Class 8

In this article, we will share MP Board Class 8th Special English Solutions Chapter 2 The Making of the Missile Man PDF download, Class 8 English Mp Board Chapter 2, These solutions are solved subject experts from the latest edition books.

Class 8th English Chapter 2 The Making of the Missile Man Poem Question Answers MP Board

Supplementary Reader Class 8 MP Board Chapter 2 Solutions

Word Power

(A) Choose appropriate words from the box and fill in the blanks.
(Scientists, astronaut, surgeon, dentist, pilot)

1. One who flies an aeroplane is called a _____
2. The doctor who takes care of your teeth is called a _____
3. Kalpana Chawla was a famous _____
4. A person who studies and has expert knowledge of one or more natural or physical sciences is a _____
5. The doctor who specialises in surgical operation is called a _____

Answers:

1. pilot
2. dentist
3. astronaut
4. scientist
5. surgeon.

(B) Pick out the odd word and encircle it.

1. surgeon, dentist, physician, electrician
2. poet, musician, author, writer
3. SLV-3, Agni, Apple, Moon
4. English, Mathematics, Hindi, Sanskrit
5. Mother Teresa, SrinivasaRamanujan, C.V. Raman, Marie Curie

Answers:

1. electrician
2. musician
3. Moon
4. Mathematics
5. Mother Teresa.

(C) Pick out the words which do not mean the same as the given words:

1. humble : greedy, meek, unassertive, unassuming .
2. hideous : dreadful, ghastly, monstrous, pretentious
3. interfere : obstruct, incessant, hamper, impede
4. confusion : muddle, congestion, commotion, bewilderment
5. preserve : protect, engage, safeguard, conserve

Answers:

1. greedy
2. pretentious
3. incessant
4. congestion
5. engage.

(D) Write the opposite of the following words:

1. broad _____
2. sell _____
3. tolerate _____
4. close _____
5. dim _____

Answers:

1. narrow
2. purchase
3. intolerate
4. open
5. bright

Comprehension

(A) Answer these questions :

MP Board Class 8 English Chapter 2 Question 1.
Who is the Missile Man ?
Answer:
Dr. APJ Abdul Kalam is the Missile Man.

Class 8 English Chapter 2 MP Board Question 2.
What did Abdul Kalam’s family do for their living ?
Answer:
Abdul Kalam’s family was engaged in ferrying pilgrims between Dhanushkhodi and Rameshwaram in Tamil Nadu.

MP Board Class 8th English Chapter 2 Question 3.
What did Abdul Kalam do for the support of his family?
Answer:
Abdul Kalam collected tamarind seeds and sold them to a shop. He also collected a popular Tamil newspaper at Rameshwaram railway station and helped her cousin in selling it. He earned money for these services. He supported his family with this money.

English Chapter 2 Class 8 MP Board Question 4.
What religious books had he read in his childhood?
Answer:
He read the Koran and the Gita in his childhood.

MP Board Class 8 English Chapter 2 Question Answer Question 5.
How did the school celebrate the occasion when Kalam became President ?
Answer:
When Kalam became President, the school celebrated the day as a festival. Sweets were distributed among the teachers and students.

Class 8 MP Board English Chapter 2 Question 6.
One day a new teacher came to the class, he asked Abdul Kalam not to sit with Ramanandha’. How did Ramanadha’s father react to this incident ?
Answer:
Ramanandha’s father sent for the teacher. He told the teacher that what he had done would only breed communalism.

Class 8th English Chapter 2 MP Board Question 7.
How did Abdul Kalam’s sister help him when he got admission in Madras Institute of Technology ?
Answer:
His sister mortgage her ornaments to raise the required fee.

Class 8 Chapter 2 English MP Board Question 8.
How did Abdul Kalam learn that perseverance makes miracles ?
Answer:
Once Kalam was invited his teacher Mr. Subramania Iyer. Iyer’s wife was an orthodox Hindu lady. So she refused to serve food to Kalam. Iyer, without showing any adverse reaction, served the food himself. It was his patience and gravity that prompted his wife another time to serve food to Kalam. This taught Kalam the lesson of perseverance.

(B) Match the expressions in A with their correct meanings in B:

Answers:

1. – 5
2. – 4
3. – 2
4. – 1
5. – 3.

(C) Read the following sentences and tick true or false :

Class 8 English Chapter 2 The Making Of The Missile Man Question 1.
Abdul Kalam comes from a high profile family. (T/F)
Answer:
False

MP Board Solutions Class 8 English Chapter 2 Question 2.
His favourite subject were English, Maths, Science and Tamil. (T/F)
Answer:
True

The Missile Man Of India Class 8 Question 3.
Ramanandha’s father was the high priest of Rameswaram temple. (T/F)
Answer:
True

Chapter 2 English Class 8 MP Board Question 4.
Abdul Kalam got the highest award ‘Bharat Ratna’ in 1997. (T/F)
Answer:
True

MP Board Class 8 English Chapter 2 Pdf Question 5.
He is our tenth President. (T/F)
Answer:
False

Let’s Learn

(A) Study these sentences :

They hardly noticed their loss.
It is scarcely the size of a pin-head.
Note: The use of ‘scarcely’ is very similar to that of ‘hardly’.

hardly = almost not /almost none
scarcely = almost not only just)

Make sentences from B to match the ones given under A.

A

1. Mr. Mohan’s house is near the market.
2. The gathering was very small.
3. The watch is very small.
4. We must hurry up. It is nearly time for the train.
5. I cannot buy the book now.

B

Answers:

1. It is scarcely a furlong from the market.
2. These were scarcely twenty people present.
3. It is scarcely the size of a paisa.
4. We have scarcely five minutes to get to the station.
5. I have scarcely any money now.

(B) Study the sentences in the table below:

Now make sentences from B to match the ones given under A.

1. Take 100 rupees now. Send me a telegram if you happen to need more money later.
2. You can find me in the school in the evening. If by some chance I am not there, you can certainly find me at home.
3. If by some chance I cannot come and see you tomorrow, I will talk to you on the phone.
4. Try to come and see me. But if you can’t, write to me.

B

Answer:

1. Send me a telegram in case you need more money.
2. I will be at home in case you can’t find me in the school.
3. I will talk to you on the phone in case you can’t come and see me.
4. Write to me in case I can’t come and see you.

Let’s Talk

Here is a conversation between two students. Read carefully and talk with your friend as given in the example.

Student – I

Student – II

• Now talk to your friend on the childhood of the Missile Man as shown in the above example.

Clues : (his full name, his teachers in school, early education, his friends, his discoveries/inventions etc.)
Answer:

Let’s read

Indian Space Programme

Read carefully the information about our

space programme :

• The first Indian satellite “Aryabhatta” was launched on April 19, 1975 from the Soviet Union.
• “Bhaskara I” was launched on June 7, 1979 from the Soviet Union.
• “Rohini I” was launched on July 19, 1981 from India.
• “APPLE” was launched on June 19, 1981 from French Guyana.
• “Shriharikota” and “Vikram Sarabhai Space Centre” are the most important places in our space programme.

Now complete the following table :

Answer:

Let’s Write

Here are clues about our genius mathematician Shrinivasa Ramanujan. On the basis of clues write a few sentences about Ramanujan.

Answer:
Srinivasa Ramanujan was born on 22nd Dec. 1887.

The place of his birth is Erode in Tamil Nadu.
He belonged to a humble family.
He joined his first service in Madras Port Trust.
His favourite subject was Mathematics.
He did his research work in mathematics from Cambridge University (London).
He was awarded Trinity College Fellowship.
He died on 20th April, 1920.

The Making of the Missile Man Word Meanings

Page 5 : Determination – firmness of purpose – दुढ इच्छाशक्ति। Virtue – inner quality – गुण। Ferrying – conveying in a boat across water, transporting – नाव से लोगों को देना। Destroyed – बर्बाद कर दिया। Pursuing – continuing – जारी रखना। Crisis – time of great difficulty or danger – संकट की स्थिति। Wished – desired – इच्छा करना। Passion – strong will – उत्कट इच्छा।

Page 6 : Dim – dull – धुधला। Tolerate – bear – सहन करना। Breed – lead to something, to cause something – उकसाना। Orthodox – holding conventional beliefs – पुराने विचारों वाला। Perseverance – continued steady efforts to achieve an aim – जुझारूपन। Miracle – surprising and welcome act or event which does not follow the laws of nature and is therefore thought to be caused by God – चमत्कार। Motto – aim, goal – उदेश। Eminent-famous – प्रसिद्द। Fascinated – attracted – अकरणित होना। Aeronauties branch – वैग्नानीक शाखा।

very big – विशाल। Martgage – a legal arrangements by which a bank lends one money – बंध की गिरवी। Required – need – आवश्यक। Compassion – feeling of pity – दया। Vowed – pledged – प्रतिगना करना। Redeem – to pay the necessary money to clear a debt – पैसा चुकाना। Commitment – a pledge – प्रतिगना। Gesture – an expressive movement or action – रुख। Below – present something as gift to somebody – प्रदान करना। Exhorted – tried hard to persuade somebody to do something – किसी को कुछ करने के लिए प्रेरित करना। Emulate – try to do well or better than somebody – कुछ अच्छा करने की कोशिश करना।

MP Board English Reader Question Answer

• Don’t Quit Class 8 Question Answer
• The Making of the Missile Man Class 8 Question Answer
• The False Flight Class 8 Question Answer
• The Narmada: The Lifeline of Madhya Pradesh Class 8 Question Answer
• Trees are the Kindest Things I know Class 8 Question Answer
• How Mowgli Joined the Pack Class 8 Question Answer
• Nothing You Can’t Do Class 8 Question Answer
• English is Amazing (Poem) Class 8 Question Answer
• To the Rescue Class 8 Question Answer
• The Sacrifice Class 8 Question Answer
• Children who are Wanted Every Hour Class 8 Question Answer
• The Cherry Tree Class 8 Question Answer
• How I Thought my Grandmother to Read Class 8 Question Answer
• The Jet Age Class 8 Question Answer
• Ahilyabai – A Great Indian Queen Class 8 Question Answer
• To the New Year Class 8 Question Answer
• Manav Sangrahalaya: One of Its Kind Class 8 Question Answer
• Handy Andy at the Post Office Class 8 Question Answer
• Bismillah Khan: The Shehnai Maestro Class 8 Question Answer
• Yoga: A Way of Life Class 8 Question Answer
• Work is Worship Class 8 Question Answer

In this article, we will share MP Board Class 8th General English Solutions Chapter 14 Azad: The Martyr PDF download, Lesson 14 Azad: The Martyr, these solutions are solved subject experts from the latest edition books.

