୋଦ୍ୟୋଗର ପ୍ରସାର ଥିଲା ଅସହଯୋଗ ଆନ୍ଦୋଳନର ମୁଖ୍ୟ ଗଠନମୂଳକ କାର୍ଯ୍ୟକ୍ରମ ।

7. ‘ତିଳକ ସ୍ଵରାଜ ପାଣ୍ଠି’ ଗଠନ କେଉଁ ଆନ୍ଦୋଳନର ଉଲ୍ଲେଖନୀୟ କାର୍ଯ୍ୟ ଥିଲା ?
Answer:
‘ତିଳକ ସ୍ଵରାଜ ପାଣ୍ଠି’ ଗଠନ ଅସହଯୋଗ ଆନ୍ଦୋଳନର ଉଲ୍ଲେଖନୀୟ କାର୍ଯ୍ୟଥିଲା ।

8. ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସ୍ଥଗିତ ନିଷ୍ପତ୍ତିକୁ ଏକ ଜାତୀୟ ବିପର୍ଯ୍ୟୟ ଭାବରେ କିଏ ବର୍ଣନା କରିଥିଲେ ?
Answer:
ସୁଭାଷ ଚନ୍ଦ୍ର ବୋଷ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସ୍ଥଗିତ ନିଷ୍ପତ୍ତିକୁ ଏକ ଜାତୀୟ ବିପର୍ଯ୍ୟୟ ଭାବେ ବର୍ଣ୍ଣନା କରିଥିଲେ ।

9. ୧୯୨୨ ମାର୍ଚ୍ଚ ୧୦ ତାରିଖରେ ଗାନ୍ଧିଜୀଙ୍କୁ ଗିରଫ କରାଯାଇ କେତେ ବର୍ଷ ପାଇଁ ଜେଲଦଣ୍ଡ ଦିଆଯାଇଥିଲା ?
Answer:
୧୯୨୨ ମାର୍ଚ୍ଚ ୧୦ ତାରିଖରେ ଗାନ୍ଧିଜୀଙ୍କୁ ଗିରଫ କରାଯାଇ ୬ ବର୍ଷ ପାଇଁ ଜେଲଦଣ୍ଡ ଦିଆଯାଇଥିଲା ।

10. ୧୯୨୨ ମସିହାରେ ତୁର୍କୀରେ କିଏ ସାଧାରଣତନ୍ତ୍ର ପ୍ରତିଷ୍ଠା କରିଥିଲେ ?
Answer:
ମୁସ୍ତାଫା କମାଲ ପାଶା ୧୯୨୨ ମସିହାରେ ତୁର୍କୀରେ ସାଧାରଣତନ୍ତ୍ର ପ୍ରତିଷ୍ଠା କରିଥିଲେ ।

11. ଭାଗୀରଥ୍ ମହାପାତ୍ର କିଏ ଥିଲେ ?
Answer:
ଭାଗୀରଥ୍ ମହାପାତ୍ର ଓଡ଼ିଶା ପ୍ରଦେଶ କଂଗ୍ରେସ କମିଟିର ପ୍ରଥମ ସମ୍ପାଦକ ଥିଲେ ।

12. ସମ୍ବଲପୁର ଜିଲ୍ଲାସ୍କୁଲର କାହା ପ୍ରେରଣାରେ ଛାତ୍ରମାନେ ଅସହଯୋଗ ଆନ୍ଦୋଳନରେ ଯୋଗଦେଲେ ?
Answer:
ପଣ୍ଡିତ ଲକ୍ଷ୍ମୀନାରାୟଣ ମିଶ୍ରଙ୍କ ପ୍ରେରଣାରେ ସମ୍ବଲପୁର ଜିଲ୍ଲାସ୍କୁଲର ଛାତ୍ରମାନେ ଅସହଯୋଗ ଆନ୍ଦୋଳନରେ ଯୋଗଦେଲେ ।

13. ସତ୍ୟବାଦୀ ବନବିଦ୍ୟାଳୟ କେବେ ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ?
Answer:
୧୯୧୯ ମସିହାରେ ସତ୍ୟବାଦୀ ବନବିଦ୍ୟାଳୟ ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ।

14. ‘ଉତ୍କଳ ସ୍ୱରାଜ୍ୟ ଶିକ୍ଷା ପରିଷଦ’ କେଉଁଠାରେ ଗଠିତ ହୋଇଥିଲା ?
Answer:
କଟକଠାରେ ‘ଉତ୍କଳ ସ୍ୱରାଜ୍ୟ ଶିକ୍ଷା ପରିଷଦ’ ଗଠିତ ହୋଇଥିଲା ।

15. କଟକରେ ସ୍ୱରାଜ ଆଶ୍ରମ କାହିଁକି ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ?
Answer:
ସ୍ଵେଚ୍ଛାସେବୀ ଓ କର୍ମୀମାନଙ୍କୁ ପ୍ରଶିକ୍ଷଣ ଦେବାପାଇଁ କଟକରେ ସ୍ୱରାଜ ଆଶ୍ରମ ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ।

