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NCERT MCQ CLASS-11 CHAPTER-15 | MATH NCERT MCQ | STATISTICS| EDUGROWN

In This Post we are providing Chapter-15 Statistics NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON STATISTICS

Question 1: What is the standard deviation of the following series ?
Measurements 0 -10 10 – 20 20 -30 30 – 40
Frequency 1 3 4 2
(a) 81
(b) 7.6
(c) 9
(d) 2.26

Answer : C

Question 2: The mean deviation from the mean of the following data :
Marks 0-10 10-20 20-30 30-40 40-50
No. of 5 8 15 16 6
Students
is
(a) 10
(b) 10.22
(c) 9.86
(d) 9.44

Answer : D

Question 3 : The standard deviation for the following data is
Wages per week upto 15 30 45 60 75 90 105 120
Number of Workers 12 30 65 107 157 202 222 230
(a) 26.75
(b) 24.46
(c) 25.89
(d) none

Answer : C

Question 4 : The mean of 13 observations is 14. If the mean of the first 7 observations is 12 and that of the last 7 observations is 16, what is the value of the 7th observation ?
(a) 12
(b) 13
(c) 14
(d) 15

Answer : C

Question 5: A batsman scores runs in 10 innings 38, 70, 48,34, 42, 55, 63, 46, 54 and 44, then the mean deviation is
(a) 8.6
(b) 6.4
(c) 10.6
(d) 9.6

Answer : A

Question 6 : Let x1, x2, …..xn be n observations. Let wi = l xi + k for i = 1, 2, … n, where l and k are constants. If the mean of xi’s is 48 and their standard deviation is 12, the mean of w’is is 55 and standard deviation of w’is is 15, the values of l and k should be
(a) l = 1.25, k = – 5
(b) l = –1.25, k = 5
(c) l = 2.5, k = – 5
(d) l = 2.5, k = 5

Answer : A

Question 7 : Standard deviations for first 10 natural numbers is
(a) 5.5
(b) 3.87
(c) 2.97
(d) 2.87

Answer : D

Question 8 : The standard deviation of some temperature data in °C is 5. If the data were converted into °F, the variance would be
(a) 81
(b) 57
(c) 36
(d) 25

Answer : A

Question 9 : The variance of the following distribution is
x₁ 2 3 11
f(x₁) 1/3 1/2 1/6
(a) 10
(b) 16
(c) 8
(d) 7.5

Answer : A

Question 10 : In an experiment with 15 observations on x, the following results were available :Sx2 = 2830, Sx =170One observation that was 20 was found to be wrong and was replaced by the correct value 30. The corrected variance is
(a) 8.33
(b) 78.00
(c) 188.66
(d) 177.33

Answer : B

Question 11 : In a series of 2 n observations, half of them equal a and remaining half equal –a. If the standard deviation of the observations is 2, then |a| equals.
(a) √2 / n
(b) √2
(c) 2
(d) 1/n

Answer : C

Question 12 : A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 gm and a standarion deviation of 2 gm. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 gm. The correct mean and standard deviation (in gm) of fishes are respectively :
(a) 32, 2
(b) 32, 4
(c) 28,2
(d) 28, 4

Answer : A

Question 13 : In a batch of 15 students, if the marks of 10 students who passed are 70, 50, 95, 40, 60, 70, 80, 90, 75, 80 then the median marks of all the 15 students is:
(a) 40
(b) 50
(c) 60
(d) 70

Answer : C

Question 14 : If f (x + y, x – y) = xy, then the arithmetic mean of f (x, y) and f(y, x) is :
(a) x
(b) y
(c) 0
(d) none of these

Answer : C

Question 15: If a variable takes discrete values x + 4, x – 7/2 , x – 5/2 , x – 3 , x – 2 , x + 1/2 , x – 1/2 , x + 5 (x is positive), then the median is :
(a) x – 5/4
(b) x – 1/2
(c) x – 2
(d) x + 5/4

Answer : A

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NCERT MCQ CLASS-11 CHAPTER-14 | MATH NCERT MCQ | MATHEMATICAL REASONING | EDUGROWN

In This Post we are providing Chapter-14 Mathematical Reasoning NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON MATHEMATICAL REASONING

Question 1.
The connective in the statement 2 + 7 > 9 or 2 + 7 < 9 is
(a) and
(b) or
(c) >
(d) <

Answer: (b) or

Question 2.
Which of the following is not a negation of the statement A natural number is greater than zero
(a) A natural number is not greater than zero
(b) It is false that a natural number is greater than zero
(c) It is false that a natural number is not greater than zero
(d) None of these

Answer: (c) It is false that a natural number is not greater than zero.

Question 3.
Which of the following is the conditional p → q
(a) q is sufficient for p
(b) p is necessary for q
(c) p only if q
(d) if q then p

Answer: (c) 6 is a natural number

Question 4.
The contra-positive of the statement If a triangle is not equilateral, it is not isosceles is
(a) If a triangle is not equilateral, it is not isosceles
(b) If a triangle is equilateral, it is not isosceles
(c) If a triangle is not equilateral, it is isosceles
(d) If a triangle is equilateral, it is isosceles

Answer: (d) If a triangle is equilateral, it is isosceles

Question 5.
Which of the following is a statement
(a) I will go tomorrow
(b) She will come today
(c) 3 is a prime number
(d) Tomorrow is Friday

Answer: (c) 3 is a prime number

Question 6.
The contra-positive of the statement if p then q is
(a) if ~p then q
(b) if p then ~q
(c) if q then p
(d) if ~q then ~p

Answer: (d) if ~q then ~p

Question 7.
Which of the following is not a statement
(a) The product of (-1) and 8 is 8
(b) All complex number are real number
(c) Today is windy day
(d) All of the above

Answer: (d) All of the above

Question 8.
If (p or q) is true, then
(a) p is true and q is false
(b) p is true and q is true
(c) p is false and q is true
(d) All of the above

Answer: (d) All of the above

Question 9.
Which of the following statement is a conjunction
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are friends
(c) Both Ram and Shyam are enemies
(d) None of these

Answer: (d) None of these

Question 10.
Which of the following is a compound statement
(a) Sun is a star
(b) I am a very strong boy
(c) There is something wrong in the room
(d) 7 is both odd and prime number.

Answer: (d) 7 is both odd and prime number.

Question 11.
Which of the following is a statement
(a) x is a real number
(b) Switch of the fan
(c) 6 is a natural number
(d) Let me go

Answer: (c) 6 is a natural number

Question 12.
Which of the following is not a statement
(a) 8 is less than 6.
(b) Every set is finite set.
(c) The sun is a star.
(d) Mathematics is fun.

Answer: (d) Mathematics is fun.

