Bhagya

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.2

निम्नलिखित प्रश्न 1 से 6 तक प्रत्येक में नाभि के निर्देशांक, परवलय का अक्ष, नियता का समीकरण और नाभिलंब जीवा की लंबाई ज्ञात कीजिए।
प्रश्न 1.
y² = 12x.
हल:
परवलय का समीकरण, y² = 12x
∴ y² = 4ax से तुलना करने पर
4a= 12 या a= 3
(i) नाभि के निर्देशांक (a, 0) या (3, 0)

(ii) परवलय का अक्ष OX
इसका समीकरण y = 0
(iii) नियता का समीकरण : x = – a अर्थात् x = – 3
(iv) नाभिलंब जीवा की लंबाई = 4a = 12.

प्रश्न 2.
x² = 6y.
हल:
परवलय का समीकरण x² = 6y
∴ 4a = 6 या a = \(\frac{3}{2}\)

इसका अक्ष y-अक्ष है जिसका
(i) समीकरण x = 0 है।
(ii) नाभि F (0, a) के निर्देशांक \(\left(0, \frac{3}{2}\right)\) है।
(iii) नियता y = – a का समीकरण y = – \(\frac{3}{2}\)
(iv) नाभिलंब जीवा की लम्बाई 4a = 6.

प्रश्न 3.
y² = – 8x.
हल:
परवलय का समीकरण y² = – 8x
∴ 4a = 8, a = 2
(i) नाभि F(- a, 0) के निर्देशांक (- 2, 0)

(ii) परवलय का अक्ष x-अक्ष
इसका समीकरण y = 0
(iii) नियता x = a का समीकरण x = 2.
(iv) नाभिलंब जीवा की लंबाई = 4a = 8.

प्रश्न 4.
x² = – 16y.
हल:
परवलय का समीकरण x² = – 16y
∴ 4a = 16 या a = 4
नियता
(i) नाभि F (0, – a) के निर्देशांक (0, – 4)
(ii) परवलय अक्ष का समीकरण x = 0.
(iii) नियता y = 0 का समीकरण y = 4.
(iv) नाभिलंब जीवा की लंबाई 4a = 16.

प्रश्न 5.
y² = 10x.
हल:
परवलय का समीकरण y² = 10x (आकृति प्रश्न 1 में देखें)
4a = 10 या a = \(\frac{5}{2}\)
(i) नाभि F (a, 0) के निर्देशांक \(\left(\frac{5}{2}, 0\right)\)
(ii) परवलय का अक्ष : x-अक्ष, समीकरण y = 0
(iii) नियता x = – a का समीकरण x = – \(\left(\frac{5}{2}, 0\right)\)
(iv) नाभिलंब जीवा की लंबाई 4a = 10.

प्रश्न 6.
x² = – 9y.
हल:
परवलय का समीकरण x² = – 9y (आकृति प्रश्न 4 में देखें)
4a = 9 या a = \(\frac{9}{4}\)
(i) नाभि (0, – a) के निर्देशांक \(\left(0,-\frac{9}{4}\right)\)
(ii) परवलय का अक्ष : y-अक्ष, समीकरण x = 0
(iii) नियता y = a का समीकरण y = \(\frac{9}{4}\)
(iv) नाभिलंब जीवा की लंबाई 4a = 9.

निम्नलिखित प्रश्न 7 से 12 तक प्रत्येक में परवलय का समीकरण ज्ञात कीजिए जो दिए प्रतिबंध को संतुष्ट करता है।
प्रश्न 7.
नाभि (6, 0), नियता x = – 6.
हल:
परवलय का अक्ष : x-अक्ष, y = 0

शीर्ष (0, 0) है, नाभि के निर्देशांक (6, 0)
परवलय का अक्ष, धन x-अक्ष के अनुदिश है।
परवलय का समीकरण y² = 24x.

प्रश्न 8.
नाभि (0, – 3), नियता y = 3.
हल:
परवलय का अक्ष y-अक्ष है।
शीर्ष (0, – 3), (0, 3) का मध्य बिन्दु (0, 0) है। नाभि (0, – 3) से स्पष्ट होता है कि परवलय की अक्ष OY के अनुदिश है।

∴ परवलय के समीकरण का रूप x² = – 4ay
यहाँ पर a = 3, ∴ 4a = 12
∴ परवलय का समीकरण x² = – 12y.

प्रश्न 9.
शीर्ष (0, 0), नाभि (3, 0) (आकृति प्रश्न 7 की देखिए)
हल:
परवलय का अक्ष OX के अनुदिश हैं।
∴ परवलय के समीकरण का रूप y² = 4ax
नाभि (3, 0) है। ∴ a = 3
4a = 4 x 3 = 12
∴ परवलय का समीकरण y² = 12x.

प्रश्न 10.
शीर्ष (0, 0), नाभि (-2, 0).
हल:
परवलय का अक्ष OX’ के अनुदिश
नाभि (- 2, 0) है तो a= 2

∴ 4a = 8
परवलय का रूप y² = – 4ax
परवलय का समीकरण y² = – 8x.

प्रश्न 11.
शीर्ष (0, 0), (2, 3) से जाता है और अक्ष, x-अक्ष के अनुदिश है।
हल:
परवलय का शीर्ष (0, 0) है और अक्ष : x-अक्ष है।
∴ परवलय के समीकरण का रूप y² = 4ax
यह बिन्दु (2, 3) से होकर जाता है
∴ 9 = 4a.2 या 4a = \(\frac{9}{2}\)
अतः परवलय का समीकरण y² = \(\frac{9}{2}\)x या 2y² = 9x.

प्रश्न 12.
शीर्ष (0, 0), (5, 2) से जाता है और y-अक्ष के सापेक्ष सममित है।
हल:
शीर्ष (0, 0), परवलय y-अक्ष के सापेक्ष सममित है।
समीकरण का रूप x² = 4ay है। यह बिन्दु (5, 2) से गुजरता है।
∴ 25 = 4a × 2
∴ 4a = \(\frac{25}{2}\)
∴ परवलय का समीकरण, x² = \(\frac{25}{2}\)y या 2x² = 25y

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 14 Ecosystem

Ecosystem NCERT Text Book Questions and Answers

Question 1.
Fill in the Blanks:

1. Plants are called as …………….. because they fix carbon dioxide.
2. In an ecosystem dominated by trees, the pyramid (of numbers) is ………………… type.
3. In aquatic ecosystem, the limiting factor for the productivity is …………………….
4. Common detritivores in our ecosystem are ………………………..
5. The major reservoir of carbon on earth is ………………………
Answer:

1. Producer
2. Inverted
3. Light
4. Earthworm
5. Ocean and Biosphere.

Question 2.
Which one of the following has the largest population in the food chain:
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers.
Answer:
(d) Decomposers.

Question 3.
The second trophic level in a lake is:
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes.
Answer:
(b) Zooplankton

Question 4.
Secondary producers are:
(a) Herbivors
(b) Producers
(c) Carnivors
(d) None of the above.
Answer:
(a) Herbivors

Question 5.
What is the percentage of photosynthetically active radiation (PAR) in the incident solar radiation:
(a) 100%
(b) 50%
(c) 1-5%
(d)2 -10%.
Answer:
(b) 50%

Question 6.
Distinguish between:
(a) Grazing food chain and detritus food chain
(b) Production and decomposition
(c) Upright and inverted pyramid
(d) Food chain and Food web
(e) Litter and Detritus
(f) Primary and Secondary productivity.
Answer:
(a) Grazing food chain and detritus food chain

(b) Production and decomposition

(c) Upright and inverted pyramid

(d) Food chain and Food web

(e) Litter and Detritus

(f) Primary and Secondary productivity.

Question 7.
Describe the components of an ecocystem.
Answer:
Ecosystem: The system resulting from the interaction between organisms and their environment is called as ecosystem.

(a) Producers: Organism, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

• Primary consumer: Animals, which feed on producers are included into this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers: Primary consumers also serve as food for water sn’akdls, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers: Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 8.
Define ecological pyramids and descirbe with examples, pyramids of number and biomass.
Answer:
Ecological Pyramid : The relation between producers and consumers in a ecosystem can be graphically represented in the form of a pyramid called ecological pyramid. The base always represent the producers or the first trophic level and the apex represents top level consumer or the lost trophic level ecological pyramids are of three types :

1. Pyramid of number
2. Pyramid of biomass
3. Pyramid of energy.

The biomass, i.e., the living weight of the organisms in the food-chain present at different trophic levels in an ecosystem forms the pyramid of biomass.

When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass.eg., pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem: The system resulting from the interaction between organisms and the environment is called as ecosystem.

Question 9.
What is primary productivity ? Give brief description of factors that affect primary productivity.
Answer:
The rate of biomass production is called productivity.
It is expressed in terms of g^-2yr^-1  or(Kcal-m^-2) yr^-1 to compare the productivity of ecosystems.
It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NFP).

Gross Primary Productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration.

Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R=NPP.
Primary productivity depends on :

• The plant species inhabiting a particular area.
• The environmental factors.
• Availability of nutrients.
• Photosynthetic capacity of plants.

Question 10.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients.
The various processes involved in decomposition are as follows:
1. Fragmentation: It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2. Leaching: It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism: It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification: The next step is humification which leads to the formation of a dark coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralization: The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization. Decomposition produces a dark coloured, nutrient-rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO₂, water, and other nutrient in the soil.

Question 11.
Given account of Explain energy flow in ecosystem.
Answer:
Energy flow: In die ecosystem, energy is transferred in an orderly sequence. The flow of solar energy from producers to consumers and to decomposers subsequently in an ecosystem is known as energy flow. Energy flow is an ecosystem is always unidirectional. Sun is the sole source of solar energy in an ecosystem. Green plants utilize this energy in photosynthesis and convert it in the form of chemical energy and store it. Plants utilize maximum part of this energy to do its biological functions. Some of it is converted into heat and released in the environment. Remaining part of the energy is stored in various components of the body.

When a consumer eats these producer plants, the energy is then transferred into its body. In any food-chain energy flows from primary producers to primary consumers, from primary consumers to secondary consumers and secondary consumers to tertiary consumers and so on. Because every organism of a trophic level continuously converts chemical energy into heat, there is always a loss of energy with each step in a food-chain. According to an estimate only 10% of the total energy obtained is transferred from one trophic level to another.

Question 12.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the earth’s crust or rocks. Nutrient elements are found in the sediments of the earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles.

Sedimentary cycles are very slow. They take a long time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long time to come out and continue circulation. Thus, it usually goes out of circulation for a long time.

Question 13.
Outline salient features of carbon cycling in an eosystem.
Answer:
Carbon constitutes 49 percent of dry weight Of an organisms.

• 71 percent of the carbon is found dissolved in oceans which is responsible for its regulations in atmosphere.
• The carbon cycle occurs through atmosphere, oceans and through living and dead organisms.
• It is estimated that 4xl013 kg of Carbon is fixed in the biosphere through photosynthesis annually.
• Carbon is returned to atmosphere as CO₂ by animals and plants through respiration and the activities of decomposers.
• Some amount of fixed carbon is lost as sediments and removed from circulation.
• Burning of wood, forest fire, volcanic activity and combustion of organicjnatter and fossil fuels are some essential sources for releasing CO₂ in the atmosphere.
• Human activities like deforestation and vehicular burning or fossil fuels has caused in increase in the amount of CO₂ in atmosphere.

Ecosystem Other Important Questions and Answers

Ecosystem Objective Type Questions

Choose the Correct Answer

Question 1.
Position of man in food chain of all ecosystem is as :
(a) Consumer
(b) Producer
(c) Producer and consumer
(d) Decomposer.
Answer:
(a) Consumer

Question 2.
Flow of energy in an ecosystem is :
(a) Unidirectional
(b) Bi-directional
(c) Three directional
(d) Quadri-directional.
Answer:
(a) Unidirectional

Question 3.
The word “ Ecosystem” was first used by :
(a) Tansley
(b) Odum
(c)Writier
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
Source of energy in an ecosystem is:
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of the above.
Answer:
(a) Solar energy

Question 5.
Pyramid of number of trees in an ecosystem is always:
(a) Inverted
(b) Upward
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 6.
Correct food chain is:
(a) Grass → Grasshopper → Frog Snake → Hawk
(b) Grass → Frog → Snake → Peacock
(c) Grass → Peacock → Grasshopper → Hawk
(d) Grass → Snake → Rabbit.
Answer:
(a) Grass → Grasshopper → Frog Snake → Hawk

Question 7.
Pyramid of biomass of an ecosystem of lake is:
(a) Upward
(b) Inverted
(c) Inverted and Upward both
(d) None of these.
Answer:
(b) Inverted

Question 8.
Man is:
(a) Producer
(b) Carnivore
(c) Herbivore
(d) Omnivorus.
Answer:
(d) Omnivorus.

Question 9.
Decomposer are:
(a) Rats
(b) Algae
(c) Bacteria and Fungi
(d) Goats.
Answer:
(c) Bacteria and Fungi

Question 10.
Man made ecosystem is:
(a) Pond
(b) Aquarium
(c) Forest
(d) Lake.
Answer:
(b) Aquarium

Question 11.
Pyramid of energy is a forest ecosystem is:
(a) Always Inverted
(b) Always upward
(c) First upward than Inverted
(d) None of the above.
Answer:
(b) Always upward

Question 12.
Study of ecosystem of species is called:
(a) Ecology
(b) Autecology
(c) Synecology
(d) None of these.
Answer:
(b) Autecology

Question 13.
Food chain starts from:
(a) Respiration
(b) Photosynthesis
(c) Decomposition
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Flow of energy and food in a food web is:
(a) Unidirectional
(b) B i-directional
(c) Multidirectional
(d) None of these.
Answer:
(c) Multidirectional

Fill in the Blanks:

1. The transitional zone between two adjacent communities is called ………………………
2 The term ecosystem was proposed for the first time by ………………………..
3. The brief source of energy in ecosystem is …………………….
4. Bacteria which fix the nitrogen is called ……………………
5. All ecosystem are dependent of energy on …………………….
6. Balance of ecosystem is called ………………………
7. Pyramid of energy is always …………………………
8. All the plant of a particular geographic area constitute ………………………. of the place.
Answer:

1. Ecotone,
2. Tansley,
3. Sunlight,
4. Rhizobuim,
5. Solar energy,
6. Ecological balance,
7. Upright,
8. Flora.

3. Match the Following:


Answer:

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

Answer:

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

Question 4.
Answer in One Word/Sentence:

1. Mention the two name of biotic components of ecosystem.
2. Who proposed the term ecosystem?
3. Which ecosystem is more stable?
4. Name the relatively loss stable ecosystem.
5. Name the ecosystem showing maximum stratification.
6. Give two examples of artificial ecosystem.
7. Give the name of any two food chains.
8. Write the name of basic unit of ecology.
9. Name the organisms found on the surface of water bodies.
10. Name the transitional zone of two adjacent communities.
11. Name the process of establishment of organisms is any new area.
12. From where xerosere started?
13. Who has proposed the term succession?
14. Name the succession started from naked rock. ‘

Answer:

1. Biotic, Aboitic
2. Tansley
3. Ocean
4. Desert
5. Aquatic
6. Aquarium, crop¬land
7. Grazing, Detritus
8. Producer
9. Benthos
10. Ecotone
11. Succession
12. Rocks
13. Hult(1885)
14. Lithosere.

Ecosystem Very Short Answer Type Questions

Question 1.
Name the process of soil formation.
Answer:
Pedogenesis.

Question 2.
What is the orientation of pyramid of energy?
Answer:
Upright.

Question 3.
Which ecosystem is most established?
Answer:
Complex ecosystem.

Question 4.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 5.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 6.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro consumer or Decomposer.

Question 7.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10 %.

Question 8.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 9.
Name the word which is similar to ecosystem given by Pro. R. Mishra.
Answer:
Ecocosm.

Question 10.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 11.
Who gave the word transformer for producer?
Answer:
By E. J. kormondy.

Question 12.
Name any two sedimentary cycle.
Answer:

1. Phosphorus cycle
2. Sulpher cycle.

Question 13.
The energy pyramids are always.
Answer:
Upright.

Question 14.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 15.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 16.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (No₃^–).

Question 17.
Name two types of food chain.
Answer:

1. Grazing food chain,
2. Detritus food chain.

Ecosystem Short Answer Type Questions

Question 1.
Differentiate between food chain and food web.
Answer:

Question 2.
Write the names of important components of atmosphere.
Answer:
(A) Abiotic components:

• Energy: Light, temperature, energy of chemical substances.
• Materials: Water, soil, salts etc.

(B) Biotic components:

• Producers: e.g., green plants.
• Consumers: Primary, secondary and tertiary.
• Decomposers: Bacteria, fungi.

Question 3.
Write the name and ratio of different components of biosphere.
Answer:
Name and ratio of different components of biosphere is :

Some gases are also found in biosphere e.g., Helium, Neyon and Crypton are found in less quantity.

Question 4.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, e.g., earth worm. And decomposer are organisms which by secreting enzyme’s break down complex organic matter into in organic substance e.g. some bacteria and fungi.

Question 5.
Explain consumers of ecosystems.
Answer:

1. Producers: All the green plants.
2. Consumers: Depends on others for food.
   (i) Primary consumer: Depends on plants called herbivores.
   (ii) Secondary consumers: Depends on herbivores for food.
   (iii) Tertiary consumers: Depends on secondary consumers.
3. Decomposers: They decomposed dead organic matter.

Question 6.
Differentiate between biome and ecosystem.
Answer:
An ecosystem is the interaction of living and non-living things in an environment. A biome is a specific geographic area notable for the species living there.

Question 7.
Draw a pyramid of energy of grass land in ecosystem.
Answer:

Question 8.
Explain nitrogen cycle in nature.
Answer:

Question 9.
Explain sulphur cycle by diagram.
Answer:

Question 10.
Explain the effect of light on plants.
Answer:
Effect of light on plants: Light is the source of energy. It is essential for life. It is an important factor of an ecosystem. The existence of life on earth is because of light obtained from sun. Sunlight is essential for photosynthesis, help in preparation of food for the whole living world. Light effects biological activities of plants by its intensity, period and duration. Plants are classified into following two categories on the basis of requirement of light intensity:

• Heliophytes: Plants, which can grow better in bright light are called heliophytes.
• Sciophytes: Plants, which require relatively less of light and they can grow better in shades are called sciophytes.

Question 11.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web: In nature, food-chains are not isolated sequences, but are interrelated and interconnected with one another. When many food-chains operate simultaneously and interlock such pattern is termed as food web. Thus, the food web is a description of feeding connections between the organisms which make up a community. Energy passes through one trophic level to next via these food web links, e.g., a rat feeds on various kinds of grains, fruits, stems, roots, etc.

A rat in its turn is consumed by a snake which is eaten by a falcon. The snakes feed on both frogs and rats. Thus, a network of food-chains exists and this is called food web. The food web gets more complicated because of variability in taste and preference availability and compulsion and several other factors at each level. For example, tigers normally do not feed on fishes or crabs but in Sunderbans they are forced to eat them.

Question 12.
Explain Calcium cycle with well labelled diagram.
Answer:

Question 13.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 14.
What do you mean by ecosystem? Describe the important components of a pond ecosystem.
Or
Write about role of decomposers in an ecosystem with example.
Answer:
Ecosystem: The system resulting from the interaction between organisms and their environment is called as ecosystem.

Components of a pond ecosystem:
A pond ecosystem should contain all components of ecosystems like:

(a) Producers: Organism, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

• Primary consumer: Animals, which feed on producers are included into this category e.g., Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers: Primary consumers also serve as food for water sn’akdls, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers : Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

(c) Decomposers: All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of then- bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 15.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the food-chain present at different trophic levels in an ecosystem forms the pyramid of biomass.

When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass. eg. pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem: The system resulting from the interaction between organisms and the environment is called as ecosystem.

Ecosystem Long Answer Type Questions

Question 1.
What are ecological pyramids? Explain various types of ecological pyramids.
Answer:
Trophic level: In an ecosystem, the producer consumer arrangement is a kind of structure known as trophic structure and each food level in the food-chain is called as trophic level or energy level. In other, words each level of food in food-chain is called its trophic level. The first trophic level (T,) in an ecosystem is occupied by producers. Herbivores (primary consumers) form second trophic level (T2), secondary consumers form third trophic level (T3), tertiary consumers form fourth trophic level (T4) and decomposers form fifth trophic level (T5) in an ecosystem. .

Food or Ecological Pyramids: If we express the organisms of various trophic levels according to their number, biomass and ratio of energy stores within it, then we obtain ac6ne or pyramid like structure which is known as food or ecological pyramid. Ecological pyramids represent the trophic structure and function of an ecosystem. In base and successive trophic levels the tiers which make up the apex. Ecological pyramids are of the following three types :

1. Pyramid of Biomass
2. Pyramid of number
3. Pyramid of energy.

1. Pyramid of Biomass : Biomass is the dry weight of living organisms per unit of space. The ecological pyramid, which shows the quantitative relationship of the standing crop at each trophic level. The pyramid of biomass shows gradual reduction in biomass at each trophic level from base to apex. The pyramid of biomass may be:

• Upright: e.g., all terrestrial ecosystems.
• Inverted: e.g., all aquatic ecosystems.

2. Pyramid of Number : The ecological pyramid which shows the number of individual organisms at each trophic level. It represents numerical relationship between different trophic level of a food-chain. In this pyramid more abundant species from the first trophic level and from the base of pyramid and the less abundant species remain near the top. The pyramid of number may be:

• Upright: e.g. grassland, pond, forest ecosystem.
• Inverted :e.g. ecosystem of tree.