Class 8 English Chapter 14 Azad: The Martyr Question Answers MP Board

MP Board Class 8th English Lesson 14 Azad: The Martyr Questions And Answers

Read and Learn
(पढ आर याद:)
Answer:
Do Yourself.

Word Power
(शब्द सामर्थ्य):

A. The letters in the following words are jumbled. Rearrange them to make meaningful words:
(निम्नलिखित शब्दों के अक्षरों को अस्त-व्यस्त कर दिया गया है। उन्हें अर्थपूर्ण शब्द बनाने के लिये पुनः व्यवस्थित करें:)
Answer:

B. Make antonyms of the following words by using the prefix ‘dis’:
(उपसर्ग लगकर नम्नलिखित शब्दो के विलोम लिखें:)

1. appoint
2. courage
3. appear
4. able
5. agree
6. comfort
7. connect

Answer:

1. disappoint
2. discourage
3. disappear
4. disable
5. disagree
6. discomfort
7. disconnect.

Now complete the following sentences using the correct form of the words from the box:
(अब बॉक्स में से शब्दों के सही रूप लेकर निम्नलिखित वाक्यों को पूरा करोः)

1. His father ……….. to his joining politics.
2. The mice ……. as soon as they saw the cat.
3. Meghna was ……. to see her marks in English.
4. He was …….. after the car accident.
5. They ………….. the phone because people were misusing it.
6. One should not …….. the children.
7. Although he comes from a royal family, he put up with all the ………।

Answer:

1. disagreed
2. disappeared
3. disappointed,
4. disabled
5. disconnected
6. discourage
7. discomfort.

C. Use the following pair of words in sentences to show the difference in their meanings as given in the example:
(उदाहरण में दिये के अनुसार निम्नलिखित शब्दों के जोड़ों को उनके अर्थों में भिन्नता दिखाने के लिए वाक्यों में प्रयोग करें:)
1. Sun – The sun sets in the west.
Son-I have one son.

2. New – My pink dress is new.
Knew – Ram knew my address.

3. Wait – Please wait for a minute.
Weight – My weight is 40 kg.

4. See – We all see with our eyes.
Sea – The sea water is salty.

5. Pray – We should all pray to God.
Prey – The lion was killing its prey.

6. Hear – We hear with our ears.
Here Please come here.

7. Hair – I have short hair.
Hare – Hare is a swift animal.

Comprehension
(बोध प्रश्न):

Class 8 English Chapter 14 Azad The Martyr Question 1.
When and where was Chandra Shekhar born ?
(व्हेन एण्ड व्हेयर वॉज़ चन्द्रशेखर बॉर्न ?)
चन्द्रशेखर का जन्म कब और कहाँ हुआ था ?
Answer:
Chandra Shekhar was born on 23rd July, 1906 in village Bhabra in Jhabua district.
(चन्द्रशेखर वॉज़ बॉर्न ऑन 23 जुलाई, 1906 इन विलेज भाबरा इन झाबुआ डिस्ट्रिक्ट।)
चन्द्रशेखर का जन्म 23 जुलाई, 1906 को भाबरा गाँव, जिला झाबुआ में हुआ था।

Mp Board Class 8 English Chapter 14 Question 2.
Where did he receive his early education?
(व्हेअर डिड ही रिसीव हिज़ अर्ली एज्यूकेशन ?)
उन्होंने अपनी प्राथमिक शिक्षा कहाँ प्राप्त की।
Answer:
He received his early education in Bhabra.
(ही रिसीव्ड हिज़ अर्ली एज्यूकेशन इन भाबरा।)
उन्होंने अपनी प्राथमिक शिक्षा भाबरा में प्राप्त की।

Class 8 English Chapter 14 Mp Board Question 3.
In 1920-21 which movement attracted Chandra Shekhar ?
(इन 1920-21 विच मूवमेण्ट अट्रैक्टिड चन्द्र शेखर ?)
सन् 1920-21 में चन्द्र शेखर को किस आन्दोलन ने आकर्षित किया ?
Answer:
He was attracted by the non-cooperation movement of 1920-21 under the leadership of Mahatma Gandhi.
(ही वॉज़ अट्रैक्टिड बॉय द नॉन कोऑपरेशन मूवमेंट ऑफ 1920-21 अन्डर द लीडरशिप ऑफ महात्मा गाँधी.)
वो महात्मा गाँधी के नेतृत्व में होने वाली, नॉन-कोऑपरेशन मूवमेन्ट जो 1920-21 में हुई, से आकर्षित हुए।

Class 8 English Chapter 14 Question Answer Question 4.
One of Chandra Shekhar’s statement provoked the magistrate. What was that statement?
(वन ऑफ चन्द्र शेखर्स स्टेटमेन्ट प्रोवोक्ड द मैजिस्ट्रेट। व्हॉट वॉज़ दैट स्टेटमेन्ट ?)
चन्द्र शेखर के एक कथन ने मजिस्ट्रेट को भड़का दिया था। वह कथन क्या था ?
Answer:
He gave his name as ‘Azad’, his father’s name as ‘Swatantra, and his residence as ‘Prison’. This provoked the magistrate.
(ही गेव हिज नेम ऐज़ ‘आजाद’, हिज़ फादर्स नेम ऐज़ ‘स्वतन्त्र’, एण्ड हिज़ रेसिडेन्स ऐज़ ‘प्रिज़न’. दिस प्रोवोक्ड द मेजिस्ट्रेट.)
उन्होंने अपना नाम ‘आज़ाद’ बताया, अपने पिता का नाम ‘स्वतन्त्र’ व घर ‘जेल’ बताया। इससे मजिस्ट्रेट भड़क गए।

Chandrashekhar Azad Question Answer Question 5.
Why was British police able to surround him?
(व्हाई वाज ब्रिटिश पुलिस ऐबिल टू सराउन्ड हिम ?)
ब्रिटिश पुलिस उनको चारों ओर से घेरने में क्यों समर्थ थी ?
Answer:
On February 27, 1931 Chandra Shekar Azad met two of his comrades at Alfred Park in Allahabad. He was betrayed by an informer to British Police. So he was surrounded by British Police.
(ऑन फैब्रुऐरी 27, 1931 चन्द्रशेखर आजाद मेट टू हिज कॉमरेड्स ऐट अल्फ्रेड पार्क इन अलाहाबाद. ही वाज बिट्रेड बाइ ऐन इनफोरमर टू ब्रिटिश पुलिस. सो ही वाज सराउन्डड बाइ ब्रिटिश पुलिस.)
27 फरवरी, 1931 को चन्द्रशेखर आजाद इलाहाबाद के अल्फ्रेड पार्क में अपने दो साथियों से मिले। किसी एक मुखबिर ने ब्रिटिश पुलिस को सूचना देकर उनके साथ विश्वासघात कर दिया। अतः वह ब्रिटिश पुलिस द्वारा घेर लिये गये।

Chandra Shekhar Azad Lesson 10 Question 6.
How did Chandra Shekhar prove his name Azad?
(हाउ डिड चन्द्रशेखर प्रूव हिज़ नेम आजाद ?)
चन्द्रशेखर ने अपना नाम आज़ाद कैसे सिद्ध किया ?
Answer:
Once he was surrounded by police in a park. He didn’t surrender but shot himself. Thus he proved his name Azad by not being caught alive.
(वन्स ही वॉज़ सराउन्डिंड बाय पुलिस इन अ पार्क. ही डिडन्ट सरेन्डर बट शॉट हिमसेल्फ. दस ही प्रूव्ड हिज़ नेम आज़ाद बाय नॉट बीइन्ग कॉट अलाइव।)
एक बार पुलिस ने एक पार्क में उन्हें घेर लिया था। उन्होंने समर्पण नहीं किया, पर स्वयं को गोली मार ली। इस तरह उन्होंने जिन्दा न पकड़े जाने से अपना नाम आजाद सिद्ध कर दिया।

Let’s Learn
(आओ याद करें):

You meet Manju and she tells you about her grandmother. Now tell your friend Anil what Manuja told you.
(तुम मंजू से मिलती हो और वह अपनी दादी के बारे में तुम्हें बताती है। अब अपने मित्र अनिल को जो अंजू ने कहा, बताओ।)
Answer:

1. She told that her grandmother was widely travelled.
2. She told that her grandmother had recently bought a new dress.
3. She told that her grandmother had written number of books.
4. She told that her grandmother was very interesting company.
5. She told that her grandmother had recently broken her left foot.

