Category: Class 12

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 1.
One of the possible early sources of energy was/were
(a) CO₂
(b) chlorophyll
(c) green plants
(d) UV rays and lightning.
Answer:
(d) UV rays and lightning.

Question 2.
Abiogenesis theory of origin supports
(a) spontaneous generation
(b) origin of life from blue-green algae
(c) origin of life is due to pre-existing organisms
(d) organic evolution is due to chemical reactions.
Answer:
(a) spontaneous generation

Question 3.
Who proposed that the first form of the could have come from pre-existing non-living organic molecules ?
(a) S.L. Miller
(b) Oparin and Haldane
(c) Charles Darwin
(d) Alfred Wallace
Answer:
(b) Oparin and Haldane

Question 4.
According to one of the most widely accepted theories, earth’s atmosphere before origin of life was
(a) oxidising
(b) oxidising along with H₂
(c) reducing with free O₂ in small amount
(d) reducing with oxygen absent in O₂ form.
Answer:
(b) oxidising along with H₂

Question 5.
According to Oparin, which one of the following was not present in the primitive atmosphere of the earth ?
(a) Methane
(b) Oxygen
(c) Hydrogen
(d) Water vapour
Answer:
(b) Oxygen

Question 6.
The correct sequence for the manufacture of the compounds on the primitive earth is
(a) NH_3, CH₄, protein and carbohydrate
(b) protein, carbohydrate, water and nucleic acid
(c) NH₃, CH₄, carbohydrate and nucleic acid
(d) NH₃, carbohydrate, protien and nucleic acid.
Answer:
(d) NH₃, carbohydrate, protien and nucleic acid.

Question 7.
Early at mosphere contained methane and other hydrocarbons. They have been now replaced by
(a) nitrogen
(b) oxygen
(c) carbon dioxide
(d) hydrogen.
Answer:
(c) carbon dioxide

Question 8.
The first life originated
(a) on land
(b) in air
(c) in water
(d) all of these.
Answer:
(c) in water

Question 9.
Coacervates are
(a) colloid droplets
(b) nucleoprotein containing entities
(c) microspheres
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 10.
First life from on earth was a
(a) cyanobacterium
(b) chemoheterotroph
(c) autotroph
(d) photoautotroph.
Answer:
(b) chemoheterotroph

Question 11.
The ship used by Charles Darwin during the sea voyages was
(a) HMS Beagle
(b) HSM Beagle
(c) HMS Eagle
(d) HSM Eagle.
Answer:
(a) HMS Beagle

Question 12.
Fitnes according to Darwin refers to
(a) number of species in a community
(b) useful variation in population
(c) strength of an individual
(d) reproductive fitness of an organism.
Answer:
(d) reproductive fitness of an organism.

Question 13.
Alfred Wallace worked in
(a) Galapagos Island
(b) Australian Island Continent
(c) Ma lay Archipelago
(d) none of these.
Answer:
(c) Ma lay Archipelago

Question 14.
The theory of natural selection was given by
(a) Lamarck
(b) Alfred Wallace
(c) Charles Darwin
(d) Oparin and Haldane.
Answer:
(c) Charles Darwin

Question 15.
The preserved fossil remains of Archaeopteryx show that
(a) it was a flying reptile from the Permian period
(b) reptiles gave rise to birds during Jurassic period
(c) it was a flying reptile in the Triassic period
(d) reptiles gave rise to birds during Permian period.
Answer:
(b) reptiles gave rise to birds during Jurassic period

Question 16.
Which of the following isotopes is used for Finding the fossil age maximum about 35,0000 years ?
(a) ²³⁸U
(b) ¹⁴C
(c) ³H
(d) ²⁰⁶Pb
Answer:
(b) ¹⁴C

Question 17.
Which of the following statements is True ?
(a) Wings to birds and insects are homologous organs.
(b) Human hands and bird’s wings are analogous organs.
(c) Human hands and bat’s wings are analogous organs.
(d) Flipper of penguin and dolphin are analogous organs.
Answer:
(d) Flipper of penguin and dolphin are analogous organs.

Question 18.
Replacement of the lighter-coloured variety of peppered moth (Biston betularia) to its darker variety (Biston carbonaria) in England is the example of
(a) natural selection
(b) regeneration
(c) genetic isolation
(d) temporal isolation.
Answer:
(a) natural selection

Question 19.
Phenomenon of ‘industrial melanism’ demonstrates
(a) geographical isolation
(b) reproductive isolation
(c) natural selection
(d) induced mutation.
Answer:
(c) natural selection

Question 20.
Which one of the following phenomena supports Darwin’s concept of natural selection in organic evolution ?
(a) Development of transgenic animals
(b) Production of “Dolly’, the sheep by cloning
(c) Prevalence of pesticide resistant insects
(d) Development of organs from ‘stem cells’ for organ transplantation.
Answer:
(c) Prevalence of pesticide resistant insects

Question 21.
The phenomenon ‘ontogeny repeats phylogeny’ is explained by
(a) recapitualtion theory
(b) Inheritcance theory
(c) mutation theory
(d) natural selection theory.
Answer:
(a) recapitualtion theory

Question 22.
The presence of gill slits, in the embryos of vertebrates, supports the theory of
(a) metamorphosis
(b) biogenesis
(c) organic evolution
(d) recapitulation
Answer:
(d) recapitulation

Question 23.
Which is not a vestigial organ in man ?
(a) Nictitating membrane
(b) Tail vertebrae
(c) Vermiform appendix
(d) Nails
Answer:
(d) Nails

Question 24.
Which one is not a vestigia organ ?
(a) Wings of kiwi
(b) Coccyx in man
(c) Pelvic girdle of python
(d) Flipper of seal
Answer:
(d) Flipper of seal

Question 25.
By the statement ‘survival of the fittest’, Darwin meant that
(a) the strongest of all species survives
(b) the most intelligent of the species survives
(c) the cleverest of the species survives
(d) the species most adaptable to changes survives.
Answer:
(d) the species most adaptable to changes survives.

Question 26.
Which of the following are the two key concepts of Darwinian theory of evolution ?
(a) Genetic drift and mutation
(b) Adaptive radiation and homology
(c) Mutation and natural selection
(d) Branching descent and natural selection
Answer:
(d) Branching descent and natural selection

Question 27.
According to Lamarckism, long necked giraffes evolved because
(a) nature selected only long necked ones
(b) humans preferred only long necked ones
(c) short necks suddenly changed into long necks
(d) of stretching of necks over many generations by short necked ones.
Answer:
(d) of stretching of necks over many generations by short necked ones.

Question 28.
Which of the following evidences does not favour the Lamarckian concept of inheritance of acquired characters ?
(a) Lack of pigment in cave-dwelling animlas
(b) Melanisation in peppered moth
(c) Absence of limbs in snakes
(d) Presence of webbed toes in aquatic birds
Answer:
(b) Melanisation in peppered moth

Question 29.
“Human population grows in geometric ratio while food materials increase in arithmetic proportion”. It is a statement from
(a) Darwin
(b) Bateson
(c) AmartyaSen
(d) Malthus.
Answer:
(d) Malthus.

Question 30.
Which one of the following scientist’s name is correctly matched with the theory put forth by him ?
(a) de Vries – Theory of natural selection
(b) Darwin – Theory of pangenesis
(c) Weismann – Theory of continuity of germplasm
(d) Pasteur – Theory of inheritance of acquired charcters
Answer:
(c) Weismann-Theory of continuity of germplasm

Question 31.
Single step large mutation leading to speciation is also called
(a) founder effect
(b) saltation
(c) branching descent
(d) natural selection
Answer:
(b) saltation

Question 32.
At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium ?
(a) 0.36
(b) 0.16
(c) 0.24
(d) 0.48
Answer:
(d) 0.48

Question 33.
Hardy-Weinberg equilibrium is known to be affected by gene flow, genetic drift mutation, genetic recombination and
(a) evolution
(b) limiting facrtors
(c) saltation
(d) natural selection.
Answer:
(d) natural selection.

Question 34.
The Hardy-Weinberg principle cannot operate if
(a) a population does not migrate for a longtime to a new habitat.
(b) frequent mutations occur in the population
(c) the population has no chance of interaction with other populations
(d) free interbreeding occurs among all members of the population.
Answer:
(b) frequent mutations occur in the population

Question 35.
Genetic drift operates only in
(a) larger populations
(b) Mendelian populations
(c) island populations
(d) smaller populations.
Answer:
(d) smaller populations.

Question 36.
Which of the following is most important for speciation ?
(a) Seasonal isolation
(b) Reproductive isolation
(c) Behavioural isolation
(d) Tropical isolation
Answer:
(b) Reproductive isolation

Question 37.
The factors involved in the formation of new species are
(a) Isolation and competition
(b) gene flow and competition
(c) competition and mutation
(d) isolation and variation.
Answer:
(d) isolation and variation.

Question 38.
Stabilising selection favours
(a) both extreme forms of a trait
(b) intermediate forms of a trait
(c) environmental differences
(d) one extreme form over the other extreme form and over intermediate forms of a trait.
Answer:
(b) intermediate forms of a trait

Question 39.
The different forms of interbreeding species that live in different geographical regions are called
(a) sibling species
(b) sympatric species
(c) allopatric species
(d) polytypic species.
Answer:
(c) allopatric species

Question 40.
Which of following represents correct order of evolutiton ?
(a) Amoeba → Leucosolenia → Hydra → Ascaris
(b) Leucosolenia → Hydra → Amoeba → Ascaris
(c) Ascaris → Amoeba → Leucosolenia → Hydra
(d) None of these
Answer:
(a) Amoeba → Leucosolenia → Hydra → Ascaris

Question 41.
Presence of gills in the tadpole of frog indicated that
(a) fishes were amphibious in the past
(b) fishes evolved from frog-like ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled ancestors.
Answer:
(d) frogs evolved from gilled ancestors.