3. Pyramid of Energy : It indicates the total amount of energy at each trophic level of the food-chain. At each producer level, the total energy available is relatively more than at the higher trophic levels because of the loss of the energy at each trophic level. Thus, there is a gradual loss of energy at each trophic level. The pyramid of energy of each types of ecosystem is always upright.

Question 2.
What is meant by terrestrial biomes? What are its types? Explain any one biomes in detail.
Answer:
Terrestrial Biome : Large area occupying ecosystems in nature are called biomes. If biomes are on land than they are called Terrestrial biomes.
Terrestrial biomes may be:
(a) Forest Biomes : They may be as below;

• Topical rain forest
• Cold tropical forest
• Taiga forest.

(b) Grassland biomes : They may be as below:

• Tropical rain forest
• Cold tropical forest.

(c) Desert Biomes
(d) Tundra Biomes.
Grassland Biome or Ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

(a) Abiotic component: All organic, inorganic substances and climatic factors together form abiotic component.

(b) Biotic component :

• Producers: Grasses, herb, shrubs.
• Primary consumers: Herbivore like cow, buffalo, goats, sheep, deer, rabbit, rat insect.
• Secondary consumers: Carnivore animals which eat primary consumers-like snake, birds, foxes, jackal etc.
• Tertiary consumers: These organism which eat secondary consumers because no other one eats them like Hawk, Peacock, etc.
• Decomposers: Micro-Fungus, Bacteria, actinomycetes are decomposer of grassland biomes and recycle the material back to soil and used by producers.

Question 3.
What are Biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
All living organisms get matter from the biosphere components i.e„ lithosphere, hydrosphere and atmosphere. Essential elements or inorganic substances are provided by earth and are required by organisms for their body building and metabolism, they are known as Biogeochemicals or biogenetic nutrients.

Sulphur Cycle : Producers (green plants) need sulphur in the form of sulphates from soil or from water (aquatic plants). The animals get sulphur through food. Some animals get sulphur from water also. Sulphur is found in three amino acids hence, sulphur is component of most proteins, some vitamins and enzymes. Plants pick up sulphur in the form of sulphates. They are converted to organic form mostly as component of some amino acids. It is found in nature as element and also as sulphates in soil, water and rocks. After the death of plants and animals, they are decomposed by microbes like Asperigillus, Neurospora and Escherichia releasing hydrogen sulphide.

Calcium cycle : Calcium is slowly released from the rocks by water and wind action. These are either blown into the air or absorbed by plants through their roots. Animals obtain it directly as compounds and also from plants. Calcium is released from plant and animal bodies by decomposition after death. Molluscs and corals deposit a large quantity of calcium in their shells and skeletons making it unavailable for quick cycling.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 11 शंकु परिच्छेद Ex 11.1

निम्नलिखित प्रश्न 1 से 5 तक प्रत्येक में वृत्त का समीकरण ज्ञात कीजिए :
प्रश्न 1.
केंद्र (0, 2) और त्रिज्या 2 इकाई।
हल:
यहाँ h = 0, k = 2 तथा r = 2 रखने पर,
वृत्त का समीकरण, (x – 0)² + (y – 2)² = 2²
या x² + y² – 4y + 4 =4
अतः वृत्त का अभीष्ट समीकरण, x² + y² – 4y = 0.

प्रश्न 2.
केंद्र (- 2, 3) और त्रिज्या 4 इकाई।
हल:
∴ वृत्त का समीकरण (x + 2)² + (y – 3)² = 4²
या (x² + 4x + 4) + (y² – 6y + 9) = 16
या x² + y² + 4x – 6y – 3 = 0.

प्रश्न 3.
केंद्र \(\left(\frac{1}{2}, \frac{1}{4}\right)\) और त्रिज्या \(\frac{1}{12}\) इकाई।
हल:
यहाँ h = \(\frac{1}{2}\), k = \(\frac{1}{4}\), तथा r = \(\frac{1}{12}\) हो, तब
वृत्त का समीकरण,

प्रश्न 4.
केंद्र (1, 1) और त्रिज्या -2 इकाई।
हल:
यहाँ h = 1, k = 1 तथा r = \(\sqrt{2}\) हों, तब
वृत्त का समीकरण,
(x – 1)² + (y – 1)² = (\(\sqrt{2}\))²
या (x² – 2x + 1) + (y² – 2y + 1) = 2
या x² + y² – 2x – 2y= 0.

प्रश्न 5.
केंद्र (- a, – b) और त्रिज्या \(\sqrt{a^{2}-b^{2}}\) इकाई।
हल:
वृत्त का समीकरण,
(x + a)² + (y + b)² = \((\sqrt{a^{2}-b^{2}})\)²
या x² + 2ax + a² + y² + 2by + b² = a² – b².
या x² +y² + 2ax + 2by + 2b² = 0.

निम्नलिखित प्रश्न 6 से 9 तक में प्रत्येक वृत्त का केन्द्र और त्रिज्या ज्ञात कीजिए :
प्रश्न 6.
(x + 5)² + (y – 3)² = 36.
हल:
वृत्त (x + 5)² + (y – 3)² = 36 की (x – h)² + (y – k)² = r² से तुलना करने पर,
– h = 5, – k = – 3, r² = 36
∴ h = – 5, k = 3, r= 6
∴ केन्द्र (- 5, 3), त्रिज्या = 6.

प्रश्न 7.
x² +y² – 4x – 8y – 45 = 0.
हल:
(x² – 4x) + (y² – 8y) = 45
या (x² – 4x + 4) + (y² – 8y + 16) = 45 + 4 + 16 = 65
(x – 2)² + (y – 4)² = 65
∴ h = 2, k = 4, r = \(\sqrt{65}\)
⇒ केन्द्र (2, 4) और त्रिज्या = \(\sqrt{65}\)
दूसरी विधि : 2g = – 4, 2f= – 8, c = – 45
g = – 2, f = – 4, r = \(\sqrt{g^{2}+f^{2}-c}\)3
r = \(\sqrt{4+16+45}\)
= \(\sqrt{65}\)
केन्द्र (- g, – f) अर्थात् (2, 4)
त्रिज्या = r = \(\sqrt{65}\).

प्रश्न 8.
x² + y² – 8x + 10y – 12 = 0.
हल:
(x² – 8x) + (y² + 10y) = 12
या (x² – 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25
(x – 4)² + (y + 5)² = 53.
∴ केन्द्र (4, – 5), त्रिज्या= \(\sqrt{53}\).

प्रश्न 9. 2x² + 2y² – x= 0.
हल:

प्रश्न 10.
बिन्दुओं (4, 1) और (6, 5) से जाने वाले वृत्त का समीकरण ज्ञात कीजिए जिसका केन्द्र रेखा 4x + y = 16 पर स्थित है।
हल:
वृत्त का व्यापक समीकरण
x² + y² + 2gx + 2fy + c = 0
बिन्दु (4, 1) इस पर स्थित है
∴ 16 + 1 + 8g + 2f + c = 0
∴ 8g + 2f + c = – 17 …..(1)
बिन्दु (6, 5) वृत्त पर स्थित है
∴ 36 + 25 + 12g + 10f + c = 0
∴ 12g + 10f + c = – 61 …..(2)
केंद्र (- g, – 1) रेखा 4x + y = 16 पर स्थित है
∴ – 4g – f = 16
या 4g + f = – 16 …..(3)
समीकरण (1) को (2) में से घटाने पर
4g + 8f = – 44
समीकरण (3) को (4) में से घटाने पर
7f = – 44 + 16 = – 28
f = – 4
समीकरण (3) में f का मान रखने पर
4g – 4 = – 16 या 4g = – 12
∴ g = – 3
f और g का मान समी (1) में रखने पर
– 24 – 8 + c = – 17
c = 32 – 17 = 15
अतः वृत्त का समीकरण
x² + y² – 6x – 8y + 15 = 0.

प्रश्न 11.
बिन्दुओं (2, 3) और (- 1, 1) से जाने वाले वृत्त का समीकरण ज्ञात कीजिए जिसका केंद्र रेखा x – 3y – 11 = 0 पर स्थित है।
हल:
मान लीजिए वृत्त का समीकरण
x² + y² + 2gx + 2fy + c = 0 ….(1)
इस पर बिन्दु (2, 3) स्थित है।
∴ 4 + 9 + 4g + 6f + c = 0
या 4g + 6f + c = – 13 ……(2)
इसी प्रकार (- 1, 1) भी वृत्त (1) पर स्थित है।
जब p = – 2, वृत्त का समीकरण
(x + 2)² + y² = 25
या x² + y² + 4x – 21 = 0
जब p = 6, वृत्त का समीकरण
(x – 6)² + y² = 25
x² + y² – 12x + 36 – 25 = 0
या x² + y² – 12x + 11 = 0
∴ वृत्त के अभीष्ट समीकरण
x² + y² + 4x – 21 = 0 और x² + p² – 12x + 11 = 0^π

प्रश्न 13.
(0, 0) से होकर जाने वाले वृत्त का समीकरण ज्ञात कीजिए जो निर्देशांक्षों पर a और b अंत: खण्ड बनाता है।
हल:
वृत्त मूल बिन्दु से होकर जाता है और अक्षों पर अंत:खण्ड a, b बनाता है।
OA = a, ∴ A के निर्देशांक (a, 0)
OB = b, ∴ B के निर्देशांक (0, b)

या x² + y² – ax – by = \(\frac{a^{2}+b^{2}}{4}-\frac{a^{2}+b^{2}}{4}\)
∴ वृत्त का अभीष्ट समीकरण
x² + y² – ax – by = 0.

प्रश्न 14.
उस वृत्त का समीकरण ज्ञात कीजिए जिसका केंद्र (2, 2) हो तथा (4, 5) से जाता है।
हल:
वृत्त की त्रिज्या = केंद्र (2, 2) और बिन्दु (4, 5) के बीच की दूरी

प्रश्न 15.
क्या बिन्दु (- 2.5, 3.5) वृत्त x² + y² = 25 के अंदर, बाहर या वृत्त पर स्थित है।
हल:
वृत्त का केंद्र O(0, 0) है।
दिया हुआ बिन्दु P(- 2.5, 3.5) है।
OP= \(\sqrt{(-2.5)^{2}+(3.5)^{2}}\)
= \(\sqrt{6.25+12.25}\)
= \(\sqrt{18.50}\)
= 4.25 (लगभग)
यह त्रिज्या जो 5 इकाई से कम है
अतः बिन्दु (- 2.5, 3.5) वृत्त के अंदर स्थित होगा।

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 4 Reproductive Health

Reproductive Health NCERT Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Significance of reproductive health in a society are :

• Control over the transmission of STDs.
• Less death due to reproduction related diseases like-AIDS, cancer of reproductive tract.
• Control in population explosion.
• Not only this reproductive health of men and women affects the health of the next generation.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Special attention need to be given to the following aspect :

• Introduction of sex education in school that to helps in eradicating myths and mis-conceptions regarding sex-related aspects.
• Proper information about reproductive organs, safe and hygienic sexual practices and sexually transmitted diseases.
• Awareness of problems due to uncontrolled population growth, social evils like sex abuse and sex-related crimes etc.
• Strong infrastructural facilities, professional expertise and material support to provide medical assistance and care to people in reproduction related problems.
• Educating people about available birth control options, care of pregnant mothers, post-natal care of mother and child, importance of breast feeding, equal opportunities for the male and female child.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, sex education is necessary in schools because :

• It will provide proper information about reproductive organs, adolescence, safe, hygienic sexual practices and Sexually Transmitted Diseases (STDs).
• It will provide right information to avoid myths and misconceptions about sex related queries.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 year’s ? If yes, mention some such areas of improvement
Answer:
Yes, Reproductive health in our country has improved in the last 50 year’s.

Some areas of improvement are :

• Better awareness about sex-related matters.
• Increased number of medically assisted deliveries and better post-natal care of child and mother leading to decreased maternal and infant mortality rates.
• Increased number of couples with small families.
• Better direction and cure of STDs and overall increased medical facilities for all sex-related problems.

Question 5.
What are the suggested reasons of population explosion?
Answer:
The situation where population exceeds productive capacity is known as population explosion. In nature, the amount of resources are limited hence, if the population increases in the present rate and increasing beyond the limit, they (Resources) would get exhausted.
Reasons of population explosion : Following are the reasons of population explosion:

• Increasing birth rate.
• Decreasing death rate.
• Higher rate of reproduction.
• Medical services have brought down mortality due to fatal diseases and epidemics.
• Lack of predator in the civilized world today, the only predator of man is man himself.

Question 6.
Is the use of contraceptives justified ? Give reasons.
Answer:
Yes, use of contraceptive is justified because it helps to control the rapid growth of human population. It will also help in preventing unwanted pregnancies and STDs. Contraceptive also help in controlling the population growth rate.

Question 7.
Removal of gonads can not be considered as a contraceptive option, why?
Answer:
Removal of gonads not only stops the production of gametes but will also stop the secretions of various important hormones, which are important for bodily functions. This method is irreversible and thus, can not be considered as a contraceptive method.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment
Answer:
Yes, the ban is necessary because amniocentesis is misused for determining the sex of the fetus and then aborting the child of it is a female. This process is illegal it causes harm to the fetus as well as mother it can also disturb the sex ratio.

Question 9.
Suggest some method to assist infertile couples to have children.
Answer:
If the couples are enable birth the children and corrections are not possible, the couples could be assisted to have children through certain special techniques, commonly known as Assisted Reproductive Technologies (ART). Some methods are given as :

1. In Vitro Fertilization (IVF) : In this method, ova from the female and the sperm from the male are collected and induced to form zygote under stimulated conditions in the laboratory. This process is called In Vitro Fertilization (IVF). Some method given as follows :

(i) Zygote Intrafallopian Transfer (ZlFT) : The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.

(ii) Intra Uterine Transfer (IUT) : Embryo with more than 8 blastomeres is transferred into the uterus in females who cannot conceive embryos formed by fusion of gametes in another female are transfered.

(iii) Test tube baby : In this method, ova from the donor (female) and sperm from the donor (male) are collected and are induced to form zygote under stimulated conditions in the laboratory. The zygote could then be transferred into the fallopian tube and embryos transferred into the uterus, to complete its further development. The child born from this method is called test tube baby.

(2) Gamete Intra Fallopian Transfer (GIFT) : It is the transfer of an ovum collected from a donor into the fallopian tube 8 another female who cannot produce one, but can provide suitable environment for fertilization and further development of the embryo.

(3) Intra Cytoplasmic Sperm Injection (ICSI) : It is a procedure to form an embryo HI* the laboratory by directly injecting the sperm into an ovum.

(4) Artificial Insemination (AI) : In this method, the semen collected either from the husband or a healthy donor is artificially introduced into the vegina or into the uterus (Intra Uterine Insemination, IUI). This technique is used in cases where the male is unable to inseminate sperms in the female reproductive tract or due to very low sperm counts in the ejaculation.

(5) Host Mothering : In this process, the embryo is transferred from the biological mother to a surrogate mother. The embryo then develops till it is fully developed or partially developed. It is then transferred to the biological mother or into any other. This technique is useful for females in which embryo forms but is not able to develop.

Question 10.
What are the measures one has to take to prevent from contracting STDs ?
Answer:
STDs can be prevented by the following methods :

• Avoid sex with unknown partners/multiple partners.
• Always use condoms during coitus.
• Always contact a qualified doctor for any doubt in early stage of infection and get complete treatment if diagnosed with disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
Answer:
False, Abortion does not happen under normal conditions. It happens accidently or under the will of Parents.

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
Answer:
False, Sterility always does not occur due to female sometimes. Males are also responsible for this.

(c) Complete lactation could help as a natural method of contraception. (True/False)
Answer:
True, Menstrual cycle does not occur after parturition which can act as natural
contraception but this method is functional for a period of six months from parturition.

(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of die people. (True/False)
Answer:
True, this creates better reproductive health among people.

Question 12.
Correct the following statements :
(a) Surgical methods of contraception prevent gamete formation.
Answer:
It prevents the transportation of gametes not their formation.

(b) All sexually transmitted diseases are completely curable.
Answer:
Hepatitis-Band AIDS are not curable.

(c) Oral pills are very popular contraceptives among the rural women.
Answer:
Oral pills are not popular among rural women. They require sex education.

(d) In E.T. techniques, embryos are always transfered into the uterus.
Answer:
In E.T. techniques 8-called blastomere is transfered into the fallopian tube. While more than 8-celled blastomere is transfered into the uterus.

Reproductive Health Other Important Questions and Answers

Reproductive Health Objective Type Questions

1. Choose the Correct Answer
Question 1.
Removal of Fallopian tube in females are called :
(a) Vasectomy
(b) Tubectomy
(c) Ovariectomy
(d) Castration (Gonadectomy).
Answer:
(b) Tubectomy

Question 2.
Main region ofthe population growth are :
(a) Less death rate
(b) Growth of birth rate
(c) Drought
(d) Less war.
Answer:
(a) Less death rate

Question 3.
The causes of popülatîon explosion in large cities is :
(a) Chances of education
(b) Available facilities
(c) Sources of income
(d) All of these.
Answer:
(d) All of these.

Question 4.
Population density is more in :
(a) USA
(b) India
(c) China
(d) Japan.
Answer:
(c) China

Question 5.
AIDS disease caused by :
(a) From bacteria
(b) From protozoa
(c) From virus
(d) From fungus.
Answer:
(c) From virus

Question 6.
Sex transmitted diseases are caused by :
(a) Virus
(b) Bacteria
(c) Protozoa
(d) All of these.
Answer:
(a) Virus

Question 7.
Population calculated in India :
(a) In 1891
(b) In 1947
(c) In 1950
(d) In 1961.
Answer:
(a) In 1891

Question 8.
The technique of the control of birth rate is called :
(a) TUD
(b) GIFT
(c) MIF
(d) IVEET.
Answer:
(a) TUD

Question 9.
Alcohol affected organ is :
(a) Liver
(b) Cerebrum
(c) Cerebellum
(d) Heart
Answer:
(a) Liver

Question 10.
Population growth rate of world is :
(a) 2-4%
(b) 2%
(c) 3%
(d) 4%.
Answer:
(b) 2%

2. Fill in the Blanks :

1. Statistical study of population is called ……………….
2. According to MALTHUS, Population grows ……………….whereas the means of its subsistences grows ……………….
3. The entry and exit of member in any population is called ……………….
4. Cutting and ligating ends of segments of vas deferens is called ……………….
5. The population density of terrestrial animals, the area is represented by ……………….
6. Birth rate and migration ………………. the population density.
7. Revised birth fate is always greater than ………………. birthrate.
8. Vital index = ………………. /Mortality x 100.
9. Population ………………. with time and place.
10. From the population point of view, every ………………. person in the world, is an Indian.

Answer:

1. 1. Demography
   2. Geometrically
   3. Migration
   4. Vasectomy
   5. Meter2 (m2)
   6. Growth
   7. Unrevised
   8. Natality
   9. Changes
   10. Sixth.

3. Match the Following :

Answer:

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

Answer:

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

III. Answer in One Word/Sentence:

1. Write two such causes by which unchecked growth takes place. (SSCE 1992)
2. Write the full name of IUCD. (SSCE 1992!)
3. What we called the study of human population?
4. What type of chemical contraceptive is found in Mala-D and N?

Answer:

1. Early marriage, illiteracy
2. Intra-Uterine Contraceptive Devices
3. Demography
4. Oestrogen and Progesteron hormone.

Reproductive Health Very Short Answer Type Questions

Question 1.
What is the other name of family planning?
Answer:
The other name of family planning is “Parivar Kalyan”.

Question 2.
Where is Central Drug Research Institute located ?
Answer:
Central Drug Research Institute (CDRI) is located in Lucknow (Uttar Pradesh).

Question 3.
Write one benefit of condom.
Answer:
Condom provided protection from Sexually Transmitted Diseases (STDs).

Question 4.
What is the name of contraceptive pills which is taken only in once in a week ?
Answer:
“Saheli.”

Question 5.
Write the full form of IUCD.
Answer:
Intra Uterine Contraceptive Device.

Question 6.
Write the full form of STDs.
Answer:
Sexually Transmitted Disease.

Question 7.
What is the legal age for marriage of male and female in India ?
Answer:
For male 21 Year and for female 18 years.

Question 8.
Write the name of two STDs which are spread through infected blood.
Answer:
AIDS, Hepatitis-B.

Question 9.
Write the name of two disease which are caused by sexual contact
Answer:
Gonorrhoea, Syphilis.

Question 10.
Write the full form of HIV and AIDS.
Answer:
HIV : Human Immunodeficiency Virus.
AIDS : Acquired Immuno Deficiency Syndrome.

Question 11.
Write the full form of ZIFT.
Answer:
Zygote Intra Fallopian Transfer.

Question 12.
What is the process for surgical sterilization in males called ?
Answer:
Vasectomy.

Reproductive Health Short Answer Type Questions

Question 1.
What is test-tube baby ?
Answer:
In some cases, a woman is unable to have a normal fertilization to bear the child. In such cases, test-tube technique may be successful. In this technique, the unfertilized eggs of such woman is isolated in aseptic condition and fertilized it in test-tube by the sperms of her husband. The fertilized egg or blastocyst can be maintained in vitro till it gets 32 celled stage. It can be implanted in the uterus of the female. The female remains under the supervision of doctor till completion of gestation period of 280 days. The baby produced in such a way is called a test-tube baby.

Question 2.
What is GIFT ? Explain in brief.
Answer:
GIFT (Gametes Intra Fallopian Transfer): It is a latest technique to produce a child. In this technique, the gametes are kept separately in a catheter and injected directly into the fallopian tube of the woman using laproscopy. Thus, in this case, fertilization occurs inside the body of woman. Prior to injection of gametes, the mother would be given hormones for about a week to stimulate follicle formation. This causes development of several eggs i.e,, super ovulation. The first GIFT in India was done in Mumbai and twins were born in August, 1990.