Question 13.
Which of the following is true
(a) A prime number is either even or odd
(b) √3 is irrational number.
(c) 24 is a multiple of 2, 4 and 8
(d) Everyone in India speaks Hindi.

Answer: (d) Everyone in India speaks Hindi.

Question 14.
If (p and q) is false then
(a) p is true and q is false
(b) p is false and q is false
(c) p is false and q is true
(d) all of the above

Answer: (d) all of the above

Question 15.
The converse of the statement p ⇒ q is
(a) p ⇒ q
(b) q ⇒ p
(c) ~p ⇒ q
(d) ~q ⇒ p

Answer: (b) q ⇒ p

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CLASS 12TH CHAPTER -16 Environmental Issues |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

All chapter class 12 Biology NCERT Solutions are prescribed by the subject teachers based on the NCERT textbook questions and explained all solutions in a better way for easy understanding. In-depth knowledge of the biology subject can be attained by the students with the help of detailed NCERT Solutions of Class 12 Biology

NCERT Solutions for Class 12 Biology Chapter :16 Environmental Issues

Question 1.
What are the various constituents of domestic sewage ? Discuss the effects of sewage discharge on a river.
Solution:
The domestic sewage contains every-thing that goes down the drain into the sewer of the house. The various constituents of domestic sewage are suspended solids, colloidal particles, pathogenic contaminants and dissolved materials. Suspended solids are sand and silt. Colloidal particles include clay, faecal matter, fine fibres of paper and cloth. Pathogenic contaminants are eggs of coliforms and enterococci. Dissolved materials includes inorganic nutrients such as nitrates, phosphates, ammonia, sodium and calcium. Effects of sewage discharge on a river :

Water becomes unfit for bathing and drinking and also for domestic or industrial use as it becomes colored, turbid with a lot of particulate matter floating on water.
The domestic sewage adds nitrates and phosphates into the river. These nitrates and phosphates encourage a thick bloom of blue green algae, which depletes the oxygen content of the water during night. This suffocates the fish and other aquatic life. Consequently river become highly polluted.

Question 2.
List all the wastes that you generate, at home, school, or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Solution:
Wastes generated at home include plastic bags, paper napkins, 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 plastic bags with biodegradable jute bags can reduce wastes generated at home, school or during trips. Domestic sewage can be reduced by optimizing the use of water while bathing, cooking, and other household activities.

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

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Solution:

Global warming is a rise in the mean temperature of the lower atmosphere and the earth’s surface. Causes – increase in the quantity of radioactively active greenhouse gases CO₂, CH₄, N₂O, CFCs. They allow heat waves to reach the surface and prevent their escape.
They are produced by combustion of fossil fuels, biomass [CO₂]; burning of nitrogen-rich fuels [N₂O]; paddy fields, fermentation in cattle and wetlands [CH₄]; refrigerators, aerosols, drying, cleaning [CFCs].
Effects: Heating of earth surface [mean temperature is increased] Climatic changes e.g.: El Nino effect.
Increased melting of polar ice caps and Himalayan snowcaps. Increased sea levels and coastal areas will submerge.
Measures – Decreased use of fossil fuels, improve the efficiency of energy usage, Reduce deforestation, plant trees Control of man-made sources of greenhouse gases like vehicles, aerosol sprays.

Question 4.
Match the items given in Column A and B
column A Column B
(a) Catalytic converter (i) Particulate matter
(b) Electrostatic precipitator (ii) Carbon monoxide and nitrogen oxides
(c) Earmuffs (iii) High noise level
(d) Landfills (iv) Solid wastes
Solution:
(a) – (ii); (b) – (i); (c) – (iii); (d) – (iv).

Question 5.
Write critical notes on the following :
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Solution:
(a) Eutrophication: The natural aging process of lakes by nutrient enrichment of their water. In young lake water is cold and clear and supports only little life. With time, streams introduce nutrients into lake which increases lakes’ fertility and encourages aquatic growth. Over centuries silts and organic debris pile up, and lake becomes shallow and warmer. It supports plants and later gets converted into land. Lakes span depends on the climate, size of lake.

(b) Biological magnification: Industrial wastes released into water contain toxic substances, such as arsenic, cadmium, lead, zinc, copper, mercury, and cyanides, besides some salts, acids, and alkalies. All these materials can prove harmful for our health.

They may reach the human body directly with contaminated food or indirectly by way of plants and other animals. The concentration of the toxic materials increases at each trophic level of a food chain. This is called biological magnification. River water may have a very low concentration of DDT, but the carnivorous fish in that river may contain a high concentration of DDT and become unfit for eating by man. Mercury discharged into rivers and lakes is changed by bacteria to the neurotoxic form called methyl mercury. The latter is highly poisonous and may be directly absorbed by fish.

(c) Groundwater depletion and ways for its replenishment: Groundwater depletion is defined as long-term water-level decline caused by sustained groundwater pumping. The volume of ground water in storage is decreasing in many areas of the world in response to pumping. Some of the negative effects of groundwater depletion include increased pumping costs, deterioration of water quality, reduction of water in streams and lakes.
Some ways for water replenishment are:

Reduction in consumption: Sprinkler and subsurface irrigation techniques reduce the amount of water used in irrigation.
Rain water harvesting: Rain water collected over roofs is allowed to pass into the ground through deep water pipes.

Question 6.
Why does the ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Solution:
Ozone hole forms over Antarctica where no one lives and no pollution is present but not over Newyork, Bangalore etc., (polluted cities). It is because CFCs and ozone-depleting substances (ODS) released worldwide accumulates in the stratosphere and drifts towards, Antarctica in winters (July – August) when temperatures is -’85° C in Antarctica.

In winters polar ice clouds are formed over Antarctica. It provides a catalytic surface for (CFCs and other ODS to release CL and other free radicals that breakdown ozone layer forming an ozone hole during spring in presence of sunlight. In summer, the ozone hole disappears due to mixing of air worldwide.

Ozone holes allow UV radiations (UV_A & UV_B) to reach earth’s surface. Which was earlier reflected by the ozone layer. UV_B damages DNA, skin cells and causes mutations and skin cancers respectively. UV_B even causes corneal damage (Snow Blindness).

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Solution:
Forest Conservation and Management:
It is time to think deeply and act seriously in order to protect this vital natural resource. Some of the measures of conservation are

Social forestry programme: It was started in 1976 and involves the afforestation on public and common lands for fuel, fodder, timber for agricultural equipment and fruits. These are mainly meant for rural people.
Agroforestry programme: It involves the multiple use of same land for agriculture, forestry and animal husbandary. Taungya System and Jhum are examples.
- Taungya System: It involves growing agricultural crops between planted trees.
- Jhum (Slash and burn agriculture): It involves felling and burning of forests, followed by the cultivation of crops for a few years. Later the cultivation is abandoned for the growth of forests. It is a traditional agroforestry system.
Urban forestry programme: It involves afforestation in urban land areas e.g. along the roads, big parks, big compounds etc. with ornamental and fruit trees.
Commercial forestry: It involves planting of fast-growing trees on available land to fulfill commercial demand.
Conservation forestry: It involves protection of degraded forest to allow recoupment of their flora and fauna.

