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MP Board Class 7th Maths Solutions Chapter 12 बीजीय व्यंजक Ex 12.2

प्रश्न 1.
समान पदों को संयोजित (मिलान) करके सरल कीजिए:
(i) 21b – 32 + 7b – 20b
(ii) -z² + 13z² – 5z + 7z³ – 15z
(iii)p – (p – q) – q – (q – p)
(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
हल:
(i) 21b – 32 + 7b – 20b
= 21b + 7b – 20b – 32
= (21 + 7 – 20) b – 32
= 8b – 32

(ii) – z² + 13z² – 5z + 7z³ – 15z
= 7z³ – z² + 13z² – 5z – 15z
= 7z³ + (-1 + 13)z² + (-5 -15) z
= 7z³ + 12z³ – 20z

(iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
= p – p + p + q – q – q
= (1 – 1 + 1)p + (1 – 1 – 1)q
= p – q

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= (3a – a – a) + (-2b + b + b) + (-ab -ab + 3ab)
= (3 – 1 – 1)a + (-2 + 1 + 1) b + (- 1 – 1 + 3)ab
= (1)a + (0) b + (1) ab = a + ab

(v) 5x² y – 5x² + 3yx² – 3y² + x² – y² + 8xy²
= (5x²y + 3yx²) + (8xy²) + (-5x² + x²) + (-3y² – y² – 3y²)
= (5 + 3) x²y + 8xy² + (- 5 + 1)x² + (- 3 – 1 – 3) y²
= 8x²y + 8xy² – 4x² – 7y²

(vi) (3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4
= (3y² + y²) + (5y – 8y) + (- 4 + 4)
= (3 + 1) y² + (5 – 8) y + (- 4 + 4)
= 4y² – 3y + 0 = 4y² – 3y

प्रश्न 2.
जोड़िए:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz – z, z – 1
(iii) -7mn + 5,12mn + 2, 9mn – 8, -2 mn – 3
(iv) a + b – 3,b – a + 3,a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy,4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, – 3xy², -5xy², 5x²y
(viii) 3p²q² – 4pq + 5, – 10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1 – y² – 1 – x², 1 – x² – y²
हल:
(i) अभीष्ट योग
= 3mn + (-5mn) + 8mn + (-4mn)
= [3+ (-5) + 8 + (-4)] mn
= [11 – 9] mn = 2mn

(ii) अभीष्ट योग
= (1 – 8tz) + (3tz – z) + (z – t)
= t – 8tz + 3tz – z + z – t
= (t – t) + (-z + z) + (-8tz + 3tz)
= (1 – 1) t + (- 1 + 1) z + (- 8 + 3) tz
= (0) t + (0) z + (-5)tz = – 5tz

(iii) अभीष्ट योग = (-7mn + 5) + (12mn + 2) + (9mn – 8) + (-2mn – 3)
= -7mn + 5 + 12mn + 2 + 9mn – 8 – 2mm – 3
= (-7mm + 12mn + 9mn – 2mn) + (5 + 2 – 8 – 3)
= (-7 + 12 + 9 – 2) mn + (7 – 11)
= (21 – 9) mn + (-4) = 12mn – 4

(iv) अभीष्ट योग = (a + b – 3) + (b – a + 3) + (a – b + 3)
= a + b – 3 + b – a + 3 + a – b + 3
= ( a – a + a) + (b + b – b) + ( – 3 + 3 + 3)
= (1 – 1 + 1)a + (1 + 1 – 1) b + (- 3 + 6)
= (1) a + (1) b + (3)
= a + b + 3

(v) अभीष्ट योग
= (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy
= 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
= (14x – 7x) + (10y – 10y) + (- 12xy + 8xy + 4xy) + (- 13 + 18)
= (14 – 7) x + (10 – 10)y (- 12 + 8 + 4) xy + (5)
= (7)x + (0)y + (- 12 + 12) xy + (5)
= 7x + (0) y + (0) xy + 5
= 7x + 5.

(vi) अभीष्ट योग
= (5m – 7n) + (3n – 4m + 2) + (2m – 3mn – 5)
= 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= (5m – 4m + 2m) + (-7n + 3n) -3mn + (2 – 5)
= (5 – 4 + 2) m + (- 7 + 3) n – 3 mn – 3
= 3m – 4n – 3mn – 3

(vii) अभीष्ट योग
= 4x²y + (-3xy²) + (-5xy²) + 5x²y
= 4x²y – 3xy – 5xy² + 5x²y
= (4 + 5) x²y + (- 3 – 5) xy²
= 9x²y – 8xy²

(viii) अभीष्ट योग
= (3p²q² – 4pq + 5) + (-10 p²q²) + (15 + 9pq + 7p²q²)
= 3p²q² – 4pq + 5 – 10p²q² + 15 + 9pq + 7p²q²
= 3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5 + 15
= (3 – 10 + 7)p²q² + (- 4 + 9) pq + (5 + 15)
= (0)p²q² + 5pq + 20
= 5pq + 20

(ix) अभीष्ट योग
= (ab – 4a) + (4b – ab) + (4a – 4b)
= ab – 4a + 4b – ab + 4a – 4b
= ab – ab – 4a + 4a + 4b – 4b
= (0) ab + (0) a + (0) b
= 0 + 0 + 0 = 0

(x) अभीष्ट योग
= (x² – y² – 1) + (y² – 1 – x²) + (1 – x² – y²)
= x² – y² – 1 + y² – 1 – x + 1 – x² – y²
= x² – x² – x² – y² + y² – y² – 1 – 1 + 1
= (1 – 1 – 1) x² + (- 1 + 1 – 1) y² + (- 1 – 1 + 1)
= – x² – y² – 1

प्रश्न 3.
घटाइए:
(i) y² में से – 5y²
(ii) – 12xy में से 6xy
(iii) (a + b) में से (a – b)
(iv) b (5 – a) में से a (b – 5)
(v) 4m² – 3mn + 8 में से – m² + 5mn
(vi) 5x – 10 में से – x² + 10x – 5
(vii) 3ab – 2a² – 2b² में से 5a² – 7ab + 5b²
(viii) 5p² + 3q² – Pq में से 4pq – 5q² – 3p²
हल:
(i) अभीष्ट अन्तर
y2 – (-5y2)
= y2 + 5 = 6y2

(ii) अभीष्ट अन्तर = – 12xy – 6xy = -18xy

(iii) अभीष्ट अन्तर
= (a + b) – (a – b) = a + b – a + b
= (1 – 1)a + (1 + 1) b = 2b

(iv) अभीष्ट अन्तर
= b (5 – a) – a (b – 5)
= 5b – ab – ab + 5a
= 5a + 5b + ( – 1 – 1) ab
= 5a + 5b – 2ab

(v) अभीष्ट अन्तर
= (4m² – 3mn + 8) – (- m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= (4 + 1)m + (- 3 – 5)mn + 8
= 5m² – 8mn + 8

(vi) अभीष्ट अन्तर
= (5x – 10) – (- x² + 10x – 5)
= 5x – 10 + x² – 10 x + 5
= x² + (5 – 10) x + (- 10 + 5)
= x² – 5x – 5

(vii) अभीष्ट अन्तर
= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= -2a² – 5a² – 2b² – 5b² + 3ab + 7ab
=(- 2 – 5) a² + (- 2 – 5) b² + (3 + 7) ab
= -7a – 7b² + 10ab

(viii) अभीष्ट अन्तर
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – pq – 4pq
= (5 + 3)p² + (3 + 5) q² + (- 1 – 4)pq
= 8p² + 8q² – 5pq

प्रश्न 4.
(a) 2x² + 3xy प्राप्त करने के लिए x² + xy + ya में क्या जोड़ना चाहिए?
(b) -3a + 7b + 16 प्राप्त करने के लिए 2a + 8b + 10 में से क्या घटाना चाहिए ?
हल:
(a) अभीष्ट व्यंजक
= (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= 2x² – x² – y² + 3xy – xy
= (2 – 1) x² – y² + (3 – 1)xy
= x² – y² + 2xy

(b) अभीष्ट व्यंजक
= (2a + 8b + 10) – (- 3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= (2 + 3)a + (8 – 7)b + (10 – 16)
= 5a + b – 6

प्रश्न 5.
-x² – y² + 6xy + 20 प्राप्त करने के लिए, 3x² – 4y² + 5xy + 20 में से क्या निकाल लेना चाहिए?
हल:
अभीष्ट व्यंजक = (3x² – 4y² + 5xy + 20) -(- x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
= (3 + 1)x² + (- 4 + 1) y² + (5 – 6) xy + (20 – 20)
= 4x² – 3y² – xy + 0
= 4x² – 3y² – xy

प्रश्न 6.
(a) 3x – y + 11 और – y – 11 के योग में से 3x – y – 11 को घटाइए।
(b) 4 + 3x और 5 – 4x + 2x² के योग में से 3x² – 5x और – x² + 2x + 5 के योग को घटाइए।
हल:
(a) 3x – y + 11 और – y – 11 का योग
= (3x – y + 11) + (- y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
अब 3x – 2y में से 3x – y – 11 को घटाने पर,
अभीष्ट अन्तर = (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= (3x – 3x) + (-2y + y) + 11
= 0 – y + 11
= – y + 11

(b) 4 + 3x और 5 – 4x + 2x² का योग
= 4 + 3x + 5 – 4 x + 2x²
= (4 + 5) + (3x – 4x) + 2x²
= 9 – x + 2x²
3x² – 5x और = x² + 2x + 5 का योग
= 3x² – 5x – x² + 2x + 5
= 3x² – x² – 5x + 2x + 5
= (3 – 1)x² + (- 5 + 2)x + 5
= 2x² – 3x + 5
अब, प्रश्नानुसार
अभीष्ट अन्तर = (9 – x + 2x²) – (2x² – 3x + 5)
= 9 – x + 2x² – 2x² + 3x – 5
= (9 – 5) + (- x + 3x) + (2x² – 2x²)
= 4 + (- 1 + 3)x + (2 – 2) x²
= 4 + (2)x + (0) x²
= 4 + 2x ⇒ 2x +4
पाठ्य-पुस्तक पृष्ठ संख्या # 258