Question 3.
Explain the following:

1. Tubectomy
2. Vasectomy.

Answer:

1. Tubectomy : The surgical removal of a segment of oviduct and then ligating the cut end is called tubectomy. It is applied in females to check the pregnancy.
2. Vasectomy : Surgical cutting and ligating ends of segments of vas deferens is called vasectomy.

Question 4.
What is amniocentesis ? Write its ill effects and two advantages.
Answer:
Amniocentesis : It is prenatal diagnostic technique in which amniotic fluid of uterus is isolated by a surgical needle and foetus cells are cultured on a culture medium and chromosomes are examined. This technique is used to understand the following things or Advantages :

• Chromosomal abnormalities like that of Down’s syndrome, Philadelphia syndrome and Edward’s syndrome.
• Metabolic disorders like that of PKV, Cretinism, Alkaptonuria, etc.
• Sex of the embryo can be examined by this technique.
• Effects of amniocentesis: Due to this type of test, the female embryos are being eradicated. This leads into the
• decrease in number of females which may cause a serious problem.

Question 5.
Explain the social causes of human population growth.
Answer:
Social causes of human population growth rate are:

• Illiteracy in society.
• Low level of society.
• Various orthodox tradition like son leads to progeny.
• Social barriers.
• Early child marriage presuming that after mid-age marriage reproductive capacity degenerates.
• Social backwardness.
• Social set-up that “Putra Ratan se Moksh milta hai”.

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ विविध प्रश्नावली

प्रश्न 1.
k के मान । ज्ञात कीजिए जब कि रेखा (k – 3)x – (4 – k²)y + K² – 7k + 6 = 0
(a) x-अक्ष के समान्तर है।
(b) y-अक्ष के समान्तर है।
(c) मूल बिन्दु से जाती है।
हल:
(i) x-अक्ष के समान्तर y = a
∴ प्रश्न में दिए गए समीकरण में x का गुणांक = 0 या k – 3 = 0 अर्थात् k = 3.
(ii) xy-अक्ष के समान्तर रेखा x =q
दिए गए समीकरण में y का गुणांक = 0 या 4 – k² = 0 अर्थात् k = ± 2.
(iii) यदि रेखा मूल बिन्दु से जाती है तो (0, 0) इसके समीकरण को संतुष्ट करेगा।
0 – 0 + K² – 7k + 6 = 0 या (k – 6) (k – 1) = 0 या k = 1, 6.

प्रश्न 2.
θ और p के मान ज्ञात कीजिए यदि समीकरण x cos θ + y sin θ = p रेखा \(\sqrt{3}\)x + y + 2 = 0 का लम्ब रूप है।
हल:

प्रश्न 3.
उन रेखाओं के समीकरण ज्ञात कीजिए जिनके अक्षों से कटे अंतःखण्डों का योग और गुणनफल क्रमशः 1 और – 6 हैं।
हल:
मान लीजिए अक्षों पर कटे अतः खण्ड a और b हैं।
दिया है : a + b = 1, ab = – 6
b = 1 – a
∴ a(1 – a) = – 6
या a – a₂ = – 6
a₂ – a – 6 = 0
या (a – 3) (a + 2) = 0
∴ a = 3, – 2
∴ b = – 2, 3
3, – 2 अंत:खण्ड वाली रेखा का समीकरण,
\(\frac{x}{3}+\frac{y}{-2}\) = 1
या 2x – 3y = 6.
और – 2, 3 अंत:खण्ड वाली रेखा का समीकरण,
\(\frac{x}{-2}+\frac{y}{3}\) = 1
या – 3x + 2y = 6.

प्रश्न 4.
y-अक्ष पर कौन से बिन्दु ऐसे हैं, जिनकी रेखा \(\frac{x}{3}+\frac{y}{4}\) = 1 से दूरी 4 इकाई है।
हल:
मान लीजिए y-अक्ष पर बिन्दु (0, y₁) है।
(0, y₁) की रेखा 4x + 3y = 12 से दूरी

प्रश्न 5.
मूल बिन्दु से बिन्दुओं (cos θ, sin θ) और (cos ϕ, sin ϕ) को मिलाने वाली रेखा की लांबिक दूरी ज्ञात कीजिए।
हल:
(cos θ, sin θ), (cos ϕ, sin ϕ) को मिलाने वाली रेखा का समीकरण,

प्रश्न 6.
रेखाओं x – 7y + 5 = 0 और 3x + y = 0 के प्रतिच्छेद बिन्दु से खीची गई और y-अक्ष के समान्तर रेखा का समीकरण ज्ञात कीजिए।
हल:
दी गयी रेखाएँ
x – 7y + 5 = 0 …(1)
3x + y = 0 …(2)
समी (2) से,
y = – 3x
y का मान समी (1) में रखने पर,
x – 7(- 3x) + 5 = 0
या 22x + 5 = 0
x = \(\frac{-5}{22}\)
अब y = – 3x = – \(3\left(\frac{-5}{22}\right)\) = \(\frac{15}{22}\)
वह रेखा का जो \(\left(-\frac{5}{22}, \frac{15}{22}\right)\) से जाती है और y-अक्ष के समान्तर है, समीकरण x = – \(\frac{5}{22}\) या 22x + 5 = 0.

प्रश्न 7.
रेखा \(\frac{x}{4}+\frac{y}{6}\) = 1 पर लंब उस बिन्दु से खींची गई रेखा का समीकरण ज्ञात कीजिए जहाँ यह रेखा y-अक्ष से मिलती है।
हल:
रेखा AB का समीकरण,
\(\frac{x}{4}+\frac{y}{6}\) = 1 या 3x + 2y = 12,
2y = – 3x + 12
y = \(\frac{-3}{2} x+\frac{12}{2}\)

∵ रेखा y-अक्ष पर प्रतिच्छेद करती है, इसलिए बिन्दु B(0, 6) है।
∴ BC रेखा का समीकरण
y – 6 = \(\frac{2}{3}\)(x – 0)
या 3y – 18 = 2x
या 2x – 3y + 18 = 0.

प्रश्न 8.
रेखाओं y – x = 0, x + y = 0, और x – k = 0 से बने त्रिभुज का क्षेत्रफल ज्ञात कीजिए।
हल:
y – x = 0 और y + x = 0 बिन्दु (0, 0) पर मिलते हैं।
x = k को y – x = 0 में रखने से, y – k = 0 या y = k
x – k = 0 और y – x = 0 बिन्दु (k, k) पर मिलते हैं।
x = k को y + x = 0 में रखने से,
y + k = 0 या y = – k
x = k और y + x = 0 बिन्दु (k, – k) पर मिलते हैं।
अब बिन्दु (0, 0), (k, k) और (k, – k) से बने त्रिभुज का क्षेत्रफल
= \(\left|\frac{1}{2}[0 .(-2 k)+k(-k)+k(-k)]\right|\)
= \(\left|\frac{1}{2}\left(-k^{2}-k^{2}\right)\right|\)
= k₂ वर्ग इकाई।

दूसरी विधि : त्रिभुज OPQ का क्षेत्रफल
= 2 × क्षेत्रफल ∆OAP
= 2 × [\(\frac{1}{2}\) × k × k] = k₂ वर्ग इकाई।

प्रश्न 9.
p का मान ज्ञात कीजिए जिससे तीन रेखाएँ 3x + y – 2 = 0, px + 2y – 3 = 0 और 2x – y – 3 = 0 एक बिन्दु पर प्रतिच्छेद करें।
हल:
दी गयी रेखाएँ
3x + y = 2 …(1)
2x – y = 3 …(2)
समी (1) और (2) को जोड़ने पर,
5x = 5 या x = 1
∴ y = 2 – 3x = 2 – 3 = – 1
∴ समी (1) और (2) वाली रेखाएँ बिन्दु (1, – 1) पर प्रतिच्छेद करती हैं।
तीसरी रेखा px + 2y – 3 = 0 भी इसी बिन्दु से जाती है इसलिए (1, – 1) इस समीकरण को संतुष्ट करेगा।
p × 1 + 2 × (- 1) – 3 = 0
p – 2 – 3 = 0
∴ p = 5
अतः दी गयी रेखाएँ संगामी हैं यदि p = 5.

प्रश्न 10.
यदि तीन रेखाएँ जिनके समीकरण y = m₁x + c₁, y = m₂x + c₂ और y =m₃x + c₃ हैं, संगामी हैं, तो दिखाइए कि m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0.
हल:

प्रश्न 11.
बिन्दु (3, 2) से जाने वाली उस रेखा का समीकरण ज्ञात कीजिए जो रेखा x – 2y = 3 से 45° का कोण बनाती है।
हल:
माना रेखा AB का समीकरण : x – 2y = 3
या y = \(\frac{1}{2}\)x – 3
तब रेखा AB की ढाल = \(\frac{1}{2}\)

+ve चिन्ह लेने पर, 1 = \(\frac{2 m-1}{m+2}\)
या 2m – 1 = m + 2
∴ m = 3
2m -1
– ve चिन्ह लेने पर, – 1 = \(\frac{2 m-1}{m+2}\)
या 2m – 1 = – m – 2
∴ 3m = – 1
या m = \(\frac{-1}{3}\)
अतः रेखा PA का समीकरण जहाँ बिन्दु P = (3, 2) हो और m = \(\frac{-1}{3}\) हो।
y – 2 = – \(\frac{-1}{3}\)(x – 3)
3y – 6 = – x + 3
या x + 3y – 9 = 0
अब जब ढाल m = 3 हो, तब बिन्दु P(3, 2) से रेखा का समीकरण,
y – 2 = 3(x – 3)
y – 2 = 3x – 9
या 3x – y – 7 = 0.

प्रश्न 12.
रेखाओं 4x + 7y – 3 = 0 और 2x – 3y + 1 = 0 के प्रतिच्छेद बिन्दु से जाने वाली रेखा का समीकरण ज्ञात कीजिए जो अक्षों से समान अंतः खण्ड बनाती हैं।
हल:
4x + 7y = 3 …(1)
2x – 3y = – 1 …(2)
समी (2) को 2 से गुणा करने पर,
4x – 6y = – 2 …(3)
समी (3) को (1) में से घटाने पर
13y = 5
∴ y = \(\frac{5}{13}\)

जो रेखा अक्षों पर बने अंतः खण्ड समान हैं तो वह धन x-अक्ष के साथ 45° या 135° का कोण बनाती हैं। इसलिए उसकी ढाल ±1 होगी।
∴ PA और PB रेखाओं के समीकरण
y – y₁ = m(x – x₁)
(i) जब m = 1 हो, तब y – \(\frac{5}{13}\) = 1 \(\left(x-\frac{1}{13}\right)\)
या 13y – 5 = 13 x – 1 या 13x – 13y + 4 = 0.
(i) जब m = – 1 हो, तब y – \(\frac{5}{13}\) = 1 \(\left(x-\frac{1}{13}\right)\)
13y – 5 = – 13x + 1
∴ 13x + 13y – 6 = 0.

प्रश्न 13.
दर्शाइए कि मूल बिन्दु से जाने वाली और रेखा y = mx + c से θ कोण बनाने वाली उस रेखा का समीकरण \(\frac{y}{x}=\pm \frac{m+\tan \theta}{1-m \tan \theta}\) हैं।
हल:
रेखा PA का समीकरण y = mx + c है
यह रेखा OP के साथ कोण θ बनाती है।
रेखा PA की ढाल = m
मान लीजिए OP की ढाल = m₁ है।


या \(\frac{y}{x}\) = m₁
∴ अभीष्ट रेखाओं का समीकरण
\(\frac{y}{x}\) = \(\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\)

प्रश्न 14.
(- 1, 1) और (5, 7) को मिलाने वाली रेखाखण्ड को रेखा x + y = 4 किस अनुपात में विभाजित करती है ?
हल:
मान लीजिए बिन्दु P रेखाखण्ड AB को k : 1 के अनुपात में विभाजित करता है। जबकि A और B के क्रमशः (- 1, 1) और (5, 7) निर्देशांक हैं।

अतः बिन्दु P रेखाखण्ड AB को 1 : 2 के अनुपात में विभाजित करता है। .

प्रश्न 15.
बिन्दु (1, 2) से रेखा 4x + 7y + 5 = 0 की 2x – y = 0 के अनुदिश दूरी ज्ञात करो।
हल:
माना रेखा PC का समीकरण, 2x – y = 0 जिस पर बिन्दु P(1, 2) स्थित है।

रेखा AB का समीकरण 4x + 7y + 5 = 0 और 2x – y = 0 को हल करने पर,
∴ x = \(\frac{-5}{18}\)
और y = \(-\frac{5}{9}\)

प्रश्न 16.
बिन्दु (- 1, 2) से खींची जा सकने वाली उस रेखा की दिशा ज्ञात कीजिए जिसका रेखा x + y = 4 से प्रतिच्छेदन बिन्दु दिए बिन्दु से 3 इकाई की दूरी पर है।
हल:
मान लीजिए अभीष्ट रेखा PQ की ढाल m है
रेखा PQ जो बिन्दु P(- 1, 2) से होकर जाती है और ढाल m है, का समीकरण
y – y₁ = m(x – x₁)
y – 2 = m(x + 1)
या mx – y + m + 2 = 0 …(1)

रेखा AB का समीकरण x+ y = 4
∴ y = 4 – x
y का मान समी (1) में रखने पर,
mx – (4 – x) + m + 2 = 0
या (m + 1) x + m – 2 = 0
∴ x = – \(\frac{m-2}{m+1}\)
अब y = 4 – x = 4 + \(\frac{m-2}{m+1}\)

अतः रेखा PQ की ढाल 0 है अर्थात् रेखा x-अक्ष के समांतर है।

प्रश्न 17.
समकोण त्रिभुज के कर्ण के अंत्य बिन्दु (1, 3) और (- 4, 1) हैं। त्रिभुज के पाद (leg) (समकोणीय भुजाओ) का एक समीकरण ज्ञात कीजिए।
हल:
माना त्रिभुज ABC एक समकोणीय त्रिभुज है जिसका कर्ण AB है। A और B के निर्देशांक क्रमशः (1, 3) और (- 4, 1) हैं।
मान लीजिए BC की ढाल m है।

AC ⊥ BC
∴ AC की ढाल = – \(\frac{1}{m}\)
रेखा BC का समीकरण,
y – y₁ = m(x – x₁)
y – 1 = m(x + 4)
या mx – y + 4m + 1 = 0 …(1)
रेखा AC का समीकरण
y – 3 = – \(\frac{1}{m}\)(x – 1)
या my – 3m = – x + 1
या x + my – 3m – 1 = 0 …(2)
यह दोनों रेखाएँ m के दिए मान से इन का समीकरण ज्ञात कर सकते हैं। यदि BC भुजा x-अक्ष के समांतर हो तो m = 0.
BC का समीकरण, y – 1 = 0
या y = 1
∴ AC, y-अक्ष के समांतर हो और यह A(1, 3) से जाती है। अतः AC का समीकरण x = 1
अत: BC और AC के समीकरण y = 1 और x = 1 हैं।

प्रश्न 18.
किसी बिन्दु के लिए रेखा को दर्पण मानते हुए बिन्दु (3, 8) का रेखा x + 3y = 7 में प्रतिबिंब ज्ञात कीजिए।
हल:
माना रेखा AB का समीकरण x + 3y = 7 है और बिन्दु P के निर्देशांक (3, 8) हैं।
y = – \(\frac{1}{3} x +\frac{7}{3}\)
बिन्दु P का प्रतिबिंब Q होगा यदि PQ ⊥ AB, PQ और AB बिन्दु M पर इस प्रकार प्रतिच्छेद करते हैं कि
PM = QM

रेखा AB की ढाल = – \(\frac{1}{3}\)
और PQ की ढाल = 3
∴ PQ रेखा का समीकरण,
y – 8 = 3(x – 3)
= 3x – 9
या 3x – y = 1 …(1)
AB का समीकरण x + 3y = 7 …(2)
समी (1) को 3 से गुणा करके समी (2) में जोड़ने पर,
10x = 10 या x = 1
समी (1) से y = 3x – 1 = 3 – 1 = 2
∴ बिन्दु M के निर्देशांक (1, 2) हैं।
मान लीजिए Q के निर्देशांक (x₁, y₁) हैं
बिन्दु M रेखाखण्ड PQ का मध्य बिन्दु है
∴ जबकि P(3, 8) है।
∴ \(\frac{x_{1}+3}{2}\) = 1 या x₁ = – 1
\(\frac{y_{1}+8}{2}\) = 2 या y₁ = – 4
∴ P का प्रतिबिंब (- 1, – 4) हैं।

प्रश्न 19.
यदि रेखाएँ y = 3x + 1 और 2y = x + 3, रेखा y = mx + 4 पर समान रूप से आनत हो तो m का मान ज्ञात कीजिए।
हल:
रेखा AB का समीकरण, y = 3x + 1 की ढाल = 3
रेखा BC का समीकरण, y = mx + 4 को ढाल = m


अतः m का अभीष्ट मान = \(\frac{1 \pm 5 \sqrt{2}}{7}\).

प्रश्न 20.
यदि एक वर बिन्दु P(x, y) की रेखाओं x + y – 5 = 0 और 3x – 2y + 7 = 0 से लांबिक दूरियों का योग सदैव 10 रहे तो दर्शाइए कि P अनिवार्य रूप से एक रेखा पर गमन करता है।
हल:
P(x, y) से रेखा x + y – 5 = 0 की दूरी
= \(\frac{x+y-5}{\sqrt{2}}\)
P(x, y) से रेखा 3x – 2y + 7 = 0 की दूरी
= \(\frac{3 x-2 y+7}{\sqrt{9+4}}\)
दोनों दूरियों का योग = 10 (दिया है)
∴ \(\sqrt{13}\)(x + y – 5) + \(\sqrt{2}\)(3x – 2y + 7) = 10\(\sqrt{26}\)
या \((\sqrt{13}+3 \sqrt{2}) x+(\sqrt{13}-2 \sqrt{2}) y-5 \sqrt{13}+7 \sqrt{2}-10 \sqrt{26}\) = 0
जो कि एक सरल रेखा का समीकरण है।
अत: P एक अनिवार्य रूप से एक रेखा पर गमन करता है।

प्रश्न 21.
समांतर रखाओं 9x + 6y – 7 = 0 और 3x + 2y + 6 = 0 से समदूरस्थ रेखा का समीकरण ज्ञात कीजिए।
हल:
दी गयी समांतर रेखाएँ
9x + 6y – 7 = 0 …(1)
और 3x + 2y + 6 = 0
या 9x + 6y + 18 = 0 …(2)
एक रेखा जो इसके समांतर है, उसका समीकरण
9x + 6y + c = 0 …(3)
रेखा (1) और (3) के बीच दूरी
\(\frac{(c+7)}{\sqrt{81+36}}=\frac{(c+7)}{\sqrt{117}}\) …(4)
रेखा (2) और (3) के बीच दूरी
\(\frac{(c-18)}{\sqrt{81+36}}=\frac{c-18}{\sqrt{117}}\) ….(5)
दूरियाँ (4) और (5) आपस में समान हैं।
\(\frac{(c+7)}{\sqrt{117}}=\pm \frac{c-18}{\sqrt{117}}\) [+ve चिन्ह मान्य नहीं है।]
c + 7 = – (c – 18)
= – c + 18
2c = 18 – 7 = 11
या c = \(\frac{11}{2}\)
c का मान समी (3) में रखने पर,
9x + 6y + \(\frac{11}{2}\) = 0
या 18x + 12y + 11 = 0.

प्रश्न 22.
बिन्दु (1, 2) से होकर जाने वाली एक प्रकाश किरण x-अक्ष के बिन्दु A से परावर्तित होती है और परावर्तित किरण बिन्दु (5, 3) से होकर जाती है। A के निर्देशांक ज्ञात कीजिए।
हल:
मान लीजिए BC, x-अक्ष के अनुदिश उस बिन्दु के निर्देशांक A (a, 0) है। AN इस पर लंब है। PA एक आपतित किरण है और AQ परावर्तित किरण है।

⇒ आपतित कोण PAN = परावर्तित कोण NAQ
⇒ ∠PAB = ∠QAC
⇒ यदि QA का झुकाव 0 हो तो PA का झुकाव 180 – θ होगा।
QA की ढाल जबकि Q(5, 3) और A(a, 0) हो, तो

प्रश्न 23.
दिखाइए कि (\(\sqrt{a^{2}-b^{2}}\), 0) और (-\(\sqrt{a^{2}-b^{2}}\),0) बिन्दुओं से रेखा \(\frac{x}{a}\)cosθ + \(\frac{y}{b}\)sinθ = 1 पर खींचे गए लम्बों की लंबाइयों का गुणनफल b₂ है।
हल:

प्रश्न 24.
एक व्यक्ति समीकरणों 2x – 3y + 4= 0 और 3x + 4y – 5 = 0 से निरूपित सरल रेखीय पथों के संधि बिन्दुओं (junction/crosing) पर खड़ा है और समीकरण 6x – 7y + 8 = 0 से निरूपित पथ पर न्यूनतम समय में पहुँचना चाहता है। उसके द्वारा अनुसरित पथ का समीकरण ज्ञात कीजिए।
हल:
AB और BC दो रेखीय पथ हैं। AB व BC रेखाओं के समीकरण
2x – 3y + 4 = 0 …(1)
और 3x + 4y – 5 = 0 …(2)

AB और BC बिन्दु B पर मिलते हैं।
समी (1) को 3 से तथा समी (2) को 2 से गुणा करने पर
6x – 9y = – 12 ….(3)
6x + 8y = 10 ….(4)
समी (3) को समी (4) में से घटाने पर,
17y = 10 + 12 = 22
∴ y = \(\frac{22}{17}\)
y का मान समी (1) में रखने पर,

B से AC तक न्यूनतम समय में पहुंचने के लिए कम से कम दूरी BD ( BD ⊥ AC) तय करनी है।
रेखा AC का समीकरण, 6x – 7y + 8= 0 की ढाल = \(\frac{6}{7}\)
BD की ढाल = – \(\frac{7}{6}\)
BD बिन्दु B \(\left(\frac{-1}{17}, \frac{22}{17}\right)\) से होकर जाती है।
∴ रेखा BD का समीकरण
y – y₁ = m(x – x₁)
\(y-\frac{22}{17}=-\frac{7}{6}\left(x+\frac{1}{17}\right)\)
102 से गुणा करने पर,
102y – 132 = – 119x – 7
119x + 102y – 125 = 0 .
अतः B से AC तक पहुँचने के लिए BD पथ अपनाना है जिसका समीकरण 119x + 102y – 125 = 0 है।

MP Board Class 11th Maths Solutions

MP Board Class 12th Biology Solutions Chapter 15 Biodiversity And Conversation

Biodiversity and Conversation NCERT Text Book Question and Answers

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.
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 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses 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.

