Prasanna

MP Board Class 12th Biology Important Questions Chapter 16 Environmental Issues

Environmental Issues Important Questions

Environmental Issues Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Disease caused due to mercury pollution is:
(a) Methenamine
(b) Minimata
(c) Asbestosis
(d) Liver inflammation.
Answer:
(b) Minimata

Question 2.
Plants are natural air purifiers because they exhibit :
(a) Respiration
(b) Photosynthesis
(c) Transpiration
(d) Drying.
Answer:
(b) Photosynthesis

Question 3.
Metal found in atmosphere is :
(a) Cadmium
(b) Lead
(c) Mercury
(d) Zinc.
Answer:
(b) Lead

Question 4.
Burning of fossil fuel results in:
(a) SO₂ pollution
(b) NO₂ pollution
(c) N₂O pollution
(d) NO pollution.
Answer:
(a) SO₂ pollution

Question 5.
Increase of the concentration of which gas results greenhouse effect :
(a) CO₂
(b) CO
(c) O₂
(d) Nitrogen oxide.
Answer:
(a) CO₂

Question 6.
Gas responsible for depletion of O₃ in atmosphere is :
(a) CFC
(b) NO₂
(c) CO₂
(d) SO₂
Answer:
(a) CFC

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

Question 8.
Effect of air pollution is seen in :
(a) Leaves
(b) Flower
(c) Forests
(d) None of these.
Answer:
(b) Flower

Question 9.
Shrinking of leaves occurs by :
(a) SO₂
(b) O₂
(C) H₂S
(d) CO.
Answer:
(a) SO₂

Question 10.
Example of greenhouse gas is :
(a) CO₂
(b) CH₄
(c) CFC
(d) All of these.
Answer:
(d) All of these.

Question 11.
Gas produced during combustion of fossil fuels is :
(a) CO₂
(b) CH₄
(c) CFC
(d) N₂O
(a) CO₂

Question 12.
Gas used in refrigeration is :
(a) Freon
(b) N₂O
(c) CH₄
(d) CO₂.
Answer:
(a) Freon

Question 13.
Radiation emitted due to heating of earth is:
(a) UV
(b) Infrared
(c) Gamma
(d) None of these.
Answer:
(b) Infrared

Question 14.
Gas responsible for depletion of atmospheric ozone layer is:
(a) CO₂
(b) CH₄
(c) CFC
(d) O₂
Answer:
(c) CFC

Question 2.
Fill in the blanks:

1. ………………….. is a chief green house gas.
2. ………………….. is the measurement unit of water pollution.
3. ………………….. and ………………….. gases are responsible for acid rain.
4. ………………….. gas is essential for life. (MP 2009 Set B)
5. The substances which produce pollution are called ………………….. (MP 2013)
6. Strong, loud, unpleasant or untolerable sound is called ………………….. (MP 2013)
7. We can ………………….. the quantity of CO₂ from forestation.
8. CO₂, CH₄, N₂O, CFCs are………………….. gases.
9. ………………….. is discharged in Bhopal gas disaster.
10. In human DDT is reached by …………………..
11. ………………….. is also known as chemical weeds.
12. Pollution is the …………………. changes which occurs in atmosphere.
13. Deficiency of soil nutrients is called …………………..
14. The percentage of CO₂ of atmosphere can be ………………….. by forestation. (MP 2009 Set D)

Answer:

1. CO₂
2. BOD
3. SO₂ and NO₂
4. O₂
5. Pollutant
6. Noise
7. Lost
8. Green house
9. Methyl isocyanate
10. Food chain
11. Ozone
12. Unwanted
13. Negative pollution
14. Reduced.

Question 3.
Match the followings:
I.

Answer:

1. (c)
2. (f)
3. (e)
4. (b)
5. (d)
6. (a)
7. (g)
8. (h)

II.

Answer:

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

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

1. Write full name of BOD.
2. What is the hearing capacity of man?
3. Name the fish used to control mosquito.
4. What is the concentration of CO₂ in environment?
5. When is World Environment Day celebrated?
6. Write the names of two main gases which pollute air.
7. Write the name of highly polluted river of India.
8. Name the term used, for the increase in atmospheric temperature, due to increase in CO₂ concentration.
9. What is the amount of oxygen taken up by microorganisms, that decompose waste matter?
10. Write full name of CFC.
11. Write the name of a chief greenhouse gas.
12. Write the name of gas responsible for ozone layer depletion.
13. Who is the greatest enemy of nature and organism? (MP 2013)
14. When we celebrated ozone day?

Answer:

1. Biological Oxygen Demand
2. 10 – 12 decibel
3. Gambussia
4. 0 03%
5. 5th June
6. SO₂and CO
7. Ganga
8. Green House effect
9. B. O. D
10. Chlorof – luorocarbon
11. CO₂
12. CFC
13. Global warming
14. 16December.

Environmental Issues Very Short Answer Type Questions

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

Question 2.
What are the essential components of biodiversity?
Answer:
Essential components of biodiversity are:

1. Genetic diversity.
2. Species diversity.
3. Ecological diversity.

Question 3.
When and how photochemical smog is formed?
Answer:
Photochemical smog is the chemical reaction of sunlight, nitrogen oxides and volatile organic compounds in the atmosphere, which leaves air borne particles and ground level ozone. Peroxyacetyl Nitrate (PAN) is present in the photochemical smog. It causes damage to respiratory system and environment system. (The word smog is derived from smoke and fog.)

Question 4.
Name two plants which uses carbon monoxide (CO).
Answer:
Two plants which uses CO are:

1. Daucus carota,
2. Ficus variegata.

Question 5.
Name the types of water pollution based on source of water.
Answer:
Types of water pollution based on source of water:

1. Underground water pollution
2. Surface water pollution
3. Lake water pollution and
4. River water pollution.

Question 6.
How does fertilizers pollute water?
Answer:
Excess fertilizers used in the field are drained into the pond and river, where they proliferate growth of algae, which spread all over the surface of the water and cut off oxygen supply for lower aquatic organisms. Due which lower aquatic organisms dies. This is known as eutrophication.

Question 7.
Give harmful effects of cutting of forest.
Answer:
Harmful effects of cutting of forest are:

1. It increases landscape.
2. It decreases fertility of the soil.
3. Rainfall reduces and become uncontrolled.

Environmental Issues Short Answer Type Questions

Question 1.
What is acid rain? Write any two effects of acid rain on human beings.
Answer:
Acid rain:
SO₂ gas is released into the environment by burning of fossil fuels containing sulphide from industries, example 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 2.
Write the effect of air pollution on plants.
Answer:
Effect of air pollution on plants:

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

Question 3.
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 puiposes gives out gases like CO₂, CO, SO₂ and consumes oxygen from the air.

Question 4.
Write preventive measures to control air pollution.
Answer:
Preventive measures to control air pollution:

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

Question 5.
Write any four measures to control greenhouse effect.
Answer:
Following important measures can be sited as the base steps towards controlling greenhouse effect:

1. 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.
2. Disposing of the greenhouse gases as they are formed elsewhere than in the atmosphere.
3. By recovering greenhouse gases present already in the atmosphere and disposing them off elsewhere.
4. Prevention of deforestation and planting of more trees.

Question 6.
Write short note:

1. Bio – magnification.
2. UV rays.
3. Biodegradable 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. Biodegradable pollutants:
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.

4. 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.

Question 7.
What is the effect of SO₂ in the environment?
Answer:

Question 8.
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 9.
Write a note on the effect of water pollution on aquatic organisms.
Answer:
Effect of water pollution on aquatic plants:

1. 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.

2. 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.

3. The number of microorganisms increases in polluted water.

4. Siltation occurs in water.

5. Temperature of water increases and O₂ 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:

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

Question 10.
Write the measures for control and prevention of water pollution.
Answer:
Measures for control and prevention of water pollution are:

Preventive measures:

1. Use of harmful pesticides and weedicides must be stopped completely.
2. Discharge of effluents into rivers, lakes and sea should be strictly prohibited without treatment.
3. Oil spill should be prevented.
4. Proper disposal of sewage so, that it does not find its way into water bodies.
5. 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, example Dung of cow and buffalo can be used for the production of gobar gas a cheap source of fuel and also as manure.

Question 11.
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 SO₂, CO₂, CO, H₂S and hydrocarbons together with dust and smoke. These are produced by the burning of coal and petroleum. The chemical industries releases HC₁, Cl₂, nitrogen oxide, oxides of copper, zinc, lead, arsenic etc. Industrial pollutants causes air pollution and water pollution.

Question 12.
Write the effects of noise pollution.
Answer:
Effects of Noise pollution:

1. The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to. deafness.
2. Interference with speech communication.
3. Noise pollution leads to neurosis, anxiety hypertension, cardiovascular disease, hepatic stress, giddiness.
4. Annoyance leading to ill – temper, bickering, mental disorientation and violent behaviour.
5. 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 consequence rise in blood pressure. This ultimately results in tension and nervousness.
6. Affects different metabolic activities.

Question 13.
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 14.
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 15.
Write preventive measures of sound pollution.
Answer:
Preventive measures of noise pollution:

1. 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.
2. Noise producing industries should be installed away from residential areas.
3. Use of public addressing systems should be restricted at a fixed intensity and hours of the day.
4. Stuffing of cotton plugs in the ear or use of earmuffs and minimise the danger of occupational exposure of noise.
5. Trees absorb sound vibrations and thus, reduce the noise pollution. Thus, plantation of trees along highways, establishment of parks help in reducing noise pollution.
6. By use of silencer in all automobiles and engines.

Question 16.
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 characteristics of soil due to addition of unwanted substances which adversely affects productivity of soil is termed as soil pollution.

Question 17.
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 the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 18.
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 19.
What are the causes 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.

Environmental Issues Long Answer Type Questions

Question 1.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming:
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 wanning:

• 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 2.
Write critical notes on the following:

1. Eutrophication,
2. Biological magnification,
3. Groundwater depletion and ways for its replenishment.

Answer:
1. 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.

2. 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 was 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 have the highest concentration of DDT.

3. 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 3.
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 + O_{3 →} ClO + O₂
• ClO + O → 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 4.
Discuss briefly the following:

1. Radioactive wastes,
2. Defunct ships and e-wastes,
3. Municipal solid wastes.

Answer:
1. 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.

2. 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. Thus, they 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.

3. 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 5.
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 qualify 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.

1. Phasing out of old vehicles
2. Use of unleaded petrol.
3. Use of low – sulphur petrol and diesel.
4. Use of catalytic converters.
5. Application of stringent pollution-level norms for vehicles.
6. 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 6.
Discuss briefly the following:

1. Greenhouse gases
2. Catalytic converter
3. Ultraviolet – B

Answer:
1. Greenhouse 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 radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are released back into the atmosphere. These radiations 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.

2. 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 unbumt 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).

3. 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 humAnswer: 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 comeal cataract in human beings.

Question 7.
Describe the diseases caused due to radioactive pollution.
Answer:
Following diseases are caused by radioactive pollution:

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

Question 8.
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 caused 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 9.
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 affect our eyes, lungs, skins, heart and brain and produce various disorders in them.

Sources of air pollution : Following are the chief sources of air pollution:

1. Human sources:

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

2. 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.

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.
• RA.N. inhibits Hill reaction and thus, decreases photosynthetic production of an ecosystem.

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 10.
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 haemoglobin 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 11.
Write down the methods to control soil pollution.
Answer:
Methods to control soil pollution:
By following methods soil pollution can be prevented:

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

Question 12.
What are the sources of soil pollution?
Answer:
Sources of soil pollution:

1. Acid rainwater and water from mines are main sources of soil pollution.
2. Mixing of debris, waste products with the soil cause soil pollution.
3. By excessive use of pesticides and fertilizers, pollute the soil.
4. When industrial wastes are discharged in the soil, they also pollute the soil.
5. Heavy metals example 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.
6. Bones of dead animals, paper putrified flesh and food, iron, copper, mercury etc., pollute the soil.
7. 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.
8. 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 13.
What is visible spectrum? Write short note on spectrum of solar light and also write effect of UV rays on the living organisms.
Answer:
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. Electromagnetic radiation in this range of wavelength is called visible light or simply light. It ranges from 390 nm to 760 nm wavelength.

Different spectrum of sunlight (solar light):
Spectrum of sunlight can be divided into three parts :

1. Ultraviolet spectrum – The UV spectrum covers the wavelength range 100 – 390 nm and is divided into three bands. It is invisible.
2. Visible spectrum – It include light of ranges from 390 to 760 nm wavelength.
3. Infrared spectrum – Light ranges 760 nm above is called as infrared ray. It is also invisible.

Effect of UV rays in the living organisms:

1. It make DNA abnormal thus, protein sythesis in the body get effected.
2. It may cause skin cancer.
3. It supress immune system of the body.
4. It causes sunburn and premature ageing of skin.
5. It causes formation of harmful photochemical smog which causes respiratory diseases.
6. It damage eyes.

Question 14.
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 and 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:

1. 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 in high dose.

2. 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 15.
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 :

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

Effects of noise pollution:

1. The more acute and immediate effect of noise pollution is impairing of hearing leading to auditory fatigue and may finally lead to deafness.
2. Interference with speech communication.
3. Noise pollution leads to neurosis, anxiety, hypertension, cardiovascular disease, hepatic stress, giddiness.
4. Annoyance leading to ill – temper, bickering, mental disorientation and violent behaviour.
5. 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.
6. Affects different metabolic activities.

Question 16.
What is Greenhouse effect? Write any four impacts of Greenhouse effect
Answer:
Greenhouse effect:
The Greenhouse 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 Greenhouse 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 Greenhouse gases does not increase the 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 Greenhouse 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 amount 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 nitric 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 Important Questions

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

Biodiversity and Conservation Important Questions

Biodiversity and Conservation Objective Type Questions 

Question 1.
Choose the correct answer:

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

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

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

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

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

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

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

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

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

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

Question 2.
Fill in the blanks :

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

Answer:

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

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

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

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

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

Answer:

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

Biodiversity and Conservation Very Short Answer Type Questions

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

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

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

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

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

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

Biodiversity and Conservation Short Answer Type Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Biodiversity and Conservation Long Answer Type Questions

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

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

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

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

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

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

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

The organisms which are near extinction are of following types:

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

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

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

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

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

Five national parks of India are:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

6. Carnivorous animals – Some Indian wild carnivores are:

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

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

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

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

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

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

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

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

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

MP Board Class 12th Biology Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 6 Thermodynamics

Thermodynamics Important Questions

Thermodynamics Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
In adiabatic expansion of an ideal gas always:
(a) Increase in temperature
(b) ∆H = 0
(c) q = 0
(d) W = 0
Answer:
(c) q = 0

Question 2.
For a reversible process, free energy change at equilibrium:
(a) More than zero
(b) Less than zero
(c) Equal to zero
(d) None of these.
Answer:
(c) Equal to zero

Question 3.
In isothermal expansion of an ideal gas:
(a) 9 = 0
(b) AE = 0
(c) W = 0
(d) dV = Q
Answer:
(b) AE = 0

Question 4.
Hess’s law is an application of the following:
(a) First law of Thermodynamics
(b) Second law of Thermodynamics
(c) Entropy change
(d) Free energy change.
Answer:
(a) First law of Thermodynamics

Question 5.
When the value of heat of neutralization of an acid with a base is 13.7 kcal, then:
(a) Acid and base both are weak
(b) Acid and base both are strong
(c) Acid is strong and base is weak
(d) Acid is weak and base is strong
Answer:
(b) Acid and base both are strong

Question 2.
Fill in the blanks:

1. Enthalpy is an …………………….. property.
2. Nicely closed thermos flask is an example of an ……………………….
3. Value of heat of combustion (∆H) is always ………………………..
4. Extensive property depend on the ………………………. of matter.
5. Value of heat of neutralization is always …………………………. kilocalorie.

Answer:

1. Extensive
2. Isolation
3. Negative
4. Amount
5. – 13.7

Question 3.
Answer in one word/sentence:

1. Tell the value of heat of neutralization of strong acid and strong base?
2. Give two examples of state function?
3. What is entropy?
4. Among NaCl, H₂O_(s) and NH_3(g) value of whose entropy is higher?
5. Expect the system, what is the remaining part of the universe called?
6. System in which changes occur spontaneously and by which entropy of the system increases, what are they called?
7. What is the type of heat of combustion?

Answer:

1. – 57 kJ
2. Enthalpy, Entropy
3. Measure of disorder
4. NH₃
5. Surroundings
6. Spontaneous process
7. Exothermic.

Thermodynamics Very Short Answer Type Questions

Question 1.
Give two examples of state function?
Answer:
Enthalpy and Entropy.

Question 2.
Whose entropy is greater among NaCl, H₂O_(s) and NH_3(g)?
Answer:
NH_3(g).

Question 3.
Who gave the equation ∆G = ∆H – T∆S?
Answer:
Gibbs – Helmholtz.

Question 4.
Write the equation of first law of thermodynamics?
Answer:
∆E = q + W.

Question 5.
What is the value of entropy when ice melts?
Answer:
Entropy increases.

Question 6.
What is Closed system?
Answer:
System which can exchange energy only and not matter with the surroundings.

Question 7.
The value of which enthalpy is always negative?
Answer:
Enthalpy of combustion.

Question 8.
Heat of neutralization of strong acid and strong base is equal to?
Answer:
-13.7 kcal or -57.1 kJ.

Question 9.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process.

Question 10.
What is Enthalpy?
Answer:
Heat change at constant pressure is known as enthalpy.

Question 11.
The process in which pressure remains constant is called?
Answer:
Isobaric process.

Question 12.
In Exothermic reaction the value of ∆H is?
Answer:
Negative.

Question 13.
Unit of specific heat capacity is?
Answer:
joule per kelvin per gm.

Question 14.
The relation between ∆G, ∆S and ∆H is given by?
Answer:
∆G = ∆H – T∆S.

Question 15.
Which type of property is heat capacity?
Answer:
Extensive property.

Question 16.
What type of properties are temperature, pressure and surface tension?
Answer:
Intensive property.

Question 17.
What is the unit of molar heat capacity?
Answer:
joule kelvin^-1 mol^-1.

Question 18.
For which process dq = 0?
Answer:
Adiabatic process.

Question 19.
The efficiency of any fuel is measured by which value?
Answer:
Calorific value.

Question 20.
What is entropy?
Answer:
The measurement of degree of disorder or randomness of the molecule of the system.

Question 21.
Write the relation between standard free energy change ∆G° and equilibrium constant (K)?
Answer:
∆G° = -RTlnK.

Question 22.
What is the value of ∆G for Spontaneous process?
Answer:
∆G < 0.

Question 23.
“The electricity obtained from an electrochemical ceil is equivalent to decrease in free energy”. Write expression for this sentence?
Answer:
∆G° = -nE°F.

Thermodynamics Short Answer Type Questions – I

Question 1.
What is System?
Answer:
System:
A specified portion of the universe which is selected for experimental or theoretical investigations is called the system. (MPBoardSolutions.com) In the system, the effects of certain properties such as pressure, temperature, etc. are observed. A system is said to be homogeneous if it consists of only one phase. On the other hand, it is heterogeneous if it consists of more than one phase.

Question 2.
What is process and what are its kinds?
Answer:
The operation which brings about change in the state of a system is called a thermodynamics process.
Thermodynamics process may be further classified as follows:

1. Isothermal process
2. Adiabatic process
3. Isobaric process
4. Isochoric process
5. Reversible process
6. Irreversible process
7. Cyclic process.

Question 3.
Explain Exothermic reaction with example?
Or, Explain why the value of enthalpy change negative for exothermic reaction?
Answer:
The reactions in which heat is emitted are called exothermic reactions. In such reactions total heat content of products is less than that of reactants. Hence, ∆H is negative.
Example: 2NO_(g) → N_2(g) + O_2(g); ∆H = – 180.5 kJ/mol.

Question 4.
Explain extensive and intensive properties?
Answer:
1. Intensive properties:
The properties of the system which are independent of the amount of matter present in it, are called intensive properties.
Example: Temperature, viscosity, surface tension, refractive index, specific heat, density, etc.

2. Extensive properties:
The properties of the system which depend upon the amount of matter present in it, are called extensive properties.
Example: Mass, volume, energy, etc.

Question 5.
State and Explain Zeroth law of Thermodynamics?
Answer:
According to this law, “Two bodies which are separately in thermal equilibrium with a third body are also in thermal equilibrium with each other”.

Question 6.
Explain Enthalpy of neutralization with example?
Answer:
The enthalpy of neutralization is defined as:
“Change in enthalpy when one gram equivalent of an acid is neutralized with one gram equivalent of base in dilute solution at constant temperature.
Example: NaOH_(aq) + HCl_(aq) ⇄  NaCl_(aq) + H₂O_(l); ∆H =- 57.1 kJ

Question 7.
The enthalpy of neutralization of weak acid and weak base is less than enthalpy of neutralisation of strong acid and base. Why?
Answer:
If either acid or base weak then its ionisation in solution remains incomplete. As a result a part of energy liberated during combination of H+ and OH” ion is used up for the ionisation of weak acid and weak base. Therefore, the value of enthalpy of neutralisation of weak acid with strong base or vice – versa is numerically less than – 57.1 kJ.

Question 8.
What is Bond enthalpy and bond dissociation energy?
Answer:
It is a well known fact that during the formation of a chemical bond, energy is required. Therefore the breaking of a bond energy is to be supplied. Thus, the energy required to break a particular bond in a gaseous molecule is called bond dissociation energy.
Example: 2HCl_(g) → H_2(g) + Cl_2(g)

Question 9.
What is the first law of thermodynamics? Write its mathematical form?
Or, Write the first law of thermodynamics and derive the mathematical expression of it?
Answer:
First law of thermodynamics is the law of conservation of energy .The common statement of this law is:
“Energy can neither be created nor be destroyed but it can be converted from one form to another form.” Let internal energy of the system is E₁ and q calorie heat is supplied to the system. E₂ is the energy of the final stage and work done is W. Therefore,
E₂ – E₁ = q + W
or ∆E = q + W.

Question 10.
Define the term Entropy?
Answer:
A change that brings about disorder or randomness is more likely to occur than one that brings about order. To account for the randomness or disorder of a system a state function called entropy was introduced. It is (MPBoardSolutions.com) defined as the measure of degree of disorder or randomness of the molecule of the system. Entropy is represented by symbol S. It is easier to define entropy change than entropy of a system.

Question 11.
Whose entropy is more: Water vapour, water or ice, why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are completely arranged, therefore its entropy is minimum and in gas the molecules move randomly in all directions, so the value of entropy is more. So
S_(ice) < S_(water) < S_{(water vapour)}.

Question 12.
Among NaCl, H₂O and NH₃ whose entropy will be maximum and why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are regularly arranged, so entropy is minimum whereas in gas the molecules move randomly in all directions, so the entropy is maximum. So, in the above example, NaCl is a solid, H₂O is liquid and NH₃ is a gas. So, the entropy of NH₃ is maximum and entropy of NaCl is minimum.

Question 13.
Prove that: P∆V = ∆ntRT?
Answer:
From ideal gas equation
P V = nRT
If the volume of gas at initial state is V₁ and the number of moles of gas is n₁ then,
PV₁ = ∆n₁RT …………….. (1)
If at final state, the volume of gas is V₂ and number of moles of gas is n₂, then,
PV₂ = ∆n₂RT …………. (2)
From eqn. (i) and (2),
P(V₂ – V₁) = (n₂ – n₁)RT
or P∆V = ∆nRT.

Question 14.
What is relation between ∆H and ∆U?
Answer:
If H be the enthalpy of any system and U is the internal energy, then the relation will be
H = U + PV
For enthalpy change, ∆H = ∆U + P∆V
We know that P∆V = ∆nRT
On putting the value,
∆H = ∆U + ∆nRT.

Question 15.
What do you mean by specific heat capacity?
Answer:
It is the heat required to raise the temperature of 1 gram of substance by 1°C. It is denoted by C_s.
C_s = \(\frac{C}{m}\)
Where, C = Heat capacity, C_s = Specific heat capacity, m = Mass of substance.
Its S.I. unit is joule K^-1g^-1.

Question 16.
What do you mean by molar heat capacity?
Answer:
It is the heat required to raise the temperature of 1 mole of substance by 1°C.
Molar heat capacity = \(\frac{C}{n}\) \(\frac{q}{∆T × n}\)
Where C = Heat capacity (Absorbed heat), ∆T = Increase in temperature, n = Number of moles.
Its S.I. unit is joule K^-1 mol^-1.

Question 17.
Write Hess’s law?
Answer:
In 1840, G.H. Hess formulated a law known as Hess’s law. According to law, “The enthalpy change in a physical or chemical change is same whether the process is carried out in one step or in several steps.”

Question 18.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process. This process mainly occurs in isolated system. For this type of process dq = 0.

Question 19.
What is standard enthalpy of reaction?
Answer:
The enthalpy change takes place at standard state i.e., at 298 K temperature (25°C), 1 atm pressure (760 mm) condition is called standard enthalpy of reaction. It is denoted by ∆H°_r or ∆_rH°.

