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MP Board Class 12th Biology Important Questions Chapter 15 Biodiversity and Conservation

Biodiversity and Conservation Important Questions

Biodiversity and Conservation Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Tera biodiversity is used for the first time by :
(a) V. G. Rosseu
(b) Linnaeus
(c) Odum
(d) Theophrastus.
Answer:
(a) V. G. Rosseu

Question 2.
When does wild life conservation Act come in the force in India :
(a) 1883
(b) 1972
(c) 1973
(d) 1972
Answer:
(b) 1972

Question 3.
When does IBWL is started :
(a) 1952
(b) 1981
(c) 1973
(d) 1972
Answer:
(a) 1952

Question 4.
Where is NBPR is situated :
(a) Delhi
(b) Kolkata
(c) Lucknow
(d) Mumbai
Answer:
(a) Delhi

Question 5.
The number of biosphere reserves in India is :
(a) 73
(b) 7
(c) 416
(d) 23
Answer:
(b) 7

Question 6.
Which National Park is associated with white tiger:
(a) Kanger
(b) Satpura
(c) Bandhavgarh
(d) Kanha
Answer:
(c) Bandhavgarh

Question 7.
First National Park of M. P. is :
(a) Shivpuri
(b) Bandhavgarh
(c) Kanha
(d) Kanger
Answer:
(c) Kanha

Question 8.
Where is “Plant Fossil” National Park is situated:
(a) Shivpuri
(b) Mandala
(c) Bastar
(d) Bhopal
Answer:
(b) Mandala

Question 9.
The National Park of M.P. which is marked as biosphere reserve is :
(a) Kanha
(b) Shivpuri
(c) Bandhavgarh
(d) Satpura.
Answer:
(b) Shivpuri

Question 10.
First Tiger Project of M.P. is :
(a) Bandhavgarh
(b) Kanha
(c) Sidhi
(d) Shivpuri.
Answer:
(b) Kanha

Question 2.
Fill in the blanks :

1. FRI is situated is …………………….
2. Kanjiranga sanctuary is associated with the conservation of …………………….
3. Red data book is related with the ……………………. conservation.
4.  ……………………. is the fust biosphere reserve of India.
5. List of the endangered species are available in ……………………. data book.
6.  ……………………. is the important componant of air for life.
7. Mineral is the example of ……………………. sources.
8. ……………………. percent part of the earth has water.
9. Biodiversity is the source of different types of …………………….
10. Species for which there are no living representative is called …………………….
11. Farming which is develop in fresh water is called …………………….
12. Botanical garden is an ……………………. method of plant conservation.
13. ……………………. conservation is the on site conservation.
14. ……………………. percent part of the mangrove plant are found in India.
15. Gir national park in India is famous for …………………….

Answer:

1. Dehradun
2. Rhinoceros
3. Endangered species
4. Nilgiri
5. Red
6. Oxygen
7. Non – renewable
8. 71
9. Germplasm
10. Extinct
11. Aquatic culture
12. Ex – situ
13. In – situ
14. 5
15. Asiatic lion.

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

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

1. Where does Kanha National Park situated?
2. From which person Chipko movement is associated?
3. When does Wild life Conservation Act is emenced in force?
4. How many percent contribution of India is in world’s biodiversity?
5. How many species are found in India?
6. Which are known as the lungs of the earth?
7. How many national park in India?
8. How many wild 1 ife sanctuary in India?
9. What is the basic component of biodiversity?
10. Write the full name of IUCN.
11. Which are most helpful in nature balancing?
12. How many species of bamboo found in India?
13. Name any one book associated with the conservation of wild life.
14. Upper surface of the land is called.

Answer:

1. M.P
2. Sunder Lai Bahuguna
3. 1972
4. 8.1%
5. 1200 species
6. Amazon – rain forest
7. 90
8. 448
9. Gene
10. International Union for conservation of nature and natural resourses
11. Wild animal
12. 100 (about)
13. Red data book
14. Soil.

Biodiversity and Conservation Very Short Answer Type Questions

Question 1.
Name the three important components of biodiversity.
Answer:
Three important components of biodiversity are:

1. Genetic diversity.
2. Species diversity.
3. Ecosystem diversity.

Question 2.
What is forest?
Answer:
A large area of land covered with trees, shrubs and grasses.

Question 3.
As which day 21st March is observed?
Answer:
21st March is observed as “World Forest Day”.

Question 4.
What are extinct species?
Answer:
Species which are not found at present in the earth but survived in the past in this world are called as extinct species.
Example : Dinosaur.

Question 5.
What are mentioned in the Red Data Book of IUCN?
Answer:
The IUCN Red Data Book contains list of threatened species. It was founded in 1964. Correction was done in 2004.

Biodiversity and Conservation Short Answer Type Questions

Question 1.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the earth is very vast.According to an estimate by researchers, it is about seven millions. The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 2.
The species diversity of plants (22 %) is much less than that of animals (72 %). What may be the reason that animals achieved greater diversifications?
Answer:
Animals have achieved greater diversification than plants due to following reasons:

1. They are mobile and thus, can move away from their predators or unfavourable, environments. On the other hand plants are fixed and have fewer adaptations to obtain optimum amount of raw materials and sunlight therefore, they show lesser diversity.

2. Animals have well developed nervous system to receive stimuli against external factors and thus, can respond to them. On the other hand, plants do not exhibit any such mechanism, thus they show lesser diversity than animals.

Question 3.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease causing microbes that we deliberately want to eradicate from the earth. Since, these microorganisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease causing microbes.

Question 4.
Explain conservation of biodiversity.
Answer:
Conservation of biodiversity:
India is one of the richest (among the 12 mega centres of the world) countries in biological diversity. This rich biodiversity is due to a variety of climatic conditions prevailing on different ecological habits ranging from tropical, subtropical, temperate, alpine to desert. These varied conditions harbour a plethora of organisms, which forms an important natures wealth, responsible for socio – economic development of life in our country. But the biodiversity of organisms are under serious threat and there is an urgent need for biodiversity conservation on war foot level.

Question 5.
What is social forestry?
Answer:
The planning of social forestry started in India since 1976, which is related with the conservation of forests. This project ;s useful for local people in various ways, such as it fulfil their requirements, provides work to unemployed, use of wasteland and help to maintain O₂ and CO₂ balance in the atmosphere, etc. Thus, project is started by Indian government, the chief objectives of this project are as follows:

1. Plantation of useful plants in the forest.
2. Development of forests on personal lands by the cooperation of government.
3. To prevent the harmful effects of pollution by development of artificial forest.
4. Preservation of endangered wild animals.

Question 6.
Write importance of forests.
Answer:
Importance of forests:

1. Forests play a vital role in the life and economy of all tribes living in forests, by providing food, medicines and other products of commercial value.

2. Forests are large biotic communities. It provides shelter and sustenance to a larger number of diverse species of plants, animals and microorganisms.

3. Forests prevent soil erosion by wind and water. The trees provide shade which prevents the soil from drying during summer. Trees reduce the velocity of raindrops or wind striking the ground so that dislodging of the slil partiles is reduces. The root system of plants firmly binds the soil.

Question 7.
Write any five features of Indian forests.
Answer:
Indian forests are characterized by:

1. Indian forests are mainly tropical forests.
2. Himalayan forests are characterized by the presence of coniferous trees.
3. In few parts of our country having temperate forests.
4. Our forests contain a large number of useful varieties of plants and animals.
5. A great variations are present in Indian forests.

Question 8.
Enumerate any five reasons for the destruction of wildlife (animals).
Answer:
The main reason of destruction of wildlife (animals) are:

1. Entertainment, personal profit, earning money by wrong methods are the human illattitudes, unkindness toward wild animals has brought the animals at endanger level.
2. Huge reduction in the natural habitat of wild animal so it has reduced their living area due to urbanization, industrialization and deforestation.
3. Exhorbitant extraction and consumption is harmful for wild animals like skin of animals, Teeth of elephant etc.
4. Various types of pollution has force the reduction of wild animals.
5. Very loose and unpunishable wildlife act which has increase poaching, hunting of wild animals.

Question 9.
How is biodiversity important for functioning of ecosystem?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods. If an ecosystem is rich in biodiversity, then the ecological balance would not get affected.

Various tropic levels are connected through food chains. If any one organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food. If all deer’s are dead, soon the tigers will also die.

Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 10.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are forest patches for worship in several parts of India. All the trees and wildlife in them are again treated and given total protection. They are found in khasiandjointia hills in Meghalaya. Western Ghat regions of Karnataka and Maharashtra, etc. Tribals do not allow anyone to cut even a single branch of tree in these sacred groves thus, sacred groves have been free form all types of exploitations.

Question 11.
Among the ecosystem services are control of floods and soil erosion, how is this achieved by the biotic components of the ecosystem?
Answer:
Control of soil erosion:
Plant roots hold the soil particles tightly and do not allow the top soil to be drifted away be winds or moving water. Plants increase the porosity and fertility of the soil.

Control of floods:
It is carried out by retaining water and preventing run off rain water. Litter and humus of plants function as sponges thus, retaining the water which percolates down and get stored as underground water. Hence, the flood is controlled.

Question 12.
Give three hypothesis for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypothesis proposed by scientists for explaining species richness in the tropics:

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Biodiversity and Conservation Long Answer Type Questions

Question 1.
What is the significance of the slope of regression in a species – area relationship?
Answer:
The slope regression (z) has a great significance in order to find a species – area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis done in larger areas, then the slope of regression is much steeper.

Question 2.
What are the major causes of species loss in a geographical regions?
Answer:
The following are the major causes for the loss of biodiversity around the world:

1. Habitat loss and fragmentation:
Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

2. Over – exploitation:
Due to over – hunting and over – exploitation of various plants and animals by humans, many species have become endangered of extinct (such as; the tiger and the passenger pigeon).

3. Alien species invasions:
Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

4. Co – extinction:
In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 3.
What do you understand by Threatened species? Explain its types.
Answer:
Species which have been greatly reduced in their number or whose natural habitats have been, disturbed due to which these are near extinction and may become extinct if the causative factors continue are called threatened species. It is estimated that about 25,000 plant species and 1,000 vertebrate species and subspecies and many invertebrate species are threatened with extinction. It is believed that at least 10% of the living species are in danger.

The organisms which are near extinction are of following types:

1. Endangered (E) species:
The species which are facing danger of extinction and whose survival is unlikely if the causal factors continue to operate. These are the species whose number have been reduced to a critical level or whose habitats have been so drastically reduced that they are deemed Lto be in immediate danger of extinction. For example, Indian rhinoceros, Asiatic lion and the great Indian bustard, snow leopard etc.

2. Vulnerable (V) species:
These are the species having sufficient number of individuals in their natural habitats. However, in the near future, they might represent the category of endangered species if unfavourable factors in the environment continue to operate, e.g., Musk deer, black buck, golden langur, etc.

3. Rare (R) species:
These are species with small population in the world. At present these are not endangered and vulnerable but are at risk. These species are usually localize within geographical areas or habitats or are thinly scattered over a more extensive range, e.g., Indian elephant, Asiatic wild ass, gharial, wild yak etc.

4. Threatened (T) species:
The species which do not fall under the endangered or vulnerable categories but indications are available that such species may come under any of these two categories if appropriate measures are not taken to protect them.

Question 4.
What are National Parks? Explain any five National Parks found in India.
Answer:
National Park:
Natinal park is an area which is strictly reserved for the betterment of the wildlife and where activities like forestry, grazing or cultivation are not permitted.

Five national parks of India are:

1. Shivpuri national park – It is located at Shivpuri near Gwalior (Madhya Pradesh). Wildlife present in this park tiger, cheetal, sambhar.
2. Guindy deer national park – It is located near Chennai (Madras) in Tamil Nadu. Wildlife found here ore Alvino deer, black buck, cheetal and famous snake park.
3. Betla national park – It is located in Palamu at Bihar. Wildlife found in this national park are elephant, tiger.
4. Dachigham national park – It is located at Srinagar in Jammu and Kashmir. Wildlife leopard, black beer, brown bear, musk deer, hangul, scrow, etc.
5. Bandhavgarh national park – It is located at Shahdol in Madhya Pradesh. Wildlife such as white tiger, panther, cheetal, bison, nilgai, barking deer, wild boar, etc. are found here.

Question 5.
Write in brief, the reasons necessary for conservation of wild species.
Answer:
Necessity for wildlife conservation : The conservation of wildlife is required for the following reasons :

1. To maintain balance in nature:
The wildlife helps us in maintaining the balance of nature. Once this equilibrium is disturbed it leads to many problems. The destruction of carnivores or insectivores often leads to an increase in the herbivores which in turn affects the forest vegetation or crop.

2. Economic value:
The wildlife can be used commercially to earn money example Animal products like hides, ivory, fur etc. are of tremendous economic value. The collection and supply of dead or living specimens of wildlife for museums and zoos fetches good amount of money. Wildlife can increase our earning of foreign exchange if tourism is promoted properly.

3. Scientific value:
The preservation of wildlife helps many naturalists and behaviour biologists to study morphology, anatomy, physiology, ecology and behaviour biology of the wild animals under their natural surroundings.

4. Recreational value:
The wildlife of any country provides best means of sports and recreation. Bird – watching is a hobby of many people all over the world. A visit to the parks and sanctuaries is an enjoyable proposition for children as well as for adults.

5. Cultural value:
The wildlife of India is our cultural asset and has deep rooted impact on Indian art, sculpture, literature and religion. Indus valley civilization shows the use of animals symbols in their seals.

6. Preservation of human race:
The destruction of wildlife in an area may eventually lead to the end of human civilization.

Question 6.
Describe the National and International efforts prescribed for the conservation of forests.
Answer:
The forest conservation is started in India on national level since British government. In 1856, Lord Dalhousi had formulated a policy for the conservation of forest in Burma. In 1894, Indian government also prepared a forest policy on national level. The main points of this policy are:

1. Forest management
2. Proper use of forest land
3. Polity for, protected forests
4. Improved forest production

Indian government established national parks, sanctuaries and zoological parks, The F.A.O. of United Nations is also functioning on forest conservation on international level. This organization also provides financial help for this purpose. In 1952, Indian government also prepared India’s New National Forest Policy under the direction of F.A.O. Forest policy has been planned for :

1. Prevention of deforestation of hill plants.
2. Reforestation of grazing land.
3. Development of grazing land.
4. Plantation of economically useful forest trees.
5. Increase in the profit of government from forests.

Question 7.
Write a short note on wild animals in India.
Answer:
India as a country has a diverse range of wildlife. India is home for many species of wild animals. More than 25% land are dense forest in India and around 400 national parks. Some of the most important and popular wild animals in India are as follows:

(A)
Animals : In Indian forest, below mentioned animals are found:

1. Deer – Its many species are found in India example Musdeer, Sambhar deer, chital, etc.

2. Antelope – These are same as deer example Nilgai, Barasingha (Swamp deer), four homed antelope (Chousingha) etc.

3. Elephant – Elephants are large mammals of the family elephantidae. It is found in Kerela and North India.

4. Rhinoceros – It is found in Himalaya region and in the forest of Bengal and Assam. Humans are the biggest threat to the Indian rhinoceros as the have been hunted to the brink of extinction for their horns.

5. Wild Ass – Wild asses are not found in any part of the world. Now in India, it is found in the little Rann of Kutch in the Gujarat state of India.

6. Carnivorous animals – Some Indian wild carnivores are:

• Indian lion (Asiatic lion) : Now it is confined to forests in the fall.
• Cheetah : It is on the verge of extinction.
• Lion : Lion is a national animal, at present its population in our country are more than 3,000.
• Leopard : It is similar to cheetah but smaller than cheetah.

(B)
Birds:
Peacock, wild fowl, many types of duck, stork, pigeons, partridge, quail, vulture, kite, piquant, owl, indian paradise flycatcher (dudhraj) are found in forests of our country.

(C)
Reptiles:
Crocodiles, alligators, tortoise, lizards, snakes and other reptilians are found in Indian forest. Many vertebrates and invertebrates are also found in Indian forest.

Question 8.
What are the main rules of Indian forest act?
Answer:
The Indian Forest Act, 1927 was largely based on previous Indian forest acts implemented under the British. Things which are included in this act are as follows:

1. Forest arrangement – Due to this Act, give the protection and arrangement to forest and Midlife.
2. Appropriate use of forest land – Uses of extra land for forest animals and grow some plant these are useful in wild animals.
3. Act for protection of forest – This act, stop passion of deforestation and must be conservation of forest and wild animals.
4. Increasing of forest product – Due to this act try to increasing of forest product and discovered new information which are better for wildlife.

Question 9.
Write an essay on measures of forest conservation.
Answer:
Forest conservation:
Forest conservation and management are essential to maintain the forests in their natural state and also to prevent the depletion of wildlife and forest wealth. For the success of conservation it is necessary to know the cause of depletion and destruction of forests. Forests are generally destroyed by fire, improper cutting of trees and by animals.

Essentiality of forest conservation:
Forest is a complex system which is responsible for the ecological balance in nature. Deforestation causing natural imbalance and affects the biotic components of the environment resulting floods, drought, epidemics, environmental pollution. Many ecologically important species of plants and animals are lost due to which economically important substance like wood, medicines, resin, lac and various food materials will not be available for us.

Measures of forest conservation : The following measures or efforts are prescribed for reforestation:

1. Establishment of conserved forests and their conservation in proper way.
2. Reforestation on deforested land. Old and damaged plants would be replaced by new plants.
3. Proper management of forests.
4. By promoting public awareness about forests.
5. Replacement of burnt off areas of the forests.
6. Plantation of trees that increase forest productivity.
7. Forestation of plants on hills and wastelands and prevention of grazing by cattle.
8. Prevent forests from fire, diseases and insects.
9. Providing basic protection for all forests by law.
10. Regulating human activity in the forest such as grazing by cattle and collection of firewood and fodder etc.
11. Provide special attention for the conservation of endangered plant and animal species under the inspection of specialists.
12. Removal of undesirable trees and vegetation for the better growth of desirable species.
13. Forestation of industrially useful plants.
14. The government will arrange the management of useful forests.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 9 Strategies for Enhancement in Food Production

Strategies for Enhancement in Food Production Important Questions

 Strategies for Enhancement in Food Production Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Triticum aestivum wheat is:
(a) Haploid
(b) Diploid
(c) Tetraploid
(d) Hexaploid.
Answer:
(c) Tetraploid

Question 2.
Man – made cereal is :
(a) Potato
(b) Triticale
(c) Triticum
(d) Sugarcane.
Answer:
(b) Triticale

Question 3.
In cattles, anthrax is disease is caused by :
(a) Bacteria
(b) Fungi
(c) Virus
(d) Ticks.
Answer:
(a) Bacteria

Question 4.
The causal organism of haemorrhagic septicaemia as :
(a) Brucella avartus
(b) Bacillus sp.
(c) Pasturella boviseptica
(d) Clostridium.
Answer:
(c) Pasturella boviseptica

Question 5.
Foot and Mouth disease of cattles is caused by :
(a) Fungi
(b) Bacteria
(c) Virus
(d) Mycoplasma.
Answer:
(c) Virus

Question 6.
The disease caused after rainy season is :
(a) Black fever
(b) Haemorrhagic septicaemia.
(c) Ponkni
(d) Anthrax.
Answer:
(c) Ponkni

Question 7.
The vaccination for galghontu is given to animal in :
(a) January – February
(b) March – April
(c) May – June
(d) October – Nov ember.
Answer:
(c) May – June

Question 8.
The cause of plague disease of animals is :
(a) Fungi
(b) Bacteria
(c) Virus
(d) Mycoplasma.
Answer:
(c) Virus

Question 9.
Example of cereal plant is :
(a) Wheat
(b) Rice
(c) Maize
(d) All of these.
Answer:
(d) All of these.

Question 10.
Botanical name of common wheat is :
(a) Triticum aestivum
(b) T. vulgare
(c) T. durum
(d) T. monococcum
Answer:
(b) T. vulgare

Question 11.
Wheat and rice belong to family
(a) Graminae
(b) Papillionaceae
(c) Euphorbiaceae
(d) Compositae
Answer:
(a) Graminae

Question 12.
Padma and Jaya are the improved varieties of:
(a) Wheat
(b) Rice
(c) Gram
(d) Groundnut
Answer:
(b) Rice

Question 13.
Botanical name of Ratanjot is :
(a) Pongamia pinnata
(b) Ricinus communis
(c) Jatropha curcas
(d) None of these
Answer:
(c) Jatropha curcas

Question 14.
Fertilizer elements are:
(a) Nitrogen
(b) Phosphorus
(c) Potassium
(d) All of these
Answer:
(d) All of these

Question 15.
Fertilizer that supplies least % of nitrogen is :
(a) Urea %
(b) (NH₄) SO₄
(c) (NH₄) NO₃
(d) Organic nitrogen fertilizer.
Answer:
(d) Organic nitrogen fertilizer.

Question 16.
Fertilizer that supplies both nitrogen and phosphorus to soil:
(a) Ammonium sulphate
(b) Ammonium nitrate
(c) Urea
(d) Super phosphate
Answer:
(d) Super phosphate

Question 17.
Nitrogen fixing organism is :
(a) Rhizobium
(b) Azolla
(c) Azotobacter and Azospirillum
(d) All of these
Answer:
(d) All of these

Question 18.
VAM is :
(a) Bacteria
(b) Yeast
(c) Fungi
(d) Virus.
Answer:
(c) Fungi

Question 2.
Fill in the blanks :

1. The process of separating animals with desired characters for breeding purpose is called ………………………..
2. Breeding between two genetically different animals is called ………………………..
3. Anthrax fever of cattles is caused by bacterium ………………………..
4. Ranikhet disease of fowls was first observed in ……………………….. district.
5. Fowl pox disease is caused by ………………………..
6. No. of chromosomes in Triticum vulgare is ………………………..
7. Groundnut belongs to family ………………………..
8. Botanical name of Karanj is ………………………..
9.  ……………………….. is the first genetically engineered food.
10. Rhizobium, Azolla and Anabaena are the examples of ………………………..
11. Insecticides azadirachtin is obtained from leaves of ……………………….. plants.
12. Insects harming plants are called ………………………..
13.  ……………………….. bacterium is found in the roots of leguminous plants. (MP 2015)
14. Nostoc and Anabaena are called ………………………..
15. Bacteria responsible for N₂ fixation is ……………………….. (MP 2009 Set A)

Answer:

1. Selection
2. Outbreeding
3. Bacillus anthraxis
4. Ranikhet (Kumau)
5. Virus
6. 42
7. Papillionaceae
8. Pongamiapinmta
9. Flavr – Savr Tomato
10. Biofertilizers
11. Neem
12. Pest
13. Rhizobium
14. Cyanobacteria
15. Rhizobium.

Question 3.
Match the followings :
I.

Answer:

1. (c)
2. (d)
3. (e)
4. (a)
5. (b).

II.

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b).

III.

Answer:

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

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

1. Name the type of vegetative propagation in the plants having bulb.
2. Name the product of hybridization of two different species.
3. Name the first man – made crop.
4. Give one example of millet.
5. Write botanical name of coconut. (MP 2013,17)
6. What is called breeding between two closely related animals?
7. In which animal Ranikhet disease occur ? (MP 2015)
8. Name the causal organism of anthrax disease of cattles.
9. Write the name of galghontu disease of cattles.
10. Which hormone is used for production of milk in cows? (MP2016)
11. Give the botanical name of any one cereal plants.
12. Name the soil best suited for rice cropping.
13. Give the botanical name of any oil seed crop.
14. Give the botanical name of Ratanjot.
15. Name the technique by which genetically modified plants are produced.

Answer:

1. Scaling
2. Hybrid
3. Triticale
4. Rye
5. Coccus nucifera
6. Inbreeding
7. Fowls
8. Bacillus anthraxis
9. Pasturella boviseptica
10. Stilbrestol hormone,
11. Triticum aestivum (Wheat)
12. Clay loam
13. Arachis hypogea (Groundnut)
14. Jatropha curcas
15. Transgenesis.

Strategies for Enhancement in Food Production Short Answer Type Questions

Question 1.
What is meant by the term ‘breed’? What are the objective of animal breeding?
Answer:
Breed referes to the group of animals having same ancestral characters, general appearance, size, etc.

Objectives of animal breeding:

1. To increase the quantity of yield.
2. To improve the desirable qualities of the produce.

Question 2.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Answer:
The methods employed in animal breeding are :

1. Natural methods: These can be carried out by inbreeding and outbreeding methods.
2. Artificial methods : These are carried out by artificial insemination and Multiple Ovulation Embryo Transfer (MOET).

The artificial method of animal plant breeding is best as it ensures good quality of progeny. It is also economic and time saving process to obtain the desirable progeny.

Question 3.
What is apiculture? How is it important in our lives?
Answer:
Apiculture is the practice of bee – keeping for the production of various products such as honey-bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. Other commercial products obtained from honeybees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes and is even used inseveral medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee – keeping on a large scale. It has become an income generating activity for farmers since, it requires a low investment and is labour intensive.

Question 4.
Discuss the role of fishery in enhancement of food production.
Answer:
Fishery is an industry which deals with catching, processing and marketing of fishes and other aquatic animals that have a high economic value. Some commercially important aquatic animals are Prawns, Crabs, Oysters, Lobsters and Octopus. Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein. A fishery is an employment generating industry especially for people staying in the coastal areas.

Question 5.
Briefly describe various steps involved in plant breeding.
Answer:
The major steps in breeding a new genetic variety of a crop are as follows:

1. Collection of variability.
2. Evaluation and selection of parents.
3. Cross – hybridization among the selected parents.
4. Selection and testing of superior recombinants.
5. Testing, release and commercialization of new cultivars.

Question 6.
Explain what is meant by biofortification.
Answer:
Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals and micro – nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

Question 7.
Which part of the plant is best suited for making virus free plant and why?
Answer:
The terminal bud having apical meristem are the best suited parts of plant for making virus free plant because they are not infected by virus.

Question 8.
What are the major advantages of producing plants by nticropropagation?
Answer:
Major advantages of producing plants by micropropagation are:

1. Large number of plants can be grown in short – time
2. Disease – free plants can be obtained.
3. Plants that have lost the capacity to produce seeds can be grown
4. The plants where sexual reproduction is absent, may be hybridised by tissue culture.
5. Plants produced are genetically similar to the parent and have all its characteristics.

Question 9.
Find out what are the various components of the medium used for propagation of an explant in vitro are?
Answer:
The major components of the medium for in-vitro propagation are :

1. Water
2. Agar – agar
3. Sucrose
4. Inorganic salts
5. Vitamins
6. Amino acids
7. Growth hormones like Auxin, Cytokinins.

Question 10.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:

1. Cauliflower varieties – Pusa shubhra and Pusa snowball K – 1
2. Brassica varieties – Pusa swamim (Karan rai)
3. Wheat varieties – Himgiri
4. Rice varieties – Jaya and Ratna
5. Chilli varieties – Pusa Sadabahar.

Question 11.
What is scaling? What are its importance?
Answer:
Scaling:
It is a method of vegetative propagation which is applicable for growing plants having bulbs, (example Onion, Garlic). In this method, all the scales of bulb are separated and planted in the field (Soil) having all the necessary conditions for their growth. Here scales develop to produced small bulblets. About 3 to 5 bulblets developing from one scale. This method is applicable for the plants belonging to the family Liliaceae. example Garlic, Onion, Lilium, etc.

Question 12.
What is tissue culture? What are its objectives?
Answer:
Tissue culture:
“Tissue culture is an experimental process through which a mass of cells (callus) is produced from an explant tissue and the callus produced in this way can be used directly either to regenerate plantlets or to extract to some primary and secondary metabolites.”

When appropriate culture conditions were provided, cell masses could then proceed along various developmental pathway, to regenerate shoot and root organs and eventually whole plants.

Aims of Plant Tissue Culture:

1. To develop new plants from the plant organ other than seeds.
2. To produce hybrid varieties of plants.
3. To produce disease free plants from diseased plants.
4. To reduce the period of reproductive cycle.
5. Development of haploid plants.
6. To develop stress resistance plants.

Question 13.
What is callus culture? Give its technique.
Answer:
Callus culture:
Callus is the unorganised and undifferentiated mass of tissue. The plant body of higher plants is made of multicellular, well organised differentiated structures like root, stem, leaves etc. Sometimes the cells of these differentiated structures proliferate to form large mass of unorganized and undifferentiated cells which is called as callus and this process is called callus culture. In this methods, the isolated plant cells tissues or organs are cultured in nutrient medium in glass culture tubes or in petridishes under aseptic conditions. The cultured part is called as explant. The growth responses depend upon the source of the explant, composition of the nutrient medium and the suitable growth conditions.

Technique of callus culture : Callus culture involves the following steps :

1. First it is necessary to sterilize the plant organ from which an explant is taken. Sterilizing agents commonly used are mercuric chloride solution (0.1 w/v), sodium hypochloride (1.6 available chlorine) and a solution of bromine in water (1% w/v). Explants can be taken from seedlings (cotyledon, hypocotyl or root) or mature organs or from wood stem parts.

2. Wash the explant with distilled water, cut small pieces of it and transfer it on suitable culture medium. Agar agar is used for making culture medium. The culture media was supplemented with amino acids, vitamins, kinetin or other growth factor. An auxin and usually a cytokinin promote high growth rate of callus.

Question 14.
Why aeration is essential in the process of tissue culture?
Answer:
For any living being, respiration is a must. Through respiration only it can perform various activities. Oxygen is necessary for respiration.Tissue culture also needs oxygen, then only it can grow. Aeration provides oxygen to the plant parts, thus aeration is essential in the process of tissue culture.

Question 15.
What is inbreeding ? Give its advantages.
Answer:
The mating of more closely related individuals within the same breed for 4 – 6 generation is called inbreeding. It may lead to inbreeding depression i.e., the loss of fertility, vigour and productivity of the hybrid.

Question 16.
List the components which are present in honey. Name the three species of honeybee and write the chemical compositions of honey.
Answer:
There are three species of honeybee:

1. Apis indica
2. Trigona species
3. Malinopa species.

Uses of honey:

1. In the form of medicine.
2. In the form of nutritive food material.

Chemical composition :

Fructose – 41%
Glucose – 35%
Sucrose – 1.9%
Dextrine – 1.5%
Protein – 0.18%
Mineral salts – 3.3%
Water – 17.25%

Some amount of vitamins B, B6, Coline, Vitamin C and D.

Question 17.
What precautions should be kept in the process of incubation?
Answer:
Incubatory period is 21 days in fowl, precautions of incubation of hens are as follows:

1. The eggs chosen should be of good quality.
2. Egg size should be medium.
3. Colour of egg should be white.
4. Eggs should be washed in water.
5. Eggs should not be shaken.
6. In summers, eggs should not be kept more than 30 days.
7. At night, the fowl should be fed before incubation. Incubation of egg should be done by Indian fowl.

Question 18.
Give three significance of poultry farming.
Answer:

1. Poultry provide humans with companionship, food and fiber in the form of eggs, meat and feathers.
2. Poultry rearing and poultry farming is a good source of income.
3. It provides additional income and job opportunities.

Question 19.
Give the characteristics of meat providing chickens and give two examples of these.
Answer:
Characteristics of meat providing chickens are as follows:

1. Size is big.
2. They intake large quantity of nutrition.
3. Theii feathers are loose.
4. Their growth rate is low.
   Example: Australorps, Sussex.

Question 20.
Write the examples of three freshwater and three marine water fishes and give its significance.
Answer:
Examples of three freshwater fishes :

1. Rohu
2. Catla,
3. Seenghala.

Significance of Freshwater fishes :

1. Rohu (Labio rohita) : Its brain is rich in phospho protein which improve eye sight.
2. Catla {Catla catla) : Its brain is rich in phospho protein which improve eye sight.
3. Seenghala {Mystus seenghala) : It is rich in iron and copper which is good for circulatory system.

Examples of three marine water fishes:

1. Hilsa
2. Pomfret
3. Bombayduck.

Significance of Marine water fishes :

1. Hilsa (Ilishaa species): It is rich in omega – 3 fatty acid.
2. Pomfret {Sfromatus niger): High in vitamin D
3. Bombay duck {Harpodon): It is the source of calcium.

Question 21.
Fish meat is better than other animals, why?
Answer:

1. Fish meat is better because it is rich in protein.
2. Iodine is found in it which protects from goiter disease.
3. It has low fat so, it protects our heart.
4. Fat soluble vitamin A and D are found in much quantity.
5. It is digestible. So, it is better for children.

Question 22.
Name the developed breed of cow and buffalo. Explain why buffalo milk is better than cows milk.
Answer:
Developed varieties of Cow:

1. Holstein friesian
2. Jersey
3. Ayer Shayer and
4. Brown swiss.

Developed varieties of Buffalo:

1. Murra
2. Surti
3. Bhadavari and
4. Nagpuri.

Buffalo’s milk is better than Cows milk because :

1. Buffalo gives triple qunatity of milk than cow.
2. Buffalo milk is rich in fat.
3. It has much resistance power.

Question 23.
What is the significance of sheep and goat? Give the name of three Indian species of each.
Answer:
Zoological name of sheep is Ovis aries. It gives wool, meat and leather. Its Indian species are:

1. Lohi: It gives carpet quality of wool and meat production.
2. Bakharwal.
3. Patanwadi.

Zoological name of goat is Capra aegagrus. It gives us both meat and milk. The three, species are as follows :

1. Cashmere goat.
2. Sirohi.
3. Jamunapari.

Question 24.
Name the five species of chicks.
Answer:

1. Rhode Island Red
2. New Hampshired
3. Light Sussex
4. Australorps
5. White Leghorn.

Question 25.
How foot and mouth disease of cattles spreaded? Write symptoms and control measures of this disease.
Answer:
Foot and mouth disease is a highly contagious viral disease caused by Foot and Mouth Disease Virus (FMDV). It affects clovenhoofed animals including cattle, buffalo, camels, sheep, goats, deer and pigs. Since, it is highly infectious, can be spread by infected animals through contact with contaminated animals.

Symptoms of disease:

1. High fever lasting two to five days.
2. Animals become dull, eat less due to painful lesions in the mouth.
3. Weight loss.
4. Ulcer in mouth, foot and teats and causes wound thus, parts become painful and it fails to walk properly.

Question 26.
Give characteristics of egg laying chicken.
Answer:
Characteristics of egg laying chicken are as follows:

1. Their skin is soft.
2. A gap of 3 – 4 fingers is there between pubic and keel bones.
3. Their body is heavy in weight.
4. Their vent is soft and moist.
5. Their romb is well developed, red in colour and soft.

Strategies for Enhancement in Food Production Long Answer Type Questions

Question 1.
Write the name of causal organism, symptoms of disease and control measures of ranikhet disease of fowls.
Answer:
1. Ranikhet Disease : This disease was first discovered from the hills of Ranikhet (Kumau), hence called Ranikhet disease.

2. Causal organism: Virus.

3. Symptoms:

• Difficulty in breathing.
• Diarrhoea.
• Head, neck and legs are paralyzed.
• Loss of appetite.
• Increase in the temperature of the body.
• Secretion of mucilagenous substance from the mouth and nostrils.
• The colour of the body become voilet.
• Laying of egg is inhibited and eggs become ruptured.

4. Control and Treatment:

1. Isolation.
2. Died individuals should be immediatly burned.
3. Water should be disinfected.
4. Vaccination by ranikhet vaccine :

• F strain vaccine given to 1 to 3 days old chicks.
• Freeze dried chick embryo vaccine is given to 6 to 8 months old chicks.

