Class 9

Class 9th English The Rainbow Workbook Chapter 13 Stopping by Woods on a Snowy Evening Question Answers

Stopping by Woods on a Snowy Evening Class 9 Questions and Answers

Stopping by Woods on a Snowy Evening Vocabulary and Pronunciation

A. Learn with the help of a dictionary the different usage of the following words:
know, deep, see.
Answer:
Do yourself.

B. Say the following words and notice the difference in the vowel sounds:
bell, bill, ball, bull, bail, be, bee, boil.
Answer:
Do yourself.

Listening Skill

Listen to the following poem twice.

See workbook pages 97-98.

Fill in the blanks using the missing lines.

Tell me not in mourful numbers,
Answer:
‘Life is but an empty dream’.
For the soul is dead that slumbers,
Ans. And the things are not what they seem.
Life is real life is earnest! ;
And the grave is not its goal,
Ans. Oust then art to dust returnest.
Was not spoken of the soul,
Ans. Not enjoyment and not sorrow.
Is our destined end or way.
But to act, that each tomorrow Finds us further than today.
In the world’s broad field of battle,
In the bivouac of life.
Answer:
Be not like dumb driven cattle;
Be a hero in the strife!

Speaking Skill

A The poem “Stopping by the Woods on a Snowy Evening” has many rhyming words. Now you add more words to the list:

B. Repeat the following lines giving attention to the rhyming words:
See workbook page 99.
Answer:
Do yourself.

Reading Skill

Read the given poem and answer the questions that follows: See workbook pages 99-100.

Question 1.
Where does the brook originate from ?
Answer:
The brook originates from the haunts of coot.

Question 2.
How does it sparkel ?
Answer:
It sparkles out among the fern to bicker down the valley.

Question 3.
What does the word ‘bicker’mean in this context ?
Answer:
‘Bicker’ means the dim light down the vally.

Question 4.
Describe the courses of the brook as suggested in the poem.
Answer:
The brook flows down from the haunts of coot making different sounds movement. It flows continuously.

Question 5.
What does the expression ‘hurty down’ suggest about the movement of the river ?
Answer:
It suggest that the flow is downwards. Its movement is very fast.

Question 6.
What does the ‘netted sunbeam’ mean? How does it dance?
Answer:
The ‘netted sunbeam’ means the sunnay that falls upon the water coming through the leaves and branches of the tree. It appess to dancing when the leaves shake with winds.

Question 7.
Who does ‘I’ refer to in the given lines ?
Answer:
‘I’ refers to the brook.

Question 8.
How does it chatter ?
Answer:
It chatter with its flowing sounds.

Writing Skill

A. Imagine you are going on a walk in woods. On the way you come across bushes, ferns, trees, streams, rocks, grasslands, slopes etc.
Write a paragraph using the above words. (50 words)
Answer:
I walk along in woods. It is really a unique experience. I come across bushes, thorns, ferns, trees, streams, rocks, grasslands, slopes etc. They all delight me in their own way. Bushes denour white the thorns and firms prick. Streams give pleasure and grasslands, fill with nature’s delight. It is really a bliss to have such experience.

B. ‘Planting saplings today is ensuring greenery tomorrow’. Write a short notice on the importance of preserving the environment through plantation drive in your village/city. (150 words)
Answer:
Our life is a part of nature. Every object of nature has its contribution in the maintenance of eco system. Trees and plants plays their measure rute. They add greenery to nature. They make wind blow and rains to rain to make the land fertile and soothe our life.
We get oxygen from them. They help and contribute to waliv conservation. They protect us from flood. But we by cutting trees, damage our own life. If we plant a new sapling todays it ensures greenery tomorrow and also guarantees our better future. It is our duty to save nature. Planting a new plant is our effort to maintain ; eco system.

MP Board Class 9th English Solutions

The Rainbow Workbook Special English Class 9th Solutions

• Bharat Our Land Class 9 Questions And Answers
• The Victory Class 9 Questions And Answers
• Little Girls Wiser than Men Class 9 Questions And Answers
• Past and Present Class 9 Questions And Answers
• Dead Man’s Riddle Class 9 Questions And Answers
• Arise, Awake! Class 9 Questions And Answers
• The World is Too Much with Us Class 9 Questions And Answers
• The Goal not Scored Class 9 Questions And Answers
• The Mission-Agni Class 9 Questions And Answers
• Polonious Advice Class 9 Questions And Answers
• Grandpa Fights an Ostrich Class 9 Questions And Answers
• The Poet and the Pauper Class 9 Questions And Answers
• Stopping by Woods on a Snowy Evening Class 9 Questions And Answers
• Old Blockhead repairs his House Class 9 Questions And Answers
• How it all began Class 9 Questions And Answers
• Where the Mind is without Fear Class 9 Questions And Answers
• On Saying Please Class 9 Questions And Answers
• The Never-Never Nest Class 9 Questions And Answers

Class 9th English The Rainbow Chapter 11 Grandpa Fights an Ostrich Question Answers

Grandpa Fights an Ostrich Class 9 Questions and Answers

Grandpa Fights an Ostrich Textual Exercises

Grandpa Fights an Ostrich Vacabulary

A. Refer to the dictionary and find out the meanings of the following. Use them in sentences. You can use the sentences given in the dictionary as models.
Strange, instant, passion, emerge, unqeual, strides elude, dodging, terrific, desperate
Answer:

Word	Meanings	Usage
Strange Instant Passion Emerge Unequal Strides Elude	odd a short while strong emotion to come out not at par with big steps to baffle	Her behaviour was quite strange. I can beat my rival in an instant. Fashion breeds passion. Our team emerged victorious. You have undertaken an unequal task. A camel runs with strides. Don’t try to elude or delude me.

B. Use the following expressions in your own words.
needless to say strange though it may seem
a glimpse of
in an instant
suddenly.
Answer:

1. Needless to say that he is a cheat.
2. Strange though it may seem yet it is true.
3. I had a glimpse of her face.
4. I can solve this sum in an instant.
5. Suddenly the cat sprang upon the rat.

C. Pick out from this lesson some words that suggest :
1. movement .
2. surprise
3. anger.
Answer:

1. go, walk run, climbed, started, gave chase, dating, emerge, bounding, expanding, rushed, circled, charged, step, sprawling, made off.
2. Startled, dazed, wondered.
3. aggressive, infuriated, enraged, spiteful.

D. Choose the correct word and fill in the blanks.
1. I was ……………… by tire maddening behaviour of the clerk at the post office, (infuriated, delighted, admired).
2. She’s made a …………….. recovery, (strange, miraculous, shocking)
3. I had to go to the …………………… (work-site work-sight, work-cite).
4. The dog was jumping around me, my face and hands …………….. (licking, liking, leaking)
5. My horse had a accident …………………. (feeble, small, slight)
Answer:

1. infuriated
2. miraculous
3. work-site
4. licking
5. slight.

Comprehension

A. Answer each of the following questions in about 25 words.
1. Why did Grandpa decide to go through the ostrich camp?
2. Why did he feel quite safe in such a dangerous situation?
3. What was the only chance to keep him safe during the chase?
4. Why was the huge bird frightened?
5. Describe the unexpected withdrawal of the ostrich.
Answer:
1. Grandpa’s horse had met with a minor accident. He had no other mode of transport. He was a great walker. The passage through the hills was shorter by six miles. Therefore, he decided to go through the ostrich camp.

2. It was dangerous to cross an ostrich farmering the breeding season. The mate ostriches are violent then. They can trample a man to death. Grandpa had his pet dog with him. Even the biggest ostriches are scared even of a small dog. Therefore, he felt quite safe in such a dangerous situation.

3. The grandpa had nothing to defend himself. He turned and ran towards the fence. But it was an unequal race between a man and an ostrich. There was only one chance to keep him safe during the chase. It was to get behind a large bush and try to elude the bird. A dodging game was his only chance.

4. The huge bird came upon grandpa when he had fallen down. But it did not strike. By this time Grandpa’s dog had come there. Ostriches are mightily scared of the dogs. Therefore, the huge bird got frightened. On hearing the dog’s bark, he turned and ran off.

5. Grandpa had fallen down. The ostrich had come upon him. Grandpa was afraid that his end was near. He put up his hand to save his face. To his amazement, the ostrich did not strike. The ostrich heard the dog’s bark. It was afraid that the dog would bring about its end. In order to save its own life, the ostrich withdrew unexpectedly.

B. Answer each of the following questions in about 50 words.

1. Why did Grandpa dare to cross the ostrich farm?
2. Describe the nature and behaviour of ostriches as known to Grandpa.
3. There was an unequal race between Grandpa and the ostrich. Describe it.
4. What traits of character do you notice in Grandpa?
Answer:
1. Grandpa’s horse had met with an accident. So he had to come back on foot from his workplace. An ostrich camp lay on L the way. It was their breeding season. The grandpa’s dog was with him. He knew it well that he was quite safe in the company of his dog. The ostriches were mightily scared of even a small dog. The journey through the hills could save the grandfather a distance of six miles. Therefore, he dared to cross the ostrich farm. He was over confident that no harm could come to him so long as his dog was with him.

2. Grandpa was fairly familiar with the ways of ostriches. He knew that male birds were very violent in breeding season. They could attack on the slightest provocation. He also knows that they , are mightily scared of a dog. It was quite strange that even the biggest ostrich would run away at the sight of even a small dog. It would try to save its own life and never try to fight with even the pet dug.

3. Suddenly, a big male ostrich emerged from a prickly bush about a hundred yards away. He came bounding towards the grandpa. The grandpa turned and ran towards the fence. However, it was an unequal race. Grandpa’s steps were of the length of two or three feet. Hie ostrich’s steps were like strides of sixteen or twenty feet. It means the ostrich’s speed was eight times of the grandpa’s speed.

4. Grandpa had a bundle of good qualities. He was adventurous. He had the courage to go through the hills knowing well that there was an ostrich camp on the way. The male ostriches could even kill him during die breeding season. He was over confident of his dog’s help. He started a dodging game when an ostrich ran towards him. He caught-the ostrich by die wing. It was an act of fearlessness and courage. He covered his face with his hands when the ostrich was upon him. He was careful enough not to come directy in front of his deadly kick.

Grandpa Fights an Ostrich Grammar

A . Study these sentences.

• Now and then I caught a glimpse of birds.
• He began to turn or rather waltz.
• All the while the ostrich kept opening and shutting his beak with loud snaps.
• I don’t know whether it was the dog’s bark or my own shouting but what I was most anxious to avoid immediately happened.

The underlined words are connectors. In the first sentence ‘aid’ connects words, in the second sentence ‘or’ connects phrases in the third sentence ‘and’ connects clauses and in the fourth sentence ‘or’ connects phrases, ‘but’ and ‘whether’ connect clauses.

B. Fill in the blanks with appropriate connectors given in brackets.
1. He roamed the whole world over to find a real princess …………………….. there was always something wrong, (and, but)
2. He may offer either Mathematics ………………. Physics (nor, or)
3. The frock was spalshed ……………….. so were. Akoulya’s eyes ……………… nose, (nor, and but; or)
4. Some patients had died ………………. the doctor arrived, (before,after)
5. Their game plan was almost final …………. they were beginning to play well as a team, (but, and)
Answer:

1. but
2. or
3. and, and
4. before.
5. and.

Speaking Activity

A. Discuss with your friends in the class and find out main characteristics of Ostriches.
Answer:
Class-room discussion.
The following are the main characteristics of ostriches. Almost all birds have four toes but an ostrich has only two toes. Its feet resemble those of the cattle and antelopes. Its inner toes have disappeared. The ostrich can fly as well as run. It can run at a speed of 34 km per hour.