MP Board Class 7th Maths Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 11 Economic Development Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 11 Economic Development

MP Board Class 8th Social Science Chapter 11 Text Book Exercise

Choose the most suitable alternative:

Question 1.
In India the total percent of working people in agriculture is:
(a) 50 percent
(b) 60 percent
(c) 70 percent
(d) None of the above
Answer:
(b) 60 percent

Question 2.
The total contribution of agriculture in. Indian G.D.P. is:
(a) 26 percent
(b) 36 percent
(c) 42 percent
(d) 100 percent
Answer:
(a) 26 percent

Question 3.
The industrial development:
(a) lessens dependence on agriculture
(b) improves life style
(c) Both of the above
(d) None of the above
Answer:
(b) improves life style

Fill up the blanks:

1. India, Nepal, China, Pakistan are ……………. countries and Japan, Singapore, Britain and U.S. are …………. countries.
2. The industrial development brings ………….. for the country and citizens.
3. The modern iron and steel plant was set up ……………
4. The cotton cloths from India are exported to prominent countries …………..
5. The slow pace of economy increases ……………..

Answer:

1. developing, developed
2. Prosperity
3. At Porto-nova in Tamil Nadu in 1830
4. U.S., Britain, Russia, France, Eastern Countries of Europe, Nepal, Singapore, Sri Lanka and African countries
5. Poverty.

MP Board Class 8th Social Science Chapter 11 Very Short Answer Type Questions

Question 1.
What is poverty?
Answer:
Poverty is the inability to secure minimum requirements of life, health and efficiency. One is unable to get minimum requirements for life such as food, clothing, housing, education and health due to poverty. Poverty leaves bad effect on the life and health of man.

Question 2.
What is unemployment?
Answer:
Unemployment is a situation in which people are willing to work for income but they are unable to find work.

Question 3.
Which are mineral-based industries?
Answer:
The industries based on raw material like iron, steel, cement, chemicals are called mineral-based industry.

MP Board Class 8th Social Science Chapter 11 Short Answer Type Questions

Question 1.
How was our economy in ancient times?
Answer:
In the ancient times Indian economy was rich and developed. India was independent in economy. The industry of pottery, metal, buildings were in advanced stage during the Indus Valley Civilization. In the medieval time Indian silk was in so high demand that Roman people were ready to give gold equal to the weight of Indian silk, Indian trade of cotton, silk, cotton clothes and spices was on the peak and our economy was rich in ancient times.

Question 2.
What do you mean by small scale industry?
Answer:
Cottage and small scale industries is the traditional industry as it needs little money and goods are made by the family members. The farmers when they have no work they earn money by this industry. These industries produce various traditional products like baskets, small toys, bides, ropes, envelopes, papad, mats, bade, spices and weaving items etc.

Question 3.
What is sustainable agriculture?
Answer;
Most of the farmers in India still use traditional tools for farming. They cannot make use of refined seeds, fertilizers and pesticides due to’poor economy and it also affects production. Their food production is consumed by their family members.

Question 4.
Write the names of Cottage industries.
Answer:
Some of the Cottage industries are:

• Small toys making
• Carpet weaving
• Papad making
• Basket making
• Bidi making
• Rope making
• Badi making
• Mat weaving
• Paper envelope making
• Spices preparing etc.

Question 5.
What is Gramin Rozgar Guarantee Programmer?
Answer:
The Central Government has passed the “Gramin Rozgar Guarantee Yojna” Bill, 2005. Under this bill a provision has been made to provide to one adult person of every family a 100 days job in the periphery of 5 Km near his residence. This scheme will be in force in 600 districts of the country by 2010.

MP Board Class 8th Social Science Chapter 11 Long Answer Type Questions

Question 1.
How agriculture is helpful in economic growth?
Answer:
The whole country depends on agriculture for its food requirements. Through agriculture, we get cereals, vegetables and fruit; it also supplies raw materials to various industries. Many industries, in India, depend on agriculture for their raw material for example, jute, sugar, cotton, cotton textile. Good crops raise the farmer’s income. With increase of income, the farmers purchase various items of daily life manufactured in industry. Agriculture is the backbone of our economic growth.

Question 2.
Describe the mineral-based industry.
Answer:
Such industries based on raw material like iron, steel, cement, chemicals are called mineral-based industries. The first iron and steel plant was set up at Porto nova in Tamil Nadu in 1830. Modern Iron and Steel plant was set up in Jamshedpur in 1907. India has chemical industries under which fertilizers, artificial fibers, rubber, plastic goods, paints and medicines are made. Transport equipment like rail engine, motor cars, bus, truck, motorcycle, aeroplane and ships industries have been set up in India. These industries provide employments in large number to our people.

Question 3.
Describe the major economic problems.
Answer:
The major economic problems are:

1. Pulsation growth
2. Poverty
3. Unemployment
4. Price List
5. Corruption

1. Pulsation growth:
The rapid growth of population in India is the major problem in the development. In 1951 the population was 36 crore which increased to 102 crore by 2002. The main causes of population growth are poverty, illiteracy, desire to love boy to run family and child marriages. The population increase impedes in the economic growth of nation. This leaves impact oil the lifestyle of people, the expansion of food grains and industries are not possible at the rate of increase in population.

2. Poverty:
Poverty is a very complex problem and requirements include minimum needs in respect of food, clothing, housing, education and health. Poverty leaves bad effect on the life and health of man. The working capacity decreases and poverty continues. Some people are living below the poverty line. Poverty is an obstacle in economic growth.

3. Unemployment:
A person who is not gainfully employed is called an unemployed person. In other words, unemployment is a situation in which a person mentally and physically is willing to work for income but is
Project Work unable to find it. Even the educated and skilled persons are unemployed in our country. Its main cause is slow growth of our economy.

4. Price List:
The continuous and uncontrolled price rise leads the economy to crises and poor became poorer. The price rise increases economic inequality and poverty, with the sincere efforts of govt, and citizens the price rise could be brought under control.

5. Corruption:
To get own’s word done in favor people pay either money or other things to a person or any agency is called corruption. Stealing of taxation also corruption. The corruption creates economic inequality. All the above problems are obstacles in our rapid economic growth.

project work:
Fill in the given table with the prices of 10 goods used in your home.

MP Board Class 8th Social Science Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 5 Our National Goals (a) Achievement of National Goals (b) Democracy and Citizen Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 5 Our National Goals (a) Achievement of National Goals (b) Democracy and Citizen

MP Board Class 8th Social Science Chapter 5 Text Book Exercise

Choose the correct option of the following

Question 1.
Our national goals?
(a) Democracy
(b) Kingship
(c) imperialism
(d ) Dictatorship
Answer:
(a) Democracy

Question 2.
Who said need peace in the world?
(a) Sardar vallab bai Patel
(b) Pandit Jawaharlal Nehru
(c) Lal Baliadur Shastre.
(d) Mahatma Gandhi
Answer:
(b) Pandit Jawaharlal Nehru

Question 3.
Why is disarmament a must?
(a) for the world peace
(b) for war
(c) for forming government
(d) for atomic arms
Answer:
(a) for the world peace

Question 4.
Voting on the basis of caste and religion makes democracy ?
(a) Strong
(b) weak
(c) Protects our rights
(d) All the above
Answer:
(b) weak

Question 5.
Age is fixed for Adult Franchise?
(a) 14 years
(b) 18 years
(c) 21 years
(d) 25 years
Answer:
(b) 18 years

Fill up the blanks:

1. in the constitution total …………… Fundamental rights, have been conferred
2. Democracy is the rule by…………
3. There is freedom of ………….. in secularism
4. Non-Alignment means to be ……………… from armed lobbies
5. Democracy , chief factors are ……………. equality and ………..

Answers:

1. Six
2. people
3. religion
4. separation
5. Democracy Freedom

MP Board Class 8th Social Science Chapter 5 very Short Answer Type Questions

Question 1.
Answer:
Individual freedom civil freedom. political economic, social and Religious freedom

Question 2.
Write meaning of equality
Answer:
The equal opportunity being made available by the state for all development of citizen is called equality

Question 3.
What is representative democracy?
Answer:
The democracy in which the people elect their representative to form govt is Know representative democracy

Question 4.
Why each citizen should be literate in democracy
Answer:
So that they can make use of their voting rights correctly

MP Board Class 8th Social Science Chapter 5 Short Answer Type Questions

Question 1.
Define Adult Franchise.
answer:
Indian constitution has provided voting right to the citizens Every citizen who has attained the age of 18 has the right to vote This right is available to all this is Adult Franchise.

Question 2.
What are our national goals
Answer:
National goals are defined and deter-mused. There are necessary for path of progress The national goals are democracy socialism Secularism and Internationale peace.

Question 3.
Write the true meaning of just-ire
Answer:
The justice indicates about correct genuine and logical facts Justice is that state in democratic society where individual rights are protected and law and order is maintained in the Society.

MP Board Class 8th Social Science Chapter 5 Long Answer Type Questions 

Question 1.
Why International peace and cooperation are important?
Answer:
International peace and co-operation is necessary for the citizens of the country. This works for the welfare of the people. This reduces arms and ammumitions procurement. The country can spend the major part of resources on social and economic developments. The progress and prosperity of the country depends upon the international peace and co-operation

Question 2.
What are the conditions required for the success of democracy?
Answer:
In a democratic system, the people play important role in electing the govt. The citizen should be aware of his duties so that govt, could work in the interest of the people. Every citizen should have equal rights with freedom. All the section of the society get equal opportunities to make progress without any discrimination. The citizen should follow the constitution and show respect to its ideals, institutions, national flag and national anthem.