B. ଗୋଟିଏ ଶବ୍ଦରେ ଉତ୍ତର ଲେଖ ।

1. ଗୋପବନ୍ଧୁ ଦାସଙ୍କଦ୍ଵାରା ପ୍ରତିଷ୍ଠିତ ସମ୍ବାଦପତ୍ରର ନାମ କ’ଣ ଥିଲା ?
Answer:
‘ସମାଜ’

2.ପ୍ରଥମ ବିଶ୍ଵଯୁଦ୍ଧ ପରେ ତୁର୍କୀ ସହିତ କେଉଁ ଚୁକ୍ତି ସ୍ୱାକ୍ଷରିତ ହୋଇଥିଲା ?
Answer:
ସେଭର୍ସ ଚୁକ୍ତି

3. ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ ଜଗତ୍‌ସିଂହପୁରଠାରେ କେଉଁ ଆଶ୍ରମ ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ?
Answer:
ଅଳକା ଆଶ୍ରମ

4. କେଉଁ ଅନୁଷ୍ଠାନ ‘ସ୍ବରାଜ୍ୟ ସମାଚାର’ ପତ୍ରିକା ପ୍ରକାଶ କରୁଥିଲେ ?
Answer:
ସ୍ଵରାଜ ମନ୍ଦିର

5. ସମ୍ବଲପୁରର ମିଶ୍ର ପ୍ରେସ୍ କେଉଁ ପ୍ରାଚୀରପତ୍ର ଛାପି ୨୫ ଟଙ୍କା ଜୋରିମାନା ଦେଇଥିଲା ?
Answer:
‘ସ୍ଵରାଜ ସଙ୍ଗୀତ’

6. ମୁସଲମାନମାନଙ୍କଦ୍ଵାରା ଗଠନ କରାଯାଇଥିବା ସଙ୍ଗଠନକୁ କ’ଣ କୁହାଯାଉଥିଲା ?
Answer:
ମୁସଲିମ୍ ଲିଗ୍

7. କାହା ନେତୃତ୍ୱରେ ‘ସ୍ଵରାଜ ମନ୍ଦିର’ ଗଠିତ ହୋଇଥିଲା ?
Answer:
ହରେକୃଷ୍ଣ ମହତାବ

8. କେବେ ପୁରୀଠାରେ ବିଦେଶୀ ବସ୍ତ୍ରଗୁଡିକରେ ଅଗ୍ନିସଂଯୋଗ କରାଯାଇଥିଲା ?
Answer:
୧୯୨୧ ଅଗଷ୍ଟ ୩

9. ଲୋକମାନ୍ୟ ବାଲ ଗଙ୍ଗାଧର ତିଳକଙ୍କ ସ୍ମୃତିରେ ଗଠିତ ପାଣ୍ଠିର ନାମ କ’ଣ ଥିଲା ?
Answer:
‘ତିଳକ ସ୍ବରାଜ ପାଣ୍ଠି’

10. ୧୯୧୯ ନଭେମ୍ବର ମାସରେ ନିଷ୍ଫଳ ଭାରତ ଖୋଲାଫତ୍ ସମ୍ମିଳନୀରେ ସଭାପତି ଭାବେ କିଏ ନିର୍ବାଚିତ ହୋଇଥିଲେ ?
Answer:
ମହାତ୍ମା ଗାନ୍ଧି

11. ମୌଲାନା ଆବୁଲ୍ କାଲାମ୍ ଆଜାଦ୍ କେବେ ମରଣୋତ୍ତର ଭାରତ ରତ୍ନ ସମ୍ମାନରେ ଭୂଷିତ ହୋଇଥିଲେ ?
Answer:
୧୯୯୨

12. ଛାତ୍ରମାନଙ୍କୁ ଜାତୀୟତାବାଦୀ ଶିକ୍ଷା ଦେବା ନିମନ୍ତେ ଗୋପବନ୍ଧୁ କେଉଁ ବିଦ୍ୟାଳୟ ପ୍ରତିଷ୍ଠା କରିଥିଲେ ?
Answer:
ସତ୍ୟବାଦୀ ବନବିଦ୍ୟାଳୟ

13. ଇଂରେଜ ସରକାରଙ୍କ ବିରୋଧରେ ଗାନ୍ଧିଜୀଙ୍କଦ୍ବାରା ପରିଚାଳିତ ପ୍ରଥମ ଗଣ-ଆନ୍ଦୋଳନର ନାମ କ’ଣ ?
Answer:
ଅସହଯୋଗ ଆନ୍ଦୋଳନ