Question 20.
What is enthalpy of solution? Explain with example?
Answer:
Enthalpy change taking place during the dissolution of one mole of a substance in excess of a solvent such that further addition of solvent does not produce any heat change is known as enthalpy of solution.
Example: KCl_(s) + aq → KCl_(aq)

Question 21.
What is enthalpy of hydration? Explain with example?
Answer:
Enthalpy change taking place when one mole of anhydrous salt combines with the required number of moles of water to form hydrated salt is called enthalpy of hydration.
Example: CuSO_4(s) + 5H₂O_(l) → CuS0₄.5H₂O ∆H = – 78.2kJ.

Thermodynamics Short Answer Type Questions – II

Question 1.
What is law of energy conservation? Write its mathematical expression. Or, What is the first law of thermodynamics? Deduce its mathematical expression?
Answer:
This law was first expressed by Meyer and Helmholt. According to this law, “Energy cannot be created or destroyed although one form and vice – versa”.
Or
Whenever a certain quantity of energy in one form disappear, an equivalent amount of energy in another form reappear.

Mathematical form:
Let us suppose that system has internal energy equal to U₁. If it absorbs heat energy 17 from the surroundings then internal energy will increase and becomes U₁ + q. If the work is done on the system then its final internal energy will become
U₂ = U₁ + q + w
U₂ – U₁ = q + w,
∆U – q + w, (∴ U₂ – U₁ = ∆U
If the work done on the system is w, then
∆U = q + w
If work done by the system
∆U = q – w

Question 2.
What is Enthalpy of fusion? Explain with example?
Answer:
Enthalpy change taking place during the conversion of 1 mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
Example: Enthalpy of fusion of ice at 273 K is 6.0 kJ.
H₂O_(s) → H₂O_(l);
i.e 6.0 kJ of energy is absorbed for the conversion of 1 mole of ice into water. ∆H = +6.0 KJ

Question 3.
For an isolated system if ∆U = 0, what will be ∆S?
Answer:
For an isolated system, AU = 0 and for a spontaneous process, the change in entropy should be positive. For example: For a closed container, which is an isolated system, two gases A and B are diffused (MPBoardSolutions.com) Both gases A and B are separated by a kinetic separator. When the separator is removed, the gases starts fusing with each other and randomness increases in the system. For this process, ∆S > 0 and ∆U = 0.
Again ∆S = \(\frac { q_{ rev } }{ T } \) = \(\frac { \Delta H }{ T } \) = \(\frac { \Delta U+P\Delta V }{ T } \) = \(\frac { P\Delta V }{ T } \) [∴∆U = 0]
So, T∆S or ∆S > 0.

Question 4.
On the basis of the following equations, write a note on thermodynamic stability of NO_(g):
\(\frac{1}{2}\) N_2(g) + \(\frac{1}{2}\) O_2(g) → NO_(g); ∆_rH° = 90kJ mol^-1
NO_(g) + \(\frac{1}{2}\) O_2(g) → NO_2(g); ∆_rH° = 94kJ mol^-1
Answer:
NO_(g) is unstable, as the preparation of NO is an endothermic reaction. But preparation of NO_2(g) occurs because it is an exothermic reaction (energy evolved). So unstable NO_(g) is converted into NO_2(g).

Question 5.
At equilibrium state, whose value will be zero ArG or ArG°?
Answer:
∆_rG = ∆_rG° + RTlnK
At equilibrium, 0 = ∆_rG° + RTlnK
or ∆_rG° = – RTlnK
∆_rG° = 0 (When K = 1)
For other values K, ∆_rG° will be zero.

Question 6.
What is the meaning of calorific value of any fuel? Explain with example?
Answer:
The heat or energy evolved in joule or calorie by the (combustion) burning of 1 gram food or fuel is known as calorific value of fuel.
C₂H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H_2(g); ∆H = -2840 kJ
In this process from 1 mole glucose or 180 gm = 2480 kJ energy obtained
So, energy obtained from 1 gm glucose = \(\frac{2480}{180}\) = 15.78 kJ/gm.
So, calorific value of glucose is 15.78 kJ/gm.

Question 7.
Explain heat of vapourization and heat of reaction?
Answer:
Heat or Enthalpy of vapourization : Enthalpy change taking place during the conversion of 1 mole of liquid into vapours at its boiling point and 1 atm pressure is called enthalpy of vapourization.
Example:
H₂O ⇄ H₂O_(g); ∆H = +40.7 kJ

Heat of reaction:
Enthalpy change taking place when number of moles of reactants as represented by the chemical equation have completely reacted is known as enthalpy of reaction. It is denoted by ∆H_f.
Example: C_(s) + O_2(g) → CO_2(g); ∆H_f = – 393.5 kJ/mol

Question 8.
Explain enthalpy of fusion and enthalpy of sublimation with example?
Answer:
Enthalpy of fusion:
Enthalpy change taking place during the conversion of I mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
H₂O_(s) ⇄ H₂O_(l); ∆H = +6.01 kJ
Enthalpy of Sublimation:
Enthalpy change taking place when one mole of solid changes into vapours without passing into intermediate liquid state at a temperature below its melting point is known as enthalpy of sublimation.
I_2(s) ⇄  I_2(g). ∆H = +62.4 kJ

Question 9.
How the change in entropy takes place in vapourisation process?
Answer:
Entropy of vapourisation:
Change in entropy when one mole of a liquid at its boiling point changes to the vapour state at the same temperature.
∆S_vapour = S_(vapour) – ∆S_(liquid) = \(\frac { \Delta H_{ vap } }{ T_{ b } } \)
Here, ∆_vap H is the latent heat of vapourisation (enthalpy of vapourisation) and T_b is the boiling point when liquid is converted into vapour, entropy of the system increases. Thus, ∆S_vap is +ve.

Question 10.
What is Internal Energy? Is its absolute value can be determined?
Answer:
Physical and chemical process occurs by some energy change. This energy may appear in the form of light, heat and work. (MPBoardSolutions.com) This evolution and absorption of energy clearly shows that every’ substance or system is associated with some definite amount of inherent energy’. The actual value of this inherent energy depends upon:

1. Chemical nature of substance
2. Conditions, like temperature, pressure and volume, and
3. Composition of the substance.

Thus, “The energy stored within a substance is called internal energy or intrinsic energy”.
Actually, internal energy is the sum of various forms of energy such as; electronic energy E_e, nuclear energy E_n, chemical bond energy E_c, potential energy E_p and kinetic energy E_k . Kinetic energy is the sum of translational energy (E_t), vibrational energy (E_v) and rotational energy (E_r).
It is represented by the symbol ‘U’,
U = E_e + E_n + E_c + E_p + E_k
It may be noted that, absolute value of internal energy cannot be determined, because it is not possible to determine the exact values for the constituent energies, such as: translational, vibrational, rotational energies etc.

Question 11.
Expansion of any gas in vacuum is called free expansion. 1 L of an ideal gas expands isothermally upto 5 L, then determine the change in internal energy and work done?
Solution:
Work done, W = – P_(external) (V₂ – V₁)
When P_(external) = 0,
So, W = – 0 (5 – 1) = 0
For isothermal expansion,
∆U = 0
So, ∆T = 0.

Question 12.
An ideal gas filled in a cylinder (according to figure.) is compressed in a single step with external pressure P_(external) Then what will be the work done on the gas? Explain with graph?
Solution:
Let the initial volume of the gas is V_i and pressure of cylinder is P. On compressing the gas by pressure P the final volume of gas is V_f.
So, change in volume ∆V = (V_f – V_i)
If W is the work done by the piston on the system
W = P_(external) (- ∆V)
W = P_(external) (V_f – V_i)
This can be shown in the figure by (P – V) graph, The work done is equal to ABV_fV_i. The positive sign shows that the work is done on the system.

Thermodynamics Long Answer Type Questions – I

Question 1.
What is heat capacity? Deduce the expression C_p – C_v = R?
Answer:
Heat Capacity: It is equal to the amount of heat required to raise the temperature of the system through 1°C. Its unit is JK^-1.
Relationship between C^p and C^v:
At constant volume: q_v = C_v∆T = ∆U
At constant pressure: q_p = C_p∆T = ∆H
∆H and ∆U are related to each other as
∆H = ∆U + ∆n_gRT
or ∆H = ∆U + ∆n_g(PV)
For 1 mole of ideal gas PV = RT
∴ ∆H = ∆U + ∆(RT)
or ∆H = ∆U + R∆T
on putting the value of ∆H and ∆U
C_p∆T = C_v∆T = C_v∆T + R∆T
Dividing whole equation by ∆T,
C_p = C_v + R
C_p – C_v = R
Thus, value of C_p is always more than C_v and the difference between them is about 2 calories or 8.314 joule.
This relationship is known as Meyer’s relationship.
The ratio C_p/C_v:
The ratio of molar heat capacities at constant pressure (C_p) to that at constant volume (C_v) is represented by γ. Value of γ gives information about the atomicity of the gas. Thus,
For monoatomic gases γ = 1.67
For diatomic gases γ = 1.40
For triatomic gases γ = 1.30

Question 2.
Explain Enthalpy of combustion? Write its uses also?
Answer:
Enthalpy of combustion:
The enthalpy change taking place when one mole of substance is completely oxidised or burnt in presence of excess of oxygen is known as enthalpy of combustion. For example, the enthalpy of combustion of methane is 890.4 kJ.
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g) ∆Hc = – 890.4 kJ

Carbon on the other hand is oxidised to carbon monoxide and carbon dioxide.
C_(s) + \(\frac{1}{2}\) O_2(g) → CO_(g); ∆H = – 110.5kJ
and C_(s) + O_2(g) → CO_2(g); ∆H_c = – 393.5 kJ

In this case enthalpy of combustion of carbon is – 393.5 kJ and not – 110.5 kJ as formation of CO is a result of incomplete combustion of carbon.
Uses of Enthalpy of combustion:

1. To determine the calorific value of fuel.
2. To determine the enthalpy of reaction of compounds.
3. Determination of structure of compounds.
4. To calculate the calorific value of food.

Question 3.
Prove that qr = ∆H_p?
Or, Prove that at constant pressure and constant temperature the heat of reaction is equal to the change in enthalpy of the system?
Answer:
Suppose enthalpy, internal energy and volume of a system in initial state are H₁, U₁ and V₁ respectively and after gaining heat these values becomes H₂, U₂ and V₂ respectively, then according to definition.
H₁ = U₁ + PV₁, (in initial state) ……………… (1)
H₂ = U₂ + PV₂, (in final state) …………….. (2)
Subtracting eqn. (1) from eqn eqn. (2),
H₂ – H₁ = U₂ – U₁ + P(V₂ – V₁)
or ∆H = ∆U + P∆V
Where ∆H is enthalpy change, ∆U is change in internal energy and ∆V is change in volume. Therefore at constant pressure enthalpy change is equal to sum of internal energy change and expansion type of mechanical work.
According to first law of thermodynamics,
∆U = q – P∆V
q = ∆U + P∆V
= (U₂ – U₁) + P(V₂ – V₁)
= (U₂ + PV₂) – (U₁ + PV₁)
= H₂ – H₁
= ∆H
∴ ∆H = q_p
Thus, enthalpy change represents the heat change occurring at constant temperature and pressure. It is noteworthy that though q is path dependent, q_p is not because ∆H is a state function.

Question 4.
Differentiate between Reversible and Irreversible processes?
Answer:
Differences between Reversible and Irreversible process:
Reversible process:

1. It is carried out infinitesimally slowly i.e., the difference between driving force and the opposing force is very very small.
2. It is an ideal process requiring infinite time for completion.
3. Equilibrium is not disturbed at any stage during the process.
4. Work obtained is maximum.
5. It is an imaginary process and cannot be realised in actual practice.

Irreversible process:

1. This process is carried out rapidly /.e.,the difference between driving force and the opposing force and the opposing force is quite large.
2. It is a spontaneous process requiring finite time for completion.
3. Equilibrium may occur only after the completion of the process.
4. Work obtained is not maximum.
5. It is a natural process which occurs in
6. particular direction under given set of conditions.

Question 5.
What are the factors affecting enthalpy of reaction?
Answer:
Factors on which enthalpy of a reaction (∆H) depends: Enthalpy of a reaction i.e., ∆H depends upon the following factors:
1. Physical state of reactants and products:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

2. Quantities of the reactants involved:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

3. Allotropic modifications:
Allotropes of an element may have different enthalpies. For example, enthalpy change during combustion of graphite and diamond is – 393.5 kJ/mol^-1 and – 395.4 kJ/mol^-1 respectively.
C_(graphite) + O_2(g) → CO_2(g); ∆H = – 393.5 kJ
C_(diamond) + O_2(g) → CO_2(g); ∆H = – 395.4 kJ

4. Temperature:
Value of enthalpy of reaction is dependent on the temperature at which the reaction is carried out. For example, at 25°C enthalpy of formation of HCl_(g) is 184.6 kJ while at 75°C it is 184.4 kJ.
H_2(g) + Cl_2(g) → 2Hl_(g); ∆H = 184.6 kJ at 25°C
H_2(g) + Cl_2(g) → 2HCl_(g); ∆H = 184.4 kJ at 75°C

Question 6.
Prove that at constant volume q_v = ∆U?
Answer:
When a reaction occurs at constant volume, then no work is done by the system. So W = 0.
So, ∆U = q + W
On putting the value, ∆U = q
So, at constant volume the energy is absorbed which increases the internal energy of the system. So,
∆U = q + W = q + p∆V, (W = P∆V)
Since, the reaction occurs at constant volume. So, ∆V = 0.
On putting vallue, ∆U = q_v

Question 7.
From following data determine the heat of reaction of CH₄ or enthalpy, ∆H:
C_(s) + O_2(g) → CO₂; ∆H = – 97k cal ………… (1)
2H_2(g) + O_2(g) → 2H₂O_(g) ∆H = – 136k cal …………. (2)
CH₄ + 2O_2(g) → CO_2(g) + 2H₂O_(g); ∆H = – 212k cal ………………… (3)
Solution:
To determine
C_(s) + 2H_2(g) → cH_4(g); ∆H = ?
On adding eqn.(1) and (2),

Thermodynamics Long Answer Type Questions – II

Question 1.
What is Hess’s Law of constant heat summation? Explain with an example?
Answer:
In 1840, G.H. Hess gave an important law of constant heat summation according to which, “The enthalpy change in a particular reaction is always constant and does not depend on the path in which reaction takes place”.
Or
“The enthalpy change in a physical or chemical process is the same whether the process is carried out in one or in several steps.”
This law is based on the law of conservation of energy. Suppose that the conversion of substance A to (MPBoardSolutions.com) substance Z takes place in a single step by first method and through several steps in second method. In single step by first method and through several steps in second method. In single step:
A → Z + Q₁

Where, Q₁ is the energy in several steps:
A → B + q₁
B → C + q₂
C → Z + q₃
Total energy evolved in several steps = q₁ + q₂ + q₃
= Q₂ calories (suppose)
Acoording to Hess’s law,
Q₁ = Q₂
Suppose Hess’s law is incorrect and Q₂ > Q₁. In this stage if we convert A to Z by several steps and then Z directly to A, then heat equal to (Q₂ – Q₁) is produced. By repeating this cyclic process several times, an (MPBoardSolutions.com) unlimited amount of heat (energy) may be produced in an isolated system. But this is against the law of conservation of energy.
Practically also, Hess’s law is proved to be true.
Example: Carbon can be directly burnt to produce C0₂ or in the second method it is first converted to carbon monoxide and then oxidized to carbon dioxide.

Energy evolved by both the methods is nearly same. Different of 0.3 kcal is due to experimental error.
∴ ∆H = ∆H₁ + ∆H₂

Question 2.
Prove that ∆H = ∆U + P∆V?
Or, Explain the relationship between ∆H and ∆U?
Answer:
Relation between AH and AU: In case of solids and liquids, the difference between ∆H and ∆U is not significant but in gases it is significant. Let us consider a reaction involving gases. Let the process be isothermal and carried out at constant pressure (P). If V_A is the total volume of the gaseous reactants and V_B be the total volume of gaseous products, also n_A be the number of moles of gaseous reactants and n_B be the number of moles of gaseous products. Then
PV_A = n_ART ……………………. (1)
and PV_B = n_BRT ………………….. (2)
Substarcting eqn. (1) from eqn. (2) we get
PV_B – PV_A = (n_B – n_A)RT
or P(V_B – V_A) = (n_B – n_A)RT
P∆V = ∆nRT
For gaseous reactants PV_R = n_RRT …………………… (3)
and for gaseous products PV_p = n_pRT …………………… (4)
Substracting eqn. (3) from eqn. (4), we get
P(V_p – V_R) = (n_p – n_R) RT
or P∆V = ∆n_gRT ……………………. (5)
But enthalpy change ∆H = ∆U + P∆V …………………….. (6)
Substracting the value of P∆V from eqn. (5) into eqn. (6), we get
∆H = ∆U + ∆n_gRT ………………. (7)
Thus, using above eqn. (7) ∆H can be converted into ∆U or vice – versa.eqn. (7) can be wriien as
q_p = q_v + ∆n_gRT ………………….. (8)
Because ∆H = q_p and ∆U = q_v.
Conditions:

1. If the number of moles of products is greater than that of reactants than ∆n will be +ve and ∆H is greater than ∆U. So,
∆H = ∆U + ∆nRT

2. If the number of moles of reactants is greater than the number of moles of products then ∆n will be – ve and the value of ∆H is less than ∆U.
∆H = ∆U – ∆nRT

3. If the No. of moles of reactants is equal to No. of moles of products then ∆n = 0, then in this condition ∆H = ∆U.

Question 3.
Deduce an expression for PV work done?
Answer:
Let us, consider a cylinder, fitted with a weightless, frictionless piston having a cross – sectional area A, filled with 1 mole of an ideal gas. The total volume of gas is V_i and pressure inside the cylinder is P_in.
Suppose, external pressure on the gas is P_ex which is slightly greater than the internal pressure of the gas. (MPBoardSolutions.com) Due to this difference in pressure the gas is compressed till the pressure inside becomes equal to P_ex Suppose, the change is achieved in one single step and the final volume of the gas is V_f The gas is compressed and suppose the piston moves a distance l. Work done during compression is
W = Force × Displacement = F × l

or F = P × ABV
∴W = P × A × large
or W = – P∆V, [∴ A × l = ∆V]

The negative sign in the expression is required to obtain conventional sign W. In compression, V_f < V_i and therefore, (V_f – V_i) or ∆V is – ve.
Hence, W will come out to be +ve from the above expression. In expansion type work V_f > V_i and value of ∆V is positive, therefore, work done will be negative.

This expression is useful for all types of PVwork and for irreversible flow.
Now, we will calculate, the work done during expansion of ideal gas in a reversible manner and in isothermal condition.

Question 4.
What is free energy? Derive its mathematical form and write Gibbs – Helmholtz equation?
Answer:
The free energy of a system is defined as the maximum amount of energy of the system which can be converted into useful work.
Free energy is related to enthalpy (H), entropy (S) and absolute temperature (T) as,
G = H – TS
Since, we know H = E + PV
∴ G = E + PV – TS
Change in free energy may be given as,
∆G = ∆E + ∆(PV) – ∆(TS)
If the process is carried out at constant temperature and pressure, then
∆(TS) = T∆S and ∆(PV) = P∆V
∆G_Tp = ∆E + P∆V – T∆S or
∆G = ∆H – T∆S
The above equation is called Gibbs – Helmholtz equation and it helps to predict the spontaneity of a process.
Free energy change and spontaneity:
For a system which is not isolated with surroundings
∆S_total = ∆S_system + ∆S_surrounding …………………… (1)
when reaction takes place at constant temperature and constant pressure, heat is supplied to surrounding.
∆S_surrounding = \(\frac { -q_{ p } }{ T } \) = \(\frac { -\Delta H }{ T } \), (q_p = ∆H at constant pressure) ………………………. (2)
From eqns. (1) and (2),
∆S_total = ∆S_system – \(\frac { -\Delta H }{ T } \) …………….. (3)
Since, all the quantities on the right – hand side are system properties, the subscript ‘system’ is not used in equations.
Multiplying both sides by T, we get
T∆S_total = T∆S – ∆H, (Where, ∆S = ∆S_system)
or -T∆Sc = ∆H – T∆S …………………… (4)
For Gibbs free energy (G),
G = H – TS
∆G = ∆H – T∆S – S∆T ………………… (5)
or ∆G = ∆H – T∆S_total
For the process taking place at constant temperature and constant pressure, eqn. (5) will be
So that, ∆G = ∆H – T∆S ……………………… (6)
Comparing eqns. (4) and (6),
∆G = – T∆S_total ………………….. (7)
So that, ∆G = – ve(for sontaneous chnages)
We know that, for spontaneous chemical change ∆S_total is positive. Eqn. (7) shows that the spontaneity of a change can be predicted on the basis of the value of ∆G.
Three special cases may be considered according to eqn. (7):

1. If AG is negative, the change is spontaneous.
2. If AG is zero, the system is in equilibrium.
3. If AG is positive, the change is non – spontaneous.

Conditions for spontaneity of a process (conditions for ∆G to be – ve):

1. If AH is negative and AS is positive, AG would certainly be negative and the process will be spontaneous.
2. If AH is negative and AS is also negative, then AG would be negative if AH is greater than TAS in magnitude.

Question 5.
Explain the determination of internal energy ∆U by Bomb calorimeter under the following heads:

1. Labelled diagram of the apparatus,
2. Explanation of the process
3. Calculations.

Answer:
Experimental determination of change in internal energy:
The change in internal energy in a chemical reaction is determined with the help of an apparatus called bomb calorimeter. (MPBoardSolutions.com) It is made up of steel so that it can bear high pressure developed during the chemical reaction taking place in the calorimeter. The inner side of the steel vessel is coated with some non – oxidizable metal like Pt or Au. It is also fitted with a pressure tight screw – cap. The two electrodes are connected to each other through a platinum wire dipped in a platinum cup.

A small known mass of the substance under investigation is taken in the platinum cup. The bomb is filled with excess of oxygen under a pressure of 20 – 25 atm and sealed. Now it is kept in an insulated water – bath which contains a known amount of water. The water – bath is also provided with a thermometer and mechanical stirrer.

The initial temperature of water is noted and the reaction (i.e., combustion of the sample) is started by passing an electric current through the Pt wire. (MPBoardSolutions.com) The heat evolved during the chemical reaction raises the temperature of water which is recorded by the thermometer. When rise in temperature and the heat capacity of the calorimeter are known, the amount of heat evolved in the chemical reaction can be calculated. This will be equal to the change in internal energy (∆E) of the reaction.

Calculation: Let W = Mass of calorimeter in gm, w = Water equivalent of calorimeter, bomb stirrer, etc. t°C = Rise in temperature, x = Mass of compound ignited in gm and m = Molecular mass of the compound.
Heat produced by x gm compound = (W + w) t calories
∴ Heat of combustion of the compound at constant volume,
∆U or ∆E = \(-\frac { m }{ x } \) (W + w) t calorie/mol.
Heat is evolved so negative sign is used.
Using the equation ∆H = ∆U + ∆ng RT heat of combustion ∆H at constant pressure can be caluculated.

MP Board Class 11 Chemistry Important Questions

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

Electrostatic Potential and Capacitance Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

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

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

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

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

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

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

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

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

Question 2.
Fill in the blanks:

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

Answers:

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

Question 3.
Match the Column:

Answers:

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

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

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

Answers:

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

Electrostatic Potential and Capacitance Important Questions Very Short Answer Type Questions

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

Question 2.
Write the name of the physical quantity whose SI unit in J/C. Is it a scalar or vector ?
Answer:
Electric potential, it is a vector quantity.

Question 3.
Draw a equi potential surface for a unit charge.
Answer:

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

Question 5.
On going in direction of electric lines of force, electric potential decreases or increases.
Answer:
Electric potential decreases.

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

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

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

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

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

Question 11.
Why condenser are used in computer’s ?
Answer:
Condenser are used as memory chip in computer.

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

Electrostatic Potential and Capacitance Important Questions Short Answer Type Questions

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

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

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

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

Question 5.
What would be the work done if a point charge +q is taken from a point A to point B on the circumference of a circle with another point charge +q at the center:
Answer:
The points A and B are at same distance from the charge + q at the center, so V_A = V_B So, work done, W= q₀ (V_A – V_B) = 0.

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

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

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

Question 9.
Why must electrostatic field be normal to the surface at every point of a charged conductor ?
Answer:
If electric field is randomly directed, then it can be resolved, into two components. The horizontal component on this surface is E sin θ.
For electrostatic situation
£ sin θ = 0
⇒ sin θ = 0
⇒ θ = 0°

So, the electric field is normal to surface.

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

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

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

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

Question 14.
In the shown figure what will be the work done to bring a z point charge from the point X to Y to Z ?
Answer:
There the point Z and Y are situated on same equaipotential surface. Therefore work done to bring a point charge from A’to Zand from X to Z will be same.
.’. W_y= W_z.

Question 15.
Derive an expression for electric potential due to a point charge. Is it scalar or vector and why ?
Answer:
Consider a point charge q placed at origin O. Potential at P has to be found out. Let the medium between charge ‘q’ and P has dielectric constant E_r.

Electric field at P due to charge q is
E = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r^{2}}\)
The electric field \(\vec{E}\) points away from the charge q. A force \(\vec{F}\) = -q₀ \(\vec{E}\) has to be applied on the charge so that it can be brought near to q. The small work required to move the test charge q₀ from P to Q through a small distance dr is given by dW = Fdr
= -q₀ Edr
The total work done in moving the charge q0 from infinity to point P will be obtained by integrating the above equation as –

But electric potential is defined as work done per unit test charge.