Writing Activity

A. How will you save yourself if you are chased by a dog?
Answer:
It is a common sight that dogs chase human beings in dark and lonely places. If a man gets scared and starts running, he is sure to be bitten. If I am chased by a dog I shall call the dog as ‘Montu’ very affectionately. I will try to pat the dog. The best thing to avoid a chasing dog is to offer it a piece of bread. I shall also pick up some stone lying nearby and show it to the dog. The dog will definitely retreat because it is always careful about its own safety.

B. Make an entry in you diary of a day when you missed your school bus. (150 words)
Answer:
New Delhi. 7th March, 2007
I was appearing at the Secondary School’s Boards examination. My examination was scheduled to start on 6th March, 2007. I got ready to catch the school bus. Unforunately, I happened to miss it. I felt like a fish out of water. I was alone at the bus stop. I had not brought my mobile phone lest it should be lost in the school. I was sweating profusely. My school was located at a distance of ten kilometres. I had only twenty rupees in my pocket. No three wheeler could be hired with this petty amount. I found myself in a tight comer. Just then I noticed a neighbour coming towards me driving his car. His daughter too had to appear at the examination. He gave me a lift. He reached the school in time. Life revived in me due to that good Samaritan, my neighbour.

Think it over

A. An ostrich is a large bird. There is a proverb on the habit of this bird. Why shouldn’t we adopt ‘ostrich policy?’ Think.
Answer:
An ostrich is a very large African bird with a long neck and long legs. Ostriches cannot fly. But they can very fast. They are known for their ‘self-deluding’ methods. We should never adopt ‘ostrich policy’. We should rather obey the voice of our inner conscience. He who disobeys his inner voice loses confidence and “ultimately suffers.

B. When a person encounters a dangerous situation, he prepares himself to face it. Experience helps him? How?
Answer:
Life is full of obstacles. Everyone has to encounter a dangerous situation some time or the other. Those who get nervous have to pay through the nose. One should learn from one’s own and others’ experiences not-to lose heart but show his wit and manliness at such times. One must always remain prepared for the worst. One day I had picked up five oranges from a cart. My pockets were bulging with them. A policeman caught me. He locked me in a cabin. He went away to bring another policeman. In the mean while. I ate the oranges with fibre, barks and pips. The policemen arrived. They found no evidence against me. They released me with a cheerful face. A thief’s experience had helped me.

Things to do

We read about ‘disaster management.’ Collect information about the situation described below :
Answer:

Disaster	What should be done
1. Fire breaks out in the school	The firebrigade should be in­formed, sand and water should be thrown on the fire.
2. There is an earthquake, the students are in the classes.	They should either fun out in the open. They can also hide themselves under desks or tables.
3. There is a bus accident.	The police should be informed. First aid should be given to the injured.
4. A building collapse 3 in your neighbourhood.	The police should be informed. To call up military. Garbage should be removed The injured should be given first aid.

Grandpa Fights an Ostrich Additional Questions

Short-Answer Type Questions (In About 25 words)

Question 1.
What’ did grandpa do before joining Indian Railway? What was memorable for him during those days?
Answer:
Before joining Indian Railway Grandpa worked for sometime on the East African Railway. Grandpa’s encounter with an ostrich is the memorable event for him during those days.

Question 2.
Describe the attack of an ostrich on Grandpa.
Answer:
Grandpa was attacked by an ostrich. Grandpa began running towards the fence. But his steps of two or three feet were nothing in comparison to the ostrich’s strides of sixteen to twenty feet.

Question 3.
How did the Grandpa escape?
Answer:
Grandpa adopted the game of dodging with the ostrich. When the ostrich attacked Grandpa, Grandpa jumped sideways and it saved him. Moreover, the ostrich saw Grandpa’s dog. An ostrich is always frightened of a dog. So he ran away leaving him lying on the ground.

Question 4.
When did Grandpa think his end had come? What happened then?
Answer:
The Grandpa caught the ostrich’s wings. The ostrich’s turn and waltz quickly loosened Grandpa’s hold and he fell down. The ostrich came upon Grandpa. He stood with one foot raised to rip grandpa open. The Grandpa thought his end had come. He put up his hands to save his face.

Long Answer Type Question

Question 1.
Describe the Ostrich’s attack on Grandpa.
Answer:
Grandpa’s dog started barking and the Grandpa started shouting. It startled the ostriches and they began darting to and fro. Just then the dog saw a hare and chased him Suddenly a male ostrich emerged from a thicket about a hundred yards away. He stood still and stared Grandpa. For a moment it began spreading its wings. It also erected its tail and started bounding towards Grandpa. It pouhed on the Grandpa who held its wings tightly. A dodging game took place between the two. Ultimately the Grandpa fell down on the ground. His life was saved when his dog appeared on the scene.

Grandpa Fights an Ostrich Summary in English

Ruskin Bond’s grandpa had worked for some time on the East African Railway before he joined the Indian Railways. He had an encounter with an ostrich during his service in Africa.The author’s grandpa was working in the laying of a new railway line. His workplace was twelve miles away. He was returning on foot one day- To save half the distance, he took a short route through the hills. There lay an ostrich ‘camp’ on this way. This was the breeding season. The male birds are very dangerous then. But the grandpa felt safe with his dog. Even the biggest ostrich is afraid of a dog.

The grandpa got through the wire fencing of the farm. The ostriches were feeding some distance away. His dog chased a hare. He called the dog in vain. Just then he saw a big male os .rich com mg out of a bush about a hundred yards away. He stood still staring at the grandpa. Soon he spread his wings and came jumping towards the grandpa. Grandpa ran towards the fence. But the ostrich soon overtook him. There was a strange encounter. Grandpa took care to avoid his kick. He was soon breathless and helpless. He circled a big bush. He was likely to drop from tiredness. Somehow he held the ostrich’s wing. The frightened ostrich began to dance round and round. Grandpa kept clinging to the ostrich’s wing.

Suddenly the ostrich went into reverse. Grandpa fell down and the ostrich was upon him. The ostrich did not strike him. He was ready to rip grandpa open. Soon the ostrich jumped back and ran fast. Grandpa heard his dog bark. The dog and the grandpa were clear of the camp.

Grandpa Fights an Ostrich Summary in Hindi

भारतीय रेलवे में भर्ती होने से पहले रस्किन बांड के दादा ने कुछ समय के लिए पूर्वी अफ्रीका रेलवे में नौकरी की थी। अफ्रीका में नौकरी करने के दौरान उसकी एक शुतुर्मुर्ग-से भिड़न्त हो गई थी।
लेखक के दादा, नई रेलवे लाइन बिछाने का काम कर रहे थे। उसका कार्यस्थल बारह मील की दूरी पर था। एक दिन वे पैदल जा रहे थे। आधी दूरी बचाने के लिए उन्होंने पहाड़ियों के बीच छोटा रास्ता अपनाया। इस रास्ते पर शुतुर्मुर्गों का एक कैम्प था। यह प्रजनन की ऋतु थी। उस समय नर पक्षी बड़े भयानक होते हैं। परन्तु अपने कुत्ते के साथ होते हुए दादा सुरक्षित महसूस कर रहे थे। बड़े-से-बड़ा शुतुर्मुर्ग भी कुत्ते से डरता है। दादा, फार्म की बाड़ के बीच से गुजरे। कुछ दूरी पर शुतुर्मुर्ग भोजन खा रहे थे।

उसका कुत्ता एक खरगोश का पीछा करने लगा। उसने कुत्ते को बुलाया परन्तु व्यर्थ रहा। तभी उसने लगभग सौ गज की दूरी पर एक झाड़ी के बीच से आते हुए एक नर शुतुर्मुर्ग को देखा। वह शान्त खड़ा होकर दादा को घूरने लगा। तुरन्त उसने अपनी पंखें फैलाईं और उछलता-फांदता हुआ दादा की तरफ आया। दादा, बाड़ की तरफ दौड़े। परन्तु शुतुमुर्ग ने तत्काल उन पर काबू कर लिया। एक विचित्र मुठभेड़ (भिड़न्त) हुई। दादा ने उसकी ठोकर से बचने की चौकसी रखी। शीघ्र ही उसका सांस फूल गया और वह लाचार हो गया। वह एक बड़ी झाड़ी का चक्कर लगाने लगा। वह थककर गिरने ही वाला था। किसी तरह से उसने शुतुर्मुर्ग के पंख पकड़ लिये। भयभीत शुतुर्मुर्ग ने गोलाई में नाचना शुरू कर दिया। दादा, शुतुर्मुर्ग के पंख के साथ चिमटा

रहा। अचानक शुतुर्मुर्ग पीछे की तरफ मुड़ा। दादा नीचे गिर गए और शुतुर्मुर्ग उनके ऊपर चढ़ गया। शुतुर्मुर्ग ने उनके ऊपर प्रहार नहीं किया। वह दादा को चीर देने के लिए तैयार था। तुरन्त शुतुर्मुर्ग पीछे की तरफ उछला और तेज दौड़ने लगा। दादा ने अपने कुत्ते के भौंकने की आवाज सुनी। दादा और कुत्ता शिविर में से सुरक्षित निकल आए।

Grandpa Fights an Ostrich Word Meanings

MP Board Class 9th English Solutions

The Rainbow Textbook Special English Class 9th Solutions

• Bharat Our Land Class 9 Questions And Answers
• The Victory Class 9 Questions And Answers
• Little Girls Wiser than Men Class 9 Questions And Answers
• Past and Present Class 9 Questions And Answers
• Dead Man’s Riddle Class 9 Questions And Answers
• Arise, Awake! Class 9 Questions And Answers
• The World is Too Much with Us Class 9 Questions And Answers
• The Goal not Scored Class 9 Questions And Answers
• The Mission-Agni Class 9 Questions And Answers
• Polonious Advice Class 9 Questions And Answers
• Grandpa Fights an Ostrich Class 9 Questions And Answers
• The Poet and the Pauper Class 9 Questions And Answers
• Stopping by Woods on a Snowy Evening Class 9 Questions And Answers
• Old Blockhead repairs his House Class 9 Questions And Answers
• How it all began Class 9 Questions And Answers
• Where the Mind is without Fear Class 9 Questions And Answers
• On Saying Please Class 9 Questions And Answers
• The Never-Never Nest Class 9 Questions And Answers

MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 10.5

Question 1.
In the figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution:
We have a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°

∴ ∠AOB + ∠BOC = ∠AOC
∠AOB = 60° + 30° = 90°
Now, tne arc ABC subtends ∠AOC = 90° at the centre and ∠ADC at a point D on the circle other than the arc ABC.
∴ ∠ADC = \(\frac{1}{2}\) [∠AOC]
∠ADC = \(\frac{1}{2}\) (90°) = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle = 60
∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ABC at a point on the minor arc of the circle.

∠ACB = \(\frac{1}{2}\) [reflex ∠AOB]
= \(\frac{1}{2}\) [300°] = 150°
Similarly, ∠ADB = \(\frac{1}{2}\) [∠AOB]
= \(\frac{1}{2}\) x [60°] = 30°
Thus, the angle subtended by the chord on the minor arc = 150° and on the major arc = 30°.