Question 2.
Write the method and advantages of artificial breeding in cattles.
Or
What is artificial insemination? Write its importance.
Answer:
Artificial Hybridization:
It is a method in which sperms of male are injected into the reproductive tract of female ones. Where they fertilize egg to produce individuals with new characters.

Artificial hybridisation involves the following steps:

1. Selection of Parents:
It is the first step of natural hybridization in which male and female individuals with desired characters are selected. Usually healthy animals with desired characters are selected.

2. Collection of Semen:
It is the second step of artificial hybridization. In this step male individuals are stimulated mechanically or electrically so, that they release semen. This semen is collected in vials.

3. Preservation of Semen:
Collected semen is diluted with appropriate diluting liquid and is stored in refrigerators.

4. Introduction of Semen into vagina (= Insemination):
In this step, the desired semen is injected into the vagina of female individuals when they are heated. This technique was used for the first time by Spallanzani in 1780 in dogs. In India, this technique was used for the first time by Animal Research Institute, Eta Nagar (U.P.). More than 10 – 70% species of cattles are developed by this technique.

Precaution taken during insemination:

1. Insemination should be done in appropriate period.
2. Only high quality semen should be used.
3. Correct technique of insemination should be used.
4. The animals should be healthy during insemination.

Advantages of artificial insemination:

1. Few semen of healthy males are used for inseminating many female individuals.
2. The transportation of semen in ampules is easier.
3. The problem of availability of suitable males is solved by the development of this technique.
4. Cattles with desired characters should be used.

Question 3.
How can we provide high living standards for poultry farming?
Answer:
The steps for providing high living standards are as follows:

1. Protection
2. Provide proper places
3. Necessary facility
4. Cheap and comfortable
5. Clean
6. Facility of water and light
7. Nearness from market
8. Protection from parasites and other insects.

While the construction of poultry farms see that they are made of low prices and keep in mind the following things:

1. There is no moisture left in the farm.
2. Farm should be in a place where there is proper sunlight because microorganism can not grow in this situation.
3. There should be proper aeration facility because there are no sweat glands present in fowls as they release moisture by
4. the process of breathing.
5. There should be cleanliness in farm.
6. Fowl is a feable bird who has lots of enemies this is the reason they should be defended.

Question 4.
What is aquaculture? Describe various steps of aquaculture.
Answer:
The culture of useful aquatic organisms specially animals is called as aquaculture.

Steps involved in aquaculture (Fishery):
Four types of ponds are required for the culture of fished at commercial level.

1. Hatchery tank : For fish culture first of all fish seeds are sown in hatchery tank.
2. Nursing tank : After hatching, small fishes are transferred to the nursing tank.
3. Rearing tank : When fishes attain a size of 2 – 3 inches than they are transferred into the rearing tank.
4. Storage or Stocking tank : After attaining Suitable size, these fishes are transferred into the storage tank where these fishes are stored and supplied to the market.

Question 5.
What is single cell culture? Explain paper raft technique and its application.
Answer:
Single Cell Culture :
The culturing of a single cell on suitable sterilized culture media under controlled and aseptic conditions is called single cell culture. Establishment of a single cell culture provides an excellent opportunity to investigate the properties and potentialities of plant cells. Several workers have successfully isolated single cell division and even raised complete plants from single cell cultures.

Paper raft technique of single cell culture:
1. In this technique, the single cells are isolated with the help of micropipette and transferred on to the upper surface of filter paper resting on nurse callus. These active nurse callus and the culture medium piovides growth factors to the single cells.

2. The single cells then divides and as a result of proliferation in them, the colonies develop.

3. Colonies formed in nutrient medium are transferred to the subcultured fresh nutrient media. The callus developing from such single cells are called as single cell clones.

Question 6.
There is a scope for better prospects through aquaculture in the rural areas of Chhattisgarh. Discuss.
Answer:
In the village area of Chhattisgarh ponds are developed. In all villages and towns of Chhattisgarh many ponds can be seen. Canals, dams and water reservoirs are constructed for development purpose. These all sources can be used for progress of the state by scientific way.

Production of useful aquatic organisms is known as aquaculture. Various algae and aquatic animals such as fishes, prawn, crab, oyster etc. are cultured in pond Pisciculture, Prawn culture and Pearl culture are various main culture of aquaculture. Pisciculture is rearing and management of fishes. Prawn culture is production of prawn. Pearl culture is rearing and management of oyster for pearl production. All these productions has commercial importance.

Question 7.
How can crop varieties be made disease resistant to overcome food crisis in India? Explain, Name some disease resistance variety in India. Give the method of breeding for disease resistant variety.
Answer:
Crop varieties can be made disease – resistant by conventional breeding methods or by mutation breeding. The germplasm is screened for resistance source or mutations are introduced followed by hybridization of selected parents. The resulting hybrids are evaluted and tested finally, disease resistant varieties are released.

Disease resistant variety of:

1. Wheat to leaf and stripe rust – Himgiri.
2. Brassica to white rust – Pusa swamim.

Methods to breeding for disease resistance:

1. Hybridization.
2. Selection.

These steps are involved it:

1. Selections of breed from resistant source.
2. Hybridization of selected breed.
3. Selection of Hybrid.
4. Evaluation.
5. Testing of new variety and its production.

By hybridization and selection some fungi, bacteria and viruses can be disease resistant. These varieties given below :

Table : Disease resistant variety of crops.

Question 8.
How is plant breeding helpful to provide resistance to pests?
Answer:
Plant breeding for developing resistance to insect pests:

1. The host crop plants may be resistant to insect pests due to the morphological, biochemical or physiological characteristics.

2. Some characteristics that lead to pest resistance are :

Hairy leaves in plant example resistance to Jassids in cotton and cereal leaf beetle in wheat.
Solid stem in wheat exhibits non – preference by stem sawfly.
In cotton, smooth leaf and absence of nectar repel bollworms.
In maize, high aspartic acid, low nitrogen and sugar content protects them from stem borers.

3. The steps of breeding method is same as for the other agronomic traits.

4. Some varieties developed by hybridization and selection are as follows:

Question 9.
Explain in brief the role of animal husbandry in human welfare.
Answer:
1. Animal husbandry evolves new techniques and technologies for the management of livestock like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc., that are useful to humans.

2. These methods can also be applied rearing animals like bees, silkworm, prawns, crabs, fishes birds, pigs, cattle, sheep and camels for their products like milk, eggs, meat, wool, silk, honey, etc.

Question 10.
If your family owned a diary farm, what measures would you undertake to improve the quality and quantity of milk production?
Answer:
Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities and regular cleaning of cattle. Choosing improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are produced for improved productivity.

Therefore, it is essential that hybrid cattle breeds should have a combination of various desirable genes such, as high milk production and high resistance to diseases, Cattle should also be given healthy and nutritious food consisting of roughage, fibre concentrates and high levels of proteins and other nutrients. Cattle’s should be housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from harsh weather conditions such as heat, cold and rain. Regular baths and proper brushing should be ensured to control diseases. Also, time – to – time check – ups by a veterinary doctor for symptoms of various diseases should be undertaken.

MP Board Class 12th Biology Important Questions

MP Board Class 11th Chemistry Solutions Chapter 13 हाइड्रोकार्बन

हाइड्रोकार्बन NCERT अभ्यास प्रश्न

प्रश्न 1.
मेथेन के क्लोरीनीकरण के दौरान ऐथेन कैसे बनती है ? आप इसे कैसे समझाएँगे?
उत्तर:
मेथेन का क्लोरीनीकरण मुक्त मूलक क्रियाविधि द्वारा होता है। मेथिल मुक्त मूलक (CH₃ ) श्रृंखला समापन पद के दौरान ऐथेन में रूपान्तरित हो जाते हैं।

प्रश्न 2.
निम्नलिखित यौगिकों के IUPAC नाम लिखिए –
(i) CH₃ CH = C(CH₃)₂
(ii) CH₃ = CH – C ≡ C – CH₃

उत्तर:

प्रश्न 3.
निम्नलिखित यौगिकों, जिनमें द्विआबन्ध तथा त्रिआबन्ध की संख्या दर्शायी गई है, के सभी सम्भावित स्थिति समावयवों के संरचना-सूत्र एवं IUPAC नाम दीजिए –
(a) C₄H₈ (एक द्विआबन्ध)
(b) C₅H₈ (एक त्रिआबन्ध)।
उत्तर:
(a) C₄H₈ के समावयवी जो द्विआबंध रखते हैं –

(b) C₅H₈ के समावयवी जो त्रिआबंध रखते हैं –

प्रश्न 4.
निम्नलिखित यौगिकों के ओजोनी-अपघटन के पश्चात् बनने वाले उत्पादों के नाम लिखिए –
(1) पेन्ट-2-ईन
(ii) 3, 4-डाइमेथिलहेप्ट-3-ईन
(iii) 2- एथिलब्यूट-1-ईन
(iv) 1-फेनिलब्यूट-1-ईन।
उत्तर:

प्रश्न 5.
एक ऐल्कीन ‘A’ के ओजोनी अपघटन से पेन्टेन-3-ओन तथा ऐथेनल का मिश्रण प्राप्त होता है। A का IUPAC नाम तथा संरचना दीजिए।
उत्तर:
ओजोनी अपघटन के फलस्वरूप बने उत्पादों की संरचना इस प्रकार लिखते हैं जिससे उनके ऑक्सीजन परमाणु एक दूसरे की ओर इंगित रहे। दोनों ऑक्सीजन परमाणु को निकालने के पश्चात् दोनों सिरों की द्विआबंध से जोड़कर, एल्कीन ‘A’ की संरचना प्राप्त करते है।

प्रश्न 6.
एक ऐल्कीन A में तीन C – C, आठ C – H सिग्मा आबन्ध तथा एक C – C पाई आबन्ध है। A ओजोनी अपघटन से दो अणु ऐल्डिहाइड, जिनका मोलर द्रव्यमान 44 है, देता है। A का आई.यू.मी.ए.सी. नाम लिखिए।
उत्तर:
मोलर द्रव्यमान 44u वाला ऐल्डिहाइड, ऐथेनल (CH₃CHO) है। अतः इसके दो अणु की संरचनाओं से –

अत: एल्कीन ‘A’ अर्थात् ब्यूट-2-ईन में तीन C-C, आठ C- HO – आबंध तथा एक C – C  2-आबंध उपस्थित है। जो संख्या प्रश्नों में दी गई है।

प्रश्न 7.
एक एल्कीन, जिसके ओजोनी अपघटन से प्रोपेनल (Propanal) तथा पेन्टेन-3-ओन प्राप्त होते हैं, का संरचनात्मक सूत्र क्या है ?
उत्तर:
प्रोपेनल तथा पेन्टेन-3-ओन में से ऑक्सीजन अलग होकर ये द्वि-बंध द्वारा जुड़ जाते हैं।

प्रश्न 8.
निम्नलिखित हाइड्रोकार्बनों के दहन की रासायनिक अभिक्रिया लिखिए –
(i) ब्यूटेन
(ii) पेन्टीन
(iii) हेक्साइन
(iv) टॉलुईन।
उत्तर:
सभी हाइड्रोकार्बन पूर्ण दहन पर CO₂ और H₂O देते है।

प्रश्न 9.
हेक्स-2-ईन की समपक्ष (सिस) तथा विपक्ष (ट्रान्स ) संरचनायें बनाइए। इनमें से कौनसे समावयव का क्वथनांक उच्च होता है और क्यों ?
उत्तर:
हेक्स-2-ईन (CH₃CH₂CH₂CH = CHCH₃) की समपक्ष तथा विपक्ष समावयव की संरचनाएँ निम्न हैं –

सिस समावयवी में क्वथनांक अपेक्षाकृत उच्च होगा क्योंकि इसके द्विध्रुव-द्विध्रुव अन्योन्य क्रिया का परिणाम, ट्रान्स समावयवी की अपेक्षा अधिक होता है।

प्रश्न 10.
बेंजीन में तीन द्वि-आबन्ध होते हैं, फिर भी यह अत्यधिक स्थायी है, क्यों?
उत्तर:
अनुनाद तथा इलेक्ट्रॉनों के विस्थानीकरण के कारण बेन्जीन अणु अत्यधिक स्थायी होता है।

संकरित संरचना संकरित संरचना में बिन्दु वृत्त, बेन्जीन वलय के छ: कार्बन परमाणु पर विस्थानीकृत छ: इलेक्ट्रॉनों को दर्शाता है। अतः विस्थानीकृत छः इलेक्ट्रॉनों की उपस्थिति के कारण बेन्जीन अत्यधिक स्थायी होता है।

प्रश्न 11.
किसी निकाय द्वारा ऐरोमैटिकता प्रदर्शित करने के लिए आवश्यक शर्ते क्या हैं ?
उत्तर:
किसी निकाय का ऐरोमैटिक गुण प्रदर्शित निम्न शर्तों पर निर्भर करता है –

• यौगिक संरचना समतलीय होनी चाहिये।
• यौगिक अनुनादी संरचनायें प्रदर्शित करें।
• यौगिक हकल नियम का पालन करें।
• यौगिक इलेक्ट्रॉन स्नेही अभिक्रियाएँ प्रदर्शित करें

प्रश्न 12.
इनमें से कौन-से निकाय ऐरोमैटिक नहीं है ? कारण स्पष्ट कीजिए –

उत्तर:

(a) यह संरचना समतलीय नहीं है क्योंकि इसमें एक C परमाणु sp’ संकरित अवस्था में उपस्थित है।
(b) यह संरचना इलेक्ट्रॉन स्नेही प्रतिस्थापी अभिक्रियाएँ प्रदर्शित नहीं करती।

(ii)
यह संरचना ऐरोमैटिक गुण प्रदर्शित नहीं करता क्योंकि
(a) इसकी संरचना समतलीय नहीं है (इसमें एक C परमाणु का संकरण अवस्था sp’ है।)
(b) यह हकल नियम का पालन नहीं करता।
(c) यह अनुनादी संरचनायें भी प्रदर्शित नहीं करता।
(d) यह इलेक्ट्रॉन स्नेही प्रतिस्थापी अभिक्रियाएँ नहीं देती।

(ii)
यह संरचना ऐरोमैटिक गुण प्रदर्शित नहीं करता क्योंकि
(a) इसकी संरचना समतलीय नहीं है।
(b) यह संरचना हकल नियम का पालन नहीं करती।।
(c) यह इलेक्ट्रॉन स्नेही प्रतिस्थापी अभिक्रियाएँ प्रदर्शित नहीं करती।

प्रश्न 13.
बेंजीन को निम्नलिखित में कैसे परिवर्तित करेंगे –

1. p – नाइट्रोब्रोमोबेंजीन
2. m – नाइट्रोक्लोरोबेंजीन
3. p – नाइट्रोटॉलुईन
4. ऐसीटोफिनोन।

उत्तर:
ये रूपान्तरण इस तथ्य पर आधारित होते हैं कि –

• क्लोरीन की प्रकृति ऑर्थों और पैरा निर्देशी होती है।
• नाइट्रो समूह की प्रकृति मेटा निर्देशी होती है।

1. बेंजीन से p – नाइट्रोनोमोबेंजीन –

2. बेंजीन से m – क्लोरोनाइट्रोबेंजीन –

3. बेंजीन से p – नाइट्रोटॉलुईन –

4. बेंजीन से p – एसीटोफिनोन –

प्रश्न 14.
ऐल्केन H₃C – CH₂ – C(CH₃)₂ – CH₂– CH(CH₃)₂ में 1,2° तथा 3° कार्बन परमाणुओं की पहचान कीजिए तथा प्रत्येक कार्बन से आबन्धित कुल हाइड्रोजन परमाणुओं की संख्या भी बताइए।
उत्तर:

प्रश्न 15.
क्वथनांक पर ऐल्केन की श्रृंखला के शाखन (Branch) का क्या प्रभाव पड़ता है ?
उत्तर:
श्रृंखला में शाखन में वृद्धि के फलस्वरूप पृष्ठ सतह घटता है। अतः एल्केन की श्रृंखला के शाखन में वृद्धि के फलस्वरूप वाण्डर वाल्स आकर्षण बलों में कमी के कारण, क्वथनांक घटता है।

प्रश्न 16.
प्रोपीन पर HBr के संकलन से 2-ब्रोमोप्रोपेन बनता है, जबकि बेन्जॉयल परॉक्साइड की उपस्थिति में यह अभिक्रिया 1-ब्रोमोप्रोपेन देती है। क्रियाविधि की सहायता से इसका कारण स्पष्ट कीजिए।
उत्तर:
प्रोपीन (असममित ऐल्कीन) पर HBr योग मार्कोनीकॉफ नियम के अनुसार धनायनिक क्रियाविधि से संपादित होती है। HBr इलेक्ट्रॉन स्नेही, H’ उत्पन्न करता है जो कार्बधनायन से द्विबंध पर आक्रमण करता है।
(i) H – Br → H⁺ + Br^–

इस -2° कार्बधनायन पर Br आयन आक्रमण करके उत्पाद बनाता है।

ब्रोमो प्रोपेन (मुख्य उत्पाद) बेन्जॉयल परॉक्साइड की उपस्थिति में प्रोपीन पर HBr का योग मार्कोनीकॉफ नियम के विपरीत होता है। यह मुक्त मूलक की क्रिया से संपादित होता है। इस अभिक्रिया को प्रति मार्कोनीकॉफ योग या परॉक्साइड प्रभाव या खराश प्रभाव कहते हैं। यह केवल HBr के साथ होता है, HCl तथा HI के साथ नहीं। क्रियाविधि निम्न हैं –

प्रश्न 17.
1, 2-डाइमेथिलबेंजीन (o – जाइलीन) के ओजोनी अपघटन के फलस्वरूप निर्मित उत्पादों को लिखिए। यह परिणाम बेंजीन की केकुले संरचना की पुष्टि किस प्रकार करता है ?
उत्तर:

इसी प्रकार,

ये तीनों उत्पाद दोनों कैकुले संरचनाओं में से किसी एक द्वारा ही नहीं प्राप्त किए जा सकते है। इससे प्रदर्शित होता है कि o-जाइलीन, वास्तव में दो कैकुले संरचनाओं ग्लाइऑक्सल (I तथा II) का अनुनादी रूप है।

प्रश्न 18.
बेंजीन, n – हेक्सेन तथा एथाइन को घटते हुए अम्लीय व्यवहार के क्रम में व्यवस्थित कीजिए और इस व्यवहार का कारण बताइए।
उत्तर:
अम्लीय सामर्थ्य का घटता हुआ क्रम है –

अम्लीय अभिलक्षण S-अभिलक्षण की प्रतिशतता से सम्बन्धित होता है। 5-अभिलक्षण जितना अधिक होता है, कार्बन परमाणु की वैद्युत ऋणात्मकता उतनी ही अधिक होती है तथा उसका अम्लीय अभिलक्षण भी उतना ही अधिक होगा।

प्रश्न 19.
बेंजीन इलेक्ट्रॉनस्नेही प्रतिस्थापन अभिक्रियाएँ सरलतापूर्वक क्यों प्रदर्शित करती है, जबकि उसमें नाभिकस्नेही प्रतिस्थापन कठिन होता है ?
उत्तर:
बेंजीन में द्विआबन्धों को प्रदर्शित करने वाले तीन T – इलेक्ट्रॉनों युग्मों के कारण इलेक्ट्रॉन घनत्व उच्च होता है। यद्यपि इलेक्ट्रॉन आवेश अनुनाद के कारण बहुत अधिक विस्थानीकृत हो जाता है, तथापि इसमें इलेक्ट्रॉनस्नेही प्रतिस्थापन कराने वाला इलेक्ट्रॉन-स्नेही आक्रमण सम्भव होता है। किन्तु, बेंजीन नाभिकस्नेही प्रतिस्थापन के प्रति अनुक्रियाशील नहीं होता है क्योंकि नाभिकस्नेही प्राथमिकता के तौर पर कम इलेक्ट्रॉन घनत्व के केन्द्र पर आक्रमण करता है।

प्रश्न 20.
आप निम्नलिखित यौगिकों को बेंजीन में कैसे परिवर्तित करेंगे –

1. एथाइन
2. एथीन
3. हेक्सेन।

उत्तर:
1. एथाइन से बेंजीन –

2. एथीन से बेंजीन –

3. हेक्सेन से बेंजीन –

प्रश्न 21.
उन सभी एल्कीनों की संरचनाएँ लिखिए, जो हाइड्रोजनीकरण करने पर 2-मेथिलब्यूटेन देती हैं।
उत्तर:
उपर्युक्त प्रश्न को हम निम्न प्रकार से मालूम करेंगे –
2-मेथिलब्यूटेन प्राप्त करने के लिये कौन-सी एल्कीन लेनी होगी –

उपर्युक्त संरचना को हमें एल्कीन से प्राप्त करनी है। उपर्युक्त संरचना में पास-पास वाले C परमाणुओं से H परमाणु हटा कर द्विबन्ध जोड़ दें।

प्रश्न 22.
निम्नलिखित यौगिकों को उनकी इलेक्ट्रॉनस्नेही (E’) के प्रति घटती सापेक्षिक अभिक्रियाशीलता के क्रम में व्यवस्थित कीजिए –
1. क्लोरोबेंजीन, 2-4-डाइनाइट्रोक्लोरोबेंजीन, p-नाइट्रो-क्लोरोबेंजीन
2. टॉलुईन, P – H₃C – C₆H₄ – NO₂, P – O₂N – C₆H₄ – NO₂
उत्तर:
1. इलेक्ट्रॉनस्नेही प्रतिस्थापन के प्रति अभिक्रियाशीलता का घटता हुआ सही क्रम है –
क्लोरोबेंजीन >p-नाइट्रोक्लोरोबेंजीन >2,4-डाईनाइट्रोक्लोरोबेंजीन
नाइट्रो समूह (NO₂) एक विसक्रियकारी समूह है। बेंजीन वलय पर इसकी उपस्थिति इसे इलेक्ट्रॉनस्नेही आक्रमण के प्रति विसक्रियित करेगी, क्योंकि इलेक्ट्रॉनस्नेही उच्च इलेक्ट्रॉन घनत्व के केन्द्र की खोज करता है। अतः नाइट्रो समूहों की संख्या जितनी ही अधिक होगी, इलेक्ट्रॉनस्नेही प्रतिस्थापन के प्रति यौगिकों की अभिक्रियाशीलता उतनी ही कम होगी।

2. घटती हुई अभिक्रियाशीलता का सही क्रम है – .
टॉलुईन > p – नाइट्रोटॉलुईन > p – डाईनाइट्रोबेंजीन
मेथिल समूह एक सक्रियकारी समूह है, जबकि नाइट्रो समूह की प्रकृति विसक्रियकारी होती है। इस तथ्य के प्रकाश में इलेक्ट्रॉनस्नेही आक्रमण के प्रति अभिक्रियाशीलता का घटता हुआ यह क्रम तर्कसंगत सिद्ध होता है –
टॉलुईन > बेंजीन > m-डाईनाइट्रोबेंजीन

प्रश्न 23.
बेंजीन, m – डाइनाइट्रोबेंजीन और टॉलुईन में से किसका नाइट्रीकरण आसानी से होता है। और क्यों ?
उत्तर:
बेंजीन के नाइट्रीकरण में वलय पर \(\mathrm{NO}_{2}^{\ominus}\) (नाइट्रोनियम आयन) का इलेक्ट्रॉनस्नेही आक्रमण होता है। चूंकि CH₃ समूह में +I प्रभाव होता है, इसलिए यह वलय को सक्रिय कर देता है और इसका इलेक्ट्रॉनस्नेही प्रतिस्थापन आसानी से होता है। दूसरी ओर, नाइट्रो समूह की विसक्रियकारी प्रकृति के कारण यह m-डाइनाइट्रोबेंजीन में सबसे कठिनाई से होता है। इस प्रकार टॉलुईन का नाइट्रीकरण सबसे आसानी से होगा। टॉलुईन > बेंजीन > m – डाईनाइट्रोबेंजीन

प्रश्न 24.
बेंजीन के एथिलीकरण में निर्जल ऐल्युमिनियम क्लोराइड के स्थान पर कोई दूसरा लुईस अम्ल सुझाइए।
उत्तर;
निर्जल फेरिक क्लोराइड (FeCl₅) वह अन्य लुईस अम्ल है जिसे प्रयुक्त कर सकते हैं। यह इलेक्ट्रॉनस्नेही (C₂H₅ई) पैदा करने में सहायक होता है। स्टेनिक क्लोराइड (SnCl₄) एवं बोरॉन ट्राइफ्लुओराइड (BF₃) . का भी प्रयोग किया जा सकता है।

प्रश्न 25.
क्या कारण है कि वुर्ट्ज अभिक्रिया से विषम संख्या कार्बन परमाणु वाले विशुद्ध ऐल्केन बनाने के लिए प्रयुक्त नहीं की जाती? एक उदाहरण देकर स्पष्ट कीजिए।
उत्तर:
कार्बन परमाणुओं की विषम संख्या वाले ऐल्केन के निर्माण में दो भिन्न-भिन्न हैलोऐल्केन आवश्यक होते हैं, जिनमें से एक में कार्बन परमाणुओं की विषम संख्या हो और दूसरे में सम संख्या हो। उदाहरण के लिए ब्रोमोएथेन और 1-ब्रोमोप्रोपेन इस अभिक्रिया के फलस्वरूप पेन्टेन देंगे।

परन्तु जब इस अभिक्रिया में भाग लेने वाले सदस्य अलग-अलग अभिक्रिया करेंगे तो पार्श्व उत्पाद भी बनेंगे। उदाहरण के लिए, ब्रोमोएथेन, ब्यूटेन देता है और 1- ब्रोमोप्रोपेन, हेक्सेन देता हैं।

इस प्रकार ब्यूटेन, पेन्टेन और हेक्सेन का मिश्रण बनेगा। इस मिश्रण से इनके घटकों को पृथक् करना काफी कठिन होगा।

हाइड्रोकार्बन अन्य महत्वपूर्ण प्रश्न

हाइड्रोकार्बन वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
मेथेन में H – C – H बन्ध कोण है –
(a) 100°5′
(b) 109°
(c) 109°28′
(d) 180°
उत्तर:
(c) 109°28′

प्रश्न 2.
C = C में है –
(a) 30 बन्ध
(b) एक सिग्मा और तीन पाई बन्ध
(c) 37 बन्ध
(d) एक सिग्मा और दो पाई बन्ध
उत्तर:
(d) एक सिग्मा और दो पाई बन्ध

प्रश्न 3.
एथीन में है –
(a) पाँच सिग्मा और एक पाई बन्ध
(b) छ: सिग्मा बन्ध
(c) चार सिग्मा और दो पाई बन्ध
(d) पाँच सिग्मा बन्ध।
उत्तर:
(a) पाँच सिग्मा और एक पाई बन्ध

प्रश्न 4.
किसकी क्रियाशीलता सर्वाधिक है –
(a) C₂H₂
(b) CH₄
(c) C₂H₄
(d) C₂H₆.
उत्तर:
(a) C₂H₂

प्रश्न 5.
ज्यामितीय समावयवता प्रदर्शित करता है –
(a) ब्यूट-2-ईन
(b) ब्यूट-2-आइन
(c) ब्यूटेन-2-ऑल
(d) ब्यूटेनल।
उत्तर:
(a) ब्यूट-2-ईन

प्रश्न 6.
बेंजीन में कार्बन द्वारा उपयोग में लाई गई प्रसंकरित ऑर्बिटल (कक्षक) होती है –
(a) sp³
(b) sp²
(c) sp
(d) dsp³
उत्तर:
(b) sp²

प्रश्न 7.
HC ≡ C – CH = CH, के C – C एक आबन्ध के कार्बन परमाणुओं का संकरण है –
(a) sp³ – sp³
(b) sp² – sp³
(c) sp³ – sp
(d) sp – sp²
उत्तर:
(d) sp – sp²

प्रश्न 8.
ब्यूट-1-इन में सिग्मा बन्धों की संख्या है –
(a) 8
(b) 10
(c) 11
(d) 12.
उत्तर:
(c) 11

प्रश्न 9.
एथिलीन में उपस्थित दो परमाणुओं के मध्य विद्यमान द्विबन्ध होता है –
(a) लम्बवत् दो सिग्मा-बन्ध
(b) एक सिग्मा (σ) तथा एक पाई (π) बन्ध
(c) लम्बवत् दो पाई-बन्ध
(d) 60 अंश का कोण बनाते हुए दो पाई-बन्ध।
उत्तर:
(b) एक सिग्मा (σ) तथा एक पाई (π) बन्ध

प्रश्न 10.
निम्न में से कौन-सा अभिकर्मक एथिलीन और ऐसीटिलीन में विभेद करता है –
(a) जलीय क्षारीय KMnO₄
(b) CCl₄ में Cl₂ विलेय
(c) अमोनियामय Cu₂ Cl₂
(d) सान्द्र H₂SO₄
उत्तर:
(c) अमोनियामय Cu₂Cl₂

प्रश्न 11.
इंजन में अपस्फोटी ध्वनि उत्पन्न होती है, जब ईंधन –
(a) धीरे जलता है
(b) तेजी से जलता है
(c) पानी होता है
(d) मशीन तेल मिश्रित होता है।
उत्तर:
(b) तेजी से जलता है

प्रश्न 12.
ईंधन का अपस्फोटक गुण बढ़ाने के लिये मिलाया जाता है –
(a) PbBr₂
(b) ZnBr₂
(c) Pbo
(d) TEL (Tetraethyl lead).
उत्तर:
(d) TEL (Tetraethyl lead).

प्रश्न 13.
मेथेन बनाने के लिए निम्न में से किस विधि का उपयोग किया जा सकता है –
(a) वु अभिक्रिया
(b) कोल्बे अभिक्रिया
(c) ऐल्किल हैलाइड का अपचयन
(d) ऐल्कीन का हाइड्रोजनीकरण।
उत्तर:
(c) ऐल्किल हैलाइड का अपचयन

प्रश्न 14.
जब ऐसीटिलीन HCl की उपस्थिति में HgCl₂ से अभिक्रिया करती है, तो बनता है
(a) मेथिल क्लोराइड
(b) ऐसीटैल्डिहाइड
(c) वाइनिल क्लोराइड
(d) फॉर्मेल्डिहाइड।
उत्तर:
(b) ऐसीटैल्डिहाइड

प्रश्न 15.
जब प्रोपाइन को HgSO₄ की उपस्थिति में जलीय H₂ SO₄ से प्रतिकृत किया जाता है, तो मुख्य उत्पाद बनता है –
(a) प्रोपेनल
(b) प्रोपिल हाइड्रोजन सल्फेट
(c) ऐसीटोन
(d) प्रोपेनॉल।
उत्तर:
(c) ऐसीटोन

प्रश्न 16.
प्रोपीन तथा प्रोपाइन में विभेद करने के लिए प्रयुक्त अभिकर्मक है –
(a) ब्रोमीन
(b) क्षारीय KMnO₄
(c) अमोनियामय AgNO₃
(d) ओजोन।
उत्तर:
(b) क्षारीय KMnO₄

प्रश्न 17.

(a) वुज अभिक्रिया
(b) कोल्बे अभिक्रिया
(c) साबात्ये सेण्डेरेन्स अभिक्रिया
(d) कार्बिल-ऐमीन अभिक्रिया
उत्तर:
(c) साबात्ये सेण्डेरेन्स अभिक्रिया

प्रश्न 18.
अमोनियाकृत सिल्वर नाइट्रेट विलयन से अभिक्रिया करके C₂ H₂ बनाता है –
(a) सिल्वर दर्पण
(b) सिल्वर ऑक्साइड
(c) सिल्वर फॉर्मेट
(d) सिल्वर ऐसीटिलाइड।
उत्तर:
(d) सिल्वर ऐसीटिलाइड।

प्रश्न 19.
एक अज्ञात यौगिक A का अणु सूत्र C₄ H₆ है । जब Aकी ब्रोमीन की अधिक मात्रा से अभि क्रिया करायी जाती है, एक नया पदार्थ B बनता है जिसका अणुसूत्र C₄ H₆ Br₄ है। A अमोनिया युक्त AgNO₃ के साथ सफेद अवक्षेप देता है। A हो सकता है –
(a) ब्यूट-1-आइन
(b) ब्यूट-2-आइन
(c) ब्यूट-1-ईन
(d) ब्यूट-2-ईन।
उत्तर:
(b) ब्यूट-2-आइन

प्रश्न 20.
क्रियाशीलता बढ़ाने वाले समूह होते हैं –
(a) o-,p-निर्देशकारी समूह
(b) m-निर्देशकारी समूह
(0) NO₂
(d) इनमें से कोई नहीं।
उत्तर:
(a) o-,p-निर्देशकारी समूह

प्रश्न 21.
निम्नलिखित में से कौन-सा एल्केन, वु अभिक्रिया द्वारा प्राप्त नहीं होता है –
(a) CH₄
(b) C₂H₆
(c) C₃H₈
(d) C₄H₁₀
उत्तर:
(a) CH₄

प्रश्न 22.
निम्न में प्रति ऐरोमैटिक है –
(a) बेंजीन
(b) साइक्लोऑक्टाडाईन
(c) ट्रोपोलियम धनायन
(d) साइक्लोपेन्टाडाइनील धनायन।
उत्तर:
(d) साइक्लोपेन्टाडाइनील धनायन।

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिये –

1. कैरोसीन तेल …………. का मिश्रण है।
2. ………… में कार्बन-कार्बन बन्ध लम्बाई न्यूनतम होती है।
3. प्रोपीन तथा प्रोपाइन में विभेद करने के लिए ………… अभिकर्मक प्रयुक्त किया जाता है।
4. टेफ्लॉन …………. का बहुलक है।
5. एथेनॉल के निर्जलीकरण से ………… बनता है।
6. बेंजीन H₂ से योग करके …………. बनाता है।
7. ज्यामितीय समावयवता ……………… में पायी जाती है।
8. सर्वाधिक स्थायी संरूपी …………… रूप होता है।
9. विपक्ष समावयवी, समरूप समावयवी से …………. स्थायी होता है।
10. ऐल्युमीनियम कार्बाइड की जल से अभिक्रिया कराने पर …………. प्राप्त होता है।

उत्तर:

1. एल्केन
2. एथाइन
3. बेयर अभिकर्मक (क्षारीय KMnO₄)
4. टेट्रा फ्लोरो एथिलीन
5. एथीन
6. साइक्लोहेक्सेन
7. ऐल्कीन
8. सांतरित
9. अधिक
10. CH₄

प्रश्न 3.
उचित संबंध जोडिए –

उत्तर:

1. (b) कोल्बे अभिक्रिया
2. (d) वु अभिक्रिया से
3. (e) मेथेन।
4. (a) इलेक्ट्रॉन स्नेही प्रतिस्थापन अभिक्रिया
5. (c) sp – संकरण

उत्तर:

1. (e) पार्श्व अतिव्यापन
2. (a) इलेक्ट्रॉन स्नेही प्रतिस्थापन
3. (1) 0, p समूह
4. (b) असंतृप्त हाइड्रोकार्बन
5. (c) द्रवीभूत पेट्रोलियम गैस
6. (g) कम्प्रेस्ड नेचुरल गैस
7. (d) समपक्ष-विपक्ष समावयवी

प्रश्न 4.
एक शब्द / वाक्य में उत्तर दीजिए –
1. T.E.L. का पूरा नाम है।
2. एथिलीन डाइ ब्रोमाइड को Zn चूर्ण के साथ गर्म करने पर क्या बनता है?
3. L.P.G. में दुर्गन्ध पैदा करने वाले पदार्थ का नाम बताइये।
4. पोटैशियम एसीटेट के जलीय विलयन के विद्युत् अपघटन से एथेन बनता है, इस विधि का नाम बताइए।
5.
6.