Let’s Talk
(आओ बात करें):

The timetable for class VIII is given below. Use the information given in the timetable to complete the conversation between Abdul and Meena. (Work in Pairs)
(कक्षा VIII की समय-सारणी नीचे दी गई है। अब्दुल और मीना के बीच हुई बातचीत को समय-सारणी में ही सूचना का प्रयोग कर पूरा करो।)

Abdul: Hello, Meena! When did you come back? (I had a lovely time. It was a long journey to Kolkata.)
Meena: Hello, Abdul! Glad to see you. I came back yesterday. My journey was enjoyable. I was in Delhi for 2 weeks.
Abdul: Now you have to study again according to your timetable. Have you seen it?
Meena: Yes’ have you seen it?
Abdul: No, that’s why I want to know something about it.
Abdul: Who teaches us EVS?
Meena: Mr. R. Khan teaches us EVS.
Abdul: How long is the lunch break?
Meena: The Lunch break is 35 minutes long.
Abdul: When does the science class begin?
Meena: The science class begins at 2:30 p.m.
Abdul: In which period do we have sports?
Meena: We have sports in 7th period.
Abdul: What does Mr. Mathew teach?
Meena: Mr. Mathew teaches us Maths.
Abdul: What is taught in the fifth period?
Meena: Sanskrit is taught in the fifth period.
Abdul: At what time does the school close?
Meena: Our school closes at 3.40 p.m.

Let’s Read
(आओ पढ़ें):

Read the given passage carefully:
Now answer the following questions:
(दिये गये गद्यांश को पढ़ें। अब निम्नलिखित प्रश्नों के उत्तर दें।
(A) Say whether the following statements are true or false.

1. Bhagat Singh’s name echoed all over the country.
2. His name did not inspire the youth of India.

Answer:

1. True
2. False.

(B) 1. “Bhagat Singh became a symbol of freedom.” Who said this?
Answer:
Pandit Jawahar Lal Nehru said this.

2.. Arrange the following words in an alphabetical order
Answer:
amazing, brought, hero, symbol.

3. Supply a suitable title to the passage.
Answer:
Suitable title to the passage is –
“Bhagat Singh: A Real Hero.”

Let’s Write
(आओ लिखो):

Write a paragraph about Chandra Shekhar Azad by putting the given sentences in the correct order:
(दिये गये वाक्यों को सही क्रम में रखकर चन्द्रशेखर आजाद के बारे में एक पैराग्राफ लिखोः)
Answer:
Chandra Shekhar was born in village Bhabra in Jhåbua district. For higher education he went to Sanskrit Pathshala at Varanasi. As a revolutionary, he adopted the surname ‘Azad’. He was deeply hurt by the Jallianwala Bagh massacre in Amritsar in 1919. A special award will be presented at Balidan Diwas.

Let’s do it
(आओ इसे करें):

(1) Collect the pictures of any five patriotic leaders. Paste them in your notebook and write five sentences on each:
(किन्हीं पाँच देशभक्त नेताओं के चित्र एकत्रित करो। उन्हें अपनी नोट-बुक में चिपकाओ और प्रत्येक पर पाँच वाक्य लिखोः)

(2) Class game/Activity
Divide the class into two teams. The first team mentions three things about a person, an object or a thing that everybody known is the class. The other team guesses who the person or what the object or thing is. One example has been given.
(कक्षा खेल/क्रिया कलाप)
(कक्षा को दो टीमों में बाँट दो। पहली टीम एक व्यक्ति, एक उद्देश्य या वस्तु के बारे में तीन बात बताती है जो कक्षा में सभी जानते हैं। दूसरी टीम अनुमान लगाती है कि वह व्यक्ति, उद्देश्य या चीज कौन-सी है। एक उदाहरण दिया गया है।)
Answer:
(1) Students can collect the pictures of patriots such as Bhagat Singh, Chandra Shekhar “Azad’, Mahatma Gandhi, Lal Bahadur Shastri and Bal Gangadhar Tilak. They can themselves write five sentences on each
(2) Students can do this in their class themselves.

Azad: The Martyr Word Meanings

Receive (रिसीव) – पाना; Ardent (आर्डेन्ट) – प्रबल, तीव्र; Followers (फॉलोअर्स) – शिष्य, अनुयायी; Disguised (डिज़गाइज़्ड) – छद्म वेश; Betray (बिट्रे) – धोखा देना, विश्वासघात करना; Escape (ऐस्केप) – बचाव; भाग – निकलना; Priest (प्रीस्ट) – पुजारी; Clutches (क्लचिस) – पकड़, चंगुल; Valiantly (वैलिअन्टली) – बहादुरी से; Provoke (प्रोवोक) – उकसाना, भड़काना; Pledge (प्लेज) – प्रतिज्ञा, वचन; Mentor (मेन्टर) – परामर्शदाता; Memorial (मेमोरियल) – यादगार, स्मारक; Flog (फ्लॉग)-कोड़े, लगाना; Claim (क्लेम) – दावा करना; Fierce (फिअर्स) – भयानक; Patriotism (पेट्रीऑटिज्म) – देशभक्ति; Revolutionary (रेवॉल्यूशनरी) – क्रान्तिकारी; Avenge (अवेन्ज) – बदला लेना; Involve (इनवॉल्व) – शामिल होना; Massacre (मेसैकर) – जनसंहार; Inspire (इन्सपायर) – प्रेरणा देना; Aggressive (अग्रेसिव) आक्रामक; Homage (होमेज) – श्रद्धांजलि; Capture (कैप्चर) – पकड़ना; Favorite (फेवरिट) – मनपसन्द।

Azad: The Martyr Summary, Pronunciation & Translation

1. Chandra Shekhar, son of Pandit Sita Ram Tiwari and Jagrani Devi, was born in village Bhabra in Jhabua district (Present Madhya Pradesh) on 23rd July 1906.

चन्द्रशेखर, सन ऑफ पंडित सीताराम तिवारी एण्ड जगरानी देवी, वाज़ बॉर्न इन विलेज भाबरा इन झाबुआ डिस्ट्रिक्ट (प्रेजेन्ट मध्य प्रदेश) ऑन ट्वन्टी थर्ड जुलाई, 1906.

अनुवाद:
पंडित सीताराम तिवारी और जगरानी देवी के पुत्र चन्द्रशेखर का जन्म झाबुआ जिला (वर्तमान में मध्य प्रदेश)। के भाबरा गाँव में 23 जुलाई, 1906 को हुआ था।

2. He received his early education in Bhabra. For higher education he went to Sanskrit Pathshala at Varanasi. He was an ardent follower of Hanuman and once disguised himself as a priest in a Hanuman temple to escape the clutches of the British police. Young Chandra Shekhar was attracted by the great national upsurge of the non-violent non-cooperation movement of 1920 – 21 under the leadership of Mahatma Gandhi. He joined it and was arrested and produced before the magistrate. He gave his name as Azad’, his father’s name as ‘Swatantra’, and his residence as ‘Prison. This provoked the magistrate, who sentenced him for fifteen lashes of a flog.

ही रिसीव्ड हिज़ अर्ली एजुकेशन इन भाबरा. फॉर हायर एजूकेशन ही वैन्ट टु संस्कृत पाठशाला, ऐट वाराणसी. ही वाज़ ऐन ऑर्डेन्ट फालोअर ऑफ हनुमान एण्ड वन्स डिसगाइज्ड हिमसैल्फ ऐज़ अ पीस्ट इन अ हनुमान टेम्पिल टु ऐस्केप द क्लचिस ऑफ द ब्रिटिश पुलिस. यंग चन्द्रशेखर वाज अट्रेक्टिड बाइ द ग्रेट नेशनल अपसर्ज ऑफ द नान-वायलेंट नॉन-कोऑपरेशन मूवमेंट ऑफ. 1920-21 अण्डर द लीडरशिप ऑफ महात्मा गाँधी. ही जॉइन्ड इट एण्ड वाज़ अरैस्टिड एण्ड प्रोड्यूस्ड बिफोर द मजिस्ट्रेट. ही गेव हिज़ नेम ऐज़ ‘आजाद’, हिज़ फादर्स नेम ऐज़ ‘स्वतन्त्र’, एण्ड हिज़ रेजीडेन्स एज़ ‘प्रिज़न’. दिस प्रोवोक्ड द मजिस्ट्रेट, हू सेन्टेन्स्ड हिम फॉर फिफ्टीन लैशेस ऑफ अ फ्लॉग।

अनुवाद:
उनकी प्रारम्भिक शिक्षा भाबरा में हुई। उच्च शिक्षा उन्होंने वाराणसी के संस्कृत पाठशाला में प्राप्त की वह हनुमान के अत्यधिक भक्त थे और एक बार ब्रिटिश पुलिस के चंगुल से बचने के लिए उन्होंने एक हनुमान मन्दिर में पुजारी बनने का स्वांग रचा। युवा चन्द्रशेखर महात्मा गाँधी के नेतृत्व में 1920-21 के अहिंसक असहयोग आन्दोलन की महान राष्ट्रीय लहर से अत्यधिक प्रभावित थे। उन्होंने इसमें भाग लिया और गिरफ्तार हुए और मजिस्ट्रेट के सामने पेश हुए। उन्होंने अपना नाम ‘आजाद’,अपने पिता का नाम ‘स्वतन्त्र’ और अपना निवास-स्थान जेल को बताया। इससे मजिस्ट्रेट भड़क गया और उसने उन्हें 15 कोड़ों की सजा दे दी।

3. As a revolutionary, he adopted the surname ‘Azad’ which means ‘free’. He once claimed that as he was named “Azad’ he would never be taken alive by the police. Chandra Shekhar Azad was a great Indian freedom fighter. His fierce patriotism and courage inspired others to enter the freedom struggle. Chandra Shekhar was the mentor of Bhagat Singh, who was also a great freedom fighter. Both Chandra Shekhar and Bhagat Singh actively participated in revolutionary activities. He was deeply troubled by the Jallianwala Bagh massacre in Amritsar in 1919. He was also involved in the Kakori Case.