Question 3.
In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution:
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

∴ reflex ∠POR = 2∠PQR,
But ∠PQR = 100°
reflex ∠POR = 2 x 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
∠POR + 200° = 360°
∠POR = 360° – 200°
∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180° [Sum of the angles of a triangle = 180°]
∠OPR + ∠OPR + 160° = 180° [∴ ∠OPR = ∠ORP]
2∠OPR = 180° – 160° = 20°
∠OPR = \(\frac{20^{\circ}}{2}\) =10°

Question 4.
In the Figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
We have, in ∆ABC,
∠ABC = 69° and
∠ACB = 31°

But ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° +31° + ∠BAC = 180°
∠BAC = 180° – 69° – 31°
= 80°
Since, angles in the same segment are equal.
∴ ∠BDC = ∠BAC
∠BDC =80°

Question 5.
In the figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Solution:
In ∆CDE,
Exterior ∠BEC – ∠AED = 130°

130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
∠EDC = 130° – 20° = 110°
∠BDC = 110°
Since, angles in the same segment are equal.
∠BAC = ∠BDC
∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Given:
∠BAC = 30°, ∠DBC = 70°
To find:
∠BCD and ∠ECD.
∠BDC = ∠BAC = 30° (∠s on the same segment of a circle are equal)

In ∆BCD, 70° + 30° + ∠C = 180° (ASP)
100° + ∠C = 180°
∠C = 80°
AB = BC (Given)
∠BAC = ∠BCA = 30°
∠BCD = ∠BCA + ∠ECD
80° = 30° + ∠ECD
∠ECD – 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given
ABCD is a cyclic quadrilateral in which AC and BD are diagonals.
To prove:
ABCD is a rectangle.

Proof:
BD is the diameter of the circle.
∠BAD = 90°(angle in a semicircle)
Similarly ∠BCD = 90°
AC is the diameter of the circle
∠ABC = 90° (angle in a semicircle)
Similarly ∠ADC = 90°
In quadrilateral ABCD, ∠A = ∠B = ∠C = ∠D = 90°
ABCD is a rectangle.

Question 8.
If the non-parallel sides of h trapezium are equal, prove that it is cyclic.
Solution:
Given:
AD = BC, AB ∥ DC.
To prove:
ABCD is cyclic quadrilateral.
Construction:
Draw AE and BF As on DC.

Proof:
In ∆ADE and ∆BCF
∴ AE = BF
(∴ Distance between two parallel lines are equal)
∠E = ∠F (Each 90°)
AD = BC (Given)
∆ADE = ∆BCF (By RHS)
and so ∠D = ∠C (By CPCT)
AB ∥ DC and AD is the transversal.
∴ ∠BAD + ∠ADC =180° (CIA’s)
⇒ ∠BAD + ∠BCD = 180° (∠ADC = ∠BCD)
∴ ABCD is cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are J drawn to intersect the circles at A, D and Q respectively (see Fig.). Prove that ∠ACP = ∠QCD.

Solution:
Given
C (O, r) and C (O₁, r₁) are two circles. Two lines ABD and PBQ are drawn which intersect at B.
To prove:
∠1 = ∠3

Proof:
∠1 = ∠2 …(1) (Z s on the same segment of a circle are equal.)
∠3 = ∠4 …..(2) (Z s on the same segment of a circle are equal)
∠2 = ∠4 …(3) (OA’s)
From (1), (2) and (3), we get
∠1 = ∠3 …(2)

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given
C (O, r) and C (O₁, r₁) are two Circles in whichAB andAC are diameter. These circles intersect at point A and D.
To prove:
BDC is a line.
Construction: JoinAD.
Proof:
∠ADB = 90° (∠s in a semicircle)
∠ADC) = 90° (∠s in a semicircle)

Adding (1) and (2), we get
∠ADB + ∠ADC = 90° + 90°
∠BDC = 180°
∴ BDC is a line and hence D lies on the third side.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD – ∠CBD.
Solution:
Given:
ABC and ADC are two right ∆’s on common base AC. ∠B – 90° and ∠D = 90°.
To prove: ∠CAD = ∠CBD
Proof:
∠ABC + ∠ADC = 90° + 90°
= 180°

ABCD is a cyclic quadrilateral.
∠CAD and ∠CBD are angles on the same segment of a circle.
∴ ∠CAD = ∠CBD.

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: ABCD is a ∥^gm
To prove: ABCD is a rectangle.

Proof:
∠A = ∠C
∠A + ∠C =180°
∠A + ∠A = 180°
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.
∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ ABCD is a rectangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 10 Gravitation

Gravitation Intext Questions

Gravitation Intext Questions Page No. 134

Question 1.
State the universal law of gravitation.
Answer:
Suppose there are two objects having mass M and m respectively.
The distance between their centres is equal to d.
The force of attraction is F.
Thus, F ∝ M . m … (i)
and, F ∝ 1/d² … (ii)
Joining equation (i) and (ii)
we get F ∝ M . m/d²
⇒ F = G . M . m/d² … (iii)
where, G is the proportionality constant and called Universal Gravitation Constant.
The expression (iii) is called expression for Universal Law of Gravitation.
The universal law of gravitation is represented as:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
Where, G is the universal gravitation constant given by:
G = 6.67 × 10^-11 Nm² kg^-2.

Question 2.
Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the earth.
Answer:
Let M_e be the mass of the Earth and m be the mass of an object on its surface.
And say R is the radius of the Earth,
then according to the universal law of gravitation,
the gravitational force (F) acting between the Earth and the object is given by the relation:
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
F = GM_em/R

Gravitation Intext Questions Page No. 136

Question 1.
What do you mean by free fall?
Answer:
Its a phenomenon of gravity. When an object falls from any height under the influence of gravitational force only, it is said to have a free fall. In the case of free fall, no change in direction takes place but the magnitude of velocity changes because of acceleration.

Question 2.
What do you mean by acceleration due to gravity?
Answer:
Change in velocity due to variation in height produces acceleration which is due to gravity in the object and is known as acceleration due to gravity denoted by letter g. The value of acceleration due to gravity is g = 9.8 m/s².

Gravitation Intext Questions Page No. 138

Question 1.
What are the differences between the mass of an object and its weight?
Answer:

Mass	Weight
Mass is a measurement of the amount of matter something has.	Weight is the measurement of the pull of gravity on an object.
Mass is a constant quantity.	Weight is not a constant quantity. It is different at different places.
It is a scalar quantity.	It is a vector quantity.
Its SI unit is kilogram (kg).	Its SI unit is the same as the SI unit of force, i.e., Newton (N).

Question 2.
Why is the weight of an object on the moon \(\frac { 1 }{ 6 }\)th its weight on the earth?
Answer:
The mass of moon is \(\frac { 1 }{ 100 }\) times and its radius \(\frac { 1 }{ 4 }\) times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is \(\frac { 1 }{ 6 }\)th of its weight on the earth.

Gravitation Intext Questions Page No. 141

Question 1.
Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Answer:
It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

Question 2.
What do you mean by buoyancy?
Answer:
The upward force exerted by a liquid on an object that is immersed in it is known as buoyancy.

Question 3.
Why does an object float or sink when placed on the surface of water?
Answer:

1. An object sinks in water if its density is greater than that of water.
2. An object floats in water if its density is less than that of water.

Gravitation Intext Questions Page No. 142

Question 1.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer:
When we weigh our body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

Question 2.
You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?
Answer:
The cotton bag is heavier than the iron bar. The cotton bag experiences larger up – thrust of air than the iron bar. So, the weighing machine indicates a smaller mass for cotton bag than its actual mass.

Gravitation NCERT Textbook Exercises

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass M and m is proportional to the product of their masses and inversely proportional to the square of distance r between them. So, force F is given by
F = G\(\frac { M\times m }{ { r }^{ 2 } } \)
Now, when the distance ‘r’ is reduced to half then force between two masses becomes
F’ = G\(\frac { M\times m }{ { (\frac { r }{ 2 } ) }^{ 2 } } \)
Or
F’ = 4F
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Answer:
All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3.
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m).
Answer:
Given that,
Mass of the body, m = 1 kg
Mass of the Earth, M = 6 × 10²⁴ kg
Radius of the Earth, R = 6.4 × 10⁶ m
Now, magnitude of the gravitational force (F) between the Earth and the body can be given as,
F = G\(\frac { M\times m }{ { r }^{ 2 } } \) = \(\frac { 6.67 × 10 × 6 × 10 × 1 }{ (6.4 × 6.4) }\)
= \(\frac { 6.67 × 6 × 10 }{ (6.4 × 6.4) }\) = 9.8N (approx.)

Question 4.
The Earth and the Moon are attracted to each other by gravitational force. Does the Earth attract the Moon with a force that is greater or smaller or the same as the force with which the Moon attracts the Earth? Why?
Answer:
According to the Universal Law of Gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the Moon with an equal force with which the Moon attracts the Earth.

Question 5.
If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Answer:
The Earth and the Moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the Moon. Hence, it accelerates at a rate lesser than the acceleration rate of the Moon towards the Earth. For this reason, the Earth does not move towards the Moon.

Question 6.
What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Answer:
(i) From Universal Law of Gravitation, force exerted on an object of mass at by Earth is given by
F = G\(\frac { M\times m }{ { R }^{ 2 } } \) ….1
When nws of the object say ne is doubled than
F’ = G\(\frac { M\times 2m }{ { R }^{ 2 } } \)  = 2F
So as the mass of any one of the object is doubled the force is also doubled,

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects es doubled, then the gravitational force of attraction between them is reduced to one fourth of its original value, Similarly, if the distance between two objects is tripled. then the gravitational force of attraction becomes one ninth of its original value.

(iii) Again from Universal Law of Attraction, from equation 1, force ‘F’ is directly proportional to the product of both the masses, So, if both the masses are doubled then, the gravitational force of attraction become four times the original value.

Question 7.
What in the importance of Universal Law of Gravitation?
Answer:
Universal Law of Gravitation is important because it tells us about:

1. the force that is responsible for binding us to Earth.
2. the motion of Moon around the Earth.
3. the motion of planets around the Sun.
4. the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both Sun and Moon on the Earth.

Question 8.
What in the acceleration of free fall?
Answer:
Acceleration of free tall is the acceleration produced when a body falls under the influence of the force of gravitation of the Earth alone. It is denoted by ‘g’ and its value on the surface of the Earth is 9.8 ms^-2.

Question 9.
What do we call the gravitational force between the Earth and an object?
Answer:
Gravitational force between the Earth and an object is known as the weight of the object.

Question 10.
Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator}.
Answer:
Weight of a body on the Earth is given by:
W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator.
Therefore, gold at the equator weighs less than at the poles.
Hence, Amit’s friend will not agree with the weight of the gold bought.

Question 11.
Why does a sheet of paper fell slower than one that is crumpled into a ball?
Answer:
When a sheet of paper is crumpled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12.
Gravitational force on the surface of the Moon is only \(\frac { 1 }{ 6 }\) as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:
Weight of an object on the Moon = \(\frac { 1 }{ 6 }\) × Weight of an object on the Earth.
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the earth = 10 × 9.8 = 98 N
And, weight of the same object on the Moon = 1.6 × 9.8 = 16.3 N.

Question 13.
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises.
(ii) the total time it takes to return to the surface of the earth.
Answer:
According to the equation of motion under gravity:
-u² = 2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s.
During upward motion, g = – 9.8 ms^-2.
(i) Let ‘h’ be the maximum height attained by the ball.
Hence,
0 – 49² = 2 × 9.8 × h
h = \(\frac { 49 × 49 }{ 2 × 9.8}\) = 122.5 m

(ii) Let ‘t’ be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
= 49 + t × (- 9.8)
9.8 t = 49
t = \(\frac { 49 }{ 9.8}\) = 5 s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14.
A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Answer:
According to the equation of motion under gravity: v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 ms^-2
∴ v² – 0² = 2 × 9.8 × 19.6
v² = 2 × 9.8 × 19.6 = (19.6)²
v = 19.6 ms^-1
Hence, the velocity of the stone just before touching the ground is 19.6 m s^-1.

