MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3

MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3

Class 7 Maths Chapter 2 Exercise 2.3 Solutions Hindi Medium प्रश्न 1.
ज्ञात कीजिए :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1a
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1b
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 1c

Class 7 Maths Chapter 2 Exercise 2.3 Solutions In Hindi प्रश्न 2.
गुणा कीजिए और न्यूनतम रूप में बदलिए (यदि सम्भव है) :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 2
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 2a

MP Board Solutions

MP Board Class 7 Maths Solutions English Medium प्रश्न 3.
निम्नलिखित भिन्नों को गुणा कीजिए-
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3a
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 3b

MP Board Class 7 Maths Solutions प्रश्न 4.
कौन बड़ा है:
(i) \(\frac { 3 }{ 4 }\) का \(\frac { 2 }{ 7 }\) अथवा \(\frac { 5 }{ 8 }\) का \(\frac { 3 }{ 5 }\)
(ii) \(\frac { 6 }{ 7 }\) का \(\frac { 1 }{ 2 }\) अथवा \(\frac { 3 }{ 7 }\) का \(\frac { 2 }{ 3 }\)
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 4

MP Board Solutions

Class 7 Maths Chapter 2.3 In Hindi प्रश्न 5.
सैली अपने बगीचे में चार छोटे पौधे एक पंक्ति में लगाती है। दो क्रमागत छोटे पौधों के बीच की दूरी – मीटर है। प्रथम और अन्तिम पौधे की बीच की दूरी ज्ञात कीजिए।
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 5
दो क्रमागत पौधों के बीच की दूरी = \(\frac { 3 }{ 4 }\) मीटर
∴ प्रथम और अन्तिम (चौथे) पौधे के बीच की दूरी = 3 x \(\frac { 3 }{ 4 }\) मीटर
= \(\frac { 3\times3 }{ 4 }\) मीटर = \(\frac { 9 }{ 4 }\) मीटर = 2\(\frac { 1 }{ 4 }\) मीटर

Class 7 Maths Exercise 2.3 Solutions In Hindi प्रश्न 6.
लिपिका एक पुस्तक को प्रतिदिन 1\(\frac { 3 }{ 4 }\) घण्टे पढ़ती है। वह सम्पूर्ण पुस्तक को 6 दिनों में पढ़ती है। उस पुस्तक को पढ़ने में उसने कुल कितने घण्टे लगाए ?
हल:
प्रतिदिन पढ़ने का समय = 1\(\frac { 3 }{ 4 }\) घण्टे = \(\frac { 7 }{ 4 }\) घण्टे
वह पुस्तक को पढ़ती है = 6 दिनों में
= \(\frac { 6\times7 }{ 4 }\) = \(\frac { 42 }{ 4 }\)
उसने पुस्तक पढ़ने में 10\(\frac { 1 }{ 2 }\) घण्टे लगाए।

Class 7 Maths 2.3 In Hindi प्रश्न 7.
एक कार 1 लीटर पैट्रोल में 16 किमी दौड़ती है। 2\(\frac { 3 }{ 4 }\) लीटर पैट्रोल में यह कार कुल कितनी दूरी तय करेगी?
हल:
∵ कार 1 लीटर में जाती है = 16 किमी
∴ कार द्वारा 23 लीटर में चली गई दूरी
= 16 x 2\(\frac { 3 }{ 4 }\) किमी
= 16 x \(\frac { 11 }{ 4 }\) = 4 x 11 = 44 किमी

MP Board Solutions

Class 7 Math 2.3 In Hindi प्रश्न 8.
(a) (i) बक्सा □ में संख्या लिखिए, ताकि \(\frac{2}{3}\) x □ = \(\frac{10}{30}\)
(ii) □ में प्राप्त संख्या का न्यूनतम रूप….. है।

(b) (i) बक्सा □ में संख्या लिखिए, ताकि \(\frac{3}{5}\) x □ = \(\frac{24}{75}\)
(ii) □ में प्राप्त संख्या का न्यूनतम रूप… है।
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 8
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 8a

पाठ्य-पुस्तक पृष्ठ संख्या # 44

ज्ञात कीजिए-
3 ÷ \(\frac { 1 }{ 2 }\) और 3 x \(\frac { 2 }{ 1 }\)
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 9

भिन्न का व्युत्क्रम

रिक्त स्थानों की पूर्ति कीजिए-
हल :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 9a
ऐसी शून्येतर संख्याएँ जिनका परस्पर गुणनफल 1 है, एक-दूसरे के व्युत्क्रम कहलाते हैं।

ऐसे पाँच और युग्मों को गुणा कीजिए।
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10
सोचिए, चर्चा कीजिए एवं लिखिए

MP Board Solutions

प्रश्न
(i) क्या एक उचित भिन्न का व्युत्क्रम भी उचित भिन्न होगी?
(ii) क्या एक विषम भिन्न का व्युत्क्रम भी एक विषम भिन्न होगा?
(ii) इसलिए हम कह सकते हैं कि
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10a
हल:
(i) नहीं, एक उचित भिन्न का व्युत्क्रम उचित भिन्न नहीं होगा।
(ii) नहीं, एक विषम भिन्न का व्युत्क्रम भी विषम भिन्न नहीं होगा।
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 10b

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 45
प्रयास कीजिए

Class 7th Bhinn Ka Guna Question प्रश्न 1.
ज्ञात कीजिए:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 11
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 11a

Kaksha 7 Prashnawali 2 Point 3 प्रश्न:
हल कीजिए:
(i) 4 ÷ 2\(\frac { 2 }{ 5 }\) = 4 ÷ \(\frac { 12 }{ 5 }\) = ?
(ii) 5 ÷ 3\(\frac { 1 }{ 3 }\) = 5 ÷ \(\frac { 10 }{ 3 }\) = ?
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 12

प्रयास कीजिए

Class 7 Maths MP Board प्रश्न 1.
ज्ञात कीजिए:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 13
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 13a

पूर्ण संख्या से भिन्न को भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 14

MP Board Solutions

मिश्रित भिन्न को पूर्ण संख्या से भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 15

एक भिन्न को दूसरी भिन्न से भाग
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 16

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 46
प्रयास कीजिए

MP Board Class 7th Maths प्रश्न 1.
ज्ञात कीजिए :
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17a
हल:
MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.3 17b

MP Board Class 7th Maths Solutions

MP Board Class 6th Special English Revision Exercises 2

MP Board Class 6th Special English Solutions Revision Exercises 2

Word Power

(a) Match the Following

MP Board Class 6th Special English Revision Exercises 2 img-1
Answer:
1. → 3
2. → 1
3. → 2
4. → 6
5. → 4
6. → 5

MP Board Solutions

Fill in the blanks with help of given words:

(friendship, favour, treasure, wasteful, righteousness, justice, praises, virtues, leader, outlaws)

  1. Saladin, the great, found one day that he had spent all his …………… in wars and ……………….. living.
  2. To the end of his days, the Jew enjoyed Saladin’s ………….. and ……………
  3. The king ruled the Kingdom with …………… and ……………
  4. He only heard …………….. of his ……………. from everyone.
  5. Robin Hood was the ………………. of a company of ………………. who lived in Sherwood Forest.

Answer:

  1. treasure, wasteful
  2. friendship, favour
  3. justice, righteousness
  4. praises, virtues
  5. leader, outlaws.

(c) Make sentences with the given words and phrases.
(inherit, willingly, countryside, virtue, passing through, ignorant, put on)

Inherit                 : He has inherited huge property from his ancestors.
Willingly              : I do my work willingly.
Countryside        : The natural scene of the countryside is very beautiful.
Virtue                  : Everyone praises him for his virtues.
Passing through : I got afraid while I was passing through a dense forest.
Ignorant              : A child is ignorant of the ways of the worlds.
Put on                 : Small children put on new clothes on festive occasions.

Comprehension

Answer these questions.

Class 6 English Revision Exercise 2 Question 1.
Why did Saladin think of a trick to get some money from the Jew?
Answer:
Saladin thought of a trick to get some money from the Jew because he was so miserly that he would never lend money willingly.

Class 6th English Revision Exercise 2 Question 2.
Why was it not possible to settle the questions of inheritance?
Answer:
It was not possible to settle the question of inheritance because all the three rings were so similar that no one could tell which was the true one.

Class 5 English Revision Exercise 2 Question 3.
What was Brahmadatta’s object in leaving the city in disguise?
Answer:
Brahmadatta left the city in disguise to search a person who could find any fault in him.

Revision Exercise 2 Class 6 Question 4.
Why were the two chariots not able to move on?
Answer:
The two chariots were not able to move on because the track was narrow and neither of the charioteers was willing to let the other pass.

Class 5 English Revision Exercise 2 MP Board Question 5.
Has anyone seen the wind?
Answer:
No, no one has seen the wind.

Class 5 English Revision Exercise 2 MP Board Solutions Question 6.
What do the leaves do when the wind is passing through?
Answer:
The leaves hang trembling when the wind passing through.

Class 6 Revision Exercise 2 Question 7.
Why was Robin Hood a terror to the rich people?
Answer:
Robin Hood used to rob the rich people to help the poor with the booty. Hence they were afraid of him (Robin Hood).

Head Make Sentence For Class 1 Question 8.
When did Robin Hood let the Sheriff go?
Answer:
Robin Hood allowed the Sheriff to leave only after snatching all the money he had.

MP Board Solutions

Grammar

(a) Combine the following pairs of sentences using “So ……. that.”
1. This book is not very difficult.
We can read it easily.

2. This bag is not very heavy.
I can lift it.

3. These pens are not very expensive.
We can buy them.

4. That bus is not very crowded.
We can travel in it comfortably.
Answer:
1. This book is not so difficult that we cannot read it easily.
2. This bag is not so heavy that I cannot lift it.
3. These pens are not so expensive that we cannot buy them
4. That bus is not so crowded that we cannot travel in it comfortably.

(b) Change the voice.

  1. They are writing the letters.
  2. He is playing cricket.
  3. He has posted the letters.
  4. She has answered the questions satisfactorily.

Answer:

  1. The letters are being written by them.
  2. Cricket is being played by him.
  3. The letters has been posted by him.
  4. The questions have been answered satisfactorily by her.

(c) Combine the following pairs of sentences using neither ………. nor.

1. Bholu didn’t play cricket yesterday.
Bholu didn’t play hockey yesterday.

2. Ritu didn’t take coffee.
Ritu didn’t take tea.
Answer:
1. Bholu played neither cricket nor hockey yesterday.
2. Ritu took neither coffee nor tea.

Let’s Write

Move Sentence For Class 2 Question 1.
Write a paragraph on “The Story of the True Ring.” What lesson does it teach?
Answer:
A rich man had a very beautiful ring. It was declared that before he died, he would pass the ring on to the worthiest of his sons. The son who received the ring, would inherit his property and be the head of the family after his death.

In this way the ring passed from father to son through several hundred years. At last it came into the hands of a man who had three sons.

The man loved his sons equally. He promised the ring to each of them. So he got two rings made. The three rings were alike. When the man was dying, he secretly gave each of the sons a ring. So after his death each claimed the inheritance. In order to prove their claim, each showed a ring. Everyone found the rings so similar that no one could tell which was the true ring. So the question of inheritance could not be solved. It is clear from the story that all the religions are true.

MP Board Solutions

Revision Test 2 Class 6 Question 2.
Describe in a paragraph how Brahmadatta ruled his kingdom.
Answer:
Brahmadatta was a just, impartial and righteous ruler. He gave fearless judgements. The people were satisfied under his rule. They settled their disputes without violence. People remained peaceful. He asked his ministers and the courtiers about his faults, but no one pointed his faults, as he had none. The people praised him highly for his virtues.

Revision Exercises Question 3.
Write a letter to your friend congratulating him/her on his/her birthday.
Answer:
F-20, Patel Nagar
Indore …….
May 15, 2005

Dear friend,
Accept my heartiest congratulation on your birthday. I had wished to join you on this occasion. But I could not came. Anyway hope you will manage everything in a grand manner. Your parents are very enthusiastic to arrange for such occasion, your brothers and sisters are there to enjoy and give you company. Once again I congratulate you and wish all my best for the occasion.

Yours
Ashish

MP Board Class 6th English Solutions

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I

In this article, we will share MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I

Columbus Discovers America-I Text Book Exercise

Word Power

(a) Fill in the blanks with the words given below:
MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I img-1

  1. The workers have gone on strike. They are …………… higher wages.
  2. My brother has failed six times in the Intermediate examination. He won’t take the examination again. He has lost all …………. of passing it.
  3. Chennai can be attacked by ships because it is on the sea ……………
  4. A crowd had collected at the scene of the accident. A policeman came up and asked one of the men in the crowd, “Can you …………… how the accident happened?”
  5. “I want to sew these buttons on this coat. Can you get me a …………… and some thread?”
  6. Marco Polo first went to China by land ………….. across Central Asia, but he returned home by sea.
  7. Drinking too much brought about the …………… of Mr. Bharti’s health.
  8. Iron is hard, but silk is ……………….
  9. You must ……………. a thing carefully before you describe it.

Answer:

  1. demanding
  2. hope
  3. coast
  4. describe
  5. needle
  6. route
  7. ruin
  8. soft
  9. observe.

MP Board Solutions

(b) Fill in the blanks with suitable words given below.

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I img-2

  1. The bite of snake is often ……………… It kills a man within a few minutes.
  2. The boy was …………… when a big dog rushed at him barking.
  3. My son left for Delhi ten days ago. I am worried because I have not heard from him …………… .
  4. “What you read in this book is a true story. It is not an …………….. one.”
  5. Our school football team this year is strong. It is sure to win the inter school cup this year. Its defeat is …………..
  6. Our school football team was playing against the Town Club. By half time the Town Club had scored two goals and our players were beginning to lose heart. “Don’t give up,” we all shouted “Fight on!” Our players ………….. again, played well, and won the match.
  7. Mr. Sampath was unwilling to become the Chief Minister. But all the M.L.A.s met him in a body and tried to …………… him to take up the office. At first they did not …………. , but they did not give up. They met him again and again. They …………. to give him every cooperation and to act according to his wishes ………….. , Mr. Sampath agreed to become the Chief Minister.
  8. The people …………… against the king only when his rule became unbearable.
  9. No one was able to recognise the policeman who was disguised as a sanyasi. He wore a ………….. beard and yellow clothes.

Answer:

  1. deadly
  2. terrified
  3. so far
  4. imaginary
  5. unthinkable
  6. took heart
  7. persuade, succeed, undertook, at last
  8. rebelled
  9. false.

(c) Match the words under A with the meaning under B.

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I img-3
Answer:
1. → (c)
2. → (f)
3. → (e)
4. → (a)
5. → (g)
6. → (h)
7. → (d)
8. → (b)

MP Board Solutions

Comprehension

(a) Say true (T) or false (F) in the given box.

  1. The Sargasso Sea was covered with seaweed.
  2. People were eager to travel with Columbus
  3. The Spanish King and Queen gave two ships to Columbus.
  4. Columbus’s first destination was the Canary Islands.
  5. The ship, Pinta lost its sail.
  6. Columbus set sail on Friday, the 3rd of August 1492.

Answer:

  1. (T)
  2. (F)
  3. (F)
  4. (T)
  5. (T)
  6. (T)

(b) Answer the following questions:

Class 6 English Chapter 12 Question Answer Question 1.
What was the plan of Vaso-de-Gama?
Answer:
The plan of Vasco-de-Gama was to find a sea-route to India with the help of the Portuguese sailors.

Class 6 English Chapter 12 Questions And Answers Question 2.
Why did Columbus think that one could find a sea route to India by sailing west from Europe?
Answer:
The scientists had told that the earth was spherical in shape. Hence, Columbus thought that he could reach India by sailing west from Europe.

Lesson 12 Class 6 English Question 3.
What stopped Columbus from sailing at once after getting the ships?
Answer:
Columbus got the ships from the king and queen of Spain. He could not start his voyage immediately, because the sailors were not ready for-it. They were afraid of various dangers of the unknown sea, especially the deadly snakes.

Class 6 English Lesson 12 Question Answer Question 4.
What kind of people made up Columbus’s crew?
Answer:
Columbus had to train the beggars and prisoners as the members of his crew.

Class 6 English Chapter 12 Question 5.
What dangers did people imagine in the seas beyond the Canary Island?
Answer:
The people had mis-beliefs like the presence of doorway to hell in and beyond the Canary Islands, sea snakes of the dangerous sea and many other things. So they did not accept the offer of Columbus.

Class 6 English Lesson 12 Question 6.
Why did Columbus stop for some time at the Canary Islands?
Answer:
One of the ships of Columbus had lost her rudder on the way to the Canary Islands. It was due to the mischief of the unwilling crew. Columbus could not sail forward. Therefore, he had to stop there for some time to get the ship repaired.

Class 6 Chapter 12 English Question 7.
“This hope turned out to be false.”
(a) What was the hope?
(b) How did the sailors come to have the hope?
Answer:
(a) Their hope was that the land was quite near to them.
(b) The sailors hoped so, because they had seen two birds flying in the air over the sea. The sight of birds made the presence of the land nearby sure.

Class 6 Lesson 12 English Question 8.
At what point during the voyage did the sailors demand that the ships should turn back home?
Answer:
When the compass started pointing to the North-West, instead of the North, the sailors were confused. Being terribly afraid they demanded that they should drop the idea of going and return to their homeland.

MP Board Solutions

Grammar in Use

1. Read these sentences.

Columbus promised his sailors rich rewards, Yet not many came forward to sail with him.
The meaning can also be expressed thus:
Though Columbus promised his sailors rich rewards, not many came forward to sail with him.
The clause beginning with though is an Adverbial clause of Concession. You can use although in place of though.

Rewrite the sentences given below using though/although and underline the Adverbial Clause.

  1. He is only thirteen years old yet he has passed the matriculation examination
  2. My watch is a cheap one; yet it has been keeping good time for the last ten years.
  3. My servant is lazy; yet I keep him because of his honesty.
  4. She was very tired; yet she kept working.
  5. I have read this poem several times; yet I can’t under stand it.
  6. Some people are very rich; yet they are not happy.
  7. There was no sign of land; yet they sailed on and on.

Answer:

  1. Though he is only thirteen years old, he has passed the matriculation examination.
  2. Though my watch is a cheap one, it has been keeping good time for the last ten years.
  3. Although my servant is lazy, I keep him because of his honesty.
  4. Although she was tired, she kept working.
  5. Though I have read this poem several times, I can’t understand it.
  6. Though some people are very rich, they are not happy.
  7. Though there was no sign of land, they sailed on and on.

Let’s Talk

1. You want to go on a trip of Pachmarhi. Talk to your partner and prepare the list of items you need for the trip.

Talk like this
Monu : Sonu, I’ll take a blanket and a torch. What are you taking?
Sonu : I’ll take a raincoat, few fruits and a knife.
Monu :
Answer:
Do yourself.

MP Board Solutions

Let’s Write

1. From the table below make six sentences referring to events in the stories you have read. Here is one of the sentences given as an example: Columbus expected to reach India by sailing west.

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I img-4
Answer:

  1. He decided to ask the Jew.
  2. Akbar agreed to send Birbal to heaven.
  3. The courtiers wished to bring about Birbal’s death.
  4. Kassim promised to give Ali ten marbles.
  5. Columbus expected to reach India by sailing west.
  6. Columbus tried to discover a new sea route to india.

Columbus Discovers America-I Word Meanings

MP Board Class 6th Special English Solutions Chapter 12 Columbus Discovers America-I img-6

MP Board Class 6th English Solutions

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1

MP Board Class 8 Maths Chapter 8 प्रश्न 1.
निम्नलिखित का अनुपात ज्ञात कीजिए –

(a) एक साइकिल की 15 km प्रति घण्टे की गति का एक स्कूटर की 30 km प्रति घण्टे की गति से।
(b) 5 m का 10 km से
(c) 50 पैसे का ₹ 5 से।

हल:
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-1

Rashiyon Ki Tulna Class 8 प्रश्न 2.
निम्नलिखित अनुपातों को प्रतिशत में परिवर्तित कीजिए –

(a) 3 : 4
(b) 2 : 3

हल:
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-2

Class 8 Maths Chapter 8 Exercise 8.1 In Hindi प्रश्न 3.
25 विद्यार्थियों में से 72% विद्यार्थी गणित में अच्छे हैं। कितने प्रतिशत विद्यार्थी गणित में अच्छे नहीं हैं?
हल:
25 में से 72% विद्यार्थी गणित में अच्छे हैं। ऐसे विद्यार्थी जो गणित में अच्छे नहीं हैं –
= (100 – 72)% = 28%
अत: 28% विद्यार्थी गणित में अच्छे नहीं हैं।

MP Board Solutions

MP Board Class 8 Maths Chapter 8 Exercise 8.1 प्रश्न 4.
एक फुटबॉल टीम ने जितने मैच खेले उनमें से 10 में जीत हासिल की। यदि उनकी जीत का प्रतिशत 40 था तो उस टीम ने कुल कितने मैच खेले?
हल:
माना कि फुटबॉल टीम ने कुल x मैच खेले।
अब, क्योंकि कुल मैचों का 40% = 10 है।
∴ x का 40% = 10
या x × \(\frac{40}{100}\) = 10
या x = \(\frac{10×100}{40}\) = 25
अतः कुल 25 मैच खेले।

MP Board Class 8 Maths Solutions English Medium प्रश्न 5.
यदि चमेली के पास अपने धन का 75% खर्च करने के बाद ₹ 600 बचे तो ज्ञात कीजिए कि उसके पास शुरू में कितने ₹ थे?
हल:
माना कि चमेली के पास शुरू में x रुपये थे।
चमेली द्वारा खर्च किए ₹ = x का 75%
खर्च करने के बाद बचे ₹
= x का (100 – 75)%
= x का 25%
∴ प्रश्नानुसार, x का 25% = ₹ 600
या x × \(\frac{25}{100}\) = 600
या x = \(\frac{600×100}{25}\) = ₹ 2400
25 अत: चमेली के पास शुरू में ₹ 2400 थे।

Rashiyo Ki Tulna Class 8 प्रश्न 6.
यदि किसी शहर में 60% व्यक्ति क्रिकेट पसन्द करते हैं, 30% फुटबॉल पसन्द करते हैं और शेष अन्य खेल पसन्द करते हैं। यदि कुल व्यक्ति 50 लाख हैं, तो प्रत्येक प्रकार के खेल को पसन्द करने वाले व्यक्तियों की यथार्थ संख्या ज्ञात कीजिए।
हल:
कुल व्यक्तियों की संख्या = 5000000
∴ क्रिकेट पसन्द करने वाले व्यक्तियों की संख्या = \(\frac{60}{100}\) x 5000000
= 3000000 = 30 लाख
फुटबॉल पसन्द करने वाले व्यक्तियों की संख्या \(\frac{30}{100}\) x 5000000
= 1500000 = 15 लाख
अन्य खेल पसन्द करने वाले व्यक्ति = 100 – (60 + 30)%
= (100 – 90)% = 10%
∴ अन्य खेल पसन्द करने वाले व्यक्तियों की संख्या = \(\frac{10}{100}\) x 5000000
= 500000 = 5 लाख

पाठ्य-पुस्तक पृष्ठ संख्या # 129 – 130

प्रयास कीजिए (क्रमांक 8.2)

MP Board Solutions

Class 8 Maths 8.1 Hindi Medium प्रश्न 1.
एक दुकान 20% बट्टा देती है। निम्नलिखित में से प्रत्येक का विक्रय मूल्य क्या होगा?