Question 42.
The character that proves that frogs have evolved from fishes is
(a) their ability to swim in water
(b) tadpole larva in frogs
(c) similarity in the shape of the head
(d) their feeding on aquatic plants.
Answer:
(b) tadpole larva in frogs

Question 43.
Identify the correct arrangement of periods of Palaeozoic era in ascending order in geological time scale.
(a) Cambrian → Devonian → Ordovician → Silurian → Carboniferous → Permian
(b) Cambrian → Ordovician → Silurian → Devonian→Carboniferous → Permian
(c) Cambrian → Ordovician → Devonian → Silurian → Carboniferous → Permian
(d) Silurian → Devonian → Cambrian → Ordovician → Permian → Carboniferous
Answer:
(b) Cambrian → Ordovician → Silurian → Devonian → Carboniferous → Permian

Question 44.
Which is the correct order of increasing geological time scale for a hypothetical vertebrate evolution ?
(a) Cenozoic, Mesozoic, Palaeozoic, Proterozoic
(b) Cenozoic, Palaeozoic, Mesozoic, Proterozoic
(c) Proterozoic, Cenozoic, Palaezoic, Mesozoic
(d) Proterozoic, Palaeozoic, Mesozoic, Cenozoic
Answer:
(d) Proterozoic, Palaeozoic, Mesozoic, Cenozoic

Question 45.
The ‘Devonian period’ is considered to be as
(a) age of fishes
(b) age of amphibians
(c) age of reptiles
(d) age of mammals.
Answer:
(a) age of fishes

Question 46.
Amphibians were dominant during _______ period.
(a) Carboniferous
(b) Silurian
(c) Ordovician
(d) Cambrian
Answer:
(a) Carboniferous

Question 47.
The primate which existed 15 mya was
(a) Homo habilis
(b) Australopithecus
(c) Ramapithecus
(d) Homo erectus
Answer:
(c) Ramapithecus

Question 48.
The extinct stone ancestor, who ate only fruits and hunted with stone weapons was
(a) Ramapithecus
(b) Australophecus
(c) Dryopithecus
(d) Homo erectus
Answer:
(b) Australophecus

Question 49.
One of the oldest, best preserved and most complete hominid fossil commonly known as ‘Lucy’ belongs to the genus
(a) Austraolopithecus
(b) Oreopithecus
(c) Dryopithecus
(d) Pithecanthropus
Answer:
(a) Austraolopithecus

Question 50.
The brain capacity of Homo eractus was about
(a) 650 c.c.
(b) 900 c.c.
(c) 1500 c.c.
(d) 1400 c.c.
Answer:
(b) 900 c.c.

Question 51.
The extinct human who lived 1,00,000 to 40,000 years ago, in East and Central Asia, used hides to protect their bodies and had brain capacity of 1400 c.c. were
(a) Homo habilis
(b) Neanderthal man
(c) Cro-Magnon man
(d) Ramapithecus.
Answer:
(b) Neanderthal man

Question 52.
Which of the following statements is correct ?
(a) Australopithecus has large brain around 900 c.c.
(b) Neanderthal man lived in East Africa and the fruits.
(c) Homo erectus had brain capacity 900 c.c.
(d) Homo sapiens arose in Central Asia and moved to other continuents and developed into distinct races.
Answer:
(c) Homo erectus had brain capacity 900 c.c.

Question 53.
Which of the following statements js correct regarding evolution of mankind ?
(a) Homo erectus is preceded by Homo habilis.
(b) Neanderthal man and Cro-Magnon man were living at the same time.
(c) Australopithecus was living in Australia.
(d) None of these
Answer:
(a) Homo erectus is preceded by Homo habilis.

Question 54.
The cranial capacity was largest among the
(a) Peking man
(b) Java ape man
(c) African man
(d) Neanderthal man.
Answer:
(d) Neanderthal man.

Question 55.
The most apparent change during the evolutionary history oiHomo sapiens is traced in
(a) loss of body hair
(b) walking unright
(c) sortening of the jaws
(d) remarkable, increase in the brain size.
Answer:
(d) remarkable, increase in the brain size.

Question 56.
What kind of evidences suggested that man is more closely related with chimpanzee than with other hominoid apes ?
(a) Evidence from DNA of sex chromosomes only
(b) Comparison of chromosome morphology and number
(c) Evidence from fossil remains, and the fossil mitochondrial DNA alone
(d) Evidence from banding pattern of chromosome 3 and 6
Answer:
(d) Evidence from banding pattern of chromosome 3 and 6

Question 57.
Which of the following eras, in geological time scale, corresponds to the period when life had not orginated upon the earth ?
(a) Azoic
(b) Palaeozoic
(c) Mesozoic
(d) Archaeozoic
Answer:
(a) Azoic

Question 58.
Homo sapiens arose during which epoch ?
(a) Pleistocene
(b) Pliocene
(c) Oligocene
(d) Holocene
Answer:
(d) Holocene

Question 59.
Study the human evolution is called
(a) archaeology
(b) anthropology
(c) pedigree analysis
(d) chronobiology.
Answer:
(b) anthropology

Question 60.
Which of the following is used as an atmospheric pollution indicator ?
(a) Lepidoptera
(b) Lichens
(c) Lycopersicon
(d) Lycopodium
Answer:
(b) Lichens

Question 61.
The theory of spontaneous generation stated that
(a) life arose from living forms only
(b) life can arise from both living and non-living
(c) life can arise from non-living things only
(d) life arises spontaneusly, neither from living nor from the non-living.
Answer:
(c) life can arise from non-living things only

Question 62.
Animal husbandry and plant breeding programmes are the examples of
(a) reverse evolution
(b) aritifical selection
(c) mutation
(d) natural selection.
Answer:
(d) natural selection.

Question 63.
Palaentological evidences for evolutaion refer to the
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs.
Answer:
(c) fossils

Question 64.
The bones of forelimbs of whale, bat, cheeah and man are similar in structure, because
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function.
(d) the have biochemical similarities.
Answer:
(b) they share a common ancestor

Question 65.
Analogous organs arise due to
(a) divergent evolution
(b) artificial selection
(c) genetic drift
(d) convergent evolution.
Answer:
(d) convergent evolution.

Question 66.
(p+q)² = p² + 2pq + q² = 1 represents an equation used in
(a) population genetics
(b) Mendilian genetics
(c) biometircs
(d) molecular genetics.
Answer:
(a) population genetics

Question 67.
Appearnace of antibiotic-resistant bacteria is an example of
(a) adaptive radiation
(b) transduction
(c) pre-existing variation in the population
(d) divergent evolution.
Answer:
(c) pre-existing variation in the population

Question 68.
Fossils are generally found in
(a) sedimentasry rocks
(b) igneous rocks
(c) metamorphic rocks
(d) any type of rock.
Answer:
(a) sedimentasry rocks

Question 69.
Which type of selection is industrial melanism observed in moth, Biston betularia?
(a) Stabilising
(b) Directional
(c) Disruptive
(d) Artificial
Answer:
(b) Directional

Question 70.
Which of the following is an example for link species ?
(a) Lobe fish
(b) Dodo bird
(c) Seaweed
(d) Chimpanzee
Answer:
(a) Lobe fish

Question 71.
Variations during mutations of meiotic recombinations are
(a) random and directionless
(b) random and directional
(c) random and small
(d) random small and directional
Answer:
(a) random and directionless

Bihar Board 12th Entrepreneurship Model Papers

Bihar Board 12th Entrepreneurship Model Question Paper 1 in English Medium

Time: 3 Hours 15 Min
Marks: 70

Instructions for the candidates:

1. Candidates are required to give their answers in their own words as far as practicable.
2. Figures in the right-hand margin indicate full marks.
3. 15 minutes of extra time has been allotted for the candidate to read the questions carefully.
4. This question paper has two sections: Section – A and Section – B.
5. In Section – A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with black/blue ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/ Liquid/Nail on OMR Sheet, otherwise, the result will be treated as invalid.
6. In Section – B, there are Non-objective type questions. There are 18 Short answer type questions, out of which any 10 questions are to be answered. Each question carries 2 marks. Apart from this, there are 6 Long answer type questions, out of which any 3 of them are to be answered. Each question carries 5 marks.
7. Use of any electronic device is prohibited.

Objective Type Questions

Question No. 1 to 35 has four options provided out of which only one is correct. You have to mark, your selected option, on the OMR-Sheet. Each question carries 1 (one) mark. (35 × 1 = 35)

Question 1.
The term ‘Fund’ as used in Fund Flow Analysis means:
(a) Cash only
(b) Current Assets
(c) Current Liability
(d) Excess of current Assets over Current Liabilities.
Answer:
(d) Excess of current Assets over Current Liabilities.

Question 2.
Networking capital means:
(a) C.A – C. L.
(b) C.A. + C.L.
(c) C. L. – C.A.
(d) None of the above
Answer:
(a) C.A – C. L.

Question 3.
A. E. P. = _________
(a) \(\frac { FixedCost }{ \frac { P }{ V } Ratio }\)
(b) \(\frac { FixedCost }{ \frac { P }{ V } Ratio } \times 100\)
(c) \(\frac { \frac { P }{ V } Ratio }{ FixedCost }\)
(d) None of these
Answer:
(b) \(\frac { FixedCost }{ \frac { P }{ V } Ratio } \times 100\)

Question 4.
Risk Capital Foundation was established in
(a) 1970
(b) 1975
(c) 1986
(d) 1988
Answer:
(b) 1975

Question 5.
Capital Intensive technique is favoured due to:
(a) Rapid economic growth
(b) Social influence
(c) Increase in employment opportunities
(d) All of the above
Answer:
(d) All of the above

Question 6.
Management is an art of:
(a) doing work himself
(b) taking work from others
(c) both for doing work himself and taking work from others
(d) All
Answer:
(c) both for doing work himself and taking work from others

Question 7.
Out of the following which is the method of quality control?
(a) Inspection method
(b) Statistical Quality Control Method
(c) Both A and B above
(d) Neither A nor B above
Answer:
(c) Both A and B above

Question 8.
Dividend paid on cumulative preference share:
(a) In the year of profit
(b) In the year of loss
(c) In the year of profit or loss
(d) None of these
Answer:
(a) In the year of profit

Question 9.
The best example of variable cost is:
(a) Interest on capital
(b) Material cost
(c) Wealth tax
(d) Rent
Answer:
(b) Material cost