Project Work

Question 1.
Make a chart of the Key Points of National Goals.
Answer:

• Democracy
• Freedom
• Equality
• Justice
• Secularism
• International Peace and Cooperation
• Non – Aligned Movement
• Disarmament

MP Board Class 8th Social Science Solutions

MP Board Class 8th Maths Solutions Chapter 7 घन और घनमूल Intext Questions

MP Board Class 8th Maths Chapter 7 पाठान्तर्गत प्रश्नोत्तर

पाठ्य-पुस्तक पृष्ठ संख्या # 117

भूमिका हार्डी-रामानुजन संख्या

प्रश्न 1.
1729 सबसे छोटी हार्डी-रामानुजन संख्या है। इस प्रकार की अनेक संख्याएँ हैं : उनमें से कुछ हैं 4104 (2,16; 9,5), 13832 (18, 20; 2,024)। कोष्ठकों में दी हुई संख्याएँ लेकर इसकी जाँच कीजिए।
हल:
जाँच –

• 4104 = 4096 + 8 = 16³ + 2³ और
• 4104 = 3375 + 729 = 15³ + 9³
• 13832 = 5832 + 8000 = 18³ + 20³
• और 13832 = 13824 + 8 = 24³ + 2³

घन –

प्रश्न 1.
1 सेमी भुजा वाले कितने घनों से 2 सेमी भजा वाला एक घन बनेगा?
हल:
2 सेमी भुजा वाला एक घन बनाने के लिए 1 सेमी भुजा वाले 2 x 2 x 2 = 8 घनों की आवश्यकता होगी।

प्रश्न 2.
1 सेमी भुजा वाले कितने घनों से 3 सेमी भुजा वाला एक घन बनेगा?
हल:
3 सेमी भुजा वाला एक घन बनाने के लिए 1 सेमी भुजा वाले 3 x 3 x 3 = 27 घनों की आवश्यकता होगी।

पाठ्य-पुस्तक पृष्ठ संख्या # 118

प्रश्न 1.
क्या आप बता सकते हैं कि इनको ये नाम क्यों दिए गए हैं?
हल:
हाँ, बता सकते हैं। इनको ये नाम इसलिए दिए गए हैं क्योंकि इसमें एक संख्या को स्वयं उसी से तीन बार गुणा किया जाता है।

प्रश्न 2.
नीचे 1 से 10 तक की संख्याओं के घन दिए गए हैं:

पूर्ण कीजिए।
हल:

प्रश्न 3.
यहाँ 1 से 1000 तक दस पूर्ण घन हैं। (इसकी जाँच कीजिए), 1 से 100 तक कितने पूर्ण धन हैं?
हल:
जाँच –

• 1 = 1 x 1 x 1
• 8 = 2 x 2 x 2
• 27 = 3 x 3 x 3
• 64 = 4 x 4 x 4
• 125 = 5 x 5 x 5
• 216 = 6 x 6 x 6
• 343 = 7 x 7 x 7
• 512 = 8 x 8 x 8
• 729 = 9 x 9 x 9
• 1000 = 10 x 10 x 10.

यहाँ स्पष्ट है कि संख्या को उसी संख्या से 3 बार गुणा करने पर संख्याएँ 1, 8, 27, 64, 125, 216, 343, 512, 729 और 1000 प्राप्त होती है।
∴ 1, 8, 27, 64, 125, 216, 343, 512, 729 और 1000 पूर्ण घन संख्याएँ हैं।
यहाँ 1 से 100 तक 1, 8, 27 और 64, 4 पूर्ण घन हैं।

प्रश्न 4.
सम संख्याओं के घनों को देखिए। क्या ये सभी सम हैं? आप विषम संख्याओं के घनों के बारे में क्या कह सकते हैं?
हल:
हाँ, सम संख्याओं के सभी घन सम हैं। विषम संख्याओं के घन विषम हैं।

पाठ्य-पुस्तक पृष्ठ संख्या # 119

प्रश्न 1.
ऐसी कुछ संख्याओं पर विचार कीजिए जिनकी इकाई का अंक 1 है। इनमें से प्रत्येक संख्या का घन ज्ञात कीजिए। उस संख्या के घन के इकाई के अंक के बारे में आप क्या कह सकते हैं, जिसकी इकाई का अंक 1 है।?
इसी प्रकार, उन संख्याओं के घनों की इकाई के अंकों के बारे में पता कीजिए, जिनकी इकाई के अंक 2,3,4 इत्यादि हैं।
हल:
1, 11, 21, 31,41,… आदि कुछ ऐसी संख्याएँ हैं जिनके इकाई का अंक 1 है। इन संख्याओं के घन हैं –

• 1³ = 1
• 11³ = 1331
• 21³ = 9261
• 31³ = 29791
• 41³ = 68921

आदि यहाँ यह स्पष्ट है कि ऐसी संख्याएँ जिनके इकाई का अंक 1 है उन संख्याओं के घनों का इकाई अंक भी 1 है।
इन संख्याओं के घन जिनके इकाई अंक 2, 3, 4, …… आदि हैं –

• 2 → 23 = 8 12³ = 1728 – 22³ = 10648
• 3 → 33 = 27 13³ =2197 – 23³ = 12167
• 4 → 43 = 64 14³ = 2744 – 24³ = 13824
• 5 → 53 = 125 15³ = 3375 – 25³ = 15625
• 6 → 63 = 216 16³ =4096 – 26³ = 17576
• 7 → 73 = 343 17³ = 4913 – 27³ = 19683
• 8 → 83 = 512 18³ = 5832 – 28³ = 21952
• 9 → 93 = 729 19³ = 6859 – 29³ =24389
• 10 → 103 = 1000 20³ = 8000 – 30³ = 27000 .. इत्यादि।

यहाँ यह स्पष्ट है कि जिन संख्याओं के इकाई अंक 2, 3, 4, 5, 6, 7, 8, 9, 0 हैं उनके घनों के इकाई अंक क्रमशः 8, 7, 4, 5, 6, 3, 2, 9 और 0 हैं।

प्रयास कीजिए (क्रमांक 7.1)

प्रश्न 1.
निम्नलिखित संख्याओं में से प्रत्येक के घन के इकाई अंक ज्ञात कीजिए:

1. 3331
2. 8888
3. 149
4. 1005
5. 1024
6. 77
7. 5022
8. 53

हल:
संख्याओं के घन के इकाई अंक –

1. 3331 → 1³ = 1 x 1 x 1 = 1; इकाई अंक =1
2. 8888 → 8³ = 8 x 8 x 8 = 512; इकाई अंक = 2
3. 149 → 9³ = 9 x 9 x 9 = 729; इकाई अंक = 9
4. 1005 → 5³ = 5x5x5 = 125; इकाई अंक = 5
5. 1024 → 4³ = 4 x 4 x 4 = 64; इकाई अंक = 4
6. 77 → 7³ = 7 x 7 x 7 = 343; इकाई अंक = 3
7. 5022 → 2³ = 2 x 2 x 2 = 8; इकाई अंक = 8
8. 53 → 3³ = 3 x 3 x 3 = 27; इकाई अंक = 7

कुछ रोचक प्रतिरूप

क्रमागत विषम संख्याओं को जोड़ना

विषम संख्याओं के योगों के निम्नलिखित प्रतिरूप को देखिए –

• 1 = 1 = 1³
• 3 + 5 = 8 = 2³
• 7 + 9 + 11 = 27 = 3³
• 13 + 15 + 17 + 19 = 64 = 4³
• 21 + 23 + 25 + 27 + 29 = 125 = 5³

प्रश्न 1.
क्या यह रोचक नहीं है? योग 10³ प्राप्त करने के लिए कितनी क्रमागत विषम संख्याओं की आवश्यकता होगी?
हल:
हाँ, यह रोचक है। उपर्युक्त प्रतिरूप से स्पष्ट है कि योग 103 प्राप्त करने के लिए 10 क्रमागत विषम संख्याओं की आवश्यकता होगी।

प्रयास कीजिए (क्रमांक 7.2)

प्रश्न 1.
उपर्युक्त प्रतिरूप का प्रयोग करते हुए, निम्नलिखित संख्याओं को विषम संख्याओं के योग के रूप में व्यक्त कीजिए –

1. 6³
2. 8³
3. 7³

हल:

1. 6³ = 31 + 33 + 35 + 37 + 39 + 41 = 216
2. 8³ = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512
3. 7³ = 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343

प्रश्न 2.
निम्नलिखित प्रतिरूप को देखिए:

• 2³ – 1³ = 1 + 2 x 1 x 3
• 3³ – 2³ = 1 + 3 x 2 x 3
• 4³ – 3³ = 1 + 4 x 3 x 3

उपर्युक्त प्रतिरूप का प्रयोग करते हुए, निम्नलिखित के मान ज्ञात कीजिए:

1. 7³ – 6³
2. 12³ – 11³
3. 20³ – 19³
4. 51³ – 50³

हल:
उपर्युक्त प्रतिरूप का प्रयोग करते हुए –

1. 7³ – 6³ = 1 + 7 x 6 x 3 = 1 + 126 = 127
2. 12³ – 11³ = 1 + 12 x 11 x 3 = 1 + 396 = 397
3. 20³ – 19³ = 1 + 20 x 19 x 3 = 1 + 1140 = 1141
4. 51³ – 50³ = 1 + 51 x 50 x 3 = 1 + 7650 = 7651

पाठ्य-पुस्तक पृष्ठ संख्या # 120

प्रश्न 1.
यदि किसी संख्या के अभाज्य गुणनखण्ड में प्रत्येक गुणनखण्ड तीन बार आता है, तो क्या वह संख्या एक पूर्ण घन होती है?
हल:
यदि किसी संख्या के अभाज्य गुणनखण्डन में प्रत्येक गुणनखण्ड तीन बार आता है, तो वह संख्या एक पूर्ण घन होती

प्रश्न 2.
क्या 729 पूर्ण घन है?
हल:
∴ 729 = 3 x 3 x 3 x 3 x 3 x 3
प्रश्न 3.
क्या आपको याद है कि a^m x b^m = (a x b)^m होता है?
हल:
हाँ, याद है कि a^m x b^m = (a x b)^m

प्रयास कीजिए (क्रमांक 7.3)

प्रश्न 1.
निम्नलिखित में से कौन-सी संख्याएँ पूर्ण घन हैं?