  1. ₹ 120 अंकित वाली एक पोशाक।
  2. ₹ 750 अंकित वाले एक जोड़ी जूते।
  3. ₹ 250 अंकित मूल्य वाला एक थैला।

हल:
1. अंकित मूल्य = ₹ 120, बट्टा = 20%
∴ बट्टा = ₹ 120 का 20%
= ₹ 120 x \(\frac{20}{100}\) = ₹ 24
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹ 120 – ₹ 24
= ₹ 96

2. अंकित मूल्य = ₹ 750, बट्टा = 20%
∴ बट्टा = ₹ 750 का 20%
= ₹ 750 x \(\frac{20}{100}\) = ₹ 150
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹750 – ₹ 150
= ₹ 600

3. अंकित मूल्य = ₹ 250, बट्टा = 20%
∴ बट्टा= ₹ 250 का 20%
= ₹ 250 x 20 = ₹ 50
∴ विक्रय मूल्य = अंकित मूल्य – बट्टा
= ₹ 250 – ₹ 50
= ₹ 200

MP Board Class 8 Maths Solutions प्रश्न 2.
₹15,000 अंकित मूल्य वाली एक मेज ₹ 14,400. में उपलब्ध है। बट्टा और बट्टा प्रतिशत ज्ञात कीजिए।
हल:
अंकित मूल्य = ₹ 15,000, विक्रय मूल्य = ₹ 14,400
बट्टा = अंकित मूल्य – विक्रय मूल्य
= ₹ 15000 – ₹ 14,400 = ₹600
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-3

Class 8 Maths 8.1 In Hindi प्रश्न 3.
एक अलमारी 5% बट्टे पर ₹ 5,225 में बेची जाती है। अलमारी का अंकित मूल्य ज्ञात कीजिए।
हल:
बट्टा प्रतिशत = 5%,
विक्रय मूल्य = ₹ 5,225
माना कि अलमारी का अंकित मूल्य = ₹ x है।
∴ बट्टा = ₹ x का 5%
= ₹ x × \(\frac{5}{100}\) = ₹ \(\frac{5x}{2}\)
∴ बट्टा = अंकित मूल्य – विक्रय मूल्य
₹ \(\frac{5x}{100}\) = ₹ x – ₹ 5225
5x = 100x – 5225 x 100
या 100x – 5x = 5225 x 100
या 95x = 5225 x 100
या x = \(\frac{5225×100}{95}\) = ₹ 5,500
अतः अलमारी का अंकित मूल्य ₹ 5,500 है।

MP Board Solutions

पाठ्य-पुस्तक पृष्ठ संख्या # 130

प्रतिशत में आकलन

Class 8 Rashiyon Ki Tulna प्रश्न 1.
यदि दुकान पर आपका बिल ₹ 577.80 है।

  1. इस बिल राशि का 20% बट्टे से आकलन करने का प्रयास कीजिए।
  2. ₹ 375 का 15% ज्ञात करने का प्रयास कीजिए।

हल:
1. बिल को ₹ 577.80 की निकटतम दहाई में पूर्णांकित करने पर
∴ ₹ 577.80 = ₹ 580
अतः बट्टा = ₹ 580 का 20%
= ₹ 580 x \(\frac{20}{100}\) = ₹ 116
∴ बिल की राशि का सन्निकट मान = ₹ 580 – ₹ 116
= ₹ 464

2. ₹ 375 का 15% = ₹ 375 x = \(\frac{15}{100}\)
= ₹ 56.25

पाठ्य-पुस्तक पृष्ठ संख्या # 131

प्रयास कीजिए (क्रमांक 8.3)

MP Board Class 8 Maths प्रश्न 1.
यदि लाभ की दर 5% है, तो निम्नलिखित का विक्रय मूल्य ज्ञात कीजिए –

  1. ₹ 700 की एक साइकिल जिस पर ऊपरी खर्च ₹ 50 है।
  2. ₹ 1,150 में खरीदा गया एक घास काटने का यन्त्र जिस पर ₹ 50 परिवहन व्यय के रूप में खर्च किए गए हैं।
  3. ₹ 560 में खरीदा गया एक पंखा जिस पर ₹ 40 मरम्मत के लिए खर्च किए गए हैं।

हल:
1. साइकिल का क्रय मूल्य = ₹ 700
ऊपरी खर्च = ₹50
∴ साइकिल का कुल क्रय मूल्य = ₹ 700 + ₹ 500
= ₹ 750 लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-4
= ₹ 787.50

2. घास काटने के यन्त्र का क्रय मूल्य = ₹ 1,150
परिवहन व्यय = ₹ 50
∴ अब, यन्त्र का क्रय मूल्य = ₹ 1,150 + ₹ 50
=₹ 1,200
लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-5
= ₹ 1,260

3. पंखे का क्रय मूल्य = ₹ 560
ऊपरी खर्चा = ₹40
∴ अब, यन्त्र का क्रय मूल्य = ₹ 560 + ₹ 40 = ₹ 600
लाभ = 5%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-6
= ₹ 630

पाठ्य-पुस्तक पृष्ठ संख्या # 132

प्रयास कीजिए (क्रमांक 8.4)

Class 8 Maths Chapter 8.1 In Hindi प्रश्न 1.
एक दुकानदार ने दो टेलीविजन सेट ₹ 10,000 प्रति सेट की दर से खरीदे। उसने एक को 10% हानि से और दूसरे को 10% लाभ से बेच दिया। ज्ञात कीजिए कि कुल मिलाकर, उसे इस सौदे में लाभ हुआ अथवा हानि?
हल:
पहले टेलीविजन के लिए,
क्रय मूल्य = ₹ 10,000 तथा लाभ = 10%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-7
= ₹ 110 x 100
=₹ 11,000
दूसरे टेलीविजन के लिए,
क्रय मूल्य = ₹ 10,000, हानि = 10%
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-8
= ₹ 90 x 100 = ₹ 9000
दोनों टेलीविजन का कुल क्रय मूल्य
= ₹ 10000 + ₹ 10000 = ₹ 20000
कुल विक्रय मूल्य = ₹ 11000 + ₹ 9000 = ₹ 20000
∴ विक्रय मूल्य = क्रय मूल्य है
अतः दुकानदार को न तो लाभ हुआ और न ही हानि।

पाठ्य-पुस्तक पृष्ठ संख्या # 133

प्रयास कीजिए (क्रमांक 8.5)

Maths Class 8 MP Board प्रश्न 1.
निम्नलिखित वस्तुओं को खरीदने पर यदि 5% बिक्री कर जुड़ता है तो प्रत्येक का खरीद (विक्रय) मूल्य ज्ञात कीजिए –

  1. ₹ 50 वाला एक तौलिया।
  2. साबुन की दो टिकिया जिनमें से प्रत्येक का मूल्य ₹ 35 है।
  3. ₹ 15 प्रति किलोग्राम की दर से 5 kg आटा।

हल:
1. तौलिया का मूल्य = ₹ 50
बिक्री कर दी दर = 5%
बिक्री कर = ₹ 50 x \(\frac{5}{100}\) = ₹ 2.50
∴ तौलिया का खरीद मूल्य = ₹ 50 + ₹ 2.50
= ₹ 52.50

2. साबुन की दो टिकियाओं का मूल्य = ₹ 35 x 2 = ₹ 70
बिक्री कर = ₹ 70 का 5%
= ₹ 70 x \(\frac{5}{100}\) = ₹ 3.50
∴ साबुन की दो टिकियाओं का खरीद मूल्य
= ₹ 70 + ₹ 3.50
= ₹ 73.50

3. 5 kg आटे का मूल्य = ₹ 15 x 5 = ₹ 75
बिक्री कर = ₹ 75 का 5%
= ₹ 75 x \(\frac{5}{100}\) = ₹ 3.75
∴ 5 kg आटा का खरीद मूल्य = ₹ 75 + ₹ 3.75
= ₹ 78.75

MP Board Solutions

Class 8 Maths Chapter 8 Exercise 8.1 Solutions In Hindi प्रश्न 2.
निम्नलिखित वस्तुओं के मूल्य में यदि 8% वैट सम्मिलित है तो वास्तविक मूल्य ज्ञात कीजिए –

  1. ₹ 14,500 में खरीदा गया एक टेलीविजन
  2. ₹ 180 में खरीदी गई शैम्पू की एक शीशी।

हल:
1. माना कि टेलीविजन का वास्तविक मूल्य = ₹ x है।
8% की दर से ₹ x पर वैट = ₹ \(\frac{8}{100}\) × x
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-9
= ₹ 13,425.93
अतः टेलीविजन का वास्तविक मूल्य = ₹ 13,425.93

2. माना कि शैम्पू की शीशी का वास्तविक मूल्य = ₹x है।
8% की दर से ₹ x पर वैट = ₹x × \(\frac{8}{100}\) = ₹ \(\frac{8x}{100}\)
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-10
= ₹ 166.67
अतः शैम्पू की शीशी का वास्तविक मूल्य = ₹ 166.67

सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 8.1)

Math Class 8 MP Board प्रश्न 1.
किसी संख्या को दुगुना करने पर उस संख्या में 100% वृद्धि होनी है। यदि हम उस संख्या को आधा कर दें तो कितना प्रतिशत ह्रास होगा?
हल:
माना कि कोई संख्या x है, तब इसका आधा = \(\frac{1}{2}\)x
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-11
= 50%

MP Board Class 8 Maths प्रश्न 2.
₹ 2,400 की तुलना में ₹ 2,000 कितना प्रतिशत कम है ? क्या यह प्रतिशत उतना ही है, जितना ₹ 2,000 की तुलना में ₹ 2,400 अधिक है?
MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.1 img-12
= 20%
नहीं, दोनों प्रतिशत समान नहीं हैं।

MP Board Class 8th Maths Solutions

MP Board Class 7th General English Solutions Chapter 13 One Way Ticket-II

MP Board Class 7th General English Solutions Chapter 13 One Way Ticket-II

One Way Ticket-II Textual Exercises

Read and Learn (पढ़ो और याद करो) :
Students should do themselves.
(छात्र स्वयं करें।)

Comprehension (बोध प्रश्न) :

Answer the questions given below :
(निम्न प्रश्नों के उत्तर दीजिए)

One Way Ticket 2 Class 7 Question 1.
What did the.children like most when they reached the fair?
(व्हॉट डिड द चिल्ड्रन लाइक मोस्ट व्हेन दे रीच्ड् द फेयर?)
बच्चे जब मेले में पहुँचे तो उन्हें सबसे ज्यादा क्या पसंद आया?
Answer:
Children liked the lights in the fair, the most.
(चिल्ड्रन लाइक्ड् द लाईट्स् इन द फेयर द मोस्ट।) बच्चों को मेले में रोशनी सबसे ज्यादा पसंद आई।

Class 7 English Chapter 13 One Way Ticket Question 2.
Why did the children decide to stay together?
(व्हॉइ डिड द चिल्ड्रन डिसाइड टू स्टे दुगैदर?) बच्चे एक साथ रहने का फैसला क्यों करते हैं?
Answer:
The children decided to stay together so that they are not lost in the crowd.
(द चिल्ड्रन डिसाइडेड टू स्टे टूगेदर सो दैट दे आर नॉट लॉस्ट इन द क्राउड।)
बच्चे एक साथ रहने का फैसला करते हैं जिससे वे भीड़ में खो न जाएँ।

MP Board Class 7 English Chapter 13 Question 3.
Why do you think jalebies and bhajias are filling ?
(व्हाइ डू यू थिंक जलेबीज़ एण्ड भजीयाज़ आर फिलिंग?)
तुम्हें क्यों लगता है कि जलेबी और भजिया पेट भरते हैं?
Answer:
I think jalebies and bhajias are filling because they are heavy and take long to digest.
(आइ थिंक जलेबीज एण्ड भजियाज़ फिलिंग बिकॉज दे आर हैवी एण्ड टेक लॉन्ग टू डाइजेस्ट।)
मैं सोचता हूँ कि जलेबी और भजिया से पेट भर जाता है क्योंकि ये भारी होते हैं और इनके पाचन में समय लगता है।

Class 7 English Chapter 13 MP Board Question 4.
Why do you like a caterpillar train ride?
(व्हाइ डू यू लाइक अ कैटरपिलर ट्रेन राइड?)
तुम्हें एक इल्ली रेलगाड़ी वाली सवारी क्यों पसंद है?
Answer:
I like the caterpillar train ride because it is a joy ride and it takes us round the fair too.
(आइ लाइक द कैटरपिलर ट्रेन राईड बिकॉज़ इट इज़ अ जॉय राईड एण्ड इट टेक्स अस राउण्ड द फेयर टू।)
मुझे इल्ली रेलगाड़ी वाली सवारी पसंद है क्योंकि यह एक मजेदार सवारी है और यह हमें पूरा मेला घुमा देती है।

One Way Ticket 1 Class 7 Question 5.
What kind of seats does a merry-goround have ?
(व्हॉट काइण्ड ऑफ सीट्स् डज़ अ मैरी-गो-राउण्ड हैव?)
मैरी-गो-राउण्ड की सीटें किस प्रकार की हैं?
Answer:
The merry-go-round has decorated wooden horses as seats.
(द मैरी-गो-राउण्ड हैज़ डैकोरेटेड वुडन हॉर्सेज़ एज़ सीट्स्।)
मैरी-गो-राउण्ड में सजाये हुए घोड़ों वाली सीटें हैं।

Lesson 13 One Way Ticket Question 6.
Why did ‘Lambu’ want to go on a ‘merry go-round’?
(व्हाइ डिड’लम्बू’वॉन्ट टू गो ऑन अ’मैरी-गो-राउण्ड’?)
लम्बू मैरी-गो-राउण्ड पर क्यों जाना चाहता था?
Answer:
Lambu wanted to go on a merry-go-round as on it they could ride a horse too.
(लम्बू वॉन्टेड् टू गो ऑन अ मैरी-गो-राउण्ड एज़ ऑन इट दे कुड राइड अ हॉर्स टू।)
लम्बू मैरी-गो-राउण्ड पर जाना चाहता था क्योंकि उस पर वे घुड़सवारी भी कर सकते हैं।

Class 7 English Chapter 13 Question Answer Question 7.
Why is the story called “One way Ticket” ?
(व्हाइ इज़ द स्टोरी कॉल्ड ‘वन वे टिकट’?)
कहानी को ‘एकमार्गी टिकट’ के नाम से क्यों पुकारा गया
Answer:
All the five friends collected their money they had and then only they were able to buy a one way ticket for the return journey.
(ऑल द फाइव फ्रेन्ड्स कलेक्टेड देअर मनी दे हैड एण्ड दैन ओनली दे वर एबल टू बाय अ वन वे टिकट फॉर द रिटर्न जी)
सभी पाँच मित्रों ने, जो भी उन पर धन था उसे इकट्ठा किया, तब वे एकमार्गी टिकट अपनी लौटती यात्रा के लिए खरीद सके।

MP Board Solutions

Word Power (शब्द-सामर्थ्य):

(A) Use the phrases given below in sentences given after them :
(निम्नलिखित वाक्यांशों को उनके बाद में दिये वाक्यों में प्रयोग कीजिए।)
trace (anyone, anything), opt for, get lost, find out, joy rides, get dizzy, pool money, go round.

  1. I have put my pen somewhere. I am not able to ………………
  2. What will you ……………… after matriculation Arts, Science or Commerce ?
  3. The child held his mother’s hand tightly so that he may not ………………
  4. ……………… the correct solution of this sum.
  5. There were several ……………… at the fair.
  6. If you stand under the hot sun you will. ………………
  7. The principal like to ……………… the school to see all the classes.

Answer:

  1. trace it
  2. opt for
  3. get lost
  4. Find out
  5. joy rides
  6. get dizzy
  7. go round.

(B) Given below are some words which have similar sounds but different meanings, match these words with their meanings in each pair:
(नीचे कुछ शब्द दिए गए हैं जिनकी ध्वनि समान है लेकिन उनके अर्थ भिन्न हैं। इन शब्दों का उनके अर्थों से मिलान कर जोड़े बनाइए।)

  1. fare – (a) opposite of ‘there’
  2. here – (b) receiving sounds by ear
  3. fair – (c) money paid for a ticket
  4. hear – (d) gathering of shops etc. for public entertainment.

Answer:

  1. → (c)
  2. → (a)
  3. → (d)
  4. → (b)

MP Board Solutions

Grammar in Use (व्याकरण प्रयोग):

(A) Combine the following sentences using who, what, when, which, or where.
(निम्न वाक्यों को Who,what,when,which या where का प्रयोग कर जोड़िए)

Class 7 English Lesson 13 Question Answer Question 1.
I entered the room. Sheila was sleeping. (when)
Answer:
I entered the room when Sheila was sleeping.

Class 7 English Chapter 12 One Way Ticket Question 2.
Mohan has gone. I do not know. (where)
Answer:
I do not know where Mohan has gone.

Class 7 English Lesson 13 Question 3.
The doctor came. The patient had died. (when)
Answer:
The doctor came when the patient had died.

Class 7th English Chapter 13 Question 4.
Mother kept the milk on the shelf. The cat could not reach it. (where)
Answer:
Mother kept the milk on the shelf where the cat could not reach.

Lesson 13 English Class 7 Question 5.
You want something. I do not know. (what)
Answer:
I do not know what you want.

Class 7th English Chapter 13 Questions Question 6.
The police caught the thief. He had stolen the necklace. (who)
Answer:
the police caught the thief who had stolen the necklace.

(B) Rewrite the following sentence using ‘tool:
(‘too’ का प्रयोग कर निम्नलिखित वाक्यों को लिखिए)

Class 7 Lesson 13 English Question 1.
This exercise is so difficult that I can’t do it.
Answer:
This exercise is too difficult for me to do.

Class 7 Chapter 13 English Question 2.
It is so hot that I can’t walk outside.
Answer:
It is too hot for me to walk outside.

Class 7 Lesson 13 English Question Answer Question 3.
It is so dark that one can’t see.
Answer:
It is too dark for to see.

Lesson 13 Class 7 English Question 4.
She is so young that she can’t do such hard work.
Answer:
She is too young to do such hard work.

One-Way Ticket Meaning In Hindi Question 5.
The bed is so hard that no one can sleep on it.
Answer:
The bed is too hard to sleep on it.

(C) Rewrite the following sentences using fullstops (.), capital letters, question marks (?), commas (,), inverted commas (” “) and exclamation marks (!) wherever necesary:
(निम्न वाक्यों को फुल स्टॉप (.), केपीटल अक्षर, प्रश्नवाचक चिन्ह्न (?), कोमा (,), उद्धरण चिह्न (” “) और विस्मयादि बोध (!) चिन्ह्न लगाइए।)
Answer:
Once Bishni, Kishto, Gittu and Lambu went to see a Bal Mela in the nearby high school. There, they saw many science models made by the students. Bishni said, “What is this model about ?” Kishto explained, “This tells you about how water is pumped up with the help of electric motor.” Bishni, “I can see water coming out of the tap.” Lambu said, “Such models are called working models.”

MP Board Solutions

Let’s Talk (आओ बात करें) :

Talk to your friends about various sweets/ dishes they like to eat and the swings/rides they enjoy in a fair and write them in the table. You can choose from these words.
swings and rides : (merry-go-round, giant wheel, horse ride, see-saw, train ride, car-rides, motor cycle rides)
sweets and dishes : (jalebi, bhajia, samosa, barfi, kulfi, papad, dosa, chat, pani-puri).
MP Board Class 7th General English Chapter 13 One Way Ticket-II 1
Ask four of your friends these questions to complete your table. Your friends will make the same chart and ask some questions.

  1. What is your name?
  2. Which sweets and dishes do you like most? (one sweet/one dish)
  3. Which swings do you prefer?
  4. What rides did you take in the fair you visited last?

अपने मित्रों से विभिन्न प्रकार के मिठाइयों/पकवानों के बारे में बात करो जो उन्हें पसन्द हों और वे झूले जो वे मेले में पसन्द करते हों, उसकी एक सारिणी बनाओ। इन प्रश्नों को अपने चार मित्रों से पूछो और सारिणी पूरी करो। तुम्हारे मित्र भी ऐसी ही एक सारिणी बनाकर तुमसे वे ही प्रश्न पूँछेगे।

  1. आपका नाम क्या है?
  2. आप कौन-सी मिठाइयाँ और पकवान अधिक पसन्द करते हो?
  3. आप किस झूले को प्राथमिकता देते हो? 4. मेले में अंतिम बार आपने कौन-सी सवारी की?

Let’s Write (आओ लिखें):

(A) Read the given telephonic conversation and then write in the following paragraph what Sonam will have to do the following morning:
(नीचे दिए टेलीफोन वार्तालाप को पढ़िए और सोनम को अगली सुबह क्या करना है, इस पर एक अनुच्छेद लिखिए।)
Answer:
Sonam will have to go to receive Montu the next morning at 9.30 a.m. at Habibganj station and take him to Professor’s Colony. He is having a computer test at NIIT, Arera Colony the next day so she will have to leave his luggage at home, give him some breakfast and take him for the test. She will have to pick him up at 5 p.m. from there when he finishes the test and put him on the train for Delhi at 7.30 p.m. (B) Fill in the blank spaces in the paragraph below using the words given in the box below:
(कोष्ठक में दिये गए शब्दों का प्रयोग कर अनुच्छेद के रिक्त स्थानों करे भरो।
(receive, bring, leave, computer centre, Habibganj, will, then, I’ll, luggage, Bhopal, test, 5 P.M., Professor’s Colony, have)
Answer:
I’ll receive, Montu at Habibganj railway station. Then I’ll bring Montu to our house at Professor’s Colony. We’ll then have breakfast and leave, for NIIT computer centre. Montu will then take his test. He will be free by 5 p.m. I’ll then take him home. We will pick up his luggage and then go to the main station at Bhopal.