प्रश्न 7.
एल्किल हैलाइड व सोडियम से ऐल्केन बनाने के लिए प्रयुक्त अभिक्रिया का नाम लिखिए।
उत्तर:
1. टेट्रा एथिल लेड
2. एथिलीन
3. एथिल मर्केप्टेन
4. कोल्बे अभिक्रिया
5. साबात्ये सेण्डरेन्स अपचयन अभिक्रिया
6. (A) CH = CH, (B) CH₃ – CHO, (C) CH₃– CH₂– OH
7. वु फिटिग अभिक्रिया।

हाइड्रोकार्बन अति लघु उत्तरीय प्रश्न

प्रश्न 1.
ऐल्काइन के द्रव अमोनिया में सोडियम द्वारा अपचयन के फलस्वरूप विपक्ष ऐल्कीन बनती है। क्या 2-ब्यूटाइन के इस प्रकार हुए अपचयन के फलस्वरूप ब्यूटीन ज्यामितीय समावयवता प्रदर्शित करेगी?
उत्तर:

2-ब्यूटाइन के अपचयन से प्राप्त विपक्ष-2-ब्यूटीन ज्यामितीय समावयवता प्रदर्शित करती है।

प्रश्न 2.
-1 प्रभाव के बावजूद, हैलोजन, हैलोएरीन यौगिकों में 0 – (ऑर्थो) तथा p – (पैरा) निर्देशी होती है। कारण स्पष्ट कीजिए।
उत्तर:
हैलोजन अत्यधिक निष्क्रिय समूह है। इनके प्रबल – I प्रभाव के कारण बेन्जीन वलय पर इलेक्ट्रॉन घनत्व घटता है। परन्तु अनुनाद के कारण o – तथा p – स्थानों पर इलेक्ट्रॉन घनत्व m – स्थान की अपेक्षा अधिक होता है। जिसके कारण ये o – तथा p – निर्देशी समूह होते हैं।

प्रश्न 3.
एल्केनों से एल्कीन की अधिक क्रियाशीलता का क्या कारण है ?
उत्तर:
एल्केनों में C – C के मध्य केवल आबन्ध होता है। जबकि एल्कीन में C = C के मध्य एक – आबन्ध और एक आबंध होता है। पार्वीय अतिव्यापन के कारण π बंध, σ बंध से दुर्बल होता है। इस कारण ऐल्कीन, ऐल्केनों से अधिक क्रियाशील है। 1 आबंध की आबन्ध ऊर्जा (251 kJ/mol) σ आबंध की आबंध ऊर्जा (347kJ/mol) से कम होने के कारण भी इनकी क्रियाशीलता ऐल्केन से अधिक होती है।

प्रश्न 4.
असममित कार्बन किसे कहते हैं ?
उत्तर:
वह कार्बन परमाणु जिससे चार भिन्न समूह या परमाणु जुड़े रहते हैं। असममित कार्बन परमाणु कहलाता है। इस असममित कार्बन की उपस्थिति के कारण यौगिक असममिति दर्शाता है तथा प्रकाश घूर्णकता का गुण दर्शाता है।

प्रश्न 5.
ब्यूटेन के ग्रसित तथा सांतरित संरूपों का न्यूमैन प्रक्षेप आरेख दीजिए।
उत्तर:
ब्यूटेन का संरचना सूत्र –

प्रश्न 6.
किरेलता से क्या समझते हो?
उत्तर:
वे अणु जो अपने दर्पण प्रतिबिम्ब पर अध्यारोपित नहीं होते हैं किरेल अणु कहलाते हैं तथा इस गुण को किरेलता कहते हैं। किरेल अणु ध्रुवण घूर्णक होते हैं। अणु में उपस्थित किरेलता असममित कार्बन परमाणु की उपस्थिति के कारण होती है। जिन अणुओं में किरेलता होती है वे ध्रुवण घूर्णक होते हैं लेकिन असममित कार्बन परमाणु युक्त अणु में ध्रुवण घूर्णकता हो यह जरूरी नहीं है। वे ध्रुवण घूर्णक हो भी सकते हैं और नहीं भी।

प्रश्न 7.
ऐल्केन किसे कहते हैं ? तथा इनमें किस प्रकार के बंध होते हैं ?
उत्तर:
ऐल्केन संतृप्त हाइड्रोकार्बन हैं। इन्हें पैराफिन भी कहते हैं। ये अत्यन्त कम क्रियाशील होते हैं तथा अधिक स्थायी होते हैं। इनका सामान्य सूत्र – C_nH_2n+2होता है। ऐल्केन में कार्बन – कार्बन के मध्य एकल बंध व 6 बंध होता है तथा कार्बन-हाइड्रोजन बंध भी बंध होता है। ऐल्केन में प्रत्येक कार्बन sp3 संकरित अवस्था में होता है।
उदाहरण – मेथेन CH₄एथेन C₂H₆

प्रश्न 8.
ऐल्कीन किसे कहते हैं ? इनमें कार्बन किस संकरित अवस्था में होता है ?
उत्तर:
ऐल्कीन असंतृप्त हाइड्रोकार्बन होते हैं। इन्हें ओलीफिन भी कहते हैं। ये ऐल्केन की तुलना में अधिक क्रियाशील होते हैं तथा इनका स्थायित्व ऐल्केन की तुलना में कम होता है। ये सरलता से योगात्मक अभिक्रिया देते हैं। इनका सामान्य सूत्र C_nH_{2n – 2} होता है। ऐल्कीन में कार्बन-कार्बन के बीच द्विबंध होता है तथा कार्बन sp संकरित अवस्था में होता है। उदाहरण – एथिलीन CH₂ = CH₂ प्रोपीलीन CH₃ – CH = CH₂

प्रश्न 9.
ऐल्काइन किसे कहते हैं ? तथा इनमें किस प्रकार के बंध होते हैं
उत्तर:
ऐल्काइन असंतृप्त हाइड्रोकार्बन होते हैं। ये अत्यधिक क्रियाशील होते हैं तथा अत्यधिक अस्थायी होते हैं। इनका सामान्य सूत्र C_nH_2n+1 होता है।
कार्बन – कार्बन के मध्य त्रिबंध होता है। जिसमें एक बंध व दो 7 बंध होते हैं तथा दोनों कार्बन sp संकरित अवस्था में होते हैं।
उदाहरण – एसीटिलीन CH = CH, प्रोपाईन CH₃ – C ≡ CH₂

प्रश्न 10.
समपक्ष-विपक्ष समावयवता क्या होती है ? उदाहरण सहित समझाइये।
उत्तर:
समपक्ष-विपक्ष समावयवता को ज्यामितीय समावयवता भी कहते हैं, जो युग्म बंध युक्त ऐसे यौगिकों के द्वारा दर्शायी जाती है, जिनमें युग्म बन्धित कार्बन परमाणु दो भिन्न परमाणुओं या समूहों द्वारा जुड़े रहते हैं। जब समान समूह या हाइड्रोजन परमाणु द्विबंध के एक ही ओर स्थित होते हैं तो प्राप्त यौगिक समपक्ष समावयवयी तथा जब विपरीत ओर स्थित होते हैं, तो प्राप्त यौगिक विपक्ष समावयवयी कहलाता है।
उदाहरण – CH₃ CH = CH – CH₃

प्रश्न 11.
ऐसीटिलीन से नाइट्रोबेन्जीन आप किस प्रकार प्राप्त करोगे?
उत्तर:

प्रश्न 12.
इलेक्ट्रॉन स्नेही तथा नाभिक स्नेही क्रमशः इलेक्ट्रॉन न्यून तथा इलेक्ट्रॉन समृद्ध अभिक्रिया मध्यवर्ती है। अतः इनकी प्रवृत्ति क्रमशः इलेक्ट्रॉन समृद्ध तथा इलेक्ट्रॉन न्यून नाभिकों पर आक्रमण करने की होती है। निम्नलिखित स्पीशीज को इलेक्ट्रॉन स्नेही तथा नाभिक स्नेही के रूप में वर्गीकृत कीजिए –
(i) H₃CO^–
(iii) Cl
(iv) Cl₂C:
(v) (CH₃)₃C⁺
(vi) Br^–
(vii) CH₃OH,
(viii) R – NH – R
उत्तर:
इलेक्ट्रॉन स्नेही, इलेक्ट्रॉन न्यून स्पीशीज होते हैं। ये उदासीन या धनावेशित हो सकते हैं।
(iii) Cl,
(iv) Cl₂C:
(v) (CH₃)₃C
नाभिक स्नेही, इलेक्ट्रॉन समृद्ध स्पीशीज होते है। ये उदासीन या ऋणावेशित हो सकते हैं।
(i) CH₃O

प्रश्न 13.
साइक्लोहेक्सेन की तुलना में साइक्लोप्रोपेन अधिक क्रियाशील होता है, क्यों ?
उत्तर:
साइक्लोप्रोपेन में C – C – C बंध कोण 60° का होता है, जिसके कारण वलय तनाव में होता है इसलिये इसका स्थायित्व कम तथा क्रियाशीलता अधिक होती है। जबकि साइक्लोहेक्सेन में C – C – C बंध कोण 109°28 के पास होता है, जिसके कारण वलय का तनाव कम होता है, जिसके कारण इसका स्थायित्व अधिक और क्रियाशीलता कम होती है।

प्रश्न 14.
क्रियात्मक समावयवता को उदाहरण सहित समझाइये।
उत्तर:
जब दो या दो से अधिक यौगिकों में अलग-अलग क्रियात्मक समूह उपस्थित हों पर उनका अणुसूत्र एक ही हो तो इस प्रकार उत्पन्न समावयवता को क्रियात्मक समावयवता कहते हैं।
उदाहरण – C₂H₆O
CH₃ – O – CH₂ ⇔ C₂H₅ – OH.

प्रश्न 15.
वियोजन क्या है ?
उत्तर:
उपयुक्त विधियों द्वारा रेसेमिक मिश्रण को d या ! प्रतिबिम्ब समावयवयी में पृथक् करने की विधि को वियोजन कहते हैं। इसके लिये यौगिक, जीव रासायनिक एवं रासायनिक विधियों द्वारा वियोजन किया जा सकता है।

प्रश्न 16.
एरीन में कौन-सी समावयवता पायी जाती है ? उदाहरण सहित स्पष्ट कीजिए।
उत्तर:
एरीन में स्थिति समावयवता पायी जाती है। टॉलुईन के अतिरिक्त सभी बेंजीन के यौगिकों में स्थिति समावयवता होती है।
उदाहरण – C₈H₁₀ चार समावयवयी रूपों में मिलता है

प्रश्न 17.
ऐल्काइन की अम्लीय प्रकृति को दर्शाने के लिये एक अभिक्रिया लिखिए।
उत्तर:
वे ऐल्काइन जिनके अणु के अंतिम छोर में त्रिबंध होता है, वे दुर्बल अम्ल की तरह कार्य करते हैं । ऐसे ऐल्काइन की अभिक्रिया Na, Ca जैसी क्रियाशील धातुओं से कराने पर ये हाइड्रोजन मुक्त करते हैं तथा इनके व्युत्पन्न प्राप्त होते हैं, जिन्हें एसीटिलाइड कहते हैं।

प्रश्न 18.
बेयर अभिकर्मक क्या है, तथा इससे असंतृप्तता का परीक्षण कैसे करते हैं ?
उत्तर:
क्षारीय KMnO₄ को बेयर अभिकर्मक कहते हैं। असंतृप्त हाइड्रोकार्बन की अभिक्रिया बेयर अभिकर्मक से कराने पर क्षारीय KMnO₄ का रंग उड़ जाता है। इसलिये इसका उपयोग यौगिक की असंतृप्तता के परीक्षण के लिये करते हैं।

प्रश्न 19.
ऐल्कीन पर बेयर अभिकर्मक की अभिक्रिया को समीकरण सहित स्पष्ट कीजिए।
उत्तर:
एथिलीन कमरे के ताप पर 1% क्षारीय KMnO₄ अर्थात् बेयर अभिकर्मक के साथ अभिक्रिया कराने पर एथिलीन ग्लाइकॉल बनाता है तथा इस योगात्मक अभिक्रिया के कारण KMnO₄ का गुलाबी रंग उड़ जाता है।

प्रश्न 20.
परॉक्साइड प्रभाव क्या है ? एक उदाहरण दीजिए।
उत्तर:
यदि कार्बनिक परॉक्साइड जैसे बेंजॉयल परॉक्साइड की उपस्थिति में असममित एल्कीन या एल्काइन पर असममित अभिकर्मक का योग कराया जाता है तो अभिक्रिया मार्कोनीकॉफ नियम के विपरीत होती है। इस अभिक्रिया को प्रति मार्कोनीकॉफ नियम या परॉक्साइड प्रभाव या खराश प्रभाव कहते हैं।

प्रश्न 21.
एथेन, एथिलीन तथा एसीटिलीन में विभेद कैसे करते हैं ? परीक्षण सहित समझाइए।
उत्तर:
एथेन, एथिलीन तथा एसीटिलीन में विभेद –

प्रश्न 22.
बर्थलोट संश्लेषण क्या है ?
उत्तर:
दो कार्बन इलेक्ट्रोड के बीच विद्युत् आर्क में से हाइड्रोजन गैस की धारा प्रवाहित करने पर एसीटिलीन प्राप्त होता है। इसे बर्थलोट संश्लेषण कहते हैं।

प्रश्न 23.
वेस्ट्रॉन तथा वेस्ट्रॉसोल क्या है ?
उत्तर:
(1) ऐसीटिलीन पर CCl₄, विलायक की उपस्थिति में हैलोजन की योगात्मक अभिक्रिया से बना .. यौगिक वेस्ट्रॉन कहलाता है।

(2) वेस्ट्रॉन को ऐल्कोहॉलिक KOH के साथ गर्म किया जाता है तो विहाइड्रो – हैलोजनीकरण के कारण वेस्ट्रॉसोल प्राप्त होता है।

प्रश्न 24.
प्रोटोट्रॉफी किसे कहते हैं ?
उत्तर:
प्रोपाइन पर जलयोजन होने पर इनॉल व कीटो रूप प्राप्त होते हैं। यह चलावयवता समावयवता को प्रदर्शित करते हैं तथा इस प्रकार की समावयवता उन यौगिकों में होती है, जिनमें कम-से-कम एक हाइड्रोजन होता है तथा एक स्थिति से दूसरे तक प्रोटॉन के स्थानांतरण के कारण उत्पन्न होती है इसे प्रोटोट्रॉफी कहते हैं।

प्रश्न 25.
संरूपण समावयवता किसे कहते हैं ?
उत्तर:
कार्बन-कार्बन एकल बंध के मध्य घूर्णन के कारण जो विभिन्न आकाशीय व्यवस्थाएँ उत्पन्न होती हैं वे संरूपी कहलाती हैं तथा संरूपियों से सम्बन्धित आण्विक ज्यामिति को संरूपण समावयवता कहते हैं।

प्रश्न 26.
बहुलीकरण अभिक्रिया को उदाहरण सहित समझाइये।
उत्तर:
वह रासायनिक अभिक्रिया जिसमें दो या दो से अधिक सरल इकाई जिन्हें एकलक कहते हैं। परस्पर संयोजन कर एक जटिल अणु का निर्माण करते हैं बहुलक कहलाते हैं तथा इस अभिक्रिया को बहुलीकरण कहते हैं।
उदाहरण – एथिलीन अणुओं के बहुलीकरण से पॉलीएथिलीन बनता है।

प्रश्न 27.
एल्कीन इलेक्ट्रॉन स्नेही योगात्मक अभिक्रियाएँ प्रदर्शित करती है, जबकि ऐरीन इलेक्ट्रॉनस्नेही प्रतिस्थापन अभिक्रियाएँ प्रदर्शित करती है। कारण स्पष्ट कीजिए।
उत्तर:
एल्कीन ढीले बँधे π – इलेक्ट्रॉनों के धनी स्रोत होते है, जिसके कारण ये इलेक्ट्रॉनस्नेही योगात्मक अभिक्रियाएँ प्रदर्शित करता है। ऐल्कीनों की इलेक्ट्रॉन स्नेही योगात्मक अभिक्रियाओं में ऊर्जा में अत्यधिक बदलाव आता है। जिसके कारण ये इलेक्ट्रॉन स्नेही प्रतिस्थापन अभिक्रियाओं की अपेक्षा ऊर्जात्मक रूप से अधिक प्रभावी होती है। एरीन में इलेक्ट्रॉन स्नेही योगात्मक अभिक्रिया के दौरान बेंजीन वलय की ऐरोमैटिक प्रकृति नष्ट हो जाती है जबकि इलेक्ट्रॉन स्नेही प्रतिस्थापन अभिक्रिया में यह स्थिर रहती है। ऐरीन की इलेक्ट्रॉन स्नेही प्रतिस्थापन अभिक्रियाएँ ऊर्जात्मक रूप से इलेक्ट्रॉन स्नेही योगात्मक अभिक्रियाओं की अपेक्षा अधिक प्रभावी होती है।

प्रश्न 28.
ज्यामितीय समावयवता की आवश्यक शर्ते क्या हैं ?
उत्तर:
ज्यामितीय समावयवता के लिये निम्नलिखित शर्ते आवश्यक हैं –

• इसमें कम-से-कम एक कार्बन-कार्बन द्विबंध होना चाहिए।
• द्विबंधित कार्बन परमाणु से जुड़े परमाणु या समूह भिन्न-भिन्न होने चाहिए।

प्रश्न 29.
सेटजेफ नियम क्या है ? उदाहरण सहित समझाइये।
उत्तर:
यदि ऐल्किल हैलाइड में से हाइड्रोजन हैलाइड का विलोपन किया जाता है तो यह भिन्न-भिन्न प्रकार से हो सकता है तथा ऐल्कीन प्राप्त होता है। इस नियम के अनुसार उस ऐल्कीन के बनने की संभावना अधिक होगी, जिसमें द्विबंध से जुड़े कार्बन परमाणु अधिक ऐल्किलीकृत होते हैं। अर्थात् हाइड्रोजन का विलोपन उस कार्बन से होता है जहाँ हाइड्रोजन की संख्या कम होती है।

प्रश्न 30.
ऑक्टेन संख्या से आप क्या समझते हैं ?
उत्तर:
विभिन्न ईंधनों के तुलनात्मक अपस्फोटरोधी गुण को व्यक्त करने के लिए एडगर ने सन् 1927 में ऑक्टेन अंक शब्द प्रयुक्त किया, जिसके अनुसार किसी ईंधन की ऑक्टेन संख्या आवश्यकतानुसार आइसो ऑक्टेन के मिश्रण में उपस्थित होती है, जिसका किसी मानक (Standard) इंजन में अपस्फोटन दिये हुए ईंधन के अपस्फोटन के बराबर होता है, अर्थात् किसी ईंधन की ऑक्टेन संख्या, हेप्टेन तथा आइसो ऑक्टेन के मिश्रण में आइसो ऑक्टेन का प्रतिशत होती है।

प्रश्न 31.
पैराफीन्स, ओलिफीन्स एवं ऐसीटिलीन की निम्न बिन्दुओं में तुलना कीजिए –
1. IUPAC नाम
2. सामान्य सूत्र
3. क्रियाशीलता
4. ब्रोमीन जल से क्रिया।
उत्तर:

हाइड्रोकार्बन लघु उत्तरीय प्रश्न

प्रश्न 1.
बेन्जीन, n – हेक्सेन तथा एथाइन को घटते हुए अम्लीय व्यवहार के क्रम में व्यवस्थित कीजिए और इस व्यवहार का कारण बताइए।
उत्तर:
दिये गये यौगिकों में कार्बन की संकरण अवस्था निम्न है –

अम्लीय व्यवहार, कक्षक के S – गुण में वृद्धि के साथ-साथ बढ़ता है। अतः बेन्जीन, n-हेक्सेन तथा एथाइन के अम्लीय व्यवहार का घटता हुआ क्रम निम्न है –
ऐथाइन > बेन्जीन > n – हेक्सेन

प्रश्न 2.
किसी निकाय द्वारा ऐरोमैटिकता प्रदर्शित करने के लिए आवश्यक शर्ते क्या हैं ?
उत्तर:
किसी निकाय द्वारा ऐरोमैटिकता प्रदर्शित करने के लिए आवश्यक शर्ते निम्न हैं –

• अणु समतलीय होना चाहिए।
• अणु चक्रीय होना चाहिए जिसमें एकान्तर क्रम में एकल तथा द्विबंध होना चाहिए अर्थात् वलय में T – इलेक्ट्रॉनों का संपूर्ण विस्थानीकरण होना चाहिए।
• अणु के वलय (ring) में (4n + 2) π इलेक्ट्रॉन होने चाहिए। जहाँ n = 0,1,2,3 …….. (ह्यूकेल नियम)। यदि कोई अणु उपरोक्त में से किसी एक अथवा अधिक दशाओं को संतुष्ट नहीं कर पाता है, तो उसे अ-ऐरोमैटिक कहते हैं।

प्रश्न 3.
मध्यावयवता समावयवता क्या है ? उदाहरण सहित समझाइये।
उत्तर:
जब दो या दो से अधिक यौगिकों के अणुसूत्र समान हो और उनमें उपस्थित क्रियात्मक समूह भी समान हो लेकिन उनसे जुड़े एल्किल मूलक में भिन्नता हो तो इस प्रकार उत्पन्न समावयवता को मध्यावयवता कहते हैं।
उदाहरण –

प्रश्न 4.
चलावयवता समावयवता क्या है ? उदाहरण सहित समझाइये।
उत्तर:
यह एक विशिष्ट प्रकार की क्रियात्मक समावयवता है, जिसमें दोनों समावयवयी साम्य अवस्था में होते हैं तथा सरलता से एक-दूसरे में परिवर्तित हो सकते हैं। इस प्रकार की समावयवता तब उत्पन्न होती है, जब हाइड्रोजन या कोई भी अन्य समूह द्विबंध या त्रिबंध के दोनों तरफ दोलन करता है।
यह दो प्रकार की होती है –
1. द्विक प्रणाली – इस प्रणाली में हाइड्रोजन का दोलन दो बहु-संयोजकीय परमाणुओं के बीच होता है।
उदाहरण – H – C = N – H – N – C
2. त्रिक प्रणाली – इस प्रणाली में हाइड्रोजन का दोलन तीन बहु-संयोजकीय परमाणुओं के बीच होता है।

इनॉल रूप इसे कीटो-इनॉल समावयवता भी कहते हैं।

प्रश्न 5.
न्यूमैन प्रक्षेपण सूत्र क्या है ?
उत्तर:
यदि दो कार्बन परमाणु के मध्य एकल बंध उपस्थित हो और दोनों कार्बन को एक वृत्त के रूप में दर्शाया जाये तथा इनसे जुड़े हाइड्रोजन को C – H बंध द्वारा उनके केन्द्र से जुड़ा हुआ दर्शाया जाये। यदि दोनों कार्बन एक-दूसरे के पीछे व्यवस्थित हो तो उनमें से केवल सामने वाला कार्बन दिखाई देता है तथा उससे जुड़े हुए तीनों हाइड्रोजन उसके केन्द्र से जुड़े दिखाई देते हैं तथा पीछे वाले कार्बन के हाइड्रोजन पहले कार्बन की परिधि पर जुड़े हुए दिखाई देते हैं।

प्रश्न 6.
मार्कोनीकॉफ नियम क्या है ? उदाहरण सहित समझाइये।
उत्तर:
जब सममित हाइड्रोकार्बन पर असममित अभिकर्मक का योग होता है तो आक्रमणकारी समूह का एक भाग द्विबंध या त्रिबंध वाले एक कार्बन परमाणु पर तथा दूसरा भाग दूसरे कार्बन परमाणु पर जुड़ता है।

लेकिन जब असममित ऐल्कीन पर असममित अभिकर्मक का योग होता है तो सामान्यतः अभिक्रिया मार्कोनीकॉफ नियम के अनुसार होती है। इस नियम के अनुसार युग्म बंध पर जुड़ने वाले आक्रमणकारी समूह का ऋणात्मक भाग द्विबंध से जुड़े उस कार्बन परमाणु पर जुड़ता है जिस पर हाइड्रोजन की संख्या न्यूनतम होती है।

प्रश्न 7.
रेसेमिक मिश्रण को उदाहरण सहित समझाइये।
उत्तर:
दो प्रतिबिम्ब समावयवी की समान मात्राओं का मिश्रण ध्रुवण घूर्णकता नहीं दर्शाता क्योंकि मिश्रण में उपस्थित दोनों प्रतिबिम्ब समावयवी की ध्रुवण घूर्णकता बराबर किन्तु विपरीत दिशा में होती है तथा एक – दूसरे के कारण उत्पन्न ध्रुवण घूर्णकता को विनष्ट कर देती है। ऐसे मिश्रण को रेसेमिक मिश्रण कहते हैं तथा इसे dl या ± चिन्ह द्वारा दर्शाते हैं। उदाहरण – रेसेमिक लेक्टिक अम्ल – यह d तथा l रूप का समअणुक मिश्रण है। इसमें 50% d रूप तथा 50% l रूप होता है। यह रूप ध्रुवण अघूर्णक होता है तथा इसे d तथा l रूप में वियोजित किया जा सकता है। इसे dl या ± द्वारा दर्शाते हैं।

प्रश्न 8.
ऐल्केन की प्रतिस्थापन अभिक्रिया को उदाहरण सहित समझाइये।
उत्तर:
वह अभिक्रिया जिसमें किसी यौगिक में उपस्थित परमाणु किसी अन्य परमाणु या मूलक से इस प्रकार प्रतिस्थापित होता है कि यौगिक की आंतरिक संरचना में कोई परिवर्तन नहीं आता प्रतिस्थापन अभिक्रिया कहलाती है। ऐल्केन में होने वाली हैलोजनीकरण, नाइट्रीकरण, सल्फोनीकरण अभिक्रिया इसके उदाहरण हैं। हैलोजनीकरण अभिक्रिया – ऐल्केन की अभिक्रिया क्लोरीन के साथ सूर्य प्रकाश की उपस्थिति में कराने पर ऐल्केन में उपस्थित हाइड्रोजन का प्रतिस्थापन क्लोरीन द्वारा होता है तथा क्रमशः मोनो, डाई, ट्राई तथा टेट्रा हैलोएल्केन का मिश्रण प्राप्त होता है।

प्रश्न 9.
B.H.C. क्या है ? इसका उपयोग बताइये।
उत्तर:
बेंजीन की अभिक्रिया सूर्य प्रकाश अथवा पराबैंगनी प्रकाश की उपस्थिति में क्लोरीन से कराने पर बेंजीन हेक्साक्लोराइड प्राप्त होता है।
यह आठ त्रिविम समावयवयी रूपों में मिलता है। लेकिन इसके केवल चार समावयवयी रूप α β γ या δ रूप को शुद्ध अवस्था में प्राप्त किया जा सकता है। इसके विभिन्न समावयवयी रूप हाइड्रोजन तथा क्लोरीन की विभिन्न व्यवस्थाओं के कारण उत्पन्न होते हैं। इसका γ समावयवयी रूप सबसे अधिक स्थायी तथा प्रबल कीटनाशी है। इसको लिण्डेन या गैमेक्सेन या 666 के नाम से भी जाना जाता है। इसका उपयोग कीटाणुनाशक के रूप में होता है।

प्रश्न 10.
निम्नलिखित परिवर्तन कैसे करोगे केवल समीकरण दीजिए –

1. मेथेन से एथेन
2. एथेन से मेथेन
3. ऐसीटिलीन से बेंजीन।

उत्तर:
1. मेथेन से एथेन –

2. एथेन से मेथेन –

3. ऐसीटिलीन से बेंजीन –

प्रश्न 11.
कैसे प्राप्त करोगे (केवल समीकरण दीजिए) –

1. बेंजीन से B.H.C.
2. बेंजीन से ऐसीटोफिनोन
3. क्लोरोएथीन से पी.वी.सी.
4. टेट्राफ्लोरो एथीन से टेफ्लॉन।

उत्तर:
1. बेंजीन से B.H.C –

2. बेंजीन से ऐसीटोफिनोन –

3. क्लोरोएथीन से –

4. टेट्राफ्लोरो एथीन से टेफ्लॉन

प्रश्न 12.
ज्यामितीय समावयवता किस प्रकार के यौगिकों में पायी जाती है ? उदाहरण सहित समझाइये।
उत्तर:
यह एक प्रकार की त्रिविम समावयवता है। जो युग्म बंध युक्त ऐसे यौगिकों द्वारा प्रदर्शित की जाती है, जिनमें युग्म बंधित कार्बन परमाणु से जुड़े दो परमाणु या समूह भिन्न-भिन्न प्रकार के हो सकते हैं। जब समान परमाणु या समूह द्विबंध के एक ओर स्थित होते हैं तो प्राप्त यौगिक समपक्ष समावयवयी कहलाता है तथा जब विपरीत परमाणु या समूह द्विबंध के एक ओर स्थित हो तो प्राप्त यौगिक विपक्ष समावयवयी कहलाता है। इसलिये इसे समपक्ष-विपक्ष समावयवता भी कहते हैं।

उदाहरण – 2 – butene

प्रश्न 13.
ह्यूकेल का नियम क्या है ?
उत्तर:
ह्यूकेल ने अनेक ऐरोमैटिक यौगिकों का अध्ययन करने के पश्चात् एक नियम प्रतिपादित किया जिसे ह्यूकेल का नियम कहते हैं। इस नियम के अनुसार वे सभी समतलीय चक्रीय यौगिक ऐरोमैटिक गुण प्रदर्शित करते हैं। जिनके चक्र में (4n + 2) π इलेक्ट्रॉन होते हैं, यहाँ n पूर्णांक है जिसका मान 0, 1, 2 ………. होता है। अत: जिन चक्रीय यौगिक में 6(n = 1), 10 (n = 2), 14(n=3) इलेक्ट्रॉन होते हैं, उनमें ऐरोमैटिक गुण होता – है। ऐरोमैटिक यौगिक दो प्रकार के होते हैं –
1. बेंजीनोइड्स – वे यौगिक जो एक या अधिक बेंजीन वलय युक्त होते हैं, बेंजीनोइड्स कहलाते हैं।

2. नॉन – बेन्जीनोइड्स – ऐसे यौगिक जिनमें बेंजीन रिंग नहीं होती है।

प्रश्न 14.
ऐल्कीन सरलता से योगात्मक अभिक्रिया प्रदर्शित करता है, क्यों ?
उत्तर:
ऐल्कीन एक असंतृप्त हाइड्रोकार्बन है। इसमें कार्बन – कार्बन के बीच द्विबंध होता है तथा इस द्विबंध में एक σ बंध एक π बंध होता है। बंध – बंध की तुलना में दुर्बल होता है। इस π बंध की उपस्थिति के कारण एल्कीन सरलता से योगात्मक अभिक्रिया दर्शाता है। इस योगात्मक अभिक्रिया के दौरान एक π बंध टूटता है तो दो नये σ बंध का निर्माण होता है।
CH₂ = CH₂ + HCl → CH₃ – CH₂Cl

प्रश्न 15.
फ्रीडल क्राफ्ट्स अभिक्रिया को समीकरण सहित दर्शाइये।
उत्तर:
एल्किलीकरण:
बेंजीन की अभिक्रिया निर्जल AlCl₃ की उपस्थिति में CH₃Cl के साथ कराने पर टॉलुईन प्राप्त होता है। इस अभिक्रिया के दौरान बेंजीन में से हाइड्रोजन का प्रतिस्थापन एल्किल समूह द्वारा होता है इसलिये इसे एल्किलीकरण कहते हैं।

एसिलीकरण – बेंजीन की अभिक्रिया निर्जल AlCl₃ की उपस्थिति में एसीटिल क्लोराइड

के साथ कराने पर एसीटोफिनोन प्राप्त होता है। इस अभिक्रिया के दौरान बेंजीन में से हाइड्रोजन का प्रतिस्थापन एसीटिल या एसिल समूह द्वारा होता है। इसलिये इसे एसिटलीकरण या एसिलीकरण अभिक्रिया कहते हैं।

प्रश्न 16.
ज्यामितीय समावयवियों के नामकरण हेतु E व z संकेत क्या है ?
उत्तर:
जब द्विबंधित कार्बन परमाणु पर सभी चारों समूह अलग – अलग हों तो ज्यामितीय समावयवियों को E व Z संकेत से दर्शाते हैं। ये निम्न अनुक्रम नियम पर आधारित हैं –

नियम I:
सभी परमाणु और समूह जो द्विबंधित कार्बन परमाणु से जुड़े होते हैं। उनके प्रथम परमाणु के परमाणु क्रमांक के बढ़ते क्रम के आधार पर उनकी प्राथमिकता निर्भर करते हैं। कम परमाणु क्रमांक वाले परमाणु या समूह को निम्नतम प्राथमिकता देते हैं तथा अधिक परमाणु क्रमांक वाले परमाणु या समूह को उच्चतम प्राथमिकता देते हैं । यदि इन समूह के प्रथम परमाणु समान हों तो द्वितीय परमाणु अथवा तृतीय परमाणु के परमाणु क्रमांक द्वारा प्राथमिकता का अनुक्रम निर्धारित करते हैं।

नियम II:
जब द्विबंध से बँधे दोनों कार्बन परमाणु पर उच्च प्राथमिकता वाले परमाणु या समूह विपरीत ओर एक ही दिशा में हो तो इस विन्यास को E रूप से दर्शाते हैं। जब यदि द्विबंध से बँधे दोनों कार्बन परमाणु पर उच्च प्राथमिकता वाले परमाणु या समूह एक ही दिशा में हो तो इस विन्यास को Z रूप से दर्शाते हैं।
उदाहरण –

प्रश्न 17.
डाइईन्स किसे कहते हैं ? डाइईन्स कितने प्रकार के होते हैं ? उदाहरण सहित समझाइये।
उत्तर:
डाइईन्स असंतृप्त हाइड्रोकार्बन है, इनमें कार्बन श्रृंखला में कार्बन – कार्बन के मध्य दो द्विबंध होते हैं तथा द्विबंधों की सापेक्षिक स्थिति के आधार पर डाइईन्स तीन प्रकार के होते हैं –
1. विलगित डाइईन्स:
ऐसे डाइईन्स जिसमें दो द्विबंध परस्पर एक से अधिक एकल बंधों द्वारा अलग होते हैं।
CH₂ = CH – CH₂ – CH = CH₂

2. संयुग्मित डाइईन्स – ऐसे डाइईन्स में दो द्विबंध संयुग्मित या एकान्तर स्थिति पर होते हैं।

3. संचयी डाइईन्स – इन डाइईन्स में दो द्विबंध समीपस्थ स्थिति पर होते हैं।
CH₂ = C = CH – CH₃
CH₃ – CH = C = CH₂

प्रश्न 18.
डील्स एल्डर अभिक्रिया को समीकरण सहित लिखिए।
उत्तर:
जब संयुग्मित डाइईन्स को किसी एल्कीन या प्रतिस्थापित एल्कीन के साथ गर्म करते हैं तो एक छः सदस्यीय चक्रीय यौगिक बनता है इसलिये इस अभिक्रिया को चक्रीय संकलन अभिक्रिया भी कहते हैं। इसमें एक 47 इलेक्ट्रॉन निकाय से जुड़ता है। इस तरह की अभिक्रिया को डील्स एल्डर अभिक्रिया कहते हैं।

प्रश्न 19.
लिण्डलार उत्प्रेरक क्या है ? तथा इसके उपयोग लिखिए।
उत्तर:
बेरियम सल्फेट या कैल्सियम कार्बोनेट पर निक्षेपित पैलेडियम उत्प्रेरक का मिश्रण लिण्डलार उत्प्रेरक कहलाता है। यहाँ सल्फर या क्विनोलीन उत्प्रेरक विष का कार्य करता है और केवल एल्कीन अवस्था तक ही एल्काइन का अपचयन करता है।
उपयोग – एल्काइन का अपचयन हाइड्रोजन द्वारा लिण्डलार उत्प्रेरक की उपस्थिति में कराने पर एल्कीन प्राप्त होता है। यह उत्प्रेरक आगे एल्कीन को एल्केन में अपचयित नहीं होने देता।

प्रश्न 20.
बेंजीन की संरचना के लिये कैकुले सूत्र के पक्ष में चार प्रमाण दीजिए।
उत्तर:
बेंजीन की संरचना के लिये कैकुले सूत्र के पक्ष में प्रमाण –
1. बेंजीन तीन अणु हाइड्रोजन के साथ संयोग कर चक्रीय यौगिक साइक्लोहेक्सेन बनाता है, जिससे बेंजीन में तीन युग्म बंध की पुष्टि होती है।

2. बेंजीन तीन अणु क्लोरीन से संयोग कर B.H.C. बनाता है। अतः यह भी तीन युग्म बंध की पुष्टि करता है।

3. बेंजीन ओजोन के साथ संयोग कर ट्राइ ओजोनाइड बनाता है जो जल अपघटित होकर ग्लाईऑक्सेल के तीन अणु बनाता है।

4. एसीटिलीन के तीन अणु बहुलीकरण के द्वारा बेंजीन का निर्माण करते हैं। ..