ऐज़ अ रिवॉल्यूशनरी, ही अडॉप्टिड द सरनेम आज़ाद ‘व्हिच मीन्स ‘फ्री’. ही वन्स क्लेम्ड दैट ऐज़ ही वाज़ नेम्ड ‘आज़ाद’ ही वुड नैवर बी टेकन अलाइव बाइ द पुलिस. चन्द्रशेखर आजाद वाज़ अ ग्रेट इण्डियन, फ्रीडम फाइटर, हिज फीअर्स पेट्रियोटिज्म एण्ड करेज इन्सपायर्ड अदर्स टु एण्टर द फ्रीडम स्ट्रगल. चन्द्रशेखर वाज़ द मेंटर ऑफ भगत सिंह, हू वाज़ ऑल्सो अ ग्रेट फ्रीडम फाइटर । बोथ चन्द्रशेखर एण्ड भगत सिंह ऐक्टिवली पार्टीसिपेटिड इन रिवॉल्यूशनरी ऐक्टिविटीज़. ही वाज़ डीपली ट्रबल्ड बाइ द जलियाँवाला बाग मैसेकर इन अमृतसर इन 1919. ही वाज़ ऑल्सो इन्वॉल्वड् इन द काकोरी केस.

अनुवाद:
एक क्रान्तिकारी के रूप में उन्होंने अपना उपनाम ‘आजाद’ रख लिया जिसका अर्थ ‘स्वतन्त्र’ है। उन्होंने यह दावा किया था कि अपने नाम के अनुरूप वह पुलिस द्वारा कभी जीवित नहीं पकड़े जाएँगे। चन्द्रशेखर आज़ाद महान भारतीय स्वतन्त्रता सेनानी थे। उनकी प्रचण्ड राष्ट्रभक्ति एवं साहस ने अन्य लोगों को स्वतन्त्रता आन्दोलन से जुड़ने को प्रेरित किया। चन्द्रशेखर भगत सिंह के गुरु थे, जो एक महान् स्वतन्त्रता सेनानी थे। चन्द्रशेखर एवं भगत सिंह दोनों ने क्रान्तिकारी गतिविधियों में बढ़-चढ़ कर भाग लिया। वह 1919 में अमृतसर के जलियाँवाला बाग जनसंहार से बहुत अधिक दुःखी (परेशान) हुए। वह काकोरी के काण्ड में भी सम्मिलित थे।

4. After the suspension of the non-cooperation movement he was attracted towards more aggressive revolutionary ideals.Chandra Shekhar was a terror for the British police. He was on their hit list and the British police badly wanted to capture him dead or alive. On February 27, 1931, Chandra Shekhar Azad met two of his comrades at the Azad Park (Alfred Park) in Allahabad. He was betrayed by an informer to the British police. The police surrounded the park and ordered Chandra Shekhar Azad to surrender. Azad fought valiantly and killed three policemen. But finding himself surrounded, he shot himself. Thus he kept his pledge of not being caught alive. He used to recite his favourite Hindustani couplet.

आफ्टर द सस्पेंशन ऑफ द नॉन-कोऑपरेशन मूवमेंट ही वाज़ अट्रैक्टिड टुवर्ड्स मोर अग्रेसिव रिवॉल्यूशनरी आइडियल्स. चन्द्रशेखर वाज़ अ टेरर फॉर द ब्रिटिश पुलिस. ही वाज़ ऑन देअर हिट लिस्ट एण्ड द ब्रिटिश पुलिस बैंडली वॉन्टेड टु कैप्चर हिम डैड ऑर अलाइव. ऑन फेब्रुअरी 27, 1931, चन्द्रशेखर आजाद मैट टू ऑफ हिज़ कॉमरेड्स ऐट द आज़ाद पार्क (अल्फ्रेड पार्क) इन इलाहाबाद. ही वाज़ ब्रिट्रेड बाइ ऐन इन्फॉर्मर टु द ब्रिटिश पुलिस द पुलिस सराउन्डिड द पार्क एण्ड ऑर्डर्ड चन्द्रशेखर आजाद टु सरेन्डर. आज़ाद फॉट वैलिएण्ट्ली एण्ड किल्ड थ्री पुलिसमैन. बट फाइंडिंग हिमसैल्फ सराउन्डिड, ही शॉट हिमसैल्फ. दस ही कैप्ट हिज़ प्लैज ऑफ नॉट बीइंग कॉट अलाइव। ही यूज्ड टु रिसाइट हिज़ फेवरिट हिन्दुस्तानी कप्लेट.

अनुवाद:
असहयोग आन्दोलन के थम जाने के बाद वे और अधिक आक्रामक क्रान्तिकारी आदर्शों की ओर आकर्षित हुए। चन्द्रशेखर ब्रिटिश पुलिस के लिए एक आतंक थे। वह उनके निशाने पर थे और ब्रिटिश पुलिस उन्हें किसी भी तरह जीवित या मृत पकड़ना चाहती थी। 27 फरवरी, 1931 को चन्द्रशेखर आजाद इलाहाबाद के आजाद पार्क (अल्फ्रेड पार्क) में अपने दो साथियों से मिलने आये। ब्रिटिश पुलिस के एक मुखबिर द्वारा उनके साथ विश्वासघात किया गया। पुलिस ने पार्क को चारों ओर से घेर लिया और चन्द्रशेखर आजाद से समर्पण करने को कहा। आज़ाद बहादुरी के साथ लड़े और उन्होंने तीन पुलिसकर्मियों को मार दिया। लेकिन स्वयं को चारों ओर से घिरे हुए देखकर उन्होंने स्वयं को गोली मार ली। इस प्रकार उन्होंने स्वयं को जीवित न पकड़े जाने के प्रण को निभाया। वह अपने प्रिय हिन्दुस्तानी दोहे को गाते रहते थे।

5. Dushman ki goliyon ka hum samna karenge Azad hee rahe hain, Azad hee rahenge. To pay homage to such a revolutionary hero of Bharat Mata the State Government has decided to institute ‘Shaheed Chandra Shekhar Azad Memorial Award’. The Award will carry an amount of Rs. 1-50 Lakhs and be presented at a special function to be organized at Bhopal on Balidan Diwas of Chandra Shekhar Azad.

दुश्मन की गोलियों का हम सामना करेंगे आजाद ही रहे हैं, आजाद ही रहेंगे. टु पे होमेज टु सच अ रिवॉल्यूशनरी हीरो ऑफ भारत माता, द स्टेट गवर्नमेंट हैज़ डिसाइडिड टु इन्सटिट्यूट “शहीद चन्द्र शेखर आजाद मैमोरियल अवार्ड”. द अवार्ड विल कैरी ऐन अमाउन्ट ऑफ रुपीज़ 1.50 लैक्स एण्ड बी प्रेजेन्टिड ऐट अ स्पेशल फंक्शन टु बी ऑर्गनाइज्ड ऐट भोपाल ऑन बलिदान दिवस ऑफ चन्द्रशेखर आजाद.

अनुवाद:
दुश्मन की गोलियों का हम सामना करेंगे आज़ाद ही रहे हैं, आजाद ही रहेंगे। भारत माता के ऐसे क्रान्तिकारी नायक को श्रद्धांजलि देने के लिए राज्य सरकार ने ‘शहीद चन्द्रशेखर आजाद स्मृति पुरस्कार’ स्थापित करने का निर्णय किया है। पुरस्कार की राशि ₹ 1.50 लाख होगी और यह चन्द्रशेखर आज़ाद के बलिदान दिवस पर आयोजित एक विशेष समारोह में प्रदान किया जाएगा।

MP Board Class 8 English Question Answer

• Another Chance Class 8 Question Answer
• Thrills of Kanha Kisli Class 8 Question Answer
• The Ungrateful Jeweller Class 8 Question Answer
• Trees: Our Saviours Class 8 Question Answer
• The Miser and his Gold Class 8 Question Answer
• Measure for Measure Class 8 Question Answer
• Water Water Everywhere Class 8 Question Answer
• The Moon Class 8 Question Answer
• Kalpana: The Star Class 8 Question Answer
• The Balloon Man Class 8 Question Answer
• Save Water Save Life Class 8 Question Answer
• Know More About Class 8 Question Answer
• Virtue Pays Class 8 Question Answer
• Azad: The Martyr Class 8 Question Answer
• A Secret Class 8 Question Answer

In this article, we will share MP Board Class 8th Special English Solutions Chapter 5 Trees are the Kindest Things I know PDF download, Mp Board Class 8 Chapter 5 English, These solutions are solved subject experts from the latest edition books.

Class 8th English Chapter 5 Trees are the Kindest Things I know Poem Question Answers MP Board

Supplementary Reader Class 8 MP Board Chapter 5 Solutions

Word Power

(A): (i) The following words tell us about the different parts of a tree. Add some more words.
roots, trunk, branches, ________, ________, ________
(ii) Write five words which have relation to trees :
forests, wood, ________,________
Answers:
(i) leaves, fruits, flowers, buds.
(ii) fruit, oxygen, timber, furniture.