Question 15.
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Answer:
According to the equation of motion under gravity:
v² – u² = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = -10 ms^-2
Let h be the maximum height attained by the stone.
Therefore,
0 – (40)² = 2 × h × (-10)
h = \(\frac { 40 × 40 }{ 20 }\) = 80 m
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (-80) = 0.

Question 16.
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Answer:
According to question,
M_Sun = Mass of the Sun = 2 × 10³⁰ kg
M_Earth = Mass of the Earth = 6 × 10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m.
From Universal Law of Gravitation,
F = G\(\frac { M\times m }{ { R }^{ 2 } } \)
Therefore, putting all the values given in question in above equation we get
F = 6.67 × 10^-11 \(\frac { (6\times { 10 }^{ 24 })\times (2\times { 10 }^{ 30 }) }{ { (1.5\times { 10 }^{ 11 }) }^{ 2 } } \) = 3.56 × 10²² N.

Question 17.
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let be the point at which two stones meet and let ‘h’ be their height from the ground. It is given in the question that height of the tower is H = 100 m
Now, first consider the stone which falls from the top of the tower.
So, distance covered by this stone at time ‘t’ can be calculated using equation of motion:
x – x₀ = u₀t + \(\frac { 1 }{ 2 }\)gt²
Since, initial velocity u = 0,
so we get
100 =  x \(\frac { 1 }{ 2 }\)gt² ………… (1)
The distance covered by the same stone that is thrown in upward direction from ground is
x = 25t –\(\frac { 1 }{ 2 }\)gt²
In this case intitial velocity is 25 m/s.
So, x = 25t – \(\frac { 1 }{ 2 }\)gt²  ………… (2)
Adding equations (1) and (2) we get,
100=25t
or,
t = 4s
Putting value in equation (2).
x = 25 × 4 – \(\frac { 1 }{ 2 }\) × 9.8 × (4)²
= 100 – 78.4
= 21.6 m.

Question 18.
A ball thrown up vertically returns to the thrower after 6 s. Find:
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer:
(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey. Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0 Acceleration due to gravity, g = -9.8 m s^-2
Equation of motion, v = u + gt
will give,
0 = u + (-9.8 × 3)
u = 9.8 × 3
= 29.4 ms^-1
Hence, the ball was thrown upwards with a velocity of 29.4 ms^-1.

(b) Let the maximum height attained by the ball be ‘h’
Initial velocity during the upward journey, u = 29.4 ms^-1
Final velocity, v = 0
Acceleration due to gravity, g = -9.8 ms^-2
From the equation of motion, s = ut + \(\frac { 1 }{ 2 }\) at²
h = 29.4 × 3 + \(\frac { 1 }{ 2 }\) × – 9.8 × (3)² = 44.1 m.

(c) Ball attains the maximum height after 3 s.
After attaining this height, it will start falling downwards.
In this case, Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in
4 s – 3 s = 1 s.
Equation of motion, s = ut + \(\frac { 1 }{ 2 }\) g²
will give,
s = 0 x t + \(\frac { 1 }{ 2 }\) x 9.8 x 12 = 4.9 m
Total height  = 44.1 m.
This means that the ball is 39.2 m (44.1 m – 4.9 m) above the ground after 4 seconds.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
An object immersed in a liquid experiences buoyant force in the upward direction.

Question 20.
Why does a block of plastic released under water come up to the surface of water?
Answer:
For an object immersed in water two forces act on it:

1. Gravitational force which tends to pull object in downward direction
2. Buoyant force that pushes the object in upward direction.

Here, in this case buoyant force is greater than the gravitational pull on the plastic block. This is the reason the plastic block comes up to the surface of the water as soon as it is released under water.

Question 21.
The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^-3, will the substance float or sink?
Answer:
If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.

= \(\frac { 50 }{ 20 }\)
= 2.5 g cm^-3
The density of the substance is more than the density of water (1 g cm^-3).
Hence, the substance will sink in water.

Question 22.
The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^-3? What will be the mass of the water displaced by this packet?
Answer:
Density of the 500 g sealed packet

= \(\frac { 500 }{ 350 }\)
= 1.428 g cm^-3
The density of the substance is more than the density of water (1 g cm^-3). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.

Gravitation Additional Questions

Gravitation Multiple Choice Questions

Question 1.
Two objects of different masses falling freely near the surface of moon would __________ .
(a) have same velocities at any instant.
(b) have different accelerations.
(c) experience forces of same magnitude.
(d) undergo a change in their inertia.
Answer:
(c) experience forces of same magnitude.

Question 2.
The value of acceleration due to gravity __________ .
(a) is same on equator and poles.
(b) is least on poles.
(c) is least on equator.
(d) increases from pole to equator.
Answer:
(c) is least on equator.

Question 3.
The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become __________ .
(a) \(\frac { F }{ 4 }\)
(b) \(\frac { F }{ 2 }\)
(c) F
(d) 2 F.
Answer:
(a) \(\frac { F }{ 4 }\)

Question 4.
A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone __________ .
(a) will continue to move in the circular path.
(b) will move along a straight line towards the centre of the circular path.
(c) will move along a straight line tangential to the circular path.
(d) will move along a straight line perpendicular to the circular path away from the boy.
Answer:
(c) will move along a straight line tangential to the circular path.

Question 5.
An object is put one by one in three liquids having different densities. The object floats with 1 2 3, and 9 11 7 parts of their volumes outside the liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following statement is correct?
(a) d₁ > d₂ > d₃
(b) d₂₁ > d₂ < d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃.
Answer:
(d) d₁ < d₂ < d₃.

Question 6.
In the relation F = GM m/d₂, the quantity G __________ .
(a) depends on the value of ‘g’ at the place of observation.
(b) is used only when the Earth is one of the two masses.
(c) is greatest at the surface of the Earth.
(d) is universal constant of nature.
Answer:
(d) is universal constant of nature.

Question 7.
Law of gravitation gives the gravitational force between __________ .
(a) the Earth and a point mass only.
(b) the Earth and Sun only.
(c) any two bodies having some mass.
(d) two charged bodies only.
Answer:
(c) any two bodies having some mass.

Question 8.
The value of quantity G in the law of gravitation __________ .
(a) depends on mass of Earth only.
(b) depends on radius of Earth only.
(c) depends on both mass and radius of Earth.
(d) is independent of mass and radius of the Earth.
Answer:
(d) is independent of mass and radius of the Earth.

Question 9.
Two particles are placed at some distance. If the mass of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be __________ .
(a) 14 times
(b) 4 times
(c) 12 times
(d) unchanged.
Answer:
(b) 4 times

Question 10.
The atmosphere is held to the earth by __________ .
(a) gravity
(b) wind
(c) clouds
(d) Earth’s magnetic field.
Answer:
(a) gravity

Question 11.
The force of attraction between two unit point masses separated by a unit distance is called __________ .
(a) gravitational potential.
(b) acceleration due to gravity.
(c) gravitational field.
(d) universal gravitational constant.
Answer:
(d) universal gravitational constant.

Question 12.
The weight of an object at the centre of the Earth of radius R is __________ .
(a) zero.
(b) infinite.
(c) R times the weight at the surface of the Earth.
(d) \(\frac { 1 }{ { R }^{ 2 } } \) times the weight at surface of the Earth.
Answer
(a) zero.

Gravitation Very Short Answer Type Questions

Question 1.
Why Moon revolves around the Earth?
Answer:
Gravitational force of Earth.

Question 2.
What is the SI unit of gravitational force?
Answer:
Newton (N).

Question 3.
Which law of physics is represented by the statement every object attract other object in universe towards itself’?
Answer:
Universal Law of Gravitation.

Question 4.
State the relation between gravitational force and distance among objects.
Answer:
Inversely proportional.

Question 5.
In which conditions free fall of an object occur?
Answer:
When an object falls from a height under the influence of gravity and no other force, it is said to have a free fall.

Question 6.
What is the SI unit of gravitational constant?
Answer:
Nm²kg^-2.

Question 7.
Express the relation between thrust and pressure.
Answer:
Pressure = thrust / area.

Question 8.
What kind of force is exerted by a liquid?
Answer:
Equal and unidirectional.

Question 9.
In what condition an object sinks?
Answer:
If the weight of the object is more than 9.8 N, then the object will sink.

Question 10.
Which material is taken as standard to calculate any object’s relative density?
Answer:
Water.

Gravitation Short Answer Type Questions

Question 1.
What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
Answer:
Gravitational force. This force depends on the product of the masses of the planet and Sun, and the distance between them.

Question 2.
On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?
Answer:
Both the stones will take the same time to reach the ground because the two stones fall from the same height.

Question 3.
Suppose gravity of Earth suddenly becomes zero, then in which direction will the Moon begin to move if no other celestial body affects it?
Answer:
The Moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of Moon is due to centripetal force provided by the gravitational force of Earth.

Question 4.
Two identical packets are dropped from two Aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming all conditions are identical, will those packets take same time to reach the surface of Earth. Justify your answer.
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Question 5.
The weight of any person on the Moon is about \(\frac { 1 }{ 6 }\) times that on the Earth. He can lift a mass of 15 kg on the Earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the moon?
Answer:
The value of ‘g’ at the equator of the Earth is less than that at poles. Therefore, the packet falls slowly at equator in comparison to the poles. Thus, the packet will remain in air for longer time interval, when it is dropped at the equator.

Gravitation Long Answer Type Questions

Question 1.
State ‘Archimedes’ Principle and write its two applications.
Answer:
‘Archimedes’ Principle:
Force exerted by liquid on wholly or partly immersed object is equal to the weight of the fluid displaced by the object. Applications based on ‘Archimedes’ principle are:

1. designing of water transport vehicles.
2. hydrometers used for determining the density of liquids

Question 2.
What is relative density? What is the density of water?
Answer:
Relative Density (RD) or Specific Gravity (SG) is the ratio of either densities or weights. Hence, when we compare or divide value of an objects’ density with water’s density, it is called Relative Density of a substance to water.

1. In SI units, the density of water is (approximately) 1000 kg/m³ Or 1 g/cm³.

Question 3.
Mass of a rectangular copper solid piece is 300 g. With dimensions 5 × 2 × 5 cm³, what should be its specific gravity, calculate? Will the bar float or sink in water?
Answer:

Given:
Mass of copper = 300 g
5 × 2 × 5 = 50 cm³
Density of copper = mass / volume
\(\frac { 300 }{ 50 }\) = 6 g / cm³
Density of water, = 1 g/cm³
Specific gravity of iron = \(\frac { 6 }{ 1 }\) = 6.
Hence the bar will sink.

Question 4.
How does the weight of an object vary with respect to mass and radius of the earth. In a hypothetical case, if the diameter of the earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the earth be affected?
Answer:
We know, weight of an object is directly proportional to the mass of the earth and inversely proportional to the square of the radius of the earth, i.e.,.
Weight of a body ∝ \(\frac { M }{ { R }^{ 2 } } \)
Original weight, W₀  = mg = mG\(\frac { M }{ { R }^{ 2 } } \)
When hypothetically M becomes 4 M and R becomes \(\frac { R }{ 2 }\) then weight becomes
W₀ = mG \(\frac { 4M }{ { (\frac { R }{ 2 } ) }^{ 2 } } \) = (16 m G) M
R² = 16 × W₀
The weight will be 16 times heavier.