Question 10.
Generally, diversification is classified in
(a) Two ways
(b) Three ways
(c) Four ways
(d) Five ways
Answer:
(c) Four ways

Question 11.
Is it necessary to give due consideration on internal resources
(a) Yes, it is necessary
(b) No, not necessary
(c) Necessary for external resources
(d) None of the above
Answer:
(a) Yes, it is necessary

Question 12.
Incentives are not concerned with:
(a) Rebate
(b) Exemption from Tax
(c) Provision of seed capital
(d) Lump-sum payment
Answer:
(c) Provision of seed capital

Question 13.
Which of the following is known as Market Demand?
(a) Demand Forecasting
(b) Real Demand
(c) Supply
(d) None of the above
Answer:
(a) Demand Forecasting

Question 14.
Which of the following factors are to be considered while selecting a product or service?
(a) Competition
(b) Cost of production
(c) Profit possibility
(d) All of the above
Answer:
(d) All of the above

Question 15.
Selection of an enterprise depends on:
(a) Sole Trading
(b) Right of Entrepreneur
(c) Self Ability of Entrepreneur
(d) None of the above
Answer:
(d) None of the above

Question 16.
An entrepreneur prefers which form of an organization if production is to be undertaken on a small scale basis?
(a) Sole Trader
(b) Partnership
(c) Company
(d) None of the above
Answer:
(a) Sole Trader

Question 17.
Project identification deals with:
(a) Viable Product idea
(b) Logical opportunity
(c) Effective demand
(d) None of these
Answer:
(a) Viable Product idea

Question 18.
The techno-economic analysis deals with the identification of the
(a) Supply potential
(b) Demand potential
(c) Export potential
(d) Import potential
Answer:
(b) Demand potential

Question 19.
Project appraisal is an/a
(a) Export Analysis
(b) Expert Analysis
(c) Profitability Analysis
(d) None of the above
Answer:
(c) Profitability Analysis

Question 20.
Various public utility undertaking has to invest heavily in:
(a) Current Assets
(b) Fixed Assets
(c) Fictitious Assets
(d) None of the above
Answer:
(b) Fixed Assets

Question 21.
Pay-back period deals with:
(a) The period required for-profit earning process
(b) The period required for the cost of investment recovery
(c) The period required for fixed cost recovery
(d) None of the above
Answer:
(b) The period required for the cost of investment recovery

Question 22.
The project life cycle is not concerned with the following:
(a) Pre-investment stage
(b) Constructive stage
(c) Normalisation stage
(d) Stabilisation stage
Answer:
(d) Stabilisation stage

Question 23.
Which of the following is the main problem connected with business?
(a) Profit
(b) money
(c) Sales
(d) Risk Management
Answer:
(d) Risk Management

Question 24.
What creates perfection in .the market which ultimately increases the volume of sales & profit?
(a) Innovation
(b) Promotion
(c) Marketing
(d) None of the above
Answer:
(c) Marketing

Question 25.
Short-term forecast involves a period of how many months?
(a) 12 months
(b) 24 months
(c) 18 months
(d) 36 months
Answer:
(a) 12 months

Question 26.
Business Regulatory framework is concerned with what
(a) Direction of Business
(b) Volume of Business
(c) Regulation
(d) None of the above
Answer:
(b) Volume of Business

Question 27.
The product which is more in demand are more
(a) Less profitable
(b) Possible
(c) More profitable
(d) None of the above
Answer:
(c) More profitable

Question 28.
Out of the following formulating of the general plan of a business depends?
(a) Project Report
(b) Plant and Product Planning
(c) Marketing Planning
(d) Financial Planning
Answer:
(d) Financial Planning

Question 29.
What break-even point shows:
(a) Profit
(b) Loss
(c) Neither Profit nor Loss
(d) None of these
Answer:
(c) Neither Profit nor Loss

Question 30.
The best example of Fixed cost is:
(a) Interest on capital
(b) Material cost
(c) Wealth Tax
(d) None of these
Answer:
(a) Interest on capital

Question 31.
What is the thing that enterprise do? The paths they follow and the decision they take in order to reach certain points and levels of success:
(a) Production
(b) Distribution
(c) Marketing
(d) Strategies
Answer:
(d) Strategies

Question 32.
Out of the following, what is essential to study in the feasibility study?
(a) Cost
(b) Price
(c) Operation
(d) All of the above
Answer:
(d) All of the above

Question 33.
What is included in the nature of marketing?
(a) Product Planning
(b) Classification of product
(c) Consumer
(d) Customer
Answer:
(c) Consumer

Question 34.
Social behaviour is not concerned with:
(a) Production of goods for public
(b) Avoidance of unethical behaviour
(c) Fulfilment of social obligations.
(d) Profit earning process
Answer:
(d) Profit earning process

Question 35.
Business opportunity relates to:
(a) Commercially feasible project
(b) Personal feasible projects
(c) Neither A nor B above
(d) None of the above
Answer:
(a) Commercially feasible project

Non-Objective Type Questions

Short Answer Type Questions

Question No. 1 to 18 are short answer type. Answer any 10 questions. Each question carries 2 marks. (10 × 2 = 20)

Question 1.
Why is entrepreneurship regarded as a creative activity?
Answer:
Entrepreneurship is creative in the sense that it involves the creation of value. By combining the various factors of production, entrepreneurs produce goods and services that meet the needs and wants of society. Every entrepreneurial act results in income and wealth generation. Even when innovations destroy the existing industries.

Entrepreneurship is creative also in the sense that it involves innovation introduction of new products, the discovery of new market and sources of supply of inputs, technological break thoughts as well as the introduction of newer organisational forms for doing things better, cheaper, faster and in the present context, in a manner that causes the least harm to the environment. It is possible that entrepreneurs in developing countries may not be pioneering innovative in introducing path-breaking, radical innovations.

Question 2.
What do you understand by Micro Environment?
Answer:
The microenvironment is the first pillar to build a business empire. All marketing plans, strategies and objectives are carried out through these components. It is, therefore, the executive arm of business where the practical implementation of ideas, thoughts, and concepts are done and based on the responses of these components, a business either moves forward or may step back.

Question 3.
What do you mean by market environment?
Answer:
Marketing Environment is the combination of external and internal factors and forces which affect the company’s ability to establish a relationship and serve its customers. The marketing environment of a business consists of an internal and external environment.

Question 4.
What is the object of project appraisal?
Answer:

• To extract relevant information for determining the success or failure of a project.
• To apply standard yardsticks for determining the rate of success or failure of a project.
• To determine the expected costs and benefits of the projects.
• To arrive at specific conclusions regarding the project.

Question 5.
What is factoring?
Answer:
Factoring is a transaction in which a business sells its invoices, or receivables, to a third-party financial company known as a “Factor”.
Factoring is known in some industries as “accounts receivables financing”.

Question 6.
Give the definitions of Fund Flow Statement.
Answer:
A fund flow statement is a statement prepared to indicate the increase in the cash resources and the utilisation of such resources of business during the accounting period.

Question 7.
What do you mean by Ratio Analysis?
Answer:
Ratio analysis is the process of determining and interpreting numerical relationships based on financial statements. A ratio is a statical yardstick that provides a measure of the relationship between variables or figures. This relationship can be expressed as a percent.

Question 8.
How is break-even point determined?
Answer:
Break-even analysis related to the study of cost and return investment in relation to the sales of a business unit. That point or level of sales at which the business makes no profit and no loss is termed as the break-even point. Break-even point can be applied from two viewpoints- Narrow viewpoint and broad viewpoint, from the narrow viewpoint break-even analysis is a technique which determines that level of work at which the total return is equal to total expenditure. That is why this is known as the point of no profit no loss.
Break-point is the point at which the following characteristics are observed:

• The firm is neither earning any profit nor undergoing any loss i.e. it is in a state of no profit no loss.
• The total return of the firm is equal to the total cost.
• Amount of contribution (Sales-variable cost is) equal to the fixed cost.

Because of the above characteristics of the break-even point, it is also known as ‘Equilibrium point’ or ‘balancing point’ ore ‘critical point’.

Question 9.
Who is an Entrepreneur?
Answer:
The term ‘entrepreneur’ was first introduced in economics by the early 18th-century French economist Richard Cantillon. The entrepreneur as the agent who buys means production at certain prices in order to sell the produce at uncertain prices in the future. “ Since then a perusal of the usage of the term in economics shows, that entrepreneurship implies risk uncertainty bearing coordination of productive resources; introduction of innovation; and the provision of capital.

Question 10.
State the two objectives of identification of business opportunities.
Answer:
Following are the objectives of identifying business opportunities:

• Identify the sources of business ideas.
• Explain methods for generating business ideas.
• Identify various entrepreneurial opportunities.
• Understand the nature of the occupational and geographical mobility of entrepreneurs.

Question 11.
Give any two objectives of environmental scanning.
Answer:
Two objectives of environment scanning are mention below.

• Environmental scanning is the process of gathering information about the event and their relationship within an organisation’s internal and external environments.
• Environmental scanning is to help management determine the future direction of the organisation.

Question 12.
Mention any two objects of the project report.
Answer:
Two objectives of project report are following.

• To evaluation of environment opportunity.
• It is necessary to present the report for taking financial help.

Question 13.
What do you’re mean by the cycle of working capital?
Answer:
Availability of enough working capital is indicative of liquidity of business and capacity to make prompts payment. Working capital is needed for making payment of day to day expenditure for meeting current liabilities and for availing benefit of cash discount.

Question 14.
Why time boundness is essential in a project?
Answer:
Project is a written account of various activities undertaken by a firm or entrepreneurs and their technical, commercial and social feasibilities. Project is made after analysis of essential document therefore there is a time limit for making project because of the project is made for particulars time and subject.

Question 15.
What is the marketing mix?
Answer:
Marketing mix implies a combination of all marking elements or ingredients so that the objective of the enterprise may be realised. The marketing mix is the appointment of effort the combination the designing and the integration of the elements of marketing into a programme or mix which on the basis of the marketing forces will best achieve the objectives of an enterprise at a given time.

Question 16.
List any two problems faced by first-generation entrepreneurs?
Answer:
List of problems faced by first-generation entrepreneurs.

• Business ideas issues.
• Lack of finance.
• Inability to market their business.
• Not knowing how to plan the business.
• Legal question.