प्रश्न 21.
कारण दीजिए –
1. शाखायुक्त यौगिकों के क्वथनांक सीधी श्रृंखला युक्त यौगिकों की अपेक्षा कम होते हैं, क्यों ?
2. विषम संख्या कार्बन परमाणु वाले यौगिक के गलनांक सम संख्या कार्बन परमाणु वाले यौगिकों से कम होते हैं, क्यों?
उत्तर:
1. रेखीय शृंखला वाले कार्बनिक यौगिकों का पृष्ठीय क्षेत्रफल अधिक होता है और इनमें अंतरअणुक आकर्षण बल अधिक होता है जबकि कार्बन परमाणु की श्रृंखला में शाखायुक्त हो जाने पर अणु में स्थित विभिन्न परमाणु पास-पास आ जाते हैं, जिससे उनके बीच अंतर-अणुक आकर्षण बल कम होता है इसलिये श्रृंखला वाले यौगिकों का क्वथनांक शाखायुक्त यौगिकों से अधिक होता है।

2. कार्बन परमाणु की विषम संख्या युक्त ऐल्केन कार्बन श्रृंखला के सिरों पर स्थित कार्बन परमाणु समान पक्ष में होते हैं। जबकि कार्बन परमाणु की सम संख्या वाले ऐल्केन में कार्बन शृंखला के सिरों पर स्थित कार्बन विपरीत पक्ष में होते हैं। सम संख्या वाले यौगिकों के विषम कार्बन संख्या वाले यौगिकों की तुलना में पैकिंग अधिक सघन होती है। इसलिए सम कार्बन संख्या वाले यौगिकों के क्वथनांक अधिक होते हैं।

प्रश्न 22.
निम्नलिखित पर टिप्पणी लिखिए –

1. सबात्ये और सेण्डरेन्स अभिक्रिया
2. बुर्दज अभिक्रिया
3. इयमा अभिक्रिया
4. स्वार्ट अभिक्रिया।

उत्तर:
1. सबात्ये और सेण्डरेन्स अभिक्रिया – ऐल्कीन की अभिक्रिया Ni या Pt की उपस्थिति में हाइड्रोजन के साथ कराने पर हाइड्रोजनीकरण के पश्चात् ऐल्केन प्राप्त होता है।

2. बुर्दज अभिक्रिया – दो अणु ऐल्किल हैलाइड की अभिक्रिया सोडियम के साथ शुष्क ईथर की उपस्थिति में कराने पर ऐल्केन प्राप्त होता है। इस प्रकार प्राप्त ऐल्केन में ऐल्किल हैलाइड की तुलना में दो गुना ज्यादा कार्बन होते हैं।

3. ड्यूमा अभिक्रिया – मोनोकार्बोक्सिलिक अम्ल के सोडियम लवण की अभिक्रिया सोडालाइम के साथ कराने पर विकार्बोक्सीकरण के पश्चात् एल्केन प्राप्त होता है।

4. स्वार्ट अभिक्रिया – जब एल्किल हैलाइड की अभिक्रिया मरक्यूरिक फ्लोराइड के साथ कराने पर फ्लोरो एल्केन प्राप्त होता है। इस अभिक्रिया के दौरान एल्किल हैलाइड के हैलोजन का प्रतिस्थापन फ्लोरीन द्वारा होता है।
2C₂H₅ – I + HgF₂ → 2CH₃ → 2C₂H₅ + F + HgI₂
प्रश्न 23.
कोल्बे की वैद्युत अपघटन अभिक्रिया को समीकरण सहित लिखिए।
उत्तर:
मोनो कार्बोक्सिलिक अम्ल के सोडियम या पोटैशियम लवण के सान्द्र जलीय विलयन का वैद्युत अपघटन करने पर एनोड पर एल्केन प्राप्त होता है।

एनोड पर – एसीटेट आयन इलेक्ट्रॉन त्यागकर उदासीन हो जाता है और विघटित होकर एल्केन देता है।
2CH₃COO^– – 2e^– → 2CH₃ COO → C₂H₆ + 2CO₂

कैथोड पर – हाइड्रोजन आयन इलेक्ट्रॉन ग्रहण कर हाइड्रोजन देता है।
2Na⁺ + 2HOH + 2e^– → 2NaOH + H₂

प्रश्न 24.
निम्नलिखित अभिक्रियाओं के केवल समीकरण दीजिए –

1. कैल्सियम कार्बाइड की जल से।
2. एथिलीन पर ब्रोमीन जल की अभिक्रिया।
3. एथिलीन को क्षारीय KMnO₄के साथ गर्म करने पर।
4. बेंजीन को सान्द्र HNOJ तथा सान्द्र H2SO, के साथ गर्म करने पर।
5. बेंजीन को मेथिल क्लोराइड के साथ निर्जल AICI, की उपस्थिति में गर्म करने पर।

उत्तर:
1. कैल्सियम कार्बाइड की जल के साथ अभिक्रिया –
CaC₂ + 2HOH → CH = CH + Ca(OH)₂
2. एथिलीन पर ब्रोमीन जल की अभिक्रिया –

3. एथिलीन को क्षारीय KMnO, के साथ गर्म करने पर

4. बेंजीन को सान्द्र HNOJ तथा सान्द्र H,SO, के साथ गर्म करने पर –

5. बेंजीन को CHICI के साथ निर्जल AICI की उपस्थिति में गर्म करने पर

प्रश्न 25.
निम्नलिखित को कैसे प्राप्त करोगे

1. एसीटिलीन से एसिटैल्डिहाइड
2. एथिलीन से मस्टर्ड गैस
3. ग्रिगनार्ड अभिकर्मक से एथेन
4. एसीटिलीन से क्यूप्रस एसीटिलाइड
5. ऐल्युमिनियम कार्बाइड से मेथेन।

उत्तर:
1. एसीटिलीन से एसिटैल्डिहाइड –

2. एथिलीन से मस्टर्ड गैस –

3. ग्रिगनार्ड अभिकर्मक से एथेन –

4. एसीटिलीन से क्यूप्रस एसीटिलाइड –
CH ≡ CH + Cu₂Cl₂ + 2NH₄OH D Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O
5. ऐल्युमिनियम कार्बाइड से मेथेन –
Al₄C₃ + 12H₂ O → 3CH₄ + 4Al(OH)₂

प्रश्न 26.
किसी प्राथमिक ऐल्किल हैलाइड की वु अभिक्रिया कराने पर एकमात्र उत्पाद के रूप में C₈H₁₈ प्राप्त होता है। इस ऐल्केन का मोनोब्रोमीकरण करने पर तृतीयक ब्रोमाइड का एकल समावयव प्राप्त होता है। ऐल्केन तथा तृतीयक ब्रोमाइड की पहचान लिखिए।
उत्तर:
चूँकि ऐल्केन C₈H₁₈ मोनोब्रोमीकरण के पश्चात् तृतीयक ब्रोमाइड का एक समावयव बनाता है अतः ऐल्केन में तृतीयक हाइड्रोजन उपस्थित होनी चाहिए। यह तभी सम्भव है जबकि प्राथमिक ऐल्किल हैलाइड (जो वु अभिक्रिया में भाग लेता है) के पास तृतीयक हाइड्रोजन हो।

प्रश्न 27.
2 – मेथिल प्रोपेन के मोनो क्लोरीनीकरण में बनने वाले मध्यवर्ती हाइड्रोकार्बन मूलकों को लिखिए। इनमें से कौन-सा अधिक स्थायी होगा। कारण भी दीजिए।
उत्तर:
2-मेथिल प्रोपेन निम्न दो प्रकार के मूलक देता है, जो निम्न है –

मूलक (I) अधिक स्थायी है क्योंकि यह 3° मुक्त मूलक है तथा नौ अति संयुग्मित संरचनाओं को स्थायी करता है (क्योंकि इसमें 9a -हाइड्रोजन है।)
मूलक (II) कम स्थायी है क्योंकि यह 1° मुक्त मूलक है तथा यह केवल एक अति संयुग्मित संरचना को स्थायी करता है। (क्योंकि इसमें केवल 1a- हाइड्रोजन है।)

प्रश्न 28.
विहाइड्रोहैलोजनीकरण तथा विहैलोजनीकरण अभिक्रिया को उदाहरण सहित समझाइये।
उत्तर:
विहाइड्रोहैलोजनीकरण – किसी कार्बनिक यौगिक में हाइड्रोजन हैलाइड के अणु का विलोपन विहाइड्रोहैलोजनीकरण कहलाता है। इस अभिक्रिया के दौरान β स्थिति से हाइड्रोजन का विलोपन होता है। इसलिये इसे β विलोपन अभिक्रिया भी कहते हैं।

उदाहरण – नॉर्मल प्रोपिल क्लोराइड को एल्कोहॉलिक KOH के साथ गर्म करने पर एल्कीन प्राप्त होता है।
CH₃ – CH₂ – CH₂ – Cl + KOH → CH₃ – CH = CH₂ + KCl + H₂O

विहैलोजनीकरण – बिस डाइहैलाइड को Zn चूर्ण के साथ मेथिल एल्कोहॉल की उपस्थिति में गर्म करने पर एल्कीन प्राप्त होता है, इस अभिक्रिया के दौरान केवल हैलोजन का विलोपन होता है इसलिए इसे विहैलोजनीकरण अभिक्रिया कहते हैं।

प्रश्न 29.
क्या होता है जब

1. सोडियम एसीटेट को सोडालाइम के साथ गर्म करने पर।
2. 2 – प्रोपेनॉल को एलुमिना के साथ 300°C ताप पर गर्म करने पर।
3. एसीटिलीन को अमोनियामय AgNO₃ के विलयन में प्रवाहित करने पर।
4. एथिल आयोडाइड को सोडियम के साथ गर्म करने पर।
5. एथिलीन पर हाइपोक्लोरस अम्ल की अभिक्रिया कराने पर।

उत्तर:
1. सोडियम एसीटेट को सोडालाइम के साथ गर्म करने पर –

2. 2-प्रोपेनॉल को एलुमिना के साथ 300°C ताप पर गर्म करने पर –

3. एसीटिलीन को अमोनियामय AgNOJ के साथ गर्म करने पर –
CH ≡ CH + 2AgNO₃ + 2NH₄OH → Ag – C ≡ C – Ag + 2NH₄NO₃ + 2H₂O
4. एथिल आयोडाइड को सोडियम के साथ गर्म करने पर –

5. एथिलीन पर हाइपोक्लोरस अम्ल की अभिक्रिया कराने पर –

प्रश्न 30.
निम्नलिखित पर टिप्पणी लिखिए –
1. हाइड्रोबोरोनीकरण
2. एपॉक्सीकरण।
उत्तर:
1. हाइड्रोबोरोनीकरण – एल्कीन पर जब डाइबोरेन से अभिक्रिया करायी जाती है तब एल्कीन के द्विबंध पर योगात्मक क्रिया होती है एवं ट्राइऐल्किल बोरेन बनता है, जो जल अपघटन पर एल्कोहॉल देता है यह हाइड्रोबोरोनीकरण अभिक्रिया कहलाती है।

2. एपॉक्सीकरण – एल्कीन ऑक्साइड को एपॉक्साइड कहते हैं। एल्कीन के ऑक्सीकरण से एपॉक्साइड बनने की क्रिया एपॉक्सीकरण कहलाती है।
उदाहरण – निम्नतर एल्कीन 200 – 400°C ताप पर सिल्वर उत्प्रेरक की उपस्थिति में ऑक्सीजन से संयोग करके एपॉक्साइड बनते हैं।

प्रश्न 31.
वुज अभिक्रिया को विषम कार्बन संख्या वाले ऐल्केन बनाने के लिये क्यों उपयुक्त नहीं माना जाता है ?
उत्तर:
एल्किल हैलाइड के दो अणु सोडियम के साथ शुष्क ईथर की उपस्थिति में गर्म करने पर ऐल्केन प्राप्त होता है।

इस अभिक्रिया द्वारा विषम कार्बन संख्या वाले ऐल्केन का निर्माण नहीं किया जा सकता है। क्योंकि इसके लिये दो अलग-अलग एल्किल हैलाइड को सोडियम के साथ गर्म करने पर ऐल्केनों का मिश्रण प्राप्त होता है जिन्हें पृथक् करना बहुत कठिन होता है। उदाहरण के लिये यदि मेथिल ब्रोमाइड तथा एथिल ब्रोमाइड को शुष्क ईथर की उपस्थिति में सोडियम के साथ गर्म करने पर एथेन, प्रोपेन तथा ब्यूटेन का मिश्रण प्राप्त होता है और इसमें से प्रोपेन को पृथक् करना अत्यन्त कठिन होता है।

प्रश्न 32.
HI, HBr तथा HCl की प्रोपीन के साथ क्रिया में बना मध्यवर्ती कार्बधनायन समान रहता है। HCl, HBr तथा Hl की आबंध ऊर्जाएँ क्रमशः 430.5 kJmol^– 363.7 kJmol^– तथा 296.8 kJmol^– होती है। इन हैलोजन अम्लों की क्रियाशीलता का क्रम क्या होगा?
उत्तर:
हैलोजन अम्लों का ऐल्कीन के साथ योग एक इलेक्ट्रॉन स्नेही योगात्मक अभिक्रिया है।

चूँकि प्रथम पद मंद है अतः यह दर निर्धारक पद है। इस पद की दर, प्रोटॉन की उपलब्धता पर निर्भर करती है। जो पुनः H – X अणुओं की आबंध वियोजन ऊर्जा पर निर्भर करती है। H – X अणु की आबंध वियोजन ऊर्जा जितनी कम होती है, हैलोजन अम्ल की क्रियाशीलता उतनी ही अधिक होती है। अतः हैलोजन अम्लों की क्रियाशीलता का घटता हुआ क्रम निम्न है
HI(296.8kJ) > HBr(363.7kJ) > HCl(430.5kJ)

प्रश्न 33.
भंजन से क्या समझते हैं ? भंजन तथा इसके उपयोग पर टिप्पणी लिखिए।
उत्तर:
उच्च ताप के द्वारा या उत्प्रेरक की उपस्थिति में उच्चतर और जटिल हाइड्रोकार्बन का सरल हाइड्रोकार्बन में होने वाला अपघटन भंजन कहलाता है। दूसरे शब्दों में, कम वाष्पशील तथा उच्च हाइड्रोकार्बन का अधिक वाष्पशील तथा निम्न हाइड्रोकार्बन में तापीय अपघटन भंजन कहलाता है।

भंजन दो प्रकार का होता है –
1. ताप भंजन – जब भंजन उच्च ताप व दाब की परिस्थिति में उत्प्रेरक की अनुपस्थिति में कराया जाता है, तो इसे तापीय भंजन कहते हैं।
2. उत्प्रेरक भंजन – जब उत्प्रेरक की उपस्थिति में भंजन कराया जाता है, तो उसे उत्प्रेरक भंजन कहते हैं। उत्प्रेरक की उपस्थिति में भंजन तापीय भंजन की तुलना में कम ताप तथा कम दाब पर कराया जाता है।

उपयोग –

• मिट्टी के तेल से गैस बनाना।
• पेट्रोल के भंजन से पेट्रोलियम गैस बनाना।

प्रश्न 34.
एक हाइड्रोकार्बन A वाष्प घनत्व (14) बेयर अभिकर्मक को रंगहीन कर निम्नानुसार रासायनिक क्रिया करता है –

प्राप्त उत्पादों A, B, C और D के नाम और सूत्र लिखिये।
उत्तर:

(A) एथीलीन, (B) 1, 2 डाई ब्रोमो एथेन या एथीलीन ब्रोमाइड ,(C) एसीटिलीन, (D) एसीटेल्डिहाइड।

हाइड्रोकार्बन दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
एल्कीन बनाने की प्रयोगशाला विधि का चित्र सहित वर्णन कीजिए।
उत्तर:
प्राथमिक एल्कोहॉल को सान्द्र H₂SO₄ के आधिक्य के साथ 170°C ताप पर गर्म करने पर एल्कोहॉल में से एक जल का अणु निकल जाता है और निर्जलीकरण के पश्चात् एल्कीन प्राप्त होता है।

एक गोल पेंदी वाले फ्लास्क में एल्कोहॉल तथा सान्द्र H2SO लेते हैं। इस मिश्रण में निर्जल Al(SOA)3 तथा रेत मिलाते हैं। AICSO), तथा रेत का मिश्रण झाग उत्पन्न होने नहीं देता तथा अभिक्रिया 140°C ताप पर ही सम्पन्न हो जाती है। आसवन फ्लास्क में या गोल पेंदी वाले फ्लास्क में एक थर्मामीटर, बिन्दुपाती कीप तथा निकास नली लगाते हैं। निकास नली का दूसरा सिरा दो वाष्प बोतलों से जुड़ा रहता है।

जिसमें सान्द्र H SoA तथा KOH भरा रहता है। बिन्दुपाती कीप की सहायता से एल्कोहॉल तथा सान्द्र H2SOA के मिश्रण को गिराते हैं। इसे बालू उष्मक पर 150°C ताप पर गर्म करते हैं । बनने वाली एथिलीन गैस में SO, तथा CO, की अशुद्धि भी होती है जो क्रमशः H,SO, तथा KOH में अवशोषित हो जाती है और शुद्ध एथिलीन गैस को गैस जार में जल के ऊपर एकत्रित कर लेते हैं।

प्रश्न 2.
एसीटिलीन बनाने की प्रयोगशाला विधि का चित्र सहित वर्णन कीजिए।
अथवा
एसीटिलीन बनाने की प्रयोगशाला विधि का वर्णन निम्नांकित बिन्दुओं के आधार पर कीजिए

1. विधि एवं रासायनिक समीकरण
2. नामांकित चित्र।

उत्तर:
प्रयोगशाला में एसीटिलीन गैस को CaC₂पर जल की अभिक्रिया द्वारा बनाया जाता है। इस विधि से प्राप्त एसीटिलीन में NH₃ तथा PH₃ की अशुद्धि होती है जिन्हें दूर करने के लिये इसे अम्लीय CuSO₄ में प्रवाहित करते हैं।
CaC₂ + 2H₂O → CH == CH + Ca(OH)₂

एक कोनिकल फ्लास्क में थोड़ी – सी रेत लेकर उसके ऊपर CaC₂ के टुकड़े रखते हैं तथा फ्लास्क में उपस्थित वायु को तेल गैस प्रवाहित कर विस्थापित कर देते हैं। कोनिकल फ्लास्क में बिन्दुपाती कीप तथा निकास नली लगी होती है। बिन्दुपाती कीप की सहायता से CaC₂ पर धीरे-धीरे H₂O की बूंदें डालते हैं जिससे एसीटिलीन गैस प्राप्त होती है। जिसे अम्लीय CuSO₄ विलयन में प्रवाहित कर – एसीटिलीन विरंजक चूर्ण के निलंबन में प्रवाहित करते ram हैं तथा एसीटिलीन गैस को गैस जार में जल के ऊपर एकत्रित कर लेते हैं।

सावधानी:

• प्रयोग के पूर्व बालू फ्लास्क की वायु को तेल गैस द्वारा विस्थापित कर देते हैं क्योंकि एसीटिलीन अम्लीकृत विरंजक चूर्ण Cuson का निलंबन वायु के साथ विस्फोटक मिश्रण बनाती है।
• CaC₂ के ऊपर धीरे-धीरे जल डाला जाता है क्योंकि अभिक्रिया बहुत तीव्र गति से होती है।
• फ्लास्क की तली में पहले रेत रखकर CaC₂ के टुकड़ों को रखा जाता है क्योंकि अभिक्रिया में बहुत अधिक मात्रा में ऊष्मा प्राप्त होती है।

प्रश्न 3.
मेथेन बनाने की प्रयोगशाला विधि का सचित्र वर्णन कीजिए।
उत्तर:
वसीय अम्लों के सोडियम लवणों को सोडा लाइम के साथ 630°K ताप पर गर्म करने पर एल्केन प्राप्त होता है। कॉस्टिक सोडा NaOH तथा अनबुझा चूना Cao के मिश्रण को सोडा लाइम कहते हैं। यहाँ अभिक्रिया में केवल NaOH भाग लेता है Cao, NaOH को शुष्क रखता है और NaOH की तीव्रता को कम करता है जिससे काँच पर अभिक्रिया नहीं हो पाती।

एक कठोर काँच की नली में कडे काँच की नली सोडियम एसीटेट व सोडालाइम का मिश्रण लेते हैं तथा कॉर्क की सहायता साड से इसमें निकास नली लगाते हैं। जिसका सोडालाइम दूसरा सिरा जल से भरे गैस जार में लगा रहता है। सोडियम एसीटेट व सोडा लाइम के मिश्रण को परखनली में बुन्सन बर्नर पर गर्म करते हैं तथा बनने वाली मेथेन गैस को जल से भरे गैस जार में जल के ऊपर एकत्रित कर लेते हैं।

प्रश्न 4.
एक असंतृप्त हाइड्रोकार्बन ‘A’ दो अणु H, को जोड़ता है तथा अपचयित ओजोनीकरण के पश्चात् ब्यूटेन-1,4 डाइअल, एथेनल तथा प्रोपेनोन देता है। यौगिक ‘A’ की संरचना तथा IUPAC नाम लिखिए। प्रयुक्त अभिक्रियाओं को समझाइए।
उत्तर:
हाइड्रोकार्बन ‘A’ में दो अणु H₂ का योग हो सकता है। अतः ‘A’ एल्केन, डाइईन या ऐल्काइन होना चाहिए। अपचयित ओजोनीकरण के पश्चात् ‘A’ तीन स्पीशीज देता है जिनमें से एक डाइऐल्डिहाइड है। इसका अर्थ है कि अणु दोनों सिरों से टूटता है। अतः ‘A’ में दो द्विआबंध होने चाहिए। यह निम्नलिखित तीन स्पीशीज देता है।

अतः यौगिक ‘A’ की संरचना निम्न है –

प्रश्न 5.
एसीटिलीन की अम्लीय प्रकृति को उदाहरण सहित समझाइये।
उत्तर:
ऐसे एल्काइन जिनके अणु के अंतिम छोर में = बंध होता है वे दुर्बल अम्लीय प्रकृति दर्शाते हैं। इनकी अम्लीय प्रकृति को संकरण के आधार पर समझाया जा सकता है। ऽ कक्षक में उपस्थित इलेक्ट्रॉन pad कक्षक में उपस्थित इलेक्ट्रॉन की तुलना में नाभिक से अधिक दृढ़ता से बँधे होते हैं। क्योंकि ऽ कक्षक, p व d कक्षक की तुलना में नाभिक के अधिक पास स्थित होता है। दूसरे शब्दों में sp संकरित कक्षक में 5 गुण (50%), sp² (33:3%) तथा sp³ संकरित कक्षक (25%) की तुलना में अधिक होता है।

जिसके कारण sp संकरित कार्बन परमाणु sp²तथा sp³ संकरित कार्बन की तुलना में अधिक ऋणविद्युती होता है। जिसके कारण C – Hबंध के इलेक्ट्रॉन कार्बन की ओर विस्थापित होने लगता है। जिसके फलस्वरूप कार्बन पर आंशिक ऋणावेश व हाइड्रोजन पर आंशिक धनावेश आ जाता है और इस प्रकार यह आयनित होकर धनावेशित H⁺ को प्रोटॉन के रूप में दान कर सकता है। इसलिये यह दुर्बल अम्ल के रूप में कार्य करता है।
उदाहरण –

(1) AgNO₃ से अभिक्रिया – एसीटिलीन गैस को अमोनियामय सिल्वर नाइट्रेट के विलयन में प्रवाहित करने पर सिल्वर एसीटीलाइड का श्वेत अवक्षेप प्राप्त होता है। H – C ≡ C – H + 2NH₄OH + 2AgNO₃ → Ag – C = C – Ag + 2NH₄NO₃ + 2H₂O

(2) अमोनियामय कॉपर क्लोराइड के साथ अभिक्रिया-एसीटिलीन गैस को अमोनियामय क्यूप्रस क्लोराइड के विलयन में प्रवाहित करने पर लाल रंग का क्यूप्रस एसीटीलाइड का अवक्षेप प्राप्त होता है। H – C ≡ C – H + Cu₂Cl₂ + 2NH₄ OH ≡ Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

प्रश्न 6.
संरूपण से आप क्या समझते हैं ? एथेन के संरूपण का विस्तारपूर्वक वर्णन कीजिए।
उत्तर:
कार्बन-कार्बन एकल बंध के मध्य घूर्णन के कारण जो विभिन्न आकाशीय व्यवस्थाएँ उत्पन्न होती हैं, जिन्हें संरूपी कहते हैं तथा संरूपियों से संबंधित आण्विक ज्यामिति को संरूपण समावयवता कहते हैं। एथेन में संरूपण समावयवता-एथेन में दो कार्बन परमाणु एकल बंध की सहायता से जुड़े रहते हैं तथा प्रत्येक कार्बन पर तीन हाइड्रोजन परमाणु द्वारा जुड़े रहते हैं। अर्थात् एथेन अणु में दो मेथिल समूह एकल बंध की सहायता से जुड़े रहते हैं।

यदि इसमें से एक कार्बन को स्थिर रखते हुए दूसरे कार्बन परमाणु को एकल बंध के चारों तरफ मुक्त घूर्णन करने दिया जाये तो विभिन्न त्रिविम व्यवस्थाएँ संभव हैं। सामान्यतः यदि एक कार्बन को स्थिर रखते हुये दूसरे कार्बन को एकल बंध के चारों तरफ 60° के कोण से मुक्त घूर्णन करने दिया जाये तो मुख्यतः 6 त्रिविम व्यवस्थाएं संभव हैं, जिनमें से 1, 3 और 5 संरचनाएँ समान हैं। इसलिये इनकी ऊर्जा तथा स्थायित्व भी समान होगा। इसी प्रकार 2, 4 और 6 वीं संरचना समान है इसलिये इनकी ऊर्जा तथा स्थायित्व भी समान होगा।

1. सांतरित रूप:
सांतरित रूप में दोनों कार्बन परमाणुओं पर H परमाणु इस प्रकार व्यवस्थित रहते हैं कि उनके बीच दूरी अधिकतम है जिसके कारण इनके बीच लगने वाला प्रतिकर्षण बल न्यूनतम रहता है जिसके फलस्वरूप इनकी ऊर्जा कम तथा स्थायित्व अधिक रहता है।

2. ग्रसित रूप:
ग्रसित रूप में दोनों कार्बन परमाणु इस प्रकार व्यवस्थित रहते हैं कि उनसे जुड़े हाइड्रोजन परमाणु के मध्य न्यूनतम दूरी है जिसके कारण इनके बीच लगने वाला प्रतिकर्षण बल अधिकतम होता है जिसके फलस्वरूप इनकी ऊर्जा अधिकतम तथा स्थायित्व न्यूनतम होता है। एथेन में सांतरित संरूपीय ग्रसित संरूपीय की अपेक्षा 12.6 kJ/mo ऊर्जा से अधिक स्थायी है। यह ऊर्जा का अंतर बहुत कम होने के कारण एथेन में से दोनों संरूपों को पृथक् करना कठिन होता है।

प्रश्न 7.
परॉक्साइड की उपस्थिति में, प्रोपीन में HBr का योग प्रतिमार्कोनीकॉफ नियमानुसार होता है, परन्तु HCl तथा Hl परॉक्साइड प्रभाव प्रदर्शित नहीं करते हैं, क्यों?
उत्तर:
प्रोपीन में HCI तथा HI के योग में परॉक्साइड प्रभाव प्रदर्शित नहीं होता है। इसका कारण है कि H – Cl आबंध (430.5 kJ moll^-1), H – Br आबंध (363.7 kJ moll^-1) की अपेक्षा प्रबल होने के कारण मुक्त मूलक द्वारा विदलित नहीं हो पाता है। दूसरी ओर H – I आबंध (296-8 kJ mol^-1) दुर्बल होता है तथा आयोडीन मुक्त मूलक द्विआबंध में जुड़ने की अपेक्षा आयोडीन अणु से संयुक्त होता है। इसकी ऊष्मागतिकीय आँकड़ों की सहायता से भी व्याख्या की जा सकती है। हाइड्रोजन हैलाइड की मुक्त मूलक योगात्मक अभिक्रियाओं में प्रयुक्त चरणों में एन्थैल्पी परिवर्तन निम्न प्रकार है –

उपरोक्त आँकड़ों से स्पष्ट है कि HBr के योग के दोनों चरण ऊष्माक्षेपी है जो कि अभिक्रिया होने की अनुकूल दशा है। परन्तु HCl तथा HI की दशा में एक चरण ऊष्माशोषी है, जो कि किसी भी रासायनिक अभिक्रिया के होने के लिए ऊष्मागतिकी की दृष्टि से अनुकूलतम स्थिति नहीं है।

प्रश्न 8.
साइक्लोहेक्सेन में संरूपण समावयवता को समझाइये।
उत्तर:
खुली श्रृंखला वाले एल्केनों के अतिरिक्त बंद श्रृंखला वाले साइक्लोएल्केन भी संरूपण समावयवता दर्शाते हैं। साइक्लोप्रोपेन तथा साइक्लोब्यूटेन में बंध कोण क्रमश: 60° तथा 90° का होता है। जिनके कारण ये वलय तनाव में होते हैं तथा इनकी क्रियाशीलता अधिकतम होती है।

साइक्लोपेंटेन में बंध कोण 108° होता है जो कि सामान्य चतुष्फलकीय बंध कोण के निकटतम है। साइक्लोहेक्सेन में बंध कोण 109°28′ होता है। जिसके कारण वलय तनावरहित तथा अधिक स्थायी होती है। इसलिये इसमें संरूपण समावयवता दर्शायी जा सकती है। त्रिविम में विभिन्न व्यवस्थाओं के कारण यह दो समावयवी रूपों में पाया जाता है।

1. कुर्सी संरूपण – यह साइक्लोहेक्सेन का अत्यधिक स्थायी संरूपी है जिसमें सभी बंध कोण चतुष्फलकीय बंध कोण 109°28 के बराबर है। लेकिन समीपवर्ती कार्बन परमाणु के हाइड्रोजन सांतरित स्थिति में है। इसलिये यह वलय तनाव से मुक्त है और इसका स्थायित्व अधिकतम है।

2. नाव संरूपण – यह साइक्लोहेक्सेन का कम स्थायी समावयवी रूप है। इसमें भी सभी बंध कोण चतुष्फलकीय बंध कोण के बराबर है। लेकिन समीपवर्ती कार्बन परमाणु के हाइड्रोजन ग्रसित रूप में है जिसके कारण वलय तनाव में है तथा इसका स्थायित्व कुर्सी रूप की तुलना में कम है। कुर्सी रूप तथा नाव रूप में ऊर्जा का अंतर 44 kJ/mol^-1 है।
नाव संरूपण –

कुर्सी संरूपण –

प्रश्न 9.
ऐथेन के ग्रस्त तथा सांतरित रूपों के न्यूमैन तथा साँहार्स प्रक्षेप लिखिए। इनमें से कौन-सा प्रक्षेप अधिक स्थायी होगा तथा क्यों?
उत्तर:

ऐथेन का सांतरित रूप, ग्रस्त रूप की अपेक्षा लगभग 12.5 kJmol^-1 ऊर्जा द्वारा अधिक स्थायी होता है। इसका कारण है कि सांतरित रूप में कोई भी दो हाइड्रोजन परमाणु समीपवर्ती कार्बन परमाणुओं से अधिकतम दूरी पर स्थित होते हैं जबकि ग्रस्त अवस्था में ये अत्यधिक पास-पास (कभी-कभी अध्यारोपित भी हो जाते हैं) आ जाते हैं। अतः सांतरित रूप में, इनमें न्यूनतम प्रतिकर्षण बल, न्यूनतम ऊर्जा तथा अधिकतम स्थायित्व होता है।

प्रश्न 10.
ब्यूटेन में संरूपण समावयवता को समझाइये।
उत्तर:
n ब्यूटेन को एथेन का डाइ मेथिल व्युत्पन्न माना जा सकता है जिसमें प्रत्येक कार्बन का एकएक हाइड्रोजन परमाणु मेथिल समूह से प्रतिस्थापित हो गया है। ऐथेन के समान ब्यूटेन में यदि C₂ को स्थिर रखकर C₃ को घुमाया जाता है तब 360° से घुमाने पर अणु पुनः मूल स्थिति में आ जाता है। जब दोनों कार्बन परमाणु के मेथिल समूह एक-दूसरे से अधिकतम दूरी पर स्थित होते हैं तो इनमें न्यूनतम प्रतिकर्षण होता है तथा सबसे स्थायी रूप है इसे पूर्णतः सांतरित रूप (I) कहते हैं।

पूर्ण सांतरित रूप को घूमाने पर क्रमशः 120° व 240° पर III व V रूप प्राप्त होता है इसमें एक कार्बन का मेथिल समूह दूसरे कार्बन के मेथिल व हाइड्रोजन परमाणु के मध्य होता है। इसे गॉश रूप कहते हैं, यह कम स्थायी होता है। रूप I को 60° व 300° पर घुमाने से क्रमशः II व VI रूप प्राप्त होते हैं ।

ये ग्रसित रूप कहलाता है तथा 180° पर रूप IV प्राप्त होता है जो पूर्णतः ग्रसित रूप है। रूप IV सबसे कम स्थायी है। इसमें दोनों मेथिल समूह की दूरी न्यूनतम होती है। अतः प्रतिकर्षण अधिकतम होता है। पूर्णतः ग्रसित रूप IV की अपेक्षा ग्रसित रूप II व III का स्थायित्व थोड़ा अधिक होता है।

प्रश्न 11.
बेंजीन के अनुनाद सूत्र को समझाइये।
उत्तर:
बेंजीन में प्रत्येक C – Cबंध लंबाई 1.40 A होती है जो कि एकल बंध 154 A और युग्म बंध 134 A के मध्यवर्ती है। अत: बेंजीन में C – C बंध एकल बंध या युग्म बंध से न जुड़े होकर आंशिक युग्म बंध तथा आंशिक एकल बंध से जुड़े होते हैं। अर्थात् कुछ समय पश्चात् द्वि-बंध एकल बंध में तथा एकल बंध द्विबंध में परिवर्तित होते रहते हैं। ऐसे बंधों की कल्पना तो की जा सकती है। लेकिन इन्हें कागजों पर नहीं दर्शाया जा सकता है। इसलिये ऐसा माना जाता है कि बेंजीन निम्नलिखित संरचनाओं का अनुनादी सूत्र है।

उपर्युक्त संरचनाओं में पहली तथा दूसरी संरचना Kekule संरचना है तथा तीसरी, चौथी तथा पाँचवीं संरचना Dewar संरचना है। Dewar संरचना की ऊर्जा अधिक होती है Para बंध की उपस्थिति के कारण। इसलिये अनुनाद में इनका योगदान केवल 20% होता है। जबकि Kekule सूत्र समतुल्य है और अन्य सूत्र की अपेक्षा अधिक स्थायी है।

इसलिये अनुनाद में इनका योगदान 80% होता है। इसी कारण बेंजीन को इन दो संरचनाओं Kekule सूत्र का ही अनुनादी संकर माना जाता है। सामान्यतः अनुनादी संकर सूत्र को अनेक सूत्र के स्थान पर केवल एक पार्श्व सूत्र द्वारा प्रदर्शित करते हैं जिसमें खंडित वृत्त असंतृप्तता को दर्शाता है।

प्रश्न 12.
एल्केन में होने वाली हैलोजनीकरण अभिक्रिया की मुक्त मूलक क्रियाविधि को समझाइये।
उत्तर:
एल्केन में हैलोजनीकरण अभिक्रिया मुक्त मूलक क्रियाविधि द्वारा सम्पन्न होती है।
1. श्रृंखला का समारम्भ पद – उच्च ताप या प्रकाश के पराबैंगनी विकिरण की ऊर्जा से क्लोरीन के अणु का समांश विखण्डन होता है एवं क्लोरीन का मुक्त मूलक प्राप्त होता है।

2. श्रृंखला संचरण पद – उक्त पद में क्लोरीन मुक्त मूलक एल्केन अणु में हाइड्रोजन को विस्थापित कर एक एल्किल मुक्त मूलक प्राप्त होता है।

यह मुक्त मेथिल मूलक क्लोरीन अणु से क्रिया कर एल्किल हैलाइड तथा क्लोरीन मुक्त मूलक देता है।

यह मुक्त क्लोरीन मूलक पुनः एल्केन से अभिक्रिया कर मेथिल मूलक देता है। इस प्रकार अभिक्रिया की पुनरावृत्ति होती रहती है।

3. श्रृंखला का समापन पद – अंत में दोनों मुक्त मूलक परस्पर क्रिया करके श्रृंखला अभिक्रिया को समाप्त कर देते हैं।

हैलोजन का मुक्त मूलक एल्किल हैलाइड पर आक्रमण कर नया मुक्त मूलक बनाता है।

यह मुक्त मूलक हैलोजन के साथ क्रिया कर डाइ हैलोएल्केन देता है। यह प्रक्रिया धीरे – धीरे तब तक चलती रहती है जब तक सभी हाइड्रोजन परमाणु क्लोरीन द्वारा प्रतिस्थापित नहीं हो जाते हैं।

प्रश्न 13.
समूहों के दिशात्मक प्रभाव का इलेक्ट्रॉनिक स्पष्टीकरण दीजिए।
उत्तर:
यदि बेंजीन रिंग पर एक प्रतिस्थापी समूह पहले से उपस्थित है तो प्रतिस्थापन अभिक्रिया के फलस्वरूप नया समूह रिंग के किस स्थान पर प्रवेश करेगा इसका निर्धारण पहले से उपस्थित प्रतिस्थापी समूह की प्रकृति के आधार पर होता है इसे उस समूह का दैशिक प्रभाव कहते हैं।

इलेक्ट्रॉनिक सिद्धान्त के आधार पर समूह के निर्देशक प्रभाव की व्याख्या –
1. ऑर्थों व पैरा निर्देशकारी समूह – ये वे समूह होते हैं जो बेंजीन वलय पर पहले से उपस्थित होने पर नये समूह को ऑर्थो व पैरा स्थान पर निर्देशित करते हैं।
उदाहरण – OH, – OCH₃, – CH₃, – NH₂
ये समूह अपनी इलेक्ट्रॉन मुक्तकारी क्षमता के कारण बेंजीन में ऑर्थो व पैरा स्थिति में इलेक्ट्रॉन घनत्व बढ़ा देते हैं। इलेक्ट्रॉन घनत्व बढ़ जाने की वजह से बेंजीन नाभिक की क्रियाशीलता बढ़ जाती है तथा इलेक्ट्रॉन स्नेही अभिकर्मक इस स्थिति में उच्च इलेक्ट्रॉन घनत्व होने के कारण इन्हीं स्थानों पर प्रवेश करता है। .