1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648.

हल:
1.

संख्याओं के त्रिक बनाने पर 2 x 5 x 5 शेष रहता है।
अतः 400 पूर्ण घन नहीं है।

2.

यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अत: 3375 एक पूर्ण घन है।

3.

यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अतः 8000 एक पूर्ण घन है।

4.

यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अतः 15625 एक पूर्ण घन है।

5.

संख्याओं के त्रिक बनाने पर 3 x 3 शेष रहता है।
अत: 9000 पूर्ण घन नहीं है।

6.

यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अत: 6859 एक पूर्ण घन है।

7.

संख्याओं के त्रिक बनाने पर 3 x 5 x 5 शेष रहता है।
अत: 2025 एक पूर्ण घन नहीं है।

8.

यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अतः 10648 एक पूर्ण घन है।

पाठ्य-पुस्तक पृष्ठ संख्या # 121
सोचिए, चर्चा कीजिए और लिखिए (क्रमांक 7.1)

प्रश्न 1.
जाँच कीजिए कि निम्नलिखित में से कौन-सी संख्याएँ पूर्ण घन हैं –

1. 2700
2. 16000
3. 64000
4. 900
5. 125000
6. 36000
7. 21600
8. 10000
9. 27000000
10. 1000

इन पूर्ण घनों में आप क्या प्रतिरूप देखते हैं?
हल:
प्रत्येक संख्या के अभाज्य गुणनखण्ड करने पर,
1. 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5
संख्याओं के त्रिक बनाने पर 2 x 2 x 5 x 5 शेष रहता है।
अतः 2700 एक पूर्ण घन नहीं है।

2. 16000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5
संख्याओं के त्रिक बनाने पर 2 शेष रहता है।
अत: 16000 एक पूर्ण घन नहीं है।

3. 64000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5
यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अतः 64000 एक पूर्ण घन है।

4. 900 = 2 x 2 x 3 x 3 x 5 x 5
यहाँ हम त्रिक बनाकर देखते हैं, तो किसी भी संख्या का त्रिक नहीं बनता है।
अत: 900 एक पूर्ण घन नहीं है।

5. 125000 = 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5
यहाँ प्रत्येक गुणनखण्ड तीन बार आया है।
अत: 125000 एक पूर्ण घन है।

6. 36000 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 x 5
संख्याओं के त्रिक बनाने पर 2 x 2 x 3 x 3 शेष रहता है।
अतः 36000 एक पूर्ण घन नहीं है।

7. 21600 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
संख्याओं के त्रिक बनाने पर 2 x 5 शेष रहता है।
अत: 21600 एक पूर्ण घन नहीं है।

8. 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
संख्याओं के त्रिक बनाने पर 2 x 5 शेष रहता है।
अतः 10000 एक पूर्ण घन नहीं है।

9. 27000000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 5 x 5 x 5
यहाँ प्रत्येक संख्या तीन-तीन बार आयी है।
अत: 27000000 एक पूर्ण घन है।

10. 1000 = 2 x 2 x 2 x 5 x 5 x 5
यहाँ प्रत्येक संख्या तीन बार आई है।
अत: 1000 एक पूर्ण घन है।

MP Board Class 8th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 12 बीजीय व्यंजक Ex 12.1

प्रश्न 1.
निम्नलिखित स्थितियों में चरों, अचरों और अंकगणितीय संक्रियाओं का प्रयोग करते हुए बीजीय व्यंजक प्राप्त कीजिए –

1. संख्या y में से z को घटाना।
2. संख्याओं x और y के योग का आधा।
3. संख्या z को स्वयं उससे गुणा किया जाता है।
4. संख्याओं p और q के गुणनफल का एक-चौथाई।
5. दोनों संख्याओं x और y के वर्गों को जोड़ा जाता है।
6. संख्याओं m और n के गुणनफल के तीन गुने में संख्या 5 जोड़ना।
7. 10 में से संख्याओं y और z के गुणनफल को घटाना।
8. संख्याओं a और b के गुणनफल में से उनके योग को घटाना।

हल:

1. y – 2
2. \(\frac { 1 }{ 2 } \) = (x + y)
3. z²
4. \(\frac { 1 }{ 4 } \)pq
5. x² + y²
6. 3mn + 5
7. 10 – yz
8. ab – (a + b)

प्रश्न 2.
(i) निम्नलिखित व्यंजकों में पदों और उनके गुणनखण्डों को छाँटिए। पदों और उनके गुणनखण्डों को पेड़ आरेख द्वारा भी दर्शाइए :
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5x² + 7x²y
(e) – ab + 2b² – 3a²
(ii) नीचे दिए गए व्यंजकों में,पदों और उनके गुणनखण्डों को छाँटिए :
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) Pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) \(\frac { 3 }{ 4 } \)x + \(\frac { 1 }{ 4 } \)
(h) 0.12p² + 0.2q²
हल:
(i) (a) x – 3

(b) 1 + x + x²

(c) y – y³

(d) 5x² + 7x²y

(e) – ab + 2b² – 3a²

(ii) हल:

प्रश्न 3.
निम्नलिखित व्यंजकों में पदों के संख्यात्मक गुणांकों जो अचर न हों, की पहचान कीजिए :
(i) 5 – 3t²
(ii) 1 + t + t² + t³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) -p²q² + 7pq
(vi) 1.2a + 0.8b
(vii) 3.14 r²
(viii) 2(l + b)
(ix) 0.1 y + 0.01 y²
हल:

प्रश्न 4.
(a) वे पद पहचानिए जिनमें x है और फिर इनमें x का गुणांक लिखिए:
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + z
(iv) y + z + zx
(v) 1 + x + xy
(vi) 12xy + 25
(vii) 7 + xy²

(b) वे पद पहचानिए जिनमें है और फिर इनमें y² का गुणांक लिखिए :
(i) 8 – xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
हल:
(a)

(b)

प्रश्न 5.
निम्नलिखित व्यंजकों को एकपदी, द्विपद और त्रिपद के रूप में वर्गीकृत कीजिए :

1. 4y – 7z
2. y²
3. x + y – xy
4. 100
5. ab – a – b
6. 5 – 3t
7. 4p²q – 4pq²
8. 7mn
9. z² – 3z + 8
10. a² + b²
11. z² + z
12. 1 + x + x²

हल:

1. द्विपद
2. एकपदी
3. त्रिपद
4. एकपदी
5. त्रिपद
6. द्विपद
7. द्विपद
8. एकपदी
9. त्रिपद
10. द्विपद
11. द्विपद
12. त्रिपद।

प्रश्न 6.
बताइए कि दिए हुए पदों के युग्म समान पदों के हैं या असमान पदों के हैं :

1. 1,100
2. -7x, \(\frac { 5 }{ 2 } \)x
3. -29x, -29y
4. 14xy, 42yx
5. 4m²p, 4mp²
6. 12xz, 12x²z²

हल:

1. समान पद
2. समान पद
3. असमान पद
4. समान पद
5. असमान पद
6. असमान पद

प्रश्न 7.
निम्नलिखित में समान पदों को छाँटिए :
(a) -xy², -4yx², , 8x², 2xy², 7y, – 11x²,-100x, – 11yx, 20x²y, -6x²,y, 2xy, 3x.
(b) 10pq, 7p, 8q, -p²q², – 7qp, – 100q, – 23, 12q²p², -5p², 41, 2405p, 78qp, 13p²q, qp², 701p²
हल:
(a) दिए हुए पदों में समान पदों के समूह:
-xy², 2ry²; – 4yx², 20x²y; 8x², – 11x²; – 6x²; 7y, y; – 100x, 3x; -11yx, 2xy
(b) दिए हुए पदों में समान पदों के समूह –
10pq, – 7qp, 78qp; 7p, 2405p; 8q, – 100q; – p²q², 12q²p²; – 23, 41; -5p², 701p², 13p²q, qp²

पाठ्य-पुस्तक पृष्ठ संख्या # 253

प्रयास कीजिए

प्रश्न 1.
कम-से-कम ऐसी दो स्थितियों के बारे में सोचिए, जिनमें से प्रत्येक में आपको दो बीजीय व्यंजकों को बनाने की आवश्यकता पड़े और उन्हें जोड़ना या घटाना पड़े।
हल:

• राहुल की मासिक आय दीपक की आय की तीन गुनी है और मीनाक्षी की मासिक आय राहुल और दीपक की प्रति माह आय के योग से ₹ 300 अधिक है। मीनाक्षी की प्रतिमाह आय क्या है?
• दो हवाई जहाज विपरीत दिशाओं में उड़ना आरम्भ करते हैं। एक की औसत चाल, दूसरे की औसत चाल से 100 km/h अधिक है। यदि चार घण्टे बाद उनके बीच की दूरी 4000 km हो, तो उनकी औसत चाल ज्ञात कीजिए।

पाठ्य-पुस्तक पृष्ठ संख्या # 255

प्रयास कीजिए

प्रश्न 1.
जोड़िए और घटाइए:
(i) m – n, m + n
(ii) mn + 5 – 2, mn + 3
हल:
(i) योग : m – n + m + n
= m + m – n + n
= 2m + 0n = 2m
घटाना: (m – n) – (m + n)
= m – n – m – n
= m – m – n – n
= 0 – 2n = -2n

(ii) योग: mn + 5 – 2, mn +3
= mn + 5 – 2 + mn + 3
= mn + mn + 5 – 2 + 3
= 2mn + 6
घटाना: (mn + 5 – 2) – (mn + 3)
= mn + 5 – 2 – mn – 3
= mn – mn + 5 – 2 – 3
= 0 + 0 = 0

पाठ्य-पुस्तक पृष्ठ संख्या # 256

MP Board Class 7th Maths Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 4 National Integration Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 4 National Integration

MP Board Class 8th Social Science Chapter 4 Text Book Exercise

Choose the correct option of the following.