14. ହଣ୍ଟର କମିଟି କେବେ ଗଠିତ ହେଲା ?
Answer:
୧୯୧୯ ଅକ୍ଟୋବର ୧୪

15. କେବେ ନାଗପୁରଠାରେ କଂଗ୍ରେସର ବାର୍ଷିକ ଅଧୂବେଶନରେ ଅସହଯୋଗ ଆନ୍ଦୋଳନ ପ୍ରସ୍ତାବ ଅନୁମୋଦିତ ହୋଇଥିଲା ?
Answer:
୧୯୨୦ ଡିସେମ୍ବର

16. ଓଡ଼ିଶାର ପ୍ରଥମ ସମ୍ବାଦପତ୍ରର ନାମ କ’ଣ ?
Answer:
ଉତ୍କଳ ଦୀପିକା

17. ସ୍ଵାଧୀନ ଭାରତର ପ୍ରଥମ ଶିକ୍ଷାମନ୍ତ୍ରୀ କିଏ ଥିଲେ ?
Answer:
ମୌଲାନା ଆବୁଲ୍ କାଲାମ୍ ଆଜାଦ୍

18. ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ ଗୋପବନ୍ଧୁ ଦାସଙ୍କୁ କେଉଁ ଜେଲକୁ ପଠାଇ ଦିଆଯାଇଥିଲା ?
Answer:
ହଜାରିବାଗ ଜେଲ

19. ଖଲିଫା ବ୍ୟବସ୍ଥାର ଉଚ୍ଛେଦ କେବେ କରାଯାଇଥିଲା ?
Answer:
୧୯୨୨

20. ଅସହଯୋଗ ଆନ୍ଦୋଳନ ସମୟରେ କନିକାର ରାଜା କିଏ ଥିଲେ ?
Answer:
ରାଜେନ୍ଦ୍ର ନାରାୟଣ ଭଞ୍ଜଦେଓ

C. ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

1. ପ୍ରଥମ ବିଶ୍ବଯୁଦ୍ଧରେ ତୁର୍କୀ _____ ଶକ୍ତି ବିରୋଧରେ ଯୁଦ୍ଧ କରିଥିଲା ।
Answer:
ମିତ୍ରଶକ୍ତି

2. ମୁସଲମାନମାନଙ୍କ ଧର୍ମଗୁବୁଙ୍କ ______ କୁହାଯାଉଥିଲା ।
Answer:
ଖଲିଫା

3. ‘ଇଣ୍ଡିଆ ଉଇନ୍ସ ଫ୍ରିଡ଼ମ୍’ ପୁସ୍ତକ ______ ରଚନା କରିଥିଲେ ।
Answer:
ମୌଲାନା ଆବୁଲ କାଲାମ

4. ମୌଲାନା ଆବୁଲ କାଲାମ _______ ମସିହାରେ ଭାରତରତ୍ନ ପୁରସ୍କାର ପାଇଥିଲେ ।
Answer:
୧୯୯୨

5. _______ ମସିହା ନଭେମ୍ବରରେ ନିଷ୍ଫଳ ଭାରତ ଖୁଲାଫତ୍ ସମ୍ମିଳନୀର ସଭାପତିଭାବେ ନିର୍ବାଚିତ ହୋଇଥିଲେ ।
Answer:
୧୯୧୯

6. ପ୍ରଥମ ବିଶ୍ଵଯୁଦ୍ଧ ପରେ ସେଭର୍ସ ଚୁକ୍ତି ______ ଉପରେ ଲଦି ଦିଆଯାଇଥିଲା ।
Answer:
ତୁର୍କୀ

7. ୧୯୨୦ ମସିହା ଅଗଷ୍ଟ ୧ ତାରିଖରେ ଭାରତର ମହାନ୍ ନେତା _______ ଙ୍କର ଦେହାନ୍ତ ହୋଇଥିଲା ।
Answer:
ବାଲଗଙ୍ଗ।ଧର ତିଲକ

8. ________ ଆନ୍ଦୋଳନ ବ୍ରିଟିଶ୍ ସାମ୍ରାଜ୍ୟର ମୂଳଦୁଆକୁ ଦୋହଲାଇ ଦେଇଥିଲା ।
Answer:
ଅସହଯୋଗ

9. ରାଓଲାତ ଆଇନ ________ ମସିହାରେ ପ୍ରଣୟନ କରାଯାଇଥିଲା ।
Answer:
୧୯୧୯

10. ଜେନେରାଲ ଡାୟାରଙ୍କ ପାଇଁ ଇଂଲଣ୍ଡର ଜନସାଧାରଣ ________ ହଜାର ପାଉଣ୍ଡ ସଂଗ୍ରହ କରିଥିଲେ ।
Answer:
୩୦

11. ମୋତିଲାଲ ନେହେରୁ ଜଣେ ଖ୍ୟାତନାମା ______ ଥିଲେ ।
Answer:
ବାରିଷ୍ଟର

12. _______ ର ଯୁବରାଜଙ୍କ ଭାରତ ପରିଦର୍ଶନକୁ ବାସନ୍ଦ କରାଯାଇଥିଲା ।
Answer:
ୱେଲସ

13. ୧୯୨୧ ମସିହା ଡିସେମ୍ବର ମାସରେ କଂଗ୍ରେସର ଅଧ୍ଵଂଶନ ______ ଠାରେ ବସିଥିଲା ।
Answer:
ଅହମ୍ମଦାବାଦ