Potential at P is V = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r}\)
If medium between q and q₀ is vacuum then £_r = 1
Then , V =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
This is the required expression.

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

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

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

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

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

Properties:

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

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

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

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

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

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

As the negative charge is nearer therefore the potential of plate A decreases. Now, the plate B is earthed then the free positive charge will go to earth and hence the potential of A decreases by more value.
C = \(\frac { Q }{V}\)
As V decreases, C will increase. This arrangement is called capacitor or condenser.

Question 21.
Derive an expression for parallel plate capacitor.
Answer;
Let A and B be two plates of a parallel plate capacitor separated by a distance d apart. Area of each plate is A and dielectric constant of the medium between them is E_r Now, plate A is given + Q charge. Therefore, – Q charge will be induced on the nearer surface of the plate B and + Q charge on the other side. As B is connected to earth, + Q charge of B will go to earth. Let the charge density of A is cr, therefore that of B will be -σ.

Now, σ = \(\frac { Q }{A}\)
Intensity between the plates will be given by
E = \(\frac{\sigma}{\varepsilon_{0} \varepsilon_{\mathrm{r}}}\)
E = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{r}}\)
But, potential difference between the plates A and B is
V = Electric field intensity ×
Distance between to plates = Ed
V = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{s}} \cdot d\)
But, C = \(\frac{Q}{V}=\frac{Q}{\frac{Q d}{A \varepsilon_{0} \varepsilon_{r}}}\)
C = \(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}\)
This is the required relation.
For air or vacuum, E_r = 1
C = \(\frac{\varepsilon_{0} A}{d}\)

Question 22.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in series. Derive an expression for the equivalent capacitance.
Answer:
The given figure shows three capacitors of capacitances C₁ C₂ and C₃ con – nected in series. A potential difference of V is applied across the combination, charges of + Q and – Q are developed on the plates of the capacitor.
Potential difference across the individual capacitors will be

V₁ = \(\frac{Q}{C_{1}}\) , V₂ = \(\frac{Q}{C_{2}}\), V₃ = \(\frac{Q}{C_{3}}\) …….(1)
The sum of these must be equal to the applied potential difference V.
V = V₁ + V₂ + V₃ ………(2)
Let C be the equivalent capacitance of the series combination
∴ C = \(\frac { Q }{ V }\) or V = \(\frac { Q }{ C }\) ………..(3)
V₁ + V₂ + V₃ = \(\frac { Q }{ C }\) [from equ..(2)]
\(\frac{Q}{C_{1}}\) + \(\frac{Q}{C_{2}}\) + \(\frac{Q}{C_{3}}\) = \(\frac{Q}{C}\) [from equ..(1)]
Q(\(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)) = \(\frac{Q}{C}\)
\(\frac { 1 }{ c }\) = \(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)
This is the required expression.

Question 23.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in parallel. Derive an expression for the equivalent capacitance C.
Answer:
Consider three capacitor of capacitance’s C₁ C₂ and C₃ connected in parallel. A potential difference V is applied across the combination. Charges set up in the individual capacitor will be.

Q₁ = C₁V, Q₂ = C₂V,Q₃ = C₃V …(1)
Total charge stored in the parallel combination is
Q = Q₁ + Q₂ + Q₃ ……….(2)
If C is the equivalent capacitance of the combination
Then, C = \(\frac { Q }{ V }\) Q = CV …………(3)
Q₁ + Q₂ + Q₃ = CV [from eq. (2)]
C₁V + C₂V + C₃V = CV [from eq. (1)]
V(C₁ + C₂ + C₃) = CV
C = C₁ + C₂ + C₃
This is the required expression.

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

Formula derivation:
Let us consider about a conductor of capacity C which is given charge +Q due to which its potential becomes V. As the charge increases work done also increases. Let at any instant the potential of conductor be V due to charge q.
∴ C = \(\frac { q }{ v}\)
or v = \(\frac { q }{ C}\)
Now, at potential Kthe work done in giving the charge dq will be dW
∴ dw = Vdq
or dw = \(\frac { q }{ C}\)dq
Work done in charging the conductor from 0 to Q

This work done is stored as potential energy on the conductor. Energy of a charge conductor
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
But Q = CV
∴ U = \(\frac{1}{2} \frac{C^{2} V^{2}}{C}\)
or U = \(\frac{1}{2} C V^{2}\)
∴ U ∝ V² because C is constant

Question 25.
Prove that on connecting two charged conductors, charges distribute on them according to their capacities.
Answer:
When two isolated charged conductors A and B are connected by a thin wire, charge flows from the conductors at high potential to the conductor at low potential till the potential of both A and B became equal. The phenomenon involved is called distribution of charges and the total charge of the entire system remains conserved. Let the capacitance of A and B be C₁ and C₂, the charges be Q₁ and Q₂ respectively. Then the potentials are V₁ and V₂ respectively.
∴ Initially, Q₁ = C₁V₁ and Q₂ = C₂V₂

The conductors are joined by a wire of negligible capacitance, the charges flow from- the conductor at higher potential to the conductor at lower potential till the potentials on each conductor become equal.
The net charge on the system,
Q = Q₁ + Q₂
The common potential, V = \(\frac { Total charge }{ Total capacitance }\)

After the potential becomes equal let the charge on A ₁ be Q₁ and charge on A₂ be Q₂.

Dividing eqn. (2) by eqn. (3), we get

When the conductors are joined, then the charges get distributed in the ratio of their capacities.

Question 26.
Obtain an expression for potential due to a group of point charges.
Or
Derive the expression for potential energy.
Answer:
Consider a group of point charges q₁,q₂,q₃……..q_n which are situated at a dis-tance of r₁, r₂, r₃…….. _nn respectively from the point P. The potential due to these point charges is to be obtained at P. Now potential at P due to q₁ is

potential due to q₂,q₃, ……… q_netc

Total potential at P will be V = V₁ + V₂ + V₃ + ……….. + V_n

This is the required expression.

Electrostatic Potential and Capacitance Important Questions Long Answer Type Questions

Question 1.
Derive the expression for the capacity of a parallel plate capacitor, when the medium between the plates is partially filled by a dielectric medium.
Answer:
Let A and B are parallel plates of a capacitor. The distance between the plates is d and plate of thickness t and dielectric constant Er is introduced.

Now, plate A is given charge +Q.
Let the charge density be σ.
∴ σ = \(\frac { Q}{ A}\)
Intensity of field in air ,
E_{0}=\(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)
If the intensity of field inside the dielectric medium be E, then
Dielectric constant = \(\frac {Electric field in vacuum}{ Electric field in medium}\)
or \(\varepsilon_{r}=\frac{E_{0}}{E}\)
or E = \(\frac{E_{0}}{\varepsilon_{r}}=\frac{Q}{\varepsilon_{0} \varepsilon_{r} A}\)
Now, potential difference between A and B,
V=E₀ (d – t) + Et, [(d – t) is vacuum distance]

This is the required expression.
Metal is a conductor. When metal is used in place of the dielectric, it will conduct electricity and the potential difference will become zero. So, capacitor will not work.

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

• Common potential
• Loss of energy.

or
prove that when two charged conductors are connected, there will be a loss of energy
Or
In redistribution of charges, is there a loss of energy ? Deduce an expression to confirm the answer.
Answer:
Let A and B be two conductors of capacities C₁ and C₂ respectively. When charges Q₁ and Q₂ are given separately the potentials become V₁ and V₂ respectively.Total charges, Q = Q₁ + Q₂ ………..(1)
But, Q₁ = C₁V₁ and Q₂ = C₂V₂
By eqn. (1), we get
Q = C₁V₁ + C₂V₂
Total capacity, C = C₁ + C₂

(1) Common potential:
Let the conductors are connected by a wire and the common potential becomes-V.
Q₁ + Q₂ = (C₁ + C₂)V
V = \(\frac{Q_{1}+Q_{2}}{C_{1} + C_{2}}\)
V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\) ……..(2)
This is the expression for the common potential.

(2) Loss of energy: Total energy of the conductors before connection:

and total energy after connection,

Putting the value of V from eqn. (2) in eqn. (4), we get

Hence, difference of energy’,

(V₁ – V₂)² is positive, hence (V₁ – V₂) is positive. Hence, during redistribution, there will be always loss of energy.
i.e., U₁ – U₂>0 ⇒ U₁>U₂
i.e., energy before joining is greater than energy after joining.
The loss in energy,

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

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

Uses:

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

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

Answer the following questions:

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

Answer:

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

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

Electrostatic Potential and Capacitance Important Questions Numerical Questions

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

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

C₁ ,C₂ is in series, its resultant (C’) is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
or \(\frac { 1 }{ c}\) = \(\frac { 1 }{ 4}\) + \(\frac { 1 }{ 4}\) = \(\frac { 2 }{ 4}\)
or C = \(\frac { 4 }{ 2}\) = 2µF.
C and C₃ is in parallel combination,
Its resultant C is C = C + C₃
C = 2 + 4 or C = 6µF.

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

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

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

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

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

Question 4.
A test charge is moved from A to B, B to C and A to C in an electric field E as shown in the figure :

Find (1) Potential difference between A and C
(2) At which point electric potential will be high and why ?
Solution:
1. In right angled ∆ABC
AB² = AC² – BC² = 5² – 3²
∴ AB = 4 = dr
BC is perpendicular to electric field, therefore potential will be same at B and C.
V_A – V_C = V_A – V_B = -Edr = -4E
2. Therefore potential at point C will be more than potential of point A.

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

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

1. 9pF,
2. 4.5pF.

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

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

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

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

Question 9.
Two charges 5 x 10^-8C and -3 x 10^-8C are located 16 cm apart At what point, on the line joining the two charges, is the electric potential zero ? Take the potential at infinity to be zero. (NCERT)
Solution:
Case I.
Let electric potential be zero at point C lying at distance x from the positive charge.

Given, q₁ = 5 × 10^-8 C;
q₂ = -3 × l0^-8 C
AC = x cm : CB = (16 – x) cm
Now, Potential at C is zero i. e.,
V₁ + V₂ = 0

-8x + 80 = 0
8x = 80
x = 19 cm
i.e., electric potential at a distance of 10 cm from positive charge will be zero.

Case II.
The other possibility is that the point C may also lie on produced AB.

Now, V₁ + V₂ = 0

5(x – 16) – 3x = 0
5x – 80 – 3x = 0
2x – 80 = 0
x = 40 cm from the positive charge

Question 10.
Determine the equivalent capacitance between A and B in the following circuits:

Solution:
(i). Mark the junctions as C and D.

(But C will be A and D will be B)
Draw the equivalent network, which is given below

Equivalent capacitance,
C = C₁ + C₂ + C₃
or C = 1 + 1 + 1 = 3µF

(ii). To move from A to B, there are two paths P -1 and P – II.

(As A and B, the path is dislocated temporarily)
The capacitors in P -II are in series. So, the equivalent becomes

The resultant capacity of series combination is
\(\frac { 1 }{ C’ }\) = \(\frac { 1}{ 3 }\) + \(\frac { 1 }{ 3}\) + \(\frac { 1 }{3 }\)
= \(\frac { 3 }{ 3 }\) = 1µF = C’ = 1µF
The equivalent further becomes

Total capacity C = 3 +1 = 4µF.

MP Board Class 12th Physics Important Questions

MP Board Class 12th Biology Important Questions Chapter 14 Ecosystem

Ecosystem Important Questions

Ecosystem Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
The flow of energy in ecosystem: (MP2011)
(a) Unidirection
(b) Bidirectional
(c) Tridirectional
(d) Four directional
Answer:
(a) Unidirection

Question 2.
Chief source of energy in ecosystem :
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of these.
Answer:
(a) Solar energy

Question 3.
The term ‘ecosystem’ was used for the first time by : (MP 2015)
(a) Tansley
(b) Odum
(c) Reiter
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
The pyramid of number of tree ecosystem is :
(a) Inverted
(b) Upright
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 5.
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 6.
Pyramid of biomass of lake ecosystem is :
(a) Upright
(b) Inverted
(c) Upright or inverted
(d) None of these.
Answer:
(b) Inverted

Question 7.
Vegetation present between two communities :
(a) Ecad
(b) Ecotype
(c) Ecotone
(d) None of these.
Answer:
(c) Ecotone

Question 8.
Xerosere is started from :
(a) Water
(b) Naked rock
(c) Swamp
(d) All of these.
(b) Naked rock

Question 9.
Serial development of plant community is called :
(a) Attack
(b) Succession
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Succession

Question 10.
Plants starting succession in any area are called :
(a) Pioneer
(b) Sere
(c) Displacer
(d) All of these.
Answer:
(a) Pioneer

Question 11.
In forest ecosystem, the pyramid of energy is : (MP 2012,17)
(a) Always inverted
(b) Always upright
(c) First upright then inverted
(d) None of these.
Answer:
(b) Always upright

Question 12.
The study of ecology of a 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) Decomposers
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Man is:
(a) Autotrophic
(b) Carnivorous
(c) Herbivorous
(d) Omnivorous.
Answer:
(d) Omnivorous.

Question 15.
In any ecosystem, solar energy is conserved by :
(a) By producers
(b) By consumers
(c) By decomposers
(d) All of the above.
Answer:
(a) By producers

Question 16.
The two components of ecosystem are: (MP 2013)
(a) Organism and Plant
(b) Weeds and trees
(c) Frog and man
(d) Biotic and Abiotic.
Answer:
(d) Biotic and Abiotic.

Question 2.
Fill in the blanks:

1. The transitional zone present between to adjacent communities is called ………………….
2. All the plants of a particular area constitute …………………. of that place.
3. Only …………………. % energy is transferred from one trophic level to another.
4. Pyramid of …………………. is always upright. (MP 2013)
5.  …………………. and …………………. are the examples of omnivorous animals.
6. The term ‘ecosystem’ was proposed by ………………….
7. The chief source of energy ecosystems is (MP 2016)
8. N₂ fixing bacteria is called …………………. (MP 2009 Set A)
9. All ecosystems are depended for energy on …………………. (MP 2009 Set D)
10. In forest trees basically function as ………………….
11. Establishment of organisms in new habitat is called ………………….
12. A sereal development of plant community is called ………………….
13. Succession which takes place in baren rock is called ………………….
14. Succession in soil is called ………………….

Answer:

1. Ecotone
2. Flora
3. 10
4. Energy
5. Man, Pig
6. Tansley
7. Sunlight
8. Nitrifying bacteria
9. Sun (solar energy)
10. Producers
11. Ecesis
12. Succession
13. Lithosere
14. Psamosere.

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

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

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

1. What is called the study of ecology of a species?
2. What is called the process of soil formation?
3. What type of energy pyramid found in nature?
4. Give one example of an omnivorous animal.
5. Who proposed the term ecosystem?
6. Who gave the 10% rule of ecosystem?
7. In an ecosystem, producer is known as.
8. What is the direction of the flow of energy in an ecosystem?
9. When many food chains operate simultaneously and interlock such patterns is termed as?
10. Give an example of autotrophic componant.
11. Which gas is used by green plants in photosynthesis?
12. Name the special nutrient which is present in metheonine amino acid.
13. Name the cycle in which nitrogen is converted into multiple chemicals and circulated among atmosphere.
14. Due to an ecosystem fungus and bacteria are known as.
15. What is T₂?

Answer:

1. Autecology
2. Pedogenesis
3. Upright
4. Man
5. A.G. Tansley
6. Lindman
7. Biotic componant
8. Leanear
9. Food web
10. Plants
11. CO₂
12. Sulfer
13. Nitrogen cycle
14. Decomposer
15. Herbivorous animal.

Ecosystem Very Short Answer Type Questions

Question 1.
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 example Helium, Neyon and Crypton are found in less quantity.

Question 2.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, example earth worm. And decomposer are organisms which by secreting enzymes break down complex organic matter into in organic substance example some bacteria and fungi.

Question 3.
Explain consumers of ecosystem
Answer:

1. Producers – All the green plants.

2. Consumers – Depends on others for food.

• Primary consumer: Depends on plants called herbivores.
• Secondary consumers : Depends on herbivores for food.
• Tertiary consumers : Depends on secondary consumers.

3. Decomposers – They decomposed dead organic matter.

Question 4.
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 5.
What are transducers according to some ecologists?
Answer:
Some ecologists call green plants as transducers of the ecosystem as they convert solar energy into chemical energy.

Question 6.
Name four submerged plants.
Answer:

1. Hydrilla
2. Vallisneria
3. Elodea
4. Potamogeton.

Question 7.
Name the stages of xerosere.
Answer:
Stages of xerosere:

1. Crustose lichen stage
2. Foliose lichen stage
3. Moss stage
4. Herb stage
5. Shrub stage
6. Climax forest stage.

Question 8.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 9.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 10.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro – consumer or Decomposer.

Question 11.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10%.

Question 12.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 13.
Name the word which is similar to ecosystem given by Prof. R. Mishra.
Answer:
Ecocosm.

Question 14.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 15.
Who gave the word transformer for producer?
Answer:
E.J. Kormondy.

Question 16.
Name any two sedimentary cycle.
Answer:

Phosphorus cycle
Sulphur cycle.

Question 17.
The energy pyramids are always.
Answer:
Upright.

Question 18.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 19.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 20.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (NO₃^–)

Ecosystem Short Answer Type Questions

Question 1.
Draw a pyramid of energy of grassland ecosystem.

Answer:

Question 2.
Explain nitrogen cycle in nature.

Answer:

Question 3.
Explain sulphur cycle by diagram.
Answer:

Question 4.
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:

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

Question 5.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web:
In nature, foodchains are not isolated sequences, but are interrelated and interconnected with one another. When many foodchains 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, example 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 6.
Explain calcium cycle with well labelled diagram.
Answer:

Question 7.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 8.
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:

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

2. Consumers:

• Primary consumer – Animals, which feed on producers are included into this category example Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers – Primary consumers also serve as food for water snakes, 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.

3. 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 their 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 9.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the foodchain 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. example pyramid of biomass of pond ecosysyem is always inverted.

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

Question 10.
Distinguish between:

1. Grazing food chain and Detritus food chain
2. Production and Decomposition
3. Upright and Inverted pyramid
4. Food chain and Food web
5. Litter and Detritus,
6. Primary and Secondary productivity

Answer:
1. Differences between Grazing food chain and Detritus food chain :

Grazing food chain:

• Energy for the food chain comes from the sun.
• First trophic level organisms are producers.

Detritus food chain:

• Energy comes from detritus (organic matter).
• First trophic level organisms are detritivores and decomposers.

2. Differences between Production and Decomposition :

Production:

• It refers to the process of synthesis of organic compounds from inorganic substances utilising sunlight.
• Example: Plants perform the function of production of food.

Decomposition:

• It is the phenomenon of degradation of waste biomass.
• Example : Bacteria and fungi decompose dead organic matter.

3. Differences between Upright pyramid and Inverted pyramid:
Upright pyramid:
When the number of producers or theirbiomass is maximum in an ecosystem and it decreases progressively at each trophic level in a food chain, an uprightpyramid is formed.

Inverted pyramid:
When the number of individuals or their biomass at the producer level is minimum and it Increases progressively at each trophic level in a food chain, an inverted pyramid is formed.

4. Differences between Food chain and Food web:

Food chain:

• A food chain is a single pathway where energy is transferred from producers to successive orders of consumers.
• All food chains start with green plants which are the original source of all food.
• Energy flow is unidirectional.

Food web:

• A food web is a network of various food chains which are interconnected with each other like an interlocking pattern.
• It has many linkages and intercrosscs among producers and consumers.
• Energy flow in multidirectional.

5. Differences between Litter and Detritus:
Litter:
The dead remains of plants (leaves, flowers etc.) and animals excreta which falls on the surface of the earth in terrestrial ecosystems is called litter.

Detritus:
The dead remains of plants and animals constitute detritus. It is differentiated into litter fall (above ground detritus) and below ground detritus.

6. Differences between Primary and Secondary productivity:

Primary productivity:

• It is the rate at which organic matter is built up by producers.
• It is due to photosynthesis.
• Primary productivity is two types :
  Gross Primary Productivity (GPP) and Net Primary Productivity (NPP) GPP – R = NPP (R = loss in Respiration)

Secondary productivity:

• It is the rate of synthesis of organic matter by consumers.
• It is due to herbivory and predation.
• Secondary productivity is two types :
  Gross Secondary Productivity (GSP) and Net Secondary Productivity (NSP) NSP = GSP – R (Loss in Respiration)

Question 11.
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 (NPP).

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 12.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition:
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 13.
Given account of explain energy flow in ecosystem.
Answer:
Energy flow:
In the 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 14.
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.

Ecosystem Long Answer Type Questions

Question 1.
Describe various components of ecosystem.
Answer:
Ecosystem has following components:

1. Abiotic components:
The physical conditions of the ecosystem depends upon latitude, overall climatic and edaphic factors. The physical deficiencies are overcome by artificial irrigation and use of fertilizers. Thus, like natural ecosystem all organic, inorganic substances and climatic factors together forms abiotic component of the ecosystem.

• Organic substances : Such as carbohydrates, proteins, lipids, etc.
• Inorganic substances : Such as C, H, N, P, K, Ca, I, etc.
• Climatic factors : Such as temperature, light, water, humidity, wind, pH, minerals, soil structure, etc.

2. Biotic factors:
Three types are there

(a) Producers :
This type of crop (dominant species) depends upon the climate, season and the choice of the farmer.

(b) Consumers:

• Primary consumers : Insects, beetles, fishes, etc.
• Secondary consumers : Frog and fishes.
• Tertiary consumers : Snakes and cranes.

(c) Decomposers:
Bacteria, fungi etc.

Question 2.
Explain consumer components ecosystem of a pond in brief.
Answer:
Consumers:
They feed on producers directly or indirectly. It is of following categories:

1. Primary consumers:
Varied forms of zooplanktons are found in the water surface. Most of them are unicellular protists, such as Amoeba, Paramoecium whereas some are multicellular crustaceans, such as Daphnia, Cyclops, etc. Animals which are found under the surface of water are called as benthos, such as many types of fishes, crustaceans, molluscs, insects, beetles, etc. They too feed on producers.

2. Secondary consumers:
They feed on primary consumers, example big fishes, water snakes, etc.

3. Tertiary consumers:
They feed on secondary consumers, example kingfisher, cranes, omnivorous man, etc.

Question 3.
Describe carbon cycle in an ecosystem.
Answer:
Carbon Cycle:
Importance of carbon:
Carbon is considered as the basis of life. Carbon is the most important constituent of proteins, fats and nucleic acids which form the essential constituents of protoplasm.

Sources of carbon – The three sources of carbon in non – living world are:

• The carbon dioxide of the air and that which is dissolved in water.
• The rocks of the earth crust containing carbonates.
• The fossil fuel like coal and petroleum.

Recycling of carbon:
Carbon dioxide (CO₂) is the main source of carbon for the living beings. The carbon of coal, graphite, petroleum are insoluble and carbonates are not available to the organism until they are burnt or chemically changed. Most of the carbon dioxide enters the living world through photosynthesis. In this process green plants trap CO₂ from atmosphere and convert it into carbohydrates by using water and solar energy.

The amount of the carbon fixed by photosynthesis is nearly 7 x 10¹³ kg/year. The organic compounds synthesized in photosynthesis are passed from plants to the herbivores and carnivores. It is estimated that one hectare of a healthy forest produces about 10 tonnes of oxygen and absorbs 30 tonnes of carbon dioxide annually.