Question 4.
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 smallar 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 5.
What are the major causes of species losses in 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 6.
How is biodiversity important for ecosystem functioning?
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 7.
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 khasi and jointia hills in Meghalaya. Westen 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 8.
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 9.
The species diversity of plants (22 %) is much less than that of animals (72 %). What could be the explanations to how 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 10.
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 micro-organisms 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.

Biodiversity and Conversation Other Important Questions and Answers

Biodiversity And Conversation Objective Type Questions

Choose the Correct Answer:

Question 1.
Wildlife conservation act was inforced in India in:
(a) 1883
(b) 1972
(c) 1973
(d) 1982.
Answer:
(b) 1972

Question 2.
Indian board of wildlife was established in:
(a) 1952
(b) 1981
(c) 1971
(d) 1972.
Answer:
(a) 1952

Question 3.
NBPGR is situated at:
(a) Delhi
(b) Kolkata
(c) Lucknow
(d) Mumbai.
Answer:
(a) Delhi

Question 4.
Number of biosphere in our country is :
(a) 73
(b) 7
(c) 416
(d) 23
Answer:
(b) 7

Question 5.
National park related with white tiger is:
(a) Kanker
(b) Satipada
(c) Bandhavgarh
(d) Kanha.
Answer:
(c) Bandhavgarh

Question 6.
The first national park of Madhya Pradesh is :
(a) Shivpuri
(b) Bandhavgarh
(c) Kanha
(d) Kanker.
Answer:
(c) Bandhavgarh

Question 7.
Plant fossils national park is situated in :
(a) Shivpuri
(b) Mandala
(c) Kanha
(d) Kanker.
Answer:
(b) Mandala

Question 8.
Kanha national park is famous for:
(a) Rhinoceros
(b) Tiger
(c) Birds
(d) Aligator.
Answer:
(b) Tiger

Question 9.
Bandipur national park is the habitat of:
(a) Deer Project
(b) Peacock Project
(c) Elephant Project
(d) Tiger Project
Answer:
(d) Tiger Project

Question 10.
Animal which is extinct in india:
(a) Hippopotamus
(b) National Park
(c) Cheetah
(d) Wolf
Answer:
(c) Cheetah

Question 11.
Which is in-situ conservation method:
(a) Tissue culture’
(b) National park
(c) Botanical garden .
(d)Cryotest.
Answer:
(b) National park

Question 12.
Which of the following are not done in a wild sanctuary:
(a) Conservation of flora
(b) Conservation of fauna
(c) Use of soil and flora
(d) Prohibition of hunting.
Answer:
(c) Use of soil and flora

Question 13.
In India maximum area covered by forest
(a) Odisha
(b) Arunachal Pradesh
(c) Madhya Pradesh
(d) All of the above
Answer:
(b) Arunachal Pradesh

Question 14.
Forest destruction:
(a) Destruction of natural sources
(b) Environmental Pollution
(c) Genetic disorder
(d) All of the above
Answer:
(b) Environmental Pollution

Question 15.
Forest Research Institute (FRI) is situated in:
(a) Shimla
(b) Chennai
(c) Dehradun
(d) Kolkata.
Answer:
(c) Dehradun

Question 16.
World environmental day is celebrated on :
(a) 15th March
(b) 15th April
(c) 4th May
(d) 5th June.
Answer:
(d) 5th June.

Question 17.
Species which are continuously distinct:
(a) Long lasting animal
(b) Dangerous animal
(c) Simple animal
(d) None of these.
Answer:
(c) Simple animal

Question 18.
Kaziranga national park is situated in:
(a) West Bengal
(b) Kerala
(c) Assam
(d) Umariya.
Answer:
(c) Assam

Question 19.
Bandhavgarh national park is situated in:
(a) Satana
(b) Shivpuri
(c) Mandla
(d) Umariya.
Answer:
(d) Umariya.

Question 20.
Kanha national park is located in :
(a) Uttar Pradesh
(b) Rajasthan
(c) Gujrat
(d) Madhya Pradesh.
Answer:
(a) Uttar Pradesh

Question 21.
National wildlife (Protection) Act was formulated during:
(a) 1972
(b) 1974
(c) 1976
(d) 1978.
Answer:
(a) 1972

Question 2.
Fill in the Blanks:

1. Planting of trees and forests on waste land is called ……………………..
2. Wildlife protection act is come in force in …………………….
3. UNESCO has started Man and Biosphere programme in ……………………..
4. ………………….. is the first Biosphere Reserve of India.
5. Kajiranga wildlife sanctuary is famous for ………………………
6. Red Data book is related with ……………………..
7. World Environment Day is celebrated on ………………………… of every year.
8. FRI is situated at ……………………..
9. …………………….. is the main source of energy.
10. National parks are made for the conservation of ……………………….
11. ………………………….. is the main region of forest depletion.
12. Indian Lion and Cheetah is ……………………….. animal.
13. ………………………….. is called the development of forest on barren land.
14. World environment day celebrated on ………………………

Answer:

1. 1. Afforestation
   2. 1972
   3. 1971
   4. Nilgiri
   5. Rhinoceros
   6. Endangered species
   7. 5th June
   8. Dehradun
   9. Sun
   10. Threatened species

1. Growing population
2. Rare
3. Forestration
4. 5th June.

Question 3.
Match the Following:

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

Answer:
l.(e)
2. (d)
3. (a)
4. (b)
5. (c).

Question 4.
Answer in One Word/Sentence:

1. In which place of India, lion is found?
2. When does World Environment Day is celebrated?
3. When does Wildlife Protection Act was formed in India?
4. In which state Ghana bird Sanctuary is situated?
5. Where is Tiger Project situated?
6. Name any two National Parks of India.
7. Where does Kanha National Park situated?
8. Name any two threatened species.
9. Name the state having maximum forest area.
10. Name the National Animal and National Bird of India.
11. Name any one book associated with the conservation of wildlife.

Answer:

1. Gir forest, Gujarat
2. 5th June
3. 1972
4. Bharatpur, Rajasthan,
5. Kanha, M.P.
6. Rajaji Park and Gir National Park
7. M.P.
8. Lion, Wild buffalo
9. Chattisgarh (C.G.)
10. Tiger and Peacock
11. Red Data Book.

Biodiversity And Conversation Very Short Answer Type Questions

Question 1.
When does WLF-India was established?
Answer:
1962.

Question 2.
Where does Tiger project is established in Madhya Pradesh?
Answer:
Kanha national park.

Question 3.
Where does wild Ass is found?
Answer:
Runn of Kutch.

Question 4.
Where does great Indian bustards are found in India?
Answer:
Rajasthan, Punjab, Gujrat.

Question 5.
Write other name of biodiversity.
Answer:
Biological diversity.

Question 6.
Which part of the world are known as lesser biodiversity?
Answer:
Poles has lesser biodiversity.

Question 7.
What is the basic concept of biodiversity of biosphere?
Answer:
Gene.

Question 8.
Name the two plants which are worshiped any two states.
Answer:

1. Kadamba in Rajasthan.
2. Mango in Odisha.

Question 9.
Which part of the world which has more biodiversity?
Answer:
In Brazil.

Question 10.
Write the full form of IUCN.
Answer:
International Union for Conservation of Nature and Natural Resources.

Question 11.
Who perform important role in nature balancing?
Answer:
Forest animals.

Question 12.
Name the branch of science in which done the study of forest animal conservation.
Answer:
Forestry.

Question 13.
When, started the programme of social forestry?
Answer:
1976

Question 14.
Write the modern name of sanctuary.
Answer:
Shelter.

Biodiversity And Conversation Short Answer Type Questions

Question 1.
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 2.
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 is 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 :

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

Question 3.
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 4.
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 5.
Enumerate any five reasons for the destruction of wildlife (animals).
Answer:
The main reason of destruction of wildlife (animals) are:

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

Biodiversity And Conversation Long Answer Type Questions

Question 1.
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 to 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 wildass, 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 2.
What are National Parks? Explain any five National Parks found in India.
Answer:
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 park: It is located at Shivpuri near Gwalior (Madhya Pradesh). Wildlife present in this park tiger, cheetal, sambhar.
2. Gundi deer park: It is locssssated near Chennai (Madras) in Tamil Nadu. Wildlife found here are Alvino deer, black buck, cheetal and famous snake park.
3. Betia 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 3.
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 e.g., 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 andreligion. 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 4.
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:

• Forest management,
• Proper use of forest land,
• Policy for protected forests,
• 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

• Prevention of deforestation of hill plants.
• Reforestation of grazing land.
• Development of grazing land,
• Plantation of economically useful forest trees.
• Increase in the profit of government from forests.

Question 5.
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-e.g. Musdeer, Sambhar deer, chital, etc.
2. Antelope: These are same as deer e.g.-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
5. the Indian rhinoceros as the have been hunted to the brink of extinction for their horns.
6. Wild Ass: Wild asses are not found in any part of the world. Now in India, it is found in die little Rann of Kutch in the Gujrat state of India.

Carnivorous animals: Some India wild carnivores are :

(a) Indian lion (Asiatic lion): Now it is confined to forests in the fall.
(b) Cheetah: It is on the verge of extinction.
(c) Lion: Lion is a national animal, at present its population in our country are more than 3,000.
(d) Leopard: It is similar in 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 6.
What is 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 :

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

Question 7.
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:

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

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 3 Human Reproduction

Human Reproduction NCERT Textbook Questions and Answers

Question 1.
Fill in die Blanks:

1. Humans reproduce …………………….. (asexually/ sexually)
2. Humans are …………………….. (oviparous, viviparous, ovoviviparous)
3. Fertilization is …………………….. in humans, (external / internal)
4. Male and female gametes are …………………….. (diploid / haploid)
5. Zygote is …………………….. (diploid / haploid)
6. The process of release of ovum from a mature follicle is called ……………………..
7. Ovulation is induced by a hormone called ……………………..
8. The fusion of male and female gametes is called ……………………..
9. Fertilisation takes place in ……………………..
10. Zygote divides to form …………………….. which is implanted in uterus.
11. The structure which provides vascular connection between fetus and uterus is called ……………………..

Answer:

1. Sexually
2. Viviparous
3. Internal
4. Haploid
5. Diploid
6. Ovulation
7. LH and FSH
8. Fertilisation
9. Follopian tube
10. Embryo
11. Placenta.

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:

Question 4.
Write two major functions each of testis and ovary
Answer:
Functions of testis functions:

• Production of sperms by seminiferous tubules.
• Production of male sex hormone, testosterone by leydig cells.

Functions of Ovary

• Production of ova (eggs).
• Production of female sex harmones, estrogen and progesterone.

Question 5.
Describe the structure of a seminiferous tubuls.
Answer:
Seminiferous tubules are highly coiled tubes, which are lined on the inside by :

• Male germ cells called spematogonia that undergo meiotic division to form sperm celts.
• Sertoli cells provides nutrition and molecular signals to the germ cells.
  

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis : Formation of sperms in the testis is called as spermatogenesis. It involves in the following steps :

1. Multiplication phase : In this phase, sperm cells are formed in testes.
The inner layer of seminiferous tubules of testes is formed of germinal epithelium.
Some of these cells called primary germ cells divide mitotically into spermatogo¬nia which become separated in the germi¬nal layer. Other cells of this layer serve as nutrition for the dividing cells.

2. Growth phase : In this phase, spermatogonia starts growing, absorbing nutrient substances. These large cells are called primary spermatocytes.

3. Maturation phase : It is a very important phase. Primary spermatocytes divide twice. The first division is meiotic due to which the number of chromosomes is reduced to half. In this process, primary spermatocyte divides into two halves which are known as secondary spermatocytes. The second division is mitotic and no change takes place in the number of chromosomes. Thus, from two secondary spermatocytes four spermatids are formed. In this manner from one primary spermatocyte four spermatids are formed. These spermatids change into sperm cells of spermatozoa by a process called metamorphosis.

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
The hormones involved in regulation of spermatogenesis are:

• Gonadotropin releasing hormone
• Luteinizing hormone (LH)
• Follicle stimulating hormone
• Testosteron.

Question 8.
Define spcrmiogenesis and spermiation.
Answer:
Spermiogenesis : The process involving transformation of spermatid into spermatozoa is called spermiogenesis.
Spermiation: After spermiogenesis sperm heads became embedded in the sertoli cells and are finally released from the seminiferous tubules by the process called spermiation.

Question 9.
Draw a labelled diagram of human sperm and explain its defferent parts.
Answer:
Structure of sperm: A mature sperm is a delicate microscopic, motile structure. A typi¬cal mammalian sperm consists of the following three parts:

1. Head,
2. Middle piece and
3. Tail.

(1) Head: Head is knob like terminal part of the sperm. It is composed of a large nucleus and an acrosome. At die time of entry of the sperm into egg acrosome Secretes spermlysin which dis¬solve the egg membrane and thus facilitates entry of sperm into the egg or ovum.

(2) Middle Piece : It is short and lies be¬tween head and tail. It contains two granules called the proximal and distal centrioles in front side and towards posterior side cylindrical middle part of sperm. It is considered as the power house of sperm as it contains compact mass of mitochondria, which provides energy for metabolism and movement of sperm.

(3) Tail: It is situated on posterior part of sperm. It moves with the help of axial filament. The posterior part of the tail is called as end piece and it is not covered by membrane.

Question 10.
What are the major components of seminal plasma?
Answer:
The main components of the seminal plasma are fructose, calcium ion, some enzymes and prostagladines.

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The main functions of male accessory ducts and glands are as follows :

1. Functions of accessory ducts:

(a) Rete Testis : They transport sperms from seminiferour tubule to Vas efferentia.
(b) Vas efferentia : Transports sperms to epdidymis.
(c) Epididymis : Sperms are stored here. Maturation of sperms occur.
(d) Vas deference : Transports sperms from epididymis to the urethra.

(2) Functions of glands:

(a) Prostate gland : It produces milky secretion which forms considerable part of the semen. It makes sperm motile.
(b) Bulbourethral gland : Its secretion make the penis lubricated.
(c) Seminal vesicle: It secrete mucus and watery alkaline fluid which provide energy to the sperm.

Question 12.
What is oogenesis ? Give a brief account of oogenesis. .
Answer:
Oogenesis is the process of formation of mature female gametes from primordial germ cells.

• This process is initiated during embryonic developmental stage when about two million gamete mother cells (oogonia) are formed in each foetal ovary.
• Oogonia start meiotic division which gets arrested at prophases-I stage. They are referred to as primary oocytes.
• Each of these gets covered with layers of granulose cells and are then called primary follicle.
• Many of the primary follicles degenerate from birth to puberty, leaving about 60000-80000 in each ovary at puberty.
• More layers of granulose cells and another theca layer surround it and now it is called secondary follicle. Theca layer is arranged as inner theca-intema and outer theca- extema.
• Secondary follicle transforms into tertiary follicle that has a fluid filled cavity called antrum.
• The primary oocyte within the tertiary follicle grows in size and completes its first meiotic division now.
• This is an unequal division resulting in the formation of:
  (a) A large haploid cell (that keeps the majority of nutrient rich cytoplasm) called secondary oocyte.
  (b) A tiny cell, with haploid nucleus and almost no cytoplasm called first polar
  body.
• The tertiary follicle undergoes certain changes to mature into graafian follicle.
  Secondary oocyte forms a new membrane around it, called zona pellucida.
• Under the influence of LH, the Graaffian follicle now ruptures to release secondary oocyte from the ovary by a process called ovulation.

Summary of oogenesis:

Oogonia Meiosis-I initiated -4 Primary oocyte (arrested at prophase-I) → Granu¬losa layer builds → Primary follicle → More granulose and theca layer added → Secondary follicle → Fluid filled cavity develops → Tertiary follicle → Primary oocyte completes meio- sis-I → Secondary oocyte (haploid)+ polar body → Tertiary follicle transforms to Graafian follicle → Zona pellucida builds around secondary oocyte Graafian follicle ruptures → Secondary oocyte (ovum) released.
Meiosis-II will occur only at the time of penetration of sperm into the ovum.

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:

Question 14.
Draw a labelled diagram of a graafian follice.
Answer:

Question 15.
Name the function of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae.
Answer:
The functions of the following :
(a) Corpus luteum Secretes large amount of progesterone which is essential for the maintenance of endometrium of the uterus.
(b) Endometrium is necessary for the implantation of the fertillised ovum, for contrib¬uting towards making of placenta and other events of pregnancy.
(c) Acrosome is filled with enzymes that help in dissolving the outer cover of the ovum and entry of sperm nucleus.
(d) Sperm tail facilitates motility of the sperm essential for reaching the ovum to fertilise it.
(e) Fimbriae are fmgers-like projections at the mouth of fallopian tubules that help in
collection of the ovum after ovulation.

Question 16.
Identify true/ false statement to make it true.

1. Androgens are produced by sertoli cells. (True / False)
2. Spermatozoa get nutrition from sertoli cells. (True / False)
3. Leydig cells are found in ovaiy. (True / False)
4. Leydig cells synthesise androgens. (True / False)
5. Oogenesis takes place in corpus luteum. (True / False)
6. Menstrual cycle ceases during pregnancy. (True / False)
7. Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True /False)

Answer:

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

Question 17.
What is mentrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates is called menstrual cycle. The uterus linings becomes thick and spongy to receive fertilised egg. If the egg is not fertilised, this lining is not needed any longer so, it slowly breaks and comes out through vagina along with blood and mucous. This is called menstruation. It is repeated at an average interval of about 28/29 days.
Following hormones are regulate this cycle:

• Gonadotropin
• strogen
• Luteinizing hormone
• Follicular stimulating hormone
• Progesteron.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
The process of delivery of the foetus (Child birth) at the end of the pregnancy is called parturition. The signals for parturition originate from the fully developed foetus and the placenta, which trigger the release of oxytocin from the maternal pitutary. Oxytocin acts on the uterine muscles and induces stronger uterine contractions leading to expulsion of the baby. Relaxin hormone released by the ovary widens the vagina to facilitate birth.
Following hormones are involved in induction of parturition:

1. Cartisol
2. Estrogen
3. oxytocin.

Question 19.
In our society the women are often blamed for giving bird) to daughters. Can you explain why this is not correct?
Answer:
Women are blamed for giving birth to daughters. This is wrong because sex of the baby is determined by the sperm that can have either X or Y-chromosome. Women have only one type of chromosome (X) in all the ova.
If the sperm having X-chromosome fertillises the ovum (X), the resulting zygote (XX) will become a female.
If the sperm having Y-chromosome fertillises the ovum (X), the resulting zygote (XY) will become a male.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins born were fraternal?
Answer:
Only one egg is released by a human (female) ovary in a month. Only one egg is released if the mother gave birth to identical twins. Yes two or more eggs are released in case fraternal twins are born.

Question 21.
How many eggs do you think were released by the ovay of the female dog which gave birth to 6 puppies?
Answer:
Six eggs are released by the ovary of a female dog if it gave birth to six puppies.

Human Reproduction Other Important Questions and Answers

Human Reproduction Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Fertilization is related by: (CBSE PMT 2007)
(a) Realizing & gamete cells from gonad
(b) Transfer of male gametophy te to female gamete
(c) Fusion of male and female sex organs
(d) Fusion of nucleus of male gametes and nucleus of female gamete.
Answer:
(d) Fusion of nucleus of male gametes and nucleus of female gamete.

Question 2.
Cleavage is a process of fertilized egg, in which egg.
(a) Does not divide, but grows in sizes
(b) Divides continuously but does not grow in size
(c) Divided continuously and grows in size
(d) None of these.
Answer:
(b) Divides continuously but does not grow in size

Question 3.
Foetal membranes provide:
(a) Protection of embryo
(b) Nutrition to embryo
(c) Protection and nutrition to embryo
(d) None of these.
Answer:
(c) Protection and nutrition to embryo

Question 4.
Science of ageing is called:
(a) Chronology
(b) Odontology
(c) Gynecology
(d) Gerontology.
Answer:
(d) Gerontology.