14. ୧୯୨୧ ମସିହା ଅଗଷ୍ଟ ୧୧ ଓ ୧୪ ତାରିଖରେ ______ ଠାରେ ଜନସାଧାରଣ ବିଦେଶୀ ଲୁଗା ପୋଡ଼ି ଦେଇଥିଲେ ।
Answer:
କଟକ

15. ଗୌରମୋହନ ଦାସ _______ ଠାରେ ଘରୋଇ ନ୍ୟାୟାଳୟ ପ୍ରତିଷ୍ଠା କରିଥିଲେ ।
Answer:
ବାଲେଶ୍ଵର

16. _________ ସମ୍ବାଦପତ୍ର ଜାତୀୟ ଚେତନା ସୃଷ୍ଟି କରିବାରେ ପ୍ରମୁଖ ଭୂମିକା ଗ୍ରହଣ କରିଥିଲା ।
Answer:
ସମାଜ

17. ‘ସତ୍ୟ ହେଲେ ସାଂଘାତିକ’ ଲେଖା ପାଇଁ ଗୋପବନ୍ଧୁଙ୍କୁ _______ ମାସ ଜେଲ ଭୋଗିବାକୁ ପଡ଼ିଥିଲା ।
Answer:
୧

18. ‘ସ୍ଵରାଜ ସଙ୍ଗୀତ’ ପ୍ରାଚୀରପତ୍ର ଛାପିବାପାଇଁ ______ ପ୍ରେସ୍‌କୁ ୨୫ ଟଙ୍କା ଜୋରିମାନା କରାଯାଇଥିଲା ।
Answer:
ମିଶ୍ର

19. ଗୋପବନ୍ଧୁ ଦାସ ଓଡ଼ିଶା ପ୍ରଦେଶ କଂଗ୍ରେସ କମିଟିରୁ ପ୍ରଥମ ______ ଭାବେ ନିର୍ବାଚିତ ହୋଇଥିଲେ ।
Answer:
ସଭାପତି

20. ନାଗପୁର କଂଗ୍ରେସ ଅଧ୍ଵବେଶନରେ ଓଡ଼ିଶାରୁ ______ ଜଣ ପ୍ରତିନିଧ୍ଵ ଯୋଗ ଦେଇଥିଲେ ।
Answer:
୩୦

D. ଠିକ୍ ଉକ୍ତି ପାଇଁ (✓) ଚିହ୍ନ ଓ ଭୁଲ୍ ଉକ୍ତି ପାଇଁ (×) ଚିହ୍ନ ଦିଅ ।

1. ମୌଲାନା ଆବୁଲ କାଲାମ ଆଜାଦଙ୍କ ମୂଳ ନାମ ଥଲା ମୋହିଉଦ୍ଦିନ୍ ଅଲ୍ଲୀ ।
2. ୧୯୨୦ ମସିହା ଜୁନ୍ ୯ ତାରିଖରେ ଅହମ୍ମଦାବାଦ୍ ଠାରେ ସର୍ବଦଳୀୟ ବୈଠକର ଆୟୋଜନ କରାଯାଇଥିଲା ।
3. ଚୌରିଚୌରା ଦୁର୍ଘଟଣାରେ ୨୨ ଜଣ ପୋଲିସ କର୍ମଚାରୀ ଜୀବନ୍ତ ଦଗ୍ଧ ହୋଇଥିଲେ ।
4. ଚୌରିଚୌରା ଦୁର୍ଘଟଣାରେ ୨୨ ଜଣ ପୋଲିସ କର୍ମଚାରୀ ଜୀବନ୍ତ ଦଗ୍ଧ ହୋଇଥିଲେ ।
5. ବାଲେଶ୍ଵରଠାରେ ଉତ୍କଳ ସ୍ୱରାଜ୍ୟ ଶିକ୍ଷା ପରିଷଦ ସ୍ଥାପିତ ହୋଇଥିଲା ।
6. ସ୍ଵେଚ୍ଛାସେବୀ ଓ କର୍ମୀମାନଙ୍କୁ ପ୍ରଶିକ୍ଷଣ ଦେବାପାଇଁ କଟକରେ ‘ସ୍ଵରାଜ ଆଶ୍ରମ’ ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ।
7. ସମ୍ବାଦପତ୍ର ‘ସମାଜ’ ମହାତ୍ମା ଗାନ୍ଧିଙ୍କଦ୍ଵାରା ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ।
8. ୧୯୨୦ ମସିହାରେ ‘ସମାଜ’ ସମ୍ବାଦ ପତ୍ରିକା ପ୍ରତିଷ୍ଠା କରାଯାଇଥିଲା ।
9. ଇଂରେଜ ସରକାର ଓଡ଼ିଶାରେ ଆନ୍ଦୋଳନକୁ ଦମନ କରିବାପାଇଁ ଲାଠିଚାଳନା ଓ ବେତ୍ରାଘାତ କରିଥିଲେ ।
10. ‘ସ୍ଵରାଜ ସଙ୍ଗୀତ’ ଏକ କବିତା ପୁସ୍ତକ ଥିଲା ।