Question 4.
What do you mean by trophic levels? 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 (T₁) 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 a cone 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 – example all terrestrial ecosystems.
• Inverted – example 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 : example grassland, pond, forest ecosystem.
• Inverted : example 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 5.
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 :

1. Forest biomes – They may be as below :

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

2. Grassland biomes – They may be as below:

• Tropical rain forest
• Cold tropical forest

3. Desert biomes

4. Tundra biomes.

Grassland biome – Grassland biome or ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

1. Abiotic component:
All organic, inorganic substances and climatic factors together form abiotic component.

2. 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 6.
What are biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
Biogeochemical cycles:
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 Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 5 States of Matter

States of Matter Important Questions

States of Matter Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
At the time of opening the bottle of ammonia, it can be recognized from a distance because:
(a) It is very reactive
(b) It diffuses very fast
(c) It possess a pungent smell
(d) It is lighter than air
Answer:
(b) It diffuses very fast

Question 2.
Who established the relationship between density and rate of diffusion of a gas:
(a) Boyle
(b) Charles
(c) Graham
(d) Avogadro
Answer:
(c) Graham

Question 3.
The value of in Calorie (approx):
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 4.
Value of gas constant R is:
(a) 8314 × 10⁷ ergs/degree/mol
(b) 83.14 × 10⁶ ergs/degree/mol
(c) 83.14 × 10⁵ ergs/degree/mol
(d) 8.314 × 10⁷ ergs/degree/mol
Answer:
(d) 8.314 × 10⁷ ergs/degree/mol

Question 5.
Absolute temperature is:
(a) 0° C
(b) – 100° C
(c) – 273° C
(d) – 373°C
Answer:
(c) – 273° C

Question 6.
To get general gas equation which two laws are combined:
(a) Charle’s law and Dalton’s law
(b) Graham’s law and Dalton’s law
(c) Boyle’s law and Charle’s law
(d) Avogadro’s law and Dalton’s law
Answer:
(c) Boyle’s law and Charle’s law

Question 7.
Which is correct in the following:
(a) r.m.s. velocity = 0.9212 × average velocity
(b) average velocity = 0.9212 × r.m.s. velocity
(c) r.m.s. velocity = 0.9013 × average velocity
(d) average velocity = 0.9013 × r.m.s. velocity
Answer:
(b) average velocity = 0.9212 × r.m.s. velocity

Question 8.
At constant volume the pressure of monoatomic gas depends on:
(a) Thickness of the wall of the vessel
(b) Absolute temperature
(c) Atomic number of the element
(d) Number of valence electron.
Answer:
(b) Absolute temperature

Question 9.
Behaviour of real gases near to that of ideal gases if:
(a) Temperature is low
(b) Pressure is high
(c) Low pressure and high temperature
(d) Gas is monoatomic
Answer:
(c) Low pressure and high temperature

Question 10.
At mountains of high altitude, water boils at lower temperature because of:
(a) Low atmospheric pressure
(b) High atmospheric pressure
(c) Hydrogen bonding in water is more strong at height
(d) Water vapour is lighter than liquid
Answer:
(a) Low atmospheric pressure

Question 11.
If molecular masses of two gases A and B are 16 and 64 respectively ratio of rates of diffusion A and B will be:
(a) 1:4
(b) 4:1
(c) 2:1
(d) 1:2
Answer:
(c) 2:1

Question 12.
Gases deviate from ideal behavior at high pressure because:
(a) At pressure number of colloision of molecule increases
(b) Attraction between molecules increases at high pressure
(c) Size of molecule decreases at high pressure.
(d) Molecule become steady at high pressure.
Answer:
(b) Attraction between molecules increases at high pressure

Question 13.
Distance on which magnitude of energy is minimum is called:
(a) Atomic radius
(b) Lattice radius
(c) Critical distance
(d) Molecular distance
Answer:
(a) Atomic radius

Question 14.
Diffusion rate of methane is twice than that of gas x, molecular mass of gas x will be:
(a) 64
(b) 32
(c) 40
(d) 8.0
Answer:
(a) 64

Question 15.
Part of van der Waals’ equation which illustrate the inter molecule force of real gases:
(a) (V – b)
(b) RT
(c) [P+\(\frac{a}{VL}\)]
(d) (RT)^-1
Answer:
(c) [P+\(\frac{a}{VL}\)]

Question 16.
Temperature which is same in both Celsius scale and Fahrenheit scale:
(a) 0°C
(b) 32°F
(c) – 40°C
(d) 40°C
Answer:
(c) – 40°C

Question 2.
Fill in the blanks:

1. With increase in temperature viscocity ………………………..
2. S.I. unit of surface tension is ………………………….
3. ………………………… is more heavy among dry air and moist air
4. AT N.T.P., volume of gas equal to Avogrado’s numbers ………………………….
5. Unit of vander Waals’ constant ‘b’ is ………………………..
6. …………………………. has minimum value among average velocity, root mean square velocity and most probable velocity
7. Process of diffusion of air from the puncture of an autommobile tube is known as ……………………….
8. …………………………… is obtained if a graph is plotted between volume and absolute temperature
9. Apparatus used for measuring gas pressure is …………………………
10. S.I. unit of viscosity is ………………………….
11. Absolute temperature of ideal gas is ………………………… to the average kinetic theory of the molecule.
12. A liquid which is permanently super cooled is called ………………………..
13. According to kinetic theory, the average kinetic energy of a gas is directly proportional to its …………………………. temperature.
14. The kinetic energy of one mole of a gas is equal to …………………………..
15. Root mean square velocity is …………………………..

Answer:

1. Decreases
2. Nm^-1
3. Dry air
4. 22.41
5. Litre/mol
6. Most probable velocity
7. Effiission
8. Straight line
9. Monometer
10. Nm^-2S
11. Proportional
12. Glass
13. Absolute
14. \(\frac{3}{2}\) RT
15. \(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \)

Question 3.
Answer in one word/sentence:

1. Vaporization of ethanol is faster as compared to water?
2. The resistance produced in the flow of a liquid is called?
3. Unit of surface tension is?
4. Molecular mass of two gases A and B are 36 and 64 respectively. What will be the ratio of diffusion of the two gases?
5. Write S.I. unit of pressure?
6. If rate of diffusion of oxygen is r, then tell the rate of diffusion of hydrogen?
7. Compressibility factor of ideal gases is?
8. Vander Waals’ equation is?
9. Adiabatic expression of ideal gas is?
10. Number of electron in outer most orbit of crypton is?

Answer:

1. Value of molar vaporization enthalpy of water is high
2. Viscosity
3. Dyne per cm
4. 4:3
5. pascal
6. r₁ = 4r [\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { m_{ 2 } }{ m_{ 1 } } } \)]
7. 1.0
8. [P + \(\frac { an^{ 2 } }{ V^{ 2 } } \)] [V – nb] = nRT
9. q = 0
10. 8 electron

Question 4.
Match the following:
[I]

Answer:

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

[II]

Answer:

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

States of Matter Very Short Answer Type Questions

Question 1.
The resistance produced in the flow of a liquid is called?
Answer:
Viscosity.

Question 2.
Unit of surface tension is?
Answer:
Dyne per cm.

Question 3.
Write S.I. unit of pressure?
Answer:
Pascal.

Question 4.
Name the scientist who developed relation between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 5.
What is the Value of gas constant in S.I. unit?
Answer:
8.314 JK^-1 mol^-1

Question 6.
1 Pascal is equal to?
Answer:
1 Nm².

Question 7.
Who established the relationship between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 8.
Absolute temperature is?
Answer:
~273°C.

Question 9.
The value of R in calorie (approx)?
Answer:
2.

Question 10.
Behaviour of real gases is near to that of ideal gases of?
Answer:
Low pressure and high temperature.

Question 11.
At mountains or high altitude, water boils at lower temperature because of?
Answer:
Low atmospheric pressure.

Question 12.
To get general gas equation. Which two laws are combined?
Answer:
Boyle’s law and Charle’s law.

Question 13.
At constant volume the pressure of gas depends upon?
Answer:
On absolute temperature.

Question 14.
Part of van der Waals’ equation which illustrate the inter – molecular forces of real gases?
Answer:
P + \(\frac { a }{ V^{ 2 } } \).

Question 15.
Temperature which is same in both Celsius scale and fahrenheit scale?
Answer:
– 40°C.

Question 16.
What is the formula for kinetic theory of gases?
Answer:
PV = \(\frac{1}{3}\) mnv².

Question 17.
The average kinetic energy of gases is proportional to?
Answer:
Absolute temperature.

Question 18.
What is the kinetic energy of 1 mole of gas?
Answer:
\(\frac{3}{2}\) RT.

Question 19.
What is the formula of Root mean square velocity?
Answer:
\(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \).

Question 20.
Which type of Crystals are diamond and ice?
Answer:
Diamond – Covalent crystal
Ice – Ionic crystal.

Question 21.
What is Critical temperature?
Answer:
The temperature above which the gas cannot be liquefied.

Question 22.
Poise is the unit of which basic property?
Answer:
Viscosity (1 poise = dynes/cm²s).

Question 23.
What is the unit of a (volume correction), b (pressure correction) in van der Waals’ equation?
Answer:
a (Volume correction) = atm L² mol^-2
b (Pressure correction) = L mol^-1

States of Matter Short Answer Type Question – I

Question 1.
Explain Anisotropic and Isotropic?
Answer:
Anisotropic:
The crystalline solid exhibits different physical properties in the , three direction. In this way, crystalline solids are called anisotropic.

Isotropic:
The amorphous solid exhibits same physical properties in all directions. Due to this property they are called isotropic.

Question 2.
Define the term absolute zero? Write its value in centigrade scale?
Answer:
The lowest possible temperature at which all the gases are supposed to occupy zero volume is called absolute zero. The actual value of absolute zero is – 273.15°C. It is , related to temperature in centigrade scale by this relation.
t°C = t + 273 K.

Question 3.
What is unit cell?
Answer:
Smallest unit is a crystal which is formed by systematic arrangements of constituent particles as atom, molecule or ions, is called unit cell. The unit cell generates the whole lattice translation.

Question 4.
What is crystal lattice?
Answer:
Geometry or shape of any crystal in which unit cells are arranged systematically and three dimensionally is called crystal lattice.

Question 5.
How does volume of balloon used for weather study, change with height?
Answer:
At height, atmospheric pressure decreases. The volume of gas inside the balloon increases with decrease in pressure. A stage comes when due to decrease in atmospheric pressure in larger extent, volume increases and balloon bursts.

Question 6.
In winter season, a layer of ice is formed in the lake but the fishes and other organisms present in the lake remain alive. Why?
Answer:
The maximum density of water is at 4°C but below 4°C temperature the density decreases. When the temperature of lake decreases then the water present on the surface become denser and goes downward. This occurs upto the level when the temperature rises to t 4°C. (MPBoardSolutions.com) If the temperature of the upper layer is less than 4°C, the water remains on the upper surface and converts into ice slowly and the water below the surface remains as such due to high density and remains as liquid. That is why, the fishes and micro – organisms remains alive.

Question 7.
Why is the density of hot gas is less in comparison to cold gas?
Answer:
According to Charle’s law, volume of any gas of definite mass is directly proportional to absolute temperature. On increasing the temperature volume of gas also increases, but increase in volume results decrease in density.

Question 8.
Why one feel sluggish, breathlessness and headache at high altitude?
Answer:
At high altitude, the pressure is less and the corresponding volume of air is more, Thus, air becomes less dense at high altitude and the oxygen in air becomes insufficient for normal breathing. This causes what is known as altitude sickness.

Question 9.
What is Critical temperature?
Answer:
The temperature to which gas must be cooled before it can be liquefied by com-pression is known as critical temperature and is represented by Tc.
Example: Critical temperature of CO₂ gas is 31.1°C or 304.1K.

Question 10.
What is critical pressure and critical volume?
Answer:
Critical pressure:
The minimum pressure required to liquefy the gas at its critical temperature is known as critical pressure and denoted by P_c.
Example: Critical pressure of CO₂ is 72.8 atm.
Critical Volume: The volume occupied by 1 mole of gas at the critical temperature and critical pressure is known as critical volume and denoted by V_c.
Example: Critical volume of C0₂ gas is 94 cm³/mol.

Question 11.
Why are tyres of automobile inflated to lesser pressure in summer than in winter?
Answer:
As the automobiles move the temp, of tyre increases due to friction against the road. Consequently the air inside the tyre expand thereby the pressure exerted by air against the wall of tyre also increases. (MPBoardSolutions.com) In summer, there is increase in temperature hence increase in pressure is much more. This may leads bursting at tyre. In order to check bursting of tyre, the tyre are inflated with looser amount of air than in winter.

Question 12.
What would be the S.I. unit for the quantity PV²T²/n?
Answer:
\(\frac { PV^{ 2 }T^{ 2 } }{ n } \) = \(\frac { (Nm^{ -2 })(m^{ 3 })^{ 2 }(K)^{ 2 } }{ mol } \) = Nm⁴K²mol^-1.

Question 13.
Explain on the basis of Charle’s law that minimum possible temperature is – 273°C
Answer:
According to Charle’s law:
V_t = V₀ [1 + \(\frac { t }{ 273 } \)]
At t = – 273°C V_t = V₀ [ 1 – \(\frac { 273}{ 273 } \)] = 0
Therefore at – 273°C, the volume of gas becomes 0 and below this temperature the volume becomes – ve which is meaningless.

Question 14.
Why are liquid drops spherical?
Answer:
Small drops are spherical in shape:
Surface tension tries to decrease the surface area of a given liquid for a given volume. Therefore, drops of liquid are spherical because for a given volume sphere has minimum volume.

Question 15.
What is Root Mean Square velocity and Average velocity?
Answer:
1. Root Mean Square velocity:
It is defined as root of, mean of, square of, velocity of large no. of molecules of same gas. It is denoted by V.
V = \(\sqrt { \frac { v_{ 1 }^{ 2 }+v_{ 2 }^{ 2 }+v_{ 3 }^{ 2 }…..v_{ n }^{ 2 } }{ n } } \)
2. Average velocity:
It is defined as average of, velocity of all the molecules present in gas. It is denoted by V_a.
V_a = \(\frac { v_{ 1 }+v_{ 2 }+v_{ 3 }…..v_{ n } }{ n } \)

Question 16.
Explain the difference between Evaporation and Boiling?
Answer:
Differences between Evaporation and Boiling:
Evaporation:

1. Evaporation decreases spontaneously and occur at all temperatures.
2. Evaporation is a process of the liquid surface.
3. Evaporation is a slow process.

Boiling:

1. Boling takes place only when the vapour pressure of this liquid becomes equal to atmospheric pressure.
2. Boling is a process of the entire liquid and occurs in the form of bubbles inside the liquid.
3. Boling is a fast process.

Question 17.
What do you mean by compressibility factor of gases?
Answer:
The ratio of observed volume and caluculated volume of a gas at a given temperature and pressure is known as compressibility factor. It is denoted by Z.
Thus,

or Z = \(\frac { PV }{ nRT } \)
For ideal gases, PV = nRrt
∴For ideal gas, Z = 1.

Question 18.
What is the effect of pressure on melting of ice?
Answer:
By increasing pressure there occur tremendous increase in kinetic energy of molecules, due to this at low temperature, the molecules move freely, hence on increasing pressure the ice below its melting point converted into liquid.

Question 19.
Mountaineers carry oxygen cylinders with them at the lance of climbing mountains. Why?
Answer:
Atmospheric pressure is relatively low at heights. Quantity of oxygen is low in mountain and climbers feel difficulties in breathing. Therefore, they carry oxygen cylinders along with them.

Question 20.
Define viscosity of liquid. Explain the effect of temperature on viscosity?
Answer:
Resistance in flow of any liquid is called viscosity. Such resistance is produced due to internal friction of different layers of liquid. (MPBoardSolutions.com) When temperature is increased, the cohesive force, which opposes liquid flow, decreases and molecular velocity increases. Due to this, viscosity decreases.

Question 21.
On same temperature when ether and water pour on different hands, then ether seems to be more colder than water. Why?
Answer:
In ether the intermolecular attractive forces between the molecules is less in comparison to water, so ether evaporates more quickly than water and the energy required for evaporation is absorbed from hand, that is why ether seems to be colder.

Question 22.
What is surface tension? Write its S.L unit?
Answer:
It is an important property of a liquid which is related with interatomic attraction force. The molecules present inside the liquid is attracted equally by molecules present in all direction. (MPBoardSolutions.com)
But molecule present on the surface of liquid is attracted by molecules at bottom and in sides, as a result the molecules at surface are pulled downward and nature of surface is to lessen the area. Due to compactness, the surface of liquid behaves as a stretched membrane. This effect is called surface tension.

“Surface tension is a measure of work which is necessary to increase the unit cross-section of liquid.”
Its S.I. unit is Joule/metre² or Newton metre.

Question 23.
The compressibility factor Z of a gas is as follows:

1. What will be the value of Z for an ideal gas?
2. What will be the effect on Z above Boyle’s temperature for real gas?

Answer:

1. For ideal gas, compressibility factor Z = 1.
2. Above Boyle’s temperature, real gases show positive deviation. So, Z > 1.

Question 24.
What is Ideal gas? Write its characteristics?
Answer:
Ideal Gas:
The gases which obey Ideal gas equation under all conditions of temperature and pressure is called ideal gas.
Characteristics:

1. At constant temperature, product of pressure and volume of ideal gas are always constant. Therefore, horizontal line should be obtained in a graph. If graph is plotted between PV and P.
2. If an ideal gas is called at constant pressure, then its volume requestly decreased and become at – 273°C.
3. There is no force of attraction between gas molecules.
4. The compressibility factor of ideal gas is equal to one.

Question 25.
What is Real gas? What are its properties?
Answer:
Gas which does not obey Boyle’s law, Charle’s law and Ideal gas equation strictly is called Real gas.
The gases which does not follow ideal gas equation behaviour under all condtions of temperature and pressure called real gas.
Properties:

1. They do not follow gas law at low temperature and high pressure.
2. At – 273°C their volume is not zero because most of the gases converted into liquid on cooling.
3. The attractive force between gas molecule is negligible.
4. The compressibility factor of real gas is not equal to zero.

States of Matter Short Answer Type Questions – II

Question 1.
State and explain Boyle’s Law?
Answer:
According to this law: “At constant temperature, the volume of a known amount of gas is inversely proportional to the pressure.”
P ∝ \(\frac{1}{V}\) (at constant temperature)
⇒P = Constant × \(\frac{1}{V}\)
⇒ PV = Constant.
Thus, “at constant temperature, product of volume of a given mass of gas and pressure remain constant.”
At initial condition, P₁V₁ = K …………….. (1)
At final condition, P₂V₂ = K ………………. (2)
From eqn. (1) and eqn. (2).
P₁V₁ = P₂V₂

Question 2.
What is the nature of gas constant R?
Answer:
We know that,
PV = nRT
R = \(\frac{PV}{nT}\)

Question 3.
Explain concept of absolute zero from Charle’s law?
Answer:
Charle’s law: According to this law “At a constant pressure the volume of certain mass of a gas increases or decreases by of its previous volume for every 1°C changes in temperature (increases or decreases).”
Suppose V₀ is the volume of certain mass of gas at 0°C then,
Volume of the gas at 1°C temperature = V₀ [1 + \(\frac{1}{273}\)]
Volume of the gas at t°C temperature = V₀ [1 + \(\frac{t}{273}\)]
Volume of the gas at – 1°C temperature = V₀ [1 – \(\frac{t}{273}\)]
Volume of gas at – 273°C = V₀ [1 – \(\frac{273}{273}\)] = 0
Thus decrease of temperature results in the decrease in volume of the gas and ultimately the volume should become zero at – 273°C. (MPBoardSolutions.com) This lowest possible temperature at which all the gases are suppossed to occupy zero volume is called absolute zero.

Question 4.
What is Gay Lussac’s law?
Answer:
Gay Lussac law:
According to this law, “At constant volume the pressure of a given mass of gas is directly proportional to its absolute temperature.”
P ∝ T (Mass and volume are constant)
⇒ P = K × T
⇒ \(\frac{P}{T}\) = K
Suppose at initial condition,
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = K
At final condition,
\(\frac { P_{ 2 } }{ T_{ 2 } } \) = K
From eqn. (1) and (2),
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = \(\frac { P_{ 2 } }{ T_{ 2 } } \)

Question 5.
State and explain Avogadro’s law?
Answer:
According to this law, “Equal volume of all gases under identical conditions of temperature and pressure contain equal number of molecules”.
V ∝N (at constant temperature and pressure) …………. (1)
At constant temperature and pressure number of moles of a gas n is directly proportional to number of molecules N.
Hence, N ∝n
⇒\(\frac{V}{n}\) = Constant
Suppose at initial condition volume of gas is V₁ and no. of mole of gas is n₁ hence
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = Constant …………… (2)
Similarly at final condition no. of moles and volume of gas is n₂ and V₂, hence
\(\frac { V_{ 2 } }{ n_{ 2 } } \) = Constant …………… (3)
From eqns. (2) and (3)
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \)

Question 6.
Write the applications of Graham’s law of diffusion?
Answer:
1. To determine the density and molecular weight of a gas:
If the time of diffusion and density of a gas is known and the time of diffusion of other gas is known, then the density and molecular weight of other gas can be calculated.

2. Marsh gas indicator:
The persons working in the mines get aware by the leakage of the poisonous gases by this indicator.

3. In separation of gases:
The gases can be separated easily due to difference in the rate of diffusion of gases.

4. Smell:
Bad smell and poisonous gases get separated due to diffusion in air.

Question 7.
Derive Charle’s law on the basis of Kinetic gas theory?
Answer:
According to Kinetic gas theory, kinetic energy of gases is directly proportional to absolute temperature.
K.E ∝T
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT
⇒PV = \(\frac{2}{3}\) KT,
⇒V = \(\frac{2}{3}\) \(\frac{K}{P}\).T
At constant pressure \(\frac{2}{3}\) \(\frac{K}{P}\) = constant
V = constant ∝ T
V ∝ T

Question 8.
How are rates of diffusion of different gases compared?
Answer:
Let two gases are A and B, equal volume V of both gases diffuse in times t_A and t_B respectively.
Rate of diffusion of gas A,
r _A = \(\frac { V }{ t_{ A } } \)
Rate of diffusion of gas B,
r _B = \(\frac { V }{ t_{ B } } \)
∴\(\frac { r_{ A } }{ r_{ B } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \)
From equation (3) and equation (4),
\(\frac { t_{ B } }{ t_{ A } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \).

Question 9.
The ratio between the rate of diffusion of an unknown gas (x) s&d CO₂ Is 40:45 Find out the molecular mass of unknown gas (x).
Solution:
According to Graham’s law of diffusion,
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \)
Given, r₁:r₂ = 40:45 M₂(CO₂) = 44, M₁ = ?
⇒\(\frac{40}{45}\) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
⇒ \(\frac { (40)^{ 2 } }{ (45)^{ 2 } } \) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
M₁ = \(\frac { 44\times 45\times 45 }{ 40\times 40 } \) = 55.68.

Question 10.
If relative density of chlorine is 36, diffusion of 25 volume of hydrogen takes 40 sec. under the condition how much time will be taken for the diffusion of 30 volume of chlorine?
Solution:
Hydrogen d₁ = 1, r₁ = \(\frac{25}{40}\), Chlorine d₂ = 36, r₂ = \(\frac{30}{t}\)

States of Matter Long Answer Type Questions

Question 1.
State and explain Graham’s law of diffusion?
Answer:
Graham’s Law of Diffusion:
The rate of diffusion of gas under similar condition of temperature and pressure is inversely proportional to the square roots of their density.”
Thus, Rate of diffusion ∝\(\frac { 1 }{ \sqrt { density } } \)
⇒r ∝\(\frac { 1 }{ \sqrt { d } } \)
If the rate of diffusion of two gases are r₁ and r₂ and their density are d₁ and d₂ respectively.
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 1 } } } \)
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 2 } } } \)
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \)
∵ M. Mass = 2 × Vapour density = \(\frac { Molecular\quad mass }{ 2 } \) = Vapour density
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \).

Question 2.
Write difference between Real and Ideal gas?
Answer:
Differences between Ideal gas and Real gas:
Ideal gas:

1. Ideal gas obeys die equation, PV = RT at all temperature and pressure.
2. There is no ideal gas, they are hypothetical.
3. Total volume of gas molecules is supposed to be negligible in comparison to total volume.
4. There is no attraction force between gas molecules.

Real gas:

1. Real gas obeys PV = RT only at low pressure and high temperature.
2. All existing gases are real gases and show deviation from ideal gas behaviour, positive or negative.
3. Volume of gas molecules are not negligible in comparison to total volume.
4. Gas molecules attract each other.Therefore, total pressure is less than ideal gas.

Question 3.
What are the main differences between Crystalline and Amorphous solids?
Answer:
Differences between Crystalline and Amorphous solids:
Crystalline:

1. In it the constituent particles are arranged in regular manner.
2. Melting point of crystalline solid is fixed.
3. They are an anisotropic i.e. some of their physical properties are different in different directions.
4. They are rigid and their shape is not distorted by mild distorting tone.
5. They have a definite heat of fusion.
6. They are true solid in real meaning.

Amorphous solids:

1. In it the constituent particles are arranged in irregular manner.
2. Melting point of amorphous solid is not fixed.
3. They are Isotropic i.e. their physical properties are same in all directions.
4. They are not very rigid, they can be distorted easily.
5. They do not have definite heat of fusion.
6. They are super cooled liquid.

Question 4.
Using state equation clarify that the density of a gas at given temperature is proportional to pressure of gas?
Answer:
PV = nRT
PV = \(\frac{m}{M}\) RT,
Or P = \(\frac{mRT}{VM}\),

Or P = \(\frac{dRT}{M}\)
Or d = \(\frac{PM}{RT}\)
if T = known constant
∴d ∝P

Question 5.
Derive Ideal gas equation on the basis of kinetic gas equation?
Answer:
According to Kinetic gas theory, “ The average kinetic energy of gas molecules is directly proportional to absolute temperature.”
Average Kinetic energy = \(\frac{1}{2}\) mnv²
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT,
⇒\(\frac{PV}{T}\) = R
⇒ PV = RT.

Question 6.
Explanation of Dalton’s law on the basis of Kinetic gas theory?
Answer:
If the volume of container V litre and no. of moles of gas A is n₁ ar.d mass of each particles in m₁ R.M.S. velocity is V₁ then,
P_A = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \)
Kinetic gas equation PV = \(\frac{1}{3}\) mnv²
For B gas number of moles = n₂
Mass = m₂
R.M.S. Velocity = v₂
P_B = \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
If both the gases are kept in same container total pressure of mixture
P = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \) + \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
⇒ P = P_A + P_B
so in general, for more than two gases
P = P_A + P_B + P_C + …………………..
It is Daltons law of partial pressure.