Question 5.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water? Give reason for each case.
(b) A ball weighing 4 kg of density 4000 kg m^-3 is completely immersed in water of density 10³ kg m^-3. Find the force of buoyancy on it. (Given: g = 10 ms^-2.)
Answer:
(a)

1. The cube will experience a greater buoyant force in the saturated salt solution because the density of the salt solution is greater than that of water.
2. The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.

(b) Buoyant force = weight of the liquid displaced = density of water x volume of water displaced xg 4
= 1000 × \(\frac { 4 }{ 4000 }\) × 10 = 10N.
4000

Gravitation Higher Order Thinking Skills (HOTS)

Question 1.
How will the weight of a body of mass 250 g of changes, if it is taken from equator to the poles? Give reasons.
Answer:
As we move from equator to poles, acceleration due to gravity increases. It is because radius of earth is less at poles than at equator. Therefore, its weight will increase.

Question 2.
Aman tried to immerse an empty plastic bottle in a bucket of water. But each time he fails. Why does this happen?
Answer:
When Aman tried to immerse an empty plastic bottle in a bucket of water, it comes above the surface of water. It is due to the upward force (upthrust or buoyant force). The upthrust exerted by water on the bottle is greater than its own weight.

Gravitation Value Based Question

Question 1.
Rashmi was wearing a high heel shoes for a beach party. Her friend told her to wear flat shoes as she will be tired soon with high heel and will not feel comfortable.

1. What is the reason of one’s feeling tired with high heel shoes on a beach?
2. Name the unit of pressure.
3. What value of Rashmi’s friend is reflected in the above act?

Answer:

1. Because the high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. Therefore, large amount of force will be required to walk with high heels.
2. Pascal.
3. Rashmi’s friend showed the value of being intelligent, concerned and helpful.

MP Board Class 9th Science Solutions

MP Board Class 9th General English The Spring Blossom Solutions Objective Type Questions [Based on Textual Lessons]

Multiple Choice Questions
(बहु विकल्पीय प्रशन)

Question 1.
The real aim of Miss Beam’s school was to:
(A) Be practical
(B) Teach thought
(C) Teach thoughtfulness
(D) Increase memory.
Answer:
(C) Teach thoughtfulness

Question 2.
Those who prepare the matter for advertisements and publicity are called:
(A) Dramatists
(B) Playwrights
(C) Script writers
(D) Copywriters.
Answer:
(D) Copywriters.

Question 3.
J. C. Bose was born in:
(A) Faridabad
(B) Faridkot
(C) Firozpur
(D) Faridpur.
Answer:
(D) Faridpur.

Question 4.
Kasturba Gandhi was married at an age of:
(A) thirteen
(B) eighteen
(C) twenty one
(D) twenty two.
Answer:
(A) thirteen

Question 5.
The poem ‘Am I a child’ is about:
(A) Child’s disputes
(B) Childhood worries
(C) An adolescent child’s situation
(D) Childhood fantasies.
Answer:
(C) An adolescent child’s situation

Question 6.
For Sudhir what mattered most was that he was:
(A) Back in the field
(B) Winner
(C) Loser
(D) In the team.
Answer:
(A) Back in the field

Question 7.
Chaturbhuj Babu had passed:
(A) M. A.
(B) B.A.
(C) Ph.D.
(D) Inter.
Answer:
(A) M. A.

Question 8.
In the opinion of the wise saint a best friend on the earth is:
(A) His own good sense
(B) His courage
(C) His physical power
(D) His smartness.
Answer:
(A) His own good sense

Question 9.
Ram Prasad Bismil was hanged on:
(A) 18th December 1927
(B) 19th December 1927
(C) 9th August 1925
(D) 19th August 1927
Answer:
(B) 19th December 1927

Question 10.
The moon walks wearing:
(A) Light shoes
(B) White shoes
(C) Silver shoes
(D) Bronze shoes.
Answer:
(C) Silver shoes

Question 11.
The tree described in the poem ‘Woodman spare that Tree’ is:
(A) A Pipal tree
(B) A Banyan Tree
(C) A Neem Tree
(D) An Oak Tree.
Answer:
(D) An Oak Tree.

Question 12.
Bose demonstrated the feelings of:
(A) Plants
(B) People
(C) Fish
(D) Animals
Answer:
(A) Plants

Question 13.
Rani Karnavati had sent Humayun:
(A) A raksha thread
(B) A gift
(C) A pearl
(D) A sword.
Answer:
(A) A raksha thread

Question 14.
Cheemi was called:
(A) Chhoti
(B) Little sparrow
(C) A bird
(D) None of these.
Answer:
(B) Little sparrow

Fill in the Blanks
(रिक्त स्थान भरो।)

1. Jumman became ……….. enemy.
2. Miss Beam’s figure was comforting to a ………. child.
3. Kasturba Gandhi led the women’s ……….. for which she was imprisoned.
4. ………… is the pen name of Ram Prasad.
5. King Vikramaditya was a ruler of ………..
6. Rani Karnavati sent a message to ……….
7. J. C. Bose was honoured with the degree of………..
8. The most deafening place on the earth is a ……….
9. Akbar was missing ………..
10. ……….. motivated Sudhir.
11. Akbar regreted ………… Birbal from the court.
12. The child will stride into a new world of ………..
13. Chaturbhuj Babu had brought a ……….. cat with him.
14. Moon is wearing ……….. shoes.

Answer:

1. Algu’s
2. homesick
3. Satyagraha,
4. Bismil
5. Ujjain
6. Humayun
7. Doctor of science
8. Gold factory
9. Birbal
10. Baljit
11. banishing
12. adult life
13. Afghan
14. silver

State True or False
(सत्य असत्य बताइए।)

1. Kasturba Gandhi was the daughter of a prosperous businessman.
2. Advertisements can be termed as the backbone of the commerical world.
3. King Vikramaditya was the ruler of Udaipur.
4. Kakori train incident took place on 18 December 1927.
5. In the second match Sudhir was to play in right in position.
6. Jumman sold his bullock to Samjhu Sahu.
7. We should do our work today.
8. During the dumb day, a child has to use his will power.
9. Parvati Kaki loved Cheemi.
10. The present day babies are too soft spoken.
11. J. C. Bose was awarded the fellowship of Royal Society in 1920.
12. J. C. Bose was also a great writer.
13. Examination for a royal scribe was to be held at Udaipur University.
14. Birbal had passed a harmless comment about Akbar’s sense of humour.

Answer:

1. True
2. True
3. False
4. False
5. True
6. False
7. True
8. True
9. False
10. False
11. True
12. False
13. False
14. True.

Match the Columns
(जोड़ी मिलाइए)

(I)

Answer:
(1) → (ii) to accept and act according to a law
(2) → (iv) an imaginary thing
(3) → (i) an easy chance to score
(4) → (v) extremely hot
(5) → (iii) filled with air

(II)

Answer:
(1) → (iii) poor orphan
(2) — (i) pen name
(3) → (v) To go from place to place to sell things
(4) → (ii) response in the living and Non-living
(5) → (iv) silver shoon

MP Board Class 9th English Solutions

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:

1. Draw a ray OA and with O as centre and any radius, draw an arc, cutting OA at B.
2. With B as centre and the same radius draw an arc cutting the arc drawn in step (i) at C.
3. With C as centre draw another arc with same radius cutting the arc drawn in step (i) at D.
4. with C as centre and the same radius draw an arc.
5. With D as centre and the same radius draw an arc, cutting the arc drawn in step (iv) at E.
6. Draw OE ∴ ∠AOF = 90°

Question 2.
Construct an angle of 45° at the initial point of a given ray justify the construction.
Solution:

1. Draw ∠AOF = 90° by following the same steps for constructing a 90° angle.
2. Draw OG, the bisector of ∠AOF.
3. ∠AOF= 45°

Question 3.
Construct the angle of the following measurements:
Solution:

1. 30°
2. 22 \(\frac{1}{2}\)°
3. 15°

1. 30°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect OA at B.
3. With B as centre and same radius draw an arc cutting the arc drawn is step (ii) at C.
4. Join OC and produce upward
5. ∠BOC = 60°
6. Draw the bisector OD of ABOC.
7. ∠BOD = ∠COD = 30°

2. 22 \(\frac{1}{2}\)°

1. Draw a ray OA.
2. With O as centre and any radius draw an arc which intersect intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii) at C.
4. With C as centre and same radius draw an arc cutting the arc drawn in step (ii) at D.
   
5. With C and D as centres and same radius draw arcs to intersect at E.
6. Join OE.
7. ∠AOG – 90°
8. Draw the bisector OE of ∠AOE, to get ∠AOF = 45°
9. Draw the bisector OG of ∠AOF.
10. ∠AOG = 22\(\frac{1}{2}\)°

3. 15°

1. Draw a ray CM.
2. With O as centre and any radius draw an arc which intersect CM at B.
3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii), at C.
4. Draw OD as the bisector of ZAOC.
5. ∠BOD = 30°
6. Draw the bisector OE of ∠AOD.
7. ∠AOE = 15°

Question 4.
Construct the following angles and verily by measuring them by a protractor.

(i) 75°
(ii) 105°
(iii) 135°

Solution:

Question 5.
Construct an equilateral triangle given its side and justify the construction.
Solution:

1. Draw a ray AY with intial point A.
2. With centred and radius equal to length of a side of the A draw an arc BY, cutting the ray AX at B.
3. With centre B and the same radius draw an arc cutting are BY at C.
4. Join AC and BC to obtain the required A.

Construction of Triangles:

Example 1:
Construction of a Triangle when its base, a base angle and sum of other two sides are given
Solution:
Step of Construction:

1. Draw the base PQ.
2. At point P draw an angle, MPQ equal to the given angle.
3. Cut a line segment PM equal to sum of sides i.e., (PR + RQ) from point P.
4. Join MQ.
5. Draw the perpendicular bisector of MQ which intersect PM at R.
6. Join QR. PQR is the required triangle.

Justification:
Mark points S as shown in Fig.
QS = MS (∴ RS is the perpendicular bisector of MQ)
RS = RS (Common)
∠QSR = ∠MSR (Each \({ 90 }^{ \underline { 0 } }\))
∆RSQ ≅ ∆RSQ (By SAS)
and so PQ = RM (By CPCT)
Now PR = PM – RM = PM – RQ
PM = PR + RQ.

Example 2:
Construct a AABC in which BC = 3.6 cm, AB + AC = 4.8 and A RSM QS – MS RS = RS cm and B = ∠60°.
Solution:
Steps of Construction:

1. Draw BC = 3.6 cm.
2. Draw ∠CBX= \({ 60 }^{ \underline { 0 } }\) at B.
3. From BX, cut off line segment BD = 4.8 cm.
4. Join DC.
5. Draw the perpcndicub: bisector ofDC meeting BD at A.
6. joinAC.
7. ABC is the required Mangle.

Justification:
‘A‘ lies on the perpendicular bisector of DC
∴ AD = AC
Now BD = 4.8 cm
⇒ BA + AD = 4.8
BA + AC = 4.8 (∴ AD = AC)

Example 3:
Construction of a Triangle when its Base Angle and the Difference of the other two sides are given.
Solution:
Case I:
Given the base BC, a base angle, say ∠B and AB – AC. Steps of Construction:

1. Draw the base BC.
2. At point B draw an angle ∠CBX equal to the given angle.
3. Cut a line segment BD = AB – AC from point B.
4. Join DC.
5. Draw the perpendicular bisector of . DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As point A lies on the perpendicular bisector of DC
∴ AD = AC
BD = AB – AD = AB – AC (∴ AD = AC)

Case II:
Given the base BC, a base angle say ∠B and (AC – AB).
Steps of Construction:

1. Draw the base SC.
2. At point S, draw an angle, ∠CBX equal to the given angle and extend the arm XB backward.
3. Cut a line segment BD equal to (AC – AB) from the extended arm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ΔABC is the required triangle.