Answer:
1. ✗
2. ✗
3. ✓
4. ✓
5. ✗
6. ✓
7. ✗
8. ✓
9. ✓
10. ✗

E. ‘କ’ ସ୍ତମ୍ଭ ସହିତ ‘ଖ’ ସ୍ତମ୍ଭର ମିଳନ କର ।

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text D: The Case against Man Textbook Activity Questions and Answers.

Class 11th Alternative English Chapter 16 The Case against Man Question Answers CHSE Odisha

The Case against Man Class 11 Questions and Answers

Activity-14

a) It means that mankind is although a living organism, it is also a thing or inanimate object.
b) The unrestrained population growth is compared with cancer. This is a good comparison because its growth will kill mankind as cancer.
c) If the present rate of population growth continues the ecology will be spoilt.
d) the thesis of the essay is increasing the birth rate and its control He waits to describe things and then concludes.
e) This is really a problem that has been shown by the author perfectly and which does not need any other way of description.
f) Interrelation and interdependence are in common among the living and nonliving things on earth.
g) The conclusion of the essay is— At the rate, we are going without birth control, then even if science serves us in an absolutely ideal way, we will reach the planetary high-rise with no animals but men, with no plants but algae, with no room for even one more person by AD. 2430.
h) The essay starts with the interrelation of the living and the nonliving and ends in control of the birth rate which will help the organism to double itself.

Activity-15

	Headings	Paragraph Numbers
a)	The thesis of the problem: “birth control”	21
b)	Reasons/causes Lowering death rate	19
c)	Examples: The number of Homo Sapiens increase	16
d)	Suggested solution: Birth control	19
e)	Special Features of the Development of the Argument (if any)	19, 20
f)	Conclusion Ready birth control without delay	20, 21, 22, 23

Activity – 16

Rather than exploiting the environment shouldn’t we be in a partnership? If we continue to waste the earth’s resources as if there were no tomorrow, there could well be no tomorrow. By the year 2010, one-third of the world’s cropland will have turned to dust, of people, will face starvation. All this is happening since our civilization has kept on expanding, on the assumption that the world’s resources are limitless. But merely stopping growth is not the answer. What we need is development that works in partnership with the environment that uses the earth’s resources more productively and after all is suitable at the same time. This is the reason why our organization Earth life exists.

Activity-17

Anita : Hello, Banita. You have a debate today, haven’t you?
Banita: Yes Anni. It’s at 2-30 p.m.
Anita: What’s it about?
Banita: It’s about population explosion.
Anita : Population explosion ! It’s a burning topic, isn’t it?
Banita: Yes, it is. But it’s a topic that needs many things to incorporate.
Anita: What’re you going to hint at?
Banita: Just the causes and consequences of population growth.
Anita: Won’t you suggest any solution?
Banita: Yes, I will.
Anita: Why’s the population on the rise now?
Banita: It’s owing to the lower death rate.
Anita: Lower death rate! Aren’t people dying now? You’ll see in the papers hundreds of people are dying every day.
Banita: No, no. People are dying but their number is eye-catching due to the high population and media network.
Anita: Do you think that the death rate has really come down?
Banita: Is there doubt about it? The death rate is very much lower than before. Thousands of people were dying of starvation, Cholera, and Smallpox in the past. But we don’t see these diseases active now. A number of villages were having mass funerals with the approach of such a disease.
Anita: You’re quite right Banita. This was a usual case that is not seen these days. Thank you.

Activity -18

Pranati: Hello. This is 250845
Minati: Can I speak to Pranati, please?
Pranati: Yes, speaking
Minati: Hi Pranati, it’s Minati here.
Pranati: Listen, Minu, We’d proposed to go to the cinema this afternoon, hadn’t we?
Minati: Yes, we had. You told me to book a pair of tickets and inform me earlier what’s about.
Pranati: I’m quite sorry. I failed to book tickets at the counter. I’d gone to do it, but I wasn’t able to.
Minati: What’s really happened?
Pranati: The counters were overcrowded. None of the counters was free to buy a ticket at.
Minati: The film has recently been released. People must be thronging to see it.
Pranati: Yes, Blakers are moving about. They are charging very high. I didn’t feel like purchasing a ticket from them.
Minati: OK. Don’t mind. We’ll see the cinema within a couple of days. The rush will be subsiding. Thank you.
Pranati: Welcome