MP Board Solutions

Let’s Do It (आओ इसे करें) :

Make a list of the things that you enjoyed in a fair. Draw a picture of the fair and write about your picture.
(मेले में आपने जिन वस्तुओं से आनन्द प्राप्त किया उनकी एक सूची बनाओ। मेले का एक चित्र बनाओ और अपने चित्र के बारे में लिखो।
Answer:
Students can write themselves according to their individual choice what they enjoy in a fair. They can then draw the picture of the fair and write about it.

Pronunciation & Translation

Gittoo: Oh! How beautiful! Look at all those lights.
(गिटू : ओह! हाउ ब्यूटीफुल! लुक ऐट ऑल दोज लाइट्स.)
अनुवाद-गिटू : ओह! कितना सुन्दर! उन सभी रोशनियों को देखो।

Krishan : Yes, yes, let us go and see all the rides first and decide which ones we really want to go to.
(कृष्ण : येस, येस, लेट अस गो एण्ड सी ऑल द राइड्स फर्स्ट एण्ड डिसाइड विच वन्स वी रीअली वान्ट टू गो टू.)
अनुवाद-कृष्ण : हाँ, हाँ, आओ चलें, और सभी सवारियों को पहले देखें और तय करें कि हम किस की ओर वास्तव में जाना चाहते हैं।

Bishni: Yes Krishan, you are absolutely right. We would like to see what rides are available and which.ones are the best. Also we should all be walking together so that we are not lost in this crowd.
(बिशनी : येस कृष्ण, यू आर एब्सोल्यूटली राइट. वी वुड लाइक टू सी व्हॉट राइड्स आर अवेलेबल एण्ड विच वन्स आर द बेस्ट. ऑल्सो वी शुड ऑल बी वॉकिंग टूगेदर सो दैट वी आर नॉट लॉस्ट इन दिस क्राउड़)
अनुवाद-बिशनी : हाँ, कृष्ण, तुम बिल्कुल सही कह रहे हो। हम देखना चाहेंगे कि कौन-सी सवारियाँ उपलब्ध हैं और | कौन-सी सबसे अच्छी हैं। हम सबको इकठे घूमना चाहिए ताकि हम भीड़ में न खो जाएँ।

Kishto : That’s true, it’ll be difficult to trace anyone who gets lost. Can we eat something before we go for the rides.
(किश्तो : दैट्स ट्र, इट ‘विल बी डिफीकल्ट टू ट्रेस एनीवन हू गैट्स लॉस्ट. कैन वी ईट समथिंग बिफोर वी गो फॉर द राइड्स.)
अनुवाद-किश्तो : यह सच है, जो भी खो जायेगा उसे ढूँढ़ पाना कठिन होगा। क्या सवारी पर जाने से पहले हम कुछ खा सकते हैं?

Krishan: Alright, let’s eat something that is really filling.
(कृष्ण : ऑलराइट, लेट्स ईट समथिंग दैट इज रीअली फिलिंग.)
अनुवाद-कृष्ण : ठीक है, आओ कुछ खायें जिससे वास्तव में पेट भर जाये।

Lambu : Well, we had our food on the way. I think we can do with some snacks. Shall we opt for jalebies, Bhajias and Kulfis?
(लम्बू : वैल, वी हैड अवर फूड ऑन द वे. आइ थिंक वी कैन डू विद सम स्नेक्स. शैल वी ऑप्ट फॉर जलेबीज, भजीयाज़ एण्ड कुल्फीस?)
अनुवाद-लम्बू : अच्छा, हमें रास्ते में खाना मिल जाता। मेरा सोचना है, हम कुछ स्नेक्स ले सकते है। क्या हमें जलेबी भाजी और कुल्फी लेना चाहिए?

Gittoo : Oh! That’s nice.
(गिटू : ओह! दैट्स नाइस.)
अनुवाद-गिट्ट : ओह! यह अच्छा है।

Kishto : Yes, that will do. (They all eat.)
(किश्तो : येस, दैट विल डू. (दे ऑल ईट.))
अनुवाद-किश्तो : हाँ, यह चलेगा। (वे सब खाते हैं।)

Bishni : Good; that was quite filling.
(बिशनी :- गुड; दैट वॉज क्वाइट फिलिंग.)
अनुवाद-बिशनी : वाह; पेट भर गया।

Krishan: Now, let’s go round the fair. Look at these train rides. That train looks like a caterpillar. It’s a joy ride and it takes us round the fair too.
(कृषण : नाउ, लेट्स गो राउण्ड द फेअर. लुक ऐट दीज ट्रेन राइड्स. दैट ट्रेन लुक्स लाइक अ कैटरपिलर. इट्स अ जॉय राइड एण्ड इट टेक्स अस राउन्ड द फेअर टू.)
अनुवाद-कृष्ण : अब, मेले में घूमें। इन रेलगाड़ियों की सवारियों को देखें। वह रेलगाड़ी देखने में कैटरपिलर की तरह दिखाई देती है। यह एक मजेदार सवारी है और यह हमें मेला भी घुमाती है।

Gittoo: Let’s us go for that. All together certainly
(गिटू : लेट्स अस गो फॉर दैट. ऑल टूगेदर सर्टेनली.)
अनुवाद-गिटू : आओ उसके लिए चलते हैं। सब एक साथ-अवश्य।

Lambu: I want to go on the ‘merry-go-round’. We can ride a horse too because all the swings on the merry-go-round have decorated horses as seats.
(लम्बू : आइ वान्ट टू गो ऑन द ‘मैरी- गो-राउण्ड.’ वी कैन राइड अ होर्स टू बिकाज ऑल द स्विंग्स ऑन द मैरी-गो-राउण्ड हैव डेकोरेटेड होर्सस एज सीट्स.)
अनुवाद-लम्बू : मैं घूमने वाले झूले में जाना चाहता हूँ। हम एक घोड़े की सवारी भी कर सकते हैं क्योंकि मैरी-गो-राउण्ड (झूले) पर सीटों जैसे घोड़ों को सजाया गया है।

Kishto: Yes, I know how to ride a horse.
(किश्तो : येस, आइ नो हाउ टू राइड अ हॉर्स.)
अनुवाद-किश्तो : हाँ मैं जानता हूँ कैसे घोड़े की सवारी की जाती है।

Bishni : But this is a wooden horse.
(बिशनी : बट दिस इज अ वुडन हॉर्स.)
अनुवाद-बिशनी : लेकिन, यह एक लकड़ी का घोड़ा है।

Gittoo : Oh! Really ?
(गिटू : ओ! रीयली?)
अनुवाद-गिटू : अरे! वास्तव में?

Bishni: It is fast indeed and gets you dizzy soon. It goes too fast and sometimes you are a little scared too.
(बिशनी : इट इज फास्ट इन्डीड एण्ड गैट्स यू डिज़ी सून. इट गोज टू फास्ट एण्ड समटाइम्स यू आर अ लिटल स्केअर्ड टू.)
अनुवाद-बिशनी : यह वास्तव में तेज है और जल्दी ही तुम्हें चक्कर आ जायेंगे। यह बहुत तेज चलता है और कभी-कभी तुम्हें थोड़ा डर भी लगेगा।

All: Let’s us ride the horses.
(ऑल : लेट्स अस राइट द हॉर्सेस.)
अनुवाद-सभी : आओ घोड़ों पर चढ़ें।

Bishni: That was fun. Now we can tell our friends what we did at the fair.
(बिशनी : दैट वाज फन. नाउ वी कैन टैल अवर फ्रेण्ड्स व्हॉट वी डिड ऐट द फेअर.)
अनुवाद-बिशनी : यह मजेदार था। अब हम अपने मित्रों को बता सकते है कि हमने मेले में क्या किया।

Gittoo : It’s almost 11.o’clock in the night. Let’s go and rest somewhere and we can leave by the first bus in the morning.
(गिटू : इट्स ऑलमोस्ट इलेवन ओ क्लॉक इन द नाइट. लेट्स गो एण्ड रेस्ट समव्हेअर एण्ड वी कैन लीव बाइ द फर्स्ट बस इन द मॉर्निग.)
अनुवाद-गिटू : अब रात के लगभग ग्यारह बजे हैं। आओ कहीं चलकर आराम करें और हम कल प्रात: पहली बस से जा सकते हैं।

Kishto: Oh my God! We don’t have enough money to buy tickets for the return journey.
(किश्तो : आह माइ गॉड ! वी डोन्ट हैव एनफ मनी टू बाय टिकट्स फॉर द रिटर्न जर्नी.)
अनुवाद-किश्तो : ओ मेरे भगवान! हमारे पास वापसी यात्रा के लिये टिकट खरीदने को पर्याप्त धन नहीं है।

Krishan: Let’s pool our money and see what we can do. There are five of us. We need Rs. 3/- for each ticket. That works out to Rs. 15/-
(कृष्ण : लेट्स पूल अवर मनी एण्ड सी व्हॉट वी कैन डू. देअर आर फाइव ऑफ अस. वी नीड रूपीज 3/- फॉर ईच टिकट. दैट वर्क्स आउट ट रुपीज 15/-.)
अनुवाद-कृष्ण : आओ अपना पैसा इकट्ठा कर लें और देखें हम क्या कर सकते हैं। हम पाँच हैं। हमें प्रत्येक टिकट के लिए तीन रुपये की जरूरत है। जो 15 रुपये होते हैं।

Gittoo: Here is all the money we have. It’s just five rupees. What shall we do now?
(गिटू. : हीअर इज ऑल द मनी वी हैव. इट ‘स जस्ट फाइव रुपीज. व्हॉट शैल वी डू नाउ?)
अनुवाद-गिट्ट : हमारे पास कुल पैसे यहीं हैं। यह केवल पाँच रुपये हैं। अब हम क्या करेंगे?

Lambu: I am very tired. I can’t walk back.
(लम्बू : आइ एम वैरी टायर्ड. आइ कॉन्ट वॉक बैक.)
अनुवाद-लम्बू-मैं बहुत थका हुआ हूँ। मैं वापस पैदल नहीं जा सकता हूँ।

Bishni : I don’t think my mother will like it if I reach home very late.
(बिशनी : आइ डोन्ट थिंक माइ मदर विल लाइक इट इफ आइ रीच होम वेरी लेट.)
अनुवाद-बिशनी : मुझे नहीं लगता है कि मेरी माँ यह पसंद करेगी यदि मैं घर बहुत देर से पहुँचूँ।

Gittoo : I don’t think it was worth spending time here if we have to walk home now. We shouldn’t have come without enough money in the first place.
(गिटू : आइ डोन्ट थिंक इट वॉज वर्थ स्पेन्डिंग टाइम हीअर इफ वी हैव टू वॉक होम नाउ. वी शुडन्ट हैव कम विदआउट एनफ मनी इन द फर्स्ट प्लेस.)
अनुवाद-गिट्टू : मुझे नहीं लगता कि यहाँ समय बेकार किया जाए यदि हमें अब घर पैदल जाना है। हमें पहले से ही पर्याप्त पैसे के बिना नहीं आना चाहिए था।

Kishto: Bapu must be looking for me.
(किश्तो : बापू मस्ट बी लुकिंग फॉर मी.)
अनुवाद-किश्तो : बापू मेरी राह देख रहे होगे।

Krishan: Well don’t worry. I have kept some money separately and have saved Rs. 10/-. That should be enough. I just found out that there is a bus. Let’s leave now and we’ll reach home by 11.30.
(कृष्ण : वैल, डोन्ट वरी. आइ हैव केप्ट सम मनी सेपरेटली एण्ड हैव सेव्ड रुपीज 10/-. दैट शुड बी एनफ. आइ जस्ट फाउण्ड आउट दैट देअर इज अ बस. लेट्स लीव नाउ एण्ड वी विल रीच होम ब्राइ 11:30.)
अनुवाद-कृष्ण : ठीक है, चिन्ता मत करो। मैंने कुछ पैसे अलग रख लिये हैं और 10 रुपये बचा लिये हैं। यह काफी होंगे। मैंने अभी-देखा कि यहाँ एक बस है। आओ। अब चलें और हम 11:30 बजे तक घर पहुँच जायेंगे।

All together: That’s wonderful. That’s why we call your Krishan, the wise. Because of you, we are able to buy our tickets for the return journey. Thank you for saving the money for our one-way tickets.
(ऑल टुगेदर : दैट्स वन्डरफुल. दैट्स व्हाइ वी कॉल योअर कृष्ण, द वाइज. बिकॉज ऑफ यू, वी ऑर एबल टू बाय अवर टिकट्स फॉर द रिटर्न जर्नी. थेंक यू फॉर सेविंग द मनी फॉर अवर वन-वे टिकट्स.)
अनुवाद–सभी एक साथ : आश्चर्यजनक है। इसीलिए हम तुम्हें कृष्ण कहते हैं यानि बुद्धिमान। तुम्हारे ही कारण हम वापिसी यात्रा के टिकिट खरीद सके हैं। हमारी एक-मार्गी टिकट के लिये पैसे बचाने के लिए तुम्हें-धन्यवाद।

MP Board Solutions

One Way Ticket-II Word Meanings

Absolutely (एब्सोल्यूटलि)-बिल्कुल; Available (अवेलेबल)-उपलब्ध; Caterpillar (कैटरपिलर)-इल्ली; झिनगा; Decorated (डैकोरेटेड)-सजा हुआ; Really (रीयली)-वास्तव में; Dizzy (डिज़ी)-चक्कर आना; Enough (इनफ)- पर्याप्त; Filing (फिलिंग)- भरना; Horse ride (हॉर्स राईड)-घुड़सवारी; Indeed (इन्डीड)वास्तव, असल में; Opt (ऑप्ट)-चुनना, पसन्द; Pool (पूल)-एकत्र करना, जमा करना; Ride (राइड)-चढ़ना,सवारी; Saved (सेव्ड्)-बचाना; Scared (स्केअर्ड)डरा हुआ, भयभीत; Separately (सेपरेट्ली)-अलग से; Snacks (स्नैक्स)-हल्का नाश्ता; Certainly (सर्टेनली)अवश्य; Swing (स्विंग)-झूला; Spend (स्पेंड)-खर्च करना; Wonderful (वण्डरफुल)-आश्चर्यजनक; Trace (ट्रेस)-खोजना; Worry (वरी)-चिन्ता; Worth (वर्थ)लायक, योग्य।

MP Board Class 7th English Solutions

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms

Respiration in Organisms Intex Questions

Question 1.
Bhoojho wants to know if cockroaches, snails, fish, earthworms, ants and mosquitoes also have lungs?
Answer:
Yes.

Table 10.1 Changes In Breathing Rate Under Different Conditions Question 2.
Boojho has seen in television programmes that whales and dolphins often come up to the water surface? They even release a fountain of water sometimes while moving upwards? Why do they do so?
Answer:
Whales and dolphins take in air during inhalation. They exhale out the air on the surface. The water vapour condenses and we see the condensed water vapour as the fountain.

MP Board Solutions

Question 3.
Paheli wants to know whether roots, which are under ground also take in oxygen? If so, how?
Answer:
Yes. Roots take up air from the air spaces preseht between the soil particles.

Activities

Activity – 1
If you try you can count your rate of breathing. Breathe in and out normally. Find out how many times you breathe in and breathe out in a minute? Did you inhale the same number of times as you exhaled? Now count your breathing rate (number of breaths/minute) after brisk walk and after running. Record your breathing rate as soon as you finish and also after complete rest. Tabulate your findings and compare your breathing rates under different conditions with those of your classmates.

Table:
Changes in breathing rate under different conditions:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 1

Activity – 2
Figure shows the various activities carried out by a person during a normal day. Can you say in which activity, the rate of breathing will be the slowest and in which it will be the fastest? Assign numbers to the pictures in the order of increasing rate of breathing according to your experience.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 2

Activity – 3
Take a deep breath. Measure the size of the chest with a measuring tape and record your observations in Table. Measure the size of the chest again when expanded and indicate which classmate shows the maximum expansion of the chest.
Answer:
Effect of breathing on the chest size of some classmates:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 3 - Copy

Respiration in Organisms Text Book Exercise

Class 7 Science Chapter 10 Table 10.1 Question 1.
Why does an athlete breathe faster and deeper than usual after finishing the race?
Answer:
During the race, the athlete has to run very fast. The demand for energy at that time increases, which increase the demand for more supply of oxygen. Thus, athlete has to breathe faster and deep to inhale more oxygen.

Question 2.
List the similarities and differences between aerobic and anaerobic respiration?
Answer:
Similarities:
Both aerobic and anaerobic respiration produce energy and give out carbon dioxide.

Differences:
Aerobic respiration require oxygen while anaerobic respiration does not require oxygen. In aerobic respiration large amount of energy is released while in anaerobic respiration small amount of energy is released.

Question 3.
Why do we often sneeze when we inhale a lot of dust laden air?
Answer:
We sneeze to get rid of the unwanted particles like dust from air body. It allows only clean and dust free air to enter our body.

MP Board Solutions

MP Board Class 7 Science Chapter 10 Question 4.
Take three test – tubes. Fill \(\frac{1}{2}\)th of each with water. Label them A, B and C. Keep a snail in test – tube A, a water plant in test – tube B and in C, keep snail and plant both. Which test – tube would have the highest concentration of CO2?
Answer:
Test – tube A.

Question 5.
Tick the correct answer:

Question (a)
In cockroaches, air enters the body through?
(a) lungs
(b) gills
(c) spiracles
(d) skin.
Answer:
(c) spiracles

Class 7 Science Chapter 10 Question (b)
During heavy exercise, we get cramps in the legs due to the accumulation of?
(a) carbon dioxide
(b) lactic acid
(c) alcohol
(d) water.
Answer:
(b) lactic acid

Question (c)
Normal range of breathing rate per minute in an average adult person at rest is?
(a) 9 – 12
(b) 15 – 18
(c) 21 – 24
(d) 30 – 33
Answer:
(b) 15 – 18

Question (d)
During exhalation, the ribs?
(a) move outwards
(b) move downwards
(c) move upwards
(d) do not move
Answer:
(b) move downwards

MP Board Class 7th Science Chapter 10 Question 6.
Match the items in Column I with those in Column II:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 4
Answer:

(a) – (iii)
(b) – (iv)
(c) – (i)
(d) – (v)
(e) – (ii)
(f) – (vi)

Table 10.1 Class 7 Science Question 7.
Mark if the statement is true and if it is false:

  1. During heavy exercise the breathing rate of a person slows down.
  2. Plants carry out photosynthesis only during the day and respiration only at night.
  3. Frogs breathe through their skins as well as their lungs.
  4. The fishes have lungs for respiration.
  5. The size of the chest cavity increases during inhalation.

Answer:

  1. False
  2. False
  3. True
  4. False
  5. True

MP Board Solutions

Question 8.
Given below is a square of letters in which are hidden different words related to respiration in organisms. These words may be present in any direction – upwards, downwards, or along the diagonals. Find the words for your respiratory system. Clues about those words are given below the square?

  1. The air tubes of insects
  2. Skeletal structures surrounding chest cavity
  3. Muscular floor of chest cavity
  4. Tiny pores on the surface of leaf
  5. Small openings on the sides of the body of an insect
  6. The respiratory organs of human beings
  7. The openings through which we inhale
  8. An anaerobic organism
  9. An organism with tracheal system

Answer:

  1. Trachea
  2. Rib
  3. Diaphragm
  4. Stomats
  5. Spiracles
  6. Lung
  7. Strils
  8. Yeast
  9. Ant.

MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 5

Class 7th Science Chapter 10 Question 10.
The mountaineers carry oxygen with them because:
(a) At an altitude of more than 5 km there is no air.
(b) The amount of air available to a person is less than that available on the ground.
(c) The temperature of air is higher than that on the ground.
(d) The pressure of air is higher than that on the ground.
Answer:
(b) The amount of air available to a person is less than that available on the ground.

Extended Learning – Activities and Projects

Question 1.
Observe fish in an aquarium. You will find flap like structures on both sides of their heads. These are flaps which cover the gills. These flaps open and close alternately. On the basis of these observations, explain the process of respiration in the fish?
Answer:
Do with the help of your subject teacher.

Question 2.
Visit a local doctor. Learn about the harmful effects of smoking. You can also collect material on this topic from other sources. You can seek help of your teacher or parents. Find out the percentage of people of your area who smoke. If you have a smoker in your family, confront him with the material that you have collected?
Answer:
Do yourself.

Class 7 Science Table 10.1 Question 3.
Visit a doctor. Find out about artificial respiration? Ask the doctor:
(a) When does a person need artificial respiration?
(b) Does the person need to be kept on artificial respiration temporarily or permanently?
(c) From where can the person get supply of oxygen for artificial respiration?
Answer:
Do yourself.

MP Board Solutions

Question 4.
Measure the breathing rate of the members of your family and some of your friends? Investigate:
(a) If the breathing rate of children is different from that of adults?
(b) If the breathing rate of males is different from that of females?
If there is a difference in any of these cases, try to find the reason?
Answer:
Do with the help of your parents.

Respiration in Organisms Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Class 7th Science Chapter 10 Question Answer Question (a)
The life processes that provide energy are?
(a) respiration
(b) nutrition
(c) both respiration and nutrition
(d) none of these.
Answer:
(c) both respiration and nutrition

Question (b)
In …………………… respiration, there is an exchange of gases between the cells and the blood?
(a) aerobic
(b) anaerobic
(c) external
(d) internal.
Answer:
(d) internal.

Class 7 Science Ch 10 Question (c)
In the cell, the food (glucose) is broken down into carbon dioxide and water using?
(a) hydrogen
(b) nitrogen
(c) oxygen
(d) none of these.
Answer:
(c) oxygen

MP Board Solutions

Question (d)
Which of the following is not a feature of respiration?
(a) involvement of enzymes
(b) occur outside the cells
(c) release of energy
(d) is a chemical process.
Answer:
(d) is a chemical process.