Justification:
As the perpendicular bisector of DC passes through A.
∴ AD – AC
BD = AD – AB
∴ BD = AC – AB

Example 4:
Construct a ∆ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw base BC = 3.4 cm.
2. Draw ∠CBX = \({ 45 }^{ \underline { 0 } }\).
3. From BX, cut line segment BD =1.5 cm.
4. Join DC.
5. Draw the perpendicular bisector of DC which intersect BX at A.
6. Join AC.
7. ABC is the required triangle.

Justification:
As A lies on perpendicular bisector of DC.
AD = AC
BD = 1.5 cm
⇒ AB – AD =1.5 cm
AB – AC =1.5 cm

Example 5:
Construct a ∆PQR in which QR = 5.8 cm, PR – PQ = 1.8 cm and ∠Q = \({ 45 }^{ \underline { 0 } }\). Justify your construction.
Solution:
Steps of Construction:

1. Draw base QR = 5.8 cm.
2. Draw ∠RQP = \({ 45 }^{ \underline { 0 } }\).
3. Produce arm XQ backward and cut a line segment QS = 1.8 cm.
4. Join SR.
5. bisector of SR which intersect QX at P.
6. Join PR.
7. PQR is the required triangle.

Justification:
As P lies on perpendicular bisector of SR
PS = PR
QS = 1.8 cm
⇒ PS – PQ = 1.8 cm
∴ PR – PQ = 1.8 cm

Example 6:
Construct a ∆ABC in which BC = 5.8 cm, ∠C = \({ 60 }^{ \underline { 0 } }\) and AB – AC = 2.5 cm.
Solution:
Steps of Construction:

1. Draw base BC = 5.8 cm
2. Draw ∠ACB = \({ 60 }^{ \underline { 0 } }\)
3. Produce arm CX backward and cut a line segment CD = 2.5 cm.
4. Join BD.
5. Draw perpendicular bisector of BD which intersect CX at A.
6. Join AB.
7. ABC is the required triangle.

Justification:
A lies on the perpendicular bisector of BD.
∴ AB =AD
CD = 2.5 cm
⇒ AD – AC = 2.5 cm
⇒ AB – AC = 2.5 cm

Example 7:
Construct a ∆ABC in which AC – AB = 3.5 cm, BC = 6.2 cm and ∠C = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw BC = 6.2 cm
2. Draw ∠ACB = \({ 45 }^{ \underline { 0 } }\)
3. Cut CD = 3.5 cm from CX.
4. Join BD and draw the perpendicular bisector of BD which intersect CX at A.
5. Join AB.
6. ABC is the required triangle.

Justification:
As A lies on the perpendicular bisector of BD.
∴ AB = AD
CD = 3.5 cm
⇒ AC – AD = 3.5 cm
∴ AC – AB = 3.5 cm

Example 8:
Construction of a Triangle when its Perimeter and Two Base Angles are given.
Solution:
Given the base angles, ∠B and ∠C and perimeter, i.e., AB + BC + AC.
Steps of Construction:

1. Draw a line segment, DE equal to AB + BC + CA.
2. Draw ∠EDM and ∠DEN equal to the base angles ∠B and ∠C respectively.
3. Draw bisectors of ∠MDE and ∠NED which intersect at A.
4. Draw the perpendicular bisectors of AD and AE which intersect DE at B and C respectively.
5. Join AE and AC.
6. ABC is the required triangle.

Justification:
As B lies on the perpendicular bisector of AD
∴ AB = DB
In ∆ADB
AB = DB
∠ADB = ∠DAB
(∴ Angles opposite to equal sides of a A are equal)
Similarly, AC = CE
and ∠CAE = ∠CEA
Now DE = DB + BC + CE
= AB + BC + AC
In ∆ABD, ∠ABC = ∠DAB + ∠ADB = ∠ADB + ∠ADB = 2∠ADB
In ∆ACE, ∠ACB = ∠CAE + ∠CEA = ∠CEA + ∠CEA = 2∠CEA

Example 9:
Construct a triangle whose perimeter is 6.4 cm and angles at the base are \({ 60 }^{ \underline { 0 } }\) and \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

1. Draw line segment DE equal to (AB + BC + CA) = 6.4 cm
2. Draw ∠EDM = \({ 60 }^{ \underline { 0 } }\) and ∠DEN = \({ 45 }^{ \underline { 0 } }\).
3. Draw AD and AE as bisectors of ∠MDE and ∠NED which inter sect at A.
4. Draw the perpendicular bisector of AD and AE which intersect DE at B and C respectively.
5. Join AB and AC.
6. ABC is the required triangle.

MP Board Class 9th Maths Solutions

MP Board Class 9th Special Hindi सहायक वाचन Solutions Chapter 7 उड़ता चल कबूतर (यात्रा वृत्तांत, रामवृक्ष बेनीपुरी)

उड़ता चल कबूतर अभ्यास

बोध प्रश्न

प्रश्न 1.
सोनभद्र नदी किस प्रदेश में बहती है?
उत्तर:
सोनभद्र नदी बिहार प्रदेश में बहती है।

प्रश्न 2.
यूरोप यात्रा पर जाने के लिए किसने पत्र लिखा?
उत्तर:
यूरोप यात्रा पर जाने के लिए सन् 1947 ई. में आचार्य नरेन्द्र देव ने पत्र लिखा।

प्रश्न 3.
जयप्रकाश जी को किस-किसकी चिन्ता थी?
उत्तर:
जयप्रकाश जी को दैनिक-पत्र ‘जनता’ की चिन्ता थी और आगामी आम चुनाव की भी चिन्ता थी।

प्रश्न 4.
बिहार से चलने के बाद लेखक का प्लेन कौन-से शहर में उतरा?
उत्तर:
बिहार से चलने के बाद लेखक का प्लेन बनारस शहर में उतरा।

प्रश्न 5.
शरीर की भंगिमा द्वारा भाव प्रकट करने वाले व्यक्ति का नाम क्या था?
उत्तर:
शरीर की भंगिमा द्वारा भाव प्रकट करने वाले व्यक्ति का नाम उदय शंकर था।

प्रश्न 6.
ननिहाल जाते समय लेखक की हाथी के साथ क्या घटना घटी?
उत्तर:
ननिहाल जाते समय लेखक एक हाथी पर चढ़कर ऊँचे आसन पर बैठकर आनन्द लेना चाहता था पर हाथी की वह सवारी भीतरी मन को जितना आनन्द नहीं दे पाई उससे ज्यादा मन में भय बैठा गई थी।

प्रश्न 7.
बर्नार्ड शा ने अंग्रेजों के बारे में क्या लिखा है?
उत्तर:
बर्नार्ड शा ने अंग्रेजों के बारे में लिखा है कि इस कौम के दिल में कोई इरादा जागता है तो वह इस तरह उसे छिपाकर रखती है कि एक दिन यह इरादा एक ज्वलन्त विश्वास में परिणत हो जाता है और इस विश्वास की अखण्ड ज्योति, इसमें इतनी शक्ति पैदा कर देती है कि वह जो चाहती है, उसे करके ही दम लेती है।

प्रश्न 8.
ब्रजनन्दन आजाद ने तार के द्वारा क्या सूचना दी?
उत्तर:
ब्रजनन्दन आजाद ने तार के द्वारा विलायत जाने की तैयारी करने की सूचना दी।

प्रश्न 9.
गोरे अंग्रेज ने हिन्दी के बारे में क्या कहा?
उत्तर:
गोरे अंग्रेज ने हिन्दी के बारे में कहा कि आजकल हिन्दी को लोग संस्कृत बना रहे हैं। प्रेमचन्द की हिन्दी कितनी अच्छी थी। पटना के हिन्दी अखबार तक ऐसी हिन्दी लिखते हैं कि उनका अर्थ समझने के लिए डिक्शनरी की जरूरत होती है।

प्रश्न 10.
दिल्ली के राजभवन में कौन-कौन से चित्र लगे थे?
उत्तर:
दिल्ली के राजभवन में वायसराय के चित्र, अंग्रेजों द्वारा बनाये गये भवनों के चित्र, उन युद्धों के चित्र जिनमें विजयी बनकर अंग्रेज भारत के शासक हुए। सुरक्षा की दृष्टि से ये चित्र राजभवन को म्यूजियम मानकर लगा दिये गये थे।

प्रश्न 11.
हवाई जहाज यात्रा के समय नीचे के दृश्यों का वर्णन लेखक ने किस प्रकार किया है?
उत्तर:
हवाई जहाज यात्रा के समय नीचे के दृश्यों का वर्णन करते हुए लेखक कहता है कि ऊपर उड़ने पर ताड़ के वृक्ष छोटे-छोटे लगते हैं। धरती की सीमाएँ जैसे सिमटी जा रही हैं। खेत ऐसे लग रहे हैं जैसे मुसल्लम धारीदार चादर हो। तेजधूप में गंगा का कछार हिमालय की सीढ़ी का पहला जीना जैसा लग रहा था। नीचे की सारी चीजें छोटी होती जा रही हैं। ताड़-खजूर आम-नीम सब छोटे-छोटे से पौधे लग रहे हैं। पृथ्वी शतरंज की एक लम्बी विसात सी लगती है। जैसे-जैसे हम ऊँचे चढ़ते जाते हैं भेदभाव मिटते जाते हैं, सीमाएँ नष्ट हो जाती हैं और एकरूपता बढ़ती जाती है।

प्रश्न 12.
अन्तर्मन और बहिर्मन का परस्पर क्या सम्बन्ध है? उदाहरण द्वारा स्पष्ट कीजिए।
उत्तर:
प्रत्येक वस्तु के दो पहलू होते हैं-आन्तरिक और बाह्य। इसी प्रकार मानव मन के भी दो पहलू होते हैं-अन्तर्मन और बहिर्मन। जब किसी वस्तु के प्रति हमारा आकर्षण अधिक होता है तो हमारा बहिर्मन उस वस्तु को प्राप्त करने का प्रयास करता है। अभिप्राय यह है कि अन्तर्मन किसी के भी बारे में सोचता रहता है, योजनाएँ बनाता रहता है पर उन योजनाओं को कार्य रूप में परिणत नहीं कर पाता है। उन योजनाओं को कार्य रूप में ‘परिणत करता है बहिर्मन। इस प्रकार हम कह सकते हैं कि अन्तर्मन का कार्य योजना बनाना है और उसे क्रिया रूप में परिणत करता है बहिर्मन। अत: इन दोनों का परस्पर अटूट सम्बन्ध है। एक के बिना दूसरे का काम नहीं चल सकता है।

प्रश्न 13.
बनारस में उदय शंकर जी से लेखक की क्या बातचीत हुई?
उत्तर:
जब लेखक बिहार से चलकर बनारस हवाई अड्डे पर उतरा तो उसने लॉन में किसी को बैठा देखा। उसे देखकर लेखक की जो बातचीत हुई, वह इस प्रकार है-“क्षमा करें, क्या आप उदय शंकर हैं?”