Reforestation: It is the process of restoring a forest that once existed, but was removed at some point of time in the past. Reforestation may occur naturally in a deforested area. The above-said methods speed up the reforestation programme.

Question 8.
What measures, as an individual, you would take to reduce environmental pollution?
Solution:
To reduce environmental pollution, we should change our habits and lifestyle so as to reduce the use of disposable materials. We should use preferably those items which can easily be recycled and also minimise the use of fossil fuels. We should also take measures to improve the quality of air by using CNG gases wherever possible instead of using diesel or petrol. We should also use the catalytic converter in our vehicles.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid wastes
Solution:
a. Radioactive waste materials are released from thermonuclear explosions. Radioactive isotopes, such as radium-226, thorium- 232, potassium-40, uranium-235, carbon-14, etc. are spread all over the world and contaminate air, soil, water, vegetation and animals.

b. Irrepairble electronic goods and computers are called electronic wastes (e-waste).
Ships that are no longer in use or that are to be dismantled are called defunct ships. Asbestos, Polychlorinated Biphenyl (PCB) produced during dismantling defunct ship cause serious health hazards especially cancer.

c. Municipal solid wastes are wastes from homes, offices, stores, schools, hospitals, etc., that are collected and disposed of by the municipality.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Solution:
Under the direction of Supreme Court of India, the State Government of Delhi took the following measures to improve the quality of air:

Switching over the entire fleet of public transport buses from diesel to CNG (Compressed Natural Gas) by the end of 2002.
Phasing out of old vehicles.
Use of unleaded petrol.
Use of low sulphur petrol and diesel.
Use of catalytic converters in vehicles.
Application of Euro II norms for vehicles.

Because of above mentioned measures adopted by the Government the air quality of Delhi has improved with a substantial fall in S0₂, CO, No_x level between 1997-2005.

Question 11.
Discuss in brief the following:
(a) Green house gases
(b) Catalytic converter
(c) Ultraviolet B
Solution:
(a) Green house gases: The gases which are transparent to solar radiation but retain and partially reflect back long wave heat radiations are called greenhouse gases. Green house gases are essential for keeping the earth warm and hospitable. They are also called radiatively active gases. They prevent a substantial part of long wave radiations emitted by earth to escape into space. Rather green house gases radiate a part of this energy back to the earth. The phenomenon is called greenhouse flux. Because of greenhouse flux, the mean annual temperature of the earth is 15°C. In its absence, it will fall to – 18°C.

However, recently the concentration of greenhouse gases has started rising to result in an enhanced greenhouse effect that is resulting in increasing the mean global temperature. It is called global warming. A regular assessment of the abundance of greenhouse gases and their impact on the global environment is being made by IPCC (Intergovernmental Panel on Climate Change). The various green house gases are CO₂ (warming effect 60%), CH₄ (effect 20%) , chlorofluorocarbons or CFCs (14%) andT nitrous oxide (N₂O, 6%). Others of minor significance are water vapors and ozone.

(b) Catalytic converter: Catalytic converters are devices that are fitted into automobiles for reducing the emission of gases. These have expensive metals (platinum – palladium, and rhodium) as catalysts. As the exhaust passes through the catalytic converters, unburnt hydrocarbons are converted into CO₂ and H₂O and carbon monoxide and nitric oxide are changed to CO₂ and N₂ respectively. Vehicles fitted with catalytic converters should be run on unleaded petrol as leaded petrol would inactivate the catalyst in the converters.

(c) Ultraviolet B – UV-B having 280-320nm wavelength. Their harmful radiations penetrate through the ozone hole to strike the earth. On earth, these can affect human beings and other animals by causing :

Skin cancer
Blindness and increased incidence of cataract in eyes, and
Malfunctioning of the immune system.
Higher number of mutations.

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CLASS 12TH CHAPTER -15 Biodiversity and Conservation |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter :15 Biodiversity and Conservation

Question 1.
Name the three important components of biodiversity
Solution:
The three important components of biodiversity are genetic diversity, species diversity and ecological diversity. These components are the basic building blocks of biodiversity. These are intimately linked and may have common elements.

Question 2.
How do ecologists estimate the total number of species present in the world?
Solution:
The diversity of living organisms present on the earth is very vast. According to an estimate by researchers, it is about seven million. The total number of species present in the world is calculated by ecologists by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 3.
Give three hypotheses for explaining why tropics show the greatest levels of species richness.
Solution:
The three hypotheses for higher species richness in tropical areas are:

Prolong evolutionary time – Temperate areas have undergone frequent glaciation in the past. It killed most of the species. No such disturbance occurred in the tropics where species continued to flourish and evolve undisturbed for millions of years.
Favourable environment – There are no unfavourable seasons in the tropics. The continued favourable environment has helped tropical organisms to gain more niche specialisation and increased diversity.
More sunlight – More solar energy is available in the tropics. This promotes higher productivity and increased biodiversity.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Solution:
The slope of regression/regression co-efficient of species-area relationship indicates that species richness decreases with a decrease in area.

The regression coefficient is between 0.1 – 0.2 regardless of taxonomic group or region, eg: Plants in Britain, Birds in California.
But in large areas like continents value is eg:- Frugivorous birds, mammals is tropical forests.

Question 5.
What are the major causes of species losses in a geographical region?
Solution:
The major causes of species losses in a geographical area are:

Habitat loss and fragmentation
Overexploitation
Alien species invasion
Co-extinctions
Disturbance and degradation
Pollution
Intensive agriculture and forestry.

Question 6.
How is biodiversity important for ecosystem functioning?
Solution:
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 to disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. Various trophic levels are connected through food chains. If anyone 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 will die due to the lack of food.