(B) Fill in the blanks with ei, ie, ea, and ee,

Answer:

Comprehensions

(A) Complete the following statements:

1. Trees are useful because ______,
2. Trees are kind because ______,
3. Trees are beautiful because ______,

Answer:

1. They give us many things, such as cool shade, fruits wood, etc.
2. They do no harm to anyone.
3. they are green and have flowers and fruits in them.

(B) Answer the following:

MP Board Class 8 English Chapter 5 Question 1.
Can you name two other things which are as kind as trees ?
Answer:
The earth and the rivers.

English Chapter 5 Class 8 MP Board Question 2.
Which lines in the poem tell us about the sun and sunlight?
Answer:
They are the first when day’s begun.
To touch the beams of morning sun.
They are the last to hold the light,
When evening changes into night.

Class 8 English Chapter 5 MP Board Question 3.
The poet says, “They are the last to hold the light’. What does ‘They’ stand for ?
Answer:
‘They’ stands for trees.

Class 8 MP Board English Chapter 5 Question 4.
Write the summary of the poem with help of the following clues:
Answer:
Trees are the kindest living beings on the earth. They provide cool shade to men and animals. They also give us fruits and wood. They are the ones who receive the light of the moon and the sun.

Lets Read

Read the following poem carefully

The world is finite, resources are scarce,
Things are bad and will be worse,
Coal is burned and gas exploded,
Wells are dry and air polluted,
Dust is blowing, trees are uprooted,
Oil is going, ores depleted,
Drains receive what is excreted
Land is sinking, seas are rising,
Man is far too enterprising,
Fires will rage with man to fan it,
Soon we will have a plundered planet.

Now Answer the following question:

(a) Choose the right meaning of the word ‘finite’:
(i) abundance
(ii) give
(iii) limited
(iv) lacking.
Answer:
iii. limited.

(b) What according to the poet, are depleting?
Answer:
According to the poet our resources are depleting.

(c) Which line in the poem tells us about industrial pollution?
Answer:
Coal is burned and gas exploded.

(d) Which line in the poem warns us that our earth will soon loose all that it has?
Answer:
Soon we will have a plundered planet.

(e) Look at the meaning of the following words:
eroded – gradually destroyed
depleted – reduced by a large amount
polluted – made harmful
plundered – looted.
Do these words suggest a bleak future for mankind ?
Answer:
Yes, the above words suggest a bleak future for mankind.

(f) Find out the nouns from the above poem which the poet uses the following adjectives for
dry ______
plundered ______
bad ______
finite ______
Answer:
Odry – wells
plundered – planet
bad – things
finite – world.

Let’s Write

(A) Complete the following lines from the poem.

Class 8th English Chapter 5 MP Board Question 1.
Trees are the kindest things I know
Answer:
They do no harm, they simply grow.

Class 8 Chapter 5 English MP Board Question 2.
They give us fruit in leaves above
Answer:
And wood to make our houses of,

MP Board Class 8th English Chapter 5 Question 3.
They are the first when day’s begun
Answer:
To touch the beams of morning sun.

MP Board Class 8 English Chapter 5 Summary Question 4.
And when a moon floats on the sky
Answer:
They hum drowsg lulla by.

(B) Mahek has written a passage on trees but she has forgotten to give suggestions on how to stop felling trees. Complete the passage by giving suggestions to stop cutting trees :

The value with which trees can be best associated is kindness. Trees do not usually cause harm to anyone. They spread their shade for rest, drop down their fruits for us to eat. They give us plenty of oxygen. Trees are selflessly devoted to the welfare of mankind. But we do not realise that we will not be able to live a happy life in their absence. If we keep. felling trees then the value of kindness shall forever be abolished from this destructive world. We must save trees by …………….
Answer:
We must save trees by promoting afforestation. People as a whole must be motivated to regard the planting and protection of trees as social duty. Educational institutions can instil in the younger generation the need for planting trees. We must keep it one our mind that our existance is bound up with the trees and therefore they must be planted especially on the roadside and near railway tracks.

(C) You are Asma residing at G-203, Palash Colony, Ratlam. Write a letter to your friend Manju, telling him the importance of trees. Use the layout as given below:

G-203
Palash Colony,
Ratlam December 15, 2006
Dear Anil,

I would like to draw your attention towards the importance of trees. We all know that trees are vital for all living beings. Without trees survival of mankind is impossible. Trees help in maintaning ecological balance. They give us fruit and leaves to plant-eating animals. Beasts of prey eat the plant -eating animals. In this way trees directly and indirectly give us food. They also provide us leaves and roots which are used in preparing many medicines. Trees keep the environment clean. They also provide us wood, shelter and shade. Trees are really very important. Hence, we mustn’t cut it. I am sure you have the same opinion about trees. Rest is fine.

Your affectionately
Wahid

Trees are the Kindest Things I know Word Meanings

Page 35 : Gather – collect – इक्टठा करना। Boughs – branches – शकए। Beams – rays – किरणों। Hum – sing with closed lips – गुनगुनाना। Lullaby – a song we sing to help a child fall asleep – लोरी।

MP Board English Reader Question Answer

• Don’t Quit Class 8 Question Answer
• The Making of the Missile Man Class 8 Question Answer
• The False Flight Class 8 Question Answer
• The Narmada: The Lifeline of Madhya Pradesh Class 8 Question Answer
• Trees are the Kindest Things I know Class 8 Question Answer
• How Mowgli Joined the Pack Class 8 Question Answer
• Nothing You Can’t Do Class 8 Question Answer
• English is Amazing (Poem) Class 8 Question Answer
• To the Rescue Class 8 Question Answer
• The Sacrifice Class 8 Question Answer
• Children who are Wanted Every Hour Class 8 Question Answer
• The Cherry Tree Class 8 Question Answer
• How I Thought my Grandmother to Read Class 8 Question Answer
• The Jet Age Class 8 Question Answer
• Ahilyabai – A Great Indian Queen Class 8 Question Answer
• To the New Year Class 8 Question Answer
• Manav Sangrahalaya: One of Its Kind Class 8 Question Answer
• Handy Andy at the Post Office Class 8 Question Answer
• Bismillah Khan: The Shehnai Maestro Class 8 Question Answer
• Yoga: A Way of Life Class 8 Question Answer
• Work is Worship Class 8 Question Answer

MP Board Class 8th Maths Solutions Chapter 15 Introduction to Graphs Ex 15.2

Question 1.
Plot the following points on a graph sheet. Verify if they lie on a line
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)
(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5).
Solution:
(a) A(4, 0), B(4, 2), C(4, 6), D(4, 2.5)

Yes, above all points lie on the same line,

(b) P(1, 1), Q(2, 2), R(3, 3), S(4, 4).

Yes, above all points lie on the same line.

(c) K(2, 3), L(5, 3), M(5, 5), N(2, 5).

No, above all points do not lie on the same line.

Question 2.
Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Solution:
After drawing a graph the line passing through (2, 3) and (3, 2), then the co-ordinates of the points which meets the x-axis and y-axis are at (5, 0) and (0, 5) respectively.

Question 3.
Write the coordinates of the vertices of each of these adjoining figures.

Solution:
After observing a graph we find that O(0, 0), A(2, 0), B(2, 3), C(0, 3), P(4, 3), Q(6,1), R(6, 5), S(4, 7), L(7, 7), M(10, 8), K(10, 5).

Question 4.
State whether True or False. Correct that are false.
(i) A point whose x-coordinate is zero and y-coordinate is non-zero will lie on the y-axis.
(ii) A point whose y-coordinate is zero and x-coordinate is 5 will lie on y-axis.
(iii) The coordinates of the origin are (0, 0).
Solution:
(i) True
(ii) False, because it will lie on x-axis.
(iii) True.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 15 Introduction to Graphs Ex 15.1

Question 1.
The following graph shows the temperature of a patient in a hospital, recorded every hour.

(a) What was the patient’s temperature at 1 p.m.?
(b) When was the patient’s temperature 38.5° C?
(c) The patient’s temperature was the same two times during the period given. What were these two times ?
(d) What was the temperature at 1.30 p.m. ? How did you arrive at your answer?
(e) During which periods did the patient’s temperature showed an upward trend?
Solution:
(a) At 1 p.m. patient’s temperature was 36.5°C.
(b) At 12 noon the patient’s temperature was 38.5°C.
(c) The patient’s temperature was 36.5°C at 1 p.m. and 2 p.m.
(d) The patient’s temperature was 36.5°C at 1.30 p.m. (because the temperature of the patient was constant from 1 p.m. to 2 p.m.). The point between 1 p.m. and 2 p.m. on the x-axis and equidistant from them will represent 1 : 30 p.m. Similarly, the point on the y-axis, between 36°C and 37°C and equidistant from them will represent 36.5° C.
(e) 9 a.m. to 10 a.m., 10 a.m. to 11 a.m. and 2 p.m. to 3 p.m.

Question 2.
The following line graph shows the yearly sales figures for a manufacturing company.
(a) What were the sales in
(i) 2002
(ii) 2006?
(b) What were the sales in
(i) 2003
(ii) 2005?
(c) Compute the difference between the sales in 2002 and 2006.
(d) In which year was there the greatest difference between the sales as compared to its previous year?

Solution:
(a) (i) In 2002 company’s sales were ₹ 4 crores.
(ii) In 2006 company’s sales were ₹ 8 crores.

(b) (i) The sales of a manufacturing company in 2003 were ₹ 7 crores.
(ii) The sales of a manufacturing company in 2005 were ₹ 10 crores.