Question 5.
The female counterpart of prostate gland in male (man) is:
(a) Bertholin’s gland
(b) Uterus
(c) Clitoris
(d) None of these
Answer:
(d) None of these

Question 6.
Humanis:
(a) Oviparous
(b) Viviparous
(c) Ovoviviparous
(d) None of these
Answer:
(b) Viviparous

Question 7.
Periodic cycle is:
(a) Of menstrual phase
(b) Of estrogen secretion
(c) Of fertilization
(d) None of these
Answer:
(a) Of menstrual phase

Question 8.
Ageing in mammals including man is due to:
(a) Adverse changes in environment
(b) Interaction between hereditary factors (genes) and the environment
(c) Malnutrition and stress
(d) All of these.
Answer:
(b) Interaction between hereditary factors (genes) and the environment

Question 9.
The part of the human body shows regeneration is:
(a) Spleen
(b) Kidney
(c) Brain
(d) Liver
Answer:
(d) Liver

Question 10.
Fertilization in mammals occurs in:
(a) Oviduct funnel
(b) Fallopian tubule
(c) Uterus
(d) Vagina.
Answer:
(b) Fallopian tubule

Question 11.
The capsule enclosing testes of mammal is called:
(a) Tunica albugenia
(b) Tunica membrana
(c) Tunica vaginalis
(d) Tunica vesculosa
Answer:
(a) Tunica albugenia

Question 12.
Sertoli cells are found in:
(a) Testes of Earthworm
(b) Testes of Frog
(c) Testes of Mammals
(d) Testes of Cockroach
Answer:
(a) Testes of Earthworm

Question 13.
The cavity of gastrula is known as:
(a) Blastocoel
(b) Coelome
(c) Archenteron
(d) Haemocoel
Answer:
(c) Archenteron

Question 14.
Implantation is the process in which
(a) Fertilization of egg
(b) Movemènt of egg
(c) Disappear of egg
(d) Blastosyst is formed by uterus.
Answer:
(a) Fertilization of egg

Question 15.
Seminiferous tubule is found in: ‘
(a) In testes
(b) In ovary
(c) In kidney
(d) In lungs.
Answer:
(a) In testes

2. Match the Following:

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

Answer:

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

3. Answer in One Word/Sentence:

1. How many sperms will be produced from 24 spermatocytes during spermatogenesis ?
2. Where does fertilization takes place in mammals ? (SSCE1993, CBSE 95)
3. Write the name of various phases of embryogenesis.
4. How many polar bodies are formed during the formation of one gamete?
5. Write die name of three layers of gastrula.
6. How many polar bodies are formed during the formation of ovum during oogenesis? (CBSE 1993)
7. Name the substance formed in sperms and which helps the sperms to enter into ovum. (AISB1991)
8. How many sperms and ova are formed from 100 primary spermatocytes and primary oocyte respectively ? (SSCE 1992)
9. Name the part of sperm secreting enzyme for the entrance of sperms into ovum. (SSCE 1992)
10. Name the organ where corpus luteum is formed. (SSCE 1995)
11. Write the duration of gestation period in human cases.
12. How many autosomes are found in human sperm ?
13. Where is corpus luteum formed ?
14. Name the developmental stage of man in which it is transplanted of uterus wall.(SSCE 1993)
15. Name the process of release of ovum from graafian follicle.

Answer:

1. 96,
2. Fallopian tube
3. Cleavage
4. Blastula and Gastrula Infant, 4.3
5. Ectoderm, mesoderm and endodenn
6. 2
7. Spermlysin
8. 400,100
9. Head
10. Ovary
11. 240 days
12. 22
13. Ovary
14. Blastula
15. Ovulation.

Human Reproduction Very Short Answer Type Questions

Question 1.
By which term, the process of the formation of morulla from zygote is referred?
Answer:
Cleavage.

Question 2.
What is called first menstrual stage or phase in the girls age 12-14 years?
Answer:
Menarche.

Question 3.
What is the example of spermatogenesis?
Answer:
Male gametogenesis.

Question 4.
What provides nourishment to foetus inside the uterus?
Answer:
Placenta.

Question 5.
How many germinal layers are there in gastrula stage?
Answer:
Three.

Question 6.
By which term, the middle germinal layer of gastrula is referred?
Answer:
Mesoderm.

Question 7.
After Fertilization, how many days are needed for the birth of human child?
Answer:
280 days (9 months and 10 days).

Question 8.
To produce the new organisms of their own kind is called by a term, name it
Answer:
Reproduction.

Question 9.
By which name, the reproduction happens by means of fusion of two different gametes?
Answer:
Sexual reproduction.

Question 10.
Which gametogenesis goes on till life span in human beings after attaining puberty?
Answer:
Spermatogenesis.

Question 11.
The stoppage of menstrual cycle in a 50 yrs old female is known as.
Answer:
Menopause.

Question 12.
What is pregnancy?
Answer:
The time period between fertilization to the birth of child is called pregnancy.

Question 13.
Where is carpora cavernosa found?
Answer:
Carpora cavernosa is found in penis.

Question 14.
What is the gestation period in elephant, dog and cat ?
Answer:
Elephant 641 days dog 58-68 days, cat 63 days.

Question 15.
Name the hormone that relaxes public symphysis during parturition.
Answer:
Relax in hormone.

Human Reproduction Short Answer Type Questions

Question 1.
Explain menarchy or menopause.
Answer:
Menopause: In every human female, puberty period starts from 12-13 yrs of age to 45-50 yrs of the age. During this period except pregnancy at every interval of a month during 26th day to 28th day. If pregnancy does not occur then the internal wall of the uterus secretes out mucilaginous liquid along with blood. Secretion continues for 3-4 days called menses. As it comes at definite period so, it is called menstruation cycle. At the age of 40-50 yrs. menses stop and females reach to a stage called menopause. Ability pregnancy also stops after attaining menopause.

Question 2.
Draw a well labelled diagram of the T.S. of human testis.
Answer:

Question 3.
Draw a well labelled diagram of Oogenesis.
Answer:

Question 4.
What is the position of fallopian tubule in female reproductive organs? What are their significance ?
Answer:
Fallopian tubules are a pair of small muscular tubes, one each on either side ol*the uterus. These tubules extend from the vicinity of the ovary to the ovary. Each tubule is about 10 cm in length. The free end of each tube lies near the ovary of its side. This end is funnel shaped and fimbriated. It is called ostium and infundibulum. Infundibulum opens in the abdominal cavity by means of abdominal ostium. The fallopian tubule is kept in position by a mesentery which is attached to the uterus.

Significance of Function of Fallopian Tubes:

• By their lashing movement of the cilia present in the lining of infundibulum and nearby area help in pulling the released ovum into fallopian tube.
• Passage of ovum into uterus is aided by muscular movement of fallopian tube as well as beating of cilia present in the lining layer of tube.
• Fertilization of ovum mostly takes place in the ampulla part of fallopian tube.

Human Reproduction Long Answer Type Question

Question 1.
Write differences between spermatogenesis and oogenesis
Answer:
Differences between Spermatogenesis and Oogenesis:

Question 2.
Describe the process of the formation and functions of corpus luteum.
Answer:
Corpus luteum is formed within ovary. It is a yellow body or structure which develops from ruptured Graafian follicle after the release of ovum. The cells become enlarged and conical. They are filled with a fluid called as luteum. It is developed under influence of L.H. of anterior pituitary. If fertilization do not occur it degenerates but if fertilization occurs, it remains for seven months during pregnancy and is responsible for the formation of some hormones which stimulates pregnancy and lactation.

Question 3.
Describe the structure and function of fallopian tubes.
Answer:
Fallopian tubules are a pair of small muscular tubes, one each on either side of the uterus. These tubules extend from the vicinity of the ovary to the ovary. Each tubule is about 10 cm in length. The free end of each tube lies near the ovary of its side. This end is funnel shaped and fimbriated. It is called ostium and infundibulum. Infundibulum opens in the abdominal cavity by means of abdominal ostium. The fallopian tubule is kept in position by a mesentery which is attached to the uterus.

Function of Fallopian Tubes:
By their lashing movement of the cilia present in the lining of infundi-bulum and nearby area help in pulling the released ovum into fallopian tube. ,
Passage of ovum into uterus is aided by muscular movement of fallopian tube as well as beating of cilia present in the lining layer of tube.
Fertilization of ovum mostly takes place in the ampulla part of fallopian tube.

Question 4.
In our society the women are often blamed for giving birth to daughters. Can you explain why this is not correct ?
Answer:
It is not correct to blame women for giving birth the daughters. The male sperm contains either X or Ychromosome whereas the female egg contains only x chromosomes. At the time of fertilisation when sperm carrying x chromosome combines with egg carring x chromosome of female Xygote z is formed which would be a female and when sperm with Y chromosome combines with egg containing X chromosome XY zygote is formed which would be a male. Thus scientifically sex of the baby is determined by the father and not by the mother as blamed in our society.

Question 5.
Describe the structure of ovary.
Answer:
These are a pair of female gonads or primary sex-organs lying one on each side of uterus. Ovary is attached to abdominal wall as well as uterus by means of ligaments. It is surrounded by a fold by peritoneum named mesovarium. Ovary is internally differentiated into four parts: germinal epithelium, tunica albuginea, cortex and medulla. Germinal epithelium is the outermost layer of cuboidal to flattened cells. Germinal epithelium is followed by a sheath of condensed connective tissues which is termed as tunica albuginea. It is followed by cortex. The central part of ovary contains medulla. A large number of groups of specialized cells are present in the cortex which are termed as ovarian follicles. These follicles are found in four development stages :

Incipient follicle : The central part of these follicles contain a large cell which is covered by smaller cells.
Primary follicle: These follicles are developed from incipient follicles.
Vascular follicle: It is formed from primary follicles. The oocyte of these follicle is covered by a many layered thick wall.

Graafian follicle: Mature ovarian follicle is termed as graafian follicle. It is covered by two sheaths derived from cortex. The follicle contains a single oocyte. A group of follicular cells surrounds the oocyte or ovum. It is called cumulus ovaricus. Another group produces membrane granulosa. Oocyte has two non-cellular membranes, inner vitelline membrane and outer zona peUucida.

Question 6.
Describe the process of fertilization and give its significance.
Answer:
Fertilization: Fertilization is the fusion of male and female gametes to produce a single diploid cell, called zygote. Fertilization in human female takes place in fallopian tube.

In a sexual mating or coitus the male ejaculates semen into the vaginal passage of the female using the copulatory organ, the penis. In a single coitus as many as 200 million sperms are introduced into the female genital tract but out of this huge number of sperms only one is destined to fuse with the ovum, provided the fallopian tube lodges of fully developed seqondary oocyte.Sperms travel through the vaginal passage and enter the uterus through the cervix.

They travel further through the uterus and finally enter the fallopian tube.The vaginal passage is highly acidic to prevent any bacterial infection but this acidic medium is not suitable for the survival of sperms. Many sperms die on the way. The liquid medium of the semen is alkaline which can neutralize the acidity of vagina to some extend and keep the. sperms alive and active. The sperms move with the speed of 1 -5 to 3 mm per minute.

The ovum gets surrounded by a large number of sperms but usually only one fuses with the ovum. The sperm penetrates the ovum using certain chemical substances of enzymatic nature.These chemicals are called spermlysins which are present in the acrosome. Certain receptors on the cell surface of the ovum enable the sperms to penetrate the wall of ovum. The ovum is surrounded by follicle cells. These cells are joined together by a glue like substance known as mucopolysaccharide, an acid, called hyaluronic acid. The sperm pro¬duces spermlysin, known as hyaluronidase. The over all changes in a sperm before the fertilization is called capacitation.

The hyaluronidase enzyme facilitates the sperm to pen¬etrate through the corona radiata (follicle cells), zona pellucida and the plasma membrane of ovum. The nucleus and the cytoplasmic components get inside the ovum, leaving the tail outside. The entry of sperm stimulates the ovum and the signal is transmitted to the egg surface incapacitating all the other cells surrounding the ovum. Nucleus of sperm move towards the nucleus of ovum and they fuses with each other to form zygote. It takes 2 \(\frac { 1 }{ 2 }\) hours to complete fertilization process. Once the ovum has been fertilized a barrier forms around it that normally prevents other sperms from entering.

Now fertilized egg reaches to the uterus and within seven days of fertilization it is transplanted to the wall of the uterus.

Significance of Fertilization
Egg become active after entry of sperm and completes its second maturation division.
Formation of fertilization membrane prevent entry of other sperms in the ovum. In human this membrane is not formed.
It restores the diploid number (2n) of chromosomes in the zygote.
It combines characters of male and female resulting in the introduction of variations. These variations make the offspring better equipped for struggle against environmental conditions to ensure the existence.
The ovum is stimulated to cleavage.
After fertilization ovum rotate inside the membrane.
Ovum do not contain centriole and obtain it from sperm during this process thus zygote continuously divides.
It is necessary for the egg to attain maturity.

Question 7.
Describe development of embryo up to the formation of three germ layer. Give the names of organs formed by three germ layer.
Or
Define cleavage. Describe the changes that occur in embryo till gastrulation.
Answer:
The term cleavage refers to a series of rapid mitotic divisions of the zygote following fertilization forming a many celled blastula. Following are the various steps of embryonic development in human up to the formation of three germ layer:

1. Formation of morula : The fertilized zygote divides. It undergoes successive quick mitotic cell divisions called cleavage. First cleavage is holoblastic, unequal and meridional. It divides the ovum completely into two unequal blastomeres. The plane of cleavage passes through animal vegetal axis, i.e., it is meridional. Large blastomere divides little earlier than the small one giving a transitional three cell stage. As a result of further cleavages, a solid mass of mulberry-shaped embryo is formed called morula.
Morula is of the size of zygote but consists of 32 cells. These cells are of two types : The outer layer of smaller cells around, inner larger cells. Within 72 hours of fertilization, morula reaches the uterus.

2. Formation of blastula : Transformation of morula into a blastula starts by the rearrangement of blastomeres. This leads to the formation of a central cavity inside the morula. The outer cells of morula absorb the nutritive fluid secreted by the uterine mjueous membrane and transform into trophoblast. The fluid absorbed by these cells collects in the central cavity called blastocoel. This cavity separates the trophoblast from the inner mass of larger cells except on one side and is termed blastocyst. The inner cell mass is pushed to one pole as a small knob. This knob gives rise to the embryo and is termed as embryonal knob.

3. Formation of gastrula : In this embryonic stage development of three germ layer occurs. In this stage morphogenic movement of cells of embryo occurs, as a result of this three germ layers are formed. A cavity develops at the centre called as archenteron which open outside through blastopore.

4. Formation of three germ layers :

Formation of endoderm : The enlargement of blastodermic vesicle is followed by the separation of few cells from the inner cell mass. These cells push out from the blastocoel to become the initial cells of the innermost layer of gastrula, the pattern of a tube within a tube. The inner tube is called primitive gut. It is differentiated into gut tract which is within the body and a yolk sac that communicates with the gut of the embryo. The remaining cells of the inner Cell mass are organized to form the embryonic disc.

Formation of mesoderm : After the formation of endoderm an increased rate of cell proliferation takes place at the caudal end of the embryonic disc. This results in the localised increase in the thickness of the disc. These cells subsequently get detached from the embryonic disc and get organized to a well demarcated mesodermal layer.
Formation of ectoderm: After the formation of endoderm and mesoderm, the remain¬ing cells of the embryonic disc arrange themselves in a layer to form ectoderm.

Fate of three germinal layer : Three germ layers differentiated at the time of gastrula- tion give rise to all the tissues and organs of the body of adult by the process of differentia¬tion and organogenesis.
Germ layers

1. Germ layers : Ectoderm
Tissue or organs of adult :

• Epidermis and its derivatives.
• Cutaneous sensory organs.
• Olfactory organs.
• Lens of eye.
• Membranous labyrinth (internal ear).
• Anterior lobe of pituitary gland.
• Lining of buccal cavity.
• Enamel of teeth.
• Endoderm
• Lining of proctodaeum.
• Brain, spinal cord, spinal and cranial nerves.

2. Germ layer : Endoderm
Tissue or organs of adult :

• Lining of alimentary canal except buccal cavity and rectum.
• Lining of larynx, trachea and lungs.
• Urinary bladder.
• Parathyroid, thyroid and thymus.
• Liver and pancreas.

3. Germ layer : Mesoderm
Tissue or organs of adult :

• Dermis.
• Vertebral column.
• Muscles of the body.
• Excretory organs.
• Reproductive tracts.
• Peritoneum.
• Circulatory system, heart, blood vessels, blood, lymph and spleen.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 16 Environmental Issues

Environmental Issues NCERT Textbook Question and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage are the waste originating from the kitchen, toilet, laundry and other sources. They contain impurities such as suspended solid (sand, salt, clay), colloidal materials (faecal matters, bacteria, plastic and cloth fiber), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia) and disease-causing microbes.

When organic wastes from the sewage enter the water bodies, they serve as a food source for microorganisms such as algae and bacteria. As a result, the population of these microorganisms in the water body increases. Here, they utilize most of the dissolved oxygen for their metabolism. This results in an increase in the levels of BOD in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate at home, school or during your trips to other places, could you very easily reduce. Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkin, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at school include waste paper, plastics, vegetable and fruit peels, food wrapping, sewage, etc. Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimized by writing on both sides of the paper and by using recycled paper.

Plastic and glass waste can also be reduced by recycling and re-using. Also, substituting plastics bags with biodegradable jute bags can reduce wastes generated at home, school or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities.

Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because microorganisms do not have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface due to greenhouse effect.

Causes of global warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane and water vapour. These gases trap solar radiation released back by the earth. Global warming is a result of industrialization, burning of fossil fuels, and deforestation.

Effects of global warming: Global warming is defined as an increase in the average temperature of the Earth’s surface. It has been observed that in the past three decades, the average temperature of the Earth has increased by 0-6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control measures for preventing global warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG, etc.
• Reforestation.
• Recycling of materials.

Question 4.
Match the items given in column A and B :

Answer:

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

Question 5.
Write critical notes on the following :
(a) Eutrophication.
(b) Biological magnification.
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication : It is the natural ageing process of lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilizers, and sewage from land which leads to an increased fertility of the lake. As a result it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting into algal blooms. Later, the decomposition of theses algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological magnification : The increase in concentration of harmful non- biodegradable substances into higher tropic level is called biological magnification. DDT used to protect the crops reach the soil and are absorbed by plants with water and minerals from the soil. Due to rain, these chemicals can also enter water sources and into the body of aquatic plants and animals. As a result, chemicals enter the food chain. Since, these chemicals cannot be decomposed, they keep on accumulating at each trophic level. The maximum concentration is accumulated at the top carnivore’s level.

The producers (phytoplankton) were found to have 0 04 ppm concentration of DDT. Since many, types of phytoplankton were eaten by zooplankton (consumers), the concentration of DDT in the bodies of zooplankton as found to be 0-23ppm. Small fish that feed on zooplankton accumulate more DDT in their body. Thus, large fish (top carnivore) that feed on several small fish have the highest concentration of DDT.

(c) Groundwater depletion and ways for its replenishment : The level of ground¬water has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers, etc.

As a result, the source of groundwater is depleting. This is because the amount or groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for replenishing groundwater:

• Preventing over-exploitation of groundwater.
• Optimizing water use and reducing water demand.
• Rainwater harvesting.
• Preventing deforestation and plantation of more trees.

Question 6.
Why ozone hole forms over Antarctica? How will enhanced ultraviolet radiations affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released ‘from Chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s release chlorine atoms by the action of UV rays on them. The release of chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion.
CF₂Cl₂ → CF₂Cl+ Cl
Cl + O₃ → ClO+O₂
ClO + O₂ → Cl+O₂
The formation of the ozone hole will result in an increased concentration of UV-B radiation on the Earth’s surface. UV-B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV-B cause corneal cataract in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.

The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the King of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and the workers went to bishnoi village. There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnoi’s showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

The Chipko movement was started in 1974, in die Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual you would take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:

Measures for preventing air pollution:

• Planting more trees.
• Use of clean and renewable energy sources such as CNG and Bio-fuels.
• Reducing the use of fossil fuels.
• Use of catalytic converters in automobiles.

Measures for preventing water pollution :

• Optimizing the use of water.
• Using kitchen waste water in gardening and other household purposes measures for controlling noise pollution:
  • Avoid burning crackers on Diwali,
  • Plantation of more trees. ‘

Measures for decreasing solid waste generation:

• Segregation of waste.
• Recycling and reuse of plastic and paper.
• Composting of biodegrable kitchen waste.
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes. ‘
(b) Defunct ships and e-wastes.
(c) Municipal solid wastes.
Answer:
(a) Radioactive wastes : Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionizing radiation such as gamma rays. These rays cause mutation in organisms which often results in skin cancer. At high dosage, these rays can be lethal. Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct ships and e-wastes : Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thu$,4hey contribute to solid wastes that are hazardous to health. E-wastes or electronic wastes generally include electronic goods such as computers, etc.

Such wastes are rich in metals such as copper, iron, silicon, gold, etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal solid wastes : Municipal solid wastes are generated from schools, offices, homes and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes and other disease-causing microbes. Hence, it is necessary to dispose municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorized as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi. Various steps have been taken to improve the quality of air in Delhi:

• Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unbumt particles.
• Phasing out of old vehicles.
• Use of unleaded petrol.
• Use of low-sulphur petrol and diesel.
• Use of catalytic converters.
• Application of stringent pollution-level norms for vehicles.
• Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.
• The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of Suspended Particulate Matter (SPM) and Respiratory
• Suspended Particulate Matter (RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Green house gases
(b) Catalytic converter
(c) Ultraviolet-B.
Answer:
(a) Green house gases: The greenhouse effect refers to an overall increases in the average temperature of the earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane and water vapour. When solar radiation reach the Earth, some of these radiation are absorbed. These absorbed radiation are released back into the atmosphere. These radiation are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival. However an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic converter: Catalytic converters are devices fitted in automobiles to reduce vehicular pollution. These devices contain expensive metals such as platinum, palladium and rhodium that act as catalysts. As the vehicular discharge passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas (respectively).