Std 7 Science Lesson No 10 Question Answer Question (e)
During heavy exercise, the breathing rate in an adult can increase upto?
(a) 25 times per minute
(b) 30 times per minute
(c) 35 times per minute
(d) none of these.
Answer:
(a) 25 times per minute

Question (f)
The percentage of oxygen and carbon dioxide in inhaled air is –
(a) 21%, 0.04%
(b) 20%, 0.10%
(c) 21%, 1.0%
(d) 20%, 1.0%.
Answer:
(a) 21%, 0.04%

Science Class 7 Chapter 10 Question Answer Question (g)
The percentage of oxygen and carbondioxide in exhalted air is –
(a) 16.4%, 4.4%
(b) 21%, 1.0%
(c) 16.4%, 3.4%
(d) 16.4%, 0.04%.
Answer:
(a) 16.4%, 4.4%

Question 1.
Fill in the blanks:

  1. All living organisms require ………………………… to perform various life process.
  2. The liver and ……………………… are found near the stomach.
  3. The exhaled air has a higher percentage of carbon dioxide as compared to the ……………………………… air.
  4. Breathing is a process at organ levels, whereas respiration is a ……………………….. process.
  5. When breakdown of glucose occurs with the use of oxygen, it is called …………………….. respiration.
  6. The taking in of air rich in oxygen into the body is called …………………………
  7. The number of times a person breathes in a minute is termed as the ………………………….
  8. Lungs are present in the ……………………….. cavity.
  9. During inhalation, ribs move up and outwards and diaphragm moves ………………………..
  10. Insects have a network of air tubes called ……………………….. for gas exchange.
  11. Like all other living cells of the plants, the root cells also need oxygen to ………………………. energy.

Answer:

  1. energy
  2. pancreas
  3. inhaled
  4. cellular
  5. aerobic
  6. inhalation
  7. breathing rate
  8. chest
  9. down
  10. tracheae
  11. generate.

MP Board Solutions

Std 7 Science Chapter 10 Question Answer Question 3.
Which of the following statements are true (T) or false (F):

  1. Respiration is a type of combustion at ordinary temperature.
  2. Breathing is a process that takes place at the cellular level.
  3. Oxygen is released during the process of respiration.
  4. During respiration the plants – take CO2 and release Or
  5. Respiration involves on exchange of gases.
  6. Cellular respiration takes place in the cells of all organisms.
  7. Anaerobic respiration do not takes places in the muscle cells to fulfill the demand of energy.
  8. The giving out of air rich in carbon dioxide is known as exhalation.
  9. A breathe means one inhalation plus one exhalation.
  10. On an average, an adult human being at rest breathes in and out 15 to 18 times in a minute.
  11. A cockroach has small openings on the sides of its body.
  12. Gills are not supplied with blood vessels for exchange of gases.
  13. The end product of anaerobic respiration are carbon dioxide and water.
  14. In earthworm, the exchange of gases occurs through the moist skin.

Answer:

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

Respiration in Organisms Very Short Answer Type Questions

Question 1.
Where does cellular respiration takes place?
Answer:
Cells of organisms.

Question 2.
Write the equation for breakdown of food in anaerobic respiration?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 6 - Copy

Up Board Class 7 Science Chapter 10 Question 3.
Write the equation for breakdown of food in aerobic respiration?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms img t

Question 4.
What are anaerobes?
Answer:
There are some organisms such as yeast that can service in the absence of air. They are called anaerobes.

Question 5.
What is yeast?
Answer:
Yeast is single – celled organisms.

Respiration In Organisms Question 6.
Write the uses of yeast?
Answer:
Yeast respire anaerobically and during this process yield alcohol. So, they are used to make beer and wine.

Question 7.
Which chemical reaction takes place in internal respiration?
Answer:
Glucose + Oxygen → Carbon dioxide + Water + Energy.

Question 8.
Give an example of an oxygen respiration?
Answer:
In human beings.

Class 7 Science Chapter 10 Short Answer Question 9.
Name the parts of digestive system of humans?
Answer:
The parts of digestive systems are mouth, oesophagus, stomach, intestine and anus.

MP Board Solutions

Question 10.
Name two processes of respiration?
Answer:
Inhalation and exhalation are the two processes of respiration.

Question 11.
Name the parts of respiratory system of human?
Answer:

  1. Nostrils
  2. Trachea
  3. Lungs with alveali, and
  4. Diaphragm.

Class 7 Science Chapter 10 Question Answer Question 12.
What waste materials are produced during respiration?
Answer:
Carbon dioxide is produced as the waste material during respiration.

Question 13.
Define respiration?
Answer:
The process of breaking down of food by using oxygen, to form carbon dioxide and release energy required for various life activities, is called respiration.

Question 14.
Name the fuels used for the production of energy during respiration?
Answer:
Glucose is oxidized to give out energy.

Class 7 Chapter 10 Science Question 15.
Which organs of plants participate in respiration?
Answer:
There is no special organ in plants for breathing.

Question 16.
Write the names of organs in human respiratory system in sequence?
Answer:
Nostrils → Nasal cavity → Pharynx → Tracheae → Lungs.

Question 17.
What is importance of hairs present in the noise?
Answer:
These small hairs present in nose act as filters. These prevent dust particles and harmful germs to enter into respiratory track.

Chapter 10 Science Class 7th Question 18.
Name the organs of the body from which blood freshly enriched with oxygen goes into the heart?
Answer:
The lungs helps the blood to get freshly enrichment of oxygen.

Question 19.
What happens during breathing?
Answer:
During breathing, oxygen enriched air is inhaled which reaches lungs. Here, oxygen centers blood and unwanted water vapour and carbon dioxide are released out during breathing.

MP Board Solutions

Question 20.
What is breathing?
Answer:
The process of taking oxygen and leaving of carbon dioxide during respiration is called breathing.

Question 21.
Write the name of gases which are involved in breathing?
Answer:
Carbon dioxide and oxygen.

Class 7 Science Lesson 10 Question Answer Question 22.
How will you prove that we exhale CO2 gas during respiration?
Answer:
Pass the exhaled air given out by us into lime water. It will turn milky in colour. We know that CO2 gas turn lime water milky This confirms that we exhale CO2 gas in respiration.

Question 23.
What do we exhale?
Answer:
We exhale air rich in carbon dioxide.

Question 24.
Do we exhale only carbon dioxide or a mixture of gases along with it?
Answer:
We exhale a mixture of gases along with carbon dioxide.

Question 25.
How do ribs and diaphragm move during inhalation?
Answer:
During inhalation, ribs move up and outwards and diaphragm moves down.

MP Board Solutions

Question 26.
Which is the respiratory organ for earthworm?
Answer:
Skin.

Question 27.
Can we survive in water?
Answer:
No.

Question 28.
How do fish breathe under water?
Answer:
Gills in fish help them to use oxygen dissolved in water. Gills are projections of the skin. Gills are well supplied with blood vessels for exchange of gases.

Question 29.
What is the function of gills?
Answer:
The fishes and other aquatic animals respire through gills or similar structure.

Question 30.
Can you guess what would happen if a potted plant is over watered?
Answer:
The roots will not get air to respire, so the roots will die and hence the whole plant will also die.

Respiration in Organisms Short Answer Type Questions

Question 1.
What is respiration?
Answer:
All the living organisms perform a number of vital activities. The energy is obtained by the oxidation of food or respiration. When the living organisms are completely at the stage of rest, even then they require some minimum amount of energy for the maintainance of cells and tissues. Thus, respiration is one of the most important process for living organisms.

Question 2.
Define respiration with the help of chemical equation?
Answer:
The process in which the oxidation of absorbed food is takes place by the CO2 which is inhaled by breathing and energy is released out is called respiration. Chemical equation of nutrition is as follows:
C6H12O2 + 6O2 → 6CO2 + 6H2O + 673 kcal (Energy)

Question 3.
Why does our body need a transporting system?
Answer:
Our body needs a transporting system to:

  1. Transport oxygen to body cells from lungs.
  2. Transport food to body cells from liver.
  3. Transport waste material from body cells to excretory organs.
  4. Maintains body temperature constant.

MP Board Solutions

Question 4.
Write difference between oxy – respiration and anoxy – respiration?
Answer:
The differences between the oxy – respiration and anoxy – respiration:
oxy – respiration:

  1. It takes place in presence of O2
  2. The end products are CO2 and H2O.
  3. The energy released is more.

Anoxy – respiration:

  1. It takes place without oxygen.
  2. The end products are ethyl alcohol and CO2.
  3. The energy released

Question 5.
Describe the various types of respiratory organs found in animals?
Answer:
In animals, there are definite respiratory organs for exchange of gases.

  1. In earthworm and leech exchange of gases takes place through moist, thin and vascular skin.
  2. In insects the trachea are the repiratory organs.
  3. In fishes the gills are the respiratory organs.
  4. Higher animals like mammal and birds including man have lungs for respiration.

Question 6.
Describe the importance of respiration in plants?
Answer:

  1. It takes place in the presence of oxygen.
  2. It is completed in cytoplasm and mitochondria of cell.
  3. It involves the complete oxidation of glucose into CO2, and
  4. H2O + C2H2O2 + 6O2 → 6CO2 + energy + 6H2O.
  5. It occurs in all the living cell of the organisms.
  6. It is day night process.
  7. Energy is released in this process.

Question 7.
Name the organs associated with the following functions:

  1. digestion
  2. absorption of minerals
  3. respiration, and
  4. excretion of carbon – dioxide in man.

Answer:

1. Digestion. Mouth, stomach, oesophagus, pharynx, small intestine, large intestine.

2. Absorption of minerals.
In Animals : Small intestine
In plants : Root hairs.

3. Respiration.
In Animals : Nose, trachaea, larynx, lung.
In plants. Stomata and lenticells.

4. Excretion of carbon – dioxide.
In Animals : Kidneys, ureters, urinary bladder, urethra.

MP Board Solutions

Question 8.
State the difference between respiration and breathing?
Answer:
Differences between Respiration and Breathing:
Respiration:

  1. It takes place inside the cells.
  2. The exchange of gases takes place between blood and the tissues of the body.
  3. In this process nutrients are oxidised to liberate energy.

Breathing:

  1. It takes place at the surface of the respiratory organs.
  2. The exchange of gases takes place between the blood and the external environment.
  3. The nutrients are not oxidised to liberate energy.

Question 9.
Name the major components of urine?
Answer:
The kidney, ureter, bladder, and urethra are the organs used in the removal of urine from the body. Renal artery carry urea and uric acid along with the large amount of water with the blood into the kindneys, when this blood enters into glomerulus, the solid wastes filter here while the water is diffused out from the network of blood capillaries into the uriniferous tubules. The ques mixture is called as urine.

Question 10.
If a person drinks very little water per day? The volume of urine decreases? In what ways does it affect the body?
Answer:
If a person drinks very little water per day. The volume of urine decreases. Large amount of water will dissolve large quantity of urea in it and large amount of urine will pass out from the body. If someone drinks lesser amounts of water, the concentration of urea in the cells and its larger quantity is very harmful for the body.

Question 11.
What is saliva? What are the functions of saliva?
Answer:
Saliva is a digestive secretion produced by three layered salivary glands, the paroted submaxillary and sunblingual present in our mouth cavity. This soften and lubricants food for easier swallowing and converts starch into reducing sugars. The salivary amylase. enzyme of saliva acts of starch in a neutral medium.

Question 12.
How does the food digested in the stomach?
Answer:
After some time from mouth the food reach inside the stomach. The gastric glands of stomach secrete the gastric juices. The gastric juice containes three enzymes. These are pepsin, renin and HCI. The HCI makes the medium acidic, and inhibites the bacterial growth and prevents the food. The renin curdiles the milk protein to be hydrololysed by pepsin. The pepsin reacts with proties and changes into peptides.

Respiration in Organisms Long Answer Type Questions

Question 1.
Define the respiratory organs in animals?
Answer:
The process of respiratory system in animals possess following organs:

  1. Nasal cavity
  2. Larynx
  3. Trachea
  4. Bronchi
  5. Lungs.

In animals some organs like gills and lungs are developed for the purpose of exchange of gases. The amount of CO2 produced after respiration cannot be utilised by animals as in plants. This is also true for the production of oxygen which is required for respiration. Mitochondria are the site of respiration in both plants and animals.

The process of break – down of glucose in the presence of oxygen and some enzymes into CO2. And water, which is accompanied with release of energy is very complex. Materials like proteins and fats are also consumed during respiration to produce energy.

MP Board Solutions

Question 2.
Write difference between photosynthesis and respiration?
Answer:
The difference between respiration and photosynthesis are:
Photosynthesis:

  1. It takes place only in green plants.
  2. It requires energy.
  3. It requires CO2 and H2O.
  4. It releases oxygen and make food.
  5. It is a building up process.
  6. It takes place in the chloroplast of the plant cell.

Respiration:

  1. It takes place in all plants and animals.
  2. It releases energy.
  3. It releases CO2 and H2O.
  4. It requires oxygen and oxidise the food.
  5. It is a breaking down process.
  6. It takes place in mitochondira of a cell.

Question 3.
Draw the labelled diagram to show respiratory system in man?
Answer:
The process of respiration is aimed to release energy
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 8

Question 4.
What is the difference in the amount of carbon choxide in the inhaled and the exhaled air? How will you test the presence of CO2 in the exhaled air?
Answer:
Excess carbon dioxide is present in exhaled air as compared inhaled air.

Test to indicate the presence of CO2 in the exhaled air:
Take two test tubes. Fill each of them half with freshly prepared lime water. Fix stoppers with two holes in both the test tubes. Insert glass tubes in both the stoppers. The lime water through which exhaled air is passed turns milky. This shows that more CO2 is present in exhaled air.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 9

Question 5.
Draw diagrams to show movements of rib and diaphragm during breathing?
Answer:
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms image 10

Question 6.
How do the following organism breathe? Amoeba, fish, frog, grasshopper, earthworm?
Answer:
Amoeba:
Amoeba breathes by diffusion of gases in between body surface and the water.

Fish:
Fish breathe with gills. The gills are special organs. They help fish to extract dissolved oxygen from the water.

Frog:
Frog can breathe through skin and lungs. In water if breathes through skin whereas in air through lungs.

Grasshopper:
The grasshopper and other insects have holes and air tubes those help them to breathe.

Earthworm:
It breathes through its moist body surface.

MP Board Solutions

Question 7.
Describe the breathing in cockroach with diagram?
Answer:
A cockroach has small openings on the sides of its body. Other insects also have similar openings. These openings are called spiracles, bisects have a network of air tubes called tracheae for gas exchange. Oxygen rich air rushes through spiracles into the tracheal tubes, diffuses into the body tissue, and reaches every cell of the body. Similarly, carbon dioxide from the cells goes into the tracheal tubes and moves out through spiracles. These air tubes or tracheae are found only in insects and not in any other group of animals.
MP Board Class 7th Science Solutions Chapter 10 Respiration in Organisms img u

MP Board Class 7th Science Solutions

MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1

MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1

MP Board Class 7 Maths Solutions Chapter 1 प्रश्न 1.
किसी विशिष्ट दिन विभिन्न स्थानों के तापमानों को डिग्री सोल्सियस (°C) में निम्नलिखित संख्या रेखा पर दर्शाया गया है:
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(a) इस संख्या रेखा को देखिए और इस पर अंकित स्थानों के तापमान लिखिए।
(b) उपर्युक्त स्थानों में से सबसे गर्म और सबसे ठण्डे स्थानों के तापमानों में क्या अन्तर है?
(c) लाहुलस्पीती एवं श्रीनगर के तापमानों में क्या अन्तर है?
(d) क्या हम कह सकते हैं कि शिमला और श्रीनगर के तापमानों का योग शिमला के तापमान से कम है? क्या इन दोनों स्थानों के तापमानों का योग श्रीनगर के तापमान से भी कम है?
हल:
(a)
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1

(b) यहाँ, सबसे गर्म स्थान बैंगलोर (22°C) और सबसे ठण्डा स्थान लाहुलस्पीती (-8°C) है।
∴ सबसे गर्म और सबसे ठण्डे स्थानों के तापमानों का अन्तर
= 22°C – (-8°C) = 22°C + 8°C = 30°C

(c) लाहुलस्पीती का तापमान = – 8°C, श्रीनगर का तापमान = -2°C
∴ अभीष्ट अन्तर = -2°C – (-8°C)
= -2°C + 8°C = 6°C

(d) शिमला और श्रीनगर के तापमानों का योग = 5°C + (-2°C) = 3°C
अतएव, हम कह सकते हैं कि
हाँ, शिमला और श्रीनगर के तापमानों का योग शिमला के तापमान से कम है।
श्रीनगर का तापमान = -2°C
∵ 3°C > -2°C नहीं, इन दोनों स्थानों के तापमानों का योग श्रीनगर के तापमान से कम नहीं है।

MP Board Solutions

MP Board Class 7th Maths Chapter 1 प्रश्न 2.
किसी प्रश्नोत्तरी में सही उत्तर के लिए धनात्मक अंक दिए जाते हैं और गलत उत्तर के लिए ऋणात्मक अंक दिए जाते हैं। यदि पाँच उत्तरोत्तर चक्करों (rounds) में जैक द्वारा प्राप्त किए गए अंक 25, -5, – 10, 15 और 10 थे, तो बताइए अन्त में उसके अंकों का योग कितना था ?
हल:
∵ पहले चक्कर में अंक = 25
दूसरे चक्कर में अंक = -5
तीसरे चक्कर में अंक = -10
चौथे चक्कर में अंक = 15
पाँचवें चक्कर में अंक = 10
∴ कुल अंक = 25 + (-5) + (-10) + 15 + 10
= 50 – 15 = 35
अतएव, जैक के अंकों का योग = 35

MP Board Class 7 Maths Solutions English Medium प्रश्न 3.
सोमवार को श्रीनगर का तापमान -5°C था और मंगलवार को तापमान 2°C कम हो गया। मंगलवार को श्रीनगर का तापमान क्या था? बुधवार को तापमान 4°C बढ़ गया। बुधवार को तापमान कितना था ?
हल:
सोमवार को श्रीनगर का तापमान = – 5°C
∵ 2°C तापमान कम हो गया।
∴ मंगलवार को तापमान = – 5°C + (-2°C)= – 7°C
बुधवार को तापमान 4°C बढ़ गय
∴ बुधवार को तापमान = – 7°C +4°C = -3°C

MP Board Class 7 Maths Solutions Hindi Medium प्रश्न 4.
एक हवाई जहाज समुद्र तल से 5000 मीटर की ऊँचाई पर उड़ रहा है। एक विशिष्ट बिन्दु पर यह हवाई जहाज समुद्र तल से 1200 मीटर नीचे तैरती हुई पनडुब्बी के ठीक ऊपर है। पनडुब्बी और हवाई जहाज के बीच की ऊर्ध्वाधर दूरी कितनी है ?
पाठ्य-पुस्तक में दिये गये चित्र के अनुसार, समुद्र तल 0 मीटर पर है और हवाई जहाज समुद्र तल से 5000 मीटर की ऊँचाई पर है।
हल:
समुद्र तल और हवाई जहाज के बीच की दूरी = 5000 मीटर
और पनडुब्बी समुद्र तल से 1200 मीटर नीचे है।
∴ समुद्र तल और पनडुब्बी के बीच की दूरी = 1200 मीटर
∴ हवाई जहाज और पनडुब्बी के बीच दूरी
= 5000 मीटर + 1200 मीटर
= 6200 मीटर

Class 7th Maths MP Board प्रश्न 5.
मोहन अपने बैंक खाते में ₹ 2000 जमा करता है और अगले दिन इसमें से ₹1642 निकाल लेता है। यदिखाते में से निकाली गई राशि को ऋणात्मक संख्या से निरूपित किया जाता है, तो खाते में जमा की गई राशि को आप कैसे निरूपित करोगे? निकासी के पश्चात् मोहन के खाते में शेष राशि ज्ञात कीजिए।
हल:
चूँकि बैंक से निकासी की राशि बैंक में जमा की गई राशि के विपरीत है, अतएव जमा की गई राशि को धनात्मक संख्या से निरूपित करेंगे।
उत्तर जमा की गई राशि = ₹ 2000
निकाली गई राशि = – ₹ 1642
अतएव निकालने के बाद शेष राशि
= ₹ 2000 – ₹ 1642
= ₹ 358

MP Board Solutions

Class 7 Math MP Board प्रश्न 6.
रीता बिन्दु A से पूर्व की ओर बिन्दु B तक 20 किलोमीटर की दूरी तय करती है। उसी सड़क के अनुदिश बिन्दु B से वह 30 किलोमीटर की दूरी पश्चिम की ओर तय करती है। यदि पूर्व की ओर तय की गई दूरी को धनात्मक पूर्णांक से निरूपित किया जाता है, तो पश्चिम की ओर तय की गई दूरी को आप कैसे निरूपित करोगे? बिन्दु A से उसकी अन्तिम स्थिति को किस पूर्णांक से निरूपित करोगे?
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
पूर्व और पश्चिम की दिशाएँ एक-दूसरे की विपरीत है।
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
यदि पूर्व की ओर चली दूरी को धनात्मक संख्या मानें, तो पश्चिम की ओर चली गई दूरी को ऋणात्मक संख्या से निरूपित करेंगे।
अब, पूर्व की ओर तय की गई दूरी = + 20 किलोमीटर
और पश्चिम की ओर तय की गई दूरी = – 30 किलोमीटर
∴ बिन्दु A से उसकी अन्तिम स्थिति = – 30 + (+ 20) = -10 किलोमीटर