Question 1.
Which feeling inspires citizen in national integration?
(a) Nationalism
(b) Religion
(c) Casteism
(d) None of the above
Answer:
(a) Nationalism

Question 2.
How many Indian languages have been mentioned in the constitution?
(a) 14
(b) 18
(c) 22
(d) 26.
Answer:
(c) 22

Fill up the blanks:

1. There are provisions of …………. fundamental rights in Indian constitution.
2. Symbol of Asoka is …………. emblem.
3. In ancient period varna system was based on ………..
4. India was divided in the year …………

Answer:

1. six
2. National
3. Karma
4. 1947

MP Board Class 8th Social Science Chapter 4 Very Short Answer Type Questions

Question 1.
What do you mean by National Integration?
Answer:
National integration is feeling of oneness among the people. irrespective of caste. religion and language.

Question 2.
what do you mean by secularism?
Answer:
In secularism every religion has the right to practice his own religion.

Question 3.
What do you mean by unity in diversity?
Answer:
Despite numerous religions. regions. languages, castes in our country India is a unit. This is unity in diversity.

MP Board Class 8th Social Science Chapter 4 Short Answer Type Questions

Question 1.
Explain the major defects of caste system?
Answer:
Major defects of caste system:

• Caste system demanded society into high and low classes.
• Upper castes exploit the lower caste people.
• Caste system creates trouble in the economic progress and unity of nation.
• Pressure of castes influence the politics

Question 2.
Which are our National Symbols?
Answer:
National Flag. National Anthem. National emblem are our national symbols

MP Board Class 8th Social Science Chapter 4 Long Answer Type Questions

Question 1.
Write notes on fundamental rights and duties?
Answer:
Indian constitution has provision of six Fundamental rights. The rights are enjoyed by all Indian without any discrimination. Provisions have been made in the constitution to provide opportunity for the development of citizens like equality, freedom and social justice etc. The interest of weaker sections of the society has been protected and safeguarded.

Fundamental Duties:
The duties have also been mentioned in the constitution it is duty of all citizens to respect the principles of the constitution. institutions, national flag and national anthem Every citizen mam-tam the the integrate and of country the citizen should project propriety and create brother hood

Question 2.
elaborate separation in detail?
Answer:
Regionalism give birth to separatism To Demand a independent state out of the country is called separatism sevent cast and languages are prevalent in country this diver-city leads to dissatisfaction and leads to a demand for separate state. the border states are often influenced by such feelings and creates a stage for separation the external force support. this in country this en-comics of country and fundamental lists choose to follow the path of violence and terrorism the feeling of separation could be avoided by social justice decentralization and balanced development.

Question 3.
Describe the chief constituents of National integration?
Answer:
There are numerous forces which lead to the promotion of national integration. Some of these forces can be summed up as under:

• Secularism is one force That unites people. a state ought not to favor any religion and should respect all  the religions alike this will ornate feeling of togetherness.
• uniformity of Law means that same law should apply on all the individual irrespective of any distinction the Whole country should be under the same law.
• national festivals and symbols together with a long liberation struggle gives the people feelings of oneness.
• climantio conditions conman economic problems, development issues, etc, etc, alike, faced by the Indian in conman.

Question 4.
mention the disruptive factions of national integration?
Answer:
the factors which hamper national integration are:
1. communal-ism:
communal-ism has always been a major threat to our country unity. it is based on ignorance and selfish interests of different sections. it leads to communal riots in the country it was communal-ism which split India into two country India and Pakistan in 194.

2. Linguism:
Linguism means love for ones own language. its but natural language that people love their language should not breed narrow mildness and hatred for the language of others. there should be not conflict between people of different state s and culture.

3. Caste-ism:
Caste feelings also Weaken country s unity people with strong caste feeling forget the interest of their country and try to serve their caste only Cast-rem is no less a danger to national unity and integrity.

MP Board Class 8th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 13 घातांक और घात Ex 13.1

प्रश्न 1.
निम्नलिखित के मान ज्ञात कीजिए :
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴
हल:
(i) 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64
(iii) 9³ = 9 × 9 × 9 = 729
(iii) 11² = 11 × 11 = 121
(iv) 5⁴ = 5 × 5 × 5 × 5 = 625

प्रश्न 2.
निम्नलिखित को घातांकीय रूप में व्यक्त कीजिए:
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
हल:
(i) 6 × 6 × 6 × 6 = 6⁴
(ii) t × t = t²
(iii) b × b × b × b = b⁴
(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³
(v) 2 × 2 × a × a = 2² × a²
(vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d

प्रश्न 3.
निम्नलिखित संख्याओं में से प्रत्येक को घातांकीय संकेतन में व्यक्त कीजिए :
(i) 512
(ii) 343
(iii) 729
(iv) 3125
हल:

प्रश्न 4.
निम्नलिखित में से प्रत्येक भाग में, जहाँ भी सम्भव हो, बड़ी संख्या को पहचानिए :
(i) 4³ या 3⁴
(ii) 5³ या 3⁵
(iii) 2⁸ या 8²
(iv) 100² या 2¹⁰⁰
(v) 2¹⁰ या 10²
हल:
(i) ∵ 4³ = 4 × 4 × 4 = 64
और 3⁴ = 3 × 3 × 3 × 3 = 81
∵ 81 > 64 ∴ 3⁴ > 4³
अतः 3⁴ बड़ा है।

(ii) ∵ 5³ = 5 × 5 × 5 = 125
और 3⁵ = 3 × 3 × 3 × 3 × 3 = 243
∵ 243 > 125 ∴ 3⁵ > 5³
अतः 3⁵ बड़ा है।

(iii) ∵ 2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 256
और 8² = 8 × 8 = 64
∵ 256 > 64 ∴ 2⁸ > 8²
अतः 28 बड़ा है।

(iv) ∵ 100² = 100 × 100 = 10000
और 2¹⁰⁰ = (2¹⁰)¹⁰
= (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)¹⁰
= (1024)¹⁰ = (1024)^5×2
= (1024 × 1024)⁵
= (1048576)⁵
∵ 1048576 > 10000
∴ (1048576)⁵ > 100²
या (2¹⁰)¹⁰ > 100² या 2¹⁰⁰ > 100²
अत: (2¹⁰)¹⁰ बड़ा है।

(v) ∵ 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 1024
और 10² = 10 × 10 = 100
∵ 1024 > 100, ∴ 2¹⁰ > 10²
अतः 2¹⁰ बड़ा है।

प्रश्न 5.
निम्नलिखित में से प्रत्येक को उनके अभाज्य गुणनखण्डों की घातों के गुणनफल के रूप में व्यक्त कीजिए:
(i) 648
(ii) 405
(iii) 540
(iv) 3600
हल:
(i) 648

∴ 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 23 × 34

(ii)405

∴ 405 = 3 × 3 × 3 × 3 × 5
= 34 × 5

(iii) 540

∴ 540 = 2 × 2 × 3 × 3 × 3 × 5
= 22 × 33 × 5

(iv) 3600

∴ 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²

प्रश्न 6.
सरल कीजिए:
(i) 2 × 10³
(ii) 7² × 2²
(iii) 2³ × 5
(iv) 3 × 4⁴
(v) 0 × 10²
(vi) 5² × 3³
(vii) 2⁴ × 3²
(viii) 3² × 10⁴
हल:
(i) 2 × 10³ = 2 × 1000 = 2000
(ii) 7² × 2² = 49 × 4 = 196
(iii) 2³ × 5 = 8 × 5 = 40
(iv) 3 × 4⁴ = 3 × 256 = 768
(v) 0 × 10² = 0 × 100 = 0
(vi) 5² × 3³ = 25 × 27 = 675
(vii) 2⁴ × 3² = 16 × 9 = 144
(viii) 3² × 10⁴ = 9 × 10000 = 90000

प्रश्न 7.
सरल कीजिए:
(i) (-4)³
(ii) (-3) × (-2)³
(iii) (-3)² × (-5)²
(iv) (-2)³ × (-10)³
हल:
(i) (-4)³ = (-4) × (-4) × (-4)
= -64

(ii) (-3) × (-2)³ = (-3) (-2) (-2) (-2)
= (-3) × (-8) = 24

(iii) (-3)² × (-5)² = (-3) (-3) (-5) (-5)
= 9 × 25 = 225

(iv) (-2)³ × (-10)3 = (-2)(-2)(-2) (-10) (-10) (- 10)
= (-8) × (-1000) = 8000

प्रश्न 8.
निम्नलिखित संख्याओं की तुलना कीजिए :
(i) 2.7 × 10¹²; 1.5 × 10⁸
(ii) 4 × 10¹⁴;3 × 10¹⁷
हल:
(i) ∵ 2.7 × 10¹² = \(\frac { 27 }{ 10 } \) × 10¹²
= 27 × 10¹¹ में 13 अंक होंगे।
और 1.5 × 10⁸ = \(\frac { 27 }{ 10 } \) × 10⁸
= 15 × 10⁷ में 9 अंक होंगे।
स्पष्ट है कि 27 × 10¹¹ > 15 × 10⁷
⇒ 2.7 × 10¹² > 1.5 × 10⁸