Question 17.
What do you mean by market assessment?
Answer:
Market Assessment: Goods are produced and facilitate for providing services are created in anticipation of their demand the ultimate aim of these activities is that these reach these for whose these are meant. The commodities are needed for consumption and used by the ultimate consumers. The services are utilized by whose Who was want to derive satisfaction out of them.

Question 18.
What is market observation?
Answer:
Market observation is a market research techniques in which highly trained researchers generally watch how people or consumers behave and interact in the market under natural conditions. It is designed to give precisely detailed and actual information on what consumers do as niche.

Long Answer Type Questions

Question No. 19 to 24 are long answer type. Answer any 3 questions. Each question carries 5 marks. (3 × 5 = 15)

Question 19.
Why the environmental study is essential for an entrepreneur?
Answer:
The process of environmental scanning and analysis is very important to an entrepreneur. The evaluation of the existing changing circumstances the forecasting and the consequent impacts need to be assessed by an entrepreneur prior to determining his policies and planning and their successful implementation. It is essential to evaluate and analyse the consequences of the changing business environment prior to giving a tangible shape to the planning and policies.

The technical innovation, new social values, problems of the community, product – market conditions etc. Must be subjected to study which is instrumental in converting strategies and planning into reality. The competitive environment and its consideration is also an integral part of environmental scanning. To plan or conceive the counter-strategies in view of the competitive threats, the environmental study and analysis is the pre-requisite in this regard by a close study and a survey of the competitive threats, the environmental.

Question 20.
How would you execute the set up of an enterprise?
Answer:
The study of these five constituents is essential for the emergence of an enterprise.

• Labour or men: In all the resources of production, the ingredient of labour is the most mobilising element that stimulates all other factors and turns even the most useless thing into a useful one.
• Raw material: If the enterprise is production oriented a prior arrangement should be emphasised with regard to availability, type, cost/price and regular supply of the raw materials.
• Machine: Whatever be the scale of the enterprise whether big or small, machines are required in either case.
• Money: Money/capital is the lifeblood of an enterprise without which the establishment is an inconceivable presumption. The capital in the long and short term and its amount and the capital resources are a pre-emptive requirements
• Market: A prior estimate is to be formed in terms of the market-consumption before the establishment of the enterprise.
• Place: It is also an important aspect and be decided where the enterprise should be set up.

Question 21.
What problems should be properly considered in the expansion of a business enterprise? Elucidate.
Answer:
(i) Cash flow management: Cash flow problems are the second most common reason why small businesses go but according to research from CB Insights. Owners have to spend money to make money during a growth period, but this concept can quickly get out of control and leave you in a precarious position.

(ii) Responding to the competition: A funny thing happens when your company is successful- others recognize the opportunity and enter the industry. Many small business owners are unprepared for the realities of fierce competition, and they quickly lose their way in an attempt to respond.

(iii) Nurturing a great company culture: Your company culture is affected by everyone involved with your organization. As you grow and more people come into your company’s orbit, it becomes more difficult to exert control over your culture and you run the risk of having it drailed.

(iv) Learning when to delegate and when to get involved: There are times when entrepreneurs need to get personally involved in specific decisions, such as big – picture strategic planning and hiring for key positions. Then there are times where it is important to delegate and trust that your managers will make the best decision for their team and the company.

(v) Keeping up with market changes: If your company operates in a sector that experiences frequent upheaval, you have to be prepared for constant change. Internalize the idea that disruption is the new normal and work on training your employees to be agile in the face of uncertainty.

(vi) Deciding when to abandon a strategy: Sometimes marketing channels that seemed full of potential don’t pan out and new product lines don’t catch on as anticipated failures are an important part of business growth and owners must train themselves to recognize where they occur, divert resources accordingly and learn from those mistakes.

Question 22.
Describe the factors affecting working capital.
Answer:
Factors affecting the working capital:
There are the following factors affecting the working capital as:
(i) Nature and size of Business: The working capital depends on the nature of the business to a great extent. If a business is engaged in the production process more working capital will be required on other hand business firms engaged in trade activities will ‘require less working’ capital.

(ii) Business cycles: The need for working capital keeps on increasing or decreasing according to the business cycle. During Boom periods the businessman tries to keep the maximum possible stock of goods in order to escape from the price increase.

(iii) Term of Buying & Selling: If businessman buy the necessary material for cash he requires more working capital. If the business unit has the policy of credit sales, more working capital will be required.

(iv) Duration of Production: The amount of working capital also depends upon the time taken in the manufacturing of a product. If the time required for the manufacture of the product is more, more working capital is needed to be.

(v) Volume & Procurement of Raw Material: If the amount to be spent on raw material is more in total investment more working capital will be required.

(vi) Manual Vs Automation: If a business unit uses automatic machines for production less working capital will be required. If the production is labour oriented like the workers engaged in tea, coffee, rubber etc. More working capital will be required.

(vii) form of Taxation: Government policy of taxation also influences the need for working capital more capital will be required if all taxes are to be paid in advance, otherwise, the need will be less.

(viii) Seasonal goods: If business units are engaged in such goods in which seasonal conditions play an important role need for working capital will be felt according to seasonal requirements.

(ix) The intensity of competition: If the market competition is intense, more amount of working capital is required as a competitive firm has to keep huge quantities of a long-range of products.

(x) Business growth: If a business is expanding it requires more money for raw materials wages etc.

Question 23.
What are the factors affecting the identification of business opportunities?
Answer:
Responsibilities of each entrepreneur of the environment are any kind of tampering with the environment in terms of filing, pollution, contamination, etc. is a serious offence which is causing great harm to the present generation. In assertion, it is raising a question mark on the fate of the coming generation. No doubt, the industrial units do cause pollution, more or less, to the environment, and it is the prime responsibility of an entrepreneur not to harm the environment in any way. The industrial waste, raw residual, etc., must not be thrown into the water, its right use should be encouraged. Personal profit should not be the sole consideration here to the extent that the environment is polluted. The government is very serious about it and does not permit, even then an entrepreneur ought to be considerate at his own in this regard.

Question 24.
Explain the meaning, definition and objective of Financial Management.
Answer:
The financial manager in order to achieve the objectives of financial management performs many functions:
(i) Estimating the financial requirements: The financial manager has to estimate the financial requirements both for long and short periods. The long term financial needs means for procuring fixed assets and part of current assets and short term requirements are meant for providing the current assets.

(ii) The decision about capital structure: After estimating the longterm financial requirements the next step is to decide about the sources from where these funds would be raised. The financial manager explores and assesses the various long-term sources in the light of requirements of the enterprise and then decides what specific sources to tap and low much to raise form each of these in the overall interest of the organisation.

(iii) Dividend Decision: It is for the finance manager to make suggestions in respect of dividend payment. He looks into the trends of earnings, share market prices the funds required for future growth, the cash flow situation, etc and recommends to the top management how much of earnings should be earmarked for payment of dividend and how much should be retained in the business.

(iv) Cash management: The finance manager has to ensure that adequate cash is available for various purposes with minimum cost, that the rate of cash inflow is accelerated and that of cash outflow decelerated, that suitable arrangements exit for meeting situation of foreseen or unforeseen shortage of cash and that adequate steps are taken to keep at the bare minimum level cash lying idle with the cashier and or at the bank.

(v) Report with investors: Many parties like financial institutions, banks, underwriters, public etc. Invest money in the organisation, the finance manager establishes close links with them, supplies them the necessary information and negotiates to get financial assistance from them.

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Time : 3 Hours 15 Min
Full Marks: 70

Instructions for the candidates

1. Candidates are required to give answers in their own words as far as practicable.
2. Figures in the right hand margin indicates full marks.
3. While answering the questions, candidate should adhere to the words limit as far as practicable.
4. 15 Minutes of extra time has been allotted for the candidates to read the questions carefully.
5. This question paper is divided into two sections : Section -A and Section-B
6. In Section A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/Blade/ Nail on OMR Sheet otherwise your result will be treated as invalid.
7. In section-B, there are 18 short answer type questions (each carrying 2 marks), out of which only 10 (ten) questions are to be answered Apart from this there are 06 Long Answer type questions (each carrying 5 marks), out of which 3 questions are to be answered.
8. Use of any electronic device is prohibited.

Objective Type Questions

In the following questions no. from 1 to 35, there is only one correct answer against each question. For each question, mark (darken) the correct answer on the OMR Sheet provided to you. (1 x 35 = 35)

Question 1.
Which type of crystals contains more than one Bravais lattice ?
(a) Hexagonal
(b) Triclinic
(c) Rhombohedral
(d) Monoclinic
Answer:
(d) Monoclinic

Question 2.
For the structure given below the site marked as S is a …………….

(a) tetrahedral void
(b) cubic void
(c) octahedral void
(d) none of these
Answer:
(c) octahedral void

Question 3.
If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is
(a) r = 0.414 R
(b) R = 0.414 r
(c) r = 2R
(d) r = \(\sqrt{2} R\)
Answer:
(a) r = 0.414 R

Question 4.
A metal crystallises into a lattice containing a sequence of layers as AB AB AB………………… What percentage of voids are left in the lattice ?
(a) 72%
(b) 48%
(c) 26%
(d) 32%
Answer:
(c) 26%

Question 5.
In ccp arrangement the pattern of successive layers can be designated as
(a) AB AB AB
(b) ABC ABC ABC
(c) AB ABC AB
(d) ABA ABA ABA
Answer:
(b) ABC ABC ABC

Question 6.
A solution is obtained by mixing 200 g of 30% and 300 g of 20% solution by weight. What is the percentage of solute in the final solution ?
(a) 50%
(b) 28%
(c) 64%
(d) 24%
Answer:
(d) 24%

Question 7.
When 1.04 g of BaCl₂ is present in 10⁵ g of solution the concentration of solution is ………….
(a) 104 ppm
(b) 10.4 ppm
(c) 0104 ppm
(d) 104 ppm
Answer:
(b) 10.4 ppm

Question 8.
In the cell, |Zn²⁺| |Cu²⁺|Cu, the negative terminals
(a) Cu
(b)Cu²⁺
(c) Zn
(d) Zn²⁺
Answer:
(d) Zn²⁺

Question 9.
What will be standard cell potential of galvanic cell with the following reaction ?