Class 7 Maths Solutions MP Board प्रश्न 7.
किसी मायावी वर्ग में प्रत्येक पंक्ति, प्रत्येक स्तम्भ एवं प्रत्येक विकर्ण की संख्याओंकायोगसमान होता है। बताइए निम्नलिखित में से कौन-सा वर्ग एक मायावी वर्ग है।
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
मायावी वर्ग (i)
अंकों का योग:
पहली पंक्ति = 5 + (-1) + (-4) = + 5 + (-5) = 0
दूसरी पंक्ति = (-5) + (-2) +7 = -7 + 7 = 0
तीसरी पंक्ति = 0 + 3 + (-3) = 3 + (-3) = 0
प्रथम स्तम्भ = 5 + (-5) + 0 = 5 + (-5) = 0
द्वितीय स्तम्भ = (-1) + (-2) + 3 = – 3 + 3 = 0
तृतीय स्तम्भ = (-4) + 7 + (-3) = – 7 +7 = 0
प्रथम विकर्ण = 5 + (-2) + (-3) = 5 + (-5) = 0
द्वितीय विकर्ण = 0 + (-2) + (-4) = – 2 – 4 = – 6
चूँकि वर्ग (i) में द्वितीय विकर्ण का योग अन्य पंक्ति, स्तम्भ एवं विकर्ण के बराबर (- 6 ≠ 0) नहीं है,
अतः वर्ग (i) मायावी वर्ग नहीं है।

मायावी वर्ग (ii)
अंकों का योग : पहली पंक्ति = 1 + (- 10) + 0 = 1 – 10 = –9
द्वितीय पंक्ति = (-4) + (-3) + (-2) = -9
तृतीय पंक्ति = – 6 + 4 + (-7) = – 13 + 4 = -9
प्रथम स्तम्भ = 1 + (-4) + (-6) = 1 + (- 10) = -9
द्वितीय स्तम्भ = (-10) + (-3) + 4 = – 13 + 4 = -9
तृतीय स्तम्भ = 0 + (-2) + (-7) = 0 – 9 = -9
प्रथम विकर्ण = 1 + (-3) + (-7)= 1 – 10 = -9
द्वितीय विकर्ण = (-6) + (-3) + 0 = – 6 – 3 = -9
मायावी वर्ग (ii) में प्रत्येक पंक्ति, स्तम्भ और विकर्ण का योग बराबर (-9) है।
अतः वर्ग (ii) एक मायावी वर्ग है।

MP Board Solutions

Class 7th Maths MP Board English Medium प्रश्न 8.
a और b के निम्नलिखित मानों के लिए a – (- b) = a + b का सत्यापन कीजिए :
(i) a = 21, b = 18;
(ii) a = 118, b = 125;
(iii) a = 75, b = 84;
(iv) a = 28, b = 11
हल:
(i) यहाँ, a = 21, b = 18
∴ L.H.S. = a – (-b)
= 21 – (-18)
= 21 + 18 = 39
R.H.S. = a + b
= 21 + 18 = 39
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(ii) यहाँ, a = 118, b = 125
∴ L.H.S. = a – (-b)
= 118 – (- 125)
= 118 + 125 = 243.
और R.H.S. = a + b
= 118 + 125 = 243
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(iii) यहाँ, a = 75, b = 84 .
∴ L.H.S. = a – (-b)
= 75 – (-84)
= 75 + 84 = 159
R.H.S. = a + b
= 75 + 84 = 159
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

(iv) यहाँ, a = 28, b = 11
∵ L.H.S. = a – (-b)
= 28 – (-11)
= 28 +11 = 391
और R.H.S: = a + b
= 28 + 11 = 39
∵ L.H.S. = R.H.S
∴ a – (-b) = a + b सत्यापित है।

MP Board Solutions

MP Board Class 7 Maths Solutions प्रश्न 9.
निम्नलिखित कथनों को सत्य बताने के लिए, बॉक्स में संकेत >, < अथवा = का उपयोग कीजिए-
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
हल:
(a) -8+ (-4) = -8 – 4 = – 12
तथा -8 – (-4) = -8 + 4 = -4
∴ (-12) < (-4)
अतः -8 + (-4) < -8 – (-4)

(b) -3 +7 – (19) = – 3 +7 – 19 = – 15
तथा 15 – 8 + (-9) = – 2
(-15) < (-2)
अतः -3 +7- (19) < 15 – 8 + (-9) (c) 23 – 41 + 11 = 34 – 41 = -7 तथा 23 – 41 – 11 = 23 – 52 = – 29 ∴ (-7) > (-29)
अतः 23 – 41 + 11 > 223 – 41 – 11

(d) 39+ (-24)- (15) = 39 – 24 – 15 = 0
तथा 36 + (-52) – (-36) = 36 – 52 + 36 = 20
∴ 0 < 20
अतः 39 + (-24) – (15) < 36 + (-52) – (-36) (e) -231 + 79 + 51 = – 231 + 130 = – 101 तथा -399 + 159+ 81 = -399 + 240 = -159 (-101) > (- 159)
अतः – 231 + 79 + 51 > -399 + 159 + 81

MP Board 7th Class Maths Book Solutions प्रश्न 10.
पानी के एक तालाब के अन्दर की ओर सीढ़ियाँ हैं। एक बन्दर सबसे ऊपर वाली सीढ़ी (यानी पहली सीढ़ी) पर बैठा हुआ है। पानी नौवीं सीढ़ी पर है।
(i) वह एक छलांग में तीन सीढ़ियाँ नीचे की ओर और अगली छलाँग में दो सीढ़ियाँ ऊपर की ओर जाता है। कितनी छलाँगों में वह पानी के स्तर तक पहुँच पाएगा?
(ii) पानी पीने के पश्चात् वह वापस जाना चाहता है। इस कार्य के लिए वह एक छलाँग में 4 सीढ़ियाँ ऊपर की ओर और अगली छलाँग में 2 सीढ़ियाँ नीचे की ओर जाता है। कितनी छलाँगों में वह वापस सबसे ऊपर वाली सीढ़ी पर पहुँच जाएगा?
(iii) यदि नीचे की ओर पार की गई संख्या को ऋणात्मक पूर्णांक से निरूपित किया जाता है और ऊपर की ओर पार की गई सीढ़ियों की संख्या को धनात्मक पूर्णांक से निरूपित किया जाता है तो निम्नलिखित को करते हुए भाग (i) और (ii) में उसकी गति को निरूपित कीजिए :
(a) – 3 + 2 – … = -8
(b) 4 – 2 +… = 8
(a) में योग (-8) आठ सीढ़ियाँ नीचे जाने को निरूपित करता है, तो (b) में योग 8 किसको निरूपित करेगा?
हल:
(i) बन्दर पहली सीढ़ी पर बैठा हुआ है।
∴ छलाँगों के बाद बन्दर की स्थिति निम्न प्रकार होगी : पहली छलाँग में वह चौथी सीढ़ी पर होगा।
दूसरी छलाँग में वह दूसरी सीढ़ी पर होगा। (∵4 – 2 = 2)
तीसरी छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 2 + 3 = 5)
चौथी छलाँग में वह तीसरी सीढ़ी पर होगा। (∵ 5 – 2 = 3)
पाँचवीं छलाँग में वह छठी सीढ़ी पर होगा। (∵ 3 + 3 = 6)
छठवी छलाँग में वह चौथी सीढ़ी पर होगा। (∵ 6 – 2 = 4)
सातवीं छलाँग में वह सातवीं सीढ़ी पर होगा। (∵ 4 + 3 = 7)
आठवीं छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 7 – 2 = 5)
नौवीं छलाँग में वह आठवीं सीढ़ी पर होगा। (∵ 5 + 3 = 8)
दसवीं छलाँग में वह छठवीं सीढ़ी पर होगा। (∵ 8 – 2 = 6)
ग्यारहवीं छलाँग में वह नौवीं सीढ़ी पर होगा। (∵ 6 + 3 = 9) जो कि पानी का स्तर है।
अत: पानी के स्तर तक वह 11 छलाँगों में पहुँच पाएगा।

(ii) यहाँ, बन्दर की स्थिति पानी के स्तर अर्थात् नौवीं सीढ़ी पर है।
∴ छलाँगों के बाद बन्दर की ऊपर वाली सीढ़ी से स्थिति निम्न प्रकार होगी :
पहली छलाँग में वह पाँचवीं सीढ़ी पर होगा। (∵ 9 – 4 = 5)
दूसरी छलाँग में वह सातवीं सीढ़ी पर होगा। (∵ 5 + 2 = 7)
तीसरी छलाँग में वह तीसरी सीढ़ी पर होगा। (7 – 4 = 3)
चौथी छलाँग में वह पाँचवीं सीढ़ी पर होगा। (3 + 2 = 5)
पाँचवीं छलाँग में वह पहली (ऊपर की) सीढ़ी पर होगा। (5 – 4 = 1)
∴ अभीष्ट छलाँगों की संख्या = 5

(iii) चूँकि बन्दर द्वारा ऊपर की ओर पार की गई सीढ़ियों की संख्या को धनात्मक पूर्णांक से निरूपित किया जाता है और नीचे की ओर पार की गई सीढ़ियों की संख्या को ऋणात्मक पूर्णांक से निरूपित किया जाता है। अतः भाग (i) में बन्दर की गति
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(a) – 3 + 2 -3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8

भाग (ii) में बन्दर की गति
MP Board Class 7th Maths Solutions Chapter 1 पूर्णांक Ex 1.1 1
(b) 4 – 2 + 4 – 2 + 4 = 8
(b) में योग 8 ऊपर की ओर 8 सीढ़ियाँ चढ़ने को निरूपित करता है।

पाठ्य-पुस्तक पृष्ठ संख्या # 06

नीचे दी हुई सारणी को देखिए और इसे पूरा कीजिए :
आप क्या देखते हैं? क्या दो पूर्णांकों का योग हमेशा एक | पूर्णांक प्राप्त होता है? क्या आपको पूर्णांकों का ऐसा युग्म मिला जिसका योग पूर्णांक नहीं है। क्या पूर्णांक योग के अन्तर्गत संवृत होते हैं।
हल:
कथन                                            प्रेक्षण
(i) 17 + 23 = 40                   परिणाम एक पूर्णांक है।
(ii) (-10) +3 =-7                  परिणाम एक पूर्णांक है।
(iii) (-75) + 18 = – 57         परिणाम एक पूर्णांक है।
(iv) 19 + (-25) = – 6           परिणाम एक पूर्णांक है।
(v) 27 + (-27) = 0              परिणाम एक पूर्णांक है।
(vi) (-20) + 0 = – 20         परिणाम एक पूर्णांक है।
(vii) (-35) + (-10) = – 45  परिणाम एक पूर्णांक है।

(a) हम देखते हैं कि किन्हीं दो पूर्णांकों का योग हमेशा एक पूर्णांक होता है।
(b) हाँ, दो पूर्णांकों का योग सदैव पूर्णांक होता है।
(c) हमें यहाँ पूर्णांकों का ऐसा युग्म प्राप्त नहीं हुआ जिनका योग एक पूर्णांक न हो।
(d) अतः पूर्णांकों का योग पूर्णांक ही होता है। इसलिए हम कहते हैं कि पूर्णांक योग के अन्तर्गत संवृत (closed) होते हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 07

निम्नलिखित सारणी को देखिए और इसे पूरा कीजिए :
आप क्या देखते हैं? क्या पूर्णांकों का कोई ऐसा युग्म है जिसका अन्तर पूर्णांक नहीं है?
क्या हम कह सकते हैं कि पूर्णांक व्यवकलन के अन्तर्गत संवृत होते हैं?
हल:
कथन                                              प्रेक्षण
(i) 7 – 9 = – 2                         परिणाम एक पूर्णांक है।
(ii) 17 – (-21) = 38               परिणाम एक पूर्णांक है।
(iii) (-8) – (-14) = 6             परिणाम एक पूर्णांक है।
(iv) (-21) – (-10) = – 11       परिणाम एक पूर्णांक है।
(v) 32 – (-17) = 49              परिणाम एक पूर्णांक है।
(vi) (-18) – (-18)= 0           परिणाम एक पूर्णांक है।
(vii) (-29 )- 0 = – 29          परिणाम एक पूर्णांक है।

(a) हम देखते हैं कि दो पूर्णांकों का व्यवकलन भी एक पूर्णांक होता है।
(b) नहीं, पूर्णांकों का ऐसा कोई युग्म नहीं है जिसका अन्तर पूर्णांक नहीं है।
(c) हाँ, हम कह सकते हैं कि पूर्णांक व्यवकलन के अन्तर्गत संवृत होते हैं।

MP Board Solutions

MP Board Maths Class 7 प्रश्न 1.
क्या पूर्ण संख्याएँ भी इस गुण को सन्तुष्ट करती हैं ?
उत्तर:
नहीं, पूर्ण संख्याएँ इस गुण को सन्तुष्ट नहीं करतीं।

Class 7 MP Board Maths प्रश्न 2.
क्या निम्नलिखित समान हैं ?
(i) (-8) + (-9) और (-9) + (-8)
(ii) (-23) + 32 और 32 + (-23)
(iii) (-45) + 0 और 0 + (-45)
पाँच अन्य पूर्णांकों के युग्मों के लिए ऐसा प्रयास कीजिए। क्या आपको पूर्णांकों का कोई ऐसा युग्म मिलता है जिसके लिए पूर्णांकों का क्रम बदल देने में उनका योग भी बदल जाता है।
हल:
(i) (-8) + (-9)= – 8 – 9 = – 17
और (-9) + (-8) = -9 – 8 = – 17
हाँ, (-8) + (-9) और (-9) + (-8) समान हैं।

(ii) (-23) + 32 = 32 – 23 = 9
और 32 + (-23) = 32 – 23 = 9
हाँ, (-23) + 32 और 32 + (-23) समान हैं।

(iii) (-45) + 0 = -45 + 0 = -45
और 0 + (-45) = 0 – 45 = – 45
हाँ, (-45) + 0 और 0 + (-45) समान हैं।
उदाहरण:
(a) (-36) + (-15) = -51
और (-15) + (-36) = – 51
∴ (-36) + (-15) = (-15) + (-36)

(b) (-10) + 6 = – 10 + 6 = -4
और 6 + (-10) = 6 – 10 = -4
∴ (-10) + 6 = 6 + (-10)

(c) (-118) + 0 = – 118 + 0 = – 118
और 0 + (-118) = 0 – 118 = – 118
∴ (-118)+ 0 = 0+ (- 118)

(d) (-12) + 10 = – 12 + 10 = – 2
और 10 + (-12) = 10 – 12 = -2
∴ (-12) + 10 = 10 + (-12)

(e) 53 + (-26) = 53 – 26 = 27
और (-26)+ 53 = – 26 + 53 = 27
∴ 53 + (-26) = (-26) + 53

पाठ्य-पुस्तक पृष्ठ संख्या # 08

पूर्णांकों के कम-से-कम पाँच विभिन्न युग्म लीजिए और इस कथन की जाँच कीजिए कि व्यवकलन पूर्णांकों के लिए क्रम-विनिमेय नहीं हैं।
हल:
(i) 100 – 86 और 86 – 100
100 – 86 = 14 और 86 – 100 = – 14.
∴ 100 – 86 ≠ 86 – 100

(ii) (-19) और 5
(-19) – 5 = – 24 और 5-(-19) = 24
∴ (-19 – 5 ≠ 5- (-19)

(iii) (-17) और (-19)
(-17) – (- 19) = – 17 + 19 = 2
और (-19) – (-17) = – 19 + 17 = – 2
∴ (-17) – (-19) ≠ (-19) – (-17)

(iv) 69 और 0
69 – 0 = 69 और 0 – 69 = – 69
∴ 69 – 0 ≠ 0 – 69

(v) 118 और (-56)
118 – (-56) = 118 + 56 = 174
और (-56)- 118 = – 174
118 – (-56) ≠ (-56) – 118
उपर्युक्त उदाहरणों से स्पष्ट है कि व्यवकलन पूर्णांकों के लिए क्रम-विनिमेय नहीं है।
अर्थात् a – b # b – a.
इसी प्रकार – 3, 1 और – 7 को लीजिए।
-3 + [1 + (-7)] = -3 + (-6) = -9
[(-3) + 1] + (-7) = – 2 + (-7) = -9
इसी प्रकार के पाँच और उदाहरण लीजिए :
उदाहरण;
(i) -9, -4 और 6
(-9) + [(-4) + 6] = -9+ 2 = -7
और [(-9) + (-4)] + 6 = – 13 + 6 = -7
अतः (-9) + [(-4) + 6] = [(-9) + (-4)] + 6

(ii) -2, 10 और 5
= 8 + 5 = 13
और (-2) + [ 10 + 5] = -2 + 15 = 13
अतः [(-2) + 10] + 5 = – 2 + [10 + 5]

(iii) 13, – 12 और -7
[13 + (-12)] + (-7) = 1-7 = – 6
और 13 + [(-12) + (-7)] = 13 – 19 = -6
अतः [13 + (-12)] + (-7) = 13 + [(-12) + (-7)]

(iv) -4, 15 और -3
[(-4) + 15] + (-3) = 11 – 3 = 8 और -4 + [15 + (-3)] = – 4 + 12 = 8
अतः [(-4) + 15] + (-3) = -4+ [15 + (-3)]

(v) – 12, 19 और 15
[(-12)+ 19] + 15 = 7 + 15 = 22
और -12 + [19 + 15] = – 12 + 34 = 22
अतः [(-12) + 19] + 15 = – 12 + [19+ 15]
अतएव पूर्णांकों के लिए योग सहचारी (Associative) है।
∴ (a + b) + c = a + (b + c)

पाठ्य-पुस्तक पृष्ठ संख्या # 09

निम्नलिखित को देखिए और रिक्त स्थानों की पर्ति कीजिए-
हल:
(i) (-8) + 0 = – 8
(ii) 0 + (-8) = -8
(iii) (-23)+ 0 = – 23
(iv) 0 + (-37) = -37
(v) 0+ (-59) = -59
(vi) 0+ (-43) = – 43
(vii) -61 + 0 = – 61
(viii) -45 + 0 = – 45
उपर्युक्त उदाहरण दर्शाते हैं कि शून्य और ऋणात्मक पूर्णांकों का योग सदैव उसी पूर्णांक के बराबर होता है। अतएव किसी पूर्णांक a के लिए शून्य योज्य तत्समक है।
a + 0 = 0 + a = a

प्रयास कीजिए

Class 7 Maths Chapter 1 Exercise 1.1 Solutions Hindi Medium प्रश्न 1.
एक ऐसा पूर्णांक युग्म लिखिए जिसके योग से हमें निम्नलिखित प्राप्त होता है
(a) एक ऋणात्मक पूर्णांक
(b) शून्य
(c) दोनों पूर्णांकों से छोटा एक पूर्णांक
(d) दोनों पूर्णांकों में से केवल किसी एक से छोटा पूर्णांक
(e) दोनों पूर्णांकों से बड़ा एक पूर्णांक।
हल:
(a) -25 और 9
योग- (-25) + 9 = -16; -16 एक ऋणात्मक पूर्णांक है।
(b) – 27 और 27
योग -(-27) + 27 = 0

(c) – 16 और -4
योग – (-16) + (-4) = – 20; – 20 पूर्णांक – 16 और -4 से छोटा है।

(d) 4 और -6
योग-4+ (-6) = -2 ; – 2 केवल 4 से छोटा है।

(e) 29 और 11
योग-29 + 11 = 40; 40 पूर्णांक 29 और 11 से बड़ा है।

MP Board Solutions

MP Board Class 7 Maths Chapter 1 प्रश्न 2.
एक ऐसा पूर्णांक युग्म लिखिए जिसके अन्तर से हमें निम्नलिखित प्राप्त होता है
(a) एक ऋणात्मक पूर्णांक
(b) शून्य
(c) दोनों पूर्णांकों से छोटा एक पूर्णांक
(d) दोनों पूर्णांकों में से केवल किसी एक से बड़ा पूर्णांक
(e) दोनों पूर्णांकों से बड़ा एक पूर्णांक
हल:
(a) 13 और – 8
अन्तर – ( – 8) – 13 = – 21; – 21 एक ऋणात्मक पूर्णांक है।
(b) – 13 और – 13
अन्तर – (-13) – (- 13) = – 13 + 13 = 0
(c) 15 और 19
अन्तर -19 – 15 = 4; 4 पूर्णांक 19 और 15 से छोटा
(d) 16 और 7
अन्तर -16 – 7 = 9; 9 पूर्णांक 7 से बड़ा है।
(e) 18 और -6
अन्तर – 18 – (-6) = 24; 24 पूर्णांक 18 और -6 से बड़ा है।

MP Board Class 7th Maths Solutions

MP Board Class 10th Maths Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2 Pdf, 5.2 Class 10 Hindi Medium, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2

Class 10 Math Chapter 5.2 In Hindi प्रश्न 1.
निम्नलिखित सारणी में रिक्त स्थानों को भरिए, जहाँ A.P. का प्रथम पद a, सार्वान्तर d और n वाँ पद an है :
हल:
MP Board Class 10th Maths Solutions Chapter 5 समान्तर श्रेढ़ियाँ Ex 5.2 1
हल:
(i) ∵ an = a + (n – 1) × d
⇒ an = 7 + (8 – 1) × 3
⇒ an = 7 + 7 × 3
⇒ an = 7 + 21 = 28

(ii) ∵ an = a + (n – 1) × d
⇒ 0 = – 18 + (10 – 1) × d
⇒ 0 = – 18 + 9d
⇒ 9d = 18
⇒ d = \(\frac { 18 }{ 9 } \) = 2.

(iii) ∵ an = a + (n – 1) × d
⇒ -5 = a + (18 – 1) (-3)
⇒ -5 = a+ 17(-3)
⇒ -5 = a – 51
⇒ a = 51 – 5 = 46.