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

Question 7.
What are sacred groves? What is their role in conservation?
Solution:
Sacred groves are forest patches around places of work. These are held in high esteem by tribal communities/state or central government. Tribals do not allow to cut even a single branch of trees in these sacred groves. Preserved over the course of many generations, sacred groves represent native vegetation in a natural or near-natural state & thus is rich in biodiversity & harbour many rare species of plants & animals. This is the reason why many endemic species flourish in these regions.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Solution:

Control of soil erosion: Plant roots hold the soil particles tightly and do not allow the topsoil to be drifted away by 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 runoff rainwater. Litter and humus of plants function as sponges thus, retaining the water which percolates down and gets stored as underground water. Hence, the flood is controlled.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations for how animals achieved greater diversification?
Solution:
Scientists recorded 22% of plant species diversity including algae, fungi, bryophytes, pteridophytes, gymnosperms, and angiosperms. But they recorded 72% of animal species (including insects, mollusks, fishes, mammals, birds etc.) diversity. Plants have the less adaptive capacity as compared to animals. Animals show locomotory movements and can move from one place to another to suit the environment and also in search of food. On the contrary, plants are fixed. Moreover, animals have well organised body structure with various organs to help adjust to the environment.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Solution:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate the 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.

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CLASS 12TH CHAPTER -14 Ecosystem |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter :14 Ecosystem

Question 1.
Fill in the blanks
(a) Plants are called as ___ because they fix carbon dioxide.
(b) In an ecosystem dominated by trees, the pyramid (of numbers) is ___ type.
(c) In aquatic ecosystem, the limiting factor for the productivity is ___
(d) Common detritivores in our ecosystem are ____
(e) The major reservoir of carbon on earth is ___
Solution:
(a) producers
(b) inverted or spindle
(c) light
(d) saprotrophs
(e) oceans

Question 2.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposer’s
Solution:
(d) decomposer’s

Question 3.
The second trophic level in a lake is
(a) phytoplankton
(b) zooplankton
(c) benthos
(d) fishes.
Solution:
(b) zooplankton

Question 4.
Secondary producers are
(a) herbivores
(b) producers
(c) carnivores
(d) none of these
Solution:
(a) herbivores

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

Question 6.
Distinguish between
(a) Grazing food chain and detritus food chain
(b) Upright and inverted pyramid
(c) Litter and detritus
(d) Production and decomposition
(e) Food chain and food web
(f) Primary and secondary productivity
Solution:
(a) Differences between grazing food chain and detritus food chain are as follows

(b) Differences between upright and inverted pyramids are as follows :

(d) Differences between production and decomposition are as follows :

(e) Differences between food chain and food web are as follows:

(f) Differences between primary productivity and secondary productivity are as follows :

Question 7.
Describe the components of an ecosystem.
Solution:
Ecosystem: The system resulting from the interaction between organisms and their environment is called an ecosystem.
(a) Producers: Organisms, which can synthesize their own food are included under producers, e.g., Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

(b) Consumers:

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

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

Question 8.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Solution:
An ecological pyramid is a graphic representation of an ecological parameter, as a number of individuals present in various trophic levels of a food chain with producers forming the base and top carnivores the tip. Ecological pyramids were developed by Charles Elton (1927) and are, therefore, also called Eltonian pyramids.

There are three types of ecological pyramids, namely,

Pyramid of numbers
Pyramid of biomass
Pyramid of energy

Pyramid of numbers: It is a graphic representation of the number of individuals per unit area of various trophic levels stepwise with producers at the base and top carnivores at the tip. In a grassland, the producers, which are mainly grasses, are always maximum in number. This number then shows a decrease towards the apex, as the primary consumers (herbivores) like rabbits, mice, etc. are lesser in number than the grasses; the secondary consumers, snakes, and lizards are lesser in number than the rabbits and mice. Finally, the top (tertiary) consumers hawks or other birds, are the least in number. Thus, the pyramid becomes upright.

Pyramid of biomass: The amount of living organic matter (fresh and dry weight) is called biomass. Here, different trophic level of the ecosystem are arranged according to the biomass of the organisms. In grassland and forest, there is generally a gradual decrease in biomass of organisms at successive levels from the producers to the top carnivores. Thus these pyramids are upright. But in pond ecosystem, it is inverted because the biomass gradually increases from the producers to carnivores.

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

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

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

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

Question 10.
Define decomposition and describe the processes and products of decomposition.
Solution:
Decomposition is the breakdown of dead or wastes organic matter by micro-organisms. Decomposition is both physical and chemical in nature. Processes involved in decomposition are – fragmentation, catabolism & leaching.

Fragmentation – The process primarily due to the action of detritus feeding invertebrate (detritivores) causes it to break into smaller particles. The detritus gets pulverized when passing through the digestive tracts of animals. Due to fragmentation, the surface area of detritus particles is greatly increased.
Catabolism – Enzyme degradation of detritus into simpler organic substances by bacteria and fungi.
Leaching – The process by which nutrients, chemicals, or contaminants are dissolved & carried away by water, or are moved into a lower layer of soil.

Various inorganic and organic substances are obtained by decomposition. Inorganic substances are obtained in the process of mineralization while organic substances are obtained in humification. A dark coloured amorphous substance called humus is formed by decomposition. Humus is highly resistant to microbial action & undergoes extremely slow decomposition. It serves as a reservoir of nutrients.

Question 11.
Give an account of energy flow in an ecosystem.
Solution:
Ecosystems require a constant input of energy as every component of an ecosystem is regularly dissipating energy.

Two laws of thermodynamics govern this flow of energy. According to the first law of thermodynamics, energy can be transferred as well as transformed but is neither created nor destroyed. According to the second law of thermodynamics, every activity involving energy transformation is accompanied by the dissipation of energy. Except for deep hydrothermal ecosystems, the source of energy in all ecosystems is solar energy. 50% of the solar energy incident over the earth is present in PAR (photosynthetically active radiation).

Energy flow in an ecosystem is always unidirectional or one way, i.e., solar radiation → producers → herbivores → carnivores. It cannot pass in the reverse direction. There is a decrease in the content and flow of energy with the rise in trophic level. Only 10% of energy is transferred from one trophic level to the next.
Producer biomass (1000 K cal) → Herbivore biomass (100 K cal) → Carnivore I biomass (10 Kcal) Carnivore II biomass (1 Kcal)

Question 12.
Write important features of a sedimentary cycle in an ecosystem
Solution:
Sedimentary Biogeochemical cycle:- It is the circulation of a biogeochemical between the biotic and abiotic compound of an ecosystem is a nongaseous being lithosphere or sediments of the earth. Sedimentary cycles occur in the case of phosphorus, calcium, magnesium, zinc, copper, etc.

Question 13.
Outline salient features of carbon cycling in an ecosystem.
Solution:
Carbon constitutes 49 percent of the dry weight of organisms and is next only to water. 71 percent of carbon is found dissolved in oceans. This ocean reservoir regulates the amount of carbon dioxide in the atmosphere. Fossil fuels also represent a reservoir of carbon. Carbon cycling occurs through the atmosphere, ocean, and living and dead organisms. 4 x 10¹³ kg of carbon is fixed in the biosphere through photosynthesis annually.