(c) Ultraviolet-B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light. It is a harmful radiation that comes from sunlight and penetrates through the ozone hole on to the earth’s surface. It induces many health hazards in humans. UV-B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV-B cause corneal cataract in human beings.

Environmental Issues Other Important Question and Answers

Environmental Issues Objective Type Questions

1. Choose the Correct Answer:

Question 1.
It is a greenhouse gas:
(a) H₂
(b) CO
(c) CO₂
(d) N₂
Answer:
(c) CO₂

Question 2.
Combustion of fissil fuel resulting in:
(a) SO₂ pollution
(b) NO₂ pollution
(c) N₂O pollution
(d) NO pollution.
Answer:
(a) SO₂ pollution

Question 3.
The causes of greenhouse effect is the increase in the concentration of:
(a) CO₂
(b) CO
(c) O₃
(d) Nitrogen oxide.
Answer:
(a) CO₂

Question 4.
Which of the following is responsible for the reduction of 03 from the atmosphere :
(a) CFC
(b) NO₂
(c) CO₂
(d) SO₂
Answer:
(a) CFC

Question 5.
Which of the following is highly dangerous radioactive pollutant:
(a) Strontium-90
(b) Phosphorus-32
(c) Sulphur-35
(d) Calcium-40.
Answer:
(a) Strontium-90

6. Causes of acid rain is:
(a) SO₂ pollutant
(b) CO
(c) Insecticide pollution
(d) Dust pollution.
Answer:
(a) SO₂ pollutant

Question 7.
Carbon monoxide is the chief pollutant of:
(a) Water
(b) Air
(c) Noise
(d) Soil.
Answer:
(b) Air

Question 8.
Plants are the purifier of air because:
(a) Respiration
(b) Photosynthesis
(c) Transpiration
(d) Drying.
Answer:
(a) Respiration

Question 9.
Bhopal gas tragedy of 1984 took place because methyl isocyanate related with:
(a) DDT
(b) Ammonia
(c) CO₂
(d) Water.
Answer:
(d) Water.

Question 10.
Maximum ozone layer depletion is caused by :
(a) CO₂
(b) CFC
(c) SO₂
(d) CH4.
Answer:
(b) CFC

Question 11.
Minimata disease is caused by the pollution of:
(a) Mercury and Lead
(b) Mercury and Cadmium
(c) Lead and Tin
(d) Lead and Strontium.
Answer:
(b) Mercury and Cadmium

Question 12.
Taj Mahal is affected by :
(a) Air pollution due to Mathura refinery
(b) Decomposition of marble
(c) Pollution of Yamuna
(d) All of the above.
Answer:
(a) Air pollution due to Mathura refinery

Question 13.
Which of the following is not a pollutant:
(a) Thermal energy plant
(b) Automatic vehicles
(c) Nuclear power energy plant
(d) Hydroelectric project.
Answer:
(d) Hydroelectric project.

Question 14.
Green house effect is increase in temperature of earth due to :
(a) High concentration of NO₂
(b) High concentration of SO₂
(c) High concentration of CO₂
(d) High concentration of CO.
Answer:
(c) High concentration of CO₂

Question 15.
Green house gases are:
(a) CO₂, CFC, CH_4, NO₂
(b) CO₂, O₂, N₂, NO₂, NH₃
(c) CH₄, N2, CO₂, NH₃
(d) CFC, CO_2, NH₃, H₂
Answer:
(d) CFC, CO₂, NH₃, H₂.

Question 16.
Which of the following is an indicator of water pollutant:
(a) BOD
(b) COD
(c) E. coli
(d) All of the above
Answer:
(a) BOD

Question 17.
Cause of water pollution is :
(a) 2,4-D and insecticides
(b) Smoke
(c) Automatic engines
(d) Aeroplanes.
Answer:
(a) 2,4-D and insecticides

Question 18.
Polluted water is treated by :
(a) Lichen
(b) Fungi
(c) Fern
(d) Phytoplankton
Answer:
(d) Phytoplankton

Question 19.
Which country produce maximum green house gas:
(a) India
(b) Brazil
(c) USA
(d) France.
Answer:
(c) USA

Question 20.
Which are not affected by acid rain :
(a) Lithosphere
(b) Plants
(c) Ozone layer
(d) Animals.
Answer:
(c) Ozone layer

Question 21.
Ozone hole causes:
(a) Global warming
(b) Reduction in the rate of photosynthesis
(c) More UV rays come to earth
(d) All of the above.
Answer:
(c) More UV rays come to earth

Question 22.
Highly polluted river of India is :
(a) Ganga
(b) Yamuna
(c) Gomti
(d) Hugh.
Answer:
(a) Ganga

Question 23.
Effect of pollution is forst marked on:
(a) Microorganism
(b) Vegetation
(c) Food crops
(d) None of the above
Answer:
(b) Vegetation

Question 24.
Ozone hole enhances:
(a) UV-radiations reaching earth
(b) Number of characters
(c) Skin cancer
(d) All of the above.
Answer:
(d) All of the above.

2. Fill in the Blanks :

1. …………………… is everything that is around us.
2. …………………. Nature of environmental study is
3. …………………. is the highest layer of the earth.
4. An interacting group of various species in a common location is called……………………..
5. …………………….. is a layer or a set of layers of gases surrounding a planet.
6. ……………………….. is electromagnetic radiation within a certain portion of the electromagnetic spectrum.
7. The is defined as a vertical section of the soil. ……………………….
8. …………………… water is available in plants easily.
9. ……………………. is essential for respiration of animals which are found in the soil.
10. …………………… is essential part of the earth for continuity of life.
11. …………………… gas is essential for life.
12. …………………….. fish is used to eat the larva of mosquitoes.
13. Quantity of atmospheric C02 is …………………… by forestation.
14. Soil pollution is caused by ……………………..
15. ………………….. rays are absorbed by ozone layer.

Answers:

1. Environment
2. Multidisciplinary
3. Biosphere
4. Bio-community
5. Atmosphere
6. Light
7. Soil profile
8. Capillary
9. Oxygen
10. Death
11. Oxygen
12. Gambusia
13. Balanced
14. Chemical fertilizers
15. Ultraviolet rays.

3. Match the Following :
(i)

Answer:

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

Answer:

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

Environmental Issues Very Short Answer Type Questions

Question 1.
The surroundings in which a person, animal or plant lives are called.
Answer:
Environment.

Question 2.
When two individuals can produce fertile offsprings, typically by sexual reproduction is called.
Answer:
Species.

Question 3.
What called, a biological community of interacting organisms and their physical environment?
Answer:
Ecosystem.

Question 4.
Name the lower part of the atmosphere which has found more than 90% gases.
Answer:
Troposphere,

Question 5.
Rays of light which are low wavelength from purple colour rays?
Answer:
Ultraviolet rays.

Question 6.
Light effect of flowering in plants.
Answer:
Photoperiodism.

Question 7.
Smallest particle which are found in soil.
Answer:
Clay.

Question 8.
Water which are absorbed from the atmosphere and held very tightly by the soil particles.
Answer:
Hygroscopic water.

Question 9.
Plants are one marked by short life-cycle.
Answer:
Ephemeral plant.

Question 10.
When an animal or plant resembles another creature or in animate object?
Answer:
Mimicry.

Question 11.
Expand BOD.
Answer:
Biological Oxygen Demand.

Question 12.
Name two gases which are caused air pollution.
Answer:
SO₂ and CO₂.

Question 13.
How much is hearing capacity of man?
Answer:
10 to 12 decibels.

Question 14.
Write the name of highly polluted river of India.
Ans.
Ganga.

Question 15.
What is called the increase in atmospheric temperature due to increased C02 concentration?
Answer:
Greenhouse effect.

Question 16.
Write the full form of DDT.
Answer:
Dichloro Diphenyl Trichloroethane.

Question 17.
How much quantity of CO₂ in atmosphere.
Answer:
0 03%.

Question 18.
Smoke which is caused eye irritation.
Answer:
No.

Question 19.
When the World Environment Day is celebrated?
Answer:
5th June.

Question 20.
Write the full form of PAN.
Answer:
Peroxy Acetyle Nitrate.

Question 21.
Write the full form of CFCs.
Answer:
Chlorofluourocarbons.

Question 22.
How much percent found of CO₂ in atmospheric temperature?
Answer:
60 %.

Environmental Issues Short Answer Type Questions

Question 1.
What is the definition of pollution?
Answer:
Pollution is defined as an undesirable change in physical, chemical or biological characteristics of air, land, water and soil.

Question 2.
What are the cause of air pollution?
Answer:
The causes of air pollution: The main causes of air pollutions are fossil fuels industries, factories and particulate matter are produced due to combustion of petrol, diesel, kerosene etc. in vehicles, houses and factories which pollute air.

In thermal power plants, steel and glass industries, paper and sugar mills etc. combustion of coal and furnace oil produce carbon monoxide, carbon dioxide, sulphur dioxide, ash, dust particles and some heavy metals which are released into atmosphere. Similarly various types of chemicals released from cloth mills, cement industries, asbesters industries, pesticides, industries etc. also causes air pollution.

Question 3.
What is acid rain? Write any two effects of acid rain on human beings.
Ans.
Acid rain: SO₂ gas is released into the environment by burning of fossil fuels containing sulphide from industries. e.g., Coal. In presence of moisture SO₂ reacts with water and forms the droplets of sulphurous and sulphuric acid in the environment. Likewise nitrogen oxides released from motor vehicles, burning materials and chemical industries . also react with water to produce the droplets of nitric acid. The droplets of sulphurous, sulphuric acid and nitric acids fall down on the earth with rainwater and thus, it is known as acid rain.
SO₂ + \(\frac { 1 }{ 2 }\)O₂ → SO₃
H₂O +SO₃ →H₂SO₄
H₂O + NO₂ → HNO₃
Effects on human beings: Skin diseases, irritation, metabolic diseases.

Question 4.
Write short note on pollution caused due to combustion activities.
Answer:
Combustion of fuel, such as wood, coal, natural gas for cooking and for some other purposes gives out gases like CO₂, CO, SO₂ and consumes oxygen from the air.

Question 5.
Write the effect of air pollution in plants.
Answer:
Effect of air pollution on plants :

• Increase in the concentration of SO₂ causes chlorosis of leaves in plants.
• The cells and chlorophyll of leaves are degraded and fall down.
• Vegetative and reproductive growth is inhibited.
• The development of plant is inhibited.

Question 6.
Write preventive measures to control air pollution.
Answer:
Measures for the control of air pollution :

• Industrial smokes must be filtered before releasing it into the atmosphere.
• Tree plantation should be increased and deforestation prevented.
• Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
• The use of crude fuels should be avoided and use of high quality fuels should be recommended.
• Nuclear explosions should be avoided.
• Legal control of air pollution.
• Plantation of air purifying plants.
• Development of parks and gardens in cities.

Question 7.
What is global warming? List four strategies for reducing global warming.
Answer:
Increase in the level of green house gases in the atmosphere causes the rise in global mean temperature called global warming. Four strategies for reducing global warming are:

1. Reducing deforestation.
2. Planting trees.
3. Slowing down the growth of human population.
4. Reduction in the emission of greenhouse gases.

Question 8.
Write any four measures to control green house effect
Answer:
Following important measures can be sited as the base steps towards controlling greenhouse effect:

• The aim is achieved to some extent by reducing the consumption of fossil fuels such as coal and petroleum. This can be achieved by depending more on non-conventional renewable sources of energy such as solar, wind, tidal, biogas and nuclear energies.
• Disposing of the greenhouse gases as they are formed elsewhere than in the atmosphere.
• By recovering greenhouse gases present already in the atmosphere and disposing them off elsewhere.
• Preventation of deforestation and planting of more trees.

Question 9.
Write short note :

1. Bio-magnification.
2. UV rays.
3. Bio digradable pollutants.
4. Non-biodegradable.

Answer:
1. Bio-magnification : Bio-magnification, also known as bio-amplification or biological magnification is the increasing concentration of a substance, such as a toxic chemical, in the tissues of organisms at successively higher levels in a food chain.

2. UV rays : Sunlight is the chief source of ultraviolet rays. The rays of sunlight having a wavelength range from 200 to 390 nm are called as UV rays.

Effect of UV rays: They have a direct effect on living cells. DNA and other chemical substances in cells are inactivated due to UV rays. Prevention of replication of the DNA molecule and its distortion cause many ill effects.

3. Non-biodegradable: The pollutants which cannot be purified by natural methods are called non-biodegradable pollutants. Plastic products, many chemicals, DDT, long chain detergents, glass aluminium, mercury salt and other synthetic products manufactured by man come under this category. These non-biodegradable pollutants not only accumulate but often “biologically magnified” as they move in biochemical cycles and along food chains.

4. Biodegradable: The pollutants which are degraded by natural factors and decomposed by natural activities are known as biodegradable pollutants, e.g., Domestic sewage, heat. The domestic sewage can be rapidly decomposed by natural processes or in engineered system (sewage treatment plant) that enhance natures great capacity to decompose and recycle.

Question 10.
What is the effect of SOz in the Environment?
Answer:

Question 11.
Write only the sources of water and air pollution.
Answer:
1. Sources of Water pollution: Following are the chief sources of water pollution:
(a) Human sources :

• Domestic sewage,
• Industrial effluents,
• Agricultural wastes,
• Oil pollution,
• Thermal and nuclear power station.

(b) Natural sources: Water pollution takes place by natural ways like soil erosion, mixing of metallic substances, plant leaves, humus and faecal matter of animals etc.

2. Sources of Air pollution: Following are the chief sources of air pollution:
(a) Human sources:

• Combustion activities,
• Industrial activities,
• Agricultural works,
• Use of solvents,
• Activities concerned with atomic energy.

(b) Natural sources: Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

Question 12.
Write a note on the effect of water pollution on aquatic organisms.
Answer:
Effect of water pollution on aquatic plants:

• Inorganic nitrates and phosphates in excess amount stimulates excessive plant growth in lakes and reservoirs. These plants deplete the oxygen contents of the water during night. This leads to suffocation of fish and other aquatic life.
• The rapid algal growth leads to the diminishing of nutrient in the medium causing rapid decay of algal filament. This increases productivity of lake and stream water brought about by nutrient enrichment is called eutrophication.
• The number of microorganisms increases in polluted water.
• Siltation occurs in water.
• Temperature of water increases and 02 ratio is reduced.

Effect of water pollution on aquatic animals: Animal life depends on aquatic plants. Aquatic animals are affected from water pollution as follows :

• Decrease in the value of B.O.D. resulting in the number of aquantic animals.
• Animals found in freshwater and killed.
• Animal diversity is also decreased and fishes suffer from various types of diseases.
• Terrestrial animals are also affected by the use of polluted water.

Question 13.
Write the measures for control and prevention of water pollution.
Answer:
Measures for control and prevention of water pollution are:
Preventive measures:

• Use of harmful pesticides and weedicides must be stopped completely.
• Discharge of effluents into rivers, lakes and sea should be strictly prohibited without treatment.
• Oil spill should be prevented.
• Proper disposal of sewage so, that it does not find its way into water bodies.
• Preventing bathing, washing cloths, throwing dead bodies and other wastes into water source.

Curative measures:
1. Adequate waste water treatment: The domestic sewage and the industrial waste should be properly treated before its disposal into water ways.
Waste water treatment involves three steps :

(a) Primary treatment: During this treatment large objects and suspended undissolved solids are removed and converted into the sludge, a valuable fertilizer.

(b) Secondary treatment: Aeration is supplied to promote bacterial decomposition, followed by chlorination to reduce its content of bacteria.

(c) Tertiary treatment: During this phase nitrates and phosphates are removed. The treated water is then released. Sewage treatment is quite expensive and in many developing countries only the first two steps are followed.

2. Treatment of industrial effluents: Industrial effluents should be properly treated to remove the pollutants. These involve neutralization of acids and alkalies, removal of harmful chemicals, coagulation of colloidal impurities, precipitation of metallic compounds and reducing the temperature of wastes to decrease thermal pollution. Chemical oxidation can be achieved by chlorination through reaction with ozone. However, there are certain chemicals which are difficult to remove.

3. Recycling: One of the best methods of prevention and control of water pollution is the recycling of the various kinds of pollutants and wastes, e.g., Dung of cow and buffalo can be used for the production of gobar gas a cheap source of fuel and also as manure.

Question 14.
Write short notes on air pollution caused due to industries.
Answer:
Industrial pollution is caused by industrial pollutants which are released by the industries such as SO_2, CO₂ , CO, H₂S and hydrocarbons together with dust and smoke. These are produced by the burning of coal and petroleum.

The chemical industries releases HCl, Cl₂, nitrogen oxide, oxides of copper, zinc, lead, arsenic etc. Industrial pollutants causes air pollution and water pollution.

Question 15.
Write preventive measures of sound pollution.
Answer:
Preventive measures of noise pollution:

• Noise pollution can be controlled by designing quieter machines, proper lubrication and better maintenance of machines. Noise producing parts of machines may be covered by suitable insulating materials.
• Noise producing industries should be installed away from residential areas.
• Use of public addressing systems should be restricted at a fixed intensity and hours of the day.
• Stuffing of cotton plugs in the ear or use of ear-muffs and minimize the danger of occupational exposure of noise.
• Trees absorb sound vibrations and thus, reduce the noise pollution. Thus, plantation of trees along highways, establishment of parks help in reducing noise pollution.
• By use of silencer in all automobiles and engines.

Question 16.
Write the effects of noise pollution.
Answer:
Effects of Noise pollution :

• The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
• Interference with speech communication.
• Noise pollution leads to neurosis, anxiety hypertension, cardiovascular disease, hepatic stress, giddiness.
• Annoyance leading to ill-temper, bickering, mental disorientation and violent behaviour.
• The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be short into blood stream with conse¬quence rise in blood pressure. This ultimately results in tension and nervousness.
• Affects different metabolic activities.

Question 17.
How many types of pollution are there ? Explain in brief.
Answer:
Pollution are of following five types :

1. Air pollution,
2. Noise pollution,
3. Water pollution,
4. Radioactive pollution,
5. Soil pollution.

1. Air pollution: Pyrotechnics’ results in the release of harmful gases like CO₂, CO and SO₂ in the environment which produces various types of diseases in living beings.

2. Noise pollution: Pyrotechnics produces the unwanted sound dumped into the atmosphere leading to health hazards.

3. Water pollution: Pyrotechnics produces some waste materials, on reaching water it causes water pollution.

4. Radioactive pollution: Pollution created due to radioactive substance is called radioactive pollution. Atomic energy centres and bombarding also causes radioactive pollution.

5. Soil pollution: Any change in physical and chemical characteristic of soil due to addition of unwanted substance which adversely affects productivity of soil it termed as soil pollution.

Environmental Issues Long Answer Type Questions

Question 1.
Describe the diseases caused due to radioactive pollution.
Answer:
Following diseases are caused by radioactive pollution:

• Leukaemia and bone cancer: In human beings and other animals, radioactive pollution causes blood and bone cancer.
• Ageing: The reproductive capacity of organisms is decreased due to radioactive
  pollution and ultimately causes ageing.
• Epidemic disease: Radioactive pollution causes the decreased rate of antitoxin production which affects the immune system and results in the production of epidemic diseases.
• Nervous system and sensory cells become irritated.
• Radioactive pollution results skin cancer.
• Mutations also take place due to radioactive pollution.
• Thyroid cancer is also produced due to radioactive iodine.

Question 2.
Burning of crackers spread pollution in the environment. Explain.
Answer:
Explosive substances are used in crackers. Crackers are burned for celebration but it causes many harms to the living organisms. Some of the harm effects of burning of crackers are as follows :

1. Air pollution: It releases harmful gases such as CO₂, CO, SO₂ etc. into the atmosphere. Which causes respiratory diseases.

2. Noise pollution: Burning of crackers produces noise which causes noise pollution which effects our hearing capacity, causes headache, increases heartbeat and blood pressure.

3. Water pollution: Waste of burning of crackers are drained into the pond, river with rainwater or through drain which causes harm to aquatic organisms.

4. Soil pollution: Burning of crackers not only produces smoke of harmful gases but its harmful chemicals mixes with soil and pollute it, which makes the soil sterile and reduced growth of plants.

Question 3.
Write an essay on air pollution.
Answer:
Air pollution : Any change in the composition of air is called pollution. The chief reason of air pollution is the releasing of harmful gas in the environment that affectdur eyes, lungs, skins, heart and brain and’produce various disorders in them. Sources of Air polution:

1. Sources of Water pollution: Following are the chief sources of water pollution:
(a) Human sources :

• Domestic sewage,
• Industrial effluents,
• Agricultural wastes,
• Oil pollution,
• Thermal and nuclear power station.

(b) Natural sources: Water pollution takes place by natural ways like soil erosion, mixing of metallic substances, plant leaves, humus and faecal matter of animals etc.

2. Sources of Air pollution: Following are the chief sources of air pollution:
(a) Human sources:

• Combustion activities,
• Industrial activities,
• Agricultural works,
• Use of solvents,
• Activities concerned with atomic energy.

(b) Natural sources: Volcano and its lava, ash, dust, smoke of forest fire, winds, cyclone. Decomposition of matters in the swamp water and liberation of methane gas and different compounds of hydrogen from forests plants, various pollen grains etc.

(A) Effects of air pollution :

• Air pollutant like SO₂ enter the soft tissues causing drying of the mouth, scratching throat and smarting eyes.
• Hydrocarbons and many other pollutants are responsible for causing cancer.
• Oxides of carbon, sulphur and nitrogen diffuse into the blood stream to combine with haemoglobin causing reduction in its oxygen carrying capacity. CO severely damages cardiovascular system and disturbs psychometry functions.
• P.A.N. inhibits Hill reaction and thus, decreases photosynthetic production of an ecosystem.