(ii) ∵ 4 × 10¹⁴ में 15 अंक होंगे
और 3 × 10¹⁷ में 18 अंक होंगे।
∴ 4 × 10¹⁴ < 3 × 10¹⁷

पाठ्य-पुस्तक पृष्ठ संख्या # 270

प्रयास कीजिए

प्रश्न 1.
सरल करके घातांकीय रूप में लिखिए :
(i) 2⁵ × 2³
(ii) p³ × p²
(iii) 4³ × 4²
(iv) a³ × a² × a⁷
(v) 5³ × 5⁷ × 5¹²
(vi) (-4)¹⁰⁰ × (-4)²⁰
हल:
(i) 2⁵ × 2³ = 2⁵⁺³ = 2⁸
(ii) p³ × p² = p³⁺² = p⁵
(iii) 4³ × 4² = 4³⁺² = 4⁵
(iv) a³ × a² × a⁷ = a³⁺²⁺⁷ = a¹²
(v) 5³ × 5⁷ × 5¹² = 5³⁺⁷⁺¹² = 5²²
(vi) (-4)¹⁰⁰ × (-4)²⁰ = (-4)¹⁰⁰⁺²⁰ = (-4)¹²⁰

पाठ्य-पुस्तक पृष्ठ संख्या # 271

प्रयास कीजिए

प्रश्न 1.
सरल करके घातांकीय रूप में लिखिए : उदाहरण के लिए, 116 + 113 = 114
(i) 2⁹ ÷ 2³
(ii) 10⁸ ÷ 10⁴
(iii) 9¹¹ ÷ 9⁷
(iv) 20¹⁵ ÷ 20¹³
(v) 7¹³ ÷ 7¹⁰
हल:
(i) 2⁹ ÷ 2³ = 2^9-3 = 2⁶
(ii) 10⁸ ÷ 10⁴ = 10^8-4 = 10⁴
(iii) 9¹¹ ÷ 9⁷ = 9^11-7 = 9⁴
(iv) 20¹⁵ ÷ 20¹³ = 20^15-13 = 20²
(v) 7¹³ ÷ 7¹⁰ = 7^13-10 = 7³

प्रयास कीजिए

प्रश्न 1.
सरल करके, उत्तर को घातांकीय रूप में व्यक्त कीजिए:
(i) (6²)⁴
(ii) (2²)¹⁰⁰
(iii) (7⁵⁰)²
(iv) (5³)⁷
हल:
(i) (6²)⁴ = 6^2×4 = 68
(ii) (2²)¹⁰⁰ = 2^2×100 = 2²⁰⁰
(iii) (7⁵⁰)² = 7^50×2 = 7¹⁰⁰
(iv) (5³)⁷ = 5^3×7 = 5²¹

पाठ्य-पुस्तक पृष्ठ संख्या # 273

प्रयास कीजिए

प्रश्न 1.
a^m × b^m = (ab)^m का प्रयोग करके, अन्य रूप में बदलिए:
(i) 4³ × 2³
(ii) 2⁵ × b⁵
(iii) a² × t²
(iv) 5⁶ × (-2)⁶
(v) (-2)⁴ × (-3)⁴
हल:
(i) 4³ × 2³ = (4 × 2)³ = (8)³
(ii) 2⁵ × b⁵ = (2 × b)⁵ = (2b)⁵
(iii) a² × t² = (a × t)² = (at)²
(iv) 56 × (-2)² = {5 × (-2)}²
= (-10)⁶ = (10)⁶
(v) (-2)⁴ × (-3)⁴ = {(-2) × (-3)}⁴ = (6)⁴

प्रयास कीजिए

प्रश्न 1.
a^m + b^m = (\(\frac { a }{ b } \))m का प्रयोग करके, अन्य रूप में बदलिए:
(i) 4⁵ ÷ 3⁵
(ii) 2⁵ ÷ b⁵
(iii) (-2)³ ÷ b³
(iv) p⁴ ÷ q⁴
(v) 5⁶ ÷ (-2)⁶
हल:

पाठ्य-पुस्तक पृष्ठ संख्या # 276

MP Board Class 7th Maths Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 8 British Policies and Administration in India After 1858 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 8 British Policies and Administration in India After 1858

MP Board Class 8th Social Science Chapter 8 Text Book Exercise

Choose the correct option of the following questions

Question 1.
The Proclamation of Queen Victoria were made in:
(a) 1757
(b) 3858
(c) 1957
(d) 1965
Answer:
(b) 3858

Question 2.
The rule of India entrusted into hands of Queen of England.
(a) By 1858 Act
(b) By 1861 Act
(c) By 1865 Act
(d) By 1876 Act
Answer:
(a) By 1858 Act

Question 3.
First Municipality in India was established:
(a) In 1865 at Madras
(b) In 1J867 in Bengal
(c) In 1868 in Uttar Pradesh
(d) None of the above
Answer:
(d) None of the above

Question 5.
The head of Indian Secretariat was called:
(a) Indian Secretary
(b) Viceroy
(c) Governor General
(d) Secretary
Answer:
(a) Indian Secretary

Fill in the blanks:

1. To help Viceroy a members …………. council was formed.
2. The British economic policies were formed to protect the interests of …………..
3. In 1876 the minimum age for Civil Services was …………..
4. To implement Wood Proposals Lord Rippon constituted ………… Commission.

Answer:

1. Four
2. Britishers
3. 19
4. Hunter

MP Board Class 8th Social Science Chapter 8 Very Short Answer Type Questions

Question 1.
By which Act Queen of England was made Empress of India.
Answer:
By Ordinance of 1858.

Question 2.
After 1858 how was the Governor General Addressed.
Answer:
The Governor General was addressed as Viceroy.

Question 3.
Who was called the father of local governance?
Answer:
Lord Rippon.

MP Board Class 8th Social Science Chapter 8 Short Answer Type Questions

Question 1.
Describe the proclamation of Queen in short.
Answer:
The rule of the company was replaced by rule of the Crown. The Company territories were now governed by English government. Indian people and rulers got some rights and privileges.

Question 2.
What changes were made by 1861 Act.
Answer:
The Legislative Council was enlarged by the addition of 6 to 12 more members and also introduced provincial councils.

Question 3.
Describe the administrative division after 1858.
Answer:
It transferred power to British crown. A minister called the Secretary of State for India was made responsible for the government of India to the British Parliament. Viceroy replaced Governor General title.

MP Board Class 8th Social Science Chapter 8 Long Answer Type Questions

Question 1.
Write about the decisions of 1858 Act.
Answer:
The decisions of 1858 Act are:
1. The control of India passed on completely to British Government. It was announced that India would be governed by and in the name of the British monarch through a Secretary of State.

2. Before 1858, there were two bodies in Britain which controlled the British policies in India, viz, the Board of Control and the Court of Directors of the East India Company. Now a minister of the British Government called the Secretary of State for India, was given complete control over the Government of India. Like other ministers of the British Government, he was responsible to the British Parliament.

3. To advise the Secretary of State, the Indian Council was created. This Council had no real powers and could only advise the Secretary of State who could ignore the advise of the Council at any time the liked.

4. Before 1858, the Governor General generally acted on his own within the framework of the general policies laid down in Britain. The advanced means of transport and communication reduced the freedom enjoyed by the earlier Governor Generals of India. Now, the Viceroy had to inform the latest developments of India to the Secretary of State thereafter he was to work according to his instructions and others.

Question 2.
Write short note on Army Organisation?
Answer:
Before 1858, the Presidencies of Bombay, Madras and Calcutta had their separate armies. Each Presidency consisted of the Indian soldiers, units of the European soldiers and the regiments of the British soldiers – the latter two were combined.

In 1859, the armies of the three Presidencies were unified and the entire army, of the British India came under the control of the commander – in – chief. The Indian soldiers were excluded from the artillery and the arsenals; The number of the European soldiers was increased. For every two Indian soldiers, one European soldier was recruited. All the officers in the army were to be the Europeans.

Question 3.
Give short Information about the local administration in brief.
Answer:
After 1857, the British government encouraged the local government units such as the municipalities and district boards. Thus, the system of administration was decentralized. The local bodies were given the tasks of providing education, health services, water supply, and the like to the people. Some important changes were made in the local government in 1882, the old system was broken down.

Matters like sanitation, roads, water supply and street lighting were neglected. After 1857, Municipal committees were set up in towns, they levied taxes to meet expenditure on local administration and works. After 1882, the district boards were set up but they consisted of only officials and not elected members. After 1882, the elected members were introduced but only with property were entitled to vote.

Question 4.
Evaluate the British economic policies in precise?
Answer:
British government reorganized the financial administration after 1858. The economic policies followed by Britishers drained the Indian resources and help the Britishers. The British government abolished the tax on Import and allowed free trade in India.

This lured lot of foreign companies to India and caused disastrous effect on Indian industries. The prominent industries that suffered due to the British financial policy were Jute, Indigo, Textiles, Woolen, Tea, Rubber, Coffee, Coal, Iron, and Steam ships. The financial policy of British government completely halted the progress of India.

Question 5.
Describe the British education policy.
Answer:
The British education policy was not for the welfare of Indians but was more for promoting Christian religion and imparting English education. Missionaries established in 1820 promoted English language. In 1854 Woods proposals suggested to promote English languages with other Indian languages.