MP Board Class 11 Chemistry Important Questions

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

Organisms and Populations Important Questions

Organisms and Populations Objective Type Questions 

Question 1.
Choose the correct answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 2.
Fill in the blanks:

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

Answer:

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

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

Answer:

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

II. (MP 2012)

Answer:

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

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

1. Name any animal which lays eggs in the nest of other animal?
2. Name the plant which has pneumatophores.
3. Name two symbiotic partners of lichen.
4. Write the type of water which is available in plants.
5. Loss of the upper surface of the soil is called.
6. When do we celebrate world environmental day?
7. Name the rootless free floating plant.
8. In which plant, root cap is absent.
9. Give an example of symbiosis.
10. Formation of soil is called.
11. Plants are dependent to other plant for shelter is called.
12. Name an amphibious plant.
13. Name the plant in which stem is modified into leaf like structure and leaves are modified into spines.
14. Give an example of sexual dimorphism.

Answer:

1. Cuckoo
2. Rhizophora
3. Algae and Fungi
4. Capillary water
5. Soil erosion
6. 5 June
7. Wolffia
8. Hydrophytes
9. Pea and Rhizobium bacteria
10. Soil fertility
11. Epiphytes
12. Ranunculus
13. Opuntia
14. Honey bee

Organisms and Populations Very Short Answer Type Questions

Question 1.
Give two examples of each:
Symbiotic, commensal, phytoplankton, zooplankton and rooted floating plant
Answer:

1. Symbiotic – Escherichia, Triconimpha
2. Commensal – Orchid and Trees, Hermit crab
3. Phytoplankton – Nostoc, Anabaena
4. Zooplankton – Paramecium, Euglena
5. Rooted floating plant – Wolffia, Lemna.

Question 2.
Name the speciality of any population.
Answer:
Any population has some speciality, these are :

1. Population density
2. Growth rate
3. Death rate
4. Age distribution
5. Biotic capacity
6. Population growth form
7. Changes of population
8. Population dispersion

Question 3.
Name any two mangrove plants.
Answer:
Rhizophora, Avicennia.

Question 4.
Name the vagetation found in sundarban delta.
Answer:
Mangrove vegetation.

Question 5.
Name the plant in which pneumatophores are found.
Answer:
Rhizophora.

Question 6.
Give the term for scientific study of human population.
Answer:
Demography.

Question 7.
What is sex ratio of India according to 2001 census?
Answer:
933 females per 1000 of njales.

Question 8.
Name any animal which lays egg in the nest of other animal.
Answer:
Koyal.

Question 9.
Give the name of larva of frog, salamander and moths.
Answer:
Tadpole, Axolotal, Catterpillar.

Question 10.
Give the reason for increase in population.
Answer:
Decrease in death rate.

Question 11.
Give example of species polymorphism.
Answer:
Honeybee.

Question 12.
Name the species found in different geographical areas.
Answer: Allopatric

Question 13.
Write the name of any saprophytic angiospermic plant
Answer:
Monotropa.

Question 14.
What is called the relationship between hermit crab and sea – anemone?
Answer:
Protocooperation.

Question 15.
Why insectivores eat insects?
Answer:
To fulfill the need of nitrogen.

Question 16.
Which are main abiotic factor?
Answer:
Temperature, water, light, soil.

Question 17.
Give the name of one insectivores plant
Answer:
Utricularia.

Question 18.
What is ecological competition?
Answer:
Ecological competition is the struggle between two organisms for the same resources within an environment.

Question 19.
A transition area between two biomes.
Answer:
Ecotone.

Question 20.
The number of live births per thousand of population per year.
Answer:
Birthrate.

Organisms and Populations Short Answer Type Question

Question 1.
How is diapauses different from hibernation?
Answer:
Diapauses is a stage of suspended development to cope with unfavorable con-ditions. Many species of zooplankton and insects exhibit diapauses to tide over adverse climatic conditions during their development. On the other hands hibernation or winter sleep is a resting stage where in animals escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2.
If a marine fish is placed in a fresh water aquarium, will the fish be able to survive? Why or why not?
Answer:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In freshwater conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment out¬side. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Give definition of :

1. Population
2. Population density
3. Biotic potential
4. Growth rate
5. Death rate

Answer:

1. Population:
A community of animals, plants or humans among whose members inbreeding occurs.

2. Population density:
Population density is a measurement of population per unit area or unit volume.

3.Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

4. Growth rate:
Human population growth is the increase in the number of individuals in a population. Global human population growth amounts to around 83 million annually.
Growth rate = \(\frac { Annually birth }{ Mid year population}\) x 100

5. Death rate:
The death rate is the number of people per thousand who die in a particular are during a particular period of time.
Death rate = \(\frac { Annually birth }{ Mid year population}\) x 100

Question 4.
What suggestions will you give for awakening people towards population control.
Answer:
Following things can be done for controlling the population:

1. Promotion of education: Misunderstanding can be avoided by promotion of education such as:

• Children are gift of god
• More income from the more children.

2. To introduce for reality of population growth.
3. Uses of family planning techniques.
4. Restriction of more than one marriage.
5. To decrease birth rate.

Question 5.
Write the differences:

1. Species and Population
2. Population growth and Population density,
3. Monospecific and Polyspecific population
4. Competition and Scattering.

Answer:
1. Species and Population:
Population on defined as organisms that belong to the identical species and identical geographical niche or area. The said area should enable these species to interbreed with each other.

2. Population growth and Population density:
In biology, Population growth is the increase in the number of individuals in a population but population density is a measurement of population per unit area, it is a quantity of type number density.

3. Monospecific or Polyspecific population:
Monospecific population is the population of individuals of only one species but poly specific or mixed population is the population of individuals of more than one species and it is generally referred to as a community.

4. Competition and Scattering:
Competition is in general a contest between two or more rivalry between two or more entities, organisms, animals, economic groups or social groups etc. Population scattering is a method which shows equilibrium by interaction of population. In population, scattering seeds away from the parent plant.

Question 6.
Give four reasons of population growth in India.
Answer:
Four reasons of population growth in India are as follows:

1. Increasing growth rate.
2. Decreasing death rate.
3. No importance of education and promotion of education is less.
4. Conservative.

Question 7.
Explain, the formation of new species.
Answer:
Although all life on earth shares various genetic similarities, only certain organisms combine genetic information by sexual reproduction and have offspring that can than successfully reproduce. Scientists call such organisms members of the same biological species.

Question 8.
What do you mean by population equilibrium?
Answer:
Population equilibrium:
When growth pattern of any population is observed then it becomes clear that any population increases at faster rate in the beginning in a new area and reaches to a maximum limit and then remain constant for long time.This phase is called as population equilibrium. At equilibrium birth and death rate of any population become equal.

Question 9.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The cooperative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honeybees, ants, wasps and termites, etc. form well – organized societies showing division of labour and polymorphism. In these social in sects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 10.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes.These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 11.
Most living organisms cannot surv ive at temperature above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Answer:
Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep sea hydrothermal vents. They are able to survive in high temperatures because their bodies have adapted to such environmental conditions. These organisms contain specialized thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures unlike other enzymes.

Most living – organism can not survive above 45°C because:

• Above 45° C enzymes get denatured.
• Protoplasm precipitates at high temperature.

Question 12.
List the attributes that populations but not individuals possess.
Answer:
A population has the following attributes that an individual does not possess:

1. Birth rates and death rates.
2. Sex ratio.
3. Population density.
4. Age distribution.
5. Population growth.

Question 13.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:
A population grow exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation :
N_t = N₀ e^rt
Where,
N_t = Population density after time t.
N_o = Population density at time zero.
r = Intrinsic rate of natural increase.
e = Base of natural logarithms (0.434).
From the above equation, we can calculate the intrinsic rate of increase (r) of a population. Now, as per the question,
Present population density =x
Then, Population density after two years = 2x
t = 3 years
Substituting these values in the formula, we get:
=> 2x = x e^3r
=> 2 = e^3r
Applying log on both sides :
=> log 2 = 3r log e
\(\frac { log 2 }{ 3 loge }\) = r
\(\frac { 0.301 }{ 3 x 0.434 }\) = r
\(\frac { 0.301 }{1.302 }\) = r
0.2311 = r
Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

Question 14.
Name important defence mechanisms in plants against herbivory.
Answer:
Several plants have evolved various mechanisms both morphological and chemical to protect themselves against herbivory.

1. Morphological defence mechanisms:

• Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.
• Sharp thorns along with leaves are present in Acacia to deter herbivores.
• In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

2. Chemical defence mechanisms :

• All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal in ingested by herbivores.
• Chemical substances such as nicotine, caffeine, quinine and opium are produced in plants as a part of self – defense.

Question 15.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:
An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 16.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The ecological principle behind the biological control method of managing with pest insects is to keep their population in check by using their natural predators and parasites.

Question 17.
Distinguish between the following:

1. Hibernation $nd Aestivation
2. Ectotherms and Endotherms.

Answer:
1. Differences between Hibernation and Aestivation :

Hibernation:

• It is called winter sleep.
• Animal rests in warm place.
• It lasts for long winter months.

Aestivation:

• It is called summer sleep.
• Animal rests in cool, moist, shady place.
• It lasts for day time as nights are cooler and animal will come out.

2. Differences between Ectotherms and Endotherms :

Ectotherms:

• These are cold – blooded animals.
• They are unable to regulate their body temperature which changes with temperature of the surrounding environment.
• They hibernate in winters and aestivate in summers.

Endotherms:

• These are warm – blooded animals.
• They are capable of maintaining their body temperature.
• They can regulate their body temperature so, they do not need to show behavioural adaptations like these.

Question 18.
List the various abiotic environmental factors.
Answer:

1. Atmospheric factors : Light, temperature, wind and water.
2. Lithosphere : Rock, soil.
3. Hydrosphere : Pond, river, lake and ocean.
4. Edaphic factors: Soil texture, soil water, soil air, soil microorganism, soil pH, minerals.
5. Topographic factors : Slope, altitude, valley.

Question 19.
Give an example for:

1. An endothermic animal,
2. An ectothermic animal,
3. An organism of benthic zone.

Answer:

1. An endothermic animals : Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such, as bears, cows, rats, rabbits etc. are endothermic animals.
2. An ectothermic animals : Fishes such as sharks, amphibians frogs, and reptiles, tortoise, snakes, and lizards are ectothermic animals.
3. An organism of benthic zone : Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 20.
Define population and community.
Answer:
Population:
Population is a group of individuals of same species, which can reproduce among themselves and occupy a particular area in a given time.

Community:
It is an assemblage of several population in a particular area and time and exhibit interaction and interdependence through trophic relationship.

Question 21.
Define the following terms and give one example for each:

1. Commensalism
2. Parasitism
3. Camouflage
4. Interspecific competition.

Answer:
1. Commensalism:
It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted. The two individuals may be physically associated. example Sucker fish and shark.

2. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

3. Camouflage:
It is a strategy adapted by prey species to escape their predators. Organisms are cryptically coloured so, that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

4. Interspecific competition:
It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 22.
Explain population dispersal in brief.
Answer:
Population dispersal:
The movement of individuals of a population from a particular place to another is known as population dispersal. Dispersal of members of population may occur for search of food, escape from enemies or for breeding purpose. There are three types of

population dispersal:

1. Immigration – Inward movement of members of other population and settle with local population. It causes increase in population.
2. Emigration – Outward movement of individuals of a population to shift in some other place. It causes decrease in population.
3. Migration – It can be observed in birds. It is periodic departure of organism and returning back to its original place, example Siberian cranes visit India during winters and go back to north during summer.

Question 23.
With the help of suitable diagram describe the logistic population growth curve.
Answer:
Logistic growth:
1. The resources become limited at certain point of time so, no population can grow exponentially.

2. This growth model is more realistic.

3. Every ecosystem or environment or habitat has limited resources to support a particular maximum number of individuals called is carrying capacity (K).

4. When N is plotted in relation to time t, the logistic growth show sigmoid curve and is also called Verhulst-Pearl logistic growth. It is given by the following equation:
\(\frac { dN }{ dt }\) = rN [ \(\frac{K-N}{K}\) ]
Where, N = Population density at time t
R = Intrinsic rate of natural increase
K = Carrying capacity.

a = When resources are not limiting the growth, plot is exponential
b = When resources are limiting the growth. Plot is logistic, k is carrying capacity.

Question 24.
Write characteristic features of aquatic plants.
Answer:
Characteristic features of aquatic plants (Hydrophytes):

1. The cuticle is either absent or very thinly deposited on the epidermis.
2. Usually stomata are absent, if present they are non – functional.
3. Many air chambers or large intercellular spaces are present inside the plant body of hydrophytes.
4. The whole body of the plant helps in the absorption of water and minerals.
5. Root system is feebly developed and usually non – branched or less branched. Some hydrophytes like Wolffia lack foots.
6. The mechanical tissues are less distributed.
7. The vascular tissues are either absent or feebly developed in the case of hydrophytes.

Question 25.
Write short note on population growth form or pattern.
Answer:
Population growth from or pattern: The way of growth of population is known as population growth pattern. Population growth shows two patterns :

1. S – shaped form : In this pattern, the increase in population is very slow in the beginning, then there is a fast growth rate and then becomes stable.

2. J – shaped form : In some organisms the population increases very fast in the beginning because of some environmental factors, the growth is checked abruptly.

Question 26.
Give only five examples of communication of informations in animals.
Answer:
Communication of information within a species:
Communication may be defined as an act which influences the activity of one individual by some behaviour of another. Most of the animal species have some mechanism of exchanging information among their members through sight, sound, touch or chemicals. The evolution of the techniques of communication in the animal kingdom is progressive but complex.

Following are few examples:

1. The honeybee gives the information to the other worker bees of the hive regarding availability of nectar in the vicinity by dancing in a special manner on the surface of the hive. Through the round dance they communicate that the availability of the food is very near to the hive and the waggle dance (by moving abdomen) shows the direction and the distance of the food source from the hive.

2. Croaking of male frog attract female frog during breeding season.

3. The dogs express their various intensions by facial expressions and movement of the tail and by making typical sound.

4. Rabbits inform their group members about any sort of danger by tapping their
tails.

5. Certain chemical compounds called pheromones are secreted by animals to transmit message to other members of the species. Pheromones are detected by smell or taste.

Question 27.
Write distinguish between:

1. Parasitism and Symbiosis,
2. Mutualism and Commensalism,
3. Hydrosere and Xerosere.

Answer:
1. Differences between Parasitism and Symbiosis:

Parasitism:

• In this case, one organism depends on other for their food.
• Only one organism benefitted.
• It produce only disease.
• example Tapeworm and Man.

Symbiosis:

• In symbiosis two organisms lives together in such way that both are benefitted by each other.
• Both organism benefitted.
• It does not produce any disease.
• example Lichens.

2. Differences between Mutualism and Commensalism:

Mutualism:

• It is an interspecific interaction in which both the species are mutually benefited.
• The two individuals may be physically or physiologically associated.
• example Rhizobium and the legume plants.

Commensalism:

• It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted.
• The two individuals may be physically associated.
• example Sucker fish and shark.

3. Differences between Hydrosere and Xerosere:

Hydrosere:

• Succession in water is called hydrosere.
• Changes are fast.
• It produce aquatic plants.

Xerosere:

• Succession in deserts is called xerosere.
• Changes are slow.
• It produce xerophytic plants.

Question 28.
Write short notes on:

1. Parasitism
2. Species dominance
3. Biotic potential.

Answer:
1. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

2. Species dominance:
Presence of any biotic community or presence of more popu¬lation of any species biosphere shown its dominance and it is called species dominance which has capacity to reduce other species in population. The effect of dominant species is more upon environment or other species.

3. Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

Question 29.
Explain population fluctuation.
Answer:
Population is generally a group of individuals of a particular species occupying a particular area at a specific time.

Population fluctuation:
Any increase or decrease in number of individuals in a population from its equilibrium state is known as population fluctuation. It may occur due to various reasons, such as due to change in climate or due to change in physical environment or due to predators.

Question 30.
Give an example for:

1. Heliophytes
2. Sciophytes
3. Viviparous.

Answer:

1. Heliophytes : Sunflower, Amranthus.
2. Sciophytes : Java moss, Lycopodium, Polytrichus.
3. Viviparous : Rhizophora, Salicomia, Sonneratia.

Organisms and Populations Long Answer Type Questions

Question 1.
Write a short note on :

1. Adaptations of deserts plants and animals
2. Adaptations of plants to water scarcity
3. Behavioural adaptations in animals
4. Importance of light to plants
5. Effect of temperature or water scarcity and the adaptations of animals.

Answer:
1. Adaptations of desert plants and animals :

(i) Adaptations of desert plants:
Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables the stomata to remain closed during the day to reduce the loss of water through transpiration.

(ii) Adaptations of desert animals:
Animals found in deserts such as desert Kangaroo rats, Lizards, Snakes, etc. are well adapted to their habitat. The Kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and bur¬row themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

2. Adaptations of plants to water scarcity :
Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration.

In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

3. Behavioural adaptations in animals:
Certain organisms are affected by temperature variations. These organisms undergo adoptions such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations.

For example, ectothermal animals and certain endosperms exhibit behavioural adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begin? to rise, the lizard burrows itself inside the sand to escape the scorching, sun.

Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against tem-perature variations.

4. Importance of light to plants:
Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need right for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses occurring in plants. Plants respond to changes in intensity of light during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for vertical distribution of plants in the sea.

5. Effects of temperature or water scarcity and the adaptations of animals:
Temperature is the most important ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermals. Those which can tolerate a narrow range of temperature are called stenothermal animals. Animals also undergo adaptations to suit their natural habitats.

Animals found in polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat. Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioral adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations.

These lizards bask in the sun during early hours when the temperature is quite low. As the temperature begins to increase, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategy is seen in other desert animals. Water scarcity is another factor that forces animals to undergo certain adaptations to suit their natural habitat. Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to stay in their habitat.

The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow in the sand as the temperature rises to escape the heat of the day. Such adaptations can be seen to prevent the loss of water.

Question 2.
List any three important characteristics of a population and explain.
Answer:
The three important characteristic of a population are:

1. Population density : The number of individuals of a species per unit area or a volume is called population density.
P_D = \(\frac { N }{ S }\)
Where, PD = Population density
N = Number of individuals in a region S = Number of unit area in a region.

2. Birth rate : It is expressed as the number of births per 1000 individuals of a population per year.

3. Death rate : It is expressed as the number of deaths per 1000 individuals of a population per year.

Question 3.
Describe the anatomical adaptation of xerophytic plants.
Answer:
The anatomical adaptations of xerophytic plants are:

1. The epidermis of xerophytic plant organs is covered by thick cuticle. It protects plants against heavy transpiration and also provides mechanical strength to some extent.

2. Epidermis may be multiple layered which also reduces the rate of transpiration.
Examples : Banyan, Nerium etc.

3. Except some monocotyledonous plants the stomata are restricted to the lower surface of the leaves. The stomata are generally sunken and surrounded by many hairs.

4. Hypodermis in xerophytes is multilayered and sclerenchymatous which checks evaporation of water.
Example: Pinus needle.

5. Mesophyll tissue is well differentiated into palisade and spongy parenchyma but palisade parenchyma is more in number than spongy parenchyma.
Examples : Pinus needle, Nerium leaf.

6. Vascular bundles are well developed and differentiated into xylem and phloem. Vascular bundles are comparatively more in number. Besides vascular bundles, mechanical tissues are well developed.

7. Leaves of some xerophytes have some rolling devices due to presence of motor cells or bulliform cells in the epidermal zone.
Examples : Poapratensis, Agropyron, Ammophila etc.

8. Stomata in certain desert plants such as Capparis spinosa and Aristida ciliata may sometimes get blocked due to deposition of resinous matter or wax. This adaptation is also known to reduce the loss of water during transpiration.

Question 4.
Describe the morphological adaptations of aquatic plants.
Or
Describe morphological adaptations found in the stem and leaves of xerophytic plants.
Or
Explain in brief, xerophytic plants along with examples.
Answer:
The morphological adaptations of aquatic plants are:

1. Adaptations in roots:

(i) Root system is feebly developed and unbranched. Some floating plants like Wolffia, Utricularia etc. and submerged plants like Ceratophyllum lack roots.

(ii) Root hairs are absent except rooted floating hydrophytes. In most of the hydro¬phytes roots are meaningless as the entire surface of the plant body is in direct contact with water and acts as absorptive surface. This may probably be the reason why roots in hydrophytes are reduced or absent.

(iii) True root caps are absent in some free floating hydrophytes and in lieu of that they develop root pockets or root sheaths which protect their tips from injuries.

(iv) If roots are present they are always of adventitious type e.g., Nymphaea, Nelumbo etc. In Jussiaea modified spongy negative geotrophic floating roots are present which pro¬vide buoyancy to the plants and also do the job of gaseous exchange.

2. Adaptations in stems:

(i) In submerged hydrophytes like Hydrilla, Potamogeton etc. stems are long, slender and flexible.

(ii) In free floating hydrophytes, stems are modified as thick, stout, stoloniferous or as offset and occur horizontal on the water surface. It bears vegetative bud at their apex and helps in vegetative propagation.

(iii) In rooted floating hydrophytes or rooted submerged hydrophytes or in marshy plants, stem may be modified as runner or rhizome.

(iv) The stem may be modified as runner, stolon, tuber to perform vegetative propaga¬tion effectively. Usually stems are of perennial nature.

3. Adaptations in leaves:

(i) Leaves are thin, long, ribbon shaped in the submerged forms.
Example: Vallisneria, Ceratophyllum etc. or they are finely dissected as found in Ranunculus aquatilis.

(ii) In rooted floating plants, like Nymphaea, Nelumbo etc. the petioles of leaves show indefinite power of growth and they keep the laminae of leaves always on the surface of water.

(iii) In free floating plants like Eichornia, Trapa etc. the petioles become charac¬teristically swollen and develop sponginess which keep the plant afloat.

(iv) Occurrence of two or more than two different types of leaves in a plant is called heterophylly. Examples are Limnophila, Heterophylla, Sagittaria, Ranunculus aquatilis etc. They bear leaves of many types viz., submerged leaves, floating leaves, aerial leaves, linear, ribbon shaped leaves or dissected leaves. Floating leaves are entire and lobed. The broad leaves found on the surface of water transpire actively and regu¬late the hydrostatic pressure in the plant body. The submerged leaves act as water absorbing organs. Heterophylly is also associated with quantitative reduction in transpiration.

(v) In the amphibious or marshy plants, the leaves that are exposed to air show typical mesophytic features. They are leathery and more tough than the submerged and floating hydrophytes.

Question 5.
What do you mean by biotic community? Describe characteristic features of any biotic community.
Answer:
Biotic community:
A biotic community is a localized association of several populations of different species living in a given geographic area of habitat. It represents heterogeneous assemblage of different groups of organisms both plants and animals. Biotic community is composed of smaller units of intimately associated members belonging to different species.

The different species of a community share common environment and their relationships are based on direct or indirect functional interactions. The nature or relationship is determined by the requirements of the members of a community.

Characteristics of a community:
Each community has its own characteristics which are not shown by its individual component species.

1. Species diversity:
Each community is made up of much different organisms Plants, animals, microbes, which differ taxonomically from each other. The number of species and population abundance in community also vary greatly.

2. Growth form and Structure:
Each community have a definite growth form. This different growth form determines the structural pattern of a community.

3. Dominance:
In each community, all the species are not equally important. There are relatively only a few of these, which determine the nature of community. These few species exert a major controlling influence on the community. Such species are known as dominants.

4. Succession:
Each community has its own development history. It develops as a result of a directional change in it with time.

5. Trophic structure (Self – sufficiency):
Nutritionally, each community, a group of autotrophic plants as well as heterotrophic animals exists as a self – sufficient, perfectly balanced assemblage of organism.

Question 6.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The co – operative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honey – bees, ants, wasps and termites, etc. form well-organized societies showing division of labour and polymorphism. In these social insects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 7.
Explain interrelationship between various species of a biotic community along with examples.
Answer:
Odum (1971) distinguished interactions into two broad categories :

1. Positive interactions and
2. Negative interactions.

1. Positive Interactions:
In this type, those type of interactions are considered in which both interacting species are mutually involved to help each other. Here, one interacting species helps the other either one way or on reciprocal terms and may be in the form of nutrition or shelter or substratum or transport. Here interaction may be obligatory or facultative. The positive interactions include three types of interactions. They are :

(i) Commensalism:
This type of interaction occurs in between two organisms of two different species in which one species benefits and the other is neither benefited nor harmed, example Lichens.

(ii) Protocooperation:
The type of interaction where both population are ben¬efited but not obligatory i.e., not essential for the survival of either population is called proto – co – operation. This is also known as non-obligatory mutualism. The relationship between hermit crab and sea anemone is an example of proto – cooperation. The crab uses the gastropod (mollusc) shell as a portable shield and the sea anemone eats the leftover food of the crab, which is protected from its predators by the stinging cells of the sea anemone.

(iii) Mutualism or Symbiosis:
When two different species grow together and are mutually benefited, the plants are known as symbiotic plants and the phenomenon is called symbiosis or mutualism. It is a sort of obligatory association. This type the organisms are dependent upon each other for survival. In this type of association, two types of species are physiologically related. This type of relationship may exist in between two plants or in between one plant and one animal or in between two animals.