Control of Air pollution:  Measures for the control of air pollution :

• Industrial smokes must be filtered before releasing it into the atmosphere.
• Tree plantation should be increased and deforestation prevented.
• Use of automobiles should be minimized which reduce the nitrogen contents in the atmosphere.
• The use of crude fuels should be avoided and use of high quality fuels should be recommended.
• Nuclear explosions should be avoided.
• Legal control of air pollution.
• Plantation of air purifying plants.
• Development of parks and gardens in cities.

Question 4.
Name the major air pollutants and their individual effects.
Answer:
Major air pollutants and their individual effects are as follows :

1. Carbon monoxide: It is highly poisonous gas. On entering blood stream, it combines with hemoglobin to block oxygen transport function. It causes laziness, headache, disturbance of psychometry functions, decrease in visual perception, serious effects on cardiovascular system etc.

2. Sulphur dioxide: The animals and human population when exposed to SO₂, suffer from respiratory diseases. It causes chest constriction, headache, vomiting and death from respiratory ailments.

3. Hydrogen sulphide: Its rotten egg like smell cause nausea, irritation in eyes and throat.

4. Nitrogen oxide: It inhibits cilia action so that soot and dust penetrate far into the lungs and finally resulting respiratory diseases in human beings and animals. In plants, it causing chlorosis.

5. Aerosoles: These are the chemicals affecting ozone layer of the atmosphere due to which UV rays enter in the atmosphere and causing various diseases in plants and animals.

6. Ammonia : It inflame upper respiratory passage in human beings. In plants, it inhibits seed germination, destruction of chloroplast and inhibition of the growth of roots and shoots.

7. Hydrogen chloride: It affects the leaves of the plants, eyes, respiratory organs of animals and human beings.

8. Hydrocarbons: It causes yellowing of plants, drying of buds, whereas in man and other animals, mucous glands of eyes and nose are irritated.

Question 5.
Write down the methods to control soil pollution.
Answer:
Control of Soil pollution: By following methods soil pollution can be prevented:

• Government should make provisions for latrines and people should not be allowed to litter in open fields.
• Solid wastes such as tin, copper, iron, glass etc., should not be dumped into soil.
• The solid wastes should be recycled to form new materials.
• Fertilizers and pesticides should be judiciously used.
• Instead of pesticides to kill pests, biological control methods should be adopted-*
• The excretory waste of man an’d cattle should be used to prepare biogas.
• Clean and well covered dustbins should be used in villages and cities.
• Closed hollow pipes should be used to collect and discharge liquid wastes.
• To prevent soil erosion, grass and small plants should be grown.
• Soil management should be adopted.

Question 6.
What are the sources of soil pollution?
Answer:
Sources of soil pollution:

• Acid rainwater and water from mines are main sources of soil pollution.
• Mixing of debris, waste products with the soil cause soil pollution.
• By excessive use of pesticides and fertilizers, pollute the soil.
• When industrial wastes are discharged in the soil, they also pollute the soil.
• Heavy metals e.g., Cadmium, Zinc, Nickel, Arsenic etc., are mixed with the soil
  from mines. These metals are harmful for plants as well as primary and secondary consumers, in a food chain. ’
• Bones of dead animals, paper putrified flesh and food, iron, copper, mercury etc., pollute the soil.
• In our villages and rural areas, where there are no latrines, people go in fields for being at ease. From their stool also the soil is polluted.
• Insecticides like D.D.T. is very dangerous substance. When these enter the body of consumers from producers, their concentration is increased because these are non-degradable substances. Moreover, these can remain in the atmosphere for upto 15.years.

Question 7.
Explain non-ionising and ionising radiation.
Answer:
Types of Radiations on the basis of their action on cells, radiations are of two types:
1. Non-ionising radiations: These include ultraviolet rays (UV rays; 100-300 nm), these have low penetration capacity. These affect only those cells which absorb them. The cause are as following :

• Sunburn: It includes rupturing of sub-cutaneous blood capillaries, blisters, reddening of skin and injury to stratum germinativum.
• Snow blindness: It damages eyesight due to damage of corneal cells.
• Inactivation of organic bio-molecules arid formation of thymin dimer in DNA.
• However they cause cancer, tumor, skin disease, etc.

2. Ionizing radiations : These include X-rays, cosmic rays and atomic radiations. These radiations have low wavelengths but high penetration power. These damage the living cells shifting the electron from one to other bio-molecule. Their harmful effects are as follows :

• Short range effects: These may appear within few days or a few weeks after exposure. The effects include loss of hair, nails, subcutaneous bleeding, metabolic changes, change in proportion of blood cells, dead tissues or death of the organisms inliigh dose.
• Long range effects: These appear in several months or even several years after their exposure. These cause tumours, cancers, mutations, genetic deformities, shorter life span and developmental defects.

Question 8.
What are sources of noise ’ pollution? Write any four effects of noise pollution.
Answer:
Sources of noise pollution: Noise is either natural such as thunder or man-made. The common sources of noise pollution are :

• Industries such as textile mills, printing-press, engineering establishments, etc.
• Defence equipment and vehicles such as tanks, artillery and rocket launching explosions, practise firing, etc.
• Transport vehicles as trains, trucks, buses, two-wheelers, jet planes, etc. mid accessory noise produced by horns, sirens.
• Other applications as dynamite blasting, use of jack hammers, pile drivers, bulldozer, lawn mowers, etc.

Effects of noise pollution:

• The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
• Interference with speech communication.
• Noise pollution leads to neurosis, anxiety, hypertension, cardiovascular disease, hepatic stress, giddiness.
• Annoyance leading to ill-temper, bickering, mental disorientation and violent behaviour.
• The high intensity of noise pollution can cause blood vessels to contract, skin becomes pale, muscles to contract and adrenaline to be shot into blood stream with consequence rise in blood pressure. This ultimately results in tension and nervousness.
• Affects different metabolic activities.

Question 9.
What is Green house effect? Write any four impacts of Green house effect
Answer:
Green house effect: The Green house effect may therefore, be defined as “The progressive warming up of the earth’s surface due to blanketting effect of man-made CO₂ in the atmosphere. It means the excessive presence of those gases blocked in the infrared radiation from earth’s surface to the atmosphere leading to an increase in temperature, which in turn would make life difficult on earth to forthcoming future generations.
Four impacts of Green house effects are :

1. Effect on global climate : The climate of earth has never been free of change. Its composition, temperature and self-cleansing ability have all varied since, the planet first formed. Yet the pace in the past two countries has been remarkable. The atmosphere’s composition in particular has changed significantly faster than it has at any time in human history.

Increase in Green house gases does not increase die temperature of the earth i.e„ it is not uniform. Temperature is normal at poles and is very less at tropics. In Iceland, Greenland, Sweden, Norway, Finland, Alaska, Siberia, temperature is more, thus at poles, ice starts melting.

2. Effect of Green house Effect on forests: When the atmospheric temperature increases, only those plants can survive which can bear high temperature. Because of this, new species of plant also come into existence, Herbaceous plant can not survive in this high temperature. Plants with high wood Will increase in number. According to a survey if the a nount of CO₂ in the atmosphere is doubled, there will be a great decrease in a green biomass.

3. Effect on crops: The positive aspect of greenhouse effect will be on agriculture. Because, CO₂ is a natural fertilizer, the plants will grow larger and faster with increasing CO₂ in the atmosphere. At first sight, the abnormal growth of plants might be expected to be beneficial because the yields of major crops might increase, however chances of depletion of characters and productivity of soil are also associated with this.

4. Effect on ozone layer : During depletion, the chlorine, fluorine, or bromine molecules of CFCs and halogens are converted into reactive free radical form by photochemical reactions from their intial non-reactive sites. Oxides of nitrogen generally inactivate Cl but the lowering of stratospheric temperature changes. NO₂ into non-reactive ritric acid. Thus, Cl or F are free to react with ozone, distintegrating it into O₂ + O.

The tiny ice particles during winter favours the conversion of chlorine into chlorine monoxide, which behaves as a catalytic compound. The abundance of chloromonoxide rich air in stratosphere continues to rise. Chlorine monoxide react with nascent oxygen and thereby gets converted into chlorine. Thus, the cycle continues destroying the ozone-level.

MP Board Class 12th Biology Solutions

MP Board Class 12th Biology Solutions Chapter 2 Sexual Reproduction in Flowering Plants

Sexual Reproduction in Flowering Plants NCERT Textbook Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Answer:
Development of male gametophyte takes place in anther and female gametophyte in ovary.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structure formed at the end of these two events.
Answer:
Differences between Microsporogenesis and Megasporogenesis:

Meiosis Occurs during micro and megasporogenesis microspores (Pollen grain) are formed at the end of micro sporogenesis and female gametophyte (enibro sac) are formed at the end of megasporogenesis.

Question 3.
Arrange the following terms in the correct developmental sequence:
Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male garnets.
Answer:
Sporogenous tissue → Pollen mother cell → Microspore → Tetrad → Pollen grain → Male garnets.

Question 4.
With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.
Answer:
Structure of Ovule: Each ovule consists of the following parts as visible in a longitudinal section:

• • A small stalk or funicle by which the ovule remains attached with the placenta of the
    ovary.
  • Hilum is the point at which it is attached with the ovule. In inverted ovule the funicle fuses with the main body of the ovule and is called as raphe.
  • The ovule is surrounded on all sides by two integuments but not at the apex where an aperture called micropyle is present. This end of the ovule is called as micropylar, while the end of the ovule opposite to it is called as chalazal end.
  • Embryo sac is situated inside the nucellus.
  • Towards the micropyle end of embryo sac one egg or oospore and two synergids are found, and towards the chalaza end of embryo sac three antipodal cells are found. At the centre secondary nuclei is found.
    

Question 5.
What is mean by monosporic development of a female gametophyte ?
Answer:
Out of theTour megaspores, three degenerate and only one remains funtional which develops in, to a female gametophyte or embryo sac. This is called monosporic development, i.e., when embryosac develops from one single megaspore it is called monosporic embryo sac.

Question 6.
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
Explanation: Nucleus of the functional megaspore undergoes mitosis resulting in 2-nuclei that move to two opposite poles forming 2-nucleate embryo sac.
Two more mitotic nuclear divisions result in 4-nucleate and later 8-nucleate stages of the embryo sac. so, far no cytokinesis (cytoplasmic division) has taken place.
Now cell walls start to build leading to the organisation of female gametophyte or embryo sac.

Six of the 8-nuclei are bound by cell wall and the remaining 2-celled polar nuclei lie below the egg apparatus in the large central cell.
Seven-celled stage
Three cells are grouped together at the micropylar end and constitute the egg appara¬tus, which is constituted of two synergids and one egg cell.
Three cells at the chalazal end are called antipodals. The large central cell has 2-polar nuclei.
Thus, a typical angiosperm embryo sac at maturity is 7-celled but 8-nucleated as the central cell has 2-nuclei

Question 7.
What are chasmogamous flowers? Can cross pollination occur in cleistogamous flower ? Give resion for your answer.
Answer:
Chasmogamous flowers are open flowers with exposed stamens and stigma which facilitate cross pollination.
No cross pollination occurs in cleistogamous flowers. As these flowers are closed and never open and thus no transfer of pollen from outside to stigma of the flower is possible. So there are no occurs cross pollination.

Question 8.
Mention two strategies evolved to prevent self-pollination in flowers.
Answer:
Two strategies evolved to prevent self-pollination are :
(i) Pollen release and stigma receptivity is not synchronized.
(ii) Anthers and stigma are placed at such positions that pollen doesn’t reach stigma.

Question 9.
What is self-incompatibility ? Why does self-pollination not lead to seed forma¬tion in self-incompatible species ?
Answer:
Self-incompatibility is a genetic mechanism to prevent self pollen from fertilizing the ovules by inhibiting pollen germination or pollen tube growth in the pistil.
In these cases, self-pollination does not lead to seed formation because fertilization is inhibited.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Answer:
It is the covering of female plant with butter paper or polythene to avoid their contamination from foreign pollens during breeding programme.

Question 11.
What is triple fusion? Where and how does it take place ? Name the nuclei involved in triple fusion?
Answer:
Triple fusion refers to the process of fusion of three haploid nuclei. It takes place in the embryo sac.
The 3-nuclei that fuse together are, nucleus of the male gamete and 2-polar nuclei of the central cell to produce a triploid primary endosperm nucleus.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilized ovule?
Answer:
The zygote is dormant in fertilized ovule for sometime because at this time, endosperm needs to develop. As endosperm is the source of nutrition for the developing embryo, the nature ensures the formation of enouth endosperm tissue before starting the process of embryogenesis.

Question 13.
Differentiate between:
(a) Hypocotyl and Epicotyl
(b) Colcoptile and Coleorhiza
(c) Integument and Testa
(d) Perisperm and Pericarp.
Answer:
(a) Differences between Hypocotyl and Epicotyl:

(b) Differences between Coleoptile and Coleorhiza:

(c) Differences Between Integement and Testa:

(d) Differences between Peris perm and Pericarp:

Question 14.
Why is apple called a false fruit? Which parts of the flower forms the fruit?
Answer:
Apple is called a false fruit because it develops from the thalamus instead of ovary (thalamus is the enlarged structure at the the base of the flower).

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Answer:
Emasculation means removal of anthers, with a forceps, from the flower bud before dehiscence.
Plant breeder employs this technique to prevent contamination of stigma with the undesired pollen. This is useful in artificial hybridisation, where desired pollen is required.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?
Answer:
Oranges, lemons, lichis could be potential fruits for inducing the parthenocarpy because seedless variety of these fruits would be much appreciated by the consumers.

Question 17.
Explain the role of tape turn in the formation of pollengrain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It produces the exine layer of the pollen grains, which is composed of the sporopollenin, the most resistant fatty, substance.
During microsporongenesis, the cells of tapetum produce various enzymes, hormones, amino acids and other nutritious material required for the development of pollen grains.

Question 18.
What is apomixis and what is its importance?
Answer:
Apomixis is the process of asexual production of seeds, without fertilisation.
The plants that grow from these seeds are identical to the mother plant.

Uses :

• It is a cost effective method for producing seeds.
• It has great use for plant breeding when specific traits of a plant have to be pre¬served.

Sexual Reproduction in Flowering Plants Other Important Questions and Answers

Sexual Reproduction in Flowering Plants Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Fertilization process is:
(a) Fusion of male gamete with egg
(b) Transference of pollengrains to stigma
(c) Fusion of polar nucleus with male gamete
(d) Formation of seed from ovule.
Answer:
(a) Fusion of male gamete with egg

Question 2.
Female gametophy te of angiosperms is:
(a) Megaspore mother cell
(b) Ovule
(c) Endosperm
(d) Nucellus.
Answer:
(c) Endosperm

Question 3.
Effect of nector demonstrate by Xenia word on:
(a) Morphological tissue
(b) Root
(c) Flower
(d) Endosperm.
Answer:
(d) Endosperm.

Question 4.
In a diploid flowering plant, no. of chromosomes are 12, then 6 chromosomes will be present in:
(a) Cotyledons cells
(b) Endospermic cells
(c) Synergids cells
(d) Leaf cells.
Answer:
(c) Synergids cells

Question 5.
Seed coats of seed after fertilization develop from:
(a) Integuments
(b) Endosperms
(c) Chalaza
(d) Ovule.
Answer:
(a) Integuments

Question 6.
Perispermis:
(a) Degenerative secondary nucleus
(b)Apomixis
(c) Peripheral part of endosperm
(d) Parthenogenesis.
Answer:
(b)Apomixis

Question 7.
Development of ovary into fruit from ovule without fertilization, the process is called:
(a) Parthenio ovule
(b) Apomixis
(c) Pathenocarpy
(d) Parthenogenesis.
Answer:
(c) Pathenocarpy

Question 8.
Mitotic cell division in an ovule takes place in:
(a) Nucellus
(b) Megaspore mother cell
(c) Macrosporangia
(d) Archisporium.
Answer:
(b) Megaspore mother cell

Question 9.
Development of sporophyte without the fusion of a gamete is called:
(a) Apomixis
(b) Apospory
(c) Apogamy
(d) Pollination.
Answer:
(c) Apogamy

Question 10.
Male gametes are found in angiospermic plants :
(a) In pollengrain
(b) In anther
(c) In ovule
(d) All of these
Answer:
(a) In pollengrain

Question 11.
The function of pollentube is:
(a) Help in pollination
(b) Protection of stigma
(c) Male gamete carrier
(d) All of these
Answer:
(c) Male gamete carrier

Question 12.
Tapetum is a part of:
(a) Male gametophyte
(b) Female gametophyte
(c) Ovary wall
(d) Anther wall.
Answer:
(d) Anther wall.

Question 13.
Cotyledons is situated in monocots:
(a) Axial
(b) Terminal
(c) Basal
(d) In Any Position
Answer:
(a) Axial

Question 14.
Which type of pollination found in sunflower:
(a) Self pollination
(b) Cross pollination
(c) All of the above
(d) None of these.
Answer:
(c) All of the above

Question 15.
Endospermic nucleus are :
(a) Haploid
(b) Diploid
(c)Triploid
(d) None of these.
Answer:
(b) Diploid

Question 16.
Double fertilization is discoverd by: :
(a) By Navaschin
(b) Leuwenhock
(c) Strasburger
(d) Hofmeister.
Answer:
(a) By Navaschin

Question 17.
The type of pollination in which genetically different pollen grains are brought to
stigma is : (AMU 2012)
(a) Xenogamy
(b) Geitonogamy
(c) Cleistogamy
(d) Dichogamy. ,
Answer:
(a) Xenogamy

Question 18.
If the number of chromosome in root cell is 14 then what will be the number of chromosome in synergid cell of in ovule of that parent (BHU2012)
(a) 14
(b) 21
(c) 7
(d) 28.
Answer:
(c)7

Question 19.
Presence of many embryo (Polyembryony) is characters tic feature of: (BHU2012)
(a) Citrus
(b) Mango
(c) Banana
(d) None of these.
Answer:
(a) Citrus

Question 20.
Sporopollenin is formed by the polymerisation of : (BHU2012)
(a) Fat and Phenol
(b) Carotenoid and Fat
(c) Fat and Ester
(d) Carotenoid and Ester.
Answer:
(a) Fat and Phenol

2. Fill in the Blanks:

1. The main function of endosperm in a seed is the storage of ………………
2. Pollination takes place by the agency of bird is called ………………
3. Apple is an example of fruit. ………………
4. Tissue on which ovules are attached is called ………………
5. ……………… types stamens are found in sunflower.
6. In plants development of fruit takes place from ………………
7. One cotyledon of maize is called ………………
8. Tetradynamous stamens are found in ………………
9. ………………is thin in endospermic seed.
10. In ovule …………….. is joined with funiculus.
11. Anther is ……………… in angiospermic plants.
12. ……………… type of pollination are found in Typha.
13. ……………… pollination is occurs in Salvia.
14. Outer wall of pollen grain is called ………………
15. Process of seedless fruit is called ………………

Answer:

1. Food material
2. Omithophilly
3. False
4. Placenta
5. Syngenecious
6. Ovary
7. Scutellum
8. Mustard
9. Cell wall
10. Hylum
11. Reproductive organ
12. Joint
13. Insect
14. Exine
15. Parthenocarpy.

3. Match the Following:

Answer:

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

4. Answer in One Word/Sentence:

1. Name the four whorls of flower.
2. Give the term for plants which produce flower and fruits many times.
3. Name the outer coat of the seed.
4. Give example of plant in which flowers are pollinated by bat.
5. Give the term used for similar calyx and corolla.
6. What parts of the plant produce seed and fruit after fertilization?
7. Name the type of pollination carried out of bat.
8. Give the term for fruit development without fertilization.

Answer:

1. Calyx, Corolla, Androecium and Gynoecium
2. Perrenial plants
3. Seed coat
4. Kadamb
5. Petalloid
6. Ovary
7. Chiropteriphilly
8. Parthenocarpy.

Sexual Reproduction in Flowering Plants Very Short Answer Type Questions

Question 1.
Development of female gametophyte occurs in which cell?
Answer:
By functional megaspore mother cell.

Question 2.
What is the other name of female gametophyte?
Answer:
Embryo sac.

Question 3.
Ovule derives nourishment from which part of the carpel?
Answer:
Ovule derives nourishment from placenta.

Question 4.
What are cleistogamous flowers ? Give an example.
Answer:
The flowers which do not open are called cleistogamous flowers, e.g. Commelina

Question 5.
What is the substance found on the exine of pollen grains?
Answer:
Sparopollenin.

Question 6.
Give the characters of wind pollinated flowers.
Answer:
White in colour, small in size and pollengrains are formed in large number.

Question 7.
What is the ploidy of angiospermic endosperm?
Answer:
Triploid.

Question 8.
Give an example of a monocotyledonous endospermic seed.
Answer:
Ricinus.

Question 9.
Give example of two false fruit.
Answer:
Apple, Jackfruit.

Question 10.
What are monocious plants?
Answer:
Plants which has both male and female flower.

Sexual Reproduction in Flowering Plants Short Answer Type Questions

Question 1.
What is parthenogenesis ?
Answer:
Parthenogenesis : It is the method of reproduction in which ovum develops into embryo and forms a new plant without fertilization. It has been observed that pollination is necessary for parthenogenesis. It stimulates the process. Ulothrix, Spirogyra and Solan- aceae and Malvaceae family plants reproduce by this method.
Sometimes in some plants ovary also develops into fruit during this process. Such fruits are called as parthenocarpic fruits.
Example: Banana, Apple, Grapes, Guava etc. shows parthenocarpic fruits.