(c) The difference between the sales in 2002 and 2006.
= Sales in 2006 – Sales in 2002
= ₹ 8 crores – ₹ 4 crores = ₹ 4 crores.

(d) In year 2005, the difference between the sales was greatest as compared to its previous year.

Question 3.
For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.

(a) How high was Plant A after
(i) 2 weeks,
(ii) 3 weeks?
(b) How high was Plant B after
(i) 2 weeks,
(ii) 3 weeks?
(c) How much did Plant A grow during the 3rd week?
(d) How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
(e) During which week did Plant A grow most?
(f) During which week did Plant B grow least?
(g) Were the two plants of the same height during any week shown here? Specify.
Solution:
(a) (i) After 2 weeks the height of plant A was 7 cm.
(ii) After 3 weeks the height of plant A was 9 cm.

(b) (i) After 2 weeks the height of plant B was 7 cm.
(ii) After 3 weeks the height of plant B was 10 cm.
(c) During the 3rd week plant A grew (9 – 7) cm = 2 cm.
(d) The plant B grew from the end of 2nd week to the end of 3rd week
= 10 cm – 7 cm = 3 cm
(e) The plant A grew most during second week.
(f) The plant B grew least during first week.
(g) Yes, at the end of second week.

Question 4.
The following graph shows the temperature forecast and the actual temperature for each day of a week.
(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?

Solution:
(a) The forecast temperature is the same as actual temperature on : Tuesday, Friday and Sunday.
(b) The maximum forecast temperature during the week was 35°C.
(c) The minimum actual temperature during the week was 15°C.
(d) On Thursday, the actual temperature differed the most from the forecast temperature.

Question 5.
Use the tables below to draw linear graphs.
(a) The number of days a hill side city received snow in different years

(b) Population (in thousands) of men and women in a village in different years.

Solution:
(a) By using a given information we draw a line graph, where the horizontal line i.e. x-axis shows the year and the vertical line i.e. y-axis shows the number of days.

(b) The horizontal line i.e. x-axis shows the year and the vertical line i.e. y-axis shows the number of men and number of women.

Question 6.
A courier-person cycles from a town to a neighbouring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph.
(a) What is the scale taken for the time axis?
(b) How much time did the person take for the travel?
(c) How far is the place of the merchant from the town?
(d) Did the person stop on his way? Explain.
(e) During which period did he ride fastest?

Solution:
(a) 4 units = 1 hour is the scale taken for the time axis.
(b) Time taken by a person for the travel = (1 +1 + 1+ \(\frac{1}{2}\)) hours = 3\(\frac{1}{2}\) hours.
(c) The place of the merchant from the town is at a distance of 22 km.
(d) Yes, this is indicated by the horizontal part of the graph from 10 a.m. -10.30 a.m. where the stability of a distance has been shown.
(e) Between 8.00 a.m. – 9.00 a.m. he rode the fastest.

Question 7.
Can there be a time-temperature graph as follows? Justify your answer.


Solution:
Figure (i) shows temperature is increasing over time.
Figure (ii) shows temperature is decreasing over time.
Figure (iii) is not a possible time – temperature graph. because, at the same time, temperature can’t be different.
Figure (iv) shows temperature is constant over time

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 14 गुणनखंडन Ex 14.1

प्रश्न 1.
दिए हुए पदों में सार्वगुणनखण्ड ज्ञात कीजिए –

1. 12x, 36
2. 2y, 22xy
3. 14pq, 28p²q²
4. 2x, 3x², 4
5. 6abc, 24ab², 12ab
6. 16x³, – 4x², 32x
7. 10pq, 20qr, 30rp
8. 3x²y³, 10x³y², 6x²y²z.

हल:
1. 12x = 12 × x
36 = 12 x 3
∴ सार्व गुणनखण्ड = 12

2. 2y = 2 x y
22xy = 2 x 11 × x × y
∴ सार्व गुणनखण्ड = 2 x y = 2y

3. 14pq = 2 x 7 x p x q
28 p²q² = 2 x 2 x 7 x p x p x q x q
∴ सार्व गुणनखण्ड = 2 x 7 x p x q = 14pq

4. 2x = 2 × x × 1
3x² = 3 × x × x × 1
4 = 2 x 2 x 1
∴ सार्व गुणनखण्ड = 1

5. 6abc = 2 x 3 x 4 x 6 x c
24ab² =2 x 2 x 2 x 3 x a x b x b
12a²b = 2 x 2 x 3 x a x a x b
∴ सार्व गुणनखण्ड = 2 x 3 x a x b = 6ab

6. 16x³ = 2 × 2 × 2 × 2 × x × x × x
4x² = (-1) × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ सार्व गुणनखण्ड = 2 × 2 × x = 4x

7. 10pq = 2 x 5 x p x q
20qr = 2 x 2 x 5 x q x r
30pr = 2 x 3 x 5 x p x r
∴ सार्व गुणनखण्ड = 2 x 5 = 10

8. 3x²y³ = 3 × x × x × y × y × y
10x³y² = 2 × 5 × x × x × x × y × y
6x²y²z = 2 × 3 × x × x × y × y × z
∴ सार्व गुणनखण्ड = x × x × y × y = xy²

प्रश्न 2.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –

1. 7x – 42
2. 6p – 12q
3. 7a² + 14a
4. -16z + 20z³
5. 20l²m + 30alm
6. 5x²y – 15xy²
7. 10a² – 15b² + 20c²
8. -4a + 4ab – 4ac
9. x²y + z + xy²z + xyz² (तीनों पदों को मिलाने पर)
10. ax²y + bxy² + cxyz

हल:
1. 7x – 42 = 7 × x – 2 × 3 × 7
= 7(x – 2 x 3)
= 7 (x – 6)

2. 6p – 12q = 2 x 3 x p – 2 x 2 x 3 x q
= 2 x 3 (p – 2 x q)
= 6 (p – 24)

3. 7a² + 14a = 7 x a x a + 2 x 7 x a
= 7 x a x (a + 2)
= 7a (a + 2)

4. – 16z + 20z³
= -2 x 2 x 2 x 2 x z + 2 x 2 x 5 x z x z x z
= 2 x 2 x z x (- 2 x 2 + 5 x z x z)
= 4z (- 4 + 57²)

5. 20l²m + 30alm
= 2 x 2 x 5 x 1 x 1 x m + 2 x 3 x 5 x a x l x m
= 2 x 5 x 1 x m (2 x 1 + 3 x a)
= 10lm (21+ 3a)

6. 5x²y – 15xy²
= 5 × x × x × y – 3 × 5 × x × y × y
= 5 × x × y × (x – 3 × y)
= 5xy (x – 3y)

7. 10a² – 15b² + 20c²
= 2 x 5 x a x a – 3 x 5 x b x b + 2 x 2 x 5 x c x c
= 5 (2 x a x a – 3 x b x b + 2 x 2 x c x c)²
=5 (2a² – 3b² – 4c)

8. -4a² + 4ab – 4ac
= – 2 x 2 x a x a + 2 x 2 x a x b – 2 x 2 x a x c
= 2 x 2 x a x (- a + b – c)
= 4a (- a + b – c)

9. xyz + xyz + xyz²
= x × x × y × z + x × y × y × z + x × y × z × z
= x × y × z (x + y + z)
=xyz (x + y + z)

10. ax²y + bxy² + cxyz
= a × x × x × y + b × x × y × y + c × x × y × z
= x × y × (a × x + b × y + c × z)
= xy (ax + by + cz)

प्रश्न 3.
गुणनखण्ड कीजिए –

1. x + xy + 8x + 8y
2. 15xy – 6x + 5y – 2
3. ax + bx – ay – by
4. 15pq + 15 + 9q + 25p
5. z – 7 + 7xy – xyz

हल:
1. x² + xy + 8x + 8y = (x² + xy) + (8x + 8y)
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)

2. 15xy – 6x + 5y – 2 = (15xy – 6x) + (5y – 2)
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

3. ax + bx – ay – by = (ax + bx) – (ay + by)
= x (a + b) – y (a + b)
= (a + b) (x – y)

4. 15pq + 15 + 9q + 25p = (15pq + 9q) + (25p + 15)
(पुनः समूहन करने पर)
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

5. z – 7 + 7xy – xyz = z – 7 – xyz + 7xy
(पुनः समूहन करने पर)
= 1 (z – 7) – xy (z – 7)
= (z – 7) (1 – xy)

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 14 गुणनखंडन Intext Questions

MP Board Class 8th Maths Chapter 14 पाठान्तर्गत प्रश्नोत्तर

पाठ्य-पुस्तक पृष्ठ संख्या # 227

प्रयास कीजिए (क्रमांक 14.1)

प्रश्न 1.
गुणनखण्ड कीजिए –

1. 12x + 36
2. 22y – 33z
3. 14pq + 35 pqr

हल:
1. 12x + 36 = (12 × x) + (12 x 3)
= 12 (x + 3)

2. 22y – 33z = (11 × 2xy) – (11 x 3 x z)
= 11 x (2xy – 3 × z)
= 11 (2y – 33)

3. 14pq + 35pqr = (7 x 2 x p x q) + (7 x 5 x p x q x r)
= 7 x p x q x (2 + 5 x r)
= 7pq (2 + 5r)

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 11 क्षेत्रमिति Ex 11.4

प्रश्न 1.
आपको एक बेलनाकार टैंक दिया हुआ है। निम्नलिखित में से किस स्थिति में आप उसका पृष्ठीय क्षेत्रफल ज्ञात करेंगे और किस स्थिति में आयत –