2. मेटा निर्देशकारी प्रभाव – ये वे समूह हैं जो बेंजीन वलय पर पहले से उपस्थित होने पर नये समूह को मेटा स्थान पर निर्देशित करते हैं।
उदाहरण- – NO₂ – COOH, – SO₃H, – CN
जब ये समूह बेंजीन वलय पर जुड़े होते हैं तो मेसोमेरिक प्रभाव के कारण बेंजीन वलय से इलेक्ट्रॉन खींच लेते हैं जिसके फलस्वरूप वलय का इलेक्ट्रॉन घनत्व कम हो जाता है तथा इलेक्ट्रॉन घनत्व मुख्यतः ऑर्थो तथा पैरा स्थिति पर घटता है जबकि मेटा स्थिति अप्रभावित रहती है अर्थात् मेटा स्थिति पर इलेक्ट्रॉन घनत्व उच्च रहता है। इसलिये इलेक्ट्रॉन स्नेही अभिकर्मक मेटा स्थिति पर प्रवेश करता है।

प्रश्न 14.
क्लोरीनीकरण के प्रति 1°,2°,3° हैलोजनों की आपेक्षिक क्रियाशीलता 1:3.8:5 है। 2 – मेथिल ब्यूटेन के सभी मोनोक्लोरीनीकृत उत्पादों की प्रतिशतता ज्ञात कीजिए।

मोनो क्लोरीनीकृत उत्पादों की आपेक्षिक मात्रा = हाइड्रोजन की मात्रा × आपेक्षिक क्रियाशीलता –
(1°) मोनो क्लोरीनीकृत उत्पाद के लिए = 9 × 1 = 9
(2°) मोनो क्लोरीनीकृत उत्पाद के लिए = 2 × 3.8 = 7.6
(3°) मोनो क्लोरीनीकृत उत्पाद के लिए = 1 × 5 = 5
मोनो क्लोरीनीकृत यौगिकों की कुल संख्या = 9 + 7.6 + 5 = 21.6
1° मोनो क्लोरीनीकृत उत्पाद की % मात्रा = \(\frac { 9 × 100 }{ 21.6 }\) = 41.67%
2° मोनो क्लोरीनीकृत उत्पाद की % मात्रा = \(\frac { 7.6 × 100 }{ 21.6 }\) = 35.18%
3° मोनो क्लोरीनीकृत उत्पाद की % मात्रा =\(\frac {5 × 100 }{ 21.6 }\) = 23.15 %

प्रश्न 15.
निम्नलिखित अभिक्रिया के फलस्वरूप कौन-सा उत्पाद प्राप्त होगा तथा क्यों ?

जब फ्रीडल – क्राफ्ट ऐल्कीकरण उच्च ऐल्किल हैलाइड जैसे n – प्रोपिल क्लोराइड के साथ होता है, तो इलेक्ट्रॉन स्नेही, n – प्रोपिल कार्बधनायन (1° कार्बधनायन) पुनर्व्यवस्थित होकर अधिक स्थायी आइसो-प्रोपिल कार्बधनायन (2° कार्बधनायन) में बदल जाता है तथा अभिक्रिया के परिणामस्वरूप मुख्य उत्पाद आइसो-प्रोपिल बेन्जीन प्राप्त होता है।

प्रश्न 16.
निम्नलिखित यौगिकों को इलेक्ट्रॉन स्नेही अभिकर्मक के प्रति उनकी घटती हुई आपेक्षिक क्रियाशीलता के क्रम में व्यवस्थित कीजिए।

उत्तर:
OCH₃ (मेथॉक्सी समूह) इलेक्ट्रॉन निर्मुक्त करने वाला समूह है। यह अनुनाद प्रभाव (+R प्रभाव) के कारण बेन्जीन नाभिक पर इलेक्ट्रॉन घनत्व बढ़ाता है। इसके कारण एनीसॉल, इलेक्ट्रॉन स्नेही अभिकर्मकों के प्रति, बेन्जीन की अपेक्षा अधिक क्रियाशील है। ऐरिल हैलाइडों की दशा में, हैलोजन अपने प्रबल – I प्रभाव के कारण अत्यधिक निष्क्रिय होते हैं जिससे बेन्जीन वलय में कुल इलेक्ट्रॉन घनत्व घट जाता है। इसके कारण पुनः प्रतिस्थापन घट जाता है।

– NO₂ समूह इलेक्ट्रॉन आकर्षित करने वाला समूह है। यह प्रबल – प्रभाव के कारण बेन्जीन नाभिक में इलेक्ट्रॉन घनत्व घटाता है। जिसके कारण नाइट्रोबेन्जीन कम क्रियाशील हो जाती है। अतः इन यौगिकों की इलेक्ट्रॉन स्नेही अभिकर्मकों के प्रति कियाशीलता का क्रम निम्न होगा –

MP Board Class 11th Chemistry Solutions

MP Board Class 12th Physics Important Questions Chapter 6 Electromagnetic Induction

Electromagnetic Induction Important Questions

Electromagnetic Induction Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The SI unit of magnetic flux is :
(a) Weber
(b) Gauss
(c) Oersted
(d) Tesla
Answer:
(a) Weber

Question 2.
The flux linked with a coil can be charged by :
(a) Keeping the coil in a time varying magnetic field
(b) Rotating the coil in a magnetic field
(c) By charging area of coil in the uniform magnetic field
(d) All the above.
Answer:
(d) All the above.

Question 3.
The direction of induced current can be determined by :
(a) Lenz’s law
(b) Flemming’s left hand rule
(c) Right hand palm rule
(d) Ampere’s swimming rule.
Answer:
(a) Lenz’s law

Question 4.
In the phenomenon of elctro magnetic induction :
(a) Electrical energy is transformed into mechanical energy
(b) Mechanical energy is transformed into electrical energy
(c) Electrical energy is transformed into thermal energy
(d) Mechanical energy is transformed into thermal energy.
Answer:
(b) Mechanical energy is transformed into electrical energy

Question 5.
The self inductance of a coil does not depend upon :
(a) Radius of coil
(b) Number of turns in the coil
(c) Lenght of coil
(d) Current through the coil.
Answer:
(d) Current through the coil.

Question 6.
If a core of soft iron is placed between the coil then their mutual inductance :
(a) Decreases
(b) Increases
(c) Remain unchanged
(d) Nothing can be said.
Answer:
(b) Increases

Question 7.
The eddy current are used in :
(a) Electrolysis
(b) Making a galvanometer dead beat
(c) Electroplating
(d) Increasing the sensitivity of galvanometer.
Answer:
(b) Making a galvanometer dead beat

Question 2.
Fill in the blanks :

1. Lenz’s law is used to find out direction of ……………
2. The SI unit of magnetic flux is ……………
3. The core of transformer is made laminated to reduce the loss of energy due to …………… currents.
4. To make the moving coil galvanometer …………… its coil is wound on the aluminium frame.
5. Magnetic flux is a …………… quantity.

Answer:

1. Induced current
2. Weber
3. Eddy
4. Dead beat
5. Scalar.

Question 3.
Match the Column :
I.

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)

II.

Answer:

1. (e)
2. (c)
3. (d)
4. (b)
5. (a)

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

1. What is the unit of magnetic flux?
2. What is unit of coefficient of self induction?
3. A straight conductor of length ‘l’ moving with a velocity of ‘V’ in a uniform magnetic field ‘B’ Perpendicular to it. Write the
4. formula for induced e.m.f. in it.
5. Write relation between Henry and Micro Henry.
6. Write the principle of alternating current generator.
7. Induced electric field is conservative field or non conservative field.
8. In which form the energy is stored in a coil?
9. By which law we determine the direction of induced e.m.f.?

Answer:

1. Weber
2. Henry
3. e = Blv
4. 1H = 10⁶pH
5. Electro magnetic induction
6. It is Non conservative field
7. Magnetic energy
8. Lenz’s law.

Electromagnetic Induction Very Short Answer Type Questions

Question 1.
When induced current flow in a closed circuit?
Answer:
When there is change in magnetic flux in a closed circuit, then induced current flow over it.

Question 2.
If a iron core is kept at the centre of a coil, then what will be the effect on coefficient of self induction?
Answer:
Coefficient of self induction will increase.

Question 3.
Lenz’s law is accordance to which law?
Answer:
Law of conservation of energy.

Question 4.
If any metal piece is kept in varying magnetic field, will eddy current generate in it?
Answer:
Yes, eddy current generate in the metallic piece.

Question 5.
Why coil with in the resistance box is double coiled?
Answer:
To avoid self – induction phenomenon.

Question 6.
When the magnitude of magnetic flux become maximum and minimum?
Answer:
When magnetic field is perpendicular to the plane, the magnetic flux is maximum and when it is parallel, its value is minimum.

Electromagnetic Induction Short Answer Type Questions

Question 1.
What is magnetic flux? Write its SI units.
Answer:
The number of magnetic lines of force passing normally through any area in a magnetic field, is called magnetic flux linked with that area. It is denoted by ϕ(phi).

Where, \(\vec { B }\) is the magnetic field, \(\vec { dS }\) is the normal to the surface and θ is the angle between \(\vec { B }\) and normal to surface. SI unit of fluxs weber.

Question 2.
Write Lenz’s law of electromagnetic induction.
Answer:
According to this law, the nature of induced e.m.f. is such that it opposes the cause which produces it. With the help of this law, the direction of induced e.m.f. can be determined.
∴ e = \(\frac {dϕ}{dt}\)
– ve sign is according to Lenz’s law.

Question 3.
A bird sitting on a wire carrying current at high voltage, flies away. Why?
Answer:
As soon as the current flows through the wire, an induced current is produced in the body of the bird, the direction of the current in the wings are opposite. Hence, they are repelled and the bird flies away.

Question 4.
In the given figure, which plate of the capacitor AB is positive and which plate is negative?
Answer:

According to Lenz’s law induced current flows from B to A, therefore plate A is positive and plate B is negative.

Question 5.
A metallic chain touching the ground is attached with the truck carrying explosive matter. Why?
Answer:
During the motion of the truck, the axle cut the magnetic flux of the earth, which causes an induced e.m.f. set up across its axle. This induced charges are earthed by the chain. Which prevent the explosive from the fire.

Question 6.
In metre bridge experiment, the battery’ key is pressed first then galvanometer. Why?
Answer:
If the galvanometer key is pressed first and then battery key, an induced current will be produced in the galvanometer circuit and galvanometer will give deflection even at the position of null point. Thus, the battery key is pressed first.

Question 7.
The resistances in the resistance box are double coiled. Why?
Answer:
In the double coil, the current flows in opposite directions, thus the magnetic fields produced cancel each other and hence induced current is not produced.

Question 8.
When any electric circuit is suddenly cut off, then sparks take place. Why?
Answer:
When the switch is off, the magnetic flux linked with the circuit becomes zero suddenly. Therefore, a strong induced current flows through the circuit in the direction of main current, which causes a high potential difference across the terminal and due to electric discharge, sparks are seen.

Question 9.
What is mutual induction?
Answer:
When the current changes in a coil, then the flux linked with another coil placed near the first changes, hence an induced current is produced in another coil. This phenomenon is known as mutual induction.

Question 10.
A coil in magnetic field is moved

1. Rapidly and
2. Slowly.

Are the e.m.f. and work done in the two cases equal or not?
Answer:
No, the e.m.f. induced in the first case will be larger due to larger rate of change of flux. Obviously, in the second case, the rate of change of flux is slow and hence induced e.m.f will be smaller.

Question 11.
Two coils A and B are kept perpendicular to each other as shown in figure. If flow of current in any one coil is change will induced current will generate in the second.
Answer:
When current is flows in use one coil, the induced flux produce will be parallel to second coil and there will be no change in magnetic flux. Hence no induced current will be produced in second coil.

Question 12.
A ring is hanging in a wall of room. When the north pole of the magnet is brought closer to the ring. Then what will be the direction of induced current if seen from the direction of magnet?
Answer:
The face of the ring which is toward the north pole of the magnet, in.it the direction of induced current will be anti clockwise. Hence this face become magnetic north pole and resist the magnetic pole to come near it.

Question 13.
Three coils are arranged as shown in the figure :

In which coil mutual induction will be maximum?
Answer:
In figure (a), the mutual induction will be maximum before the magnetic lines of force passing in figure (a) is more in comparison to other figure.

Question 14.
What is electromagnetic induction? State Faraday’s laws of electromag neticinduction.
Or
Write down the laws of electromagnetic induction. Find out the expression for the induced electromotive force.
Answer:
When a magnet and a conducting coil are in relative motion, an e.m.f. is induced in the coil. This e.m.f. is called induced e.m.f. If the coil is closed, a current flows in it, then this current is called induced current and the phenomenon is called electromagnetic induction.

Faraday’s laws of electromagnetic induction:
1. Whenever there is a change in magnetic flux linked with an electrical circuit, an e.m.f. is induced in the circuit, therefore an induced current flows in it. The induced current losts so long as there is change in flux.

2. The induced e.m.f. is directly proportional to the rate of change in magnetic flux.

Formula for induced e.m.f.:
Let Φ₁ weber be the flux linked with a closed electrical circuit at any instant of time t₁. After time ∆t i.e., at time t₂ seconds, let the flux be Φ2 weber.

This is the required formula.
Hence, the induced e.m.f. is directly proportional to the rate of change of flux.

Question 15.
Write Lenz’s law and explain how it can be used to determine the direction of induced current?
Answer:
Lenz’s law:
According to this law, the direc¬tion of induced current is such that it opposes the very cause producing it. With the help of this law, the direction of induced current can be determined as follows :

1. When N – pole of a bar magnet is brought closer to a coil, a current is induced to develop N – pole on the adjacent face of the coil to oppose the motion of N – pole towards the coil. This means the induced current must be anticlockwise as seen from the magnet, [see Fig.(a)].

2. When N – pole of the bar magnet is taken away from the coil, an induced current flows in the coil such that the face of the coil near the magnet becomes south pole which attracts the north pole and thus opposes the cause producing it (i.e., opposes the motion of the magnet away from the coil). So, the direction of the induced current in the coil must be clockwise as seen from the magnet, [see Fig. (b)].

Question 16.
Explain that Lenz’s law is in accordance with the law of conservation of eaefgy.
Answer:
When a magnetic north pole is brought near a plane of a coil, the magnetic pole induced in that plane is also north pole causing a repulsive force on the north pole of the magnet. Therefore, work will have to be done against this repulsive force to bring the magnet’s north pole closer to the coil. The work done (mechanical energy) is transformed into electri¬cal energy causing a current induced in the coil.

But, on moving the north pole of the magnet away from the coil, a south pole is developed on the face of the coil and an attractive force is developed on the face of the magnet. So, work has to be done in moving the magnet away from the coil. This results in flow of induced current in reverse direction. Thus, Lenz’s law is in accordance with the law of conservation of energy.

Question 17.
Explain self – induction and demonstrate the phenomenon by an experiment.
Answer:
Self – induction:
During the motion of the truck, the axle cut the magnetic flux of the earth, which causes an induced e.m.f. set up across its axle. This induced charges are earthed by the chain. Which prevent the explosive from the fire.

Experiment:
The circuit of the experiment is shown in the figure. It consists of an insulated copper coil L, wound on soft iron core. Cell E, rheostat Rh and a tapping key K are connected in series. A bulb B is connected in parallel with the coil. Now, the key is closed, the bulb glows slowly, then becomes bright and when it is opened, the bulb flashes bright and then goes off.

As the key is pressed, the flux linked with coil changes and an induced current flows in the opposite direction of main current, because of this the current grows slowly in the circuit.But when the key is opened, the flux suddenly decreases to zero, therefore a strong induced current flows in the same direction of circuit current makes the bulb very bright for a moment.

Question 18.
What is self – induction? What is meant by self-induction of coil? Explain. WritgTts unit.
Answer:
Self – inductance:
When the current flowing through a circuit, an induced e.m.f. is set up in the same circuit, which opposes the main current. This phenomenon is called self – induction.
When the current increases in the circuit, induced current flows in the opposite direc¬tion and when the current decreases in the circuit, the induced current flows in the same direction of circuit current.
Let ϕ be the flux linked with a circuit when a current (l) flows through it. Then,
ϕ ∝I
or ϕ = LI …(1)
Where, L is a constant, called self – inductance of the coil.
Now, if I = 1, then by eqn. (i), we get
ϕ = L
Thus, the self-inductance of a coil is numerically equal to the magnetic flux linked with the coil, produced by unit current.
By Faraday’s law, e = –\(\frac {dϕ}{dt}\)
Putting the value of ϕ from eqn. (i), e = –\(\frac {d(LI)}{dt}\)
or e = -L\(\frac {dI}{dt}\)
If \(\frac {dI}{dt}\) = 1
Then, e = – L
Therefore, the self – inductance of a coil is numerically equal to the opposing e.m.f. produced in the coil due to a unit rate of change of current in the circuit.
Unit : Unit of self-inductance is henry (H).

Question 19.
Explain mutual inductance. Give its unit. On what factors does it depend?
or
Write down the definition, unit and dimensional formula of mutual inductance.
Answer:
Mutual inductance is defined in two ways :
1. Let the flux linked with secondary coil be ϕ due to current l in primary coil.
∴ ϕ ∝ I
or ϕ = MI
Where, M is constant called mutual inductance.
If l = 1, then ϕ = M.
Hence, mutual inductance is defined by numerical value of flux linked with secondary coil, with unit current flows through primary coil.

2. When current changes in primary coil, then flux changes in secondary coil. Thus, by second law of Faraday
Induced e.m.fi, e = –\(\frac {dϕ}{dt}\)
or e = –\(\frac {d(MI)}{dt}\)
or e = ML\(\frac {dI}{dt}\)
If \(\frac {dI}{dt}\) = 1
Then, e = – M
Thus, the mutual inductance of two coils is numerically equal to the induced e.m.f. set up in secondary coil at that moment, when the rate of change of current in primary coil is unity.

Unit : Its unit is Henry in SI system.
Dimensional formula :

It depends upon:

1. Number of turns of both the coils : As the number of turns increases, mutual inductance also increases.
2. Area of cross – section of both the coils : As the area increases, the mutual inductance increases.
3. Material of core : In presence of soft – iron core, the inductance is greater than that of air.

Question 20.
Give differences between self – induction and mutual induction.
Answer:
Difference between self – induction and mutual induction :
Self – induction:

• This phenomenon happens in the same coil when the current flowing through the coil changes.
• One coil is used in this phenomenon.
• Induced current affects the main current.

Mutual induction:

• This phenomenon happens in the another coil placed nearby a coil in which the current is changing.
• Two coils are used in this phenomenon.
• The induced current produced in the second coil, therefore main current is not affected.

Question 21.
What are eddy currents? Show an experiment to demonstrate eddy currents.
Or
What are eddy currents? What are their disadvantages ? Write any two uses of eddy currents.
Answer:
Eddy currents:
When a metallic plate is moved in a magnetic field or placed in a changing magnetic field, then flux linked with the conductor changes, hence an induced current is produced in the plate. This induced current is called eddy current.

Demonstration:
A rectangular copper plate is free to oscillate about a horizontal axis, passing through O. It is placed between the pole pieces of an electromagnet NS. Oscillate the plate, when no current is flowing in the circuit, the plate will oscillate for a long time and due to air resistance finally its amplitude of oscillation decreases.

Now, oscillate and pass the current. It is observed that the oscillations of the plate will be damped soon. Just as eddies are produced on water surface, the change of flux through the metal plate develops induced current known as eddy currents.

Uses : Eddy currents are used in :

1. Making a galvanometer dead beat.
2. Induction furnace.
3. Electric break.
4. Induction motor, etc.

Disadvantages of eddy currents:
Production of eddy currents causes loss of electrical energy in the form of heat. In order to prevent this, soft-iron core is laminated in a transformer. Lamination increases resistance and decreases eddy currents. So, there is very little dissipation of electrical energy in the form of heat.

Question 22.
The self – inductance of two coils P and S are L₁ and L₂ respectively. If the coupling between them is ideal, then prove that the mutual inductance between these coils is M = \(\sqrt { { L }_{ 1 }{ L }_{ 2 } }\)
Answer:
Let coil P is having N₁ total number of turns, length of coil is l and I is the current flowing.

Question 23.
Obtain an expression for the self – inductance of two coils when they are joined in parallel.
Answer:
Consider two inductors of self – inductance L₁ and L₂and joined in parallel.
If I₁ and L₂, be the current flow through L₁and L₂ respectively, then
I = I₁ + I₂

Where 1 is the total current in the circuit differentiating both sides, we get
\(\frac {dI}{dt}\) = \(\frac { { dI }_{ 1 } }{ dt }\) + \(\frac { { dI }_{ 2 } }{ dt }\) … (2)
In parallel combination, the induced e.m.f, across each coil remains the same.

Putting the value of \(\frac {dI}{dt}\) From eqns. (1) and (3), we get

Equation (4) is the required expression

Electromagnetic Induction Long Answer Type Questions

Question 1.
Derive an expression for mutual inductance ¡n between two circular plane coils. Write the factors affecting mutual inductance between circular plane colts.
Answer:
Consider two circular plane coils P and S kept coaxially close to each other. Thenumber of turns in the primary coil P is n₁ and in the secondary coil S is n₂. The radius of primary coil P is r₁ and that of the secondary coil is r₂. Due to flow of current l in the primary coil, the magnetic field produced at its centre is
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2\pi { n }_{ 1 }I }{ { r }_{ 1 } }\) = \(\frac { { \mu }_{ 0 }{ n }_{ 1 }I }{ 2{ r }_{ 1 } }\)
The magnetic flux linked with secondary coil S due to this magnetic field.
ϕ = nBA
=n₁ \(\frac { { \mu }_{ 0 }{ n }_{ 1 }I }{ 2{ r }_{ 1 } }\) πr²₂ … (1)
But, ϕ = MI … (2)
From eqns. (1) and (2),
\(\frac { { \mu }_{ 0 }{ n }_{ 1 }{ n }_{ 2 } }{ 2{ r }_{ 1 } }\).πr²₂
If some other medium is placed inside the coils in place of air or vacuum of perme – ability µ, then
M = µ\(\frac { { n }_{ 1 }{ n }_{ 2 } }{ 2{ r }_{ 1 } }\).πr²₂ … (4)

The mutual inductance between circular plane coils depends on following factors :

1. The mutual inductance increases on increasing the number of trims in primary coil and secondary coil.
2. On increasing the radius of primary coil r¹, the mutual inductance decreases.
3. On increasing the area of secondary coil, mutual inductance increases.
4. If a medium other than air is kept between the coil Mincreases.

Question 2.
Derive an expression for self – inductance of a long solenoid. On what factors does it depend?
Answer:
Expression for self – inductance of a long solenoid:
Let the length and radius of a solenoid be l and r respectively. Also n be the number of turns per unit length.

∴ Area of cross – section, A = πr²
and total number of turns = nl = N.
If current l flows through the solenoid, then the intensity of magnetic field inside the solenoid will be given by
B = µ₀nI, [∵B = \(\frac { { \mu }_{ 0 } }{ 4\pi }\)2πnl(cos0⁰ – cos180⁰)]
(for unit length if l > >r)
Where, µ₀ is the permeability of the medium.
∴ Magnetic flux linked with the total length of solenoid.

If the permeability of soft – iron core inside the solenoid is µ_r, then
L = \(\frac { { \mu }_{ 0 }{ \mu }_{ 1 }{ N }^{ 2 }A }{ l }\) … (3)
Equation (3) is the expression for self – inductance of a long solenoid. From this equation, it is clear that the self – inductance of a solenoid depends upon the following :

1. Area of cross – section of solenoid or radius of the solenoid : The self – inductance of a solenoid increases with radius and hence with the area of cross – section.

2. No. of turns : Self – inductance of a solenoid increases with number of turns.

3. Length of the solenoid : Self – inductance of a solenoid decreases on increasing the length of the solenoid.

4. Relative permeability of the core : The self – inductance of a solenoid increases on placing a core of higher permeability. This is the reason, that self-inductance of a solenoid with soft – iron core is greater than that of air core solenoid.

Equation (3) can be written in terms of radius as follows :
L = \(\frac { { \mu }{ N }^{ 2 }\pi { { r }^{ 2 } } }{ l }\)

Question 3.
Obtain an expression for mutual inductance of two long solenoids. What are the factors on which mutual inductance depends?
Answer:
Let S₁, and S₂ be the two long coaxial solenoids such that S₁, is completely surrounded by S₂. The lengths of both the solenoids are l. n₁ and n₂ are the number of turns per unit length respectively.

Case :
1. Let S₁ be taken as primary and S₂ as secondary solenoid. If I₁ is the current flowing through S₁ then the intensity of field inside S₁ will be
B₁ = μ₀n₁I₁
∴The magnetic field B₁ will superpose on solenoid S₂ and hence, magnetic flux linked with S₂ per turn = B₁A. Where, A is the area of cross – section of S₁.
Hence, net flux associated with S₂ will be
Φ₂₁ = n₂lB₁A
Where, n₂l is the total number of turns in solenoid S₂ and Φ₂₁ is magnetic flux on S₂ due to S₁
Substituting the value of B, from eqn. (1), we get
Φ₂₁ = μ₀n₁I₁n₂A = μ₀n₁n₂lAI₁ … (2)
But, Φ₂₁ = M₂₁I₁ … (3)
From eqns. (2) and (3),
M₂₁ = μ₀n₁n₂Al
This is the equation of mutual inductance of two solenoids. … (4)

2. Let S₂ be taken as primary and S₁ as secondary. If I₂ is the current through S₂, then the intensity of field inside S₂ will be
B₂ = μ₀n₂I₂ … (5)
∴ The magnetic field of S₂ i.e., B₂ will superpose on solenoid S, and hence, magnetic flux linked with each turn of S₁ = B₂A.
Hence, the total flux associated with S₁
Φ₁₂ =n₁lBn₂A … (6)
Where, n₁l is the total number of turns of solenoid S₁, and Φ₁₂ is magnetic flux on S₁, due to S₂.
Putting the value of B₂ from eqn. (5), we get
Φ₁₂ = μ₀n₂I₂An₁l = μ₀n₁n₂lAI₂ … (7)
But, Φ₁₂ = μ₁₂I₂ … (8)
From eqns. (7) and (8), we have
M₁₂ = μ₀n₁n₂Al … (9)
From eqns. (4) and (9), we have
M₂₁ = M₁₂
This is called reciprocity theorem.
It is clear that the mutual inductance of two solenoids remains the same whether cur¬rent flows in one or the other.
Thus, mutual inductance of two long solenoids is given by.
M = μ₀n₁n₂Al .
M = μ₀μ_r\(\frac { { N }_{ 1 } }{ l } \frac { { N }_{ 2 } }{ l }\)Al
μ₀μ_r\(\frac { { N }_{ 1 } }{ l } \frac { { N }_{ 2 } }{ l }\)Al

Mutual inductance depends upon :
1. Number of turns of both the coils : As the number of turns increases, the mutual inductance also increase.

2. Area of cross – section of both the coils : As the area increases, the mutual inductance increases.

3. Material of core : If the permeabilities of the cores in the solenoid are large, then mutual inductance increases. In presence of soft – iron core, the inductance is greater than that of air.

4. Length of the solenoid : For larger solenoids, mutual inductance is less.

Electromagnetic Induction Numerical Questions

Question 1.
The magnetic flux linked with a closed loop is given by the equation ϕ_B = 6t² + 7t +1, where ϕ_B is in milliweber and t is in second. Find the electromotive force induced at time t = 2 second.
Solution:
Formula: e = – \(\frac {dϕ}{dt}\)
Given : ϕ = 6t² + 7t +1
∴ = – latex]\frac {d}{dt}[/latex](6t² + 7t +1) = – 12t – 7
At t = 2 sec,
e = – 12 x 2 – 7
= – 31 milli volt.

Question 2.
An aircraft with a wing span of 40 m flies with a speed of 1080 km hr^-1 towards East where vertical component of Earth’s magnetic field is 1.75 x 10^-5 tesla. Find the e.m.f. developed between the tip of the wings. (NCERT)
Solution:
l = 40m, B_v, B_v = 1.75 x 10^-5T V = 1080\(\frac {Km}{hr}\)
= 1080 x \(\frac {5}{18}\) = 300 m/sec
e = B_vlv = 1.75 x 10^-5 x 40 x300
= 0.21 volt

Question 3.
The flux linked with a coil changes from 1 Wb to 0.1 Wb in 0.1 sec. Calculate the induced e.m.f.
Solution:
Formula : e = – \(\frac {dI}{dt}\)
Given : dl = (0.1 – 1) and dt = 0.1.
e = – \(\frac {(0.1 – 1) }{0.1}\)
or e = \(\frac {(0.9) }{0.1}\)

Question 4.
A current is flowing through a coil having 800 turns. The flux linked with it is 1.5 x 10-s Wb. Find the self-inductance of the coil.
Sol. Formula : = LI
Flux linked with one turn = 1.5 x 10^-5 Wb,
∴ Flux linked with 800 turns = 800 x 1.5 x 10^-5Wb
Substituting the values, we get
800 x 1.5 x 10^-5 = L x 1.5
∴ L = 800×10^-5
= 8.0 x 10^-3H
= 8.0 mH.

Question 5.
An air – core solenoid with length 30 cm area of cross – section 25 cm² and number of turns 500 carries a current of 2.5A. Current is switched off for 1 m sec. Find the average back emf induced across the ends of the open switch in the circuit. (NCERT)
Solution:
l = 30 cm = 0.3 m, A = 25 cm² = 25 x 10^-4m²
N = 500, dt = 10^-3sec., dl = 0 – 2.5 = – 2.5A
Back emf

MP Board Class 12th Physics Important Questions

MP Board Class 12th Economics Important Questions Unit 9 Government Budget and Economy

Government Budget and Economy Important Questions

Government Budget and Economy Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
The duration of Government budget is :
(a) 5 years
(b) 2 years
(c) 1 year
(d) 10 years.
Answer:
(c) 1 year

Question 2.
Budget is presented in the Parliament by :
(a) Prime Minister
(b) Home Minister
(c) Finance Minister
(d) Defence Minister.
Answer:
(c) Finance Minister

Question 3.
Budget speech in Lok Sabha is given by :
(a) President
(b) Prime Minister
(c) Finance Minister
(d) Home Minister.
Answer:
(c) Finance Minister

Question 4.
Professional tax is imposed by :
(a) Central Government
(b) State Government
(c) Municipal Corporation
(d) Gram Panchayat.
Answers:
(b) State Government

Question 5.
From the following which is included in the direct tax :
(a) Income Tax
(b) Gift Tax
(c) Both (a) and (b)
(d) Excise Tax.
Answer:
(c) Both (a) and (b)

Question 6.
Who issues 1 rupee note in India :
(a) Reserve Bank of India
(b) Finace Ministry of India
(c) State Bank of India
(d) None of these.
Answer:
(b) Finace Ministry of India

Question 2.
Fill in the blanks:

1. …………………… is a document containing income and expenditure of the government.
2. Income tax is …………………… tax.
3. …………………… tax is levied on the value of the goods.
4. Service tax is levied by the ……………………
5. budget is considered good for the country.
6. Finance bill contains …………………… proposals.
7. Government budget is presented on the last day of ……………………

Answer:

1. Budget
2. Direct
3. Advalorem
4. Central
5. Deficit
6. Tax
7. February.

Question 3.
State true or false :

1. Deficit budget is not considered as a good budget.
2. Electricity tax is levied by the State Government.
3. Budget speech is given by the Finance Minister.
4. Central excise duty is direct tax.
5. Interest payment is a planned item.
6. During deflation surplus budget is made.
7. Rail budget is generally not included in the annual budget.