A considerable amount of carbon returns to the atmosphere as Co₂ through respiratory activities of the producers and consumers. Decomposers also contribute substantially to the CO₂ pool by their processing of waste materials and dead organic matter of land or oceans. Some amount of fixed carbon is lost to sediments and removed from circulation. Burning of wood, forest fire and combustion of organic matter, fossil fuels, volcanic activity are additional sources for releasing Co₂ into the atmosphere.

Human activities have significantly influenced the carbon cycle. Rapid deforestation and the massive burning of fossil fuels for energy and transport have significantly increased the rate of release of carbon dioxide into the atmosphere.

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CLASS 12TH CHAPTER -13 Organisms and Populations |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter :13 Organisms and Populations

Question 1.
How is diapause different from hibernation ?
Solution:
Diapause is different from hibernation. The table below shows the differences between them :

Question 2.
If a marine fish is placed in a freshwater aquarium/will the fish be able to survive? Why or why not?
Solution:
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 in the marine environment. In freshwater conditions, they are unable to regulate the water entering the body (through osmosis). Water enters their body due to the hypotonic environment outside. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Define phenotypic adaptation. Give one example.
Solution:
Phenotypic adaptation involves non-genetic changes in individuals such as physiological modifications like acclimatization or behavioural changes.Question 4.
Most living organisms cannot survive at temperatures above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Solution:
organisms survive at a temperature range of 0° to 40°C or less. However, there are some notable exceptions. Certain microorganisms live in hot springs and deep-sea hydrothermal vents where temperature far exceeds 100°C. They survive at the high temperature due to the occurrence of branched-chain lipids in their cell membrane that reduces the fluidity of cell membranes and the occurrence of the minimum amount of free water in their cells that provides resistance to high temperature

Question 5.
List the attributes that populations but not individuals possess.
Solution:

Natality
Mortality
Growth forms
Population density
Population dispersion
Population age distribution

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Solution:
The intrinsic rate of increase(r), can be calculated by the following exponential growth equation:

Question 7.
Name important defence mechanisms in plants against herbivory.
Solution:

Modification of leaves into thorns.
Development of spiny margins on leaves.
Development of sharp silicated edges on leaves.

Question 8.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and the mango tree?
Solution:
An orchid growing as an epiphyte on a branch of mango tree is an example of commensalism. Commensalism is the relationship between individuals of two species of which one is benefited and the other is almost unaffected, i.e., neither benefited nor harmed. A commensal may get shelter (protection), or ride, or support instead of or in addition to food. Epiphytes are space parasites, they use trees only for attachment and manufacture their own food by photosynthesis. In Vanda, an epiphytic orchid, a special kind of aerial roots (hanging roots) hang freely in the air and absorb moisture with the help of their special absorptive tissue called velamen.

Question 9.
What is the ecological principle behind the biological control method of managing pest insects?
Solution:
Predation is the means of biological control to manage pest insects where predators prey upon pests and regulate their numbers in the habitat.

Question 10.
Distinguish between the following:

Hibernation and Aestivation
Ectotherms and Endotherms

Solution:

Differences between hibernation and aestivation are as follows :
Differences between ectotherms and endotherms are as follows:

Question 11.
Write a short note on :
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioral adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.
Solution:
a. Desert plants are called xerophytes. They have adaptations for increased water absorption, reduction in transpiration and water storage. Many desert plants have a thick cuticle on their leaf surfaces and have their stomata arranged in deep pits to minimise water loss through transpiration. They also have a special photosynthetic pathway that enables their stomata to remain closed during day time. In desert plants like Opuntia, leaves are reduced to spines. Animals of dry areas may use metabolic water and reduce water loss bypassing nearly solid faeces and urine.

b. Xerophytes have special adaptations to withstand prolonged periods of drought. These are of four types – ephemerals, annuals, succulents and non-succulent perennials.

Ephemerals (drought escapers): Plants which live for a brief period and complete their life cycle during the rains.
Annuals (drought evaders): Plants which continue to live for a few
months even after rains in hot dry conditions. They have modifications to reduce transpiration.
Succulents (drought resistants): Plants have fleshy organs to store large amounts of water. They have a very thick cuticle, sunken stomata which open during night only.
Non-succulent perennials: These are true xerophytes. They have an extensive root system to absorb the maximum amount of water. They possess waxy coatings on leaves, sunken stomata, reduced leaf blades etc. to reduce transpiration.

c. The animals with variable temperatures called poikilotherms are affected by temperature variations. They are also called ectotherms. They show different adaptations like hibernation, aestivation, periodic activity, winter eggs, and migration.

d. Sun is the ultimate source of energy for most of the organisms on this earth. Light is the visible range of the electromagnetic spectrum. Light (400 nm-700nm) is effective in photosynthesis and is called photo-synthetically active radiation or PAR. The intensity of light, duration of light, etc. are also influencing the growth of plants.

e. Animals live in arid regions show two kinds of adaptations

Reducing loss of water from their bodies.
Ability to tolerate arid conditions.

Question 12.
List the various abiotic environmental factors.
Solution:
Abiotic factors are non-living factors and conditions of the environment which influence the survival, function and behaviour of organisms. Various abiotic factors are :

(i) Temperature – Temperature is one of the most important environmental factors. The average temperature varies seasonally. It ranges from subzero level in polar areas and high altitudes to more than 50°C in tropical deserts in summer and exceeds 100°C in thermal springs and deep-sea hydrothermal vents.

(ii) Water – Next to temperature, water is the most important factor which influences the life of organisms. The productivity and distribution of land plants are dependent upon the availability of water. Animals are adapted according to water availability. E.g., aquatic animals are ammonotelic while xerophytic animals excrete dry feces and concentrated urine.

(iii) Light – Plants produce food through photosynthesis for which sunlight is essential to the source of energy. Light intensity, light duration and light quality influences the number of life processes in organisms, such as – photosynthesis, growth, transpiration, germination, pigmentation, movement and photoperiodism.

(iv) Humidity – Humidity refers to the moisture (water vapour) content of the air. It determines the formation of clouds, dew and fog. It affects the land organisms by regulating the loss of water as vapour from their bodies through evaporation, perspiration and transpiration.

(v) Precipitation – Precipitation means rainfall, snow, sleet or dew. Total annual rainfall, seasonal distribution humidity of the air and amount of water retained in the soil are the main criteria that limit the distribution of plants and animals on land.

(vi) Soil – The soil is one of the most important ecological factor called the edaphic factor. It comprises of different layers called horizons. The upper weathered humus containing part of soil sustains terrestrial plant life.

Question 13.
Give an example for:

An endothermic animal
An ectothermic animal
An organism of the benthic zone.