(a) 74 V
(b) 1.14 V
(c) 34 V
(d) – 0.34 V
Answer:
(c) 34 V                         .

Question 10.
In a reaction 2HI → H₂ + I₂ the concentration of HI decreases from 0.5 mol L^-1 in 10 minutes. What is the rate of reaction during this interval ?
(a) 5 x 10^-3 M min^-1
(b) 5 x 10^-3 M min^-1
(c) 5 x 10^-2 M min^-1
(d) 5 x 10^-2 M min^-1
Answer:
(a) 5 x 10^-3 M min^-1

Question 11.
The rate of disappearance of SO, in the reaction, 2SO₂ + O₂  → 2SO₃ is 1.28 x 10^-5 Ms . The rate of appearance of SO₃ is ……………….
(a) 64 x 10^-5 M s^-1
(b) 0.32 x 10^-5 M s^-1
(c) 56 x 10^-5 Ms¹
(d) 1.28 x 10^-5  M s^-1
Answer:
(d) 1.28 x 10^-5  M s^-1

Question 12.
Which of the following is a property of physisorption ?
(a) High specificity
(b) Irreversibility
(c) Non-specificity
(d) None of these
Answer:
(c) Non-specificity

Question 13.
Which of the following statement is not correct about physisorption ?
(a) It is a reversible process
(b) It requires less heat of adsorption
(c) It requires activation energy
(d) It takes place at low temperature
Answer:
(c) It requires activation energy

Question 14.
The name of the metal which is extracted from the ore is given below. Mark the example which is not correct.
(a) Malachite – Cu
(b) Calamine – Zn
(c) Chromite – Cr
(d) Dolomite – Al
Answer:
(d) Dolomite – Al

Question 15.
Which of the followings is magnetite ?
(a) Fe₂CO₃
(b) Fe₂O₃
(c) Fe₃O₄
(d) Fe₂O₃.3H₂O
Answer:
(c) Fe₃O₄

Question 16.
Nitrogen can form only one chloride with chlorine which is NCl₃ whereas P can form PCl₃ and PCl₅. This is ……………
(a) due to absence of d-orbitals in nitrogen
(b) due to difference in size of N and P
(c) due to higher reactivity of P towards Cl than N
(d) due to presence of multiple bonding in nitrogen
Answer:
(a) due to absence of d-orbitals in nitrogen

Question 17.
On heating a mixture of NH₄Cl and KNO₂, we get
(a) NH₄NO₃
(b) KNH₄(NO₃)₂
(c) N₂
(d) NO
Answer:
(c) N₂

Question 18.
The melting point of copper is higher than that of zinc because
(a) the s, p as well as d-electrons of copper are involved in metallic bonding
(b) the atomic volume of copper is higher
(c) the d-electrons of copper are involved in metallic bonding
(d) the s as well as d-electrons of copper are involved in metallic bonding
Answer:
(c) the d-electrons of copper are involved in metallic bonding

Question 19.
Zr and Hf have almost equal atomic and ionic radii because of ………………….
(a) diagonal relationship
(b) lanthanoid contraction
(c) actinoid contraction
(d) belonging to the same group
Answer:
(b) lanthanoid contraction

Question 20.
Which of the following primary and secondary valencies are not correctly marked against the compounds ?
(a) [Cr(NH₃)₆]Cl₃, p = 3,s = 6
(b) K₂[Pt(CI₄],p = 2,s = 4
(c) [Pt(NH₃)₂Cl₂],p = 2,s = 4
(d) [Cu(NH₃)₄]SO₄,p = 4,s = 4
Answer:
(d) [Cu(NH₃)₄]SO₄,p = 4,s = 4

Question 21.
When aqueous solution of potassium fluoride is added to the blue coloured aqueous CuSO₄ solution, a green precipitate is formed. This observation can be explained as follows.
(a) On adding KF, H₂O being weak field ligand is replaced by F^– ions forming [CuF₄ ]^2- which is green in colour
(b) Potassium is coordinated to [Cu(H₂O)₄]²⁺ ion present in CuSO₄ and gives green colour
(c) On adding KF, Cu²⁺ are replaced by K⁺ forming a green complex
(d) Blue colour of CuSO₄ and yellow colour of KI form green colour on mixingAnswer:
(a) On adding KF, H₂O being weak field ligand is replaced by F^– ions forming [CuF₄ ]^2- which is green in colour

Question 22.
IUPAC name of CH₃)₂ CH — CH₂ — CH₂Br is
(a) 1-bromopentane
(b) 1-bromo-3-methylbutane
(c) 2-methyl-4-bromobutane
(d) 2-methyl-3-bromopropane
Answer:
(b) 1-bromo-3-methylbutane

Question 23.
The IUPAC name of the compound is

(a) 1 -fluoro-4-methyl-2-nitrobenzene
(b) 4-fIu6ro-lmethyl-3-nitrobenzene
(c) 4-methyl-l-fIuoro-2-nitrobenzene
(d) 2-fIuoro-5-methyl-l-nitrobenzene
Answer:
(d) 2-fIuoro-5-methyl-l-nitrobenzene

Question 24.
The C— O — H angle in ether is about
(a) 180°
(b) 190°28′
(c) 110°
(d) 105°
Answer:
(c) 110°

Question 25.
The C— O — H bond angle in alcohols is slightly less than the tetrahedral angle whereas the C— O — C bond angle in ether is slightly greater because ………
(a) of repulsion between the two bulky R groups.
(b) O atom in both alcohols and ethers is sp³-hybridised
(c) lone pair-lone pair repulsion is greater than bond pair­bond pair repulsion
(d) none of these
Answer:
(a) of repulsion between the two bulky R groups.

Question 26.
Propanone can be prepared from ethyne by
(a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C
(b) passing a mixture of ethyne and ethanol over a catalyst zinc chromite
(c) boiling ethyne with water and H₂SO₄
(d) treating ethyne with iodine and NaOH
Answer:
(a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C

Question 27.
In the following reaction, product (P) is

(a) RCHO
(b) RCH₃
(c) RCOOH
(d) RCH₂OH
Answer:
(a) RCHO

Question 28.
IUPAC name of the compound (C₂H₅)₂ NCH is
(a) 2, 2-diethylmethanamine
(b) N, N-diethylmethanamine
(c) N-ethyl-N-methylethanamine
(d) N-methylbutanamine
Answer:
(c) N-ethyl-N-methylethanamine

Question 29.
Reduction of aromatic nitro compounds using Sn and HCI gives
(a) armatic primary amines
(b) aromatic secondary amines
(c) aromatic tertiary amines
(d) aromatic amides
Answer:
(a) armatic primary amines

Question 30.
Which of the fol lowing is an example of an aldopentose ?
(a) D-Ribose
(b) Glyceraldehyde
(c) Fructose
(d) Erythrose
Answer:
(a) D-Ribose

Question 31.
Which of the following treatment will convert starch directly into glucose ?
(a) Heating with dilute H₂SO₄
(b) Fermentation by diastase
(c) Fermentation by zymase
(d) Heating with dilute NaOH
Answer:
(a) Heating with dilute H₂SO₄

Question 32.
The correct structure of monomers of buna-S is

Answer:
(c)

Question 33.
The S in buna-S refers to
(a) sulphur
(b) styrene
(c) sodium
(d) salicylate
Answer:
(b) styrene

Question 34.
The main cause of acidity in the stomach is
(a) release of extra gastric acid which decrease the pH level
(b) indidestion and pain in large intestine
(c) increase the pH level in the stomach
(d) release of extra bile juice which increase alkaline medium in stomach.
Answer:
(a) release of extra gastric acid which decrease the pH level

Question 35.
Which of the following will not act as antacid ?
(a) Sodium hydrogen carbonate
(b) Magnesium hydroxide
(c) Sodium carbonate
(d) Aluminium carbonate
Answer:
(c) Sodium carbonate

Non-Objective Type Questions

Short Answer Type Questions

In this Section, there are 18 Short answer type questions (each carrying 2 marks), out of which answer any 10 questions.  2 x 10 = 20

Question 1.
Calculate the osmotic pressure of 5% solution of urea at 272 K . (R = 0.0821 L-atm. deg^-1).
Answer:
As we know that \(\pi=\frac{n_{\text {urea }} R T}{V}\)
Number of moles in urea \(=\frac{5}{60}=0.083\)
Volume of Solution \(=\frac{100}{1000}=0.1\) litres
Hence, Osmotic Pressure, \(\pi=\frac{0.083 \times 0.0821 \times 272}{0.1}=18.53 \mathrm{atm}\)
Osmotic Pressure = 18.53atm

Question 2.
Define standard electrode potential.
Answer:
Standard Electrode Potential : The standard electrode potential is the potential difference between the electrode & electrolyte (at 1 M), at standard conditions (1 atm, 298 K). Standard electrode potential is abbreviated by (E or E°).

Question 3.
What is activation energy ? How is the rate constant of a reaction related to its activation energy ?
Answer:
Activation Energy: The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy.
Activation energy (E_a)= Threshold energy (E_T) – Average energy of the reactant (E_R). i.e. E_a = E_T – E_R
If activation energy will be less, then faster will be the reaction.
i. e . Activation energy = \(\alpha \frac{1}{\text { rate of reaction }}\)
e.g. 2NO + O₂ → 2NO₂
It is a fast reaction because of low activation energy. 2CO + O₂ → 2CO₂
It is a slow reaction because of high activation energy.

Question 4.
What is the main difference between physiorption and Chemisorption ?
Answer:
Physisorption:

• Force of attraction are Vander Waal’s forces.
• It usually takes place at low temperature and decreases with increasing temperature.
• It does not require any activation energy.

Chemisorption :

• Force of attraction are chemical bond forces.
• It takes place at high temperature.
• It requires high activation energy.