Question 2.
What is Polyembryony?
Answer:
Polyembryony : When more than one embryo develops in one seed then this condition is called as Polyembryony. It is generally found in citrus family. It is also found in Nicotiana, conifers, rice, wheat. It occurs when fertilization occurs in all embryo sacs found in the ovule.

Question 3.
Describe structure of anther with labelled diagram.
Answer:
Structure of Anther : Transverse section of anther shows that it consists of 2 lobes which are connected by connective. Each lobe contains two Pollen sacs. Innermost layer of pollen sac is called as Tapetum.

Tapetum is a layer rich in nutritive contents which supplies food material for the developing pollen grains. At first many Pollen mother cells are formed in them which divides meiotically to form haploid pollen grains.

Question 4.
Give four contrivances for self pollination.
Answer:
Contrivances for self-pollination:
(1) Bisexuality : When male and female parts are found in same flower then possibility of self-pollination increases.

(2) Cieistogamy : In Commelina benghalensis both cleistogamous and chasmogamous flowers are produced. The former are the underground flowers and the latter are the aerial ones developed on branches. In the small, inconspicuous cleistogamous flowers the pollen are shed within the closed flowers so that self-pollination is a must. This is also observed in Impatiens, Oxalis, Viola, Portulaca etc.

(3) Homogamy : Here the stamens and carpels mature at the same time. So, there is a greater chance of self-pollination as compared to cross-pollination, e.g., Mirabilis, Argemone etc.

(4) Failure of cross pollination : In some flowers generally cross pollination occurs but if they fails to do cross pollination then self-pollination occurs.

Question 5.
What do you mean by micropropagation?
Answer:
Micropropagation : It is a modern method of reproduction. By this process thousands of new plants can be obtained from few tissues of mother plant. This method is based on tissue and cell culture technique.
In this process a small part of tissue is separated from the plant and then it is grown in nutrient medium in aseptic condition. The tissue develops to form a cluster of cells which is called as callus. This callus can be preserved for long time for multiplication. A small part of the callus is transferred to nutrient medium, where it grows into a new plant. This plant is then transferred to the field. By this process Orchids, Carnations, Chrysanthemum plants can be grown successfully.

Question 6.
What are the advantages of cross-pollination ?
Answer:
Advantages of cross-pollination:

• The weaker characteristics of the race elimi¬nated and replaced by better character of nature.
• New improved varieties cap be produced by cross-pollination. ‘
• It increases the adaptability of the offsprings.
• Seeds are more vigorous.

Question 7.
What is fertilization? Draw labelled diagram showing path of pollen tube during fertilization.
Or
What do you mean by sexual reproduction? Describe the process of fertilization in plants with the help of diagram or sexual reproduction.
Answer:
Fertilization : The process of fusion of male and female gamete is called as fertilization or sexual reproduction. As a result of this process a diploid zygote is formed.

Question 8.
Differences between self and cross pollination
Answer:
Differences between Self and Cross poltination :

Sexual Reproduction in Flowering Plants Long Answer Type Questions

Question 1.
What is self-pollination ? Give advantages and disadvantages of self pollination.
Answer:
Self-pollination : When the pollen grain of one flower are transferred to the stigma of the same flower then the process is called as self-pollination.

Advantages of self-pollination:
These are the advantages of self pollination:

• Parental characters can be preserved indefinitely in several generations.
• Self-pollination helps in maintaining pure lines for experimental hybridization.
• It is most economical method of pollination. The plants do not consume their energies in the production of
• large number of pollen grains, nectar and coloured corolla.
• It ensures seed production and flowers do not take chances of the failure of fertilization.

Disadvantages of self-pollination:
These are the disadvantages of self-pollination :

• The weaker characteristics or defects of the plant can never be eliminated from the race.
• No useful characters can be introduced in the race.
• The immunity of race towards diseases falls and ultimately it falls prey to many diseases.

Question 2.
What is double fertilization?
Or
Explain double fertilization in angiosperms.
Answer:
Double fertilization : It is found in all angiospermic plants. It was first discovered by S.N. Navaschin (1899) in Lilium and Fritillaria. When pollen tube come in contact of embryo sac then the tip of the pollen tube disintegrate and release two male gametes into the embryo sac. Out of which one male gamete (x) fuses with egg cell (x) to form a diploid zygote. This process is called as fertilization. The second male gamete (x) fuses with secondary nuclei (2x) to form a triploid body.
The act of two nuclear fusion is called as double fertilization.

Question 3.
Describe development of embryo sac or Female gametophyte in angiosperms.
Answer:
Development of embryo sac or Female gametophyte in angiosperms :
During the initial stage of development of ovule primary archesporial cell gets demarcated at the apex of the nucellus below the epidermis. It divides periclinally into a outer primary parietal cell or primary wall cell and an inner primary sporogenous cell. The later acts as a megaspore mother cell which enlarges in size and divide meiotically to form a row of four megaspores. Of the four cells the upper three cells degenerate and appear as dark caps while the lowest one functions arid is called the functional megaspore. The latter greatly enlarges and forms the embryo sac.

Question 4.
Describe development of Endosperm.
Answer:
Development of Endosperm : It develops from the triploid tissue of the fertilized embryo sac after the act of double fertilization. It is of the following three types:

1. Free nuclear endosperm.
2. Cellular type of endosperm.
3. Helobial type of endosperm.

The name refers to the type of nuclear divisions of the endosperm nucleus. If triploid nucleus divides by free nuclear division the endosperm produced contains many nuclei lying freely in it and hence it is termed as free nuclear endosperm. If the nuclear division is followed by wall formation it is called as cellular type. If endosperm is intermediate between the two types it is called as helobial type.


Question 5.
What is male gametogenesis ? Describe male gametogenesis in plants.
Or
Describe development of pollen grain.
Answer:
Male gametogenesis: The process of formation of male gametes is called as male gametogenesis or microsporogenesis.

Each anther usually consists of two lobes, connected together by a connective. Each lobe contains two sacs called as Pollen sac. In the early stage of development the anther comprises of a homogenous mass of cells limited by a well demarcated epidermis. During the course of development the anther gets four lobed. Each lobe exhibits a hypodermal lining, few cells thick with distinct nuclei. These hypodermal cells constitute the archesporium. The cell of archesporium divide periclinally, cutting of parietal cells towards the periphery and sporogenous cells within it two chambers or loculi, which are termed as pollen sacs or microsporangia.

Fig. Development of Pollen grain : (A) Formation of Pollen lobes, (B) Archesporial cell, (C) and (D) Division in Parietal cell, (E) and (F) Parietal layer : Tapetum sporogenous tissue (G), (H) and (I) Pollen mother cell, (J) Pollen grain, (K) Outer layer Degenerate, (L) Four free Pollen grains.

Besides the above changes in the wall layers the sporogenous layer comprising of primary sporogenous cells also undergo several divisions to form microspore mother cells or pollen mother cells. In the initial stage the microspore mother cells are closely packed but as the anther enlarges the pollen sac becomes spacious and the pollen mother cells assume spherical shape and get loosely arranged.

In many cases some of the sporogenous cells are nonfunctional and serve only as the nutritive material for the functional microspore mother cells. Each microspore mother cell is diploid in nature and by a single meiotic division give rise to four haploid nuclei. The four nuclei so formed are arranged tetrahedrally, and called as Tetrad. Innermost layer of pollen sac is called as Tapetum. Outer to which endothecium layer is formed. In outer most layer of pollen sac epidermis is formed. Surrounding each nuclei exine and intine are formed, now each structure is called as Pollen grain. Pollen grains are liberated during the dehiscence of anthers through stomium.

Question 6.
Differenciate between:
(i) Embryo sac and Endosperm
(ii) Seed and Ovule
Answer:
(i) Difference between Embryosac and Endosperm:

(ii) Difference between Seed and Ovule

MP Board Class 12th Biology Solutions

MP Board Class 11th Maths Solutions Chapter 10 सरल रेखाएँ Ex 10.3

प्रश्न 1.
निम्नलिखित समीकरणों को ढाल अंतःखण्ड रूप में रूपान्तरित कीजिए और उनके ढाल तथा y-अंतः खण्ड ज्ञात कीजिए :
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.
हल:
(i) x + 7y = 0
∴ y = – \(\frac{1}{7}\)x + 0
∴ ढाल = – \(\frac{1}{7}\), y-अंत: खण्ड = 0.
(ii) 6x + 3y – 5 = 0,
3y = – 6x + 5
∴ y = – 2x + \(\frac{5}{3}\)
ढाल = – 2, y-अंत: खण्ड = \(\frac{5}{3}\)
(iii) y = 0
या y = 0. x + 0.
ढाल = 0, y-अंत: खण्ड = 0

प्रश्न 2.
निम्नलिखित समीकरणों को अंतःखण्ड रूप में रूपान्तरित कीजिए और अक्षों पर इनके द्वारा काटे गए अंतःखण्ड ज्ञात कीजिए :
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0.
हल:
(i) 3x + 2y – 12 = 0
या 3x + 2y = 12
12 से दोनों पक्षों में भाग देने पर
\(\frac{x}{4}+\frac{y}{6}\) = 1
अतः अंत: खण्ड 4 तथा 6 हैं।
(ii) 4x – 3y = 6
6 से दोनो पक्षों में भाग देने पर,
\(\frac{4 x}{6}-\frac{3 y}{6}\) = 1
\(\frac{x}{\frac{3}{2}}+\frac{y}{-2}\) = 1.
अत: अंत:खण्ड \(\frac{3}{2}\) तथा – 2 हैं।
3y + 2 = 0
या 3y = – 2
y = – \(\frac{2}{3}\).
अन्त: खण्ड हेतु समीकरण का रूप:
\(\frac{x}{0}+\frac{y}{-\frac{2}{3}}\) = 1
अंत:खण्ड 0 और –\(\frac{2}{3}\) हैं।

प्रश्न 3.
निम्नलिखित समीकरणों को लम्ब रूप में रूपान्तरित कीजिए। उनकी मूल बिन्दु से लांबिक दूरियाँ और लम्ब तथा धन x-अक्ष के बीच का कोण ज्ञात कीजिए :
(i) x – \(\sqrt{3}\)y + 8 = 0,
(ii) y – 2 = 0,
(iii) x – y = 4.
हल:

(ii) y – 2 = 0 या y = 2
0·x + y·1 = 2
x cos 90° + y sin 90° = 2 [∵ cos 90° = 0, sin 90° = 1]
∴ p = 2, α = 90°.
(iii) x – y =4
\(\sqrt{2}\) से भाग देने पर
\(\frac{1}{\sqrt{2}} x+\left(-\frac{1}{\sqrt{2}}\right) y\) = 2\(\sqrt{2}\)
\(\frac{1}{\sqrt{2}}\) = cos (360° – 45°) = cos 315°
और – \(\frac{1}{\sqrt{2}}\) = sin 315०
∴ x – y = 4 का लम्ब रूप
x cos 315° + y sin 315° = 4
की तुलना x cos α + y sin α = p से करने पर,
p = 2\(\sqrt{2}\) , α = 315°.

प्रश्न 4.
बिन्दु (- 1, 1) की रेखा 12(x + 6) = 5(y – 2) से दूरी ज्ञात कीजिए।
हल:
12(x + 6) = 5(y – 2).
या 12x + 72 = 5y – 10
12x – 5y + 82 = 0

प्रश्न 5.
x-अक्ष पर बिन्दुओं को ज्ञात कीजिए जिनकी रेखा \(\frac{x}{3}+\frac{y}{4}\) = 1 से दूरियाँ 4 इकाई हैं।
हल:
दिया गया समीकरण है: \(\frac{x}{3}+\frac{y}{4}\) = 1
12 से गुणा करने पर
4x + 3y – 12 = 0 …(1)
x- अक्ष पर माना कोई बिन्दु (a, 0) हो, तो बिन्दु (a, 0) से रेखा (1) की दूरी
= \(\frac{4 a+0-12}{\sqrt{16+9}}\) = ± \(\frac{4 a-12}{5}\)
∴ ± \(\frac{4 a-12}{5}\) = 4
या ± (4a – 12) = 20
+ ve चिन्ह लेकर 4a = 32 या a = 8
x-अक्ष पर अभीष्ट बिन्दु (8, 0) है।
– ve चिन्ह लेकर, –\(\frac{4 a-12}{5}\) = 4 या – 4a + 12 = 20
4a = – 8, a = – 2
दूसरा अभीष्ट बिन्दु (- 2, 0) है।

प्रश्न 6.
समान्तर रेखाओं के बीच की दूरी ज्ञात कीजिए
(i) 15x + 8y – 34 = 0 और 15x + 8y + 31 = 0
(ii) l(x +y) + p = 0 और l(x + y) – r= 0
हल:

प्रश्न 7.
रेखा 3x – 4y + 2 = 0 के समान्तर और बिन्दु (-2, 3) से जाने वाली रेखा का समीकरण ज्ञात कीजिए।
हल:
3x – 4y + 2 = 0
या 4y = 3x +2
∴ y = \(\frac{3}{4} x+\frac{2}{4}\)
∴ रेखा की ढाल = \(\frac{3}{4}\)
दिया गया बिन्दु (- 2, 3) और ढाल m = \(\frac{3}{4}\) से जाने वाली रेखा का समीकरण
y – y₁ = m(x – x₁)
y – 3 = \(\frac{3}{4}\)(x + 2)
या 4y – 12 = 3x + 6
या 3x – 4y + 18 = 0.
दूसरी विधि : कोई भी रेखा ax + by + c = 0 के समान्तर ax + by + k = 0 के रूप में लिखी जा सकती है।
∴ 3x – 4y + 2 = 0 के समान्तर रेखा 3x – 4y + k = 0 है
यह (- 2, 3) से होकर जाती है।
∴ 3 x (- 2) – 4 x 3 + k = 0 या k = 18
अभीष्ट समान्तर रेखा का समीकरण: 3x – 4y + 18 = 0.

प्रश्न 8.
रेखा x – 7y + 5 = 0 पर लम्ब और x-अन्तः खण्ड 3 वाली रेखा का समीकरण ज्ञात कीजिए।
हल:
∵ x-अंत:खण्ड = 3
∴ रेखा A(3, 0) से होकर जाती है।
रेखा PQ : x – 7y + 5 = 0
या 7y = x +5
या y = \(\frac{1}{7}\) x + \(\frac{5}{7}\)

इसलिए PQ की ढाल = \(\frac{1}{7}\)
∵ PQ ⊥ AB
∴ A से होकर जाने वाली रेखा AB की ढाल = – 7
∴ बिन्दु (3, 0) से रेखा AB का समीकरण,
y – 0 = – 7(x – 3).
= – 7x + 21
या 7x + y – 21 = 0.
दूसरी विधि : ax + by + c = 0 की लम्ब कोई रेखा bx – ay + k = 0
∴ x – 7y + 5 = 0 की लम्ब रेखा 7x + y + k = 0
यह रेखा (3, 0) से होकर जाती है।
∴ 7 x 3 + 0 + k = 0, अर्थात् k = – 21
∴ अभीष्ट रेखा का समीकरण 7x + y – 21 = 0.

प्रश्न 9.
रेखाओं \(\sqrt{3}\)x + y =1 और x +\(\sqrt{3}\)y =1 के बीच का कोण ज्ञात कीजिए।
हल:

θ = 30° = \(\frac{\pi}{6}\) रेडियन। .

प्रश्न 10.
बिन्दुओं (h, 3) और (4, 1) से जाने वाली रेखा, रेखा 7x – 9y – 19 = 0 को समकोण पर प्रतिच्छेद करती है। का मान ज्ञात कीजिए।

माना रेखा AB बिन्दु A(h, 3), B(4, 1) से जाने वाली रेखा की ढाल,

चूँकि दोनों रेखाएँ एक-दूसरे को समकोण पर प्रतिच्छेद करती हैं, ∴ m₁,m₂ = – 1
\(\frac{2}{h-4} \times \frac{7}{9}\) = – 1
14 = – 9(h – 4) = – 9h + 36
∴ 9h = 36 – 14 = 22
h = \(\frac{22}{9}\)

प्रश्न 11.
सिद्ध कीजिए कि बिन्दु (x₁, y₁) से जाने वाली और रेखा Ax + By + C = 0 के समान्तर रेखा का समीकरण
A(x – x₁) + B(y – y₁) = 0 है।
हल:
रेखा Ax + By + C = 0
या y= – \(\frac{A}{B}\)x – \(\frac{C}{B}\)
रेखा की ढाल = – \(\frac{A}{B}\)
∴ समान्तर रेखा की ढाल = – \(\frac{A}{B}\)

प्रश्न 12.
बिन्दु (2, 3) से जाने वाली दो रेखाएँ परस्पर 60° के कोण पर प्रतिच्छेद करती हैं। यदि एक रेखा की ढाल 2 है तो दूसरी रेखा का समीकरण ज्ञात कीजिए।
हल:
माना दूसरी रेखा की ढाल m है।
दोनों रेखाओं के बीच कोण
tan θ = \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)
जहाँ θ = 60°, m₁ = m और m₂ = 2
∴ tan 60 = ± \(\frac{m-2}{1+2 m}\) = \(\sqrt{3}\)

प्रश्न 13.
बिन्दुओं (3, 4) और (- 1, 2) को मिलाने वाली रेखाखण्ड के लम्ब समद्विभाजक रेखा का समीकरण ज्ञात कीजिए।
हल:
माना बिन्दुओं A(3, 4) और B(- 1, 2) को मिलाने वाले रेखाखण्ड का मध्य बिन्दु

रेखा CD बिन्दु D से होकर जाती है
∴ रेखा CD का समीकरण
y – 3 = – 2(x – 1)
= – 2x + 2
∴ 2x + y – 5 = 0.

प्रश्न 14.
बिन्दु (- 1, 3) से रेखा 3x – 4y – 16 = 0 पर डाले गए लम्बपाद के निर्देशांक ज्ञात कीजिए।
हल:
मान लीजिए रेखा AB का समीकरण, 3x – 4y – 16 = 0 …(i)
या y = \(\frac{3}{4}\)x – 4
रेखा AB की ढाल = \(\frac{3}{4}\)
बिन्दु C(- 1, 3) से AB पर डाला गया लम्ब CD है
∴ AB ⊥ CD.

अतः रेखा CD का समीकरण,
y – y₁ = m(x – x₁)
y – 3 = \(\frac{-4}{3}\)(x + 1)
या 3y – 9 = – 4x – 4
या 4x + 3y – 5 = 0 …(ii)
समी (i) को 3 से और (ii) को 4 से गुणा करने पर,
9x – 12y = 48
16x + 12y = 20
इनको जोड़ने पर
25x = 68 या x = \(\frac{68}{25}\)
x का मान (i) में रखने पर,

प्रश्न 15.
मूल बिन्दु से रेखा y = mx + c पर डाला गया लम्ब रेखा से बिन्दु (-1, 2) पर मिलता है। m और … c के मान ज्ञात कीजिए।
हल:
माना रेखा AB का समीकरण, y = mx + c
रेखा AB की ढाल = m

O से रेखा AB पर लम्ब OC डाला गया है जो बिन्दु C(- 1, 2) पर मिलता है।
∴ लम्ब रेखा OC की ढाल = –\(\frac{1}{m}\)
अब रेखा OC का समीकरण,
y – 0 = –\(\frac{1}{m}\)(x – 0)
या x + my = 0
OC की प्रवणता = \(\frac{2-0}{-1-0}\) = – 2
∴ लम्ब रेखा OC की ढाल = –\(\frac{1}{m}\)
बिन्दु C (- 1, 2) निम्न रेखा पर स्थित है :
y = mx + c
⇒ 2 = – m + c
m = \(\frac{1}{2}\) रखने पर,
2 = – \(\frac{1}{2}\) + c
∴ c = 2+ \(\frac{1}{2}\) = \(\frac{5}{2}\)
अतः m = \(\frac{1}{2}\), c = \(\frac{5}{2}\)

प्रश्न 16.
यदि p और q क्रमशः मूल बिन्दु से रेखाओं x cos θ – y sin θ = k cos 2θ और x sec θ +y cosec θ = k पर लम्ब की लंबाइयाँ हैं तो सिद्ध कीजिए कि
p₂ + 4q₂ = k₂.
हल:
मूल बिन्दु (0, 0) से x cos θ – y sin θ = k cos 2θ की दूरी,

समीकरण (1) और (2) को वर्ग करके जोड़ने पर,
k₂ = p₂ + 4q₂
अतः p₂ + 4q₂ = K₂.

प्रश्न 17.
शीर्षों A(2, 3), B(4, – 1) और C(1, 2) वाले त्रिभुज ABC के शीर्ष A से उसकी सम्मुख भुजा पर लम्ब डाला गया है। लम्ब की लम्बाई तथा समीकरण ज्ञात कीजिए।
हल:
मान लीजिए AM रेखा BC पर लंब डाला गया है
(i) रेखा BC की ढाल

रेखा AM बिन्दु A से जाती है और ढाल = 1 है।
∴ AM का समीकरण
y – y₁ = m(x – x₁)
y – 3 = 1. (x – 2)
या x – y + 1 = 0
(ii) बिन्दु B(4, – 1) और C(1, 2) से होकर जाने वाली रेखा BC का समीकरण

प्रश्न 18.
यदि p मूल बिन्दु से उस रेखा पर डाले गए लम्ब की लम्बाई हो जिसस पर अक्षों पर कटे अंत: खण्ड a और b हों, तो दिखाइए कि \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).
हल:
उस रेखा का समीकरण, जिसकी अक्षों पर कटे अंत:खण्ड a और b हों,
\(\frac{x}{a}+\frac{y}{b}\) = 1 (अंत:खण्ड समीकरण)
मूल बिन्दु (0, 0) बसे इस रेखा पर डाले गए लम्ब की लम्बाई

MP Board Class 11th Maths Solutions