“जी हाँ, आप मुझे पहचानते हैं?” जी, मैं भी एक छोटा-मोटा कलाकार ही हूँ। जिसे आप शरीर की भंगिमा द्वारा प्रकट करते हैं, उसे मैं कलम द्वारा उतारने की कोशिश करता हूँ।”

“अपने कला-केन्द्र के बारे में कुछ बता ………”
हाँ, सोच रहा हूँ, 1952 में उसे अल्मोड़ा के बदले देहरादून में खोलूँ। देहरादून बड़ी अच्छी जगह है।” मैने बताया लेखनीधारी का बेटा राइफलधारी बनने जा रहा है। “देश को उसकी भी जरूरत है। अच्छा किया है।”

प्रश्न 14.
भावार्थ स्पष्ट कीजिए-“ज्यों-ज्यों ऊँचे चढ़िए भेदभाव मिटते जाते हैं। सीमाएँ नष्ट होती जाती हैं, एकरूपता बढ़ती जाती है।”
उत्तर:
भावार्थ-लेखक कहता है कि ज्यों-ज्यों हम हवाई जहाज के द्वारा ऊपर आकाश की ओर बढ़ते जाते हैं, नीचे की सब वस्तुएँ एकसी दिखाई देने लगती हैं। इस यात्रा में सीमाएँ नष्ट होती जाती हैं और ऐसा लगता है कि सारा संसार, सारी पृथ्वी एक ही है।

MP Board Class 9th Hindi Solutions

MP Board Class 9th Special Hindi सहायक वाचन Solutions Chapter 11 यशस्वी पत्रकार दादा साहेब आप्टे (संजय त्रिवेदी)

यशस्वी पत्रकार दादा साहेब आप्टे अभ्यास

बोध प्रश्न

प्रश्न 1.
दादा साहेब आप्टे ने किस समाचार एजेंसी की स्थापना की एवं उसका संचालन किया?
उत्तर:
तमाम भारतीय भाषाओं को एक सूत्र में बाँधने के उद्देश्य से दादा साहेब आप्टे ने सन् 1948 में एक बहुभाषी समाचार एजेंसी ‘हिंदुस्तान समाचार’ की स्थापना की और सफलतापूर्वक लम्बे समय तक उसका संचालन किया। उनके द्वारा स्थापित यह समाचार एजेंसी आज भी भारतीय पत्रकारिता की सेवा कर रही है।

प्रश्न 2.
दादा साहेब आप्टे का जन्म कहाँ हुआ?
उत्तर:
यशस्वी पत्रकार दादा साहेब आप्टे का जन्म सन् 1905 ई. में गुजरात के बड़ोदरा नामक नगर में हुआ था।

प्रश्न 3.
वकालत करते समय दादा साहेब की मुलाकाल किस महापुरुष से हुई थी?
उत्तर:
अपनी प्रारम्भिक शिक्षा-दीक्षा बड़ोदरा (गुजरात) में पूर्ण कर दादा साहेब आप्टे ने मुम्बई से एल-एल.बी. की परीक्षा उत्तीर्ण की और मुम्बई उच्च न्यायालय में वकालत प्रारम्भ की। इसी दौरान आप्टे की मुलाकात उस समय के जाने-माने अधिवक्ता तथा भारतीय संस्कृति के विद्वान बैरिस्टर कन्हैयालाल माणिक लाल मुंशी से हुई। श्री आप्टे को मुंशी का मार्गदर्शन मिला तो उनकी प्रतिभा को चार चाँद लग गए।

प्रश्न 4.
किसने भारतीय भाषाओं की श्रीवृद्धि के लिए दादा साहेब आप्टे की सराहना की?
उत्तर:
भारतीय भाषाओं को एक सूत्र में पिरोने एवं उनकी श्रीवृद्धि के लिए दादा साहेब आप्टे ने अपना समूचा जीवन लगा दिया। इसके लिए उन्होंने 1948 में पहली स्वदेशी राष्ट्रीय समाचार एजेंसी ‘हिंदुस्तान समाचार’ की स्थापना की। तत्कालीन राष्ट्रपति डॉ. राजेन्द्र प्रसाद ने भारतीय भाषाओं की श्रीवृद्धि के लिए किए जा रहे इस प्रयास की सरहना की। वास्तव में, भारतीय भाषाओं की इस समाचार एजेंसी को स्थापित कर श्री आप्टे. ने राष्ट्रीय एकता और सद्भावना पैदा करने का महत्वपूर्ण काम किया।

प्रश्न 5.
स्वतंत्र भारत की पहली स्वदेशी समाचार एजेंसी कौन-सी थी?
उत्तर:
स्वतंत्र भारत की पहली स्वदेशी समाचार एजेंसी ‘हिंदुस्तान समाचार’ थी। यशस्वी पत्रकार दादा साहेब आप्टे ने। तमाम भारतीय भाषाओं को एक सूत्र में बाँधने के उद्देश्य से सन् 1948 में इसकी स्थापना की थी।

प्रश्न 6.
दादा साहेब ने पत्रकारिता की परम्पराओं और कार्य पद्धति को सीखने के लिए कहाँ काम किया?
उत्तर:
दादासाहेब आप्टे को प्रारम्भ से ही लेखन और पत्रकारिता में गहरी रुचि थी। इसलिए वे राष्ट्रसेवा में पत्रकारिता के माध्यम से योगदान करना चाहते थे। इस दौरान उन्होंने पत्रकारिता की परम्पराओं ओर कार्यपद्धति को सीखने के लिए यूनाइटेड प्रेस ऑफ इण्डिया (यू.पी.आई.) में काम किया। वहाँ अनुभव प्राप्त करने के बाद श्री आप्टे ने ‘हिंदुस्तान समाचार’ नाम से भारतीय भाषाओं की संवाद समिति गठित की।

प्रश्न 7.
आप्टे जी ने हिंदी के किस प्रख्यात कवि की कृति का मराठी में अनुवाद किया?
उत्तर:
दादा साहेब आप्टे ने हिंदी के प्रख्यात कवि जयशंकर प्रसाद की चर्चित एवं लोकप्रिय कृति ‘कामायनी’ का मराठी में अनुवाद किया।

प्रश्न 8.
भारतीय संस्कृति और सभ्यता के लिए दादा साहेब ने लगभग कितने देशों की यात्राएँ की?
उत्तर:
भारतीय संस्कृति और सभ्यता के प्रचार के लिए दादा साहेब आप्टे ने विश्व के लगभग 22 देशों की यात्रा की।

प्रश्न 9.
दादासाहेब आप्टे किन-किन भाषाओं के जानकार थे?
उत्तर:
दादा साहेब आप्टे एक यशस्वी यायावर पत्रकार थे। वे हिंदी, मराठी, गुजराती, अंग्रेजी और संस्कृत भाषा पर अपना अधिकार रखते थे।

प्रश्न 10.
दादा साहेब ने महाराष्ट्र के किस संत पर अंग्रेजी में शोध प्रबंध लिखा?
उत्तर:
दादा साहेब आप्टे ने महाराष्ट्र के सामाजिक जीवन में अत्यन्त लोकप्रिय और सम्मानित समर्थ गुरु रामदास के जीवन पर अंग्रेजी में शोध प्रबन्ध लिखकर उनके योगदान से विश्व मानवता को परिचित कराया।

MP Board Class 9th Hindi Solutions

MP Board Class 9th Special Hindi सहायक वाचन Solutions Chapter 10 जीवन दृष्टि (लघु प्रसंग)

जीवन दृष्टि अभ्यास

बोध प्रश्न

प्रश्न 1.
विक्रम साराभाई कौन थे?
उत्तर:
विक्रम साराभाई एक महान् वैज्ञानिक थे। आपने राष्ट्रीय अन्तरिक्ष अनुसन्धान संस्थान (इसरो) की स्थापना की है।

प्रश्न 2.
लेखक के अनुसार अच्छे पुरुष कौन होते हैं?
उत्तर:
लेखक के अनुसार अच्छे पुरुष उन सुन्दर फूलों के समान होते हैं जो अपनी उदारता से सुगंध और शहद देते हैं। बिना किसी चाह के प्रेम और सुन्दरता बाँटते हैं और जब अपना काम कर लेते हैं तो चुपचाप गिर जाते हैं। कहने का भाव यह है कि अच्छे पुरुष अपने कार्यों से समाज और देश में अच्छे जन उपयोगी कार्य करते हैं। इन कार्यों के करने में उनका कोई स्वार्थ नहीं होता है और जब वे अपने निर्धारित लक्ष्य को प्राप्त कर लेते हैं तो चुपचाप इस लोक से गमन कर जाते हैं।

प्रश्न 3.
विद्यालय के छात्रों को किस बात के लिए पुरस्कृत किया गया था?
उत्तर:
विद्यालय के छात्रों को उनके अच्छे आचरण एवं श्रेष्ठ चरित्र के लिए पुरस्कृत किया गया था।

प्रश्न 4.
बुढ़िया माँ कलकत्ता क्यों गई थी?
उत्तर:
बुढ़िया माँ कलकत्ता गंगा स्नान के लिए गयी थी।

प्रश्न 5.
कलकत्ता हाईकोर्ट के न्यायाधीश का नाम क्या था?
उत्तर:
कलकत्ता हाईकोर्ट के न्यायाधीश का नाम श्री गुरुदास वन्द्योपाध्याय था।

प्रश्न 6.
‘इसरो’ की स्थापना कहाँ और कैसे हुई?
उत्तर:
‘इसरो’ की स्थापना पालीथुरा, थुम्बा (केरल) में महान वैज्ञानिक प्रो. विक्रम साराभाई के अथक प्रयासों से हुई थी।

प्रश्न 7.
विक्रम साराभाई किस बात पर शोध कर रहे थे?
उत्तर:
विक्रम साराभाई अन्तरिक्ष की किरणों पर शोध कर रहे थे। आप राष्ट्रीय अनुसन्धान संस्थान की स्थापना के लिए 400 एकड़ भूमि भूमध्य रेखा के पास चाहते थे।

प्रश्न 8.
‘संसार की स्थिति भी ऐसी ही है’ के माध्यम से किस स्थिति की बात कही गई है ?
उत्तर:
‘संसार की स्थिति भी ऐसी ही है’ के माध्यम से लेखक ने बताया है कि हम आपस में बिना दूसरे की भावना को समझे अकारण ही लड़ाई-झगड़ा किया करते हैं। हमें भाषा, जाति या देश के भेदभाव के कारण आपस में टकराना नहीं चाहिए अपितु परस्पर एक-दूसरे की भावना को समझकर प्रेम और शान्ति का वातावरण बनाना चाहिए।

प्रश्न 9.
न्यायाधीश अपने आसन से क्यों खड़े हो गये?
उत्तर:
न्यायाधीश श्री गुरुदास वन्द्योपाध्याय अपनी बूढ़ी धाय माँ (पालन करने वाली माँ) को न्यायालय कक्ष के दरवाजे पर खड़े देखकर अपने आसन से खड़े हो गये।

प्रश्न 10.
गुरुदास ने बुढ़िया माँ का परिचय किस प्रकार दिया?
उत्तर:
गुरुदास ने बुढ़िया माँ का परिचय देते हुए कहा कि ये मेरी माँ हैं, इन्होंने मुझे दूध पिलाकर पाला है।

प्रश्न 11.
गुरुदास बुढ़िया माँ से किस प्रकार मिले?
उत्तर:
गुरुदास भूमि पर लेटकर उस बुढ़िया को दण्डवत् प्रणाम करके मिले।