Question 5.
Differentiate the following giving suitable example:
(i) Calcination and Roasting
(ii) Flux and Slag.
Answer:
(i) Calcination :
(a) Calcination is a thermal treatment process in absence of air applied to ores and materials to bring about a thermal decomposition, phase transition or removal of volalatile fraction It is done at temperature below the melting point of the product material.
(b)

Roasting:
(a) Roasting is reverse process of calcination. It is also thermal process in presence of air applied to ore for removal of volatile fraction under melting point of the product material is called Roasting.
(b)

Flux :
(a) It is chemical cleaning agent, which remove impurities from metal.
(b) Example – CaO + SiO₂ → CaSiO₃

Slag :
(a) Slag is partially vitereous by product of the process of smelting ores, which separates the desiredmetal fraction from the unwanted fraction. It is mixture of metal oxide
(b) Example – CaO + SiO₂ → CaSiO₃

Question 6.
Why do transition elements form coloured compounds ? Explain.
Answer:
The value of electrods potential depends upon the heat of sublimation and ionisation energy
ΔH – ΔH_sub + IE + ΔH_hyd . Due to the influence of ligand d-orbitals of transition metals divide into two unequal energy containing sets. Unpaired electrons absorb sunlight and jumps from one orbit into another orbit and one clour is reflected.

Question 7.
Predict the geometrical shapes of the following:
(a) sp³
(b)d²sp³
Answer:
(a) sp³ tetra-hedral
(b)d²SP³ – Square tetrahedral.

Question 8.
Specify oxidation numbers of the metals in the following co-ordination compounds:
(a) K₄[Fe(CN)₆]
(b) [PtCl₄]^2-
Answer:
(a) Oxidation number of (co-ordination compound)
Let oxidation number of Fe = x
Hence, 1 x 3 + (x -1 x 6 = 0
or, x – 3 = 0
x = 3 = O. N of Fe

(b) Oxidation number of (co-oridnation compaund)
Hence, x + 4 (-1) = – 2
or, x = -2 + 4 = + 2 = ON of Pt

Question 9.
Write the structural formulae of the following:
(a) 4,4 dimethyl-2-pentanol
(b) 2-butanol.
Answer:
(a) 4, 4 dimethyl, 2-Pentanol

Question 10.

Answer:

Question 11.
How will you convert the following ?
(a) Ethyl alcohol to ethylamine
(b) Ethylamine to ethyl alcohol.
Answer:

Question 12.
Explain in which of the following compounds, the chemical bond would have less ionic character: LiCl or KCl.
Answer:
LiCl > KCl – Due to large size of k, KCl compound has less ionic character. LiCl is covalent than Li⁺ has higher polarisation capacity.

Question 13.
Discuss briefly the structure of CsCl.
Answer:
Structure of CsCl.

Question 14.
The osmotic pressure of sugar solution is 2.46 atm at 27°C. Calculate the concentration of the solution.
Answer:

Question 15.
How is molarity of a solution different from molality?
Answer:
Molarity : It is defined as the number of moles of solute dissolved in one liter of solution
\(Molarity (\mathrm{M})=\frac{\text { Moles of solute }}{\text { Volume of Solution in liter }} \)

Molality : It is defined as the number of moles of solute dissolved in one kg of solvent
\(Molality (\mathrm{m})=\frac{\text { Moles of the Solute }}{\text { Mass of the solvent in } \mathrm{kg}}\)

Question 16.
Define heat of neutralisation.
Answer:
Defination of heat of neutralization : The heat reaction resulting from the neutralization of an acid or base especially the quantity produced when a gram equivalent of a base or acid is neutralized with a gram equivalent of an acid or base in dilute soliution.

Question 17.
(a) State law of mass action.
(b) What is the effect of temperature on reaction ?
Answer:
(a) Law of mass action: Law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.

(b) The effect of increasing collision frequency on the rate of the reaction is very minor increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions.

Question 18.
Discuss the following terms :
(a) Coordination number
(b) Effective atomic number.
Answer:
(a) The coordination number of a central atom in a molecure or crystal is the number of its near neighbours. This number is determined some what differently for molecules than for crystals.

(b) Effective atomic number has two different meaning; one that is the effective nuclear charge of an atom, and other that calculates the average atomic number for a compound or mixture of materials.

Long Answer Type Questions

There are 06 long answer type questions (each carrying 05 marks), out of which answer any there questions.
(5 x 3 = 15)

Question 19.
Define order to reaction and molecularity of relation. Derive a general expression for specific rate constant of first order reaction.
Answer:
It is the numbers of atom, ion, and molecule that collide with one another simultaneously so as to result into a chemical reaction
Rate constant → It is the speed at which the reactions are converted into the products at any moment of time.
Law of Mass Action → At a given temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of reacting species with each concentration term raised to the power equal to numerical co-efficient of that species in the chemical reaction.
Thus, A + B → Products
Rate of reaction, r x [A] [B] = K [A] [B]
Where, [A] and [B] are the molar concentrations of the reactants A and B respectively and K is a constant of proportionality.
∴Rate = K [A]^a [B]^b

First Order Reaction – It is the type of reaction in which the rate of reaction depends only on the first power of the concentration of a single reacting Species.
Let, A + B → Product
If, a be the initial concentration of A and after time t, the concentration of product is x, then the concentration of A left at that instant is (a – x)
A → Products
initial conc._n (a)  ….(i)
after time ‘t’ (a – x)  … (ii)

Where, C → integration constant, which can be obtained from initial condition of the reaction, when, t = 0, x = 0,
then, from (iii) In (a – 0) = K x 0 + C
or, C = In (a)  ………….. (iv)
from (iii) + (iv) – In (a -x) = Kt – In (a)
or, In (a) – In (a – x) = kt or, In (a) – In (a – x) = kt.

Question 20.
(a) Differentiate between DNA and RNA.
(b) What are Nucleic acids ?
Answer:
(a) The structural difference between DNA and RNA
DNA : DNA is double stranded helix in which two strands are coiled spirally in opposite directions.

• The sugar molecule is 2-dehydrobase.
• Nitrogeneous base uracil is not present.
• DNA molecule are very large their molecules weights may vary from 6 million to 16.
• DNA has unique property of replication.

RNA :

• It is single strainded structure.
• The sugar molecules is ribose.
• Nitrogeneous base they-mine is not present.
• RNA molecules are much smaller with molecular weight ranging from 20000 to 40000.
• RNA usually doesn’t replicate.

(b) Nucleic acid – Nucleic acid are found in all living cells in form of nucleo proteins. They constitute important class of biomolecules. Nucleic acids are genetic material of the cells and are responsible for transmission of herediatry effect from one generation to the other and also carry out the biosynthesis of proteins. Nucleic acid are biopolymers (i.e., polymers present in living system). The genetic information about in nucleic acid controls the structure of all proteins including enzymes and thus governs the entire metabolic activity in the living organism.

Question 21.
(a) Discuss the principle involved in the manufacture of sulphuric acid by contact process.
(b) How will you test for sulphate ion ?
Answer:
From following steps sulphuric acid is manufactured
(i) SO₂ gas is produced by burning sulphur of FeS₂ in excess of air
S + O₂  → SO₂
4 FeS₂ +11O₂ → 2Fe₂ O₃ + 8SO₂

(ii) SO₂ is oxidised into SO₃ in prensence of catalyst
2 SO₂ + O₂ → 2 SO₃

(iii) SO₃ on treatment with cone. H₂SO₄ oleum (H₂S₂O₇) is obtained.
SO₃ + H₂SO₄  → 2 H₂S₂2O₇

(iv) From oleum sulphuric acid is obtained of desired concentration by diluting water.
H₂S₂O₇+ H₂O → 2H₂OSO₄

(b) Test of sulphate radical (of NO₂SO₄) – We mix up Barium chloride in the salt. We get white precipate of BaSO₄ which is insoluble in dil. HCl & HNO₃.
BaCl₂ 4- Na₂SO₄ → BaSO₄ ↓ +2 NaCl
BaCl₂ + HNO₃ → In soluble

Question 22.
What are carbohydrates ? How are they classified ?
Answer:
Carbohydrates are the naturally occuring organic compounds and are a major source of energy to our body. In plants carbohydrates are formed as a result for photosynthesis.

In animal systems the carbohydrates undergo decomposition to form carbon-dioxide and water accompanied by the release of energy needed for the body.

Composition of carbohydrates – The chemical formula of the carbohydrates suggests that these are the hydrates of carbon. For example, glucose may be represented as C₆(H₂O)₆ while sucrose as C₁₂(H₂O)₁₁ . But this definition has certain limitation.

(i) Compounds like formaldehyde (CH₂O) and acetic acid (C₂H₄O₂) are the hydrates of carbon do not show characteristics of carbohydrates.

(ii) Similarly, compound like rhammose (C₆H₁₂O₆) and deoxyribose (C₅H₁₀O₄) are carbohydrates but not the hydrates of carbon.

The definition of carbohydrates has been modified and they may be defined as—The polyhydroxy aldehydes or polyhdroxy ketones are substances which generally give these on hydrolysis; contain atleast one chiral carbon and are therefore, opitcally active classification of carbohydrates—

Classification of carbohydrates are done in three way

(a) Based on molecular size – on the basis of the molecular size, carbohydrates have been classified into three types. These are

• Manosaccharides
• Oligosoccharides
• Polysaccharides

Based on taste – Carbohydrates with sweet taste are called sugar while those without a sweet taste are called non-sugars. It may be noted that all mono and oligosaccharides are sugars while polysaccharides are non­sugar.

Reducing and non-reducing sugar –  Carbohydrate which reduce Tollen’s reagent and Fehling solution are called reducing sugar while those which don’t reduce, these are called non reducing sugars. For example – glucose and fructose.

Question 23.
Disuss the following :
(a) Kolbe’s reaction
(b) Wurtz’s reaction
(c) Carbylamine reaction.
Answer:
(a) Kolbe’s reaction – The reaction of phenoxide ion with CO₂ at the high pressure and temperature followed by acidification to form salicylic acid is called Kolbe’s reaction.

(b) Wurtz reaction-When two moles of alkyl halide is heated with Na in the presence of ether goes higher alkane.

(c) Carbylamine reaction-Chloroform when heated with primary amine in presence of alcoholic KOH forms a derivative called isocyanide.