(iv) ∵ an = a + (n – 1) × d
⇒ 3.6 = 18.9 + (n – 1) × 25
⇒ 3.6 = – 18.9 + 2.5n – 2.5
⇒ 2.5n = 18.9 + 3.6 + 2.5
⇒ 2.5 n = 25.0
⇒ n = \(\frac { 25 }{ 2.5 } \) = 10

(v) ∵ an = a + (n – 1) × d
⇒ an = 35 + (105 – 1) × 0
⇒ an = 35 + 104 × 0
⇒ an = 3.5
अतः अत: an = 3.5

MP Board Solutions

Exercise 5.2class 10 In Hindi प्रश्न 2.
निम्नलिखित में सही उत्तर चुनिए और उसका औचित्य दीजिए:
(i) AP: 19,7, 4 ….. 30 वाँ पद है :
(a) 97
(b) 77
(c) -77
(d) -87

(ii) AP -3, –\(\frac { 1 }{ 2 } \), 2, …….. का 11 वाँ पद है:
(a) 28
(b) 22
(c) -38
(d) -48 \(\frac { 1 }{ 2 } \)
हल:
(i) सही उत्तर (C) -77 है, क्योंकि a = 10, d = -3, n = 30
एवं an = a + (n – 1)d ⇒ a30 = 10 + (30 – 1) (-3)
⇒ a30 = 10 – 29 × 3 ⇒ 10 – 87 = -77

(ii) सही उत्तर (B) 22 है, क्योंकि a = -3, d = 2\(\frac { 1 }{ 2 } \), n = 11
एवं an = a + (n – 1)d ⇒ an = -3 + (11 – 1) (2.5)
⇒ a11 = -3 + 10 × 2.5 = -3 + 25 = 22

MP Board Solutions

Class 10 Math 5.2 In Hindi प्रश्न 3.
निम्नलिखित सामान्तर श्रेढ़ियों में रिक्त स्थानों (boxes) के पदों को ज्ञात कीजिए।
(i) 2, [], 26
(ii) [], 13, [], 3
(iii) 5, [], [], 9\(\frac { 1 }{ 2 } \)
(iv) -4, [], [], [], [], 6
(v) [], 38, [], [], [],-22
हल:
(i) प्रश्नानुसार, a = 2, a3 = 26 एवं n = 3.
चूंकि an = a + (n – 1) (d)
⇒ 26 = 2 + (3 – 1)d
⇒ 26 = 2 + 2d ⇒ 2d = 26 – 2 = 24 ⇒ d = \(\frac { 24 }{ 2 } \) = 12
अतः रिक्त स्थान a2 = a + d = 2 + 12 = 14
अत: अभीष्ट रिक्त स्थान (box) में पद 14 होगा।

(ii) प्रश्नानुसार, a2 = 13 एवं a4 = 3
⇒ 13 = a + (2 – 1)d ⇒ a + d = 13 ….(1)
एवं 3 = a+ (4 – 1)d ⇒ a + 3d = 3 …..(2)
⇒ 2d = 3 – 13 = – 10 ⇒ d = –\(\frac { 10 }{ 2 } \) = -5
[समी. (2) – समी. (1) से]
d = -5 का मान समीकरण (1) में रखने पर,
a – 5 = 13 ⇒ a = 13 + 5 = 18
एवं a3 = a + (3 – 1) (-5) = 18 + 2(-5) = 18 – 10 = 8
अतः अभीष्ट रिक्त स्थान में पद क्रमशः 18 एवं 8 होंगे।

(iii) प्रश्नानुसार, a = 5 एवं a4 = 9 \(\frac { 1 }{ 2 } \)
⇒ 9\(\frac { 1 }{ 2 } \) = 5 + (4 – 1) (d) ⇒ 9\(\frac { 1 }{ 2 } \) = 5 + 3d
⇒ 3d = 9 \(\frac { 1 }{ 2 } \) – 5 = 4 \(\frac { 1 }{ 2 } \) ⇒ d = \(\frac { 1 }{ 3 } \) × 4\(\frac { 1 }{ 2 } \) = 1\(\frac { 1 }{ 2 } \)
⇒ a2 = 5 + 1 \(\frac { 1 }{ 2 } \) = 6\(\frac { 1 }{ 2 } \)
एवं a3 = 5 + 2 × 1 \(\frac { 1 }{ 2 } \) = 5 + 3 = 8
अत: अभीष्ट रिक्त स्थानों के अभीष्ट पद क्रमशः 6\(\frac { 1 }{ 2 } \) एवं 8 हैं।

(iv) प्रश्नानुसार, a = – 4 एवं a6 = 6
⇒ 6 = -4 + (6 – 1)d = 6 = -4 + 5d
⇒ 5d = 6 + 4 = 10 ⇒ d = \(\frac { 10 }{ 5 } \) = 2
अब a2 = -4 + 2 = -2
a3 = -4 + 2 × 2 = -4 + 4 = 0
a4 = -4 + 3 × 2 = -4 + 6 = 2
a5 = -4 + 4 × 2 = -4 + 8 = 4
अतः अभीष्ट पद क्रमशः – 2,0,2 एवं 4 हैं।

(v) प्रश्नानुसार, a2 = 38 एवं a6 = -22
⇒ 38 = a + d ⇒ a + d = 38 ….(i)
एवं -22 = a + 5d ⇒ a + 5d = – 22 ….(ii)
⇒ 4d = -60 [समीकरण (2)- समीकरण (1) से]
⇒ d = – \(\frac { 60 }{ 4 } \) = -15
d =-15 का मान समीकरण (1) में रखने पर,
a + (-15) = 38 ⇒ a = 38 + 15 = 53
अब a3 = 53 + 2 × (-15) = 53 – 30 = 23
a4 = 53 + 3 × (-15) = 53 – 45 = 8
a5 = 53 + 4 × (-15) = 53 – 60 = -7
a6 = 53 + 5 × (- 15) = 53 – 75 = – 22
अत: अभीष्ट रिक्त स्थान में पद क्रमशः 53, 23, 8, -7 होंगे।

5.2 Class 10 In Hindi प्रश्न 4.
AP : 3, 8, 13, 18, ……….. का कौन-सा पद 78 है?
हल:
प्रश्नानुसार, a = 3,d = 8 – 3 = 5, an = 78
एवं an = a + (n – 1) × d
⇒ 78 = 3 + (n – 1) × 5 = 3 + 5n – 5
⇒ 5n = 78 + 5 – 3 = 80
⇒ n = \(\frac { 80 }{ 5 } \) = 16
अत: अभीष्ट 16वाँ पद 78 है।

MP Board Solutions

10 Class Math 5.2 In Hindi प्रश्न 5.
निम्नलिखित समान्तर श्रेढ़ियों में से प्रत्येक श्रेणी में कितने पद हैं?
(i) 7, 13, 19, ………. 205
(ii) 18, 15\(\frac { 1 }{ 2 } \),13, …..(-47)
हल:
(i) चूँकि A.P : 7, 13, 19, ……. 205 (दी गयी है।)
प्रश्नानुसार, a = 7,d = 13 – 7 = 6 एवं an = 205
चँकि an = a + (n – 1) (d)
⇒ 205 = 7 + (n – 1) (6)
⇒ 205 = 7 + 6n – 6
⇒ 6n = 205 + 6 – 7 = 204
⇒ n = \(\frac { 204 }{ 6 } \) = 34
अतः श्रेणी में अभीष्ट 34 पद हैं।

(ii) चूँकि AP : 18, 15\(\frac { 1 }{ 2 } \), 13, ….., (-47) (दी गयी है)
प्रश्नानुसार, a = 18, d = 15, – 18 = -2\(\frac { 1 }{ 2 } \), एवं an = -47
चूँकि an = a + (n – 1) (d)
⇒ -47 = 18 + (n – 1) (-2\(\frac { 1 }{ 2 } \))
⇒ – 47 = 18 – 2\(\frac { 1 }{ 2 } \)n + 2\(\frac { 1 }{ 2 } \)
⇒ \(\frac { 5 }{ 2 } \)n = 47 + 18 + 2\(\frac { 1 }{ 2 } \) = 67\(\frac { 1 }{ 2 } \) = \(\frac { 135 }{ 2 } \)
⇒ n = \(\frac { 135 }{ 2 } \) × \(\frac { 2 }{ 5 } \) = 27
अतः श्रेणी में अभीष्ट 27 पद हैं।

5.2 Maths Class 10 Hindi Medium प्रश्न 6.
क्या AP. 11,8, 5, 2 ……… का एक पद — 150 है? क्यों?
हल:
प्रश्नानुसार, a = 11, d = 8 – 11 = -3, an = – 150.
चूँकि an = a + (n – 1) (d)
⇒ – 150 = 11 + (n – 1) (- 3) = 11 – 3n + 3
⇒ 3n = 150 + 11 + 3 = 164
⇒ n = \(\frac { 164 }{ 3 } \) = 54 \(\frac { 2 }{ 3 } \) जो एक पूर्णांक नहीं है।
अत: दत्त AP का कोई भी पद -150 नहीं होगा।

Class 10 Maths Chapter 5 Exercise 5.2 Hindi Medium प्रश्न 7.
उस AP का 31वाँ पद ज्ञात कीजिए जिसका 11वाँ पद 38 है और 16वाँ पद 73 है। (2019)
हल:
प्रश्नानुसार, n11 = 38 एवं n16 = 73.
⇒ 38 = a + 10d ⇒ a + 10d = 38 …..(1)
एवं 73 = a + 15d ⇒ a + 15d = 73 …..(2)
⇒ 5d = 35 [समीकरण (2) – समीकरण (1) से]
⇒ d = \(\frac { 35 }{ 5 } \) = 7
अब d = 7 का मान समीकरण (1) में रखने पर,
a + 10 × 7 = 38 ⇒ a = 38 – 70 = – 32
अब a31 = a + 30d = – 32 + 30 × 7
⇒ a31 = -32 + 210 = 178
अंतः अभीष्ट 31वाँ पद = 178 है।

MP Board Solutions

Class 10 Maths 5.2 Solutions In Hindi प्रश्न 8.
एक AP में 50 पद है, जिसका तीसरा पद 12 है और अन्तिम पद 106 है। इसका 29वाँ पद ज्ञात कीजिए।
हल:
प्रश्नानुसार, n = 50, a3 = 12 एवं a50 = 106.
चूँकि an = a + (n – 1)d
⇒ 106 = a + 49d ⇒ a + 49d = 106 …..(1)
एवं 12 = a + 2d ⇒ a + 2d = 12 …..(2)
⇒ 47d = 94 [समीकरण (2) – समीकरण (1) से
⇒ d = \(\frac { 94 }{ 47 } \) = 2
d = 2 का मान समीकरण (2) में रखने पर,
a + 2 × 2 = 12 ⇒ a = 12 – 4 = 8
अब n29 = 8 + 28 × 2 = 8 + 56 = 64
अतः अभीष्ट 29वाँ पद = 64 है।

Prashnavali 5.2 Class 10 प्रश्न 9.
यदि किसी AP के तीसरे और नौवें पद क्रमशः 4 और – 8 हैं, तो इसका कौन-सा पद शून्य होगा?
हल:
प्रश्नानुसार, a3 = 4 एवं a9 = – 8 है।
⇒ a3 = a + 2d = 4 …..(1)
एवं a9 = a + 8d = – 8 …..(2)
⇒ 6d = -12 [समीकरण (2)- समीकरण (1) से]
⇒ d = –\(\frac { 12 }{ 6 } \) = -2
d = – 2 का मान समीकरण (1) में रखने पर
चूँकि a + 2(-2) = 4 ⇒ a = 4 + 4 = 8.
अब an = a + (n – 1)d
⇒ 0 = 8 + (n – 1)(-2) ⇒ 0 = 8 – 2n + 2
⇒ 2n = 8 + 2 = 10 ⇒ n = \(\frac { 10 }{ 2 } \) = 5
अतः अभीष्ट पाँचवाँ पद शून्य होगा।

10th Class Math 5.2 In Hindi प्रश्न 10.
किसी AP का 17वाँ पद उसके 10वें पद से 7 अधिक है। इसका सार्वान्तर ज्ञात कीजिए।
हल:
प्रश्नानुसार, (a + 16d) – (a + 9d) = 7
⇒ 16d – 9d = 7 ⇒ 7d = 7 ⇒ d = \(\frac { 7 }{ 7 } \) = 1
अतः d का अभीष्ट मान = 1 है।

MP Board Class 10 Maths Solutions प्रश्न 11.
AP3 , 15, 27, 39, …………. का कौन-सा पद उसके 54वें पद से 132 अधिक होगा?
हल:
प्रश्नानुसार, a = 3, d = 15 – 3 = 12 एवं an – a54 = 132
⇒ [3 + (n – 1) (12)] – [3 + (54 – 1) (12)] = 132
⇒ (3 + 12n – 12) – (3 + 53 × 12) = 132
⇒ 12n – 12 – 636 = 132
⇒ 12n = 132 + 12 + 636 = 780
⇒ n = \(\frac { 780 }{ 12 } \) = 65
अतः अभीष्ट 65वाँ पद होगा।

Class 10 Maths Ex 5.2 Solutions In Hindi प्रश्न 12.
दो समान्तर श्रेढ़ियों का सार्वान्तर समान है। यदि इनके 100वें पदों का अन्तर 100 है, तो इनके 1000 वें पदों का अन्तर क्या होगा?
हल:
प्रश्नानुसार, दो समान्तर श्रेढ़ियाँ क्रमशः a, a + d, a + 2d, …………., a + (n – 1)d
एवं b, b + d, b + 2d, …………, b + (n – 1)d
एवं [a+ (100 – 1)d] – [b + (100 – 1)d] = 100
⇒ (a + 99d) – (b + 99a) = 100
⇒ a – b = 100 ….(1)
अब [a + (1000 – 1)d] – [b+ (1000 – 1)d]
= (a + 999d) – (b + 999d)
= a – b = 100 [समीकरण (1) से]
अतः हजारवें पदों का अभीष्ट अन्तर = 100 होगा।

MP Board Solutions

Class 10 Maths Chapter 5.2 In Hindi प्रश्न 13.
तीन अंकों वाली कितनी संख्याएँ 7 से विभाज्य हैं।
हल:
7 से विभाज्य तीन अंकों वाली संख्याओं की सूची है।
105, 112, 119, ……………., 994
जहाँ, a = 105, d = 112 – 105 = 7 एवं an = 994
चूँकि an = a + (n – 1)d
⇒ 994 = 105+ (n – 1) × 7
⇒ 994 = 105 + 7n – 7
⇒ 7n = 994 + 7 – 105
⇒ 7n = 1001 – 105 = 896
⇒ n = \(\frac { 896 }{ 7 } \) = 128
अतः 7 से विभाज्य तीन अंकों वाली कुल अभीष्ट संख्याएँ 128 हैं।

Class 10 Maths Chapter 5.2 Hindi Medium प्रश्न 14.
10 और 250 के बीच में 4 के कितने गुणज हैं?
हल:
10 और 250 के बीच 4 के गुणजों की सूची है :
12, 16, 20, 24, …………, 248
जहाँ, a = 12,d = 16 – 12 = 4 एवं an = 248
चूँकि an = a + (n – 1)d
⇒ 248 = 12 + (n – 1) (4)
⇒ 248 = 12 + 4n – 4 = 4n + 8
⇒ 4n = 248 – 8 = 240
⇒ n = \(\frac { 240 }{ 4 } \) = 60
अत: 10 और 250 के बीच 4 के गुणजों की अभीष्ट संख्या 60 है।

Class 10 Math Chapter 5.2 Solutions In Hindi प्रश्न 15.
n के किस मान के लिए दोनों समान्तर श्रेढ़ियों 63, 65,67,………….. और 3, 10, 17,……… के nवें पद बराबर होंगे?
हल:
चूँकि प्रथम AP का a = 63 एवं d = 65 – 63 = 2, एवं द्वितीय A.P. का a’ = 3 एवं d’ = 10 – 3 = 7 है, तो प्रश्नानुसार,
63 + (n – 1) (2) = 3 + (n – 1) (7)
63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n -4
⇒ 7n – 2n = 61 + 4
⇒ 5n = 65 ⇒ n= \(\frac { 65 }{ 5 } \) = 13
अतः n के अभीष्ट मान 13 के लिए दोनों श्रेढ़ियों के nवें पद बराबर होंगे।

Class 10 Maths Chapter 5 Exercise 5.2 In Hindi Medium प्रश्न 16.
वह AP ज्ञात कीजिए जिसकी तीसरा पद 16 है और 7वाँ पद 5वें पद से 12 अधिक है।
हल:
मान लीजिए कि AP का प्रथम पद a तथा सार्वान्तर d है, तो
प्रश्नानुसार, a3 = 16 ⇒ a + 2d = 16 ….(i)
एवं a7 – a5 = 12 ⇒ (a + 6d) – (a + 4d) = 12
⇒ 2d = 12 ⇒ d = \(\frac { 12 }{ 2 } \) = 6 …..(2)
d का मान समीकरण (2) से समीकरण (1) में रखने पर,
a + 2 × 6 = 16 ⇒ a + 12 = 16 ⇒ a = 16 – 12 = 4
अतः अभीष्ट AP = 4, 10, 16, 22, ……… है।

MP Board Class 10th Maths Solutions प्रश्न 17.
AP 3,8, 13, ……………, 253 में अन्तिम पद से 20वाँ पद ज्ञात कीजिए।
हल:
AP को घटते क्रम में लिखने पर,
253, 248, 243, ………… 13, 8, 3.
जहाँ a = 253 एवं d = (248 – 253) = -5
⇒ a20 = 253 + (20 – 1) (-5)
= 253 + 19 (-5) = 253 – 95
= 158
अतः दत्त AP के अन्तिम पद से अभीष्ट 20वाँ पद = 158 है।

10 Class Ka Math 5.2 In Hindi प्रश्न 18.
किसी AP के चौथे और 8वें पदों का योग 24 है तथा 6वें और 10वें पदों का योग 44 है। इस AP के प्रथम तीन पद ज्ञात कीजिए।
हल:
मान लीजिए a, a + d, a + 2d, a + 3d, ……., समान्तर श्रेढ़ी में हैं, तब प्रश्नानुसार,
∵ a4 + a8 = 24
⇒ (a + 3d) + (a + 7a) = 24
⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ……(1)
एवं a6 + a10 = 44
⇒ (a + 5a) + (a + 9d) = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …..(2)
⇒ 2d = 10 [समीकरण (2) – (1) से]
⇒ d = \(\frac { 10 }{ 2 } \) = 5
d का मान समीकरण (1) में रखने पर,
a + 5 × 5 = 12 ⇒ a + 25 = 12 ⇒ a = 12 – 25 = – 13
⇒ a2 = a + d = -13 + 5 = -8
एवं a = a + 2d = – 13 + 5 × 2 = – 13 + 10 = -3
अतः दी हुई समान्तर श्रेढ़ी के अभीष्ट प्रथम तीन पद क्रमश: -13, -8 एवं -3 हैं।

MP Board Solutions

Math 5.2 Class 10 In Hindi प्रश्न 19.
सुब्बाराव ने 1995 में ₹ 5,000 के मासिक वेतन पर कार्य प्रारम्भ किया ओर प्रत्येक वर्ष ₹200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उसका वेतन ₹ 7000 हो गया?
हल:
सुब्बाराव के प्रतिवर्ष के वेतन की सूची एक AP का निर्माण करेगी, जिसमें a = ₹ 5,000, d = ₹ 200 एवं an = ₹ 7,000 होगा।
इसलिए प्रश्नानुसार,
an = a + (n – 1)d
⇒ 7000 = 5000 + (n – 1) × 200
⇒ 7000 = 5000 + 200n – 200
⇒ 7000 = 4800 + 200n
⇒ 200n = 7000 – 4800 = 2200
⇒ n = \(\frac { 2200 }{ 200 } \) = 11
अत: सुब्बाराव का अभीष्ट वेतन 11वें वर्ष में होगा।

10th Math 5.2 In Hindi प्रश्न 20.
रामकली ने किसी वर्ष के प्रथम सप्ताह में ₹5 की बचत की और फिर अपनी साप्ताहिक बचत में ₹ 1.75 बढ़ाती गयी। यदि वें सप्ताह में उसकी बचत ₹ 20.75 हो जाती है, तो n ज्ञात
कीजिए।
हल:
रामकली के साप्ताहिक बचत की सूची एक AP का निर्माण करती है जिसमें a = ₹5 एवं d = ₹ 1.75 तथा an = ₹ 20.75, तो प्रश्नानुसार,
an = a + (n – 1)d
⇒ 20.75 = 5 + (n – 1) (1.75)
⇒ 20.75 = 5 + 1.75n – 1.75
⇒ 1.75n = 20.75 + 1.75 – 5
⇒ 1.75n = 22.50 – 5 = 17.50
⇒ n = \(\frac { 17.50 }{ 1.75 } \) = 10
अतः n का अभीष्ट मान = 10 है।

MP Board Class 12th Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

MP Board Class 12th Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Electrostatics Important Questions For Board Question 1.
The SI unit of electrical capacitance:
(a) Stat farad
(b) Farad
(c) Coulomb
(d) Stat coulomb.
Answer:
(b) Farad

Question 2.
The potential difference between the plates of a capacitor is constant. A dielectric medium is filled instead of air in between the plates. The intensity of electric field will:
(a) Decrease
(b) Remains unchanged
(c) Become zero
(d) Increase.
Answer:
(b) Remains unchanged

Buffer capacity is the measure of a buffer’s ability to resist pH change.

Class 12 Physics Important Questions Chapter 2 Question 3.
On replacing the air by an insulating material between the plates of a capacitor its capacity:
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(b) Increases

Question 4.
On increasing the separation between the plates of a parallel plate capacitor its capacitance :
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(c) Decreases

Equivalent Capacitance Questions Class 12 Question 5.
When two capacitors are joined in series each capacitor will have the same :
(a) Charge
(b) Potential
(c) Charge and potential
(d) Neither charge nor potential.
Answer:
(a) Charge

MP Board Solutions

Question 6.
When two capacitors are joined in parallel each capacitor will have the same:
(a) Charge and potential
(b) Only charge
(c) Only potential
(d) Neither charge nor potential.
Answer:
(c) Only potential

Question 7.
Two capacitors of equal capacitance first connected in parallel then connected in series. What is the ratio of their capacities in both the cases:
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4.
Answer:
(c) 4 : 1

Physics Important Questions Class 12 MP Board 2023 Question 8.
The formula of capacitance of a spherical conductor is:
(a) C = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \)
(b) C = 4πt£0R
(c) C = 4πr£0R2
(d) C = 4π£0R3
Answer:
(b) C = 4πt£0R

Question 2.
Fill in the blanks:

  1. 1 farad = one coulomb/ ……………
  2. 1 farad = …………… stat farad.
  3. Dimensional formula of capacitance is ……………
  4. is a device in which with or out changing in shape or size of a conductor its capacitance can be increased ……………
  5. On increasing the distance between the plater of a parallel plate capacitor its capacity ……………
  6. Three capacitor each of 3pF are joined in series their equivalent capacitance will be ……………
  7. The dimensional formula of electric potential is ……………
  8. The potential due to a point charge q at a distance r is given as ……………
  9. The potential difference = Intensity of electric field × ……………
  10. The increase in kinetic energy of a charge q when it is accelerated by a potential difference V is ……………
  11. Due to presence of dielectric medium the potential ……………
  12. The work done in moving a charge perpendicular to the electric field is ……………
  13. The potential of earth is considered to be ……………

Answers:

  1. 1 Volt
  2. 9 × 1011
  3. [M-1L-2T4A2]
  4. Capacitor
  5.  Decreases
  6. lµF
  7. [ML2T-3 A-11]
  8. V = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \) \(\frac { q }{ r}\)
  9. Distance between the two point
  10. qV
  11. Decreases
  12. Zero
  13. Zero.