प्रश्न 12.
धर्म और विज्ञान राष्ट्र उत्थान में सहयोगी हैं। प्रसंग के आधार पर स्पष्ट कीजिए।
उत्तर:
धर्म और विज्ञान राष्ट्र उत्थान में शत-प्रतिशत सहयोगी हैं। इसका प्रत्यक्ष उदाहरण हमें ‘इसरो’ की स्थापना में मिलता है। जब विक्रम साराभाई राष्ट्रीय अनुसन्धान संस्थान (इसरो) की स्थापना के लिए भूमध्य रेखा के निकटवर्ती क्षेत्र की भूमि चाहते थे। उन्हें यह भूमि केरल राज्य के थुम्बा क्षेत्र में दिखाई दी। इस भूमि पर उस क्षेत्र के ईसाइयों तथा बिशप परेरा का आवास था। जब विक्रम साराभाई ने राष्ट्र के निर्माण के लिए इस भू-भाग (जो लगभ 400 एकड़ में था) को माँगा तो आदरणीय फादर डॉक्टर पीटर परेरा बिशप के विज्ञान को धर्म से जोड़कर यह भूमि सहर्ष ‘इसरो’ की स्थापना के लिए दान में दे दी।

प्रश्न 13.
ओछे और अच्छे पुरूषों की क्या पहचान है? वर्णन कीजिए।
उत्तर:
ओछे लोग वे होते हैं जो सदैव अपने स्वार्थ की बातें करते हैं। वे कभी भी यह नहीं सोचते कि हमारे इन स्वार्थ पूर्ण कार्यों से अन्य व्यक्तियों, समाज और राष्ट्र को क्या हानि उठानी पड़ेगी।

इसके विपरीत जो अच्छे पुरूष होते हैं वे उस सुन्दर पुष्प के समान होते हैं जो अपनी सुगन्ध एवं मिठास रूपी सत्कार्यों से अन्य व्यक्तियों, समाज एवं राष्ट्र की नि:स्वार्थ भाव से सेवा करते हैं। वे राष्ट्रहित में अपना सर्वस्व निछावर करने में भी गर्व का अनुभव करते हैं।

प्रश्न 14.
गुरुदास की कौन-सी विशेषता ने उन्हें महान् बनाया? समझाइए।
उत्तर:
गुरुदास एक नम्र, शीलवान, विद्वान एवं मातृभक्त इन्सान थे। वे कलकत्ता हाईकोर्ट के न्यायाधीश तथा कलकत्ता विश्वविद्यालय के कुलपति थे। अंग्रेजी काल में इतना सम्मान का पद पाने वाले वे प्रथम भारतीय थे। इतना ऊँचा पद पाकर भी अहंकार से दूर थे। जब उनकी निर्धन धाय माँ अपने भीगे वस्त्रों को पहने न्यायालय के कक्ष के दरवाजे पर आ जाती है तो वे अपनी कुर्सी छोड़कर उसे दण्डवत् प्रणाम करते हैं तथा कोर्ट के सभी काम बन्द करके उसे अपने घर ले जाते हैं तथा उसकी सेवा सुश्रूषा करते हैं। उपर्युक्त विशेषताओं के कारण गुरुदास वन्द्योपाध्याय महान् बन गये थे।

MP Board Class 9th Hindi Solutions

MP Board Class 9th Special Hindi सहायक वाचन Solutions Chapter 9 प्रेरणा दीप (पौराणिक कथा संदर्भ, संकलित)

प्रेरणा दीप अभ्यास

बोध प्रश्न

प्रश्न 1.
रामायण के रचयिता कौन हैं?
उत्तर:
रामायण के रचयिता महर्षि बाल्मीकि हैं।

प्रश्न 2.
नदी के तट पर किस पक्षी का जोड़ा था?
उत्तर:
नदी के तट पर क्रौञ्ज पक्षी का जोड़ा था।

प्रश्न 3.
बाल्मीकि के शिष्य का नाम क्या था? बताइए।
उत्तर:
बाल्मीकि के शिष्य का नाम भारद्वाज था।

प्रश्न 4.
राम के वंश का नाम लिखिए।
उत्तर:
राम के वंश का नाम सूर्यवंशी इक्ष्वाकु था।

प्रश्न 5.
कौरव सेना के सेनानायक का नाम क्या था?
उत्तर:
कौरव सेना के सेनानायक का नाम भीष्म पितामह था।

प्रश्न 6.
युधिष्ठिर कौरव की सेना में क्यों गये थे?
उत्तर:
युधिष्ठिर कौरव की सेना में भीष्म पितामह एवं अन्य गुरुजनों के चरण स्पर्श करने तथा उनसे आशीर्वाद लेने गये थे।

प्रश्न 7.
बाल्मीकि ने निषाद को क्या और क्यों अभिशाप दिया?
उत्तर:
बाल्मीकि ने निषाद को यह अभिशाप दिया कि “तू अब आने वाले समय में कभी भी प्रतिष्ठा (सम्मान) को प्राप्त नहीं कर सकेगा। क्योंकि तूने कामासक्त क्रौञ्ज के जोड़े में से नर क्रौञ्च का वध कर दिया है।”

प्रश्न 8.
बाल्मीकि को ध्यान के समय ब्रह्मा जी ने दर्शन देकर क्या कहा?
उत्तर:
बाल्मीकि जब ध्यान लगाकर बैठे हुए थे तभी सृष्टि के रचयिता ब्रह्माजी ने दर्शन देकर उनसे कहा-“ऋषिवर मेरी ही इच्छा वाणी अनायास आपके मुँह से निकली और श्लोक रूप में इसलिए निकली है कि आप अनुष्टुप् छन्दों में रामचन्द्र के सम्पूर्ण चरित्र का वर्णन कीजिए। श्रीराम कथा संक्षेप में आप नारदजी से सुन ही चुके हैं। मेरे राम, लक्ष्मण, सीता और राक्षसों का गुप्त अथवा सब वृत्तान्त आपकी आँखों के सामने आ जाएगा, होगा वह भी दिखाई पड़ेगा। अतः जो आप लिखे यथार्थ और सत्य होगा। इस प्रकार आपकी लिखित रामायण इस लोक में अमर हो जायेगी।”

प्रश्न 9.
श्रीकृष्ण ने अर्जुन को अपने विराट स्वरूप में क्या दिखलाया?
उत्तर:
श्रीकृष्ण ने अर्जुन को अपना विराट स्वरूप दिखलाया जिसे देखकर अर्जुन काँप उठे। उन्होंने देखा कि एक विशाल अग्नि-ज्वाला, जिसमें चारों ओर से कीड़े-मकोड़े आते हैं और समाकर भस्म हो जाते हैं। उसी तरह समस्त कौरव भी कृष्ण के मुख में समा रहे हैं। इसे देखकर अर्जुन का मोह भंग हो गया और उन्होंने भगवान कृष्ण की वन्दना की।

प्रश्न 10.
माता सत्यवती से भीष्म ने क्या प्रतिज्ञा की थी?
उत्तर:
माता सत्यवती से भीष्म ने प्रतिज्ञा की थी कि वे राजा के पक्ष में रहेंगे तथा उसी की ओर से युद्ध करेंगे। लेकिन धर्म और सत्य की सदैव विजय होती है अतः विजयी पाण्डव ही होंगे।

प्रश्न 11.
तमसा नदी की प्राकृतिक सुन्दरता का वर्णन करो।
उत्तर:
तमसा नदी का प्राकृतिक सौन्दर्य बहुत मनोहारी था। इस समय बसन्त ऋतु का आगमन हो चुका था। नदी का जल अत्यन्त स्वच्छ एवं पवित्र था। यहाँ तक कि उसका तल भी स्पष्ट नजर आ रहा था। जल के किनारे पक्षी बैठे थे तथा उसमें मछलियाँ तैर रही थीं। नदी के किनारे हरे-भरे जंगल थे। कुछ पक्षी तो पेड़ों पर बैठे थे तथा कुछ नदी के किनारे। कुछ पक्षी नदी के जल में अपनी चोंच डुबोकर पानी पी रहे थे। इस प्रकार इस प्राकृतिक सौन्दर्य का आनन्द लेते हुए पक्षी विचरण कर रहे थे।

प्रश्न 12.
नारद जी की वाणी याद आने पर बाल्मीकि ने नारद जी से क्या पूछा ? तथा नारद जी ने क्या उत्तर दिया? लिखिए।
उत्तर:
एक बार बाल्मीकि जी निषाद को दिये गये कठोर शाप की चिन्ता में मग्न थे कि उसी समय उन्हें नारद जी की वाणी याद आयी। उन्होंने नारद जी से पूछा था-“हे देवर्षि, मुझे किसी ऐसे पुरुष का नाम बताइये जो गुणवान, बलवान एवं धर्मात्मा हो, जो सत्य पर दृढ़ रहता हो, अपने वचन का पक्का हो, सबका हित करने वाला हो, विद्वान हो और जिससे बढ़कर सुन्दर कोई दूसरा न हो।”

इस प्रश्न को सुनकर नारदजी ने कहा था कि मैं ऐसे एक ही पुरूष को जानता हूँ। वे इक्ष्वाकु वंश के राजा दशरथ के पुत्र राम हैं। वे सब तरह से गुणवान और रूपवान हैं और जब क्रोध करते हैं तब डर के मारे देवता और दानव भी काँप उठते हैं।

प्रश्न 13.
श्रीकृष्ण का कर्मयोग संक्षिप्त में समझाइए।
उत्तर:
युद्ध क्षेत्र में जब अर्जुन ने देखा कि कौरव सेना तथा पाण्डव सेना दोनों में ही उनके आत्मीय एवं परिवारीजन हैं और वे दोनों ही एक-दूसरे को राज्य प्राप्ति के लिए मारने को उद्यत हैं तो अर्जुन दु:खी हो जाते हैं और अपना गाण्डीव धनुष रथ में एक तरफ रखकर रथ के पिछले भाग में बैठ जाते हैं।

जब श्रीकृष्ण ने अर्जुन की इस मोह ग्रसित दशा को देखा तो उन्होंने अपना विराट स्वरूप दिखाकर अर्जुन का मोह भंग किया तत्पश्चात् उन्होंने अर्जुन को कर्मयोग का उपदेश देते हुए कहा कि आत्मा अजर और अमर है। इसलिए हे अर्जुन! तुम शोक मत करो तथा युद्ध करो।

प्रश्न 14.
कुरुक्षेत्र में खड़ी सेनाओं का वर्णन कीजिए।
उत्तर:
कुरुक्षेत्र में ग्यारह अक्षौहिणी सेना कौरवों की ओर थीं ओर सात अक्षौहिणी सेना पाण्डवों की ओर थीं। दोनों सेनाएँ आमने-सामने व्यूह-रचना के तरीके से खड़ी थीं। कौरव सेना के प्रधान सेनापति भीष्म पितामह थे तथा पाण्डव सेना के आगे धनुर्धारी अर्जुन थे। अर्जुन के रथ का नाम नंदिघोष था जिसके सारथी स्वयं भगवान श्रीकृष्ण थे। कौरव सेना के सेनापति भीष्म पितामह के रथ के साथ ही रथी दुःशासन था तथा थोड़ी दूरी पर मुक्ताओं से जड़ित सुन्दर रथ पर कौरवराज दुर्योधन था। पास ही गुरू द्रोणाचार्य और अश्वत्थामा थे। इस प्रकार कौरवों एवं पाण्डवों की अपार सेना एक अद्भुत रोमांचकारी दृश्य उपस्थित कर रही थी।

MP Board Class 9th Hindi Solutions