Question 24.
Give one example for each of the following :
(a) Synthetic rubber
(b) Naturally occurring amino acid
(c) Condensation polymer
(d) Additional polymer
(e) Artificial sweeteners.
Answer:
(a) Nioprin
(b) Zwitter ion
(c) NYlon-6, 6
(d) Buna-5
(e) Saccharine.

 

Bihar Board 12th Physics Model Papers

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Objective Type Questions

There are 1 to 35 objective type questions with 4 options. Choose the correct option which is to be answered on OMR Sheet. (35 x 1 = 35)

Question 1.
The electric field intensity at distance on the axis of an electric dipole is.Ej and E2 on the perpendicular bisector of dipole. The angle between E] and E2 is 6, will be
(a) 1 : 1, π
(b) 1 : 2, π/2
(c) 2 : 1, π
(d) 1 : 3 π
Answer:
(c) 2 : 1, π

Question 2.
The metre bridge is shown in figure. The value of x is
(a) 10Ω
(b) 3 ohm
(c) 9 ohm
(d) 10 ohm

Answer:
(a) 10Ω

Question 3.
The dimension of electromotive force is
(a) |ML² T^-3]
(b) [ML²T^-3I^-1]
(c) [MLT^-2]
(d) [ML²T^-3I^-1]
Answer:
(d) [ML²T^-3I^-1]

Question 4.
The power of electric circuit is
(a) V.R
(b) V2.R
(c) V2/R
(d) V2RI
Answer:
(c) V2/R

Question 5.
If a 60W and 40W bulb are joined in series the
(a) 60W bulb glow more
(b) 40 W bulb glow more
(c) Both bulb glow similar
(d) Only 60W bulb is lighted
Answer:
(b) 40 W bulb glow more

Question 6.
Kilowatt hour (kwh) is unit of
(a) Power
(b) Energy
(c) Torque
(d) None of these
Answer:
(b) Energy

Question 7.
A magnet is cut parallel to its length in n equal
parts. Magnetic moment of each part will be
\(\text { (a) } \frac{M}{n}\)
\(\text { (b) } \frac{M}{n^{2}}\)
\(\text { (c) } \frac{M}{2 n}\)
(d) None of these
Answer:
\(\text { (a) } \frac{M}{n}\)

Question 8.
The work done to rotate the magnet with 90° will be
(a) MB
(b) MB cosθ
(c) MB sin θ
(d) MB (1 – sin θ)
Answer:
(b) MB cosθ

Question 9.
S.I. unit of polar strength is
(a) Ampere metre
(b) Tesla
(c) Faraday
(d) Ampere gm²
Answer:
(a) Ampere metre

Question 10.
Which is ferromagnetic substance
(a) Mn
(b) Cr
(c) Zn
(d) Alnico
Answer:
(d) Alnico

Question 11.
Which law is based on the principle of conservation of energy?
(a) Lenz’s law
(b) Faraday’s law
(c) Ampere’s law
(d) None of these
Answer:
(a) Lenz’s law

Question 12.
The unit of reactance is
(a) mho
(b) ohm
(c) Faraday
(d) ampere
Answer:
(b) ohm

Question 13.
The phase difference between current and voltage in AC circuit is 0. Then power factor will
(a) cos θ
(b) sin θ
(c) tan θ
(d) 1θ
Answer:
(a) cos θ

Question 14.
The velocity of electro magnetic wave in air is

Answer:
(b)

Question 15.
The light waves are transverse in nature which is shown by
(a) Scattering
(b) Diffraction
(c) Interference
(d) Polarisation
Answer:
(d) Polarisation

Question 16.
When two converging lens of same total length / are placed in contact then focal length combination is
(a) f
(b) 2f
(c) f2
(d) 3f
Answer:
(c) f2

Question 17.
When the light ray passes from one medium to another medium then it is deviated, is called
(a) Dispersion
(b) Refraction
(c) Diffraction
(d) Reflection
Answer:
(b) Refraction

Question 18.
Convex lens are used for
(a) Myopia
(b) Hypermetropia
(c) Old sightedness
(d) Astigmatisn
Answer:
(b) Hypermetropia

Question 19.
The path difference for destructive interference is
(a) nλ
\(\text { (b) }(2 n+1) \frac{\lambda}{2}\)
(c) Zero
(d) infinity
Answer:
\(\text { (b) }(2 n+1) \frac{\lambda}{2}\)

Question 20.
Brewler’s law is
(a) μ = sin i_p
(b) μ = cos i_p
(c) μ = tan i_p
(d) μ = tan² i_p
Answer:
(c) μ = tan i_p

Question 21.
Light is made of vibration by
(a) Ether particle
(b) Air particle
(c) Electric and magnetic field
(d) None of these
Answer:
(c) Electric and magnetic field

Question 22.
Photo cell is based on
(a) photo electric effect
(b) chemical effect of current
(c) magnetic effect of current
(d) electromagnetic cell
Answer:
(a) photo electric effect

Question 23.
Which is uncharged
(a) α-particle
(b) β-particle
(c) photon
(d) proton
Answer:
(c) photon

Question 24.
Lymen series of Hydrogen lies in electromagnetic spectrum is
(a) X : ray
(b) Visible
(c) Infrared
(d) Ultra-violet
Answer:
(d) Ultra-violet

Question 25.
When transition takes place in hydrogen atom from higher orbit to another orbit then obtain
(a) Lymen series
(b) Balmer series
(c) Paschan series
(d) Pfund series
Answer:
(b) Balmer series

Question 26.
The average binding energy per nucleon of nucleus is
(a) 8 ev
(b) 8 Mev
(c) 8 Bev
(d) 8 Joule
Answer:
(b) 8 Mev

Question 27.
The cause of emission of energy in star is
(a) chemical reaction
(b) fusion of heavy nucleus
(c) fusion of light nucleus
(d) fission of heavy nucleus
Answer:
(c) fusion of light nucleus

Question 28.
The Boolean expression of OR gate is
(a) A + B = y
(b) A.B = y
\(\text { (c) } \bar{A}=y\)
\(\text { (d) } \mathrm{C}=\overrightarrow{\mathrm{A} \cdot \mathrm{B}}\)
Answer:
(a) A + B = y

Question 29.
The permeability of Ferromagnetic substance is
(a) μ > 1
(b) μ = 1
(c) μ < 1
(d) μ = 0
Answer:
(a) μ > 1

Question 30.
The light is emitted in forward biased junction diode, is called
(a) LED
(b) Photo diode
(c) Zener diode
(d) None of these
Answer:
(a) LED

Question 31.
Which expression is correct?

Answer:
(d)

Question 32.
The majority carriers in n-type semi conductor is
(a) proton
(b) hole
(c) a-particle
(d) electron
Answer:
(d) electron

Question 33.
Which frequency range is used for TV transmission?
(a) 30 HZ-300HZ
(b) 30 KHZ-300 KHZ
(c) 30 MHZ-300MHZ
(d) 30GHZ-300GHZ
Answer:
(a) 30 HZ-300HZ

Question 34.
Ionosphere is used for radio waves in
(a) rarer mediun
(b) denser medium
(c) free space
(d) dielectric
Answer:
(a) rarer mediun

Question 35.
The maximum distance upto which TV signal can be received from Antenna of height h is proportional to
(a) h^1/2
(b) h
(c) h^5/2
(d) h²
Answer:
(a) h^1/2

Non-Objective Type Questions!
Short Answer Type Questions

Question No. 1 to 18 are short answers type question. Each question carries 2 marks. Answer any ten (10) question.

Question 1.
Electric lines of force never intersect to each other. Why?
Answer:
If two lines of force intersect at a point then it will mean that at that point electric field has two directions. As it is not possible, the lines of force here not intersect to each other.

Question 2.
Write the limitations of Coulomb’s law.
Answer:
Coulomb’s law in electrostatics does not hold in all situation. It is applicable only in following situations :

1. The electric charges must be stationary.
2. The electric charge must be point in size. Coulomb’s law doesn’t apply to two charged bodies of finite size. It is because, the distribution of charge doesn’t remain uniform when the two bodies are brought together.

Question 3.
Prove maximum power theorem.
Answer:
It states that the output power of a source of emf is maximum, when the external resistance in the circuit is equal to the internal resistance of the source.

This is maximum power theorem.

Question 4.
Establish the relation b/w current density and drift velocity.
Answer:

Question 5.
What is Hysteresis loop? With its help explain the terms retentivity and co-ercivity.
Answer:
Hysteresis loop : Hysteresis loop is defined as the lagging of the magnetic induction B behind the corresponding magnetic field H.

When magnetic field H and the magnetic induction B are yarned from zero to maximum in one direction and then back through zero to a maximum in the opposite direction and finally back again through zero to the first maximum, a cycle of magnetisation is said to be completed. B(T) Y Using the value of H and B obtained above, we plot a graph by taking H along x-axis and B along y-axis. When the magnetic field H is increased from zero to the maximum value Og, the induction B follows the curve Oa. However, when the magnetic field is brought to zero, the.;induction does not follows back the curve Oa but a different curve ab. The same pattern of variations is seen for the rest of the curve. The B-H curve for decreasing H does not coincide with the B-H curve for increasing H.

As is clear from figure, even when the magnetising field H is zero, some magnetic induction (Ob or Oe) is still present in the speci nen. It is denoted by Br and is called remanance or retent i vity or residual magnetism.

The value of the intensity of magnetisation of a material when the magnetisi ng field is reduced to zero is called retentivity.

After the specimen has been magnetised to saturation (oh or oj) a reversed magnetising field (equal to Oc or Of) is required to reduce the magnetic induction to zero.This is called the coercivity or coercive force He. The value of the magnetising field required to reduce residual magnetism to zero is called coercivity of the material.