MP Board Solutions

Class 12 Physics Important Questions MP Board Question 3.
Match the Column:
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 1
Answers:

  1. (c) Q/V
  2. (d) \(\frac { 1}{2}\) C/V2
  3. (e) 4π£0R
  4. (a) £0A/d
  5. (b) 4π£0ab / (b – a)

Question 4.
Write the answer in one word / sentence:

  1. What is the potential of earth. Write SI units ?
  2. What will be the electric field intensity inside a hallow sphere ?
  3. In which direction of electric dipole, electric potential is zero ?
  4. What is the net charge of a charge condenser ?
  5. What quantity remains constant when the condenser are connected in series ?
  6. What quantity remain constant when the conductor are connected in parallel ?

Answers:

  1. Zero, volt
  2. Zero
  3. Broad-side-on position
  4. Zero
  5. Charge
  6. Potential difference.

MP Board Solutions

Electrostatic Potential and Capacitance Important Questions Very Short Answer Type Questions

Question 1.
What do you understand by equal potential surface ?
Answer:
The surface of the conductor where potential is in every point is called equal potential surface.

Imp Questions Of Physics Class 12 MP Board Question 2.
Write the name of the physical quantity whose SI unit in J/C. Is it a scalar or vector ?
Answer:
Electric potential, it is a vector quantity.

Question 3.
Draw a equi potential surface for a unit charge.
Answer:
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 2

Question 4.
Define farad.
Answer:
If the potential of a conductor increases by one volt when one coulomb of charges is given to it, then the capacity of the conductor is said to be one farad.

Physics Important Questions Class 12 MP Board Question 5.
On going in direction of electric lines of force, electric potential decreases or increases.
Answer:
Electric potential decreases.

Question 6.
Give an example in which electric field is non-zero but potential is zero.
Answer:
At broad-side-on position of an electric dipole electric field is non-zero and potential is zero.

MP Board Solutions

Question 7.
Does electron try to go toward high potential area or low potential area ?
Answer:
Since electron is negatively charge so it tries to go toward high potential area.

MP Board Class 12th Physics Important Questions Question 8.
Potential between two parallel surface are same. The distance between them is R. If a charge q is bought from one surface to another, then what will be the work I done to do this ?
Answer:
Amount of work done will be zero on both the surface are equipotential.

Question 9.
If area of a plate of a parallel plate condenser in made half. Will it behave as condenser.
Answer:
When area of the plate if a parallel plate condenser is made half. Its capacity become half. Therefore it will not act as condenser.

Question 10.
A capacitor of capacity C is charged with potential difference V. What will be the magnitude of electric flux passing through the surface of it ?
Answer:
Zero.

MP Board Class 12 Physics Notes In English Question 11.
Why condenser are used in computer’s ?
Answer:
Condenser are used as memory chip in computer.

Question 12.
Write one use of capacitor ?
Answer:
To accumulate electric charge.

MP Board Solutions

Electrostatic Potential and Capacitance Important Questions Short Answer Type Questions

Question 1.
What is potential ? Is it a vector or scalar quantity ?
Answer:
Work done in bringing a unit positive charge from infinity to a point in the I electric field is called potential at that point. If charge q is brought from infinity to a point and IT work is done.
∴ V = \(\frac {W}{p}\)
It is a scalar quantity.

Question 2.
Can same amount of charge be given 1 and a solid sphere of same radii, if they have same potential ?
Answer:
No, because capacities of both spheres of same radii are always equal. Therefore i both the spheres can hold same amount of charge at same potential.

Physics Important Questions Class 12 MP Board 2024 Question 3.
What is meant by capacity of a conductor ? Give its unit.
Answer:
The capacity of a conductor is defined by the charge given to the conductor, which increases its potential through unity.
Capacity = \(\frac {Charge}{Potential}\)
or C = \(\frac {q}{v}\)
Its SI unit is farad.

Question 4.
The surface of any conductor is always equipotential. Why ?
Or
The potential at every point on a charged conductor is same. Why ?
Answer:
All the points of the surface of a conductor are in electrical contact with one another. If the potential is not equal then the charges will flow from higher potential to lower potential till the potential of both the points on the surface becomes same. This will give rise to electrodynamics situations. Thus, the surface of a conductor is always equipotential.

MP Board Solutions

MP Board Class 12th Physics Imp Questions 2023 Question 5.
What would be the work done if a point charge +q is taken from a point A to point B on the circumference of a circle with another point charge +q at the center:
Answer:
The points A and B are at same distance from the charge + q at the center, so VA = VB So, work done, W= q0 (VA – VB) = 0.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 3

Question 6.
Explain the meaning of capacity of a capacitor. What will be the effect on capacity of a parallel plate capacitor if a dielectric medium of dielectric constant k is filled in between the plates ?
Answer:
The capacity of a capacitor, is equal to charge given to one of its plates which produces unit potential difference across the plates. In this case capacity increases, it becomes k times its initial value.

Class 12 Physics Chapter 2 Notes Question 7.
What will be the change in the value of charge and potential difference between the plates of a parallel plate capacitor, if after charging its battery is removed and distance between its plates is reduced ?
Answer:
Charge remains same but potential difference decreases.

Question 8.
Two equipotential surface does not intersect each other, why ?
Answer:
Electric lines of forces are always perpendicular to equipotential surface. If two equipotential surface intersect each other then at the point of intersection there will be two direction of electric fields which is impossible. Therefore they does not intersect each other.

MP Board Physics Question 9.
Why must electrostatic field be normal to the surface at every point of a charged conductor ?
Answer:
If electric field is randomly directed, then it can be resolved, into two components. The horizontal component on this surface is E sin θ.
For electrostatic situation
£ sin θ = 0
⇒ sin θ = 0
⇒ θ = 0°
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 4
So, the electric field is normal to surface.

Question 10.
The potential at any point inside the hollow conductor remains same. Why ?
Answer:
When charge is given to a hollow conductor then the distribution of charge takes place on its upper surface. Therefore the intensity of electric field inside the conductor is zero. Hence, no work is done in moving unit positive charge inside it. Therefore potential at every point inside the conductor remains same.

MP Board Solutions

Question 11.
Can the potential be zero where electric field is not zero ?
Answer:
Yes, the electric field on the equatorial line of a dipole is not zero but potential is zero.

Chapter 2 Physics Class 12 Important Questions Question 12.
What will be the effect on electric field, potential, difference, electric capacity and energy if a dielectric of dielectric constant K is filled between the plates of a capacitor ?
Answer:
The electric field will become \(\frac {1 }{ K }\)times, potential difference will become \(\frac {1 }{ K }\) times, electric field will become K time and energy will become \(\frac {1 }{ K }\) times.

Question 13.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
As we know that the formula of potential ¡s V = \(\frac { 1 }{ 4π { £ }_{ 0 } } \) \(\frac { q }{ r }\) …(1)
Given,q = lC, r = lcm = 10-2m
Putting these values in eqn. (1)
V = 9 × 109 × \(\frac{1}{10^{-2}}\) = 9 x 1011volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) =  9 × 109 in SI unit.
This value of potential is greater than barrier potential of air. Therefore IC charge cannot be given to a sphere of radius 1cm.

Chapter 2 Physics Class 12 Question 14.
In the shown figure what will be the work done to bring a z point charge from the point X to Y to Z?
Answer:
There the point Z and Y are situated on same equaipotential surface. Therefore work done to bring a point charge from A’to Zand from X to Z will be same.
.’. Wy= Wz.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 5

Question 15.
Derive an expression for electric potential due to a point charge. Is it scalar or vector and why?
Answer:
Consider a point charge q placed at origin O. Potential at P has to be found out. Let the medium between charge ‘q’ and P has dielectric constant Er.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 6
Electric field at P due to charge q is
E = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r^{2}}\)
The electric field \(\vec{E}\) points away from the charge q. A force \(\vec{F}\) = -q0 \(\vec{E}\) has to be applied on the charge so that it can be brought near to q. The small work required to move the test charge q0 from P to Q through a small distance dr is given by dW = Fdr
= -q0 Edr
The total work done in moving the charge q0 from infinity to point P will be obtained by integrating the above equation as –
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 7
But electric potential is defined as work done per unit test charge.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 9
Potential at P is V = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r}\)
If medium between q and q0 is vacuum then £r = 1
Then , V =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
This is the required expression.

MP Board Solutions

MP Board Class 12th Physics Question 16.
What are the factors affecting the potential of a charged
Answer:
The factors affecting the potential of a charged conductor are:
1. Amount of charge on conductor:
By the formula V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) it is clear that
V ∝ q, hence more is the charge, more will be the potential of charged conductor.

2. Shape of conductor (Area of conductor):
If the charge is kept constant on a conductor and its surface area is decreased then the potential of conductor increases whereas on increasing the surface area its potential decreases. So the potential of a conductor is inversely proportional to the radius.

3. Presence of other conductor near the charged conductor:
If an uncharged conductor is brought near a charged conductor then the potential of the charged conductor decreases.

4. Medium surrounding the conductor:
Due to presence of insulating medium near the charged conductor its potential will decrease.

Question 17.
Define equipotential surface. Write its properties.
Answer:
Equipotential surface:
An equipotential surface is the locus of all those points at which the potential due to distribution of charge remains same.

Properties:

  • Potentials on every point are equal
  • No work is done in moving a positive charge from one point to another
  • The electrical lines of force are normal to the equipotential surface
  • Two equipotential surfaces do not intersect each other.
  • All the points on the surface of a conductor are in electric contact. If the potentials are not same then the
  • charge will flow from higher potential to lower potential till the potential of both the points become same.
  • Thus the surface of a conductor is always equipotential.

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 10

Uniform Distribution Calculator … Calculate the probability density.

Class 12th Physics Chapter 1 Important Questions MP Board Question 18.
Obtain a relation between electric Held intensity and potential difference.
Or
Prove that E = \(\frac { dv }{ dr }\)where symbols have their usual meanings.
Answer:
Suppose A and B are two points in the electric field of charge q. The direction of electric field is radially outwards from A to B. Suppose the distance between A and B is very small (i. e., dr) then the electrie field between A and B can be taken as uniform. As the potential is inversely proportional to distance hence potential at A is more than that of B. Let the potential at B is V then that at A is V + dV.

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 11
Work done in bringing the test charge q0 from B to A is –
dW = q0dV …….(1)
Force acting on q0 will be
F = q0 E
Work done in bringing the test charge against the repulsion force will be
dW = -q0Edr
(Work = Force x Displacement) ………(2)
The negative sign shows that the direction of displacement and direction of force are opposite to each other.
From eqns. (1) and (2), we get,
q0dV = q0Edr
or dV = -Edr
E = – \(\frac { dv }{ dr }\)
This is required relation between intensity of electric field and potential difference.

Question 19.
Prove that capacity of an isolated spherical conductor is directly proportional to its radius.
Or
Derive an expression for the capacity of a spherical conductor.
Answer:
The capacity of a conductor is its ability to store electrical energy and it is equal
to that charge which increases its potential by unity.
∴ Capacity = \(\frac { Charge }{ Potential }\)

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 12
Capacity of an isolated spherical conductor:
Let us consider about a spherical conductor of radius r. The charge + Q is given to it. The charge will be distributed on its surface uni- formly. Therefore the lines of force will be emitted normally to the surface seem to becoming from its center. Hence, we can suppose that all the charges are kept at the centre.
∴ Potential on the surface V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)
But capacity C = \(\frac { Q }{ V }\)
Putting the value of V, we get
C = \(\frac{Q}{\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}}\)
C = 4πE0r
C ∝ r.
Thus, the capacity is proportional to the radius of the spherical conductor.

MP Board Solutions

Question 20.
What do you mean by a capacitor ? Explain its principle.
Answer:
The capacitor is a device by which the capacitance of a conductor is increased without changing its size or volume. Actually it stores electrical energy.
Principle of capacitor:
Let A be a charged conducting plate. Another uncharged conductor plate B is brought near to A, therefore due to induction negative charges will be induced on the front surface and positive charges on the other side of plate B.
Now, the negative charge reduces the potential while the positive charge increases.

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 13
As the negative charge is nearer therefore the potential of plate A decreases. Now, the plate B is earthed then the free positive charge will go to earth and hence the potential of A decreases by more value.
C = \(\frac { Q }{V}\)
As V decreases, C will increase. This arrangement is called capacitor or condenser.

Question 21.
Derive an expression for parallel plate capacitor.
Answer;
Let A and B be two plates of a parallel plate capacitor separated by a distance d apart. Area of each plate is A and dielectric constant of the medium between them is Er Now, plate A is given + Q charge. Therefore, – Q charge will be induced on the nearer surface of the plate B and + Q charge on the other side. As B is connected to earth, + Q charge of B will go to earth. Let the charge density of A is cr, therefore that of B will be -σ.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 14
Now, σ = \(\frac { Q }{A}\)
Intensity between the plates will be given by
E = \(\frac{\sigma}{\varepsilon_{0} \varepsilon_{\mathrm{r}}}\)
E = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{r}}\)
But, potential difference between the plates A and B is
V = Electric field intensity ×
Distance between to plates = Ed
V = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{s}} \cdot d\)
But, C = \(\frac{Q}{V}=\frac{Q}{\frac{Q d}{A \varepsilon_{0} \varepsilon_{r}}}\)
C = \(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}\)
This is the required relation.
For air or vacuum, Er = 1
C = \(\frac{\varepsilon_{0} A}{d}\)

Question 22.
Three capacitors of capacitance’s C1 C2 and C3 are connected in series. Derive an expression for the equivalent capacitance.
Answer:
The given figure shows three capacitors of capacitances C1 C2 and C3 con – nected in series. A potential difference of V is applied across the combination, charges of + Q and – Q are developed on the plates of the capacitor.
Potential difference across the individual capacitors will be
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 15
V1 = \(\frac{Q}{C_{1}}\) , V2 = \(\frac{Q}{C_{2}}\), V3 = \(\frac{Q}{C_{3}}\) …….(1)
The sum of these must be equal to the applied potential difference V.
V = V1 + V2 + V3 ………(2)
Let C be the equivalent capacitance of the series combination
∴ C = \(\frac { Q }{ V }\) or V = \(\frac { Q }{ C }\) ………..(3)
V1 + V2 + V3 = \(\frac { Q }{ C }\) [from equ..(2)]
\(\frac{Q}{C_{1}}\) + \(\frac{Q}{C_{2}}\) + \(\frac{Q}{C_{3}}\) = \(\frac{Q}{C}\) [from equ..(1)]
Q(\(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)) = \(\frac{Q}{C}\)
\(\frac { 1 }{ c }\) = \(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)
This is the required expression.

MP Board Solutions

Question 23.
Three capacitors of capacitance’s C1 C2 and C3 are connected in parallel. Derive an expression for the equivalent capacitance C.
Answer:
Consider three capacitor of capacitance’s C1 C2 and C3 connected in parallel. A potential difference V is applied across the combination. Charges set up in the individual capacitor will be.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 16
Q1 = C1V, Q2 = C2V,Q3 = C3V …(1)
Total charge stored in the parallel combination is
Q = Q1 + Q2 + Q3 ……….(2)
If C is the equivalent capacitance of the combination
Then, C = \(\frac { Q }{ V }\) Q = CV …………(3)
Q1 + Q2 + Q3 = CV [from eq. (2)]
C1V + C2V + C3V = CV [from eq. (1)]
V(C1 + C2 + C3) = CV
C = C1 + C2 + C3
This is the required expression.

Question 24.
Derive an expression for the energy of a charged conductor.
Or
Prove that energy of a charge conductor is directly proportional to its square of potential.
Answer:
The work done in charging a conductor is stored as energy in it. This energy is called electrostatic potential of conductor.

Formula derivation:
Let us consider about a conductor of capacity C which is given charge +Q due to which its potential becomes V. As the charge increases work done also increases. Let at any instant the potential of conductor be V due to charge q.
∴ C = \(\frac { q }{ v}\)
or v = \(\frac { q }{ C}\)
Now, at potential Kthe work done in giving the charge dq will be dW
∴ dw = Vdq
or dw = \(\frac { q }{ C}\)dq
Work done in charging the conductor from 0 to Q
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 17
This work done is stored as potential energy on the conductor. Energy of a charge conductor
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
But Q = CV
∴ U = \(\frac{1}{2} \frac{C^{2} V^{2}}{C}\)
or U = \(\frac{1}{2} C V^{2}\)
∴ U ∝ V2 because C is constant

MP Board Solutions

Question 25.
Prove that on connecting two charged conductors, charges distribute on them according to their capacities.
Answer:
When two isolated charged conductors A and B are connected by a thin wire, charge flows from the conductors at high potential to the conductor at low potential till the potential of both A and B became equal. The phenomenon involved is called distribution of charges and the total charge of the entire system remains conserved. Let the capacitance of A and B be C1 and C2, the charges be Q1 and Q2 respectively. Then the potentials are V1 and V2 respectively.
∴ Initially, Q1 = C1V1 and Q2 = C2V2
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 18
The conductors are joined by a wire of negligible capacitance, the charges flow from- the conductor at higher potential to the conductor at lower potential till the potentials on each conductor become equal.
The net charge on the system,
Q = Q1 + Q2
The common potential, V = \(\frac { Total charge }{ Total capacitance }\)
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 19
After the potential becomes equal let the charge on A 1 be Q1 and charge on A2 be Q2.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 20
Dividing eqn. (2) by eqn. (3), we get
MP BoardMP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 21 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 21
When the conductors are joined, then the charges get distributed in the ratio of their capacities.

Question 26.
Obtain an expression for potential due to a group of point charges.
Or
Derive the expression for potential energy.
Answer:
Consider a group of point charges q1,q2,q3……..qn which are situated at a dis-tance of r1, r2, r3…….. nn respectively from the point P. The potential due to these point charges is to be obtained at P. Now potential at P due to q1 is
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 22
potential due to q2,q3, ……… qnetc
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 24
Total potential at P will be V = V1 + V2 + V3 + ……….. + Vn
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 25
This is the required expression.

Electrostatic Potential and Capacitance Important Questions Long Answer Type Questions

Question 1.
Derive the expression for the capacity of a parallel plate capacitor, when the medium between the plates is partially filled by a dielectric medium.
Answer:
Let A and B are parallel plates of a capacitor. The distance between the plates is d and plate of thickness t and dielectric constant Er is introduced.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 26
Now, plate A is given charge +Q.
Let the charge density be σ.
∴ σ = \(\frac { Q}{ A}\)
Intensity of field in air ,
E_{0}=\(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)
If the intensity of field inside the dielectric medium be E, then
Dielectric constant = \(\frac {Electric field in vacuum}{ Electric field in medium}\)
or \(\varepsilon_{r}=\frac{E_{0}}{E}\)
or E = \(\frac{E_{0}}{\varepsilon_{r}}=\frac{Q}{\varepsilon_{0} \varepsilon_{r} A}\)
Now, potential difference between A and B,
V=E0 (d – t) + Et, [(d – t) is vacuum distance]
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 27
This is the required expression.
Metal is a conductor. When metal is used in place of the dielectric, it will conduct electricity and the potential difference will become zero. So, capacitor will not work.

Question 2.
Calculate the loss of energy, when two charged conductors are connected.
Or
The capacities of two conductors are C1 and C2, Q1 and Q2 charges are given to them so that their potentials become V1 and V2 respectively. If they are connected by a wire, then calculate the following:

  • Common potential
  • Loss of energy.

or
prove that when two charged conductors are connected, there will be a loss of energy
Or
In redistribution of charges, is there a loss of energy ? Deduce an expression to confirm the answer.
Answer:
Let A and B be two conductors of capacities C1 and C2 respectively. When charges Q1 and Q2 are given separately the potentials become V1 and V2 respectively.Total charges, Q = Q1 + Q2 ………..(1)
But, Q1 = C1V1 and Q2 = C2V2
By eqn. (1), we get
Q = C1V1 + C2V2
Total capacity, C = C1 + C2
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 28
(1) Common potential:
Let the conductors are connected by a wire and the common potential becomes-V.
Q1 + Q2 = (C1 + C2)V
V = \(\frac{Q_{1}+Q_{2}}{C_{1} + C_{2}}\)
V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\) ……..(2)
This is the expression for the common potential.

(2) Loss of energy: Total energy of the conductors before connection:
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 29
and total energy after connection,
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 30
Putting the value of V from eqn. (2) in eqn. (4), we get
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 31
Hence, difference of energy’,
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 32
(V1 – V2)2 is positive, hence (V1 – V2) is positive. Hence, during redistribution, there will be always loss of energy.
i.e., U1 – U2>0 ⇒ U1>U2
i.e., energy before joining is greater than energy after joining.
The loss in energy,
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 34

Question 3.
Explain the construction and working of Van de Graaff generator. Write its uses.
Answer:
Van de Graaff generator is a machine which produces electricity of about 107 V or more potential difference.
Construction:
It consists of a large metallic sphere S of diameter 5 m, mounted on high insulating support PP about 15 m high. An endless insulating belt made up of rubber passes over the pulleys p1 and P2. A motor rotates p1 C1 and C2 are two metallic combs called spray comb and collecting comb respectively. C1 is connected to S. To prevent the leakage of charge, the generator is put inside a large enclosure filled with gas at 15 atm. pressure. This iron enclosure is connected to earth.

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 35
Working:
The comb C1 is connected to the positive terminal of the battery, therefore the surface density of the points becomes very high which causes the wind present nearby it to get charged. Thus, the spray comb sprays the charge on the belt. Now, the electric wind moves up to the collector comb C2 When it reaches in front of the collector comb C2 opposite charge induces on the tip to neutralize the same type of charge. The negative charge wind of C2, cancels the positive charge of the belt. Thus, by the repeated actions more and more positive charge is induced on sphere, hence its potential increases to about 107volts or more.

Uses:

  • To generate high potential.
  • To accelerate the positive particles such as protons, Deuteronomy, are particle etc. and used in nuclear disintegration.