Solution:

Hedgehog
Frog
Sponges

Question 14.
Define population and community.
Solution:
Population: A population is a group of individuals of the same species, which can reproduce among themselves and occupy a particular area in a given time.

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

Question 15.
Define the following terms and give one example for each.
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Solution:

a. Commensalism is an interspecific interaction between individuals of two species where one species is benefitted and the other is not affected.
e. g. Orchid and mango tree.

b. Parasitism is an interspecific interaction between individuals of two species where generally small species is benefitted and the large species are affected, e.g. Malarial parasite and human beings.

c. Camouflage: It is the ability of the animals to blend with the surroundings or background. In this way, animals remain unnoticed for protection or aggression. An example is a stick insect.

d. Mutualism is an interspecific interaction between individuals of two species where both the interacting species are benefitted in an obligatory way. e.g. Pollination in plants by animals.

e. Interspecific competition: It is an interaction between individuals of two species where both the interacting species are affected, e.g. Monarch butterfly and Queen monarch.

Question 16.
With the help of a suitable diagram describe the logistic population growth curve.
Solution:
Logistic population growth curve or S-shaped or sigmoid growth curve is shown by the populations of most organisms. It has the following phases: lag phase, log phase, exponential phase and stationary phase. In lag phase there is little or no increase in population. In log phase increase in population starts and occurs at a slow rate in the beginning. During exponential phase, increase in population becomes rapid and soon attains its full potential rate. This is due to the constant environment, availability of food and other requirements of life in plenty, absence of predation and interspecific competition and no serious intraspecific competition so that the curve rises steeply upward. The growth rate finally slows down as environmental resistance increases.

Finally, the population becomes stable during the stationary phase because now the number of new cells produced almost equals to the number of cells that die. Every population tends to reach a number at which it becomes stabilized with the resources of its environment. A stable population is said to be in equilibrium, or at saturation level. This limit in population is a constant K and is imposed by the carrying capacity of the environment. The sigmoid growth form is represented by the following equation :

r = intrinsic rate of natural increase
N = population density at time t; K = carrying capacity.

Question 17.
Select the statement which best explains parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited
(c) One organism is benefited, other is not affected
(d) One organism is benefited, other is affected.
Solution:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Solution:
The three important characteristics of a population are:

Birth and death rate
Age structure
Sex ratio

(i) The birth rate (natality) of a population refers to the average number of young ones produced per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year. The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.

(ii) The age structure of a population is the percentage of individuals of different ages such as young, adult and old. Age structure is shown bv organisms in which individuals of more than one generation coexist. The ratio of various age groups in a population determines the current reproductive status of the population. It also indicates what may be expected in the future. The population is divided into three age groups; pre-reproductive, reproductive and post-reproductive.

(iii) The sex ratio of a population refers to the number of females per thousand male individuals. There were 933 females per 1,000 males in our country in the 2001 census. The number of females in a population is very important as it is often directly related to the number of births. The number of males may be less significant because in many species a single male can mate with several females.

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CLASS 12TH CHAPTER -12 Biotechnology and Its Applications |NCERT BIOLOGY SOLUTIONS | EDUGROWN

Class 12 biology sets a strong basis for advanced studies. Subject experts clear all your queries during exam preparation by the explanation of the solutions in a conceptual way.

NCERT Solutions for Class 12 Biology Chapter :12 Biotechnology and Its Applications

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature
(c) toxin is inactive
(d) bacteria encloses toxin in a special sac.
Solution:
(c) Toxin is inactive: In bacteria, the toxin is present in an inactive form called prototoxin. This gets converted into the active form when it enters the salivary gland of insects having an alkaline medium.

Question 2.
What are transgenic bacteria? Illustrate using any one example.
Solution:
Transgenic bacteria are one that carries a transgene or a foreign gene of interest introduced using recombinant DNA technology. e. g., bacteria carrying the genes for human insulin.

In 1983, Eli Lilly an American company prepared two DNA sequences corresponding to A and B, chains of human insulin and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted and combined by creating disulfide bonds to form human insulin.

Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Solution:
Advantages of genetically modified crops or transgenic crops are as follows :

They are resistant to pests, herbicides and diseases.
They help to reduce post-harvest losses.
They enhance the nutritional value of food, e.g., a transgenic variety of rice (golden rice) is rich in vitamin A content.
Some transgenic plants, e.g., poplar trees are used to clean up heavy metal pollution from contaminated soil.
They are efficient in mineral usage and thus prevent early exhaustion of fertility of the soil.

Transgenic crops have several disadvantages also which are mentioned below:

Bt toxins expressed in pollen grains of transgenic crops are harmful for useful varieties of insects, e.g., honey bees and butterflies.
The foods produced by transgenic crops might cause toxicity and might result in allergies.
The bacteria present in human alimentary canal can become resistant to concerned antibiotic by taking up antibiotic resistance gene present in genetically modified food and become difficult to manage.

Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Solution:
Cry proteins are a group of toxic protein which are highly poisonous to deficient types of insects. It is produced by a soil bacterium Bacillus thuringiensis. The genes controlling their formation are called cry genes eg:- Cry I Ab, Cry I Ac, Cry II Ab, The bacterium produces a protein in the crystal form of protoxin. Two cry genes have been incorporated in cotton (Bt cotton) while one has been introduced in corn (Bt corn) As a result Bt Cotton was disease resistant to bollworm and Bt corn was resistant to corn borer.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency.
Solution:
Gene therapy is the technique of genetic engineering used to replace a faulty gene with a normal, healthy functional gene. The first clinical gene therapy was given in 1990 to a 4 years old girl with adenosine deaminase deficiency (ADA deficiency). This enzyme is very important for the immune system to function. Severe combined immunodeficiency (SCID) is caused due to a defect in the gene for the enzyme adenosine deaminase. SCID patient lacks functional T-lymphocytes and, therefore, fails to fight the infecting pathogens.

To perform gene therapy, lymphocytes are extracted from the patient’s bone marrow and a normal functional copy of human gene coding for ADA is introduced into these lymphocytes with the help of a retroviral vector. The cells so treated are reintroduced into the patient’s bone marrow. The lymphocytes produced by these cells contain functional ADA genes which reactivate the victim’s immune system. But, as these lymphocytes do not divide and are short-lived, so periodic infusion of genetically engineered lymphocytes is required. This problem can be overcome if stem cells are modified at an early embryonic stage.

Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E.coli?
Solution:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Solution:
rDNA technology is a technique of genetic engineering that involves combining DNA from two different sources to produce recombined or recombinant DNA (rDNA). Oils are composed of glycerol and fatty acids. Thus, to produce oil-free seeds genes coding for glycerol or fatty acids should be identified and nucleotide sequences complementary to the sequence of these genes should be inserted adjacent to these genes in the early cells of the endosperm. During transcription, these complementary sequences will produce anti-sense RNAs to the RNAs produced by glycerol or fatty acids gene and will silence these genes. As a result, oil-free seeds will be produced.