There was proposal to train teachers and to provide financial grant to educational institutions on secular basis. In 1904 University Act was passed for affiliation of the University and to appoint professors. The 1919 Act entrusted the responsibility of education to provincial councils. The British education policy played a major role in motivating Indians towards nationality

MP Board Class 8th Social Science Solutions

MP Board Class 7th Maths Solutions Chapter 11 परिमाप और क्षेत्रफल Ex 11.4

प्रश्न 1.
एक बगीचा 90 m लम्बा और 75 m चौड़ा है। इसके बाहर चारों तरफ 5 m चौड़ा पथ बनाना है। पथ का क्षेत्रफल ज्ञात कीजिए। बगीचे का क्षेत्रफल हेक्टेअर में भी ज्ञात कीजिए।
हल:
बगीचे की लम्बाई (l) = 90 m, चौड़ाई (b) = 75 m

बगीचे का क्षेत्रफल = l × b
= 90 m × 75 m = 6750 m²
= \(\frac { 6750 }{ 10000 } \) हेक्टेअर
= 0.6750 हेक्टेअर
∵ बगीचे के चारों ओर का पथ आयत बनाता है।
∴ बाह्य आयत की लम्बाई = 90 + 5 + 5 = 100 m
व चौड़ाई = 75 + 5 + 5 = 85 m
∴ पथ सहित बगीचे का बाह्य आयत का क्षेत्रफल
= 100 × 85 = 8,500 m²
अतः पथ का क्षेत्रफल = बाह्य आयत का क्षेत्रफल – बगीचे का क्षेत्रफल
= 8500 m² – 6750 m² = 1750 m²

प्रश्न 2.
125 m लम्बाई और 65m चौड़ाई वाले एक आयताकार पार्क के चारों ओर बाहर एक 3 m चौड़ा पथ बना हुआ है। पथ का क्षेत्रफल ज्ञात कीजिए।
हल:
चित्र 11.29 में,
PQ = AB + 2 × पथ की चौड़ाई
= 125 + 2 × 3 = 125 + 6 = 131 m
व QR = BC + 2 × पथ की चौड़ाई
= 65 + 2 × 3 = 65 + 6 = 71 m

अब, पथ का क्षेत्रफल
= बाह्य आयत PORS का क्षेत्रफल – पार्क ABCD का क्षेत्रफल
= PQ × QR – AB × BC
= 131 × 71 – 125 × 65
= 9301 – 8125 = 1176m²

प्रश्न 3.
8 cm लम्बे और 5 cm चौड़े एक गत्ते पर एक चित्र की पेंटिंग इस प्रकार बनाई गई है कि इसकी प्रत्येक भुजाओं के अनुदिश 1.5 cm चौड़ा हाशिया (margin) छोड़ा गया है। हाशिये का कुल क्षेत्रफल ज्ञात कीजिए।
हल:
चित्र 11.30 में,
PQ = 8 cm, QR = 5 cm
AB = PQ – 2 × हाशिये की चौड़ाई
= 8 – 2 × 1.5 = 8 – 3 = 5 cm
BC = QR – 2 × हाशिये की चौड़ाई
= 5 – 2 × 1.5 = 5 – 3 = 2 cm

अब, हाशिये का कुल क्षेत्रफल
= गत्ते का क्षेत्रफल – पेंटिंग का क्षेत्रफल
= PQ × QR – AB × BC = 8 × 5 – 5 × 2 = 40 – 10 = 30 cm²

प्रश्न 4.
5.5m लम्बे और 4 m चौड़े कमरे के चारों ओर बाहर 2-25 m चौड़ा एक बरामदा बनाया गया है। ज्ञात कीजिए:
(i) बरामदे का क्षेत्रफल
(ii) ₹ 200 प्रति m² की दर से बरामदे के फर्श पर सीमेंट कराने का व्यय।
हल:
चित्र 11.31 में,
ABCD कमरा है तथा इसके चारों ओर 2.25 m चौड़ा बरामदा है।
PQ = AB + 2 × बरामदे की चौड़ाई
= 5.5 + 2 × 2.25 = 5.5 + 4.5 = 10 m
QR = BC + 2 × बरामदे की चौड़ाई
= 4 + 2 × 2.25
= 4 + 4.5 = 8.5 m

(i) बरामदे का क्षेत्रफल = बाह्य आयतकार भाग PQRS का क्षेत्रफल – कमरे का क्षेत्रफल
= PQ × QR – AB × CD
= 10 × 8.5 – 5.5 × 4
= 85 – 22 = 63 m²

(ii) ₹ 200 प्रति m² की दर से बरामदे के फर्श पर सीमेंट कराने का कुल व्यय = ₹200 × 63 = ₹ 12,600

प्रश्न 5.
30 m भुजा वाले एक वर्गाकार बगीचे की परिसीमा से लगा भीतर की ओर 1 m चौड़ा पथ बना हुआ है। ज्ञात कीजिए:
(i) पथ का क्षेत्रफल
(ii) ₹ 40 प्रति m² की दर से शेष भाग पर घास लगवाने का व्यय।
हल:
(i) चित्र 11.32 में,
PQRS बगीचा है। इसके भीतर की ओर 1 m चौड़ा पथ है।

AB = PQ – 2 × पथ की चौड़ाई
= 30 – 2 × 1
= 30 – 2 = 28 m.

BC = OR – 2 × पथ की चौड़ाई
= 30 – 2 × 1
= 30 – 2 = 28m
अब, पथ का क्षेत्रफल = बाह्य वर्ग PQRS का क्षेत्रफल
– अन्त:वर्ग ABCD का क्षेत्रफल
= 30 × 30 – 28 × 28
= 900 – 784
= 116 m²

(ii) बगीचे के शेष भाग का क्षेत्रफल
= ABCD का क्षेत्रफल
= (28 × 28) m²
= 784 m²
∴ ₹ 40 प्रति m² की दर से बगीचे के शेष भाग पर घास लगवाने का व्यय
= ₹40 × 784
= ₹ 31,360

प्रश्न 6.
700 m लम्बे और 300 m चौड़े एक आयताकार पार्क के मध्य से होकर जाते 10 m चौड़े दो पथ बने हुए हैं, जो एक-दूसरे पर परस्पर लम्ब और चौपड़ के आकार के हैं। इनमें से प्रत्येक पथ का क्षेत्रफल ज्ञात कीजिए तथा पार्क की भुजाओं को छोड़कर पार्क के शेष भाग का भी क्षेत्रफल ज्ञात कीजिए। उत्तर को हेक्टेयर में दीजिए।
हल:
PQ = EH = KL = KN = 10 m
लम्बाई के अनुदिश सड़क की लम्बाई = 700 m
चौड़ाई के अनुदिश सड़क की लम्बाई = 300 m

सड़कों PQRS व EFGH का क्षेत्रफल
= PQRS का क्षेत्रफल + EFGH का क्षेत्रफल – KLMN का क्षेत्रफल
(300 × 10 + 700 × 10 – 10 × 10) m²
= (3000 + 7000 – 100) m²
= 10000 m² – 100 m² = 9900 m²
= \(\frac { 9900 }{ 10000 } \) = 0.99 हेक्टेयर
अब, सड़कों को छोड़कर पार्क का क्षेत्रफल
= पार्क का क्षेत्रफल – परस्पर लम्ब सड़कों का क्षेत्रफल
= 700 × 300 – 9900
= 210000 – 9900 = 200100 m²
= \(\frac { 200100 }{ 10000 } \) हेक्टेयर = 20.01 हेक्टेयर

प्रश्न 7.
90 m लम्बाई और 60 m चौड़ाई वाले एक आयताकार मैदान में दो पथ बनाए गए हैं, जो भुजाओं के समान्तर हैं, एक-दूसरे को लम्बवत् काटते हैं और मैदान के मध्य से होकर निकलते हैं। यदि प्रत्येक पथ की चौड़ाई 3m हो, तो ज्ञात कीजिए:
(i) पथों द्वारा आच्छादित क्षेत्रफल
(ii) ₹110 प्रति m² की दर से पथ बनाने का व्यय।
हल:
(i) पथ PQRS का क्षेत्रफल = लम्बाई × चौड़ाई
= 90 × 3 = 270 m²

पथ TUVW का क्षेत्रफल = लम्बाई × चौड़ाई
= 60 × 3 = 180 m²
उभयनिष्ठ भाग EFGH का क्षेत्रफल
= 3 × 3 = 9 m²
अत: पथों का कुल क्षेत्रफल = पथ PQRS का क्षेत्रफल + पथ TUVW का क्षेत्रफल – उभयनिष्ठ भाग EFGH का क्षेत्रफल
= 270 m² + 180 m² – 9m²
= 450 m² – 9 m²
= 441 m²

(ii) ₹ 110 प्रति m² की दर से पथ बनाने का व्यय
= ₹ 110 × 441
= ₹ 48,510

प्रश्न 8.
प्रज्ञा 4 cm त्रिज्या वाले एक वृत्ताकार पाइप के चारों ओर एक रस्सी लपेटती है (जैसा दिखाया गया है) और रस्सी की आवश्यक लम्बाई को काट लेती है। इसके बाद वह उसे 4 cm भुजा वाले एक वर्गाकार बॉक्स के चारों ओर लपेटती है(दिखाया गया है)। क्या उसके पास कुछ और रस्सी बचेगी ? (π = 3.14)

हल:
वृत्ताकार पाइप की त्रिज्या (r) = 4 cm
∴ पाइप की परिधि = 2πr
= 2 × 3.14 × 4 = 25.12 cm
∴ पाइप पर लपेटी गई रस्सी की लम्बाई
= 25.12 cm
अब, वर्ग का परिमाप = 4 × a
= 4 × 4 cm = 16 cm
∴ वर्ग पर लपेटी गई रस्सी की लम्बाई = 16 m
∵ 25.12 cm – 16 cm
∴ 25.12 cm – 16 cm = 9.12 cm.
हाँ, उसके पास 9-12 cm रस्सी और बचेगी।