Answer:

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

Question 4.
Match the following

Answer:

1. (b)
2. (d)
3. (a)
4. (e)
5. (c)

Question 5.
Answer the following in one word/sentence :

1. Write meaning of surplus budget.
2. Expenditure on education is considered as?
3. Who passes the budget presented by the Finance Minister every year?
4. Which tax was levied by the government on July 2017?
5. For how many years government makes budget?
6. What is the full form of G.S.T.?
7. Land Revenue is levied by whom?
8. What is the name gives to budget?

Answer:

1. More income and less expenditure
2. Developmental
3. Parliament
4. G. S. T
5. 1 year
6. Goods and Service Tax
7. By State Government
8. Master Finacial Scheme of Government.

Government Budget and Economy Very Short Answer Type Questions

Question 1.
What do you mean by government budget? How many types are there?
Answer:
A budget is the statement of financial plan of the government for a financial year (1st April to 31st March). It indicates the revenue expenditure estimates for the next financial year of the government.

Budget is of two types:

1. Capital budget
2. Revenue budget.

Question 2.
What do you mean by primary deficit?
Answer:
Primary deficit is the difference between fiscal deficit and interest payments. It indicates how much government borrowing is going to meet expenses other than interest payments. It is often used as the basic measure of fiscal responsibility.
Primary Deficit = Fiscal Deficit – Interest payments

Question 3.
What is tax?
Answer:
A tax is a compulsory contribution which is given by the people to the government in order to meet the expenditure on the welfare of the citizens.

Question 4.
What is deficit budget?
Answer:
When the expenditure of the government is more than the income of the government, it is called deficit budget.

Question 5.
What do you mean by supplementary budget?
Answer:
Supplementary budget is prepared for the temporary period. It is prepared during the period of emergency like flood, war, earthquake etc.

Question 6.
What do you mean by zero primary deficit?
Answer:
When government has to take loan only to fulfill the liability of interest it is called zero primary deficit.

Question 7.
What do you mean by vote on account?
Answer:
Special arrangement is made under which special power is vested with lok sabha to sanction some amount as advance till the financial budget is passed for the next year.

Question 8.
What is tax evasion?
Answer:
When people do not pay tax by hiding income it is called tax evasion.

Question 9.
What do you mean by surplus budget?
Answer:
When the income of the government is more than expenditure of the government in budget, it is called surplus budget.

Question 10.
What is a balanced budget?
Answer:
Budget in which income and expenditure of the government is equal is called balanced budget.

Question 11.
Write the tax multiplier.
Answer:
Tax multiplier = \(\frac {-c}{1 – c}\)

Question 12.
What do you mean by debt trap?
Answer:
Generally developing countries take loans from foreign countries to fulfill their projects and plans. Developing countries have to pay debt along with the interest on it. To pay this amount again government has to take loan. Thus, principal amount goes on increasing. These countries take loan from one country and pay loan of other country. Thus these countries go in the clutches of debt. This is called debt trap.

Question 13.
Write the main revenue sources.
Answer:
Public revenue:
By public revenue we mean all those income of the government which are essential for government expenditure. Public revenue can be divided into two parts :

1. Revenue receipts
2. Capital receipts.

1. Revenue receipts (Items of income) are of two types :

(A) Tax revenue : It includes all direct and indirect taxes which are imposed by the central government. For example, Income Tax, Corporation Tax, Production tax.

(B) Non – tax revenue :

• Interest receipts
• Dividends and Profits
• Foreign grants
• Fiscal services, Economic services, Subsidiary assistance.

Question 14.
Give the relationship between the Revenue deficit and the Fiscal deficit.
Answer:
Revenue deficit refers to the excess of Government’s revenue expenditure over revenue receipts whereas fiscal deficit is the difference between the government’s total expenditure and its total receipts excluding borrowings.

Question 15.
Explain why public goods must be provided by the government.
Answer:
Public goods must be provided by the government as they cannot be provided through market mechanism i.e., by transactions between individual consumer and producers.

Question 16.
Explain the relation between government deficit and government debt.
Answer:
Relation between government deficit and government debt: The government deficit and government debt are closely related. The government deficit in a flow concept, but it adds to the stocked debt. If the government continues to borrow year after year there is an accumulation of debt. It implies that government has to pay more and more by way of interest. These interest payment contribute to the debt. Thus, deficit is the cause and effect of debt.

Question 17.
Explain the concept of Deficit budget.
Answer:
Deficit budget: It is the budget in which government receipts are more than government expenditure.
Formula = Expected public income < Expected public expenditure.

Government Budget and Economy Short Answer Type Questions

Question 1.
What is a government budget? Write its objectives. (Delhi, Foreign 2013)
Answer:
“A budget is the statement of the financial plan of the government for a financial year” (1st April to 31st March). It indicates the revenue expenditure estimates for the next financial year of the government. The main objectives of budget are as follows :

1. Reallocation of financial resources.
2. Removal of inequality of income and wealth.
3. Stabilization of price level.
4. Management of public enterprises.
5. Expansion of employment opportunities.

Question 2.
How inequalities in income can be removed through budget?
Or
What is the role of budget in removing income inequalities? (Delhi, All India 2011, Foreign 2012)
Answer:
Fiscal policy implies the income and expenditure policy or the budgetary policy of the government. It is a branch of public finance which deals with the types of financial statements made by any government about its probable revenue and expenditure during a given year.

Through their fiscal policies government can play a significant role in reducing inequality of income and wealth as well as inequality of opportunity. Both tax and spending policies can alter the distribution of income over both short term and medium term. Government need to reform the policies by imposing more taxes oh riches sections and to reduce the burden of tax on poor with the motive of economic welfare of the poor people.

Question 3.
Explain in brief how economic stability can be obtained by government budget? (All India 2011)
Answer:
Government budget is used to prevent business fluctuations of inflation or deflation to achieve the objective of economic stability. The government aims to control the different phases of business fluctuations through the budgetary policy. Policies of surplus budget during inflation and deficit budget during deflation helps to maintain stability of prices in the economy.

Question 4.
What is the role of government in Reallocation of Resources? Explain. (Delhi 2012)
Answer:
Through the budgetary policy, Government aims to reallocate resources in’ accordance with the economic and social priorities of the government. Government can influence allocations of resources through:

1. Tax concessions or subsidies:
To encourage investment, government can give tax concessions subsidies etc. to the producers, example Government discourages the production of harmful consumption goods through heavy taxes and encourages the use of khadi products ‘by providing subsidies.

There is disequilibrium in the balance of payments when imports are more and exports are less. With demonitization of Indian rupee imports have become costlier, efforts can be made to make balance of payments favourable. But this cannot be done again and again because by this demand of our currency will decrease.

Question 5.
Explain four types of public expenditure.
Answer:
Following are the types of public expenditure :

1. Developmental expenditure:
Under it we include those expenditure which are done for the economic development and social welfare of people. It includes education, medicine, industry agriculture, transportation roads, canals, water welfare, electricity etc.

2. Non – development expenditure:
This expenditure includes that expenditure which is done on administration, on security and legal procedure of the government. It includes salaries of police department, military department, interest on loans, pension etc.

3. Plan expenditure:
It includes expenditure to be incurred during the year on programmes under the five year plan by planning commission. In this expenditure investors and consumers both are included. For example, economic activities like agriculture, industry transportation, communication etc. It also includes social welfare activities like education, family welfare, information and communication, drinking water, cleanliness, health etc.

4. Non – plan expenditure:
It includes all those expenditure other than plan expenditure. It includes debt given to State Government and others, expenditure on maintenance of property, expenditure on purchase of shares.

Question 6.
Differentiate between Direct Tax and Indirect Tax
Answer:

Question 7.
Differentiate between Revenue Expenditure and Capital Expenditure.
Answer:
Differences between Revenue Expenditure and Capital Expenditure :

Revenue Expenditure:

• Revenue expenditure is a current expenditure incured on civil administrations, defence forces, public health and education.
• It is of recurring type of expenditure. It is incurred regularly.
• It is called non – developmental expenditure.

Capital Expenditure:

• It refers to expenditure which leads to creation of assets or reduces liabilities.
• It is a non – recurring type of expenditure.
• It is called developmental expenditure.

Question 8.
Differentiate between Developmental and Non – Developmental Expenditure.
Answer:
Differences between Developmental and Non – Developmental Expenditure:

Developmental Expenditure:

• It is incurred on economic and social development of the country.
• Expenditure on agriculture, industries, transport, etc. are included in it.

Non – Developmental Expenditure:

• Its nature is non – developmental type.
• Expenditure on administrative services like police defence, grants to government, etc. are included in it.

Question 9.
What do you mean by budget? Write its characteristics.
Answer:
Origin of the word in India:
The word ‘budget’ has been derived from the French word ‘Bougette’, which means small bag. It symbolizes a bag containing the financial proposals. In England, the chanceller used to bring economic proposals and statements in a bag; In 1773, finance minister of Britain, Robert Walpole opened his leather bag for the budget proposal to take place in Parliament. From that it becomes popular in India.

Characteristics of budget : Following are the characteristics of budget:

1. Annual plan of income and expenditure : It is a description of annual income and expenditure of the government for the next financial year. It is a detailed description.
2. Fixed period : Budget is prepared before a fixed period. It is presented before the Parliament by finance minister on the last day of the month of February. It is related to the definite fixed period.
3. Financial discussion: In budget only financial discussion takes place.Other matters are not discussed here.
4. In advance : Budget is prepared in advance for the period during which it is to operate. It is an annual financial action plan for the next financial year.
5. Balance budget : Balanced budget is the symbol of economic stability of the country.

Question 10.
Write the objectives of budget.
Answer:
The objectives of budget are as follows :
1. Reallocation of financial resources:
The government reallocates the financial resources to achieve the desired goals. It is the primary responsibility of the government to build a sound socio – economic infrastructure regarding health housing, education and raising the standard of living of people. Government imposed more taxes on those items which are harmful for people on the other hand it may reduce the taxes which are useful from social point of view.

2. Removal of inequality of income and wealth:
Every nation tries to bring equality between the various sections of the people and to bridge the gap between poor and rich. Through subsidies, taxes government can remove their inequalities. The main objective of Budget is to fill the gap of rich and poor.

3. Stabilisation of price level:
It is the responsibility of the government to stabilise the price level in the country. Unnecessary fluctuation in the prices of goods and services adversely affect the economy. Through the budget the government can effectively control the price level and bring about stability in it.

4. Management of public enterprises:
The government looks after certain departments such as railways, post and telegraph, electricity, defence etc. The budget of the government makes special provisions for such public enterprises in order to safeguard the interest of public.

5. Expansion of employment opportunities:
Government throughout budget create the employment opportunities to people. Various employments oriented and productive programs can be implemented for this purpose.

Question 11.
Differentiate between Progressive Tax and Regressive Tax.
Answer:
Differences between Progressive Tax and Regressive Tax :

Progressive Tax:

• In progressive tax the rate of the tax increses as the taxable income increases.
• The burden of it is more on rich people.
• They are justified because they reduce inequalities of income.

Regressive Tax:

• In regressive tax the rate of the tax decreases as the taxable incomes increases.
• The burden of it is on poor people.
• These are not justified because they increase inequalities.

Question 12.
Write difference betw een Revenue Receipts and Capital Receipts. (Foreign 2013)
Answer:
Differences between Revenue Receipts and Capital Receipts :

Question 13.
Write a note on Surplus budget, Balance budget and Deficit budget.
Answer:
A budget is the statement of the financial plan of the government for a financial year. It indicates the revenue expenditure estimates for the next financial year. The different types of budget are :

1. Surplus Budget – When in budget the income of the government is more than expenditure of the government, it is called Surplus Budget.
2. Balance Budget – Balance budget is said to be balance when government revenue and expenditure are balanced.
3. Deficit Budget – If the expenditure of the government is more than the income of the government, it is called the Deficit Budget.

Question 14.
Does public debt impose a burden? Explain.
Answer:
Public debt does not impose a burden all the times. Only in the following situation, it imposes a burden. When government sectors to public debt, the government transfers the burden to reduce consumption on future generation, because the government may decide to pay off deut in future by raising taxes. Taxes reduce the savings and capital formation and growth. Thus, the debt acts as a burden on future generation.

Question 15.
(a) The fiscal deficit given the borrowing requirement of the government. Explain.
Answer:
Fiscal deficit gives the borrowing requirement of the government’s total expenditure and its total receipts excluding borrowing. The fiscal deficit has to be financed through borrowing. Thus, it indicates the total borrowing requirements of the government from all sources.

(b) Are fiscal deficits inflationary? (NCERT)
Answer:
It is not correct to that all deficits are necessarily inflationary. If the fiscal deficit results in higher demand and greater output, the fiscal deficit will not be inflationary. However, if the firms are unable to produce the higher quantities that are being demanded to due fiscal deficits, prices will rise resulting the inflationary.

Question 16.
We suppose that: C= 70 + 0.70YD, I = 90, G = 100, T = 0.10 Y
(a) Find the equilibrium income, (b) What are tax revenues at equilibrium income? Does the government have a balanced budget ?
Solution:
(a) Y = \(\frac {1}{1 – 0.70}\)(70 + 90 + 100)
Y = \(\frac {1}{0.30}\)(260)
Y = \(\frac {260}{0.30}\) = 866.66

(b) Tax revenue on balance budget (T) = 0.10 Y T = 0.10 (866.66) T = 86.66.

Question 17.
Suppose that for a particular economy investment is equal to 200, government purchases are 150, net taxes i.e., lump sum taxes minus transfer in 100 and consumption is given by C = 100 + 0.75. (a) What is the level of equilibrium multiplier? (b) Calculate the value of government expenditure multiplier and the tax multiplier, (c) If government expenditure increases by 200, And the change in equilibrium income.
Answer:
(a) The government directly affects the level of equilibrium income in two specific ways : 1. Purchase of goods and services and 2. Taxes and transfers. Taxes lower the consumption expenditure and disposable income. Hence, we use the following formula for calculating the level of equilibrium income:
Formula: Y = \(\frac {1}{1-c}\)(C – CT + CTR+ 1 + G)

(b) Govt expenditure multiplier:
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{1-c}\)
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{ 1-0.75}\) \(\frac {1}{0.25}\)
Tax multiplier \(\frac { ∆Y}{∆G}\) = \(\frac {-C}{1-c}\)
\(\frac { ∆Y}{∆G}\) = \(\frac {-0.75}{ 1-0.75}\) = \(\frac {-0.75}{ 0.25}\) = – 3

(c) If income of Govt, increases by 200 :

Question 18.
In the equation given in question 9, calculate the effect on output of a 10 increase in transfers and 10 increase in lumpsum taxes. Compare the effects of two.
Solution:
Transfer multiplier, = \(\frac {C}{1-C}\)
= \(\frac {0.80}{1-0.80}\)
= \(\frac {0.80}{0.20}\)
= 4
Increase in transfer = 10%
Hence, increase in output = 10%
Tax multiplier = 4
Decrease in output = 10 x 4 = 40%
or
Y = \(\frac {1}{0.20}\)(20 – 8 + 88 + 30 + 50)
Y = 5 x 188 = 940
Increase lump sum, its effect,
or Y = \(\frac {1}{1-0.80}\)(20 – 8 + 80 + 30 + 50)
Y = 5 x 172 = 860.
Hence, we find that effected change in transfer and change in taxes are equal. It is due to fact that their size of changes as well as multiplier are equal.

Question 19.
What do you understand by Goods and Service Tax (G.S.T.)? How is G.S.T. better than old system of taxation? Explain its types.
Answer:
(a) The Goods and Services Tax (G.S.T.) is a value Added Tax (VAT) levied on most goods and services sold for domestic consumption. The G.S.T. is paid by consumers, but it is remitted to the government by the businesses selling the goods and services:

G.S.T. in comparison with old taxation system :

1. In place of old taxation system, G.S.T. will become one tax economy.
2. The tax structure will be simplified with G.S.T.
3. G.S.T. will save both time and money.
4. The growth rate of economy will show a rapid increase with G.S.T.
5. But for the time being, the G.S.T. would be expected to increase the inflation rate in comparison to the old tax system.

For Goods and Services Tax, G.S.T., U.T.G.S.T. Act and S.G.S.T. tax are passed.

Government Budget and Economy Long Answer Type Questions

Question 1.
Explain the types of budget.
Answer:
The types of budget are as follows :
1. Central budget:
Central budget is prepared by Central Government. It is the numerical statement of income and expenditure done by central government. In India Central budget is presented in two parts:

• General budget
• Railway budget.

2. State budget:
State budget is prepared by State Government. State Government takes the help from Central Government.

3. Revenue budget and Capital budget:
In revenue budget we include the expenditure and income related to revenue of the government. On the other hand in capital budget the capital expenditure and capital income is included.

4. Supplementary budget:
Supplementary budget is prepared for the temporary period. It is prepared during the period of emergency like flood, earthquake and war etc. There is no fixed time for it.

5. Balanced budget and Imbalanced budget:
Balance budget is that budget where income and expenditure of the government is equal. Imbalanced budget is that budget where expenditure is more or less than income of the government. If the expenditure is more than income it is called deficit budget and if the income is more than the expenditure it is called surplus budget.

Question 2.
Explain the procedure of the budget.
Answer:
Under the budgetary procedure in India we can study the following heads:

1. Preparation of budget : The preparation of budget involves the following steps :

• Sending estimate forms by finance minister to all ministers and their department’s heads to get the estimates of revenue and expenditure required for the next financial year.
• Preparation of estimates by departmental heads : It includes revenue and expenditure of previous year budgetary estimate for next coming year.
• Preparation of consolidated estimate by the various ministers and sending them to finance minister.
  Scruting report estimates by A.G. of India and sending the same to finance ministry.

2. Presentation of the budget : In India the budget is presented in two parts :

• General budget and
• Rail budget.

Rail budget is always present before the general budget. General budget presented on the last day of the month of February at 5 p.m. generally 28th in February.

3. Discussion on the budget:
Budget is put before the Parliament for discussion. In the processes of passing the budget, the discussions on various items continues for 3 – 4 days. Presently different committees are formed. For the discussion and finance minister gives his final reply on the budget.

4. Voting on budget:
After the general discussion the budget is put for voting. The members of parliament give their speeches for and against the budget before voting. There after voting is held for passing the budget.

5. Appropriation bill:
The finance minister presents appropriation bill. According to the constitution no amount from the Reserve fund of India can be withdrawn without passing the appropriation bill in the Parliament.

6. Financial bill:
All the financial proposals for the coming year are included in a bill which is known as financial bill. This bill is generally presented immediately after the presentation of budget in the Lok Sabha. If this bill is not passed by the Parliament f ur government L not supposed to spend any amount.

7. Vote on account:
Special arrangement is made under which special power is vested with Lok shabha to snetion some amount as advance till the final budget is passed for the next year.

8. Implementation of budget:
After the budget is passed, it is implemented by the government for which the following steps are taken :

• Collection of revenue
• Preparation of accounts in
• Audit by A.G. (Auditor General).

Audit by comptroller and Auditor General of India presents their report before the Parliament which comprise the receipt of revenue, expenditure, loans taken.

Question 3.
Write the tax revenue sources of Public Revenue:
Answer:
Following are the sources of Public Revenue :
Tax:
Taxes are neither a fees nor a penalty because it is neither taken by the government to give any additional benefits nor taken as a fine. Taxes are compulsory contribution which are given by the people to the government in order to meet the expenditure on the Welfare of the citizens.

Types of taxes : Taxes are of following types :

• Progressive tax – In progressive tax the rate of my increases as the income increases
• Proportional tax – Proportional taxes are those taxes whose impact and incidence falls equally on both the sections of the society.
• Regressive tax – The rate of taxation decreases with the increase in income. The burden of text falls more heavily on poor than on rich.
• Direct taxes – These are those taxes in which the impact of the tax and incidence of tax is on the same person.

Question 4.
What do you understand by Non – tax sources of public revenue?
Answer:
Following are the non – tax sources of public revenue:

1. Fees, licence and permit:
Government gets non – tax revenue from fees licence fee and payment made for permits.

Fees – Registration fee for land, death and birth registration fees, passport fees are included in it.

Licence and permit – Fee is that fee which is paid by people to government after giving permission by the government to the people to do something. For example: Driving licence, import licence etc.

2. Fines:
Fines are collected by the govt, for breaking rules of the country. The main aim of this is not to collect money but to teach people lesson about it.

3. Income from public enterprise: Government may raise funds from many sources like Railways, Postal department, Steel Plants fertilizer Corporations, Indian Oil department etc. Government sells products of it and gets profit from it.

4. Gifts and grants: The government may raise the funds from others. During calamities like flood, earthquake etc. citizens and some NGO’s give help to the government . For example : W.H.O., UNESCO etc. they give assistant also during this period.

5. Notes issues:
Sometimes government prints extra notes and increase the treasury. Reserve Bank of India has got the right of printing notes and minting coins.

6. Stamp, registration and land revenue: The receipts of stamps, registration and land revenue is the another source of non – tax revenue. The stamp receipts of Supreme court is also the income of the central government.

7. Interest, receipts : Interest earned by  the government from the loans given to States, Union territories. Railways, Post and Telegraph are included in it.

8. Dividends: This includes the share from profits of various undertakings in which the central government has done investments example SAIL, ONGC etc.

9. Administrative receipts:
The central government provides a number of services to people like health, medical, education etc. The government earns certain income from these services. For example; court fees, registration fees.

Question 5.
Consider an economy described by the following functions. C = 20 + 0.80 Y, I = 30, G = 50, TR = 100.
(a) Find the equilibrium level of income and the autonomous expenditure multiplier in the model,
(b) If the government expenditure increases by 30, what is the impact on equilibrium income?
(c) If a lump sum tax of 30 added to pay for increase in government purchases, how will equilibrium income change?
Answer:
C = 20 + 0.80y
I = 30
G = 50
TR = 100
(a)
Y = \(\overline { C } \) + CY – (T – TR)
Y = 20 + 0.80 (Y + TR) + I + G
Y = 20 + 0.80 (Y + 100) + 30 + 50
Y = 0.8Y + 180
Y = \(\frac {180×100}{200}\)
= 900
The equilibrium level of income is 900. Autonomous expenditure multiplier.
= \(\frac {1}{1 – C}\)
= \(\frac {1}{1-0.08}\)
= \(\frac {1}{0.20}\)
= 5
Increase in equilibrium income = AG x Expenditure multiplier = 30 x 5 = 150
Tax multiplier = \(\frac {-C}{1 – C}\) = \(\frac {-0.08}{1-0.08}\) = \(\frac {-80}{0-20}\) = -4
Decrease in equilibrium income = ∆T x Tax Multiplier
= 30 x 4 = 120.
or
Y = \(\frac {1}{0.20}\)(180)
Y = 5 x 180 = 900
autonomus multiplier \(\frac { ∆Y}{∆G}\)= \(\frac {1}{1 – C}\)
or
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{1-0.80}\) = \(\frac {1}{0.20}\) = 5

(b) Expenditure of govt, increase by 30 :
(∆Y) = \(\frac {1}{1 – C}\) ∆G
or
∆Y = \(\frac {1}{1-0.80}\) x 30
= \(\frac {1}{0.20}\) x 30 = 5 x30 = 150
New income = 900 + 150 = 1,050
So, it is clear that by 30 it become 150 to 1,050.

(c) Lump sum 30 is added then :
Change in balance income (∆Y) = \(\frac {-C}{1 – C}\) ∆T
or
∆Y = \(\frac {-0.08}{1-0.08}\) x 30
= \(\frac {-0.08}{0.20}\) x 30 = -4 x 30
= 1 – 120
= 900 – 120 = 780

MP Board Class 12th Economics Important Questions

MP Board Class 12th Maths Important Questions Chapter 9 Differential Equations

Differential Equations Important Questions

Differential Equations Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Degree of the differential equation
\(\frac { d^{ 2 }y }{ d^{ 2 }x } \) + x² \(\frac{dy}{dx}\) = e^x is:
(a) 1
(b) 2
(c) 3
(d) does not exist
Answer:
(c) 3

Question 2.
Solution of differential equation (1 + x) y dx + (1 – y) xdy = 0 is:
(a) log xy + x + y = c
(b) logxy + x – y = c
(c) log xy – x – y = c
(d) log xy – x + y = c.
Answer:
(b) logxy + x – y = c

Question 3.
The differential equation of all circles which passes through the origin whose center lie on the A – axis is:
(a) x² = y²+ xy \(\frac{dy}{dx}\)
(b) x² = y² + 3xy \(\frac{dy}{dx}\)
(c) y² = x² + 2xy \(\frac{dy}{dx}\)
(d) y² = x² – 2xy \(\frac{dy}{dx}\)
Answer:
(c) y² = x² + 2xy \(\frac{dy}{dx}\)

Question 4.
Solution of the differential equation \(\frac{dy}{dx}\) + y = e^x, y(o) = 0 is:
(a) y = e^-x(x – 1)
(b) y = xe^x
(c) y = xe^-x + 1
(d) y = xe^-x
Answer:
(d) y = xe^-x

Question 5.
The straight line which satisfies the differential equation \(\frac{dy}{dx}\) = m and cuts an intercept 3 on the positive y – axis is:
(a) y = mx + c
(b) y = mx +3
(c) y = mx – 3
(d) y = – mx + 3.
Answer:
(b) y = mx +3

Question 2.
Fill in the blanks:

1. The corresponding differential equation of the equation x² + y² = a² is ……………………………
2. The differential equation of the curve y = e^cx is …………………….. where c is arbitrary constant.
3. The integration factor of the linear differential equation \(\frac{dy}{dx}\) + Py = Q is …………………………….
4. In the linear differential equation \(\frac{dy}{dx}\) + Py = Q. P and Q are …………………………….
5. The form of differential equation (x + y + 1) dy = dx is ……………………………
6. Solution of the differential equation e^-x+y \(\frac{dy}{dx}\) = 1 is ……………………………

Answer:

1. y \(\frac{dy}{dx}\) + x = 0
2. x \(\frac{dy}{dx}\) = y log y
3. e^∫pdx
4. constant
5. linear differential equation
6. e^-y = e^-x + c

Question 3.
Write True/False:

1. Order of differential equation y = x \((\frac { dy }{ dx } )^{ 2 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) is 2.
2. Degree of differential equation \((\frac { d^{ 3 }y }{ dx^{ 3 } } )^{ 4/5 }\) – 2 \((\frac { dy }{ dx } )\) \((\frac { d^{ 2 }y }{ dx^{ 2 } } )^{ 2 }\) = 0 is 5.
3. The integration factor of linear differential equation x \(\frac{dy}{dx}\) – y = 2x² is e^-x
4. Solution of differential equation dy = sin xdx is y + cos x – c = 0.
5. Solution of differential equation ydx + (x – y³) dy = 0 is xy = \(\frac { y^{ 4 } }{ 4 } \) + c.

Answer:

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

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

1. Write the integration factor of differential equation (1 + y²) + (2xy – cot y) \(\frac{dy}{dx}\) = 0.
2. Find the solution of the differential equation (1 + x²) dy = (1 + y²) dx.
3. Find the sum of order and degree of the differential equation y = x \((\frac { dy }{ dx } )^{ 3 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)
4. Find the integration factor of the differential equation x logx \(\frac{dy}{dx}\) + y = 2 log x
5. Find he solution of the differential equation \(\frac{dy}{dx}\) + \(\frac{1}{x}\) = \(\frac { e^{ y } }{ x^{ 2 } } \)

Answer:

1. 1 + y²
2. x – y = c(1 + xy)
3. 3
4. log x
5. 2xe^-y = cx² + 1.

Differential Equations Very Short Answer Type Questions

Question 1.
Find the order and degree of \(\frac{dy}{dx}\) + y = e^-x?
Answer:
1, 1

Question 2.
Find the degree and order of \((\frac { dy }{ dx } )^{ 3 }\) = \(\sqrt { 1+(\frac { dy }{ dx } )^{ 2 } } \)?
Answer:
1, 6

Question 3.
Find the degree and order of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + \(\sqrt { 1+(\frac { dy }{ dx } )^{ 3 } } \) = 0?
Answer:
2, 2

Question 4.
Find the differential equation corresponding to equation of circle x² + y² = a²?
Answer:
y \(\frac{dy}{dx}\) + x = 0

Question 5.
Find the differential equation of line y = mx + c?
Answer:
\(\frac{dy}{dx}\) = m

Question 6.
Solve the differential equation \(\frac{dy}{dx}\) = 4y?
Answer:
y = c.e^4x

Question 7.
Solve the differential equation x² \(\frac{dy}{dx}\) = 2?
Answer:
y = c – \(\frac{2}{x}\)

Question 8.
Find the solution of differential equation dy = sinxdx?
Answer:
y + cos x = c

Question 9.
Find the solution of differential equation \(\frac{dy}{dx}\) + Px = Q?
Answer:
xe∫^pdy. dy + c

Question 10.
Solve the differential equation (1 – y²) \(\frac{dy}{dx}\) + yx = ay?
Answer:
\(\frac { 1 }{ \sqrt { 1-y^{ 2 } } } \)

Differential Equations Short Answer Type Questions

Question 1.
Solve the differential equation xlog xdy – ydx = 0?
Solution:
x log x dy – ydx = 0 (given)
⇒ x log x = ydx
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = \(\frac{1}{xlogx}\) dx
⇒ ∫\(\frac{1}{y}\) dy = ∫\(\frac{1}{xlogx}\) dx
⇒ log y = ∫\(\frac{1}{t}\) dt , (let log x = t ⇒ \(\frac{1}{x}\) dx = dt)
⇒ log y = log t + log c
⇒ log y = log log x + log c

Question 2.
Solve the differential equation \(\frac{dy}{dx}\) = e^x-y + x.e^-y?
Solution:
\(\frac{dy}{dx}\) = e^x-y + x.e^-y (given)
⇒ \(\frac{dy}{dx}\) = e^-y (e^x + x)
⇒ e^ydy = (e^x + x) dx
Integrating both sides,
∫e^ydy = ∫(e^x + x) dx
e^y = e^x + \(\frac { x^{ 2 } }{ 2 } \) + c

Question 3.
Prove that solution of y = 4sin 3x is \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0?
Solution:
y = 4 sin 3x (given)
Differentiating with respect to x,
∴\(\frac{dy}{dx}\) = 12 cos 3x
Again differentiating with respect to x,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = – 36 sin 3x = – 9 × 4 sin 3x
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -9y, [from eqn.(1)]
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0. Proved.

Question 4.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 2x?
Solution:
\(\frac{dy}{dx}\) = sec x (sec x + tan x) (given)
⇒ dy = (sec² x + sec x tan x ) dx
⇒ ∫dy = ∫sec² dx + ∫sec x tan x dx
∴y = tan x + sec x + c.

Question 5.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 3x²?
Solution:
\(\frac{dy}{dx}\) = sec² x + 3x² (given)
⇒ dy = (sec² x + 3x²) dx
⇒ ∫dy = ∫sec² xdx + 3∫x² dx
⇒ y = tan x + \(\frac { 3x^{ 3 } }{ 3 } \) + c
⇒ y = tan x + x³ + c.

Question 6.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 2x?
Solution:
Solve as Q.No . 5

Question 7.
Solve the differential equation \(\frac{dy}{dx}\) = (3x² x + 2)?
Solution:
\(\frac{dy}{dx}\) = (3x² + 2) (given)
⇒ dy = (3x² + 2) dx
⇒ ∫dy = ∫(3x² + 2) dx
⇒ y = 3 × \(\frac { x^{ 3 } }{ 3 } \) + 2x = c = x² + 2x + c.

Question 8.
Solve the differential equation x² \(\frac{dy}{dx}\) = 2?
Solution:
x² \(\frac{dy}{dx}\) = 2 (given)
⇒ dy = 2.x^-2 dx
⇒ ∫dy = 2∫x^-2 dx
⇒ y = 2 \((\frac { -1 }{ x } )\) + c

Question 9.
Solve the differential equation \(\frac{dy}{dx}\) = x³ + sin 4x?
Solution:
\(\frac{dy}{dx}\) = x³ + sin 4x (given)
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { -cos4x }{ 4 } \) + c

Question 10.
Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^3x?
Solution:
\(\frac{dy}{dx}\) + 2x = e^3x (given)
⇒ \(\frac{dy}{dx}\) = e^3x – 2x
⇒ dy = (e^3x – 2x) dx
⇒ ∫dy = ∫^3x dx – 2∫x dx
⇒ y = e^3x. \(\frac{1}{3}\) e^3x – x² + c

Question 11.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \) (given)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
⇒ ∫sec² ydy = ∫cosec² xdx
⇒ tan y = – cot x + c

Question 12.
Solve the differential equation (x² + 1) \(\frac{dy}{dx}\) = 1?
Solution:

Question 13.
Solve the differential equation \(\frac{dy}{dx}\) = sin x sin y?
Solution:
\(\frac{dy}{dx}\) = sin x sin y
⇒ cosec y dy = sin x dx
On integrating,
– log_e (cosec y + cot y) = – cos x + c
⇒ cos x – log_e (cosec y + cot y) = c

Question 14.
Solve the differential equation \(\frac{dy}{dx}\) = y sin x?
Solution:
\(\frac{dy}{dx}\) = y sin x
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = sin x
⇒ ∫\(\frac{1}{y}\) dy = ∫sin x dx
⇒ log y = – cos x + c

Question 15.
Solve the differential equation \(\frac{dy}{dx}\) = x cos x?
Solution:
Given:
\(\frac{dy}{dx}\) = x cos x
⇒ dy = x cos x dx
⇒ ∫dy = ∫x cos x dx
⇒ y = xsin x – ∫1. sin x dx + c
⇒ y = xsin x + cos x + c

Question 16.
Solve the differential equation \(\frac{dy}{dx}\) = x³ + sin 4x?
Solution:
\(\frac{dy}{dx}\) = x³ + sin 4x (given)
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { cos4x }{ 4 } \) + c

Question 17.
Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^3x?
Solution:

Question 18.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)?
Solution:
\(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
∴∫sec² ydy = ∫cosec² xdx + c
tan y = – cotx + c

Question 19.
Solve the differential equation \(\frac{dy}{dx}\) = 1 – x + y – xy?
Solution:
\(\frac{dy}{dx}\) = 1 – x + y – xy (given)
⇒ \(\frac{dy}{dx}\) = (1 – x) + y (1 – x)
⇒ \(\frac{dy}{dx}\) = (1 – x) (1 + y)
⇒ \(\frac{dy}{1+y}\) = (1 – x) dx
⇒ ∫\(\frac{dy}{1+y}\) = ∫(1 – x) dx
⇒ log_e (1 + y) = x – \(\frac { x^{ 2 } }{ 2 } \) + c

Question 20.
Solve the differential equation \(\frac{dy}{dx}\) = (1 + x) (1 + y²)?
Solution:
\(\frac{dy}{dx}\) = (1 + x) (1 + y²) (given)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = (1 + x) dx
Integrating both sides,
⇒ tan^-1y = x + \(\frac { x^{ 2 } }{ 2 } \) + c.