Activity – 19

• In Kingston, Jamaica’s capital, RSLs own cruiser is waiting to introduce them to the unique world of the Caribbean.
• Every Tuesday a British Airways flight leaves Heathrow for Jamaica.
• Like all our ships, this cruiser has been specially designed to give you maximum comfort, luxury, and enjoyment.
• For this lucky one it’s the beginning of an unforgettable air-sea holiday with the world’s leading cruiser company. The Royal Seafaring Line.
• For many of the passengers, it’s just a normal scheduled flight, but for some, it’s the start of something very special.
• Whether you choose relaxation on board or stimulation on land,’ you will have the holiday of a lifetime.
• And it’s all included in the price — Just 1,995 for 21 days.
• While you can thus spend a perfect holiday without leaving the ship, there is also the added attraction of fascinating store visits at each of our parts of the cell.
• So you can relax on the vest sun deck, bide your time with a cocktail, or dance till dawn in the nightclub or in the discotheque.
• So don’t delay- See your travel agents today.

The Case Against Man Summary in English

Section – D
Part – One

Summary:
The first mistake is to think of man as a thing in itself. It is, however, a part of an intricate problem of life. Life gets its energy from the sun. Five billion years back, the earth had undergone a vast revolution. On its first appearance, it lacked an ocean and an atmosphere. Far within the solid crust, there are slow continual changes whose hot springs, volcanoes, and earthquakes are the more noticeable manifestations here on the surface.

Portions of the surface water with solar radiation developed complicated compounds called ‘life’. It has assumed a complex proportion. But, life forms are as much part of the structure of the Earth as any inanimate portion is. It is all an inseparable part of a whole if any animal is isolated totally from other forms of life, and death by starvation will surely follow. If isolated from water, death by dehydration will follow even faster.

If isolated from air, death by asphyxiation will take place. Isolation from the sun will bring death to the animal world. The inanimate portion also suffers. The entire planet and solar system are closely interrelated. A planet is a life form made up of nonliving portions. For instance, a man is composed of 50 trillion cells of a variety of types, all interrelated and interdependent.

Part – Two

Summary:
Sometimes, the neat economy of growth within an organism such as a human being is disrupted. The growing of a group of cells is stopped. If one type of organism began to multiply without limit killing its competitors, the same thing would happen in ecology. The earth’s human population is estimated to have been 150 million al the time of Julius Caesar. This population since then has been on the rise. It is really an alarming proportion.

The current increase of the human population qualifies Homo Sapiens as ecological cancer. However, this cancerous growth must be stopped. It can be done by raising the death rate or towering the birth rate. There is no other alternative. If we do nothing, the death rate will rise fabulously. Lowering the birth rate is surely the preferable way.

Analytical outlines of the Text

• The first mistake is to think of man as a thing in itself.
• It is, however, a part of an intricate problem of life.
• Life gets its energy from the sun.
• Five billion years back, the earth had undergone a vast revolution.
• On its first appearance. it had lacked an ocean and an atmosphere.
• Far within the solid crust, there are slow continual changes.
• The hot springs, volcanoes, and earthquakes are the more noticeable manifestations here on the surface.
• Portions of the surface water with solar radiation developed complicated compounds called life.
• It has assumed a complex proportion.
• But life forms are as much part of the structure of the Earth as any inanimate portion is.
• It is all an inseparable part of a whole.
• Any animal is isolated totally from other forms of life.
• It will surely follow death by i$arvation.
• Any animal is isolated from water.
• It will follow death by dehydration.
• Any animal is isolated from the air.
• It will take place death by asphyxiation.
• Isolation from the sun will bring death to the animal world.
• The inanimate portion also suffers.
• The entire planet and solar system are closely interrelated.
• A planet is a life form made up of nonliving, portions.

• For instance, a man is composed of 50 trillion cells of a variety of types all interrelated and interdependent.
• Sometimes, the net economy of growth within an organism such as a human being is disrupted.
• The growing of a group of cells is stopped.
• One type of organism began to multiply without limit killing its competitors.
• The disruption will happen in ecology.
• The earth’s human population is estimated to have been 150 million at the time of Julius Caesar.
• This population since then has been on the rise.
• It is really an alarming proportion.
• The current increase in the human population qualifies Homo Sapiens as ecological cancer.
• However, this cancerous growth must be stopped.
• It can be achieved in two ways.
• One is by raising the death rate.
• The other is by lowering the birth rate.
• There is no other alternative.
• We have to do something.
• Otherwise, the death rate will rise fabulously.
• Lowering the birth rate is certainly the preferable way.

Meaning of difficult words

crust – the thin hard surface of the earth.
versatile – clever to do a number of things, good at doing a lot of different things.

asphyxiation – death by choking.
reefs – a line of sharp rocks, often made of coral.
quiescent – becoming quiet or silent, not developing or doing anything.
cougar – a puma, a large brown wild cat of North West America.
decimated – killed large numbers of ruined a large part of something.
predators – animals that live by killing and eating other animals.
ecology – the study of living things in their surroundings.
Homo Sapiens – the type of human beings that inhabit the earth now.
catastrophically- in a terribly destructive manner.