Since glycerol is a common component of all the oils whereas various fatty acids combine with glycerol to form oils, thus it will be easier if we silence the gene for glycerol synthesis.

Question 8.
Find out from the internet what is golden rice.
Solution:
Golden rice is a GM rice with increased vitamin A content.

Question 9.
Does our blood have proteases and nucleases?
Solution:
Proteases occur naturally in all organisms. These enzymes are involved in a multitude of physiological reactions from simple digestion of food proteins to highly-regulated cascades (e.g., the blood-clotting cascade, the complement system, apoptotic pathways, and the invertebrate prophenoloxidase activating cascade). Proteases present in blood serum (thrombin, plasmin, Hageman factor, etc.) play important role in blood clotting, as well as in lysis of the clots, and the action of the immune system. Other proteases are present in leukocytes (elastase, cathepsin G) and play several different roles in metabolic control. Nucleases, such as deoxyribonucleases and ribonucleases are found in the blood which helps in the degradation of exogenous deoxyribonucleic acid and ribonucleic acid circulating in the blood.

Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Solution:
The problem is stomach enzymes and acids. Once you orally ingest a protein, the proteases in your stomach juices (trypsin, chymotrypsin, pepsin) will cleave the holy-hell out of your therapeutic protein and the acids will denature whatever’s left beyond all recognition. This is why proteins like insulin have to be injected.

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NCERT MCQ CLASS-11 CHAPTER-13 | MATH NCERT MCQ | LIMITS AND DERIVATIVES| EDUGROWN

In This Post we are providing Chapter-13 Limits and Derivatives NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON LIMITS AND DERIVATIVES

Question 1. lim_x_→4 |x-4 | /x – 4 is equal to
(a) 1
(b) –1
(c) does not exist
(d) None of these

Answer : C

Question 2. If y = √x + 1/√x , then dy/dx at x = 1 is

(a) 1
(b) 1/2
(c) 1/√2
(d) 0

Answer : D

Question 3. lim_x_→0|Sinx | /x is
(a) 1
(b) –1
(c) does not exist
(d) None of these

Answer : C

Question 4. If y =

Class 11 Mathematics Limits And Derivatives MCQs

(a) -4x / (x² – 1)2

(b) -4x / (x² – 1)
(c) 1 – x² / 4x
(d) 4x / x² – 1

Answer : A

Question 5. lim_x_→0 tan2x – x / 3x – Sinx
(a) 2
(b) 1/2
(c) -1/2
(d) 1/4

Answer : B

Question 6. The derivative of (3x + 5)(1 + tan x) is
(a) 3x sec² x 6 sec2 x 3 + 3 tan x
(b) 3x sec² x 5 sec2 x 3 3 tan x
(c) 3x sec² x + 5 sec2 x + 3 + 3 tan x
(d) None of the above

Answer : C

Question 7. The derivative of (ax² + cot x)( p + q cos x) is
(a) ax² cot (q sin ) + ( p+ q cosx )(2ax cosec²x )
(b) ax² cot (q sin ) + ( p+ q cosx )(2ax cosec²x )
(c) ax² cot (q sin ) + ( p q cosx )(2ax cosec²x )
(d) None of the above

Answer : A

Question 8. The derivative of (x sinx cosx) 2 + 2 is
(a) x² cos x + 2x sin x sin 2x
(b) x² cos x + 2x sin x sin 2x
(c) x² cos x + x sin x sin 2x
(d) x² cos x + 2x sin x + sin 2x

Answer : B

Question 9. The derivative of sin³x cos³x xis
(a) 3 /5 sin² 2x cos 2x
(b) 3 /4 sin² 2x cos 2x
(c) 3 /4 sin 2x cos² 2x
(d) 3 /2 sin² 2x cos 2x

Answer : B

Question 10. The derivative of x⁵ cos x /sinx is

Answer : B

Question 11. Find the differentiation of x^2/3 by using first principle.
(a) 2/3 x^-1/3
(b) -2/3 x^-1/3
(c) 2/3 x^1/3(d) None of these

Answer : A

Question 12. Find the differentiation of cos (x² + 1 ) by using first principle.
(a) 2x sin (x² + 1 )
(b) 2x sin (x² + 1 )
(c) -2x sin (x² – 1 )
(d) None of these

Answer : B

Question 13. If f (x )= x-4 /2√x then f ‘(1) is

(a) 5/4
(b) 4/5
(c) 1
(d) 0

Answer : A

Question 14. Find the differentiation of x cos x by using first principle.
(a) x sin x + cos x
(b) x sin x + cos x
(c) x cos x cos x
(d) None of these

Answer : A

Question 15. If y = Sin (x+9) / Cos x , then dy/dx at x = 0 is
(a) cos 9
(b) sin 9
(c) 0
(d) 1

Answer : A

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NCERT MCQ CLASS-11 CHAPTER-12 | MATH NCERT MCQ | INTRODUCTION TO THREE DIMENSIONAL GEOMETRY| EDUGROWN

In This Post we are providing Chapter-12 Introduction to Three Dimensional Geometry NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Question 1. The coordinates of a point which is equidistant from the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c) are given by
(a) (a/2 , b/2 , c/2,)
(b) ( -a/2 , -b/2 , c/2,)
(c) (a/2 , -b/2 , c/2,)
(d) (-a/2 , b/2 , c/2,)

Answer : A

Question 2. If x2+y2 = 1, then the distance from the point (x, y, 1-x2-y2 ) to the origin is
(a) 1
(b) – 1
(c) 0
(d) 2

Answer : A

Question 3. Three vertices of a parallelogram ABCD are A(1, 2, 3), B(-1, -2, -1) and C(2, 3, 2). Find the fourth vertex D.
(a) (– 4, – 7, – 6)
(b) (4, 7, 6)
(c) (4, 7, – 6)
(d) None of these

Answer : B

Question 4. If a parallelopiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelopiped is
(a)2 √3
(b)3√2
(c) √2
(d) √3

Answer : A

Question 5. The points (5, – 4, 2),(4,– 3, 1), (7, -6, 4) and (8, – 7, 5) are the vertices of
(a) a rectangle
(b) a square
(c) a parallelogram
(d) None of these

Answer : C

Question 6. If the coordinates of the vertices of a ΔABC are A(-1, 3, 2), B(2, 3, 5) and C(3, 5, – 2), then ? ∠A is equal to
(a) 45°
(b) 60°
(c) 90°
(d) 30°