MP Board Solutions

Question 4.
When Anil opened the cap of the tap, then he found no water in coming out of it. Then he opened the cap of the water tank and found no water in the tank. To fill up water in the water tank he switch on the switch of the motor and found motor is not starting. Then he called the electric technician. The technician said him on checking that the condenser of the motor is not functioning. On replacing capacitor, the motor start working.

Answer the following questions:

  1. What values does Anil exhibits ?
  2. What is the function of condenser ?
  3. What is total charge on a charged condenser ?
  4. The capacity of a capacitor is 3pF. If it is charged up to 100 V potential difference, then what will be charged stored in it ?

Answer:

  1. Anil exhibited his presence of mind.
  2. It accumulate charge and hence it conserved energy.
  3. Net charge on a condenser is zero.
  4. C = 3µF = 3 × l0-6 F, V = 100V

∴ By formula Q = CV = 3 × l0-6 × l00
or Q = 3 × 10-4C.

Electrostatic Potential and Capacitance Important Questions Numerical Questions

Question 1.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
No.
As we know that the formula of potential is V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) …(1)
Given, q = 1C, r = 1cm = 10-2m
Putting these values in eqn. (1)
V = 9 × 109 x \(\frac{1}{10^{-2}}\) = 9 × 1011volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 109 in SI unit.
This value of potential is greater than barrier potential of air. Therefore lC charge cannot be given to a sphere of radius 1cm.

Question 2.
You are given three capacitor of 4pF each. How they will be combined to obtain resultant capacity of 6pF ?
Solution:
Given : Q = C2 = C3 = 4µF
When two capacitor is joined in series and third capacitor joined parallel with them, then resultant capacity is obtained as 6µF.

MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 36
C1 ,C2 is in series, its resultant (C’) is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
or \(\frac { 1 }{ c}\) = \(\frac { 1 }{ 4}\) + \(\frac { 1 }{ 4}\) = \(\frac { 2 }{ 4}\)
or C = \(\frac { 4 }{ 2}\) = 2µF.
C and C3 is in parallel combination,
Its resultant C is C = C + C3
C = 2 + 4 or C = 6µF.

MP Board Solutions

Question 3.
A hollow metallic sphere of radius 0-1 m is given 6pC. Calculate its potential:

  1. At the surface of sphere
  2. At the center.

Solution
Given, r = 01 m, q = 6µC = 6 x 10-6 C
Formula: V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)

1. Potential at the surface:
V = 9 × 109 × \(\frac{6 \times 10^{-6}}{0 \cdot 1}\)
V = 54 × 104
V- 5.4 × 105 volt.

2. At the center:
Inside the sphere the potential remains same and equal to that on the surface hence V = 5 . 4 x 105volt.

Question 4.
A test charge is moved from A to B, B to C and A to C in an electric field E as shown in the figure :
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 37
Find (1) Potential difference between A and C
(2) At which point electric potential will be high and why ?
Solution:
1. In right angled ∆ABC
AB2 = AC2 – BC2 = 52 – 32
∴ AB = 4 = dr
BC is perpendicular to electric field, therefore potential will be same at B and C.
VA – VC = VA – VB = -Edr = -4E
2. Therefore potential at point C will be more than potential of point A.

MP Board Solutions

Question 5.
Identical water droplets, having equal charge on each are combined to form a big drop. Compare the capacity of bigger drop with that of a small drop.
Solution:
Let radius of small droplet = r
Radius of the big drop = R
Volume of big drop = Volume of 27 droplets
\(\frac { 4 }{ 3 }\) πR3 = 27 × \(\frac { 4 }{ 3 }\) πR3
or R3 = 27r3
or R3 = (3r)3
or R = 3r
or \(\frac { R }{ r }\) = \(\frac { 3}{ 1 }\)
Since, C ∝ radius
or \(\frac{C_{1}}{C_{2}}=\frac{r_{1}}{r_{2}}\)
or \(\frac{C_{1}}{C_{2}}\) = \(\frac { 3}{ 1}\) = 3
or C1 = 3C2
The capacity of bigger drop is three times that of smaller one.

Question 6.
How three capacitor of 3pF each can be combined such that their resultant capacity is :

  1. 9pF,
  2. 4.5pF.

Solution:
1. When the three capacitor is joined in parallel, then
C = C1 + C2 + C3
= 3 + 3 + 3 = 9µF.

2. When two capacitor are joint in series, then resultant C’ is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
= \(\frac { 1 }{ 3}\) + \(\frac { 1 }{ 3}\) = \(\frac { 2 }{ 3}\)
C = \(\frac { 3 }{2}\) = 1.5 µF
Now C is joined in parallel with C3
C = C + C3 = 1.5 + 3 = 4.5µF.

Question 7.
The potential difference between two points is 10V. How much work is required to move a charge 100 pC from a point to the other ?
Solution:
Given, V= 10 volt, q = l00µC = l00 × l0-6C
Formula : w = qV
= 100 × 10-6 × 10
= 10-3 joule.

MP Board Solutions

Question 8.
Find the area of the plate of a 2F parallel plate capacitor, if the separation between the plates is 0.5 cm ?
Solution:
As C = \(\frac{\varepsilon_{0} A}{d}\)
A = \(\frac{C d}{\varepsilon_{0}}\)
Here, C = 2F, d= 0.5cm = 0.5 x 10-2m
A = \(\frac{2 \times 0 \cdot 5 \times 10^{-2}}{8.85 \times 10^{-12}}\)
= 1.13 × 109 m2 = 1130Km2

Engineering Physics MCQ Electrical Engineering.

Question 9.
Two charges 5 x 10-8C and -3 x 10-8C are located 16 cm apart At what point, on the line joining the two charges, is the electric potential zero ? Take the potential at infinity to be zero. (NCERT)
Solution:
Case I.
Let electric potential be zero at point C lying at distance x from the positive charge.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 38
Given, q1 = 5 × 10-8 C;
q2 = -3 × l0-8 C
AC = x cm : CB = (16 – x) cm
Now, Potential at C is zero i. e.,
V1 + V2 = 0
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 39
-8x + 80 = 0
8x = 80
x = 19 cm
i.e., electric potential at a distance of 10 cm from positive charge will be zero.

Case II.
The other possibility is that the point C may also lie on produced AB.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 40
Now, V1 + V2 = 0
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 41
5(x – 16) – 3x = 0
5x – 80 – 3x = 0
2x – 80 = 0
x = 40 cm from the positive charge

Question 10.
Determine the equivalent capacitance between A and B in the following circuits:
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 42
Solution:
(i). Mark the junctions as C and D.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 43
(But C will be A and D will be B)
Draw the equivalent network, which is given below
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 44
Equivalent capacitance,
C = C1 + C2 + C3
or C = 1 + 1 + 1 = 3µF

(ii). To move from A to B, there are two paths P -1 and P – II.
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 45
(As A and B, the path is dislocated temporarily)
The capacitors in P -II are in series. So, the equivalent becomes
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 46
The resultant capacity of series combination is
\(\frac { 1 }{ C’ }\) = \(\frac { 1}{ 3 }\) + \(\frac { 1 }{ 3}\) + \(\frac { 1 }{3 }\)
= \(\frac { 3 }{ 3 }\) = 1µF = C’ = 1µF
The equivalent further becomes
MP Board 12th Physics Chapter 2 Electrostatic Potential and Capacitance Important Questions - 47
Total capacity C = 3 +1 = 4µF.

MP Board Class 12th Physics Important Questions

MP Board Class 8th Sanskrit Solutions Chapter 18 सत्कर्म एव धर्मः

In this article, we will share MP Board Class 10th Sanskrit Solutions Chapter 13 महाभारते विज्ञानम् Pdf, Surbhi Class 8, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Sanskrit Solutions Surbhi Chapter 18 सत्कर्म एव धर्मः

MP Board Class 8th Sanskrit Chapter 18 अभ्यासः

Class 8 Sanskrit Chapter 18 MP Board प्रश्न 1.
एकपदेन उत्तरं लिखत(एक शब्द में उत्तर लिखो-)
(क) विक्रमादित्यः नगरभ्रमणसमये किं दृष्टवान्? (विक्रमादित्य ने नगर भ्रमण के समय क्या देखा?)
उत्तर:
रुग्णम्। (रोगी को)

(ख) विक्रमादित्यः महामन्त्रिणि राज्यभारं समर्प्य कुत्र अगच्छत्? (विक्रमादित्य महामन्त्री पर राज्यभार को समर्पित करके कहाँ गये?)
उत्तर:
वनम्। (वन में)

(ग) महात्मा कस्य समीपे तपस्यारतः आसीत्? (महात्मा किसके पास तपस्यारत थे?)
उत्तर:
विक्रमादित्यस्य। (विक्रमादित्य के)

(घ) महात्मा योगबलेन तत्र कस्य दृश्यं दर्शितवान्? (महात्मा ने योग के बल से वहाँ किसका दृश्य दिखाया?)
उत्तर:
यमलोकस्य। (यमलोक का)

(ङ) कर्मणां लेखनं कस्य पार्वे अस्ति? (कर्मों का लेखा किसके पास है?)
उत्तर:
चित्रगुप्तस्य। (चित्रगुप्त के)

(च) श्रेष्ठाचरणस्य प्रतिज्ञां कृत्वा विक्रमादित्यः कुत्रः आगतवान्? (श्रेष्ठ आचरण की प्रतिज्ञा करके विक्रमादित्य कहाँ आये?)
उत्तर:
उज्जयिनीम्। (उज्जयिनी में)

Class 8 Sanskrit Chapter 18 प्रश्न 2.
एकवाक्येन उत्तरं लिखत(एक वाक्य में उत्तर लिखो-)
(क) महात्मा विक्रमादित्यं किम् उपदिष्टवान्? (महात्मा ने विक्रमादित्य को क्या उपदेश दिया?)
उत्तर:
महात्मा विक्रमादित्यं उपदिष्टवान् यत्-“राजन्! भवान् तु यथानीति राजधर्मस्य पालनं करोतु। धर्मानुकूलं शासनम् अपि धर्म एव भवति” इति। (महात्मा ने विक्रमादित्य को उपदेश दिया कि-“हे राजन्! आप तो नीति के अनुसार राजधर्म का पालन करो। धर्म के अनुसार शासन भी धर्म ही होता है।)

(ख) कर्म-तपस्योः कः भेदः? (कर्म और तपस्या में क्या भेद है?)
उत्तर:
कर्मणः स्थानम् भिन्नम् परं तपस्या स्वर्गप्राप्तेः साधनम्।” इति कर्म-तपस्ययोः भेदः। (“कर्म का स्थान भिन्न है परन्तु तपस्या स्वर्ग प्राप्ति का साधन है।” ऐसा कर्म और तपस्या का भेद है।)

(ग) विक्रमादित्यः अन्ते किं ज्ञातवान्? (विक्रमादित्य ने अन्त में क्या जाना?)
उत्तर:
विक्रमादित्यः अन्ते ज्ञातवान् यद्-‘सदाचारः एव तपस्या, सत्कर्म एव धर्म’ इति। (विक्रमादित्य ने अन्त में यह जाना कि-“सदाचार ही तपस्या है, सत्कर्म ही धर्म है”।)

(घ) विक्रमादित्यः लोके कथं प्रसिद्धः? (विक्रमादित्य संसार में कैसे प्रसिद्ध हैं?)
उत्तर:
विक्रमादित्यः लोके सत्कर्मणा एव प्रसिद्धः। (विक्रमादित्य संसार में सत्कर्म से ही प्रसिद्धः हैं।)

(ङ) विक्रमादित्यस्य ध्येयवाक्यं किम् आसीत्? (विक्रमादित्य का ध्येय वाक्य क्या था?)
उत्तर:
विक्रमादित्यस्य ध्येयवाक्यं ‘सत्कर्म एव धर्मः’ इति आसीत्। (विक्रमादित्य का ध्येय वाक्य ‘सत्कर्म ही धर्म है’ यह था।)

(च) विक्रमादित्यस्य मनसि केन वैराग्यम् उद्भूतम्? (विक्रमादित्य के मन में किससे वैराग्य उत्पन्न हुआ?)
उत्तर:
विक्रमादित्यस्य मनसि रुग्णदर्शनेन वैराग्यम् उद्भूतम्। (विक्रमादित्य के मन में रोगी को देखने से वैराग्य उत्पन्न हुआ।)

(छ) “सत्कर्म एव धर्मः” कथायाः सारः कः? (“सत्कर्म ही धर्म है” कथा का सारांश क्या है?)
उत्तर:
अस्याः कथायाः सारः यत् ‘मनुष्यः उत्तमानि कर्माणि कृत्वा अपि परलोकं साधयितुम् शक्नोति’ इति। (इस कथा का सारांश है कि ‘मनुष्य अच्छे कार्य करके भी परलोक सिद्ध कर सकता है।)

MP Board Class 8 Sanskrit Chapter 18 प्रश्न 3.
रिक्तस्थानानि पूरयत(रिक्त स्थान भरो-)
(क) विक्रमादित्यस्य समीपे एकः ……….. तपस्यारतः आसीत्। (महात्मा/दुरात्मा)
(ख) तपस्या ……… साधनम्। (स्वर्गप्राप्ते/राज्यप्राप्तेः)
(ग)यमराजः …………. आदिष्टवान्। (विक्रमादित्यम्/दूतेभ्यः)
(घ) वने सः कठिनां …………. आरब्धवान्। (दिनचर्यां/तपस्याम्)
(ङ) प्रजापालनं धर्म …….. अस्ति। (तपस्वीनाम्/राज्ञाम्)
उत्तर:
(क) महात्मा
(ख) स्वर्गप्राप्तेः
(ग) दूतेभ्यः
(घ) तपस्याम्
(ङ) राज्ञाम्।।

Class 8 Sanskrit Chapter 18 Question Answer प्रश्न 4.
उचितं मेलयत(उचित को मिलाओ-)
MP Board Class 8th Sanskrit Solutions Chapter 18 सत्कर्म एव धर्मः 1
उत्तर:
(क) → (v)
(ख) → (iv)
(ग) → (ii)
(घ) → (iii)
(ङ) → (i)

MP Board Class 8 Sanskrit प्रश्न 5.
नामोल्लेखपूर्वकं समासविग्रहं कुरुत(नाम का उल्लेख करते हुए समास विग्रह करो-)
(क) ध्येयवाक्यम्
(ख) तपस्यारतः
(ग) प्रजापालनम्
(घ) योगबलेन
(ङ) श्रेष्ठाचरणेन।
उत्तर:
MP Board Class 8th Sanskrit Solutions Chapter 18 सत्कर्म एव धर्मः 2

Class 8 Sanskrit Chapter 17 MP Board प्रश्न 6.
नामोल्लेखपूर्वकं सन्धिविच्छेदं कुरुत(नामे का उल्लेख करते हुए सन्धि विच्छेद करो-)
(क) धर्मानुकूलम्
(ख) नेति
(ग) सदाचारः
(घ) अद्यापि
(ङ) विक्रमादित्यः।
उत्तर:
MP Board Class 8th Sanskrit Solutions Chapter 18 सत्कर्म एव धर्मः 3

Class 7 Sanskrit Chapter 18 प्रश्न 7.
रेखाङ्कितशब्दानाम् आधारेण प्रश्ननिर्माणं कुरुत (रेखांकित शब्दों के आधार पर प्रश्न निर्माण करो-)
(क) विक्रमः अवदत्। (विक्रम बोला।)
उत्तर:
कः अवदत्? (कौन बोला?)

(ख) तपस्या तु महात्मनां कर्म इति। (तपस्या तो महात्माओं का काम है।)
उत्तर:
तपस्या तु केषाम् कर्म इति? (तपस्या तो किनका काम है?

(ग) तत्रैव यमलोकस्य दृश्यं दर्शितवान्। (वहीं यमलोक का दृश्य दिखाया?)
उत्तर:
कुत्र यमलोकस्य दृश्यं दर्शितवान्? (कहाँ यमलोक का दृश्य दिखाया?)

(घ) यमराजः दूतान् पृच्छति? (यमराज दूतों से पूछते हैं।)
उत्तर:
यमराजः कान् पृच्छति? (यमराज ने किनसे पूछा?)

(ङ) राज्ञः धर्म प्रजापालनम्। (राजा का धर्म प्रजा का पालन है।)
उत्तर:
कस्य धर्मः प्रजापालनम्? (किसका धर्म प्रजा का पालन है?)

सत्कर्म एव धर्मः हिन्दी अनुवाद

एकदा विक्रमादित्यः नगरभ्रमणसमये एकम् मरणासन्नं रुग्णं दृष्टवान्। तस्य दर्शनेन मनसि उद्भूतम्। अतः मायामोहमयं संसारं ज्ञात्वा सः महामन्त्रिणि राज्यभारं समर्प्य वनम् अगच्छत्।

अनुवाद :
एक बार विक्रमादित्य ने नगर में भ्रमण के समय एक मरणासन्न (मरने के निकट) रोगी को देखा। उसको देखने से मन में वैराग्य उत्पन्न हुआ। इसलिए माया मोह से भरे संसार को जानकर वह महामन्त्री को राज्यभार सौंपकर वन चले गये।

वने सः कठिना तपस्याम् आरब्धवान्। तस्य समीपे एव एकः महात्मा अपि तपस्यारतः आसीत्। महात्मा तम् अवदत्, “राजन्! भवान् तु यथानीति राजधर्मस्य पालनं करोतु। धर्मानुकूलं शासनम् अपि धर्म एव भवति” इति स महात्मा विक्रमादित्यम् उपदिष्टवान्। विक्रमः अवदत्-“नहि महात्मन्! तपसा एव परलोकः साध्यते, कर्मणा नेति।” महात्मा अवदत्, “राजन! राज्ञः धर्मः प्रजापालनं, शासनम् एव अस्ति तपस्या तु महात्मानां कर्म इति।” इत्युक्त्वा महात्मा राजानाम् पृष्टवन्-कर्म-तपस्ययोः कः भेदः?

अनुवाद :
वन में उन्होंने कठिन तपस्या आरम्भ कर दी। उनके पास में एक महात्मा भी तपस्यारत (तपस्या में लगे) थे। महात्मा ने उनसे कहा, “हे राजन्! आप तो नीति के अनुसार राजधर्म का पालन करो। धर्म के अनुकूल शासन भी धर्म ही होता है।” इस प्रकार उस महात्मा ने विक्रमादित्य को उपदेश दिया। विक्रम ने कहा-“नहीं महात्मन्! तपस्या से ही परलोक प्राप्त होता है, कर्म से नहीं।” महात्मा ने कहा, “हे राजन्! राजा का धर्म प्रजा का पालन और शासन ही है, तपस्या तो महात्माओं का काम है।” ऐसा कहकर महात्मा ने राजा से पूछा-कर्म और तपस्या में क्या भेद है?

राजा अवदत्-कर्मणः स्थानम् भिन्नम् परं तपस्या स्वर्गप्राप्तेः साधनम्। एतच्छ्रुत्वा महात्मा हसन् अवदत्-“राजन् ! मनुष्यः उत्तमानि कर्माणि कृत्वा अपि परलोकं साधयितुम् शक्नोति।” अनन्तरम् महात्मा योगबलेन तत्रैव यमलोकस्य दृश्यं विक्रमं दर्शितवान्। दृश्ये यमराजः दूतान् पृच्छति-“एतस्य कर्म कीदृशम् ?” एकः दूतः अवदत्-“कर्मणां लेखनं तु चित्रगुप्तस्य पार्वे अस्ति।” क्षणं विचार्य यमराजः दूतान् आदिष्टवान् यत्-“यदि एतस्य कर्माणि उत्तमानि सन्ति तर्हि स्वर्गस्य द्वारम् उद्घाटयतु यदि कर्माणि अधमानि, तदा बलात् नरके पातयतु।”

अनुवाद :
राजा ने कहा-कर्म का स्थान भिन्न है परन्तु तपस्या स्वर्ग प्राप्ति का साधन है। ऐसा सुनकर महात्मा हँसते हुए बोले-“हे राजन्! मनुष्य अच्छे कर्म करके भी परलोक सिद्ध कर सकता है।” इसके बाद महात्मा ने योग के बल से वहीं यमलोक का दृश्य विक्रम को दिखाया। दृश्य में यमराज दूतों से पूछ रहे हैं-“इसका कर्म कैसा है? एक दूत ने कहा-“कर्मों का लेखा तो चित्रगुप्त के पास है।” कुछ देर विचार करके यमराज ने दूतों को आदेश दिया कि-यदि इसके कर्म अच्छे हैं तो स्वर्ग का द्वार खोल दो यदि कर्म बुरे हैं तो जबरदस्ती नरक में डाल दो।”

इदं दृश्यं दृष्ट्वा विक्रमः ज्ञातवान् यद्-‘सदाचारः एव तपस्या, सत्कर्म एव धर्म’ इति। अनन्तरं सः तम् महात्मानम् प्रणम्य, श्रेष्ठाचरणस्य प्रतिज्ञां कृत्वा राजधानीम् उज्जयिनीम् आगतवान्। आगत्य धर्मानुकूलंनीतिपूर्वकम् प्रजापालनपुरस्सरं शासनं कृतवान्। सः अद्यापि लोके सत्कर्मणा एव प्रसिद्धः। तस्य जीवनस्य ध्येयवाक्यम् आसीत्-‘सत्कर्म एव धर्मः।’

अनुवाद :
इस दृश्य को देखकर विक्रम जान गये कि-“सदाचार (अच्छा व्यवहार) ही तपस्या है और सत्कर्म (अच्छे कर्म) ही धर्म।” इसके बाद वह उन महात्मा को प्रणाम करके, श्रेष्ठ आचरण की प्रतिज्ञा करके राजधानी उज्जयिनी आ गये। आकर धर्म के अनुसार नीतिपूर्वक प्रजा पालन को प्रमुखता देते हुए शासन किया। वह आज भी संसार में अच्छे कर्म से ही प्रसिद्ध हैं। उनके जीवन का ध्येय वाक्य था-‘सत्कर्म ही धर्म है।’

सत्कर्म एव धर्मः शब्दार्थाः

बलात् = बलपूर्वक। साधयितुम् = सिद्ध करने लिए। पार्वे = पास में। प्रजापालनपुरस्सरम् = प्रजापालन को प्रमुखता देते हुए। मरणासन्न = मरने के निकट। उद्भूतम् = उत्पन्न हुआ। पातयतु = गिराओ। एतच्छ्रुत्वा = ऐसा सुनकर।