Read More:

• Text A: That Lean, Hungry Look Question Answer
• Text B: What is Art? Question Answer
• Text C: Psychobabble Question Answer
• Text D: The Case against Man Question Answer

Odisha State Board CHSE Odisha Class 11 Approaches to English Book 1 Solutions Unit 4 Text B: What is Art? Textbook Activity Questions and Answers.

Class 11th Alternative English Chapter 14 What is Art? Question Answers CHSE Odisha

What is Art? Class 11 Questions and Answers

Activity-5
Getting The Main Idea Of The Paragraph

Find out a suitable title for each of the paragraphs in Text-B (Part one)

Paragraph     Title
1                 :
2                 :
3                 :
4                 :
5                 :
6                 :
7                 :

Answer:
Paragraph -1 : Title – Defining Art.
Paragraph -2: Title – Relationship of Art.
Paragraph -3 : Title – Art Transmitting Human Thought.
Paragraph -4 : Title – Activity of Art.
Paragraph -5: Title – Man’s capacity of Receiving other’s Emotional Dimensions.
Paragraph -6 : Title – Infecting feelings.
Paragraph -7 : Title – Object of Joining Another.

Descriptive Sequence

After going through part two Text-B, read the whole of Text-B (That is, both the parts) and arrange the following items in the sequence in which they are presented in the Text.
a) Discussing what art is not.
b) Talking about defining art.
c) Speaking about the characteristics of art.
d) Citing examples that does not amount to art.
e) Comparing art with speech.
f) Arriving at a definition of art.
g) Speaking of the variety of feelings on which art is based.

Answer:
a) Talking about defining art.
b) Comparing art with speech.
c) Speaking of the characteristics of art.
d) Speaking of the variety of feelings on which art is based.
e) Arriving at a definition of art.
f) Discussing what art is not.
g) Citing examples of what does not amount to art.

Activity-7
Reacting To The Ideas In The Text

• Art is superior to speech because it transmits feelings as well as thoughts because a man transmits his thoughts to another by words but by art, he transmits his feelings.
• Tolstoy speaks of the essential elements of art in paragraph 5 but seems to contradict himself in the next paragraph. The views presented in these two paragraphs can, however, be concealed.
• The analogy between the boy who encounters- a wolf and the artist who recreates his emotions in a work of art is now appropriate because the feelings and emotions are equally infected with one another.
• The writer begins his essay by saying that art should not be considered “as a means to pleasure” but should be considered “as one of the conditions of human life”. And he has proved this in his essay taking suggestive examples from various lores of life.

Activity – 8

a) A direct approach is chosen to define the term ‘renaissance’ in passage 1, but a descriptive technique is followed in passage 2 to define the term ‘elegance’.
b) An etymological analysis of the term ‘renaissance’ finds an outlet in passage 1 but the implied meaning of the term ‘elegance’ is given in passage 2.
c) A general meaning of the word ‘renaissance’ is reflected in passage 1 whereas the views and considerations of the word ‘elegance’ have been found in passage 2.

Activity-9
Remedial Grammar

My town is an excellent place to live in, I think it is wonderful. It is an important town, because, it is the center of the district administration. It is also great because of the two very famous museums. The weather here is nice. It is hot in summer with occasional rains and is cool in winter. I like my home town very much.

What is Art? Summary in English

Section – B
Part – One
Read below the first paragraph of Leo Tolstoy’s “What is Art ?” and try to guess the writer’s purpose.
In order to define art correctly it is necessary first of all to cease to consider it as a means to pleasure and to consider it as one of the conditions of human life viewing it in this way, we can’t fail to observe that art is one of the means of intercourse between man and man. Now read part one of Text B and note how Tolstoy develops his idea of art across the paragraph.

Summary:
Leo Tolstoy defines art to cease to consider it as a means to pleasure and to consider it as one of the conditions of human life. Art is an intercourse between man and man. The receiver of every work of art enters into a certain kind of relationship both with him who produced or is producing the art and with all those who simultaneously, previously or subsequently receive the same artistic impression, speech transmitting the thoughts and experiences of man serve as a means of union among them and art serves a similar purpose.

A man communicates himself with another by means of words and by it he transmits his feelings. A man shares his feelings by listening to another man. When one man laughs, another becomes merry to hear it. But when a man cries, another feels sorry. A man is excited or irritated and another man who sees him is brought to a similar state of mind by his movements or by the sounds of his voice. A man expresses courage and determination or sadness and calmness and this state of mind passes on to others.

A man suffers expressing his suffering through groans and spasms and this suffering transmits itself to other people. A man expresses his feelings of admiration, devotion, fear, respect, or love to certain objects, persons, or phenomena and others infected by the same feelings of administration, devotion, fear, respect or love to some objects, persons or phenomena. Art begins when one person expresses his feelings by certain external indications in order to join others or others.