2. Negative Interactions:
Those interaction of two different species in which both are harmed or one organism is benefited while other is more or less harmed is referred to as negative interac¬tion. In this type of interaction one population eats the other type of population or one organism does not allow other organisms to grow near it by using the food supply of the other or producing toxic substances. The negative interactions be categorised into following types:

• Competition
• Parasitism
• Predation and
• Antibiosis.

Competition:
When interaction occurs between two species for the use of same resources and when resources are in short supply is referred to as competition. It is a relationship which involves struggle amongst the organisms for food, shelter, water, sunlight and climate.

Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria :
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

Predation:
To obtain food by hunting is predation. A predator is an animal or plant that kills and feeds on another organism, its prey. It represents a generalised carnivorous habit. Commonly predators are larger than their prey but this is not always true but predator is in any case always occur as a naturally better equipped hunter than its prey. Most of the predatory organisms are animals but there are some plants (carnivores) also especially fungi which feed upon other animals.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Physics Important Questions Chapter 1 Electric Charges and Fields

Electric Charges and Fields Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The least charge is:
(a) 1 coulomb
(b) 1 stat coulomb
(c) 1 micro coulomb
(d) Electronic charge.
Answer:
(d) Electronic charge.

Question 2.
The number of electrons in one coulomb charge is:
(a) 5.46 × 10²⁹
(b) 6.25 × l0¹⁸
(c) 1.6 × l0¹⁹
(d) 9.0 × l0²⁹
Answer:
(b) 6.25 × l0¹⁸

Question 3.
An object has – 80 µ C.What is the excess number of electrons:
(a) 5 × 10⁸C
(b) 16 × 10¹⁴C
(c) 5 × 10¹⁴ C
(d) 16 × 10⁸C.
Answer:
(c) 5 × 10¹⁴ C

Question 4.
The intensity of electric field E due to charge Q at distance r as compared to its normal state:
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Question 5.
The value of permitivity E₀ of vacuum (or free space) is :
(a) 8.85 × 10^-12 C
(b) 8.85 × 10^-12C²N^-1m^-2
(c) 9 × 10⁸ Nm²C^-2
(d) 9 × 10 C²N^-1m^-2.
Answer:
(b) 8.85 × 10^-12C²N^-1m^-2

Question 6.
The work done in rotating a dipole of dipole moment p through an angle 180° from direction of uniform electric field E is:
(a) 2pE
(b) pE
(c) \(\frac { 1 }{ p}\)pE
(d) Zero.
Answer:
(a) 2pE

Question 7.
The intensity of electric field E due to a dipole of dipole moment p at a point a distance r from center of dipole depends on r as :
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)

Question 8.
A surface S is placed in an electric field E parallel to the field. The flux linked with the surface be:
(a) ES
(b) \(\frac{ E }{S}\)
(c) Zero
(d) Infinite.
Answer:
(c) Zero

Question 9.
The total electric flux emerges out from unit positive charge in air is:
(a) E₀
(b) \(\frac { 1 }{ { E }_{ 0 } } \)
(d) \(\frac { 1 }{ { 4\pi E }_{ 0 } } \)
(d) 4πE₀
Answer:
(b) \(\frac { 1 }{ { E }_{ 0 } } \)

Question 10.
The dielectric constant of metal is:
(a) Infinite
(b) Zero
(c) One
(d) None of these.
Answer:
(a) Infinite

Question 11.
The dielectric constant of air is:
(a) 8.85 × l0^-12 C² N^-1m^-2
(b) 1
(c) Infinite
(d) None of these
Answer:
(c) Infinite

Question 2.
Fill in the blanks:

1. The dimensional formula of charge is …………….
2. The two like charges …………… each other.
3. Two unlike charges ……………. each other.
4. The SI unit of electric field is …………….
5. The dimensional formula of electric field is …………….
6. The size of an ideal dipole is …………….
7. Two charges – q and + q are situated at a distance / its dipole moment will be …………….
8. The size of an ideal dipole is …………….
9. Two charges – q and + q are situated at a distance lits dipole moment will be …………….
10. The maximum torque acting on a dipole of dipole moments p in a uniform electric field E is …………….

Answers:

1. [AT]
2. Repel
3. Attract
4. NC^-1
5. [ML^-3A^-1’]
6. Point size
7. q.I
8. pE sin G.

Question 3.
Match the Column:

Answers:
(b) newton / coulomb
(e) coulomb
(a) coulomb x metre
(d) N – m²/C
(c) coulomb²/newton x meter ²

answers:

1. (e) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)
2. (d) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { Kr }^{ 2 } } \)
3. (b) \(\frac { q }{ { E }_{ 0 } } \)
4. (c) pE(1 – cosθ)
5. (a) – pE cos θ

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

1. What type of charge get accumulated in a glass rod rubbed with silk ?
2. What type of charge get accumulated in rubber rod when rubber with fur fabric ?
3. What is the cause of quanti zation of charge ?
4. What is dielectric constant of a metal ?
5. What is the direction of electric dipole moment ?

Answers:

1. Positive charge
2. Negative charge
3. Transfer of electrons takes place in integral multiple. This is the cause of quantization
4. Infinity
5. Direction of dipole moment is from negative charge to positive charge

Electric Charges and Fields Important Questions Very Short Answer Type Questions

Question 1.
What does q₁ + q₂ = 0 signify in electrostatics ?
Answer:
The equation signifies that the electric charges are algebraically additive and here q₁ and q₂ are equal and opposite.

Question 2.
Why electric lines of force do not form closed loop ? (AH India 2014)
Answer:
Because electric lines of force start from the positive charge and end on die negative charge.

Question 3.
Why must electrostatic field at the surface of charged conductor be perpendicular to every point on it ? (Foreign 2014)
Answer:
As, electric field inside a conductor is always zero. So electric lines of force exert lateral pressure on each other and hence repulsion between like charges take place. Thus, in order to stable the spacing, the lines of force are normal to surface.

Question 4.
What quantity is electric flux ? Write its SI unit.
Answer:
It is a scalar quantity. S.I. unit of electric flux is Nm²/C 

Question 5.
Charge of on QNC is placed as centre of a cube. What will be the electric flux passing through a face of it ?
Answer:

Question 6.
Charges +q₁ -5q₁, +2q and +2q is at Gaussian surface. What will be the magnitude of electric flux passing through Gaussian surface.
Answer:
Zero, because total charges is zero as surface of Gaussian surface.

Question 7.
Fig. shows three charges + 2q, – q and+ 3q. Two charges T 2q and – q are enclosed within a surface S. What is the electric flux through the surface S ?
Answer:

Question 8.
Does charge get changes with the velocity of charged conductor ?
Answer:
No? charge does not get change with velocity of charged conductor.

Question 9.
Which scientist first discovered the charge as positive and negative ?
Answer:
Benjamin Franklin.

Question 10.
Is static electric force is Newtonian between two charges ?
Answer:
Yes, static electric force Newtonian between two charges.

Question 11.
What is the relation between dielectric constant in vacuum and medium ?
Answer:
£ = £₀ × £_r

Question 12.
What is the electric force acting upon charge q placed in electric field E ?
Answer:
qE

Electric Charges and Fields Important Questions Short Answer type Questions

question 1.
Write the limits of inverse square law ?
Answer:

• It is true only for point charge.
• It is only for that charges which is at rest.

Question 2.
Write the basic properties of electric charge ?
Answer:

• Quantization of charge i.e. q = ±ne.
• Charge cannot be created nor be destroyed.

Question 3.
What is dielectric constant of a metal ?
Answer:
The dielectric constant of a metal is ∞

Question 4.
Define intensity of electric field and write its unit.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge qo placed at a point in the electric field of intensity\(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its S.I. unit is newton/coulomb.

Question 5.
What is an electric dipole ?
Answer:
A system of two equal and opposite charges kept very close to each other is • called an electric dipole.

Question 6.
What is an electric flux ? Give its SI Units.
Answer:
Electric flux associated with electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. Its SI unit is newton meter per coulomb.

Question 7.
What do you mean by quantization of electric charge ?
Answer:
Quantization of electric charge means that the total charge (q) of a body is always ah integral multiple of basic charge (e) which is the charge on the electron.
.’. Charge on any body, q = ± ne
Where, e = 1.6 x 10^-19 coulomb and n = 0,1,2,3 In above relation, +ve sign denotes the loss of electrons and – ve sign denotes gain of electrons. certain particles are known having these charges which are called quarks.
Charge on any body will never be a fraction of e Like \(\frac { 3 }{ 2 }\)e, \(\frac { 5 }{ 2 }\)e, \(\frac { 7 }{ 2 }\)e. But, now days, certain particles are known having these charges which are called quarks.

Question 8.
Ordinary rubber are insulators but the types of an aeroplane are made a little bit conducting. Why ?
Answer:
When the aeroplane lands and takes off, the friction between types and the run-way may cause the electrification of types. Due to conducting nature of the types, the charge so produced is conducted to the earth and electrical sparking is avoided.

Question 9.
State Gauss’ theorem in electrostatics.
Answer:
The net electric flux passing through any closed surface is \(\frac { 1 }{ { E }_{ 0 } } \) times the net charge present inside it. Mathematically,

Question 10.
The mass of an isolated conductor decreases when some positive charge is given to it. Why ?
Answer:
Mass will reduce because when electrons escape out from the conductor then only it becomes positively charged.

Question 11.
Write the nature of the force acting between q₁ and q₂ if –

1. q₁ q₂ > 0
2. q₁q₂ <0
3. q₁q₂ = 0.

Answer:

1. If q₁q₂ > 0, then both charges are similar and repulsive force acts in between them.
2. If q₁ q₂ < 0, then both charges are of opposite nature and attractive force acts in between them.
3. If q₁q₂ = 0 then one of the two charges is neutral and no force acts in between them.

Question 12.
State principle of conservation of charges. Give examples.
Answer:
According to principle of conservation of charge “Net charge in any isolated system always remains constant”. In the other words “Charge can neither be created nor be , destroyed although it may be transferred from one body to another body”.

Example:
1. Pair annihilation:
When an electron and a positron come near to each other, then they get destroyed and form a y ray photon. Thus the total charge becomes zero.

2. Pair production:
A gamma ray photon splitted into a positron and an electron

Question 13.
Define electric dipole moment. Is it scalar or vector ? Write its SI unit and dimensional formula.
Answer:
Electric dipole moment is the product of magnitude of any one charge of dipole and the distance between both the charges. If + q and – q are two charges placed at a distance 2l apart, then electric dipole moment will be
\(\vec{p}\) = q. 2l

It is a vector quantity whose direction is from negative charge to positive charge.
Unit: Its SI unit is coulomb – meter.
Dimension: [p] = [q][2l] = [ATL]
|p] = [M⁰L¹T¹A¹].

Question 14.
Metallic ropes are attached with the vehicles carrying inflammable substances, connected to earth. Why ? (NCERT)
Answer:
Moving vehicles get charged due to friction. The inflammable material may catch fire due to spark produced by charged vehicle. When metallic ropes or chain is used, the charge developed on the vehicle is transferred to the ground and so the fire is prevented.

Question 15.
When the electric dipole becomes in equilibrium state when kept in electric field?
Answer:
We know that U = – pE cosθ
When the angle between electric dipole and electric field is zero then the electric dipole will be in equilibrium.

Question 16.
State Coulomb’s inverse square law of electrostatic charges. On this basis define unit charge. What is the condition for the law to be applicable ?
Answer:
Coulomb’s inverse square law:
According to it, force of attraction or repulsion between two point charges is directly proportional to the product of their magnitude and inversely proportional to the square of distance between them. Force of attraction or repulsion is the central force which acts along the line joining the two charges. Let the charges q₁ and q₂ be placed at a distance r apart from each other. So, the force acting between them according to Coulomb’s law will be
F ∝ q₁q₂ and F ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Combining the above two relations, we get

To define unit charge : Let q₁ = q₂ = q, r = lm and F = 9 x 10⁹ newton.
Putting these values in eqn. (1), we get
9 x 10⁹N = 9 × 10⁹ × \(\frac { { q }^{ 2 } }{ { 1m }^{ 2 } } \)
q = ±1 coulomb
So, if two charges of same type and equal magnitude are placed at 1 m apart in vacuum and if force of repulsion between them is 9 × 10⁹ newton, then the two charges are said to be unit charges of magnitude 1 coulomb.

Condition for the applicability of the law:

• It is applicable for the point charges which are stationary’, not for moving charges.
• It is not valid for large distances and distances less than 10^-15 meter because at distances less than 10^-15 m, nuclear forces are dominant.

Question 17.
Define intensity of electric field and write its unit Four properties of electric lines of force.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge q placed at a point in the electric field of intensity \(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) =\(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its SI unit is newton / coulomb.

Properties:
(a) Electric lines of force start from positive charge and end on negative charge.
(b) Tangent drawn at any point of the lines of force gives the direction of electric field intensity.
(c) Two lines of force will never intersect each other.
(d) Lines of force try to contract longitudinally. This demonstrates why unlike charges attract each other.
(e) They are emitted normal to the surface of conductor and end normally.

Question 18.
Why the two lines of force do not cross or intersect each other ?
Answer:
If two lines of force intersect, then there would be two tangents and hence two directions of electric field at the point of intersection, which is not possible.

Question 19.
Prove that electrostatic force is much stronger than the gravitational force.
Or
Compare the electrostatic force and gravitational force.(NCERT)
Answer:
Let a proton and an electron be placed in air or vacuum at lm distance apart, mass of proton and electron are mp and me respectively.
m_p = l.67 × 10^-27 kg
m_e = 9.1 × 10^-31 kg
Charge on proton is q_p = 1.6 ×10^-19C. and charge on electron is q_e = 1.6 x 10^-19C The electrostatic force acting between them is –
\(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ p }{ q }_{ e } }{ { Kr }^{ 2 } }\)

The gravitational force between proton and electron is –

Dividing eqn. (1) by eqn. (2), we get

F_e = 2.27 × l0³⁹ × F_G
Thus, the electrostatic force of attraction is 1039 times more than the gravitational force for the electron – proton system. Obviously the electrostatic force acting between two charged particles is very much greater than the gravitational force acting between them.

Question 20.
(a) Define electric flux. Write its SI unit and dimensional formula.
Answer:
Electric flux associated with the electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. It is denoted by Φ_E If the normal drawn to the surface dS makes an angle 0 with the electric field E, then the flux through this area is –
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\)
Unit: Its SI unit is Nm²/ C.

Dimension:

Question 20. (b)
Under what conditions flux is said to be positive and negative ?
Answer;
Electric flux:
The electric flux associated with a surface is a measure of total lines of force passing normally to the surface when held in the electric field. It is denoted by ∅_E There are two types of electric flux:

• Positive electric flux – When electric lines of force leaves any body through its surface, it is considered as positive flux.
• Negative electric flux – When electric lines of force enters into body through any surface, it is considered as negative flux.

Question 20. (c)
Show that the electric flux passing through a surface parallel to electric field is zero.
Answer:
Let \(\overrightarrow{\mathrm{E}}\) be the intensity of electric field and \(\overrightarrow{\mathrm{dS}}\) be the normal to the arbitary surface. If the normal to the surface (\(\overrightarrow{\mathrm{dS}}\)) marks an angle θ with E, then,
The electric flux can be given by
d∅_E = E. dS cos θ
If surface is parallel to the electric field
i.e., θ = 90°
d∅_E = E dS cos 90°
= E.dS × 0
d∅_E = 0

Thus flux passing through a parallel surface to an electric field is zero.

Question 21.
State the principle of electrostatic shielding.
Answer:
The electric field inside the charged hollow conductor (spherical shell) is zero. Therefore, when an object is placed inside a hollow conductor then there will be no effect of electric field on the object i.e., the hollow conductor shields the object from external electric field.

The concept of electrostatic shielding is used to shield wires carrying audio signal from external fields like atmospheric electricity or due to electric sparks. If such wires are not shielded, the audio signal will give rise to noisy sound or reproduction.

Question 22.
Derive an expression of electric field intensity on a point in axial position (end on position) of an electric dipole.
Answer:
Consider an electric dipole made up of charges + q and – q separated by a distance 2/apart and placed in vacuum. We have to find out the electric field at point P situated at a distance r from the center of dipole system. To find out the electric field intensity, imagine a unit positive test charge situated at P.

Electric field intensity due to charge (+q) situated at A will be

Electric field intensity due to charge (-q) situated at B will be 1

As AP < PB, hence repulsive effect due to charge +q will be more effect then the attractive effect of charge – q. Therefore, E₁> E₂ but their directions are opposite. Hence, the net electric field will be

This is the expression for the electric field which is directed from A to P. For small and strong dipoles,
r>>2l ⇒ r>>l ⇒ r² >>>l²
∴ r² – l² ≈ r²
Hence, eqn. (1) becomes

This is the required expression.

Question 23.
Derive the expression for the intensity of electric field on the equatorial position of the dipole.
Answer:
Equatorial position is a point on the line which lies on the perpendicular bisector of the axial line, then their position is called equatorial position or broadside on position. Let AB be an electric dipole formed by charges +q and -q. The distance between the charges is 21. P is a point on the equatorial position of the dipole, situated at a distance r from the center of the dipole. Sup- pose unit positive test charge is situated at P. Electric field at P due to +q charge will be

Electric field at P due to -q charge will be


Resolving E and E₂ into its components, we get the component of E₁ are E₁cosθ parallel to AB and E₁ sin θ along OP. Also, components of E₂ are E₂ cosθ parallel to AB and E₂ sin O along PO.
From eqns. (1) and (2), we find |E₁| = |E₂|, hence E₁ sinθ = E₂ sinθ , but their directions are opposite. So, they cancel each other. The only resultant electric field will be
E = E₁ cos θ + E₂ cos θ
= E₁ cosθ + E₁ cos θ
E = 2E₁ cos θ  …………(3)

If the charges are kept in any other medium other than vacuum, then the electric field becomes

The direction of electric field is along AB i.e., it is parallel to axial line which acts from positive charge to negative charge. For small and strong dipoles, r >> 2l ⇒ r >> l
r2 + l2 ≈ r2
Hence, eqn. (4) becomes

Question 24.
Obtain an expression for the intensity of electric field due to a charge q at a distance r from it.
Answer:
Consider a point charge q which is placed at the origin O of the generalized coordinate system. We have to find out the electric field at point P at a distance r from it. For this, imagine a test charge q₀ placed at point P. Both the charges q and q₀ have been placed in vacuum. According to Coulomb’s law, the force on charge q₀ will be

Where, r is the unit vector in the direction from q to q₀. But intensity of electric field is given by the formula,

If any other medium is present having dielectric constant then

Where, £0 = k = dielectric constant of the medium. In magnitude,

Question 25.
Derive the expression for the torque acting on the dipole when kept in a uniform electric field and hence define dipole moment.
Answer:
Consider an electric dipole made up of charges +q and -q and length 2l which is placed in a uniform electric field of strength E. The dipole moment P makes an angle Q with respect to the electric field at any instant of time. As a result, a pair of forces +qE (along the direction of electric field) and -qE (opposite to the direction of electric field) acts on the dipole. These two opposite forces are of equal magnitude. So, they apply a couple on electric dipole, which tries to bring it along the direction of electric field. This couple is called restoring couple.

So, moment of couple or torque (τ)
= Magnitude of any force x Perpendicular distance between the forces
= qE × BC …….(1)
In A ABC, sinθ = \(\frac { BC }{ AB }\)
⇒ BC = AB sinθ = 2l sinθ
Putting the value of BC in eqn. (1), we get,
τ= qE × 2l sin θ
⇒ τ = pEsin θ,
(∴q 2l = p, where p is the dipole moment) ……..(2)
In vector form,
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{p}}\) × \(\overrightarrow{\mathrm{E}}\)
From eqn. (2), if E = 1 and 0= 90°.Then,
τ = p sin 90° = p
Thus, dipole moment of an electric dipole is numerically equal to the torque acting on the dipole when it is placed perpendicular to a unit electric field.

Question 26.
Derive an expression for the potential energy of a dipole in an electric field.
Answer:
Work done in rotating any electric dipole from standard position to any angle with respect to the electric field is called potential energy. The standard position is considered to θ = i.e., perpendicular to the field. Let p be the dipole moment of the dipole, placed in an electric field of intensity E and it makes an angle θ with respect to the electric field, then
Torque, τ = p E sinθ
Also, work done in displacing the dipole through an angle dθ will be
dW = τdθ = pE sinθ dθ
Now, the work done in displacing from θ = \(\frac { π }{ 2 }\) to θ = θ is

This work done is stored in the form of potential energy,
U = – p E cosθ
In vector form, U = \(\overrightarrow{\mathrm{-p}}\).\(\overrightarrow{\mathrm{E}}\)
This is the required expression.

Question 27.
Derive the expression for the amount of work done in rotating a dipole in a uniform electric field.
Answer:
If a dipole placed in a uniform electric field is rotated from the equilibrium position, then work has to be done. Consider an electric dipole of dipole moment p(= 2 ql) which has been rotated in an electric field E by an angle 0 from the equilibrium position, then torque acting on it will be –
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{-p}}\) × \(\overrightarrow{\mathrm{E}}\) = pE sin θ, (in magnitude form)
Work done in rotating the dipole through an angle dθ is
dw = τ.dθ
Therefore, work done in rotating it by an angle θ can be obtained from the above equation under proper limits.

Above equation is the total work done in rotating any electric dipole through an angle θ

Question 28.
Write and prove Gauss’ theorem.
Or
Prove that net electric flux through any closed surface is 1 / £₀ times of total charge present inside it, where s £₀ is
permittivity of free space.
Answer:
Gauss’ law:
According to Gauss’ law, the net electric flux through any closed surface is 1 / £₀ times of the total charge present inside it.

Where, £₀ is permittivity of vacuum or air.
Proof:
Let at any point O, inside any closed surface, a charge +q is present. A point P lies at a distance r from O on surface dS. Normal to the surface dS can be represented by \(\overrightarrow{\mathrm{dS}}\) (say). Therefore, electric field at point P due to charge +q can be given by

.’. Electric flux passing through area \(\overrightarrow{\mathrm{dS}}\)will be
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\) = E dS cos0

Where, 0 is an angle between electric field f and the normal to the surface.
Net flux through the closed surface is.

But \(\frac { dscos\theta }{ { r }^{ 2 } } \) = dw is the solid angle subtended by the small surface dS at O.

As total solid angle subtended by any closed surface at any point inside it is 4π hence –

Above equation is mathematical form of Gauss’ law.

Electric Charges and Fields Important Questions Long Answer Type Questions

Question 1.
Obtain an expression for the intensity of electric field due to linear charge distribution.
Answer:
Consider a thin wire of length ‘l’ Let the charge given to it be q. So, the linear charge density will be λ = \(\frac { q }{ l }\) ⇒ q = λl
If a test charge ¡s brought near to the wire, then it will be repelled. Hence, the electric field (\(\overrightarrow{\mathrm{E}}\)) will be radially out ward. Draw a normal \(\overrightarrow{\mathrm{dS}}\) on the surface at P. The direction of \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) are the same. The electric flux passing through the curved surface of the cylinder is

= E 2πrl (∴\(\overrightarrow{\mathrm{dS}}\) = Total surface areaofcurved surface 2,πrl …(2)
But according to Gauss’ theorem,
Φ = \(\frac { 1 }{ { E }_{ 0 } } \) x (Total charge present inside the Gaussian surface) …(3)
Φ = \(\frac { q }{ { E }_{ 0 } } \)
From eqns. (2) and (3), we have
E 2πrl = \(\frac { q}{ { E }_{ 0 } } \)
E 2πrl = \(\frac { λl}{ { E }_{ 0 } } \)
[∴q = λl from eqn. (1)]

This is the expression for the electric field due to a linear charge distribution having a linear charge density λ.

Question 2.
Derive an expression for the intensity of electric field due to a uniform infinite plane sheet of charge.
Answer:
Consider an infinite plane sheet on which charge given is q. If S is the total surface area of the sheet, then the surface charge density will be σ = \(\frac {q }{ S}\) = \(\frac { charge }{ area }\) the charge will be distributed uniformly on the surface of the sheet. Hence, surface charge density (= σ) remains constant. Any test charge held near to the surface will be repelled on both the sides i.e., either to paper and the other will go inside, hence the Gaussian surface formed will be cylindrical; which is having three surfaces S₁, S₂ and S₃.

The electric field \(\overrightarrow{\mathrm{dS}}\) is directed radially outwards away from the sheet. Moreover, \(\overrightarrow{\mathrm{dS}}\) is the outward drawn normal on the Gaussian surface. Hence, the total electric flux is

The angle between normal to the curved surface S3 in the direction of electric field \(\overrightarrow{\mathrm{E}}\)] is 90°, hence

For the surfaces Sj and S2, \(\overrightarrow{\mathrm{E}}\) and\(\overrightarrow{\mathrm{dS}}\) are along the same direction, hence the angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°, so

By Guss’ theorem,
Φ_E = \(\frac { 1}{ { E }_{ 0 } } \) q ……(2)
from eqns. (1) and (2) we get
2ES = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac{q}{2 \varepsilon_{0} S}\)
Hence, E = \(\frac { σ}{ { 2E }_{ 0 } } \)
(∴ σ = \(\frac { q }{ S }\))
This is the expression for the intensity of electric field due to a uniformly charged Sheet

Question 3.
Prove that intensity of field is zero inside a spherical shell, while in outside it is such that all the charges are concentrated at the center.
Or
Determine the intensity of electric field by Gauss’ law, due to a uniformly charged spherical shell at a point

1. Outside the shell
2. On the surface of shell
3. Inside the shell.

Draw a graph showing the variation of electric field with the distance from the center.
Answer:
Intensity of electric field due to uniformly charged spherical shell:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity at any point due to this charged sphere depends on its position relative to the sphere.