(a) यह ज्ञात करने के लिए कि इसमें कितना पानी रखा जा सकता है।
(b) इसका प्लास्टर करने के लिए वांछित सीमेंट बोरियों की संख्या।
(c) इसमें भरे पानी से भरे जाने वाले छोटे टैंकों की संख्या।

उत्तर:

(a) आयतन
(b) पृष्ठीय क्षेत्रफल
(c) आयतन।

प्रश्न 2.
बेलन A का व्यास 7 cm और ऊँचाई 14 cm है। बेलन B का व्यास 14 cm और ऊँचाई 7 cm है। परिकलन किए बिना क्या आप बता सकते हैं कि इन दोनों में किसका आयतन अधिक है? दोनों बेलनों का आयतन ज्ञात करते हुए इसका सत्यापन कीजिए। जाँच कीजिए कि क्या अधिक आयतन वाले बेलन का पृष्ठीय क्षेत्रफल भी अधिक है।

हल:
बेलन B का आयतन अधिक है।
बेलन A के लिए, r = \(\frac{7}{2}\) सेमी तथा h = 14 सेमी
बेलन A का आयतन = πr²h
= \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 14 सेमी³
= 539 सेमी³
बेलन B के लिए, r = \(\frac{14}{2}\) सेमी = 7 सेमी तथा h= 7 सेमी
बलेन B का आयतन = πr²h
= \(\frac{22}{7}\) x 7 x 7 x 7 सेमी³
= 1078 सेमी³
बेलन A का पृष्ठीय क्षेत्रफल = 2πr (r + h)
= 2 x \(\frac{22}{7}\) x \(\frac{7}{2}\) (\(\frac{7}{2}\) + 14) सेमी²
= 22 (\(\frac{7+28}{2}\)) सेमी²
= 11 x 35 सेमी² = 385 सेमी²
बेलन का पृष्ठीय क्षेत्रफल = 2πr (r + h)
= 2 x \(\frac{22}{7}\) x 7 (7 + 7) सेमी²
= 44 (14) सेमी²
= 616 सेमी
हाँ, अधिक आयतन वाले बेलन B का पृष्ठीय क्षेत्रफल भी अधिक है।

प्रश्न 3.
एक ऐसे घनाभ की ऊँचाई ज्ञात कीजिए जिसके आधार का क्षेत्रफल 180 cm² और जिसका आयतन 900 cm³ है।
हल:
यहाँ, घनाभ का आयतन = 900 सेमी³ तथा आधार का क्षेत्रफल = 180 सेमी²
माना कि घनाभ की ऊँचाई = h सेमी है।
घनाभ का आयतन = आधार का क्षेत्रफल x ऊँचाई
900 सेमी² = 180 सेमी² x h
h = \(\frac{900}{180}\) = 5 सेमी
अतः घनाभ की ऊँचाई = 5 सेमी

प्रश्न 4.
एक घनाभ की विमाएँ 60 cm x 54 cm x 30 cm हैं। इस घनाभ के अन्दर 6 cm भुजा वाले कितने छोटे घर रखे जा सकते हैं?
हल:
यहाँ, l = 60 सेमी
b = 54 सेमी तथा
h = 30 सेमी
घनाभ का आयतन = l x b x h
= 60 x 54 x 30 सेमी³
घन की भुजा = 6 सेमी
घन का आयतन = (भुजा)³ = (6)³ सेमी³
= 6 x 6 x 6 सेमी³
= 216 सेमी³

अतः घनाभ में 450 घन रखे जा सकते हैं।

प्रश्न 5.
एक ऐसे बेलन की ऊँचाई ज्ञात कीजिए जिसका आयतन 1.54 m³ और जिसके आधार का व्यास 140 cm है।
हल:
यहाँ, बेलन का आयतन = 1.54 मीटर
त्रिज्या r = \(\frac{140}{2}\) सेमी = 70 सेमी
= \(\frac{70}{100}\) मीटर = 0.7 मी
माना कि ऊँचाई = h है।
अब, बेलन का आयतन = πr²h
1.54 = \(\frac{22}{7}\) x 0.7 × 0.7 x h
h = \(\frac{1.54×7}{22×0.7×0.7}\) मीटर
= 1 मीटर
अतः बेलन की ऊँचाई = 1 मीटर

प्रश्न 6.
एक दूध का टैंक बेलन के आकार का है जिसकी त्रिज्या 1.5 m है और लम्बाई 7 m है। इस टैंक में भरे जा सकने वाले दूध की मात्रा लीटर में ज्ञात कीजिए।

हल:
टैंक की त्रिज्या = 1.5 m, टैंक की लम्बाई = 7 m
टैंक की धारिता = πr²h
= \(\frac{22}{7}\) x 1.5 x 1.5 x 7 मीटर³
= 49.5 मी³
= 49.5 x 1000 लीटर
(∵ 1 मी³ = 1000 लीटर)
= 49500 लीटर
अत: टैंक में भरे जाने वाली दूध की मात्रा = 49500 लीटर

प्रश्न 7.
किसी घन के प्रत्येक किनारे को दुगुना कर दिया जाए तो –

1. इसके पृष्ठीय क्षेत्रफल में कितने गुना वृद्धि होगी?
2. इसके आयतन में कितने गुना वृद्धि होगी?

हल:
माना कि घन की भुजा = x इकाई है, तब
इसका आयतन = x³ तथा पृष्ठीय क्षेत्रफल = 6x²
अब, जबकि घन के प्रत्येक किनारे को दुगुना कर दिया जाए, तब

1. इसका पृष्ठीय क्षेत्रफल = 6(2x)² = 6 x 4x²
= 24x² = 4 x 6x²
अतः नये घन के पृष्ठीय क्षेत्रफल में चार गुना वृद्धि होगी।

2. नये घन का आयतन = (2x)²
= 8x² = 8 × x³
अतः इसके आयतन में आठ गुना वृद्धि होगी।

प्रश्न 8.
एक कुंड के अन्दर 60 लीटर प्रति मिनट की दर से पानी गिर रहा है। यदि कुंड का आयतन 108 m’ है, _ तो ज्ञात कीजिए कि इस कुंड को भरने में कितने घंटे लगेंगे?

हल:
यहाँ, कुंड का आयतन
= 108 m³
= 108 x 1000 लीटर
= 108000 लीटर
कुंड में पानी गिरने की दर = 60 लीटर प्रति मिनट
= 60 x 60 लीटर प्रति घण्टा

= 30 घण्टे
अतः कुंड को भरने में 30 घण्टे लगेंगे।

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 14 गुणनखंडन Ex 14.2

प्रश्न 1.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –

1. a² + 8a + 16
2. p² – 10p + 25
3. 25m² + 30m + 9
4. 49y² + 84yz + 36z²
5. 4x² – 8x + 4
6. 121b² – 88bc + 16c²
7. (l + m)² – 4lm
8. a⁴ + 2a²b² + b⁴

हल:
1. a² + 8a + 16 = (a)² + 2 x a x 4 + (4)²
[∴ a² + 2ab + b² = (a + b)²]
= (a + 4)²

2. p² – 10p + 25 = (p)² – 2 x p x 5 + (5)²
[∴ a² – 2ab + b² = (a – b)]
= (p – 5)²

3. 25m² + 30m + 9 = (5m)² + 2 x 5m x 3 + (3)²
= (5m + 3)²

4. 49y² + 84yz + 36z²
= (7y)² + 2 x 7y x 6z + (6z)²
= (7y + 6z)²

5. 4x² – 8x + 4 = (2x)² – 2 x 4x × 2 + (2)²
= (2x – 2)²

6. 121b² – 88bc + 16c²
= (11b)² – 2 x 11b x 4c + (4c)²
= (11b – 4c)

7. (l + m)² – 4lm = l² + 2lm + m² – 4lm
= l² – 2lm + m²
= (l)² – 2 x 1 x m + (m)² = (1 – m)

8. a⁴ + 2a²b² + b⁴ = (a²)² + 2 x a² x b² + (a)²
= (a + b)²

प्रश्न 2.
गुणनखण्ड कीजिए –

1. 4p² – 9q²
2. 63a² – 112b²
3. 49x² – 36
4. 16x⁵ – 144x³
5. (l + m)² – (l – m)
6. 9x²y² – 16
7. (x² – 2xy +y²) – z²
8. 25a² – 4b² + 28bc – 49c²

हल:
1. 4p² – 9q²
a² – b² = (a – b) (a + b)
4p² – 9q² = (2p)² – (3q)²
= (2p – 3q) (2p + 3q)

2. 63a² – 112b² = 7 (9a² – 16b²)
= 7 {(3a)² – (4b)²}
= 7 (3a – 4b) (3a + 4b)

3. 49x² – 36 = (7x)² – (6)²
= (7x – 6) (7x + 6)

4. 16x⁵ – 144x³ = 16x³ (x² – 9)
= 16x³ (x² – 3²)
= 16x³ (x – 3) (x + 3)

5. (l + m)² – (l – m) = [(l + m) – (l – m)][(l + m) – (l – m)]
= (l + m – 1 + m) (l + m + l – m)
= 2m x 2l = 4lm

6. 9x²y² – 16 = (3xy)² – (4)²
= (3xy – 4) (3xy + 4)

7. x² – 2xy + y² – z² = (x – y)² – z²
= [(x – y) – z] [(x – y) + z]
= (x – y – z) (x – y + z)

8. 25a² – 4b² + 28bc – 49c²
= 25a² – (4b² – 28bc + 49c²)
= 25a² – [(2b)² – 2 x 26 x 7c + (7c)²]
= (5a)² – (2b – 7c)²
= [5a – (2b – 7c)] [5a + (2b – 7c)]
= (5a – 25 + 7c) (5a + 2b – 7c)

प्रश्न 3.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –

1. ax² + bx
2. 7p² + 21q²
3. 2x² + 2xy² + 2xz²
4. am² + bm² + bn² + an²
5. (lm + 1) + m + 1
6. y (y + z) + 9 (y + z)
7. 5y² – 20y – 8z + 2yz
8. 10ab + 4a + 5b + 2
9. 6xy – 4y + 6 – 9x.