For instance, a boy having experienced fear of encountering a wolf relates the encounters, and in order to evoke in others the feelings he has experienced describes his conditions before the encounter, the surrounding of the world, his own lightheartedness, and then, the wolf’s appearance, its movements, the distance between himself and the wolf and so forth. If only the boy when telling the story again experiences the feelings he has lived through and infects the heart and compels them to feel that he had experienced is art.

It is also art if a man having experienced either the fear of suffering or the attraction of enjoyment expresses these feelings on canvas or in marble so that others are infected by them. It is again art of a man who feels or imagines to himself feelings of delight gladness, sorrow, despair, courage or despondency and the transition by sounds from me to another of those feelings and expresses them by sounds so that the hearers are inflected by them and experience them as they were experienced by the composer.

Analytical outlines of the text:

• According to Leo Tolstoy, art is a means to provide pleasure.
• He also considers it as one of the conditions of human life.
• Art is an intercourse between man and man.
• The receiver of every work of art enters into a certain kind of relationship both with him.
• It is a relationship with him who produced or is producing the art.
• It relates to those who simultaneously, previously or subsequently receive the same artistic impression.
• Speeches transmitting the thoughts and experiences .of men serve as a means of union, among them and art serves a similar purpose.
• A man communicates himself with another by means of words and by it he transmits his feelings.

• A man shares his feelings by listening to another.
• One man laughs, and another becomes merry to hear it.
• One man cries and another becomes feel sorry.
• A man as excited or irritated.
• Another is brought to the same state by seeing it.
• He acquires it by his movements or the sounds of his voice.
• A man expresses his courage or determination.
• This state of mind passes to another.
• A man expresses his sadness or calmness.
• This state of mind passes to another.
• A man expresses his suffering through groans or spasms.
• It transmits itself to other people.
• A man expresses his admiration, devotion, fear, respect, and love to certain objects, persons or phenomena.
• Others are infected by the same feelings.
• Art begins when one joins others with the same feelings.
• For instance, a boy experiences of fear by encountering a wolf.
• He expresses this fear in order to evoke a feeling in others.

• He provides an elaborate description of it.
• He expresses the conditions before the encounter.
• He also expresses the surrounding of the words.
• He also expresses his distance from the wolf.
• On the other hand, he also expresses the wolf’s appearance, its movement, distance from him, etc.
• When he compels others to feel his experiences, it is called art.
• Hence, to be an art, the feelings of suffering and enjoyment should be infected by them.
• When a man feels or imagines those feelings of delight, gladness, sorrow, despair, etc. it is called art.
• Therefore, art refers to the transmission of the sounds of those feelings from one man to other so that one must be infected by them and also experiences them by themselves.

Meaning of difficult words:

simultaneously – happening at the same time.
previously – formerly.
subsequently – followingly.
intercourse – deal with, interact, and communicate.
transmit – sends, communicates.
groan – moan, lamentation.
spasms – muscular contraction, stiffness of muscles.
encountering – facing, confronting, meeting
despondency – misery, sorrow, unhappiness.
transition – change, transformation, movement.

Text-B
Part – Two

Summary:
The feelings which the artist transmits to others are varied and many. Some are very strong and some are very weak, some significant and others insignificant, some very bad, and others very good. Patriotic love, self-devotion and yielding to fate or to God in drama, raptures of lovers in. a novel, voluptuousness in a picture, courage in triumphal marches, merriment in a dance, and humor in a funny story are all different forms of art. If the feelings of the author are transmitted to the spectators, they are deemed to be rightly infected.

Art is certainly a human activity that consists of external signs hands-on to other’s feelings he had lived through and that others are infected by these feelings and also experience them. The metaphysicians state that art is not the manifestation of some, mysterious idea of the beauty of God. Physiologists view a game in which man lets off his excess stored-up energy, is not man s expression of emotion by external signs. It is neither pleasure nor the production of pleasing objects.

Analytical outlines of Part Two.

• The artist’s transmission of feelings to others is varied and many.
• Some are very strong and others are very weak.
• Some are significant and others are insignificant.
• Even some are very good and others are very bad.
• There are different forms of art.
• Patriotic love and self-devotion are the same.
• Raptures of lovers in a novel, and voluptuousness in a picture are others.
• Also, courage in a triumphal march, merriment in a dance, and humor in a tunny spry are still others.
• If the feelings of the writer are transmitted to the audience, they are deemed to be rightly infected.

• Art is, certainly, a human activity.
• It consists of external signs hands-on to other’s feelings.
• Others are infected by these feelings.
• They also experience these feelings.
• The metaphysicians opine that art is not the manifestation of some mysterious idea of the beauty of God.
• Physiologists view it is a game in which man gets off his excess stored up energy.
• It is not the expression of man’s emotion by external signs.
• It is neither pleasure nor the production of pleasing objects.
• It is a means of union among men joining them together in the same feelings.

Read More:

• Text A: That Lean, Hungry Look Question Answer
• Text B: What is Art? Question Answer
• Text C: Psychobabble Question Answer
• Text D: The Case against Man Question Answer