Question 21.
Solve the differential equation:
\(\frac{dy}{dx}\) = cot² x?
Solution:
\(\frac{dy}{dx}\) = cot² x (given)
⇒ dy = cot² x
⇒ ∫dy = ∫cot² x dx
⇒ y = ∫(cosec² x – 1) dx
⇒ y = – cot x – x + c

Differential Equations Long Answer Type Questions – II

Question 1.
(A) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sec x?
Solution:
\(\frac{dy}{dx}\) + y tan x = sec x (given)
Comparing the equation with \(\frac{dy}{dx}\) + Py = Q,
P = tan x, Q = sec x
∴L.F. = e^{∫P dx} = e^{∫tanx dx} = e^{log secx}
⇒ L.F. = sec x
Applying formula
y × (1.F.) = ∫Q × (I.F.) dx + c
⇒ ∫sec² x dx + c
⇒ y sec x = tan x + c

(B) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sin x?
Solution:
Solve as Q.No.1 (A).

Question 2.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \)?
Solution:

Question 3.
Solve the differential equation 3x² dy = (3xy + y²)dx?
Solution:
3x² dy = (3xy + y²) dx (given)

Putting in eqn. (1),

⇒ – \(\frac{1}{v}\) = \(\frac{1}{3}\) log x + c
∴- \(\frac{1}{y}\) = \(\frac{1}{3}\) log x + c

Question 4.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = 4x²?
Solution:

The required solution will be:

Question 5.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x?
Solution:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x (given)
⇒ \(\frac{dy}{dx}\) + \(\frac { 2x }{ (1+x^{ 2 }) } \)
Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution of eqn. (1) is:

Question 6.
Marginal cost price of making anything is given by equation c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x. Find the total cost price c(x) for making it? [Given c (0) = 100]
Solution:
c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x (given)
Integrating both sides
∫c'(x) dx = ∫ (2 + 0.15 x) dx
c(x) = 2x + 0.15 \(\frac { x^{ 2 } }{ 2 } \) + A …………. (1)
If x = 0
c(0) = 2 × 0 + \(\frac { 0.15 }{ 2 } \) × 0² + A
⇒ c(0) = A
∴ A = 100, [∵ c(0) = 100]
Putting in eqn. (1),
c(x) = 2x + 0.075 x² + 100

Question 7.
Solve the differential equation x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0?
Solution:
x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0 (given)
⇒ \(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = – \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx
Integrating both sides,

Question 8.
Solve the differential equation (x + y + 1) \(\frac{dy}{dx}\) = 1?
Solution:
(x + y + 1) \(\frac{dy}{dx}\) = 1 (given)
⇒ \(\frac{dy}{dx}\) = x + y + 1
⇒ \(\frac{dy}{dx}\) – x = y + 1
Comparing the equation with \(\frac{dx}{dy}\) + Px = Q?
P = – 1 and Q = y + 1
∴I.F. = e^∫pdy = e^-∫dy = e^-y
Here the required solution is:
⇒ x.e^∫pdy = ∫e^∫pdy. Qdy
⇒ x.e^-y = ∫e^-y (y + 1) dy + c
⇒ x.e^-y = – (y + 1)e^-y – ∫1(-e^-y) dy + c
⇒ x = -(y + 1) -1 + ce^y
x + y + 2 = ce^y

Question 9.
Solve the differential equation sec²x tany dx + sec² y tan xdy = 0?
Solution:
sec²x tany dx + sec² y tan xdy = 0 (given)

⇒ log y = – log x + log c
⇒ log x + log y = log c
⇒ log xy = log c
⇒ xy = c

Question 10.
Solve the differential equation \(\frac{dy}{dx}\) = y tanx – 2 sin x?
Solution:
\(\frac{dy}{dx}\) – y tanx = – 2 sin x (given)
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = – tan x, Q = – 2 sin x
L.F. = e^∫pdx = e^-∫tanxdx
= e^log_ecosx = cos x
The required solution is:
y.(I.F.) = ∫Q.I.F. dx + c
⇒ ycos x = -2∫sin x cos x dx + c
⇒ ycosx = -∫sin2xdx + c
⇒ ycosx = \(\frac { cos2x }{ 2 } \) + c.

Question 11.
Solve the differential equation \(\frac{dy}{dx}\) + 2y = 4x?
Solution:
Solution:
\(\frac{dy}{dx}\) + 2y = 4x
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = 2, Q = 4x
I.F. = e^∫pdx = e^∫2dx = e^2x
The required differential solution is:
y. (I.F) = ∫Q.I.F. dx + c
⇒ ye^2x = ∫4xe^2x dx + c

Question 12.
Solve the differential equation cos² x \(\frac{dy}{dx}\) + y = 2?
Solution:
cos² x \(\frac{dy}{dx}\) + y = 2
⇒ \(\frac{dy}{dx}\) + sec² x. y = 2 sec² x
Comparing with \(\frac{dy}{dx}\) + P y = Q,
P = sec² x, Q = 2 sec² x
I.F. = e^{∫sec2 xdx} 
Required solution is:
y. (I.F) = ∫Q.I.F. dx + c
⇒ y.e^tanx = ∫2 sec² x. e^tanx dx + c
⇒ y.e^tanx = 2.∫e^t dt + c, (Let tan x = t ⇒ sec² xdx = dt)
= 2.e^tanx + c
⇒ e^tanx (y – 2) = c.

Question 13.
Solve the differential equation cos x \(\frac{dy}{dx}\) + y = sin x?
Solution:
cos x \(\frac{dy}{dx}\) + y = sin x
⇒ \(\frac{dy}{dx}\) + sec x.y = tan x
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = secx, Q = tanx
L.F. = e^∫secxdx
= e^{log_e(sec x + tan x)} dx + c
Required Solution y.I.F. = ∫Q.I.F. dx + c
⇒ y. (sec x + tan x) = ∫tan x.(sec x + tan x) dx + c
= ∫sec x tan x dx + ∫tan² xdx + c
= sec x + ∫(sec² x – 1) dx = sec x + tan x = – x + c.

Question 14.
Solve the differential equation (1 + y²) dx = (tan^-1 y – x) dy? (CBSE 2015)
Solution:
(1 + y²) dx = (tan^-1 y – x) dy (given)

The required solution is:

Question 15.
Solve the differential equation (1 + y²) + ( x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0? (CBSE 2016)
Solution:
The given equation is:


Comparing with \(\frac{dx}{dy}\) + Px = Q,

The required solution is:

Put tan^-1 y = t,

Question 16.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { x }{ 1+x^{ 2 } } \). Where, y = 0 and x = 1? (NCERT)
Solution:

Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution is:

y = 0 and x = 1 (given)
0 (1 + 1)² = tan^-1 1 + c
⇒ 0 = \(\frac { \pi }{ 4 } \) + c
⇒ c = – \(\frac { \pi }{ 4 } \)
From eqn. (2),
y. (1 + x²) = tan^-1 x – \(\frac { \pi }{ 4 } \).

Question 17.
Solve the differential equation \(\frac{dy}{dx}\) + y cot x = 4x cosec x. Given: y = 0 and x = \(\frac { \pi }{ 2 } \)? (NCERT; CBSE, 2012)
Solution:
\(\frac{dy}{dx}\) + y cot x = 4x cosec x (given)
Comparing with eqn. (1) \(\frac{dy}{dx}\) + Py = Q,
P = cot x, Q = 4x cosec x
I.F. = e^∫pdx
= e^{∫cot xdx} = e^log(sinx) = sin x
The required solution is:
y.I.F = ∫I.F. × Qdx
⇒ y.sin x = ∫sin x × 4x cosec x dx
⇒ y sin x = 4∫\(\frac { xsinx }{ sinx } \) dx
⇒ y sin x = 4∫xdx
⇒ ysin x = \(\frac { 4x^{ 2 } }{ 2 } \) + c
⇒ y sin x = 2x² + c
When x = \(\frac { \pi }{ 2 } \) and y = 0,
0 (sin \(\frac { \pi }{ 2 } \)) = 2 ( \(\frac { \pi }{ 2 } \) ) ² + c
⇒ 0 = \(\frac { 2\pi ^{ 2 } }{ 4 } \) + c
∴ c = \(\frac { -\pi ^{ 2 } }{ 2 } \)
Putting in eqn. (1),
y. sin x = 2x² – \(\frac { -\pi ^{ 2 } }{ 2 } \).

MP Board Class 12 Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives

Application of Derivatives Important Questions

Application of Derivatives Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
The rate of change of the area of a circle with respect to its radius r when r = 5 is:
(a) 10 π
(b) 12 π
(c) 8 π
(d) 11 π.
Answer:
(b) 12 π

Question 2.
Tangent line of a curve y² = 4x at y = x + 1 is:
(a) (1,2)
(b) (2, 1)
(c) (1, -2)
(d) (- 1, 2).
Answer:
(a) (1,2)

Question 3.
Approximate change in the volume of a cube of side x metre caused by increasing the side by 2%:
(a) 0.03 x³
(b) 0.02 x³
(c) 0.06 x³
(d) 0.09 x³
Answer:
(c) 0.06 x³

Question 4.
Point on the curve x² = 2y which lie minimum distance from the point (0,5):
(a) (2\(\sqrt{2}\),4)
(b) (2\(\sqrt{2}\),0)
(c) (0, 0)
(d) (2, 2).
Answer:
(a) (2\(\sqrt{2}\),4)

Question 5.
Minimum value of the function f(x) = x⁴ – x² – 2x + 6:
(a) 6
(b) 4
(c) 8
(d) None of these.
Answer:
(b) 4

Question 2.
Fill in the blanks:

1. The function f(x) = cosx, for 0 ≤ x ≤ π is …………………………..
2. The radius of circular plate is increasing at the rate of 0.2 cm /sec. when r = 10, then the rate of change of the area of the plate is ………………………………
3. The function y = x(5 – x) is maximum at x is equal to ……………………………
4. The minimum value of 2x + 3y is ………………………….. when xy = 6.
5. The maximum value of sin x + cos x is ……………………………
6. If line y = mx+ 1 is a tangent line to the curve y² = 4x. Then value of m will be ……………………………..
7. Slope of the tangent to the curve y = x² at the point (1, 1) is ………………………….
8. Using differential the approximate value of \(\sqrt{0.6}\) is …………………………….

Answer:

1. Decreasing
2. 4πcm²/sec
3. \(\frac{5}{2}\)
4. 12
5. \(\sqrt{2}\)
6. 1,
7. 2,
8. 0.8.

Question 3.
Write True/False:

1. For all real values of x, the function f(x) = e^x – e^-x is increasing function.
2. If the length of equal sides of an isosceles traingle be x then its maximum area will be \(\frac{1}{2}\) x²
3. Function f(x) = 3x² – 4x is increasing in the interval (- ∞, \(\frac{2}{3}\) )
4. Function f(x) = x – cot x is always decreasing.
5. Equation of the normal of the curve y = e^x at point (0, 1) is x + y = l.

Answer:

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

Application of Derivatives Short Answer Type Questions

Question 1.
The radius of a circle increases at the rate of 2cm/sec. At what rate the area increases when radius is 10cm?
Solution:
Given:
\(\frac { dr }{ dt } \) = 2cm/sec
Let the area of circle be A
Then,

∵ Area of circle A = πr²

= 40π sq.cm/second

Question 2.
The radius of air bubble is increasing at the rate of 1/2 cm per second. At what rate the volume of the air bubble is increasing when the radius is 1 cm?
Solution:
Let the radius of air bubble be r.
∴ Volume V = \(\frac{4}{3}\) πr³

Question 3.
The radius of a balloon is increasing at the rate of 10 cm/sec? At what rate surface area of the balloon is increasing when radius is 15cm?
Solution:
Let r be the radius of balloon at any time t and its surface be x then
A = 4πr²
Differentiating eqn.(1) w.r.t. t,

Hence, when radius of ballon is 15cm, then its area is increasing at rate of 1200π cm/sec.

Question 4.
Find those intervals in which the function f(x) = 2x³ – 15x² + 36x + 1 is increasing or decreasing?
Solution:
f(x) = 2x³ – 15x² + 36x + 1
⇒ f'(x) = 6x² – 30x + 36
= 6(x² – 5x + 6)
= 6(x – 2) (x – 3)
For increasing function of f(x),
f'(x) > 0
or (x – 2) (x – 3) > 0
⇒ x – 2 > 0 and x – 3 > 0
⇒ x > 2 and x < 3 ⇒ x > 3
Clearly the function is increasing in interval (3, ∞)
Again, (x – 2) (x – 3) > 0
or x – 2 < 0 and x – 3 < 0
⇒ x < 2 and x < 3
⇒ x < 2
Clearly the function is increasing in interval (-∞, 2)
Hence, the function is increasing in the interval (-∞, 2) ∪ (3, ∞)
Again for decreasing function of f(x),
f'(x) < 0
⇒ (x – 2) (x – 3) < 0
or x – 2 < 0 and x – 3 > 0
⇒ x < 2 and x > 3, which is impossible
or x – 2 > 0 and x – 3 < 0 ⇒ x > 2 and x < 3
⇒ 2 < x < 3
Hence, f(x) is decreasing function in the interval (2, 3).

Question 5.
(A) If x + y = 8, then find maximum value of xy?
Solution:
Let P = xy
⇒ x + y = 8
⇒ y = 8 – x
∴P = x(8 – x) = 8x – x²
⇒ \(\frac { dP }{ dx } \) = 8 – 2x
⇒ \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2
For maximum or minimum value
8 – 2x = 0
⇒ x = 4
Now, x = 4 then \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2, which is negative
∴ at x = 4, then y = 4
Maximum value of P, when x = 4, y = 4
= 4 × 4 = 16.

(B) If x + y = 10 then find maximum value of xy?
Solution:
Solve like Q.No.5(A).

Question 6.
The radius of a circle increases at the rate of 3cm/sec. At what rate the area increases when the radius is 10cm?
Solution:
Let r be the radius and A be the area of circle then,
A = πr²
\(\frac { dA }{ dt } \) = rate of change of area = ?
⇒ \(\frac { dA }{ dt } \) = 2πr \(\frac { dr }{ dt } \)
⇒ \(\frac { dA }{ dt } \) = 2π.(10).3
= 60π cm²/second.

Question 7.
The volume of a cube is increasing at the rate of 9 cm³/sec? If the edge of cube is 10cm then at what rate the surface area of cube is increasing? (NCERT)
Solution:
Let the edge of cube = a dm
Volufne = V= a³
Surface area of cube = s = 6a²

Surface area of cube

When a = 10cm, then \(\frac { ds }{ dt } \) = \(\frac { 36 }{ 10 } \) = 3.6 cm²/sec.

Question 8.
(A) A man 180 cm high walks away from a lamp post at a rate of 1.2metre per second. If the height of lamp post is 4.5 metre. Find the rate at which the length of his shadow increases?
Solution:
Let AB be the lamp post of the height 4.5 m, PQ be the position of man at time t, CQ = x and BQ = y
drrf

(B) A man of height 2 metre walking away from a lamp post at a rate of 5km/ sec. If the height of the lamp post 6m. Find the rate at which the length of his shadow is increasing? (NCERT)
Solution:
Solve like Q. No.8(A).
[Ans. \(\frac{5}{8}\) km/sec.]

Question 9.
A ladder 5 metre long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of 2 metre per second. How just is its height on the wall decreasing when the foot of the ladder is 4 metre away from the wall? (NCERT)
Solution:
Let at any time t, the bottom of the ladder be at a distance x metre from the wall and the height of the wall bey metre.
QA = xm, OB = ym, AB = 5m (given)
\(\frac { dx }{ dt } \) = 2m/sec.
In ∆OAB,

Question 10.
Find the intervals in which f(x) = 5x³ + 7x – 13 is increasing or decreasing?
Solution:
Given:
f(x) = 5x² + 7x – 13 (given)
∴ f'(x) = 10x + 7
If f(x) is increasing,
f'(x) > 0
⇒ 10x + 7 > 0
⇒ x > \(\frac{-7}{10}\)
Hence, f(x) is increasing in interval ( \(\frac{-7}{10}\) , ∞)
If f(x) os decreasing, then,
f'(x) < 0
⇒ 10x + 7 < 0
⇒ x < \(\frac{-7}{10}\)
∴ f(x) is decreasing in (-∞, \(\frac{-7}{10}\) ).

Question 11.
Find the intervals in which the function f(x) = 2x³ – 24x + 5 is increasing or decreasing?
Solution:
f(x) = 2x³ – 24x + 5
Differentiating w.r.t x,
f'(x) = 6x² – 24
⇒ f'(x) = 6(x² – 4)
⇒ f'(x) = 6(x + 2) (x-2) …………………. (2)
(A) If f(x) in is increasing, then
f'(x) > 0
6(x + 2)(x-2) > 0
∴ f(x), x ∈ (-∞,-2) ∪(2, ∞) is increasing.
(B) If f(x) is decreasing,
f'(x) < 0 ⇒ 6(x + 2) (x – 2) < 0
f(x), x ∈ (-2,2) decreasing.

Question 12.
Show that the function f(x) = x – cos x is always increasing?
Solution:
Given function f(x) = x – cos x
∴ f'(x) = 1 – (- sin x)
⇒ f'(x) = 1 + sin x
We know that -1 ≤ x ≤ 1
-1 + 1 ≤ 1 + sin x ≤ 1 + 1
0 ≤ 1 + sin x ≤ 2
Hence f'(x) = 1 + sin x is always positive for all values of x.
∴ f(x) = x – cos x is always increasing.

Question 13.
Find the least value of a such that the function given by f(x) = x² + ax + 1 is strictly increasing on (1, 2)? (NCERT)
Solution:
Given:
f(x) = x² + ax + 1
∴ f'(x) = 2x + a
When, x ∈ (1,2)
∴ 1 < x < 2
⇒ 2 < 2x < 4
⇒ 2 + a < 2x + a < A + a
⇒ 2 + a < f'(x) < 4 + a Since f(x) is increasing function ∴ f'(x) > 0
⇒ 2 + a > 0 ⇒ a > -2
Hence, the least value of a is -2.

Question 14.
Let I be any interval disjoint from [-1, 1] prove that the function f given by f(x) = x + \(\frac{1}{x}\) is strictly increasing on I? (NCERT)
Solution:
f(x) = x + \(\frac{1}{x}\)

x ∈ 1, x ∉ (-1, 1)
x < – 1 0r x > 1
⇒ x² – 1 > 0
⇒ \(\frac { x^{ 2 }-1 }{ x^{ 2 } } \) > 0
⇒ f'(x) > 0, x ∈ 1
∴ f(x) is increasing on I. Proved.

Question 15.
Find the interval in which the given function is increasing or decreasing:
f(x) = x⁴ – \(\frac { x^{ 3 } }{ 3 } \)?
Solution:
f(x) = x⁴ – \(\frac { x^{ 3 } }{ 3 } \)


Hence, the interval 0 and \(\frac{1}{4}\) on X – axis is divided in three intervals
(- ∞, 0), (0, \(\frac{1}{4}\) ), ( \(\frac{1}{4}\), ∞)
In interval (- ∞, 0)
f'(x) = x² (4x – 1),
⇒ f'(x) < 0 [x² = +ve] [4x – 1 > 0]
⇒ f'(x) < 0
∴ In interval (- ∞, 0) the function f(x) decreasing
In interval (0, \(\frac{1}{4}\) ):
f'(x) = x² (4x – 1),
⇒ f'(x) > 0 [∵ x² = positive] [4x – 1 > 0]
∴ In interval (0, \(\frac{1}{4}\) ) f(x) is decreasing
In interval ( \(\frac{1}{4}\), ∞):
f'(x) = x² (4x – 1), [∵ x² = positive] [4x – 1 > 0]
⇒ f'(x) > 0
In interval ( \(\frac{1}{4}\), ∞) in function f(x) increasing.

Question 16.
The perimeter of a rectangle is 100 cm. Find the length of sides of the rectangle for maximum area?
Solution:
Let length of rectangle be x and breadth be y.
∴Perimeter of rectangle = 2(x + y)
⇒ 2x + 2y = 100
⇒ x + y = 50
Let area of rectangle,
A = xy = x(50 – x) = 50x – x², [from eqn.(1)]
\(\frac { dA }{ dx } \) = 50 – 2x
and \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) = -2
For maximum or minimum of A,
\(\frac { dA }{ dx } \) = 0
⇒ 50 – 2x = 0 or or x = 25
For any value of x, \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) is -ve
For x = 25, area of rectangle is maximum
Put in eqn.(1),
y = 50 – x = 50 – 25 = 25.

Question 17.
Area of a rectangle is 25 sq.cm. Find its length and breadth when its perimeter is minimum?
Solution:
Let the length and breadth of rectangle be x and y units and A be the area.
xy = 25 ……………………….. (1)
Perimeter of rectangle,
P = 2(x + y)

For maximum or minimum \(\frac { dP }{ dx } \) = 0

Which is positive for x = 5
∴ For Minimum parimeter x = cm.
y = \(\frac{25}{x}\) = \(\frac{25}{5}\) = 5 cm.

Question 18.
Find the maximum value of sin x + cos x = \(\sqrt{2}\)? (NCERT)
Solution:
Given:
f(x) = sin x + cos x …………………. (1)
∴f'(x) = cos x – sin x …………………… (2)
and f”(x) = – sinx – cos x …………………….. (3)
For maxima or minima
f'(x) = 0
∴ cos x – sin x = 0
⇒ sin x = cos x
⇒ tan x = 1
∴ x = \(\frac { \pi }{ 4 } \), \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \)
Put x = \(\frac { \pi }{ 4 } \) in eqn (3)

Hence, at x = \(\frac { \pi }{ 4 } \) the given function is maximum. In the same way for x = \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \) ………………… the function is maximum.
Put x = \(\frac { \pi }{ 4 } \) in eqn.(1),
Maximum value f ( \(\frac { \pi }{ 4 } \) ) = sin \(\frac { \pi }{ 4 } \) + cos \(\frac { \pi }{ 4 } \)
= \(\frac { 1 }{ \sqrt { 2 } } \) + \(\frac { 1 }{ \sqrt { 2 } } \) = \(\frac { 2 }{ \sqrt { 2 } } \) = \(\sqrt{2}\). Proved.

Question 19.
Two positive numbers are such that x +y = 60 and xy³ is maximum. Prove (NCERT)
Solution:
Given:
x + y = 60 …………………….. (1)
Let p = xy³
⇒ P = (60 – y). y³
⇒ P = 60y³ – y⁴
⇒ \(\frac{dp}{dy}\) = 180 y² – 4y³ …………………….. (2)
Put \(\frac{dp}{dy}\) = 0
180 y² – 4y² = 0
⇒ 4y² (45 – y) = 0
∴ y = 45
Differentiating eqn. (2) w.r.t y,
\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 360y – 12y²
= 12 y(30 – y)
Put y = 45,
\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 12 × 45 (30 – 45) = – ve
Hence, at y = 45, p is maximum
From eqn.(1),
x + 45 = 60
⇒ x = 15
∴ The two numbers are 15 and 45.

Question 20.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x coordinate is 2? (NCERT)
Solution:
The equation of given curve,
y = x³ – x + 1

Slope of tangent at x = 2,

= 3 × 4 – 1 = 11.

Question 21.
Find the slope of the tangent to the curve y = \(\frac { x-1 }{ x-2 } \), x ≠ 2 at x = 10? (NCERT)
Solution:
The equation of given curve:


At x = 10 slope of tangent is:

Question 22.
Find the point at which the tangent to curve y = x³ – 3x² – 9x + 7 is parllel to X – axis? (NCERT)
Solution:
Equation of given curve y = x³ – 3x² – 9x + 7
\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) (x³ – 3x² – 9x + 7)
⇒ \(\frac { dy }{ dx } \) = 3x² – 6x – 9
= 3[x² – 2x – 3]
= 3[x² – 3x + x – 3]
= 3[x(x – 3) + 1(x – 3)]
⇒ \(\frac { dy }{ dx } \) = 3(x – 3) (x + 1)
Tangent is parllel to X – axis.
∴\(\frac { dy }{ dx } \) = 0
3 (x – 3) (x + 1) = 0
⇒ x = 3,-1
When x = 3, then y = (3)³ – 3(3)² – 9 × 3 + 7
y = 27 – 27 – 27 + 7
y = – 20
When x = 3, then y = (-1)³ – 3(-1)² – 9(-1) + 7
y = – 1 – 3 + 9 + 7
y = 12
Hence, the point at which the tangent is parllel to X – axis are (3, -20) and (-1, 12).

Question 23.
Find the equation of the tangent to the parabolas y² = 4ax at the point (at², 2at)? (NCERT)
Solution:
Given:
y² = 4ax ……………………. (1)

Equation of tangent is:
y – y₁ = \(\frac { dy }{ dx } \) (x – x₁)
Where x₁ = at², y₁ = 2at, \(\frac { dy }{ dx } \) = \(\frac { 1 }{ t } \)
y – 2at = \(\frac { 1 }{ t } \) (x – at²)
⇒ yt – 2at² = x – at²
⇒ x – ty + at² = 0.

Question 24.
Find the equation of the tangent to the curve x^2/3 + y^2/3 = 2 at point (1, 1)? (NCERT)
Solution:
Equation of curve:
x^2/3 + y^2/3 = 2


Equation of tangent at point (x₁, y₁) is:

y – 1 = – (x – 1)
⇒ y – 1 = – x + 1
⇒ x + y – 2 = 0

Question 25.
Find the equation of tangent to the curve 2y + x² = 3 at point (1, 1)? (NCERT)
Solution:
Equation of given curve:
2y + x² = 3

Equation of tangent at point (x₁, y₁) is:

⇒ y – 1 = x – 1
⇒ x – y = 0.

Question 26.
(A) Find the equation of tangent to the cure x = cos t, y = sin at t = \(\frac { \pi }{ 4 } \)? (NCERT)
Solution:
Equation of 1^st curve:
x = cos t

x = cos t = cos \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
y = sin t = sin \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
Equation of tangent at (x₁, y₁) is:

(B) Find the equation of tangent and normal to the curve 16x² + 9y² = 145 at point (x₁, y₁), where x₁ = 2 and y₁ > 0? (CBSE 2018)
Solution:
Given:
16x² + 9y² = 145 ……………………. (1)

Put x = 2 in eqn.(1), we get
16.(2)² +9y² =145
⇒ 64 + 9y² =145
⇒ 9y² = 145 – 64 = 81
⇒ y² =9, [y ≠ -3 ∵ y₁ > 0]
⇒ y = 3
At point (2, 3)
\(\frac { dy }{ dx } \) = – \(\frac { 16 }{ 9 } \). \(\frac { 2 }{ 3 } \) = – \(\frac { 32 }{ 27 } \)
Equation of tangent of curve (1) at point (2, 3),

and equation of normal of curve (1) at point (2, 3) is,

Question 27.
Use differential to approximate (25)^1/3? (NCERT)
Solution:
Let y = x^1/3
Where, x = 27 and ∆x = -2


∆y is appromimately equal to dy.

Approximately value of (25)^1/3
= 3 + ∆y
= 3 – 0.074 = 2.926.

Question 28.
Use differential to approximate \(\sqrt { 36.6 } \)? (NCERT)
Solution:
Let y = \(\sqrt{x}\), where x = 36 and ∆x = 0.6

∆y is appromimately equal to b,

Approximate value of \(\sqrt { 36.6 } \)
= ∆y + 6
= 0.05 + 6 = 6.05

Question 29.
Use differential to approximate (15)^1/4? (NCERT)
Solution:
Let y = x^1/4
Where, x = 16 and ∆x = -1

∆y is appromimately equal to dy,

Approximately value of (15)^1/4
∆y + 2 = 2 – 0.031 = 1.969.

Question 30.
Use differential to approximate (26)^1/3? (NCERT)
Solution:
Let y = x^1/3
Where, x = 27 and ∆x = -1


∆y is appromimately equal to dy

Approximate value of (26)^1/3
= ∆y + 3
= 3 – 0.037 = 2.963

Question 31.
If the radius of the sphere is measured as 9cm with an error 0.03 cm then find the approximate error in calculating its surface area? (NCERT)
Solution:
Let the radius of circle = r
Given:
r = 9 cm, ∆r = 0.03 cm
Area of sphere

= (8πr) × 0.03
= 8π × 9 × 0.03 = 2.16 π cm²
The approximate error in caluculating the surface area is = 2.16 π cm²

Question 32.
If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume?
Solution:
Let the radius of sphere = r
and ∆r be the error in measuring the radius.
r = 7m, ∆r = 0.02 m (given)
Volume of sphere V = \(\frac{4}{3}\) πr³

The approximate error in calculating the volume is 3.92π m³

Question 33.
Find the approximate change in the volume V of a cube of sides x metre caused by increasing the side by 1%? (NCERT)
Solution:
Let x be the sides of cube.
Volume of cube V = x³, ∆x = 1% of x = \(\frac{x}{100}\)

Change in volume ∆V = ( \(\frac { dV }{ dx } \) ) ∆x

Approximate change in volume = 0.03 x³m³.

Question 34.
Find the approximate change in voume V of a cube of side x metre caused by increasing the side by 2%
Solution:
Let x be the side of cube,
∆x = 2% of x = \(\frac { x\times 2 }{ 100 } \) = 0.02 x
Volume of cube V = x³
\(\frac { dv }{ dx } \) = \(\frac { d }{ dx } \) x³ = 3x²
dV = ( \(\frac { dv }{ dx } \) ) ∆x = 3x² × 0.02 x = 0.06x³ m³
Thus the approximate change in voume = 0.06 x³m³.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Biology Important Questions Chapter 13 Organisms and Populations

Organisms and Populations Important Questions

Organisms and Populations Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Example of photophilous plant is:
(a) Sunflower
(b) Abies
(c)Taxus
(d) All of these
Answer:
(a) Sunflower

Question 2.
Example of saprophytic plant is:
(a) Orobanche
(b) Monotropa
(c) Balanophora
(d) Rafflesia
Answer:
(b) Monotropa

Question 3.
Example of total parasitic plant:
(a) Balanophora
(b) Monotropa
(c) Rafflesia
(d) Drosera
Answer:
(c) Rafflesia

Question 4.
Characteristic features of aquatic plants are:
(a) Undeveloped stem with thorns
(b) Leaves with hairs
(c) Soft and flexible stem
(d) None of the above
Answer:
(c) Soft and flexible stem

Question 5.
Sunken stomata are found in:
(a) Xerophytic plants
(b) Aquatic plants
(c) Terrestrial plants
(d) Floating plants
Answer:
(a) Xerophytic plants

Question 6.
Spongy roots are present in :
(a) Vallisnaria
(b) Hydrilla
(c) Trapa
(d) Pistia
Answer:
(b) Hydrilla

Question 7.
Which soil is best for growth of plants : (MP 2016)
(a) Sandy soil
(b) Silty soil
(c) Loamy soil
(d) Clayey soil.
Answer:
(c) Loamy soil

Question 8.
Hydrophytes are characterised by : (MP 2009 Set A)
(a) Spiny less developed stem
(b) Leathery leaves
(c) Delicate and mucilage bearing stem
(d) Sunken stomata.
Answer:
(c) Delicate and mucilage bearing stem

Question 9.
Sunken stomata is found in :
(a) Xerophytes
(b) Hydrophytes
(c) Terrestrial
(d) Floating hydrophytes.
(a) Xerophytes

Question 10.
Spongy roots are found in :
(a) Jussiaea
(b) Eichhomia
(c) Trapa
(d) Pistia.
Answer:
(a) Jussiaea

Question 11.
Which one is the example of mangrove :
(a) Rhizophora
(b) Eichhomia
(c) Avicennia
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 12.
Pneumatophores are the characteristic feature of :
(a) Mangrove plants
(b) Mesophytes
(c) Hydrophytes
(d) Xerophytes
Answer:
(a) Mangrove plants

Question 13.
Vascular bundles are feebly developed in :
(a) Hydrophytes
(b) Xerophytes
(c) Psammophytes
(d) Halophytes.
Answer:
(a) Hydrophytes

Question 14.
Phyllode is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) None of these
Answer:
(c) Leaf

Question 15.
Phylloclade is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) Petiole
Answer:
(b) Stem

Question 16.
Rootless plant is :
(a) Wolffia
(b) Eichhomia
(c) Lemna
(d) Jussiaea.
Answer:
(a) Wolffia

Question 17.
Free floating plant is :
(a) Pistia
(b) Eichhomia
(c) Azolla
(d) All of these
Answer:
(d) All of these

Question 18.
Vivipary is found in :
(a) Hydrophytes
(b) Xerophytes
(c) Epiphytes
(d) Mangrove plants.
Answer:
(d) Mangrove plants.

Question 19.
Transpiration Higher among the following plants : (MP 2015)
(a) Mesophytes
(b) Hydrophytes
(c) Xerophytes
(d) Algae cells.
Answer:
(a) Mesophytes

Question 2.
Fill in the blanks:

1. Sandal plant is a …………………… parasite.
2.  …………………… and …………………… are found in lichen’s thallus.
3.  …………………… is the example of rodent.
4. Plants and animals combine togather and make the …………………… factor of the enviroments.
5. Inactivity of animals in summer season is called ……………………
6.  …………………… is prescribed the word ecology.
7. Raffesia is a …………………… plant.
8. Formation of forest ecosystem is the last stage of the ……………………
9. The study of human papulation is called ……………………
10. Respiratony root is found in ……………………  plant
11. Vivipary is the characteristics of …………………… plant
12. Group of animals which are interacted to each other is called ……………………
13. Herbivorous animals which are take the food in the plants is called ……………………
14. Plants they grow low density of light is called ……………………

Answer:

1. Partial parasite
2. Algae and fungi
3. Guinea pig
4. Biotic
5. Summer hybernation
6. Peter
7. Parasitic plant
8. Secession
9. Demography
10. Man – grove
11. Mangrove
12. Community
13. Plant eater
14. Sciophyte.

Question 3.
Match the followings:
I. (MP 2012)

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

II. (MP 2012)

Answer:

1. (c)
2. (b)
3. (a)
4. (e)
5. (d)

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

1. Name any animal which lays eggs in the nest of other animal?
2. Name the plant which has pneumatophores.
3. Name two symbiotic partners of lichen.
4. Write the type of water which is available in plants.
5. Loss of the upper surface of the soil is called.
6. When do we celebrate world environmental day?
7. Name the rootless free floating plant.
8. In which plant, root cap is absent.
9. Give an example of symbiosis.
10. Formation of soil is called.
11. Plants are dependent to other plant for shelter is called.
12. Name an amphibious plant.
13. Name the plant in which stem is modified into leaf like structure and leaves are modified into spines.
14. Give an example of sexual dimorphism.

Answer:

1. Cuckoo
2. Rhizophora
3. Algae and Fungi
4. Capillary water
5. Soil erosion
6. 5 June
7. Wolffia
8. Hydrophytes
9. Pea and Rhizobium bacteria
10. Soil fertility
11. Epiphytes
12. Ranunculus
13. Opuntia
14. Honey bee

Organisms and Populations Very Short Answer Type Questions

Question 1.
Give two examples of each:
Symbiotic, commensal, phytoplankton, zooplankton and rooted floating plant
Answer:

1. Symbiotic – Escherichia, Triconimpha
2. Commensal – Orchid and Trees, Hermit crab
3. Phytoplankton – Nostoc, Anabaena
4. Zooplankton – Paramecium, Euglena
5. Rooted floating plant – Wolffia, Lemna.