हल:
1. ax² + bx = x (ax + b)

2. 7p² + 21q² = 7 (p² + 3q²)

3. 2x² + 2xy² + 2xz² = 2x (x² + y² + z²)

4. amv + bm² + bn² + an²
= (am² + bm²) + (bn² + an²)
= m² (a + b) + n² (b+ a)
= (a + b) (m² + n²)

5. (lm + 1) + m + 1 = 1(m + 1) + 1 (m + 1)
= (m + 1) (1 + 1)

6. y (y + z) + 9 (y + z) = (y + z) (y + 9)

7. 5y² – 20y – 8z + 2yz
= (5y² – 20y) + (2yz – 8z)
= 5y (y – 4) + 2z (y – 4).
= (y – 4) (5y + 2z)

8. 10ab + 4a + 56+2
= (10ab + 5b) + (4a + 2)
= 5b (2a + 1) + 2 (2a + 1)
= (2a + 1) (5b + 2)

9. 6xy – 4y + 6 – 9x
= (6xy – 4y) – (9x – 6)
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)

प्रश्न 4.
गुणनखण्ड कीजिए –

1. a⁴ – b⁴
2. p⁴ – 81
3. x⁴ (y + z)⁴
4. x⁴ – (x – z)⁴
5. a² – 2a²b² + b⁴

हल:
1. a⁴ – b⁴ = (a²)² – (b²)²
= (a² + b²) (a² – b²)
= (a² + b²) (a + b) (a – b)

2. p⁴ – 81 = (p²)² – (9)²
= (p² + 9) (p² – 9)
= (p² + 9) (p + 3) (p – 3)

3. x⁴ – (y + z)⁴ = (x²)² – [(y + 2)²]²
= [x² – (y + z)²] [x² + (y + z)²]
= [x – (y + z)] [x + (y + z)] [x² + (y + z)²]
= (x – y – z) (x + y + z) [x² + (y + z)²]

4. x⁴ – (x – z)⁴ = (x²)² – [(x – z)²]²
= [x² – (x – z)²] [x² + (x – z)]
= [x – (x – z)] [x + (x – z)] [x² + (x² – 2x² + z²)] = (x – x + z) (x + x – z)(2x² – 2xz + z²)
= z(2x – z) (2x² – 2xz + z²)

5. a⁴ – 2a²b² + b² = (a²)² – 2 x a² x b² + (b²)²
= [a² – b²]²
= [(a – b) (a + b)]²
= (a – b)² (a + b)²

प्रश्न 5.
निम्नलिखित व्यंजकों के गुणनखण्ड कीजिए –

1. p² + 6p + 8
2. q² – 10q + 21
3. p² + 6p – 16

हल:
1. P² + 6p + 8 = p² + (4 + 2) p + 8
(∴8 = 4 x 2)
= p² + 4p + 2p +8
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

2. q² – 10q + 21 = q² – (7 + 3) q + 21
(∴ 21 = 3 x 7)
=q² – 7q – 3q + 21
= q(q – 7) – 3 (q – 7)
= q (q – 7) (q – 3)
= (q – 3) (q – 7)

3. p² + 6p – 16 = p² + (8 – 2)p – 16
(∴ 16 = 8 x 2)
= p² + 8p – 2p – 16
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)

पाठ्य-पुस्तक पृष्ठ संख्या # 234

प्रयास कीजिए (क्रमांक 14.2)

प्रश्न 1.
भाग दीजिए –

1. 24xy²3z³ को 6yz² से
2. 63a²b⁴c⁶ को 7a²b²c³ से।

हल:
1. 24xy²z³ + 6yz²

= 4xyz

2. 63a²b⁴c⁶ ÷ 7a²b²c³

= 9b²c³

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 12 घातांक और घात Ex 12.2

प्रश्न 1.
निम्न संख्याओं को मानक रूप में व्यक्त कीजिए –

1. 0.0000000000085
2. 0.00000000000942
3. 6020000000000000
4. 0.00000000837
5. 31860000000

हल:
1. 0.0000000000085 = \(\frac{85}{ 10000000000000}\)
= \(\frac { 85 }{ 10^{ 13 } } \)
= \(\frac { 8.5×10 }{ 10^{ 13 } } \) = 8.5 x 10 x 10^-13
= 8.5 x 10^-12

2. 0.00000000000942 = \(\frac{942}{ 100000000000000 }\)
= \(\frac { 942 }{ 10^{ 14 } } \)
= \(\frac { 8.42×10 }{ 10^{ 14 } } \)
= 9.42 x 10² x 10^-14
= 9.42 x 10^-12

3. 6020000000000000 = 602 x 10¹³
= 6.02 x 100 x 10¹³
= 602 x 10² x 10¹³
= 6.02 x 10¹⁵

4. 0.00000000837 = \(\frac{837 }{ 100000000000 } = \frac { 837}{ 10^{ 11 } } \)
= \(\frac { 8:37×100 }{ 10^{ 11 } } \)
= 8.37 x 10² x 10^-11
= 8.37 x 10^-9

5. 31860000000 = 3186 x 10⁷
= 3.186 x 10³ x 10⁷
= 3.186 x 10¹⁰

प्रश्न 2.
निम्न संख्याओं को सामान्य रूप में व्यक्त कीजिए –

1. 3.02 x 10^-6
2. 4.5 x 10⁴
3. 3 x 10^-8
4. 1.0001 x 10⁹
5. 5.8 x 10¹²
6. 3.61492 x 10⁶

हल:
1. 3.02 x 10^-6 = \(\frac{302}{100}\) x \(\frac { 1 }{ 10^{ 6 } } \)
= \(\frac { 302 }{ 10^{ 8 } } \)
= \(\frac{302}{100000000}\)

2. 4.5 x 10⁴ = \(\frac{45}{10}\) x 10⁴
= 45 x 10³
= 45000

3. 3 x 10^-8 = \(\frac{3}{100000000}\)
= 0.00000003

4. 1.0001 x 10⁹ = \(\frac{10001}{10000}\) x 10⁹
= 10001 x 10⁹ x 10^-4
= 10001 x 10⁵
= 1000100000

5. 5.8 x 10¹² = \(\frac{58}{10}\) x 10¹²
= 58 x 10¹² x 10^-1
= 58 x 10¹¹
= 5800000000000

6. 3.61492 x 10⁶ = \(\frac{361492}{100000}\)
= 361492 x 10⁶ x 10^-5
= 361492 x 10
= 3614920

प्रश्न 3.
निम्नलिखित कथनों में जो संख्या प्रकट हो रही है उन्हें मानक रूप में व्यक्त कीजिए –

1. 1 माइक्रॉन \(\frac{1}{100000}\) m के बराबर होता है।
2. एक इलेक्ट्रॉन का आवेश 0.000,000,000,000,000,000,16 कूलॉम होता है।
3. जीवाणु की माप 0.0000005 m है।
4. पौधों की कोशिकाओं की माप 0.00001275 m
5. मोटे कागज की मोटाई 0.07 mm है।

हल:
1. 1 माइक्रॉन = \(\frac{1}{1000000}\)m
= 1 x 10^-6m

2. एक इलेक्ट्रॉन का आवेश = 0.000,000,000,000,000,00016
= \(\frac{16}{100000000000000000000}\)
= \(\frac { 16 }{ 10^{ 20 } } \) = \(\frac { 1.6×10 }{ 10^{ 20 } } \)
= 1.6 x 10 x 10^-20
= 1.6 x 10^-19 कूलॉम

3. एक जीवाणु की माप 0.0000005 m
= \(\frac{5}{10000000}\)
= 5 x 10^-7 m

4. पौधों की कोशिकाओं की माप 0.00001275 m
= \(\frac{1275}{100000000}\)
= \(\frac { 1.275×1000 }{ 10^{ 8 } } \)
= 1.275 x 10³ x 10^-8
= 1.275 x 10^-5 m

5. एक मोटे कागज की मोटाई 0.07 mm
= \(\frac{7}{100}\)
= 7 x 10^-2 mm

प्रश्न 4.
एक ढेर में पाँच किताबें हैं जिनमें प्रत्येक की मोटाई 20 mm तथा पाँच कागज की शीटें हैं जिनमें प्रत्येक की । मोटाई 0.016 mm है। इस ढेर की कुल मोटाई ज्ञात कीजिए।
हल:
यहाँ, 1 किताब की मोटाई = 20 mm
∴ 5 किताबों की मोटाई = 5 x 20 mm
= 100mm
∴ 1 कागज की शीट की मोटाई = 0.016 mm
∴ 5 कागज की शीटों की मोटाई = 5 x 0.016 mm
= 0.080 mm
∴ ढेर की कुल मोटाई = 5 किताबों की मोटाई + 5 कागज की शीटों की मोटाई
= 100 mm + 0.080 mm
= 10008 mm
\(\frac{1.0008×10000}{100}\)
= 1.0008 x 10⁴ x 10^-2
= 1.0008 x 10² mm

MP Board Class 8th Maths Solutions