(i) When point P lies outside the spherical shell:
Consider a point P which lies outside the spherical shell of radius R. Now, imagine a sphere of radius r which passes through P. This closed surface behaves as Gaussian surface. Since, sphere is uniformly charged, so that electric field at each point on the surface is equal and points radially outward.

Net flux through the Gaussian sphere of radius r will be

But, charge enclosed inside the Gaussian surface is q. Therefore, from Gauss’ law
Φ_E = \(\frac { q }{ { E }_{ 0 } } \) …….(2)
Equating eqns. (1) and (2), we have
E. 4πr² = \(\frac { q }{ { E }_{ 0 } } \)
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \) ……..(3)

It is clear from above equation that, the electric field outside any uniformly charged spherical shell is exactly to that electric field when we take the same charge at the center of sphere.

(ii) Point P lies on the surface of spherical shell:
In this case, r = R
From eqn. (3),
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { R }^{ 2 } } \)

(iii) When point P lies inside the spherical shell:
As we know, the charge given to spherical shell spreads equally all over the surface, there is no charge present inside the sphere. Hence, by Gauss’ law,


From equs. (4) and (5),
E.4πr² =  0
⇒ E = 0
Therefore, electric field inside a spherical shell is always zero. The adjacent figure shows the variation of electric field with distance.

Question 4.
Obtain an expression for the intensity of electric field due to a solid sphere of charge using Gauss’ theorem at the following points:

1. Outside the sphere
2. On the surface of sphere
3. Inside the sphere.

Also prove that the intensity of the electric field at the center will be zero.
Answer:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity of electric field at any point depends on its position with respect to sphere. The volume charge density of sphere is
p = \(\frac { Charge }{ volume }\)

(i) When the point lies outside the sphere:
Consider a point P which lies outside the spherical shell of radius R. Now imagine a sphere of radius r which passes through P.This closed surface behaves as a Gaussian surface. The electric field at each point outside the sphere points radially outwards.

.’. Total flux through the Gaussian sphere of radius r will be –

But charge enclosed inside the Gaussian surface is q. Hence, from Gauss’ law
Φ = \(\frac { 1}{ { E }_{ 0 } } \) q …..(3)
From eqns. (2) and (3),
E 4πr² = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac { q }{ { 4\pi { E }_{ 0 }r }^{ 2 } }\)
Hence, the electric field outside any uniform charged sphere is same as if the entire charge is present at the centre.
Putting the value of q from eqn. (1) in eqn. (4), we get

(ii) When the point lies on the surface of sphere:
r = R
From eqn. (4), we have E =

and from eqn. (5), we have E =

(iii) When the point lies inside the sphere:
Suppose the point P lies at a distance r from the center of the sphere such that r < R. At radius r draw a Gaussian surface which is a sphere. Then electric field \(\overrightarrow{\mathrm{E}}\) is radially outwards. Moreover, normal to surface \(\overrightarrow{\mathrm{dS}}\) is along the direction of \(\overrightarrow{\mathrm{E}}\). Hence, angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°
The flux coming out of the Gaussian surface is –

But by Gauss’ law, the flux Φ = \(\frac { 1}{ { E }_{ 0 } } \) x (Total charge inside the Gaussian surface)
Φ = \(\frac { q}{ { E }_{ 0 } } \) q ……(2) (where q’ is the total charge inside the sphere of radius r)
On volume \(\frac { 4 }{ 3 }\) πR³charge given is q
On volume 1, charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \)
On volume\(\frac { 4 }{ 3 }\) πR³ nr charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \) x \(\frac { 4 }{ 3 }\) πR³ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
q’ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
‘Putting the value of q’ in.eqn. (2), we get
Φ = \(\frac { q}{ { E }_{ 0 } } \) \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \) …….(3)
from equs. (1) and (3) we get,

At the center of sphere, r = 0, hence E = 0.

Question 5.
Neeta return home from school. When she was taking out the sweater, she heard twinkling sound. She asked her mother but she was unable to answer.Next day she went to school and asked the science teacher, she was satisfied with her answer. Write down the answer of the question given below on the basis of above statement:

1. What values does she show ?
Answer:
She shown the value of answer, questioning.

2. What was the reason for the sound when she was taking out her sweater ?
Answer:
Due to the friction between her school dress and sweater, its acquire negative charge which get discharge. Therefore the sound was been listen by her.

3. Presence of charge can be determined by which apparatus ?
Answer:
Gold leaf electroscope.

4. Does any new charges get created on rubbing ? If yes or not than why ?
Answer:
No, new charges get created, only transfer of electron occur form one body to another. There one become positive charge and other become negative charge.

Electric Charges and Fields Important Questions Numerical Questions

Question 1.
Two positive ions which have same charges having a force of repulsion of
3.7 × 10^-9N. The distance between them is 5 Å. Find the deficiency of electron on each ions.
Solution Given:
F = 3.7 × 10^-9N, r = 5Å = 5 × 10^-10m
q₁ = q₂ = q

= 10.28 × 10³⁸
q = 3.2 × 10^-19
n = \(\frac { q }{ e }\)

There will be deficiency of two electron in each ions.

Question 2.
Calculate force of repulsion between two protons when they are separated with 4.0×10¹⁵m.
Solution:
We know that, \(\frac { 1 }{ 4\pi { E }_{ 0 } } \) = 9 × 10⁹Nm²C^-2
Given: q₁= q₂ = +1.6 × 10^-19 C,r = 4.0 × 10^-15m.
From formula, F = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)

or F = 14.4 newton.

Question 3.
Can 1 coulomb of charge be gives to a metallic sphere of radius 1cm? Give reason.
Solution:
Here, q = 1 C, r = lcm = 10^-2m
Electric field on the surface of sphere is
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \)
= 9 × 10⁹ × \(\frac { 1 }{ { \left( { 10 }^{ 2 } \right) }^{ 2 } } \) = 9 × 10³ N /C
But when intensity of electric field exceeds 3 × 10⁶ N / C air gets ionized and hence, electric discharge takes place. So l Cm of charge cannot be given to a sphere of radius 1 cm.

Question 4.
Two charge of one coulomb are separated by a distance of 1 meter from each other, find:

1. Force between the charge when was in air
2. If it is in dielectric medium, calculate its force.

Solution:
Given:
q₁= q₂ = 1 coulomb, r = l m

Question 5.
Two point charges + 9e and + e are placed 8 m apart. Where should the third charge q be placed on the line joining the two charges so that q should be in equilibrium?
Solution:
Let the charge q be placed at distance x from +9 e charge, then its distance from e will be 8 – x.
In equilibrium,


Taking + ve sign \(\frac { 3 }{ x }\) = \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = x
24 = 4x or x = 6m
Taking – ve sign, \(\frac { 3 }{ x }\) = – \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = -x
24 – 2x or x = 12 m
Since, +9e and+e are similar charges, hence q will be in equilibrium when x = 6 m i.e., q should be placed 6 m apart from +9e between the charges.

Question 6.
What is the force between two small sphere having charge of 2 × 10^-7 C and 3 × 10^-7C ? (NCERT)
Solution:
Given: q₁ = 2 × 10^-7C,
q₂ = 3 × 10^-7C,
and r = 30 cm = 30 × 10^-2
m = 0.3 m

The force will be repulsion force a both the charge are same.

Question 7.
The electrostatic force on a small sphere of charge 0.4µC due to another small sphere of charge 0.8µC in air is 2 N.

1. What is the distance between two sphere ?
2. What is the force on the second sphere due to the first ?

Solution:
Given:
q₁ = 0.4µC = 0.4 × 10^-6 C
q₂ = -0.8µC = -0.8 × 10^-6 C
F = 0.2N

= 9 × 2 × 8 × 10^-4
r = 3 × 4 × 10^-2
= 0.12 m = 12 cm.

(ii) Both the charges apply same force on each other. There force of first sphere due to second sphere will be 0.2 N. The force between them will be force of attraction.

Question 8.
A polythene piece is rubbed with woollen cloth as a result 3 x 10.7 coulomb charge is produced in it, then: .

1. How many electrons are transferred from which to which substance ?
2. Is mass transferred from woollen to polythene ?

Solution:

1. Polythene gets negative charge, so that electron is transferred from woollen to polythene.
We know that, q = ne

or n = 1.875 × 10¹² electrons.

2. As electrons are transferred from woolen to polythene, mass is also transferred in the same way.

Question 9.
A conductor has 14.4 × 10¹⁹ C positive charge. How many electrons are lacking or in excess ?
Solution:
We know that, q = ne
⇒ n =\(\frac { q }{ e}\)

Therefore, body is lacking by 9 electrons (as it has positive charge).

Question 10.
What is the net flux of uniform electric flux there through a cube of field oriented so that its faces are parallel to the coordinate plane ?
Solution:
All the faces of a cube are parallel to the coordinate axis. Therefore the Number of hold lines entering the cube is equal to the number of field lines piercing out of the cube. As a result net flux through the cube in zero.

MP Board Class 12th Physics Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Questions

Chemical Bonding And Molecular Structure Very Short Answer Type Questions

Question 1.
What type of bond is present generally in same atoms?
Answer:
Covalent bond.

Question 2.
Which hybridization is present in ammonia (NH₃)?
Answer:
sp³.

Question 3.
What is the reason for high boiling point of water?
Answer:
Presence of H – bond between molecules of water.

Question 4.
What is the dipole moment of C0₂?
Answer:
Zero.

Question 5.
What type of bonds are directional?
Answer:
Covalent bonds.

Question 6.
What is the bond angle in water?
Answer:
104°5′.

Question 7.
What is the structure of [Ni(CN)₄]^2-?
Answer:
Square planner.

Question 8.
Which bond is present in s – s overlapping?
Answer:
cr (Sigma) bond.

Question 9.
What is the structure of diamond?
Answer:
Crystal lattice (Tetrahedral).

Question 10.
Which bond is present in sidewise overlapping of p – p orbitals?
Answer:
n – bond.

Question 11.
Which type of hybridization found in PCl₅?
Answer:
sjy’d hybridization.

Question 12.
What is full form of LCAO?
Answer:
Linear Combination of Atomic Orbitals.

Question 13.
What is the structure of NH₃?
Answer:
Trigonal bipyramidal.

Question 14.
What is dipole moment (µ) of a linear covalent molecule?
Answer:
Zero.

Question 15.
What is the unit of electron gain enthalpy?
Answer:
eV per atom or kJ per mole.

Question 16.
What is the formula of bond order?
Answer:
Bond order = \(\frac{1}{2}\) (N_b – N_a).

Question 17.
What is the symbol of superoxide and peroxide?
Answer:
Superoxide – O^–₂ and peroxide – O^2-₂.

Question 18.
What do you mean by bond order?
Answer:
The number of electrons present between two atoms of molecules or ions.

Question 19.
More polarizing power and more polarisability increases which property of molecule?
Answer:
Covalent property.

Question 20.
What is determined by Born – Haber cycle?
Answer:
Lattice energy.

Question 21.
What is full form of VSEPR?
Answer:
Valence Shell Electron Pair Repulsion.

Chemical Bonding And Molecular Structure Short Answer Type Questions – I

Question 1.
What is the electronic theory of covalency? Write its main postulates?
Answer:
The electronic theory of valency and its main postulates are as follows:

1. The covalency of any element depends upon the no.of electrons present in its valence shell.
2. All the elements have the tendency to acquire the nobel gas configuration.
3. The electrons present in valence shell are called valence electrons and when the electron comes out than the vacant space is called kernel.
4. If an element is unstable, than it works for its stability for this it gives and takes the electron. On this basis the bonds are of three types:
   1. Ionic bond
   2. Covalent bond and
   3. Co – ordinate bond.

Question 2.
What do you mean by lone pair of electron?
Answer:
The electron pair which present in the valence orbital of the element and does not take part bond formation is called lone pair electron.
Example: In the H₂O atom the lone pairs of electrons on an atom:

Bond pairs = 2, Lone pair electrons = 2.

Question 3.
What do you mean by dipole moment?
Answer:
Dipole moment is defined as, the product ofthe magnitude of charge on any one of the atoms and distance between them. It is represented by Greek letter µ(mu).
Mathematically, dipole moment is expressed as µ = e × d
Where, e is charge on any one of the atoms and d is distance between the atoms.
As e is of the order of 10^-10 esu while d is of the order of 10^-8 cm µ is of the order 10¹⁸ esu cm and this unit of p is known as Debye (D). Thus,
1D = 1 × 10^-18esu cm

Question 4.
Is He₂ molecule is possible? Clearify it.
Answer:
He₂ molecule is not possible.
2He → 1s²
Two electrons are present in Is orbital of He atom. It is complete and stable orbital and so it cannot accept an extra electron. The bond order is zero. So the formation of He₂ molecule is impossible.

Question 5.
What is the total number of σ (sigma) and π (pi) bond are present in following molecules:

1. C₂H₂
2. C₂H₄

Answer:

1. C₂H₂
2. C₂H₄

Question 6.
What do you mean by hydrogen bond?
Answer:
Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of the other molecule. Hydrogen bond is of two types:

1. Intermolecular hydrogen bonding
2. Intra – molecular hydrogen bonding.

Question 7.
At normal temperature H20 is in liquid state but H2S is in gaseous state. Why?
Answer:
Hydrogen bonding affects the physical state of the molecule. For example, H₂0 and H₂S are the hydrides of group 16 elements, but H₂O is liquid whereas H₂S is gas at room temperature. Actually, H₂O fulfils the (MPBoardSolutions.com) condition of hydrogen bonding therefore, in H₂O hydrogen bonding is found and they get associated, grow in molecular size and exist in liquid state. The H₂S molecule, the magnitude of hydrogen bonding is negligible, therefore, molecules remains separated from each other and acquires gaseous state at room temperature.

Question 8.
Why, viscosity of glycerol is more than ethanol?
Answer:
In ethanol molecule, one hydroxyl gap is present whereas in every glycerol, three OH – groups are present which form hydrogen bond. Therefore, in glycerol hydrogen bond formation is more in comparison to ethanol. That is why, the viscosity of glycerol is more than ethanol.

Question 9.
Why σ – bond is stronger than π – bond?
Answer:
The strength of any bond depends upon the overlapping limit, σ – bonds are formed by axial overlapping whereas σ – bonds are formed by sidewise overlapping. So, the extent of overlapping in σ – bond is more than π – bond. So σ – bond is more stronger than π – bond.

Question 10.
Why HF molecule is more polar than HI?
Answer:
The electronegativity of F is more than I. So, the displacement of electron in covalency in HF is more than HI, resultant the deviation of charges in HF is greater than HI. So, HF is more polar than HI.

Question 11.
C – Cl bond is polar but CCl₄ is non – polar, why? Give the reason?
Answer:
In C – Cl bond the electronegativity of Cl is more than C, due to this the electrons in covalent bond shifts towards Cl, due to which partial +ve charge appears on C and partial -ve charge develops once and this bond become polar in nature. Whereas the structure of CCl₄ is symmetric, due to which the dipole moment of C – Cl bond cancelled each other. So, CCl₄ molecule is non – polar.

Question 12.
What do you mean by resonance?
Answer:
Some compounds cannot be represented by a single definite structure, rather more than one structure is required by none of them is able to explain all the known properties of the compound alone. Thus, the various structures written for a compound to explain the known properties of it completely are called resonating structure. This phenomenon is called resonance.

Question 13.
What are the conditions for resonance?
Answer:
The conditions are as follows:

1. The heat of formation of each resonating structure should be same.
2. The arrangement of atoms in each formula should be same.
3. The number of unpaired electron in each structures should be same.

Question 14.
What is resonance energy ?
Answer:
The difference between the actual energy of the resonance hybrid and the most stable one of the resonating structures is called resonance energy.

Question 15.
Determine the bond order in N₂?
Answer:
N₂ (14 electrons): The electronic configuration
σ1s², σ^*1s², σ2s², σ^*2s², (π2p_x² = π2p_y²), σ2p^z2
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (10 – 4) = 3.

Question 16.
On the basis of molecular orbital theory explain that the Be₂ atom is not formed?
Answer:
₄Be: Electronic configuration = 1s² 2s²
Be₂ (molecule) (4 + 4 = 8e^–)
Electronic configuration = σ1s², σ^*1s², σ2s², σ^*2s²
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (4 – 4) = 0.
So, the Be₂ molecule does not formed.

Question 17.
Write the definition of hybridization?
Answer:
The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbital.

Question 18.
What do you understand by lattice energy and solvation energy?
Answer:
Lattice energy:
Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three – dimensionally in a definite geometric pattern to form ionic crystal (Crystal lattice). Since, the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied (MPBoardSolutions.com) with the release of energy referred to as lattice enthalpy. Lattice enthalpy may be defined as; the amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.

Solvation energy:
The energy released when an ion get soluble in water is called solvation energy.

Question 19.
Whose boiling points are high electrovalent compounds or covalent compounds, why?
Answer:
High boiling points: The boiling points of ionic solids are very high. This is due to strong electrostatic force of attraction between the oppositely charged ions. To change the physical state of ionic compounds, high temperature is required.

Question 20.
Why BaS0₄ is insoluble in water?
Answer:
The solubility of any ionic compound depends upon the lattice energy and solvation energy. If the lattice energy of any compound is more than solvation energy, than ionic compound is insoluble in water. The solvation energy of BaS04 is less than lattice energy. So it is insoluble in water.

Question 21.
If Be – H bond is polar, the dipole moment of Be – H₂ is zero. Why?
Answer:
BeH₂ is linear. The bond moment present in opposite direction cancelled each other. That is why Be the dipole moment of BeH₂ is zero.

Chemical Bonding And Molecular Structure Short Answer Type Questions – II

Question 1.
Why ice is lighter than water? Explain?
Or, Density of ice is less than water. Why?
Answer:
Density of ice is less than water:
In ice each oxygen atom is tetrahedrally surrounded by four hydrogen atoms in which two hydrogen atoms are linked to oxgyen atom by covalent bond and other two hydrogen atoms are linked by hydrogen bond.

The molecules of H₂O are not packed closely. This H gives rise to open cage like structure for ice having a larger volume for the given mass of water. Thus, density of ice is less than water. Ice is actually hydrogen bonded crystal. (MPBoardSolutions.com) Three dimensional structure of protein and nucleic acids like biologically important substances is due to hydrogen bond. Energy of hydrogen bond is between 3.5 kJmol^-1 and 8 kJmol^-1. Thus, hydrogen bond is stronger than van der Waals force and weaker than covalent bond.

Question 2.
Write the rules of hybridization?
Answer:
Conditions for hybridization: Following are the conditions for hybridization:

1. The orbitals of one and same atom participate in hybridization. Only the orbitals and not electrons get hybridized.
2. The energy difference between the hybridizing orbitals should be small.
3. Promotion of electron is not essential prior to hybridization.
4. It is not necessary that only the half-filled orbitals may participate in hybridiza¬tion. In some cases, even the filled orbitals may participate in hybridization.

Question 3.
Explain covalent bond with example?
Answer:
Lewis – Langmuir theory:
Lewis and Langmuir suggested that, atoms may combine by sharing of electrons in their outermost shell to complete their respective octet. The shared electrons becomes the property of both the atoms. This types of linkage is known as covalent linkage or covalent bond. Thus,

1. The force which binds atoms of same or different elements by mutual sharing of electrons is called a covalent bond.

2. This type of bond is formed between two similar non-metalic elements (A, A) or (B, B) or dissimilar atoms (A and B).
Example:
Chlorine molecule:
Both the chlorine atoms (Z = 17) contain 7 electrons in their valence shells and short in one electron each. They share one electron pair in which an electron is contributed by both as shown below:

Question 4.
Differentiate between atom and ion?
Answer:
Differences between Atom and Ion:
Atom:

1. Atoms are electroneutral.
2. Not present in free state.
3. Takes part in chemical reaction.
4. No. of e^– and proton is same.

Ion:

1. Ions are charged.
2. Ions present in free state.
3. Not take part in chemical reaction.
4. No. of e^– is more than proton.

Question 5.
Differentiate between Sigma (σ) and Pi (π) bond?
Answer:
Differences between Sigma (σ) and Pi (π) bond:
Sigma (σ) bond:

1. This bond is formed by end to end or head on overlapping of orbitals along the inter nuclear axis.
2. This is formed by overlapping of s – s, s – p or p – p orbitals.
3. Overlapping is large, hence it is strong bond.
4. Free rotation about σ – bond is possible.
5. Electron cloud is symmetrical about inter nuclear axis.

Pi (π) bond:

1. This bond is formed by the sidewise overlapping of orbitals.
2. This is formed by overlapping of p – p orbitals only.
3. Overlapping is small, hence it is weak bond.
4. Free rotation about a π – bond is not possible.
5. Electron cloud of π – bond is unsymmetrical.

Question 6.
Write difference between s – and p – orbitals?
Answer:
Differences between s – and p – orbitals:
s – orbital:

1. They are oval – symmetrical.
2. Non – directional
3. 1 = 0 and m = 0

p – orbitals:

1. They are dumbelled shape and lines symmetry of axis.
2. Directional
3. l = 1 and m = -1, 0, +1

Question 7.
Explain inter molecular and intramolecular hydrogen bonds with example?
Answer:
Types of hydrogen bond:
1. Intermolecular hydrogen bonding:
When these atoms (hydrogen and electronegative atom) are of different molecules, it is called intermolecular hydrogen bonding as in H₂O, HF, C₂H₅OH etc.

Hydrogen bond is represented by dotted lines. Many molecules of HF associates and form (HF)_n. On the same way molecules of water and alcohols are linked with hydrogen bonds.

2. Intramolecular hydrogen bonding:
If these atoms (hydrogen and electronegative atoms) are present in same molecule, this type of hydrogen bonding is called intramolecular hydrogen bonding.
e.g., o – nitrophenol

Question 8.
Among NH₃ and NF₃ whose dipole moment is high. Why?
Answer:
Both these molecules have pyramidal shape with one lone pair of electron on nitrogen atom. As fluorine is more electronegative than hydrogen therefore N – F bond should be more polar than N – H bonds. Consequently, the resultant dipole moment of NF₃ should be much larger than that of NH₃. However, the dipole moment of NH₃ (µ = 1.47D) is larger than that of NF₃ (µ = 0.24D).

The anomalous behaviour can be explained due to the presence of lone pair on nitrogen. In case of NH₃, the orbital dipole due to lone pair of electron and the bond moments of three N – H bonds are in same direction. Therefore, it adds on the resultant dipole moment of the N – H bonds. On the other hand in case of NF₃, the orbital dipole moment is in the opposite direction to resultant dipole moment of three N – F bonds. Thus, the lone pair moment cancels the resultant N – F bond moments as shown in figure. Consequently, the dipole moment of NF₃ is low.

Question 9.
Is according to following equation, is the hybridization changes in B and N:
BF₃ + NH₃ → F₃B.NH₃.
Answer:
In BF₃ three bonded pair and zero lone pair electrons are present. Due to this B is sp² hybridized and in NH₃ three bonded pair and one lone pair of electron is present. So, N is sp₃ hybridized. After reaction the hybridization of B becomes sp₃ but the hybridization of N remains same as N gives its lone pair to B atom.

Question 10.
Explain the change in hybridization in A1 atom in following reaction:
AlCl₃ + cl^– → AlCl^–₄
Answer:
The electronic configuration of Al is:
At ground state = ₁₃Al = ls² 2s² 2p⁶ 3s² 3p_x¹
At excited state = 1s², 2s², 2p⁶ 3s² 3p_x¹ _y¹
In the formation of AlCl₃, Al is sp² hybridized and its geometry is trigonal bipyramidal. Whereas in the formation of AlCl₄^–, due to inclusion of 3p_z orbital. Al is sp³ hybridized and its geometry is tetrahedral.

Question 11.
What are the postulates of orbital overlap concept of covalent bond?
Answer:
According to this concept:
1. The covalent bond is formed due to partial overlap of the two half-filled atomic orbitals of the valence shells of the combining atoms. Partial overlap means that a part of the electron cloud of each of the two half – filled domic orbitals becomes common. As a result the probability of finding electrons in the region of overlap is much more at the other places. This reduces the intemuclear repulsion and hence decreases the energy.

2. The orbitals undergoing overlap must have electrons with opposite spins.

3. Greater the extent of overlapping, stronger is the bond formed.

4. Larger the size of the orbitals, less effective is the overlapping and thus weaker is the bond formed.

Question 12.
Explain that the geometry of PCl₅ is trigonal bipyramidal and that of IF₅ is pyramidal?
Answer:
PCl₅: P is central metal atom.

As P in PCl₅ in sp³d hybridized so its geometry is square pyramidal.
IF₅: Central Metal atom is I (Z = 53)

In IF₇ , I is sp³d²

Question 13.
What are σ bond and π bond? Explain with example?
Answer:
Hybridization is defined as, “The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbitals.”