Answer : A

Question 7. Find the ratio in which the YZ-plane divides the line segment formed by joining the points (– 2, 4, 7) and (3, – 5, 8).
(a) externally 2 : 3
(b) internally 2 : 3
(c) internally 3 : 2
(d) externally 3 : 2

Answer : B

Question 8. Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, -4, 6) in the ratio 2 : 3 externally.
(a) (– 8, – 17, 3)
(b) (– 8, 17, 3)
(c) (8, – 17, 3)
(d) None of these

Answer : B

Question 9. Find the length of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).
(a) 7, 7, √34
(b) 7, 8, √34
(c) 7, 9, √34
(d) None of these

Answer : A

Question 10. Find the coordinates of the points which trisect the line segment joining the points P(4 , 2, -6) and Q(10, -16, 6).
(a) (6, – 4, – 2), (8, – 10, 2)
(b) (6, 4, – 2), (8, – 10, 2)
(c) (6, – 4, – 2), (8, 10, 2)
(d) None of these

Answer : B

Question 11. Find the centroid of a triangle, the mid-point of whose sides are D(1, 2, -3), E(3, 0, 1) and F (-1, 1, -4).
(a) (1, 1, 2)
(b) (1, 1, – 2)
(c) (– 1, –1, –2)
(d) (1, –1, –2)

Answer : B

Question 12. The points A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are vertices of a
(a) square
(b) rhombus
(c) rectangle
(d) None of these

Answer : B

Question 13. If vertices of a triangle are A(1, -1, 2), B(2, 0, -1) and C(0, 2, 1), then the area of a triangle is (a) √6
(b) 2√6
(c) 3√6
(d) 4√6

Answer : B

Question 14. The point ( -2, -3, -4) lies in the
(a) first octant
(b) seventh octant
(c) second octant
(d) eight octant

Answer : B

Question 15. The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, – 1). Then, the vertices are
(a) (7, 2, 5), (3, 12, 17), (– 3, 4, – 7)
(b) (7, 2, 5), (3, 12, 17), ( 3, 4, 7)
(c) (7, 2, 5), (– 3, 12, 17), (– 3, – 4, – 7)
(d) None of the above

Answer : A

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NCERT MCQ CLASS-11 CHAPTER-11 | MATH NCERT MCQ | CONIC SECTIONS | EDUGROWN

In This Post we are providing Chapter-11 Conic Sections NCERT MCQ for Class 11 Math which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MCQ ON CONIC SECTIONS

Question 1.
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
(a) √{a² × (1 + m²)} < c
(b) √{a² × (1 – m²)} < c
(c) √{a² × (1 + m²)} > c
(d) √{a² × (1 – m²)} > c

Answer: (c) √{a² × (1 + m²)} > c
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
√{a² × (1 + m²)} > c

Question 2.
Equation of the directrix of the parabola x² = 4ay is
(a) x = -a
(b) x = a
(c) y = -a
(d) y = a

Answer: (c) y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Question 3.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is
(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8xAnswer

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Question 4.
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
(a) 7
(b) 8
(c) 9
(d) 10Answer

Answer: (a) 7
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Question 5.
The equation of a hyperbola with foci on the x-axis is
(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y² = (a² + b²)

Answer: (b) x²/a² – y²/b² = 1
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² – y²/b² = 1

Question 6.
If the line 2x – y + λ = 0 is a diameter of the circle x² + y² + 6x − 6y + 5 = 0 then λ =
(a) 5
(b) 7
(c) 9
(d) 11

Answer: (c) 9
Given equation of the circle is
x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
⇒ -3 × 2 – 3 + λ = 0
⇒ -6 – 3 + λ = 0
⇒ -9 + λ = 0
⇒ λ = 9

Question 7.
The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is
(a) 0
(b) 1
(c) 2
(d) More than 2Answer

Answer: (b) 1
Given point (1, 2) and equation of circle is x² + y² = 5
Now, x² + y² – 5 = 0
Put (1, 2) in this equation, we get
1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0
So, the point (1, 2) lies on the circle.
Hence, only one tangent can be drawn.

Question 8.
The equation of the circle x² + y² + 2gx + 2fy + c = 0 will represent a real circle if
(a) g² + f² – c < 0
(b) g² + f² – c ≥ 0
(c) always
(d) None of these

Answer: (b) g² + f² – c ≥ 0
Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0
This equation can be written as
{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²
So, the circle is real is g² + f² – c ≥ 0

Question 9.
The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is
(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0
(b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0
(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0
(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
Given focus S(3, 0)
and equation of directrix is: 3x + 4y = 1
⇒ 3x + 4y – 1 = 0
Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
Then, SP = PM
⇒ SP² = PM²
⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}²
⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25
⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0
⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
This is the required equation of parabola.

Question 10.
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
(a) 2/3
(b) 4/3
(c) 1/3
(d) 4

Answer: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 11.
The eccentricity of an ellipse is?
(a) e = 1
(b) e < 1
(c) e > 1
(d) 0 < e < 1

Answer: (d) 0 < e < 1
The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Question 12.
If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
(a) (x + 2)² + (y – 3)² = 3²
(b) (x – 2)² + (y + 3)² = 3²
(c) (x – 2)² + (y – 3)² = 3²
(d) (x + 2)² + (y + 3)² = 3²

Answer: (c) (x – 2)² + (y – 3)² = 3²
Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}
= √(4 + 9 – 4)
= √9
= 3
So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 13.
If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is
(a) 1/3
(b) 1/√3
(c) 1/√2
(d) 2√2/√3

Answer: (d) 2√2/√3
Given, the length of the major axis of an ellipse is three times the length of the minor axis
⇒ 2a = 3(2b)
⇒ 2a = 6b
⇒ a = 3b
⇒ a² = 9b²
⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}
⇒ 1 = 9(1 – e²)
⇒ 1/9 = 1 – e²
⇒ e² = 1 – 1/9
⇒ e² = 8/9
⇒ e = √(8/9)
⇒ e = 2√2/√3
So, the eccentricity of the ellipse is 2√2/√3

Question 14.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is
(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8x

Question 15.
In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is
(a) 4/5
(b) 1/√52
(c) 3/5
(d) 1/2

Answer: (c) 3/5
Given, distance between foci = 6
⇒ 2ae = 6
⇒ ae = 3
Again minor axis = 8
⇒ 2b = 8
⇒ b = 4
⇒ b² = 16
⇒ a² (1 – e²) = 16
⇒ a² – a² e² = 16
⇒ a² – (ae)² = 16
⇒ a² – 3² = 16
⇒ a² – 9 = 16
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Now, ae = 3
⇒ 5e = 3
⇒ e = 3/5
So, the eccentricity is 3/5

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