Question 2.
Name the speciality of any population.
Answer:
Any population has some speciality, these are :

1. Population density
2. Growth rate
3. Death rate
4. Age distribution
5. Biotic capacity
6. Population growth form
7. Changes of population
8. Population dispersion

Question 3.
Name any two mangrove plants.
Answer:
Rhizophora, Avicennia.

Question 4.
Name the vagetation found in sundarban delta.
Answer:
Mangrove vegetation.

Question 5.
Name the plant in which pneumatophores are found.
Answer:
Rhizophora.

Question 6.
Give the term for scientific study of human population.
Answer:
Demography.

Question 7.
What is sex ratio of India according to 2001 census?
Answer:
933 females per 1000 of njales.

Question 8.
Name any animal which lays egg in the nest of other animal.
Answer:
Koyal.

Question 9.
Give the name of larva of frog, salamander and moths.
Answer:
Tadpole, Axolotal, Catterpillar.

Question 10.
Give the reason for increase in population.
Answer:
Decrease in death rate.

Question 11.
Give example of species polymorphism.
Answer:
Honeybee.

Question 12.
Name the species found in different geographical areas.
Answer: Allopatric

Question 13.
Write the name of any saprophytic angiospermic plant
Answer:
Monotropa.

Question 14.
What is called the relationship between hermit crab and sea – anemone?
Answer:
Protocooperation.

Question 15.
Why insectivores eat insects?
Answer:
To fulfill the need of nitrogen.

Question 16.
Which are main abiotic factor?
Answer:
Temperature, water, light, soil.

Question 17.
Give the name of one insectivores plant
Answer:
Utricularia.

Question 18.
What is ecological competition?
Answer:
Ecological competition is the struggle between two organisms for the same resources within an environment.

Question 19.
A transition area between two biomes.
Answer:
Ecotone.

Question 20.
The number of live births per thousand of population per year.
Answer:
Birthrate.

Organisms and Populations Short Answer Type Question

Question 1.
How is diapauses different from hibernation?
Answer:
Diapauses is a stage of suspended development to cope with unfavorable con-ditions. Many species of zooplankton and insects exhibit diapauses to tide over adverse climatic conditions during their development. On the other hands hibernation or winter sleep is a resting stage where in animals escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2.
If a marine fish is placed in a fresh water aquarium, will the fish be able to survive? Why or why not?
Answer:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In freshwater conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment out¬side. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Give definition of :

1. Population
2. Population density
3. Biotic potential
4. Growth rate
5. Death rate

Answer:

1. Population:
A community of animals, plants or humans among whose members inbreeding occurs.

2. Population density:
Population density is a measurement of population per unit area or unit volume.

3.Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

4. Growth rate:
Human population growth is the increase in the number of individuals in a population. Global human population growth amounts to around 83 million annually.
Growth rate = \(\frac { Annually birth }{ Mid year population}\) x 100

5. Death rate:
The death rate is the number of people per thousand who die in a particular are during a particular period of time.
Death rate = \(\frac { Annually birth }{ Mid year population}\) x 100

Question 4.
What suggestions will you give for awakening people towards population control.
Answer:
Following things can be done for controlling the population:

1. Promotion of education: Misunderstanding can be avoided by promotion of education such as:

• Children are gift of god
• More income from the more children.

2. To introduce for reality of population growth.
3. Uses of family planning techniques.
4. Restriction of more than one marriage.
5. To decrease birth rate.

Question 5.
Write the differences:

1. Species and Population
2. Population growth and Population density,
3. Monospecific and Polyspecific population
4. Competition and Scattering.

Answer:
1. Species and Population:
Population on defined as organisms that belong to the identical species and identical geographical niche or area. The said area should enable these species to interbreed with each other.

2. Population growth and Population density:
In biology, Population growth is the increase in the number of individuals in a population but population density is a measurement of population per unit area, it is a quantity of type number density.

3. Monospecific or Polyspecific population:
Monospecific population is the population of individuals of only one species but poly specific or mixed population is the population of individuals of more than one species and it is generally referred to as a community.

4. Competition and Scattering:
Competition is in general a contest between two or more rivalry between two or more entities, organisms, animals, economic groups or social groups etc. Population scattering is a method which shows equilibrium by interaction of population. In population, scattering seeds away from the parent plant.

Question 6.
Give four reasons of population growth in India.
Answer:
Four reasons of population growth in India are as follows:

1. Increasing growth rate.
2. Decreasing death rate.
3. No importance of education and promotion of education is less.
4. Conservative.

Question 7.
Explain, the formation of new species.
Answer:
Although all life on earth shares various genetic similarities, only certain organisms combine genetic information by sexual reproduction and have offspring that can than successfully reproduce. Scientists call such organisms members of the same biological species.

Question 8.
What do you mean by population equilibrium?
Answer:
Population equilibrium:
When growth pattern of any population is observed then it becomes clear that any population increases at faster rate in the beginning in a new area and reaches to a maximum limit and then remain constant for long time.This phase is called as population equilibrium. At equilibrium birth and death rate of any population become equal.

Question 9.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The cooperative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honeybees, ants, wasps and termites, etc. form well – organized societies showing division of labour and polymorphism. In these social in sects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 10.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes.These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 11.
Most living organisms cannot surv ive at temperature above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Answer:
Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep sea hydrothermal vents. They are able to survive in high temperatures because their bodies have adapted to such environmental conditions. These organisms contain specialized thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures unlike other enzymes.

Most living – organism can not survive above 45°C because:

• Above 45° C enzymes get denatured.
• Protoplasm precipitates at high temperature.

Question 12.
List the attributes that populations but not individuals possess.
Answer:
A population has the following attributes that an individual does not possess:

1. Birth rates and death rates.
2. Sex ratio.
3. Population density.
4. Age distribution.
5. Population growth.

Question 13.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:
A population grow exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation :
N_t = N₀ e^rt
Where,
N_t = Population density after time t.
N_o = Population density at time zero.
r = Intrinsic rate of natural increase.
e = Base of natural logarithms (0.434).
From the above equation, we can calculate the intrinsic rate of increase (r) of a population. Now, as per the question,
Present population density =x
Then, Population density after two years = 2x
t = 3 years
Substituting these values in the formula, we get:
=> 2x = x e^3r
=> 2 = e^3r
Applying log on both sides :
=> log 2 = 3r log e
\(\frac { log 2 }{ 3 loge }\) = r
\(\frac { 0.301 }{ 3 x 0.434 }\) = r
\(\frac { 0.301 }{1.302 }\) = r
0.2311 = r
Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

Question 14.
Name important defence mechanisms in plants against herbivory.
Answer:
Several plants have evolved various mechanisms both morphological and chemical to protect themselves against herbivory.

1. Morphological defence mechanisms:

• Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.
• Sharp thorns along with leaves are present in Acacia to deter herbivores.
• In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

2. Chemical defence mechanisms :

• All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal in ingested by herbivores.
• Chemical substances such as nicotine, caffeine, quinine and opium are produced in plants as a part of self – defense.

Question 15.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:
An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 16.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The ecological principle behind the biological control method of managing with pest insects is to keep their population in check by using their natural predators and parasites.

Question 17.
Distinguish between the following:

1. Hibernation $nd Aestivation
2. Ectotherms and Endotherms.

Answer:
1. Differences between Hibernation and Aestivation :

Hibernation:

• It is called winter sleep.
• Animal rests in warm place.
• It lasts for long winter months.

Aestivation:

• It is called summer sleep.
• Animal rests in cool, moist, shady place.
• It lasts for day time as nights are cooler and animal will come out.

2. Differences between Ectotherms and Endotherms :

Ectotherms:

• These are cold – blooded animals.
• They are unable to regulate their body temperature which changes with temperature of the surrounding environment.
• They hibernate in winters and aestivate in summers.

Endotherms:

• These are warm – blooded animals.
• They are capable of maintaining their body temperature.
• They can regulate their body temperature so, they do not need to show behavioural adaptations like these.

Question 18.
List the various abiotic environmental factors.
Answer:

1. Atmospheric factors : Light, temperature, wind and water.
2. Lithosphere : Rock, soil.
3. Hydrosphere : Pond, river, lake and ocean.
4. Edaphic factors: Soil texture, soil water, soil air, soil microorganism, soil pH, minerals.
5. Topographic factors : Slope, altitude, valley.

Question 19.
Give an example for:

1. An endothermic animal,
2. An ectothermic animal,
3. An organism of benthic zone.

Answer:

1. An endothermic animals : Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such, as bears, cows, rats, rabbits etc. are endothermic animals.
2. An ectothermic animals : Fishes such as sharks, amphibians frogs, and reptiles, tortoise, snakes, and lizards are ectothermic animals.
3. An organism of benthic zone : Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 20.
Define population and community.
Answer:
Population:
Population is a group of individuals of same species, which can reproduce among themselves and occupy a particular area in a given time.

Community:
It is an assemblage of several population in a particular area and time and exhibit interaction and interdependence through trophic relationship.

Question 21.
Define the following terms and give one example for each:

1. Commensalism
2. Parasitism
3. Camouflage
4. Interspecific competition.

Answer:
1. Commensalism:
It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted. The two individuals may be physically associated. example Sucker fish and shark.

2. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

3. Camouflage:
It is a strategy adapted by prey species to escape their predators. Organisms are cryptically coloured so, that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

4. Interspecific competition:
It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 22.
Explain population dispersal in brief.
Answer:
Population dispersal:
The movement of individuals of a population from a particular place to another is known as population dispersal. Dispersal of members of population may occur for search of food, escape from enemies or for breeding purpose. There are three types of

population dispersal:

1. Immigration – Inward movement of members of other population and settle with local population. It causes increase in population.
2. Emigration – Outward movement of individuals of a population to shift in some other place. It causes decrease in population.
3. Migration – It can be observed in birds. It is periodic departure of organism and returning back to its original place, example Siberian cranes visit India during winters and go back to north during summer.

Question 23.
With the help of suitable diagram describe the logistic population growth curve.
Answer:
Logistic growth:
1. The resources become limited at certain point of time so, no population can grow exponentially.

2. This growth model is more realistic.

3. Every ecosystem or environment or habitat has limited resources to support a particular maximum number of individuals called is carrying capacity (K).

4. When N is plotted in relation to time t, the logistic growth show sigmoid curve and is also called Verhulst-Pearl logistic growth. It is given by the following equation:
\(\frac { dN }{ dt }\) = rN [ \(\frac{K-N}{K}\) ]
Where, N = Population density at time t
R = Intrinsic rate of natural increase
K = Carrying capacity.

a = When resources are not limiting the growth, plot is exponential
b = When resources are limiting the growth. Plot is logistic, k is carrying capacity.

Question 24.
Write characteristic features of aquatic plants.
Answer:
Characteristic features of aquatic plants (Hydrophytes):

1. The cuticle is either absent or very thinly deposited on the epidermis.
2. Usually stomata are absent, if present they are non – functional.
3. Many air chambers or large intercellular spaces are present inside the plant body of hydrophytes.
4. The whole body of the plant helps in the absorption of water and minerals.
5. Root system is feebly developed and usually non – branched or less branched. Some hydrophytes like Wolffia lack foots.
6. The mechanical tissues are less distributed.
7. The vascular tissues are either absent or feebly developed in the case of hydrophytes.

Question 25.
Write short note on population growth form or pattern.
Answer:
Population growth from or pattern: The way of growth of population is known as population growth pattern. Population growth shows two patterns :

1. S – shaped form : In this pattern, the increase in population is very slow in the beginning, then there is a fast growth rate and then becomes stable.

2. J – shaped form : In some organisms the population increases very fast in the beginning because of some environmental factors, the growth is checked abruptly.

Question 26.
Give only five examples of communication of informations in animals.
Answer:
Communication of information within a species:
Communication may be defined as an act which influences the activity of one individual by some behaviour of another. Most of the animal species have some mechanism of exchanging information among their members through sight, sound, touch or chemicals. The evolution of the techniques of communication in the animal kingdom is progressive but complex.

Following are few examples:

1. The honeybee gives the information to the other worker bees of the hive regarding availability of nectar in the vicinity by dancing in a special manner on the surface of the hive. Through the round dance they communicate that the availability of the food is very near to the hive and the waggle dance (by moving abdomen) shows the direction and the distance of the food source from the hive.

2. Croaking of male frog attract female frog during breeding season.

3. The dogs express their various intensions by facial expressions and movement of the tail and by making typical sound.

4. Rabbits inform their group members about any sort of danger by tapping their
tails.

5. Certain chemical compounds called pheromones are secreted by animals to transmit message to other members of the species. Pheromones are detected by smell or taste.

Question 27.
Write distinguish between:

1. Parasitism and Symbiosis,
2. Mutualism and Commensalism,
3. Hydrosere and Xerosere.

Answer:
1. Differences between Parasitism and Symbiosis:

Parasitism:

• In this case, one organism depends on other for their food.
• Only one organism benefitted.
• It produce only disease.
• example Tapeworm and Man.

Symbiosis:

• In symbiosis two organisms lives together in such way that both are benefitted by each other.
• Both organism benefitted.
• It does not produce any disease.
• example Lichens.

2. Differences between Mutualism and Commensalism:

Mutualism:

• It is an interspecific interaction in which both the species are mutually benefited.
• The two individuals may be physically or physiologically associated.
• example Rhizobium and the legume plants.

Commensalism:

• It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted.
• The two individuals may be physically associated.
• example Sucker fish and shark.

3. Differences between Hydrosere and Xerosere:

Hydrosere:

• Succession in water is called hydrosere.
• Changes are fast.
• It produce aquatic plants.

Xerosere:

• Succession in deserts is called xerosere.
• Changes are slow.
• It produce xerophytic plants.

Question 28.
Write short notes on:

1. Parasitism
2. Species dominance
3. Biotic potential.

Answer:
1. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

2. Species dominance:
Presence of any biotic community or presence of more popu¬lation of any species biosphere shown its dominance and it is called species dominance which has capacity to reduce other species in population. The effect of dominant species is more upon environment or other species.

3. Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

Question 29.
Explain population fluctuation.
Answer:
Population is generally a group of individuals of a particular species occupying a particular area at a specific time.

Population fluctuation:
Any increase or decrease in number of individuals in a population from its equilibrium state is known as population fluctuation. It may occur due to various reasons, such as due to change in climate or due to change in physical environment or due to predators.

Question 30.
Give an example for:

1. Heliophytes
2. Sciophytes
3. Viviparous.

Answer:

1. Heliophytes : Sunflower, Amranthus.
2. Sciophytes : Java moss, Lycopodium, Polytrichus.
3. Viviparous : Rhizophora, Salicomia, Sonneratia.

Organisms and Populations Long Answer Type Questions

Question 1.
Write a short note on :

1. Adaptations of deserts plants and animals
2. Adaptations of plants to water scarcity
3. Behavioural adaptations in animals
4. Importance of light to plants
5. Effect of temperature or water scarcity and the adaptations of animals.

Answer:
1. Adaptations of desert plants and animals :

(i) Adaptations of desert plants:
Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables the stomata to remain closed during the day to reduce the loss of water through transpiration.

(ii) Adaptations of desert animals:
Animals found in deserts such as desert Kangaroo rats, Lizards, Snakes, etc. are well adapted to their habitat. The Kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and bur¬row themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

2. Adaptations of plants to water scarcity :
Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration.

In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

3. Behavioural adaptations in animals:
Certain organisms are affected by temperature variations. These organisms undergo adoptions such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations.

For example, ectothermal animals and certain endosperms exhibit behavioural adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begin? to rise, the lizard burrows itself inside the sand to escape the scorching, sun.

Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against tem-perature variations.

4. Importance of light to plants:
Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need right for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses occurring in plants. Plants respond to changes in intensity of light during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for vertical distribution of plants in the sea.

5. Effects of temperature or water scarcity and the adaptations of animals:
Temperature is the most important ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermals. Those which can tolerate a narrow range of temperature are called stenothermal animals. Animals also undergo adaptations to suit their natural habitats.

Animals found in polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat. Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioral adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations.

These lizards bask in the sun during early hours when the temperature is quite low. As the temperature begins to increase, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategy is seen in other desert animals. Water scarcity is another factor that forces animals to undergo certain adaptations to suit their natural habitat. Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to stay in their habitat.

The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow in the sand as the temperature rises to escape the heat of the day. Such adaptations can be seen to prevent the loss of water.

Question 2.
List any three important characteristics of a population and explain.
Answer:
The three important characteristic of a population are:

1. Population density : The number of individuals of a species per unit area or a volume is called population density.
P_D = \(\frac { N }{ S }\)
Where, PD = Population density
N = Number of individuals in a region S = Number of unit area in a region.

2. Birth rate : It is expressed as the number of births per 1000 individuals of a population per year.

3. Death rate : It is expressed as the number of deaths per 1000 individuals of a population per year.

Question 3.
Describe the anatomical adaptation of xerophytic plants.
Answer:
The anatomical adaptations of xerophytic plants are:

1. The epidermis of xerophytic plant organs is covered by thick cuticle. It protects plants against heavy transpiration and also provides mechanical strength to some extent.

2. Epidermis may be multiple layered which also reduces the rate of transpiration.
Examples : Banyan, Nerium etc.

3. Except some monocotyledonous plants the stomata are restricted to the lower surface of the leaves. The stomata are generally sunken and surrounded by many hairs.

4. Hypodermis in xerophytes is multilayered and sclerenchymatous which checks evaporation of water.
Example: Pinus needle.

5. Mesophyll tissue is well differentiated into palisade and spongy parenchyma but palisade parenchyma is more in number than spongy parenchyma.
Examples : Pinus needle, Nerium leaf.

6. Vascular bundles are well developed and differentiated into xylem and phloem. Vascular bundles are comparatively more in number. Besides vascular bundles, mechanical tissues are well developed.

7. Leaves of some xerophytes have some rolling devices due to presence of motor cells or bulliform cells in the epidermal zone.
Examples : Poapratensis, Agropyron, Ammophila etc.

8. Stomata in certain desert plants such as Capparis spinosa and Aristida ciliata may sometimes get blocked due to deposition of resinous matter or wax. This adaptation is also known to reduce the loss of water during transpiration.

Question 4.
Describe the morphological adaptations of aquatic plants.
Or
Describe morphological adaptations found in the stem and leaves of xerophytic plants.
Or
Explain in brief, xerophytic plants along with examples.
Answer:
The morphological adaptations of aquatic plants are:

1. Adaptations in roots:

(i) Root system is feebly developed and unbranched. Some floating plants like Wolffia, Utricularia etc. and submerged plants like Ceratophyllum lack roots.

(ii) Root hairs are absent except rooted floating hydrophytes. In most of the hydro¬phytes roots are meaningless as the entire surface of the plant body is in direct contact with water and acts as absorptive surface. This may probably be the reason why roots in hydrophytes are reduced or absent.

(iii) True root caps are absent in some free floating hydrophytes and in lieu of that they develop root pockets or root sheaths which protect their tips from injuries.

(iv) If roots are present they are always of adventitious type e.g., Nymphaea, Nelumbo etc. In Jussiaea modified spongy negative geotrophic floating roots are present which pro¬vide buoyancy to the plants and also do the job of gaseous exchange.

2. Adaptations in stems:

(i) In submerged hydrophytes like Hydrilla, Potamogeton etc. stems are long, slender and flexible.

(ii) In free floating hydrophytes, stems are modified as thick, stout, stoloniferous or as offset and occur horizontal on the water surface. It bears vegetative bud at their apex and helps in vegetative propagation.

(iii) In rooted floating hydrophytes or rooted submerged hydrophytes or in marshy plants, stem may be modified as runner or rhizome.

(iv) The stem may be modified as runner, stolon, tuber to perform vegetative propaga¬tion effectively. Usually stems are of perennial nature.

3. Adaptations in leaves:

(i) Leaves are thin, long, ribbon shaped in the submerged forms.
Example: Vallisneria, Ceratophyllum etc. or they are finely dissected as found in Ranunculus aquatilis.

(ii) In rooted floating plants, like Nymphaea, Nelumbo etc. the petioles of leaves show indefinite power of growth and they keep the laminae of leaves always on the surface of water.

(iii) In free floating plants like Eichornia, Trapa etc. the petioles become charac¬teristically swollen and develop sponginess which keep the plant afloat.

(iv) Occurrence of two or more than two different types of leaves in a plant is called heterophylly. Examples are Limnophila, Heterophylla, Sagittaria, Ranunculus aquatilis etc. They bear leaves of many types viz., submerged leaves, floating leaves, aerial leaves, linear, ribbon shaped leaves or dissected leaves. Floating leaves are entire and lobed. The broad leaves found on the surface of water transpire actively and regu¬late the hydrostatic pressure in the plant body. The submerged leaves act as water absorbing organs. Heterophylly is also associated with quantitative reduction in transpiration.

(v) In the amphibious or marshy plants, the leaves that are exposed to air show typical mesophytic features. They are leathery and more tough than the submerged and floating hydrophytes.

Question 5.
What do you mean by biotic community? Describe characteristic features of any biotic community.
Answer:
Biotic community:
A biotic community is a localized association of several populations of different species living in a given geographic area of habitat. It represents heterogeneous assemblage of different groups of organisms both plants and animals. Biotic community is composed of smaller units of intimately associated members belonging to different species.

The different species of a community share common environment and their relationships are based on direct or indirect functional interactions. The nature or relationship is determined by the requirements of the members of a community.

Characteristics of a community:
Each community has its own characteristics which are not shown by its individual component species.

1. Species diversity:
Each community is made up of much different organisms Plants, animals, microbes, which differ taxonomically from each other. The number of species and population abundance in community also vary greatly.

2. Growth form and Structure:
Each community have a definite growth form. This different growth form determines the structural pattern of a community.

3. Dominance:
In each community, all the species are not equally important. There are relatively only a few of these, which determine the nature of community. These few species exert a major controlling influence on the community. Such species are known as dominants.

4. Succession:
Each community has its own development history. It develops as a result of a directional change in it with time.

5. Trophic structure (Self – sufficiency):
Nutritionally, each community, a group of autotrophic plants as well as heterotrophic animals exists as a self – sufficient, perfectly balanced assemblage of organism.

Question 6.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The co – operative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honey – bees, ants, wasps and termites, etc. form well-organized societies showing division of labour and polymorphism. In these social insects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 7.
Explain interrelationship between various species of a biotic community along with examples.
Answer:
Odum (1971) distinguished interactions into two broad categories :

1. Positive interactions and
2. Negative interactions.

1. Positive Interactions:
In this type, those type of interactions are considered in which both interacting species are mutually involved to help each other. Here, one interacting species helps the other either one way or on reciprocal terms and may be in the form of nutrition or shelter or substratum or transport. Here interaction may be obligatory or facultative. The positive interactions include three types of interactions. They are :

(i) Commensalism:
This type of interaction occurs in between two organisms of two different species in which one species benefits and the other is neither benefited nor harmed, example Lichens.

(ii) Protocooperation:
The type of interaction where both population are ben¬efited but not obligatory i.e., not essential for the survival of either population is called proto – co – operation. This is also known as non-obligatory mutualism. The relationship between hermit crab and sea anemone is an example of proto – cooperation. The crab uses the gastropod (mollusc) shell as a portable shield and the sea anemone eats the leftover food of the crab, which is protected from its predators by the stinging cells of the sea anemone.

(iii) Mutualism or Symbiosis:
When two different species grow together and are mutually benefited, the plants are known as symbiotic plants and the phenomenon is called symbiosis or mutualism. It is a sort of obligatory association. This type the organisms are dependent upon each other for survival. In this type of association, two types of species are physiologically related. This type of relationship may exist in between two plants or in between one plant and one animal or in between two animals.

2. Negative Interactions:
Those interaction of two different species in which both are harmed or one organism is benefited while other is more or less harmed is referred to as negative interac¬tion. In this type of interaction one population eats the other type of population or one organism does not allow other organisms to grow near it by using the food supply of the other or producing toxic substances. The negative interactions be categorised into following types:

• Competition
• Parasitism
• Predation and
• Antibiosis.

Competition:
When interaction occurs between two species for the use of same resources and when resources are in short supply is referred to as competition. It is a relationship which involves struggle amongst the organisms for food, shelter, water, sunlight and climate.

Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria :
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

Predation:
To obtain food by hunting is predation. A predator is an animal or plant that kills and feeds on another organism, its prey. It represents a generalised carnivorous habit. Commonly predators are larger than their prey but this is not always true but predator is in any case always occur as a naturally better equipped hunter than its prey. Most of the predatory organisms are animals but there are some plants (carnivores) also especially fungi which feed upon other animals.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Maths Important Questions Chapter Application of Integrals

Application of Integrals Important Questions

Application of Integrals Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
The value of \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x} d x\) is:
(a) π
(b) \(\frac { \pi }{ 2 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -\pi }{ 4 } \)
Answer:
(c) \(\frac { \pi }{ 4 } \)

Question 2.
Area of the best quadrant of the ellipse \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1 is:
(a) πab
(b) \(\frac{1}{2}\) πab
(c) \(\frac{1}{4}\) πab
(d) \(\frac{1}{8}\) πab
Answer:
(c) \(\frac{1}{4}\) πab

Question 3.
The value of \(\int_{2}^{4} x^{3} d x\) is:
(a) 60
(b) 50
(c) 70
(d) 256
Answer:
(a) 60

Question 4.
\(\int_{0}^{2 a} f(x) d x=0\), if:
(a) f(2a – x) = f(x)
(b) f(2a – x) = – f(x)
(c) f(x) is an even function
(d) f(x) is an odd function
Answer:
(b) f(2a – x) = – f(x)

Question 5.
The value of \(\int_{0}^{2 \pi}|\sin x| d x\) is equal to:
(a) 2
(b) \(\sqrt{3}\)
(c) 4
(d) 0
Answer:
(d) 0

Question 2.
Fill in the blanks:

Answer:

1. 0
2. 0
3. 17
4. 0
5. \(\frac { \pi }{ 4 } \)
6. πa²

Question 3.
Write True/False:

Answer:

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

Question 4.
Match the column:

Answer:

1. (b)
2. (a)
3. (d)
4. (c)
5. (f)
6. (e)

Question 5.
Write the answer in one word/sentence:

Answer:

1. 1
2. \(\frac { -\pi }{ 2 } \) log 2
3. 2
4. \(\frac { \pi }{ 2 } \)
5. -4
6. \(\frac { \pi }{ 12 } \)

Application of Integrals Long Answer Type Questions – II

Question 1.
Find the area bounded by the parabolas x² = 8y and y² = 8x?
Solution:
Equation of given parabolas:
x² = 8y ………………. (1)
and y² = 8x ………………….. (2)
Solving eqns. (1) and (2),

Coordinate of A(8, 8)
Two parabolas intersect at point O and A.
∴ Required area = Area OBALO – Area OCALO

Question 2.
Find the area bounded by the curve y = cos x, the X – axis and x = 0, x = 2π, by integration method?
Solution:
y = f(x) = cos x
When x ∈ [0, \(\frac { \pi }{ 2 } \), cos x ≥ 0;
x ∈ [0, \(\frac { \pi }{ 2 } \), [0, \(\frac { 3\pi }{ 2 } \), cos x ≤ 0;
When x ∈ \(\frac { 3\pi }{ 2 } \), 2π] cos x ≥ 0

Question 3.
(A) Find the area between the curves y² = 4x and y = 2x? (Integral method)
Solution:
Given, Curve and line are y² = 4x and – y = 2x
(2x)² = 4x
⇒ 4x² = 4x
⇒ x² – x = 0
⇒ x (x – 1) = 0
∴ x = 0 or x = 1
Then y = 2x = 0 or y = 2 × 1 = 2

The point of intersections of curve and line are (0, 0) and (1,2).
Area of shaded region = ar (OPAMO) – ar (OQAMO)

(B) Find the area bounded by the curves x² = 4y and x = 4y – 2 by integration method?
Solution:
Given, Curve and line are y² = 4x and – y = 2x
(2x)² = 4x
⇒ 4x² = 4x
⇒ x² – x = 0
⇒ x (x – 1) = 0
∴ x = 0 or x = 1
Then y = 2x = 0 or y = 2 × 1 = 2

The point of intersections of curve and line are (0, 0) and (1,2).
Area of shaded region = ar (OPAMO) – ar (OQAMO)

Question 4.
Find the area bounded by lines |x|+|y|= a by using integration method?
Solution:
The given lines are represented by |x|+|y| = a will be
x + y = a ……………. (1)
-x – y = a ……………………. (2)
x – y = a ……………………. (3)
and – x + y = a ………………… (4)
⇒ \(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1, \(\frac { x }{ -a } \) + \(\frac { y }{ -a } \) = 1
⇒ \(\frac { x }{ a } \) + \(\frac { y }{ -a } \) = 1, \(\frac { x }{ -a } \) + \(\frac { y }{ a } \) = 1

The above graph is represented by the PQ, RS, PS and QR respectively.
Hence, the area bounded by these lines
= 4 × ∆Area of OPQ

Question 5.
(A) Find the rea of circle x² + y² = a² by using integration method?
Solution:
Given:
x² + y² = a²
⇒ y² = a² – x²
⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
The circle is symmetric about the axes
Area = 4 × area of quadrant OAB

(B)
Find the area of the circle x² + y² = 25 by using integration method?
Solution:
Solve as Q. No. 5 (A) by putting a = 5.

Question 6.
Find the area included between two parabolas:
y² = 4 ax and x² = 4ay, a > 0.
Solution:
The equations of the given parabolas are
y² = 4ax.
and x² = 4ay
Putting the value of y from eqn. (2) in eqn. (1), we have

From eqn.(2), we have
y = 0, y = \(\frac { (4a)^{ 2 } }{ 4a } \) = 4a
∴The points of intersection of the given parabolas are (0, 0) and (4a, 4a).
Now, Required area = Area OBAL – Area OCAL

Question 7.
Find the area enclosed between parabolas y² = 4x and x² = 4y?
Solution:
y² = 4x ……………… (1)
x² = 4y …………………. (2)
Putting the value of y in eqn.(1), we get

∴ At O, x = 0 and at M, x = 4
Hence Required area Area OAPB O
= Area OAPMO – Area OBPMO

Question 8.
Find the area between curve y² = x and x² = y by integral method?
Solution:
Solve as Q.No. 7.

Question 9.
Find the area enclosed by the ellipse \(\frac { x^{ 2 } }{ a^{ 2 } } \) + \(\frac { y^{ 2 } }{ b^{ 2 } } \) = 1?
Solution:
Equation of the ellipse is

∴ Required area = 4 area AOB

Question 10.
Find the area of the region bounded by the parabola y² = 4ax and its latus rectum?
Solution:
Equation of the parabola is
y² = 4ax or y = ±2 \(\sqrt { ax } \)
In the figure, LSL’ is latus rectum and vertex is O (0, 0)
∴ Required area = 2 × Area OSL

Question 11.
Find the area enclosed between parabola y² = 4ax and straight line y = mx?
Solution:
Equation of the parabola y² = 4ax ……………. (1)
Equation of straight line y = mx ………………….. (2)
O is the origin (O, O). P is the point of intersection of parabola and the straight line. Solving eqns. (1) and (2) we get,
y² = 4ax
⇒ (mx)² = 4ax
⇒ m²x² – 4ax = 0
⇒ x(m² – 4a) = 0
∴ x = 0, x = \(\frac { 4a }{ m^{ 2 } } \)

Question 12.
Find the area bounded by the curve x² = 4y, y = 2, y = 4 and Y – axis in the first quadrant? (NCERT)
Solution:
Given equation of curve is:
x² = 4y
⇒ x = \(2\sqrt { y } \)
Required area = Area of ABCD

Question 13.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = \(\frac { a }{ \sqrt { 2 } } \)? (NCERT)
Solution:
Given equation of curve is:
x² + y² = a² …………………………. (1)
⇒ y² = a² – x²
⇒ y = \(\sqrt { a^{ 2 }-x^{ 2 } } \)
Equation of line is:


Point of intersection of circle and line is A \(\frac { a }{ \sqrt { 2 } } \), \(\frac { a }{ \sqrt { 2 } } \)
Coordinates of D is (a, 0).
∴ Required area = Area of ABD

Question 14.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a. Find the value of a? (NCERT)
Solution:
Given:
Equation of parabola is
x = y²
⇒ y = \(\sqrt{x}\)
Area bounded by the parabola and the line x = 4 is divided into two equal parts by the line x = a.
Area OEC = Area EFCB

Question 15.
Find the area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2? (NCERT)
Solution:
Given equation of circle is:
x² + y² = 4
⇒ y² = 4 – x²
⇒ y = \(\sqrt { (2)^{ 2 }-x^{ 2 } } \)

Question 16.
Find the area enclosed by a circle x² + y² = 32, a line y = x and axis of A in the first quadrant?
Solution:
Equation of circle is:

Putting the value of x in eqn. (2)
y = 4
The point of intersection of the line and circle are 0(0,0) and A(4,4) coordinates of B are are (4,0) and coordinates of C are (4\(\sqrt{2}\),0).
Required area = Area OACBO
= Area OAB + Area ABC

Question 17.
Find the area of triangle whose sides are y = 2x + 1, y = 3x +1 and x = 4? (NCERT)
Solution:
Let the equation of sides AB,AC and BC of & ABC ore respectively.
y = 2x + 1 …………… (1)
y = 3x + 1 ………….. (2)
and x = 4 …………….. (3)
By solving eqns. (1) and (2) we have A (0, 1), by solving eqns. (1) and (3) we have B (4, 9) and by solving eqns. (2) and (3) we have C (4, 13).
Thus required area

MP Board Class 12 Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 7B Definite Integral

Definite Integral Important Questions

Definite Integral Long Answer Type Questions – II

Question 1.
Evaluate \(\int_{0}^{4 a} \frac{f(x)}{f(x)+f(4 a-x)} d x\)?
Solution:
Let

Adding eqns.(1) and (2),

Question 2.
Prove that \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}=\frac{\pi}{12}\)? (NCERT)
Solution:

Adding eqns.(1) and (2),

Question 3.
Evaluate \(\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e^{x}(\log \sin x+\cot x) d x\)?
Solution:

Question 4.
Evaluate \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|2 \sin | x|+\cos | x|| d x\)?
Solution:
Let

Question 5.
Prove that:
\(\int_{0}^{\pi} \frac{x \tan x}{\sec x+\cos x} d x=\frac{\pi^{2}}{4}\)?
Solution:

Adding eqns. (1) and (2)

Question 6.
Prove that:
\(\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x=\frac{\pi^{2}}{16}\)?
Solution:

Adding eqns.(1) and (2)

Question 7.
Prove that:

Solution:
Let

Adding eqns. (1) and (2)

Question 8.
Prove that:

Solution:

Question 9.
Evaluate

Solution:

Question 10.
Prove that:

Solution:

Adding eqns.(1) and (2),

Question 11.
Evaluate

Solution:
Let


When x = 0, then t = tan 0 = 0
and when x = π, then t = tan \(\frac { \pi }{ 2 } \) = ∞

Question 12.
Prove that:

Solution:

Adding eqns. (1) and (2)

Question 13.
Prove that:

Solution:

Adding eqns. (1) and (2)

Question 14.
Prove that:

Solution:

Question 15.
Evaluate

Solution:

Adding eqns.(1) and (2)

Question 16.
Evaluate

Solution:
Let

MP Board Class 12 Maths Important Questions