प्रश्न 9.
संलग्न आकृति एक आयताकार पार्क के मध्य में एक वृत्ताकार फूलों की क्यारी को दर्शाती है। ज्ञात कीजिए:

(i) पूरे पार्क का क्षेत्रफल
(ii) फूलों की क्यारी का क्षेत्रफल
(iii) फूलों की क्यारी को छोड़कर, पार्क के शेष भाग का क्षेत्रफल
(iv) क्यारी की परिधि।
हल:
(i) यहाँ, पार्क की लम्बाई (l) = 10m, चौड़ाई = 5 m.
∴ पार्क का क्षेत्रफल = l × b
= 10 × 5 = 50 m²

(ii) फूलों की क्यारी की त्रिज्या (r) = 2 m
∴ फूलों की क्यारी का क्षेत्रफल = 2
= 3.14 × 2 × 2
= 12.56 m²

(iii) फूलों की क्यारी को छोड़कर शेष भाग का क्षेत्रफल
= पार्क का क्षेत्रफल – क्यारी का क्षेत्रफल
= 50 m² – 12.56 m²
= 37.44 m²

(iv) क्यारी की परिधि = 2πr
= 2 × 3.14 × 2
= 12.56 m

प्रश्न 10.
दी गई आकृति में छायांकित भाग का क्षेत्रफल ज्ञात कीजिए।

हल:
(i) आयत ABCD का क्षेत्रफल
= l × b = 18 × 10 = 180 cm²
∆ AEF का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
= \(\frac { 1 }{ 2 } \) × 10 × 6 = 30 cm²
∆ CBE का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
= \(\frac { 1 }{ 2 } \) × 8 × 10 = 40 cm²
∴ छायांकित भाग का क्षेत्रफल
= आयत ABCD का क्षेत्रफल
-(∆ AEF का क्षेत्रफल + ∆ CBE का क्षेत्रफल)
= 180 cm2 – (30 cm2 + 40 cm²)
= 180 cm2 – 70 cm = 110 cm²

(ii) वर्ग PQRS का क्षेत्रफल
= भुजा × भुजा
= 20 × 20 = 400 cm²
∆ PQT का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
\(\frac { 1 }{ 2 } \) × 20 × 10 = 100 cm²
∆ QRU का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
\(\frac { 1 }{ 2 } \) × 10 × 20 = 100 cm²
∆ TSU का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
= \(\frac { 1 }{ 2 } \) × 10 × 10 = 50 cm²
अब, छायांकित भाग का क्षेत्रफल = PQRS का क्षेत्रफल
– (∆ PQT का क्षेत्रफल + ∆ QRU का क्षेत्रफल + ∆ TSU का क्षेत्रफल)
= 400 cm² – (100 cm² + 100 cm² + 50 cm²)
= 400 cm² – 250 cm² = 150 cm²

प्रश्न 11.
चतुर्भुज ABCD का क्षेत्रफल ज्ञात कीजिए। यहाँ, AC = 22 cm, BM = 3 cm, DN = 3 cm, और BM ⊥ AC, DN ⊥ AC.

हल:
∆ ABC का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
\(\frac { 1 }{ 2 } \) × 22 × 3 = 33 cm²
∆ ACD का क्षेत्रफल = \(\frac { 1 }{ 2 } \) × b × h
= \(\frac { 1 }{ 2 } \) × 22 × 3 = 33 cm²
∴ चतुर्भुज ABCD का क्षेत्रफल
= ∆ABC का क्षेत्रफल + ∆ACD का क्षेत्रफल
= 33 cm² + 33 cm² = 66 cm²

MP Board Class 7th Maths Solutions

In this article, we will share MP Board Class 8th Social Science Solutions Chapter 7 Changing Outer Forces Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Social Science Solutions Chapter 7 Changing Outer Forces

MP Board Class 8th Social Science Chapter 7 Text Book Exercise

Choose the correct option of the following

Question 1.
By which process the materials of broken rocks assemble at its original place
(a) Weathering
(b) Transportation
(c) Erosion
(d) Deposition
Answer:
(a) Weathering

Question 2.
Where among the following winds perform its work?
(a) In deserts
(b) In Mountains
(c) In coastal areas
(d) In plains
Answer:
(a) In deserts

Question 3.
Which is vital factor that makes soil fer-tile?
(a) The original rock materials
(b) Humus
(c) Climate
(d) Man
Answer:
(b) Humus

Fill in the blanks:

1. When the rock breaks into pieces without chemical reaction is …………… weathering
2. The process of flattening of unleavened surface is called ………….
3. With the deposition …………. lake is formed.
4. On the  Orissa coast in India …………. is an example of lagoon-lake.

Answers:

1. Biological
2. Surface Balancing
3. Oxbow
4. Chilka Lake

MP Board Class 8th Social Science Chapter 7 Very Short Answer Type Questions

Question 1.
Which are the three works of river?
Answer:
Erosion, Transportation & Deposition are three works of river.

Question 2.
Which are three stages of river?
Answer:
There are three stages of river, young, matured and old age.

Question 3.
What is estuary?
Answer:
Some of the rivers become wider at their meeting point with sea and sweep away into deep sea.

Question 4.
What is geyser?
Answer:
The underground hot water is thrown up in the form of fountain is called geyser.

Question 5.
How is the Yardage formed?
Answer:
Yardage is formed by wind erosion.

Question 6.
What is Lagoon?
Answer:
The sea water enclosed between the coast and sand bar is known as lagoon.

MP Board Class 8th Social Science Chapter 7 Short Answer Type Questions

Question 1.
Write the factors that help soil formation?
Answer:
The weathering and erosion process of rocks from the soil. The helping agents in soil formation are parent rocks, surface shape, climate & vegetation.

Question 2.
What is weathering? Write the names of different types of weathering?
Answer:
Weathering is the iocal process in which graduation and split of rocks occur at their original place. The types are physical, chemical and biological weathering.

Question 3.
Differentiate between slow and rapid wasting?
Answer:
The process of surface sliding by weathering is slow wasting and the process of filling up the holes is high wasting.

Question 4.
How the V – shaped valley is formed?
Answer:
The V – shaped valley is formed when river reaches its base level and widens its banks. V – Shaped Valley

Question 5.
What are Moraines? Write its types?
Answer:
When the glacier melts, it starts depositing the sediments in different parts of the valley. The land form made is Moraines. The types are Lateral, Medial and Terminal moraines.

Question 6.
What is underground water level?
Answer:
The water which collects on the ground goes in to under the ground is called underground water. Water reaches into ground through rivers, lakes and sea.

MP Board Class 8th Social Science Chapter 7 Long Answer Type Questions

Question 1.
Discuss the surface Balancing?
Answer:
The widening and flattening of the surface is called surface Balancing. Rivers, glaciers, wind, underground water and sea waves are the key agents of leveling. The Surface Balancing can be classified in to two categories.

1. Slow wasting
2. Rapid wasting.

1. Slow wasting:
The rapid erosion cycle shifts the rocks from its original place by rubbing, cut-ting & peeling to decrease the high elevated areas.

2. Rapid wasting:
The process of deposition of weathered materials in low-lying regions or holes is called aggravation and the weathered materials fill up the holes whose height kept on growing is called Rapid wasting.

Question 2.
What is soil erosion? Write its cause and ways to conserve soil?
Answer:
The natural forces or human activities which help in removing the layer of soil is called soil erosion. The causes of soil erosion are as follows:

• When the cover of vegetation is removed the upper layer of soil gets loose.
• Heavy rains and powerful wind remove the Soil content and layer.
• Overgrazing by domestic animals.
• Cutting trees and clearing forest cover.
• Excess use of fertilize & pesticide.

The measures to conserve soil are:

• To protect forest. Trees tighten the soil material.
• Tree plantation in river valleys, barren land and mountain slopes.
• Soil rehab of the land which becomes uneven due to water can be leveled to stop erosion.
• Controlled grazing.
• Flood control measures by building stop dams.
• Crop rotation by planting crops alter – neatly.

Question 3.
How wind erosion takes place? Mention about the land forms created by the wind?
Answer:
The wind erosion generally occurs in Desert areas where wind brings pebbles and sand particles with it. These particles make friction among themselves and scratch rocks. The work of erosion is done by transportation, collision and friction. Wind erosion is affected by the speed of wind, the size, quantity and height of particles, the structure and shape of rocks, and the climate. The land forms created by the wind erosion are sand dunes and Leos.

Sand Dunes:
When some obstacle comes in the path of the sand laden wind the deposition of sand forms a dune hill which has a crest.

Loess:
Fine particles of the sand are suspended in wind. They are carried over long distance and gets deposited as Loess.

Question 4.
What is glacier? Write about glacier types with example.
Answer:
A river of moving ice is called glacier. It creeps down a valley from a snowfield above the snow line. It moves under the influence of gravity. On the basis of origin and situation glaciers can be divided into two groups.

1. Continental glacier:
These are permanent sheets of ice and snow that covers extensive land surfaces. These exist in Greenland and Antarctica.

2. Valley glacier:
These are formed in high mountains and are usually long and narrow as they occupy valleys of former streams. The biggest valley glacier is Siachen glacier.

Question 5.
Describe the function of underground water with pictures?
Answer:
The writer which collects on the ground goes into under the ground is called underground water. Water reaches into ground through rivers, lakes, and sea. Ground water partly exists on the surface like waterfall, artesian well and geyser. Ground water balances the surface and several land forms are created through erosion and transportation. Ground water dilutes the limestone by rain water entering in causes erosion which is called Karts topography

The erosion work of ground water can be classified into two parts:

1. landscape or the surface
2. underground landscape.

The land forms created on the surface are lapses and sink hole. The underground land form includes caves, stalactites stalagmites.

MP Board Class 8th Social Science Solutions