Sigma bond:
The bond formed by overlapping of two orbitals along their axis is called a sigma (σ) bond. The line joining the two nuclei of the combining atoms is called the intemuclear axis or bond axis.
Example: This type of overlapping takes place between s – s orbital, s – p orbital and pz – pz orbitals.

Pi – bond:
The bond formed by the lateral overlapping of two p – orbitals (p_x – p_x) (p_y – p_y) is called π – bond.

It is important to note that overlapping at both the lobes of the p – orbital occurs in Pi – bond whereas in case of sigma bond the overlap occurs in a single region.

Question 14.
Give the reason for the difference in properties of two allotropes diamond and graphite of carbon?
Answer:
Diamond and graphite are two allotropes of carbon. But due to difference in C arrangement their properties are different. In diamond the C atom is sp3 hybridized. Every C atom attached with four cations form tetrahedral geometry. So it forms a lattice structure and so hard and have high melting point.

In graphite every C atom is sp² hybridized, i.e. each C is surrounded by three cations and fourth valency of C is unstable. In graphite different layers are present which are joined together with weak vander Waals’ forces. That is why graphite is soft and due to presence of free electron it conducts electricity.

Question 15.
Explain the hybridized structure of acetylene by diagram?
Answer:
Formation of ethyne or acetylene (HC = CH): In the formation of acetylene molecule, each carbon atom undergoes ip – hybridization leaving two 2p – orbitals in the original unhybridized state. The two sp – hybrid orbitals of carbon atom are linear and are directed at an angle of 180°. Whereas the two unhybridized p – orbitals remain perpendicular to ip – hybrid orbital and also perpendicular to each other.

In the formation of acetylene, ip – hybrid orbital of One C – atom overlap with ip – hybrid orbital of another C – atom along the intemuclear axis forming a σ – bond. The second sp – hybrid orbital of each C – atom overlaps with the half – filled 1s – orbital of H – atom again along intemuclear axis thus forming a-bonds. (MPBoardSolutions.com) Each of the two unhybridized orbitals of both the carbon atoms overlap. Sidewise to form two π – bonds. Thus, all the carbon and hydrogen atoms are linear and there is electron cloud above and below, in the front and at the back of the C – C axis. In other words, there is electron cloud all around the intemuclear axis thus giving a cylindrical shape as represented in fig.

Question 16.
With tetrahedral geometry CH₄ molecule have a possible geometry of square plannar. In which the H atoms are present at the four corners of the square. Explain that the CH₄ molecule is not have square plannar geometry?
Answer:
The electronic configuration of C is:
In ground state ₆C = 1s² 2s² 2p_x¹2p_y¹
In excited state ₆C = 1s² 2s¹2p_x¹ 2p_y¹2p_z¹
sp³ hybridisation.
In CH₄ molecule carbon is sp³ hybridized. So its geometry is tetrahedral. For square plannar geometry dsp² hybridization is necessary. But due to absence of d – orbital in C atom. This geometry is impossible. With this according to VSEPR concept the bonded electrons in C atom is present at four comers of tetrahedron. The bond angle in tetrahedron is 109°28′ and in square plannar 90°. So in case of tetrahedral geometry the repulsion of electrons is less than in square plannar geometry.

Question 17.
On the basis of hybridization explain that the structure of BeCl₂ is linear?
Answer:
Formation of BeCl₂:
In the compound (BeF₂, BeCl₂, etc.), beryllium shows a covalency of two. In order to explain the formation of two equivalent bonds with beryllium its 25 – electron from the ground state (₄Be, Is² 2s²) is excited to 2p – orbital (1s²2s¹ 2p¹]) 2s and 2p – orbitals get mixed up to two equivalent sp – hybrid orbitals which make an angle of 180° with each other and oriented linearly. Each sp – orbitals overlap with half – filled p – orbital of chlorine (1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹) atoms to form two sigma bonds. Thus, the shape of BeCl₂ is linear.

Question 18.
Explain the formation of N₂ molecule on the basis of orbital theory as overlapping?
Answer:
π – bond formation of N₂ molecules:
Nitrogen molecule has a triple – bond consiting of one σ and two π – bonds (\(N\overset { \pi }{ \underset { \pi }{ \equiv N } } \)). Nitorgem atom has three half – filled p – orbitals.
₇N : 1s² 2s² 2p_x¹2p_y¹ 2p_z¹

When 2p_x orbital of each nitrogen atom overlaps co – axially, a σ – bond is formed. The 2p_y and 2p_z orbitals of one N atom overlap N atom to overlap laterally to form two π – bonds.

Chemical Bonding And Molecular Structure Long Answer Type Questions – I

Question 1.
Write the postulates of Valence bond theory? Write its limitation also?
Answer:
Valence bond theory was given by Heitler and London which is modified by Pauling and Slater. The postulates of the theory are:

1. The covalent bonds are formed by the partial overlapping of atomic orbitals (half filled).
2. In the orbitals taking part in overlapping electrons with opposite spin are present.
3. Strength of bond depends upon the extent of overlapping.
4. Strong directional bonds are formed between the orbitals of same stability, same energy and same symmetry.

Limitations of Valence Bond Theory:

1. According to this theory, no unpaired e^– is present in O² molecules. So the nature of O² is paramagnetic but O² is diamagnetic.
2. Not explain about the formation of coordinate bonds.
3. This theory is failed to explain the formation of H₂⁺ molecules.
4. Doesn’t give any information about resonance.

Question 2.
Write the condition for the formation of molecular orbitals by linear combination of atomic orbitals?
Answer:
Conditions for the combination of Atomic Orbitals:
Molecular orbital is formed by the linear combination of atomic orbitals. There are certain conditions for the effective linear combination of atomic orbitals. These conditions are:

1. The combining atomic orbitals should have same or nearly same energy: This means that in the formation of a homonuclear diatomic molecule Is atomic orbital of one atom will undergo linear combination with 1s atomic orbital of the other atom, but not with the 2s atomic orbital because the energy of the 2s orbital is appreciably higher than that of 1s atomic orbital. Similarly, because of the energy difference between 2s and 2p atomic orbitals, they will also not combine to form molecular orbitals.

2. There should be maximum overlap of atomic orbitals:
Greater the overlap, greater will be the charge density between the nuclei of a molecular orbital. This condition is often referred to as the principle of maximum overlap.

3. The atomic orbitals should have the same symmetry about the molecular axis:
This condition is known as symmetry condition for the combination of atomic orbitals. Taking the Z – axis as the molecular axis, the following pairs of atomic orbitals will not combine to form any molecular orbital, because of their different symmetries.

1. s – p_x pair
2. s – p_y pair
3. p_x – p_y pair and
4. ↔p_y – p_z pair.

This means that, s – s, px- px, py -py and Pz~Pz combinations are allowed because combining atomic orbitals have the same symmetry. A ,pz orbital, however, is able to combine with an s-orbital since, they have the same symmetry.

Question 3.
Differentiate between Ionic compounds and Covalent compounds?
Answer:
Differences between Ionic and Covalent compounds:

Question 4.
On what factors the formation of ionic bond depends?
Answer:
Factors influencing Ionic Bond formation:
The formation of ionic bond (Electrostatic force of attraction) depends upon the following factors:
1. Ionisation enthalpy:
One of the combining atoms (metal) must have low ionisa-tion enthalpy. So that, cation formation becomes easy. For example, alkali and alkaline earth metals of periodic table has tendency to form positive ion because they have comparatively low ionisation energy.

2. Electron gain enthalpy:
The electrons released in the formation of cation are to be accepted by the other atom taking part in the ionic bond formation. The electron accepting tendencies of an atom depends upon the electron gain enthalpy. (MPBoardSolutions.com) It may be defined as: Energy released when an isolated gaseous atom takes up an electron to form an anion. Greater the negative electron gain enthalpy, easier will be the formation of anion or negative ion. The halogen present in group 17 have the maximum tendency to form anions as they have very high negative electron gain enthalpy.

3. Lattice energy:
The amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.
A_(g)⁺ + B_(g)^– → A⁺B_(s)^– + Lattice enthalpy (∆_LH)
Thus, greater the magnitude of -ve lattice energy, more will be the stability of the ionic bond or ionic compound.

Question 5.
Explain the main points of Molecular orbital theory?
Answer:
The main points of Molecular orbital theory are:
1. In a molecule, electrons are present in new orbitals called molecular orbitals. These molecular orbitals are characterised by a set of quantum numbers just like atomic orbitals.

2. Molecular orbitals are formed by combination of atomic orbitals of equal energies (in case of homonuclear molecules) or of comparable energies (in case of heteronuclear molecules).

3. The number of molecular orbitals formed is equal to the number of atomic orbitals undergoing combination.

4. Two molecular orbitals are formed by combination of two atomic orbitals one of these two molecular orbitals has a lower energy and the other has a higher energy than either of the combining atomic orbitals. The molecular orbital with lower energy is called bonding molecular orbital and the other is called antibonding molecular orbital.

Question 6.
What is resonance? Explain with example?
Answer:
When properties of a molecule are not explained by one structure and two or more than two structures are assigned to express its characteristics, it is said that molecule is resonance hybrid of these structures and this property is known as resonance. Different resonating structures are exhibited by using sign (↔) in between these structures.
Example: Carbon dioxide (CO₂) is represented by following three structures:

Chemical Bonding And Molecular Structure Long Answer Type Questions – II

Question 1.
Show the molecular orbital energy levels of N₂ by diagram?
Answer:
N₂ molecule: Each nitrogen atom contains seven electron. Thus total 14 electrons are filled in seven orbitals of increasing energy.

Molecular orbital structure of N₂ molecule will be as follows:

Where KK represents closed shell structure (σ _1s)²(\(({ \sigma _{ 1s } })^{ 2 }\))(\(({ \sigma ^{ * }_{ 1s } })^{ 2 }\)
This structure shows that it contains 10 bonding and 4 antibonding electrons.
Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 4] = 3.
Value of bond order is more. Hence value of bond energy should also be high. Experimental value of bond energy is 945 kJ mol^-1 which proves the presence of paired electrons in nitrogen molecule. Thus, it is a diamagnetic molecule.

Question 2.
Give the important applications of dipole moment?
Answer:
Applications of Dipole moment:
1. Comparison of relative polarity:
It is possible by comparing the value of dipole moment e.g., HF (1.98 D) is more polar than HCl (1.03D).

2. Predicting the nature of molecules:
Molecules with specific dipole moments are polar in nature while those with zero value are non – polar. Thus, BeF₂ (µ – 0D) is non – polar while H₂O (µ = 1.84D) is polar.

3. Calculation of percentage ionic character:
% ionic character = \(\frac{Observed dipole moment}{Caluculated dipole moment}\) × 100 (100% ionic character)
or % I.C. = \(\frac { \mu _{ obs } }{ \mu _{ cal } } \) × 100
For example, the observed dipole moment of HCl molecule is 1.03D. For 100% ionic character i.e., complete transfer of electron charge on H⁺ and Cl^– ions would be equal to one unit (4.8 × 10^-10e.s.u.)× (1.275 × 10^-8 each. The bond length of H – Cl bond is 1.275 × 10^-8cm. Therefore, dipole moment for complete electron transfer
µ = q × d = (4.8 × 10^-10 e.s.u) × (1.275 × 10^-8cm)
= 6.12 × 10^-18 e.s.u cm = 6.12D
Observed dipole moment, µ(obs) = 1.03D
% ionic character = \(\frac { 1.03 }{ 6.12 } \) × 100 = 16.83 %

4. Dipole moment of symmetric molecules is zero, but they have two or more than two polar bonds. It is applied for the measurement of symmetry.

5. Distinction between ortho, meta and para isomers of aromatic compounds:
In general, the dipole moments follow the order: ortho > meta > para e.g., In dichlorobenzene, the dipole moments of o, m and p isomers are: 2.54 D, 1.48 D and 0 respectively.

Question 3.
What is valence shell electron pair repulsion theory? Write its limitations? Or, Explain valence shell electron pair repulsion theory with example?
Answer:
VSEPR theory:
On the basis of this theory, “When central atom is surrounded by only bonded electron pairs, in that case the geometry of the molecule will be ordinary, but if the central atom is surrounded by bonded electron pairs as well as lone pairs or non – bonded pairs, then the geometry of die molecule will become abnormal.” This (MPBoardSolutions.com) means that the repulsion between the non – bonded or lone pair of electrons and bonded electron pairs become greater than that of repulsion between only bonded electron pairs. Repulsion

Between the electron pairs is in the following order:
lone pair – lone pair > lone pair – bonded pair > bonded pair – bonded pair

Shapes of some molecules accroding to VSEPR theory:
1. Shape of CH₄:
In methane, central atom carbon is surrounded by four electron pairs and four C – H bonds. This molecule has tetrahedral geometry. Shared electrons are at the comers of tetrahedron for maximum separation. Bond angle is 109°28′.

2. Shape of H_2O:
In water molecule central oxygen atom is surrounded by four electron pairs, two of which are lone pair of electrons. Thus, Ip – lp and lp – bp repulsion exist. Due to this, bond angle reduces to 104.5° in place of 109°28′ and shape becomes V shaped.

3. Shape of NH₃:
In ammonia, central N atom is surrounded by four electron pairs, so shape of molecule is tetrahedral. According to VSEPR theory, the four groups around the central atom of ammonia should be tetrahedrally arranged at bond angle of 109°28’.

But, the measured bond angle is 107°. This is explained on the basis of repulsive effect of the lone pair of electrons on bonding electrons. In ammonia molecule there is a lone pair of electrons on the N atom. Thus, the shape of NH₃ molecule is distorted and it looks like pyramidal and it is polar in nature.

Example:

Limitations of VSEPR theory:
VSEPR theory no doubt, theoretically gives the shapes of simple molecules but could not explain them and also has limited application. To overcome these limitations, two important theories based on quantum mechanical principles are commonly used. These are:

1. Valence bond theory (VBT) and
2. Molecular orbital theory (MOt).

Question 4.
Structure of two molecules are given:

1. Among these which contain intermolecular hydrogen bond and intramolecular hydrogen bond?
2. The melting point of any compound depends upon hydrogen bonding also, on the basis of this explain which one have high M.P.
3. The solubility of any compound depends upon its tendency to formed H – bonding with water among these who will form H – bond with water easily.

Answer:
1. Compound (a) forms intramolecular H – bond. When H – bond is present between the atoms of same molecule than it is called intramolecular H – bond.
In ortho nitrophenol [(a)] H – bond is present between two O – atoms.

In compound (b) inter molecular H – bond forms.
In p – nitrophenol [(b)] between – N0₂ and – OH vacant space is present. So among H – atom of one molecule and O – atom of another molecule, H – bond is formed.

2. The melting point of compound (b) is high because many molecules are forming H – bond.

3. Due to intramolecular H – bonding, compound (a) will not form H – bond with water so it is less soluble in water. Whereas compound (b) forms H – bond easily with water so soluble in water.

Question 5.
Draw molecular orbital diagram for 0₂ molecule?
Answer:
Oxygen molecule, (0₂): Each oxygen atom has eight electrons. When two oxygen atom combine, molecular orbitals are formed. These molecular orbitals have following configuration:

Its molecular orbital diagram is:

From the above configuration we have,
N_b = 10, N_a = 6
∴ Bond order = \(\frac{1}{2}\) [N_b – N_a] = \(\frac{1}{2}\) [10 – 6] = 2
Hence, there is a double bond in oxygen molecule. Due to the presence of two unpaired electrons it is paramagnetic.

MP Board Class 11 Chemistry Important Questions

MP Board Class 12th Biology Important Questions Chapter 12 Biotechnology and its Applications

Biotechnology and its Applications Important Questions

Biotechnology and its Applications Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Restriction endonuclease is employed for cutting :
(a) A single stranded DNA
(b) Double stranded DNA
(c) RNA fregment
(d) m – RNA.
Answer:
(d) m – RNA.

Question 2.
Which enzyme is helpful in genetic engineering:
(a) RNA polymerase
(b) DNA polymerase
(c) Restriction endonuclease
(d) Nuclease.
Answer:
(c) Restriction endonuclease

Question 3.
Extrachromosomal DNA used as vector in gene cloning is :
(a) Transposon
(b) Intron
(c) Exon
(d) Plasmid.
Answer:
(d) Plasmid.

Question 4.
The most extensively used bacteria in genetic engineering is :
(a) Bacillus
(b) Clostridium
(c) Escherichia
(d) Salmonella.
Answer:
(c) Escherichia

Question 5.
Hybridomas are:
(a) Antibodies
(b) Hybrid swarms
(c) Hybrid cultures
(d) Hybrid zone.
Answer:
(c) Hybrid cultures

Question 6.
Vaccine can not protect us from :
(a) Malaria
(b) Polio
(c) Small pox
(d) Tuberculosis.
Answer:
(a) Malaria

Question 7.
Humulin is:
(a) A type of chitin
(b) New digestive enzyme
(c) Powerfull antibiotic
(d) Human insuline
Answer:
(a) A type of chitin

Question 8.
Which is not used in the form of bio weapons :
(a) Bacillus anthrasis
(b) Bacillus thuringiensis
(c) Botulinum toxin
(d) Small pox.
Answer:
(b) Bacillus thuringiensis

Question 9.
Super bug is called :
(a) Pseudomonas pulida
(b) Xanthomonas citrae
(c) Bacillus anthresis
(d) Streptococcus pneumoniae
Answer:

Question 10.
Biopiracy affected plants are :
(a) Neem
(b) Basmati rice
(c) Turmeric
(d) All of these.
Answer:
(d) All of these.

Question 11.
Golden rice will help in :
(a) Producing petrol like fuel
(b) Pest resistance
(c) Herbicide tolerance
(d) Alliviation of Vit – A deficiency
Answer:
(d) Alliviation of Vit – A deficiency

Question 12.
During gene cloning which is called as “gene taxi” :
(a) Vaccine
(b) Plasmid
(c) Bacterium
(d) Protozoa.
Answer:
(b) Plasmid

Question 13.
In transgenics, expression of transgene in target tissue is determied by.
(a) Transgene
(b) Promoter
(c) Reporter
(d) Enhancer.
Answer:
(b) Promoter

Question 2.
Fill in the blanks :

1. Bt is a ………………………..
2. Bt toxin is formed from ……………………….. by the bacillus.
3. The prooduct of Cry – 1 gene is used for ………………………..
4. RNA interpritation is the method of cellular conservation in ……………………….. genes.
5. Both peptide chains are joined by the ……………………….. in insuline.
6.  ……………………….. is the found in proinsuline which is not found in muture insulin.
7.  ……………………….. is found in transgenic cow ‘Rosi’, it is not found in commen cow’s milk.
8. Flavr savr tomato is a transgenic veriety of ………………………..
9.  ……………………….. has two polypeptide chain and 51 amino acids.
10.  ……………………….. gene which found in Rhizobium is converted from nitrogen into nitrates.
11.  ……………………….. is the reason of male infertility in maize.
12. First transgenic organism is petant in ………………………..
13. Monoclsonal antibody is produce with the help of ……………………….. technology.

Answer:

1. Bio Toxin
2. Protoxin
3. Corn bores
4. Eukaryotic
5. Three sulphaid bundles
6. Peptide ‘c’
7. Human α – heredity lectalbumen
8. Tomato
9. Insulin
10. Nif gene
11. Cytoplasmic heredity
12. 1989
13. Hybridoma.

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

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

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

1. Write the full name of RNAi.
2. Bt Toxic is encrypted by this gene.
3. Is our blood has proteoses and nucleases.
4. Who produce Bt toxin protein?
5. What is humulin?
6. What is SCID (Severe Combined Immunodeficienay)?
7. Which is controled by Trycoderma?
8. Name the organism which bear the other organism obtained gene?
9. Bt. cotton and Golden rice is the examples of.
10. Name the gene which controles the cell cycle?

Answer:

1. RNA interference
2. By Cry genes
3. No
4. Bacillus thuringiensis
5. Human insulin
6. Hereditary disease
7. Biological fungal disease
8. Transgenic organism
9. Transgenic crop
10. p53 gene.

Biotechnology and its Applications Important Questions Very Short Answer Type Questions

Question 1.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of r – DNA technology and chemistry of oil?
Answer:
To remove oil from seeds using recombinat DNA technology would involve:

1. Identifying the genes that code for oil production.
2. Deleting these genes from the seed genome.
3. Splicing back together the remaining DNA.
4. Putting it back into the cell.

It will be not be very easy because the oils are made up of fatty acids and glycerol. Since, fatty acids are important components of cell membrane system, deleting or switching off of its genes might affect the cell structure itself.

Question 2.
Find out from internet what is golden rice?
Answer:
Golden rice is a variety of rice produced through genetic engineering to biosynthesize β – carotene, a precursor of vitamin ‘A’ in the edible parts of rice. It is intended to produce a fortified food to be grown and consumed in areas with a shortage of dietary vitamin ‘A’.

Question 3.
Does our blood have proteases and nucleases?
Answer:
No.

Question 4.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).

Question 5.
Name the crops which are prepaired with the help of biotechnology.
Answer:
Bt cotton, Bt maize, paddy, tomato, potato and soyabeen.

Question 6.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.

Question 7.
By which Bt toxin is coded?
Answer:
By Cry genes Bt toxin is coded.

Question 8.
Full form ofRNAi.
Answer:
RNA interference (RNAi).

Question 9.
Name the therapy which is help of missing of defective ones in order to correct genetic disorders.
Answer:
Gene therapy.

Question 10.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.

Biotechnology and its Applications Important Questions Short Answer Type Questions

Question 1.
What is genetically modified (GM) food? Give two examples.
Answer:
Genetically Modified (GM) food:
The food substances produced from genetically modified crops or transgenic crops is called GM food. This food differ from conventionally developed varieties in the following aspects :

1. GM food contains antibiotic resistance gene itself.
2. It contains protein produced by transgene, example Cry protein in insect resistance varieties.
3. These GM foods contain enzyme produced by the antibiotic resistance gene that was used during gene transfer by recombinant DNA technology.

Examples of GM Crops, Food and Fruits:

1. Flavr Savr Tomato:
It is the first food containing genetically engineered DNA. These tomatoes contain genes for antibiotic resistance for kanamycin.

2. Maize:
GM maize has a bacterial gene which increases its resistance to pests and diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore introduction of GM maize is opposed by many European countries.

3. Rape oil seed:
It is a new type of plant that contain genes for resistance to the herbicide Basta. It has for more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also effect food chains in unpredictable ways.

Question 2.
Write down the advantage of GM Crops.
Answer:
Advantages of GM Crops:

• Genetic modification has made crops more tolerant to abiotic stresses (cold, drought, heat, salt.)
• Viral resistance can be introduced.
• Over ripening losses can be reduced. Example : Flavr Savr Tomato.
• Enhanced nutritional value of food. Example : Golden Rice.
• Reduced reliance on chemical pesticides.

Disadvantages of GM crops:

• Transgenes in crop plants can endanger native species. Example : The gene for Bt toxin expressed in pollen may end natural pollinators such as honeybees.
• Weeds also become resistant.
• Products of transgenes may be allergic or toxic.
• They cause damage to the natural environment.

Question 3.
What is perfect agriculture? How is this method better than traditional method? Explain.
Answer:
Perfect agriculture is a method of agriculture which is sustainable, perfect and harmless. Green revolution and there after the production of agricultural crops has definately increased due to use of new and high yielding varieties, development of irrigation facilities, increased irrigated area, use of fertilizers etc. but it results many problems such as loss of soil fertility, pollution of food and water and diseases. The resistance power of plants and human beings falls slowly.

Food and water borne diseases affecting the health of human beings and animals. All of these conditions and events taking place due to the modem commercial agriculture. (MPBoardSolutions.com) Therefore, it would become necessary to develop a method of agriculture which would be free from above mentioned demerits. This kind of agriculture is called to be as perfect or sustainable agriculture. Organic agriculture is the best example of perfect agriculture.

Question 4.
What is organic cropping? What are its basis?
Answer:
Organic agriculture:
Organic agriculture is a method of agriculture which does not allow the use of synthetic fertilizers, pesticides, insecticides, weedicides, plant growth regulators, substances of animal origin and genetically modified bacteria. In this method biofertilizers, crop rotation methods are used to increase crop production and biopesticides are used to control insects and weeds. Thus, organic farming is a holistic way of agriculture which tries to bridge the widening gap between man and nature.

If has the commitment of meeting production needs on one hand and sustaining resources and ecosystem function on the other hand. Thus, organic farming is an alternative agriculture production system which avoid or largely excludes the use of synthetic chemicals, fertilizers, pesticides and growth regulating hormones and live stock additives.

Basics of organic agriculture:

1. Organic agriculture is based on improvement of soil, plants, animals, man and global scinery and make it sustainable.
2. Organic agriculture is based on those ecosystems and biocycles which utilize that organisms which would be promoted.
3. It is based on the principle which are related with making pollution free environment and possibilities of life.
4. It is also based on saving environment and health of present and future generations.

Question 5.
What is eugenics? Write importance of eugenics.
Answer:
Eugenics: The branch of biology which deals with the study of improvements of human race is called eugenics.

Importance:

1. Development of selective reproduction in similar species.
2. Transfer of genetic materials in various organisms.
3. Development ofGM food and GM crops.
4. Gene cloning.
5. Gene therapy, etc.