Question 6.
Write the expression for energy stored in the
Answer:
The induced emf in coil \(e=-L \frac{d I}{d t}\)
let chasge dq passes from coil, the work done dw = e.dq
\(\begin{aligned}
&=\mathrm{L} \cdot \frac{d I}{d t} d q=\mathrm{L} \cdot d I \cdot \frac{d q}{d t} \\
d w &=\mathrm{LI} . \mathrm{dI}
\end{aligned}\)

work done after posting current O to 1
\(\omega=\mathrm{L} \int_{0}^{1} I d I=\mathrm{L}\left[\frac{I^{2}}{2}\right]_{0}^{1} ; \omega=\frac{1}{2} \mathrm{L}^{2}\)

This work done is stored in the potential energy of coil.
\(P \cdot E=\frac{1}{2} L I^{2}\)

Question 7.
What is current sensitivity and voltage sensitivity of moving coil galvanometer?
Answer:
Current sensitivity : It is defined as the deflection produced in the galvanometer, when a unit voltage is applied across its cops.
\(\frac{\alpha}{\mathrm{I}}=\frac{\mathrm{nBA}}{\mathrm{K}}\)

Voltage sensitivity : It is defined as the deflection produced in the galvanometer, when a unit voltage is applied across its COPI.
\(\frac{\alpha}{V}=\frac{n B A}{K R}\)

Question 8.
Write the expression for energy density related with electro magnetic wave and show that its ratio is equal to 1.
Answer:

Question 9.
Sketch the wave front corresponding to (a)
diverging rays and (b) converging rays.
Answer:

Question 10.
Write the differences between magnification and magnifying power.
Answer:
Magnification :

• It is a linear magnification which is equal to \(\frac{n_{2}}{n_{1}}\).
• Its value increases with increase in V.
• It may be between to -∞ to +∞.
• Under certain conditions it is equal to magnifying power.

Magnifying power :
(i) It is an angular magnification which is equal to \(\frac{\angle \beta}{\angle \alpha}\)
(ii) Its value decreases with increase in V.
(iii) Its value may be b/w \(\frac{d}{f}\) and \(\left(1+\frac{D}{f}\right)\)
(iv) It is a special condition of magnification when (v_e = D

Question 11.
Derive the expression for de-Broglie wave length.
Answer:
A/C to quantum theory E = hv ……………………….. (i)

λ is called de-Broglie wavelength.

Question 12.
What is mass defect?
Answer:
Mass defect : The difference b/w sum of the masses of neutrons and protons forming a nucleus and actual mass of the nucleus is called mass defect.
Δm = [mpz + m_n (A – Z) – M]
The unit of mass defect is amu.

Question 13.
Why does thermionic emission takes place from metal surface only?
Answer:
Thermionic emission takes place from metal surface only because metal has free electrons on its surface.This free electrons requires less energy to come out from one surface. Also sometimes, to reduce work function of metal by coating oxide layer on metal.

Question 14.
Why is semi conductor damaged by a strong current?
Answer:
A strong current, when passed through a semi conductor, heat up the semi conductor and covalent bond break up. It results in a large number of free electrons. The material behaves as a conductor. As now the semi conductor no longer possesses the property of conduction, it is said to be damaged.

Question 15.
Write the differences between Intrinsic and extrinsic semi conductor.
Answer:
Intrinsic semi conductor :

• They are the crystals of pure elements like Germanium and Silicon.
• n_e = n_h
• The electrical conductivity of Intrinsic semi conductor is low.
• Resistivity is higher.

Extrinsic semi conductor:

• When some impurity is added in Intrinsic semi conductor, we get an extrinsic semi conductor.
• n_e ≠ n_h
• Its electrical conductivity is high.
• Resistivity is lower.

Question 16.
Define mean value and root square value A.C. Write its expression.
Answer:
Mean value of Average value : It is that steady current when passes through the circuit for half time period, sends the same amount of charge of the alternating current, sends the charge in same circuit and in same time period. It is denoted by Im.
\(\begin{aligned}
&I_{m}=\frac{2 I_{0}}{\pi}\\
&I_{m}=0.636 I_{0}
\end{aligned}\)

R.M.S. value or virtual value: It is that steady current when passes through the circuit for half time period, produces the same amount of heat as the alternating current produces the heat in same time and in same circuit. It is denoted by Irms.
\(\begin{aligned}
&I_{r m s}=\frac{I_{0}}{\sqrt{2}}\\
&I_{r m s}=0.7071
\end{aligned}\)

Question 17.
What is polaroids? Write its important uses.
Answer:
It is a device which produces plane polarised light. Use of polaroids :

• Sun glasses fitted with polaroid sheets protect the eye from glass.
• Wind shields of automobiles are also made of Polaroid sheets.
• They are useful in three dimensional motion pictures.
• It is used in L.C.D., calculator, watches, T.V. etc.

Question 18.
Write the condition for sustained interference.
Answer:
Condition for sustained interference :

• The two source should emf the light wave continuously.
• The light wave should be of same wave length.
• The two light source should be narrow.
• The light waves emitted should be of the same amplitude.
• The two source of light must lie very close to each other.

Long Answer Type Questions

Question No. 19 to 24 are long answers type question. Each question of this category carries 5 marks. Answer any three (3) question. (3 x 5 = 15)

Question 19.
Explain any two of the following terms :
(i) Ground Waves
(ii) Sky Waves
(iii) Space Waves.
Answer:
(i) Ground Waves : The ground waves which progress along the surface of the earth are called ground waves or surface waves. The ground propagation is suitable for low and medium frequency, i.e., up to 2 MHz only, hence it is also called medium wave propagation: The maximum range of ground or surface wave propagation depends on :

• The frequency of the radio waves and
• Power of the transmitter.

(ii) The Sky Waves : The sky waves are the radio waves of frequency between 2 MHz to 30 MHz. The ionosphere reflects those radio waves so that they can propagate through atmosphere. The sky waves propagation is also known as ionosphere propagation.

The sky waves are used for very long distance radio communication at medium high frequencies (i.e. at medium waves and short waves). The radio waves can cover a distance approximately 400 km in a single reflection from the ionosphere.

(iii) Space Waves : The space waves are the radio waves of very high frequency (i.e. between 30 MHz to 300 MHz or more). The space waves can travel through atmosphere from transmitter antenna to receive antenna either directly or after reflection from ground in the earth’s troposphere region. It is also called Tropospherical propagation or line of sight propagation. This propagation is limited (a) to the line of sight distance, (b) by the curvature of the earth.

It is utilized in T.V. communication, Radar communication, etc.

Question 20.
Find an expression for the electric field strength at a distant point situated (i) on the axis and (ii) along the equatorial line of an electric dipole.
Answer:
Consider an electric dipole AB. The charges -q and +q of dipole are situated at A and B respectively. The separation between the charges is 21.

Electric dipole moment, p = Q 21
The direction of dipole moment is from -q to + Q

(i) At axial or end-on position : Consider a point P on the axis of dipole at a distance r from mid-point O of electric dipole.

The distance of point P from charge + q is BP = r – l and distance of point P from charge – q is, AP = r + l. Let E₁ and E₂ be the electric field strengths at point P due to charges +<7 and -q respectively. We know that the direction of electric field due to a point charge is away from positive charge and towards the negative charge. Therefore

Clearly the directions of electric field strengths \(\overrightarrow{\mathrm{E}}_{1}\) and \(\overrightarrow{\mathrm{E}}_{2}\) are along the same line but opposite to each other and E₁ > E₂ because positive charge is nearer.

∴ The resultant electric field due to electric dipole has magnitude equal to the difference of E₁ and E₂ and direction from 13 to P i.e.

If the dipole is infinitely small and point P is far away from the dipole, then r >> l, therefore equation (i) may be expressed as
\(\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p r}{r^{4}} \text { or } \mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) …………………………… (ii)

This is the expression for the electric field strength at axial position due to a short electric dipole.

Question 21.
Explain the principle and working of a Cyclotron with the help of a neat diagram. Write the expression for Cyclotron frequency.
Or,With the help of a labelled diagram, state the underlying principle of a Cyclotron. Explain clearly how it works to accelerate the charged particles.

Show that Cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.
Answer:
(a) Cyclotron : The cyclotron, devised by Lawrence and Livingston, is a device for accelerating ions to high speed by the repeated application of accelerating potentials.

Construction:
The cyclotron consists of two flat semi-circular metal boxes called ‘dees’ and are arranged with a small gap between them. A source of ions is located near the mid-point of the gap between the dees (fig). The dees are connected to the terminals of a radio frequency oscillator, so that a high frequency alternating potential of several million cycles per second exists between the dees. Thus dees act as electrodes. The dees are enclosed in an insulated metal box containing gas at low pressure. The whole apparatus is placed between the poles of a strong electro¬magnet which provides a magnetic field perpendicular to the plane of the dees.

Working :
The principle of action of the apparatus is shown in fig.The positive ions produced from a source S at the centre are accelerated by a dees which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dees, the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with high velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the deeS where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.

The function of electric field is to accelerae the charged particle and so to impart energy to the charged particle.

The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field.

Expression for Period of Revolution and Frequency :

Suppose the positive ion with charge q moves in a dees with a velocity v, then,
\(q v \mathrm{B}=\frac{m v^{2}}{r} \text { or }=\frac{m v}{q \mathrm{B}}\) …………………………… (i)

where m is the mass and r is the radius of the path of ion in the dees and B is the strength of the magnetic field. The angular velocity to of the ion is given by.
\(\omega=\frac{v}{r}=\frac{q \mathrm{B}}{m}( \text { from eq. }\) ………………… (ii)

The time taken by the ion in describing a semi-circle, i.e., in turning through an angle π is.
\(t=\frac{\pi}{\omega}=\frac{\pi m}{\mathrm{B} q}\)

Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on increasing with increase in the radius (from eQuestion i) when it moves from one dees to the other, yet it take the same time in each dees.

From eQuestion (iii) it is clear that for a particular ion, \(\frac{m}{q}\)

being known, B can be calculated for producing resonance with the high frequency alternating potential.

(b) Resonance Condition : The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particles within the dees. This is called resonance condition.

Now for the cyclotron to work, the applied alternating potential should also have the same semiperiodic time (T/2) as that taken by the ion to cross either dees, i.e.,

This is the expression for period of revolution. Obviously, period of revolution is independent of speed of charged particle and radius of circular path.

∴ Frequency of revolution of particles.
\(f=\frac{1}{T}=\frac{q B}{2 \pi m}\)

This frequency is called the cyclotron frequency. Clearly the cyclotron frequency is independent of speed of particle.

(c) Expression for K.E. attained : If R be the radius of the path and Vmax, the velocity of the ion when it leaves the periphery, then in accordance with eq (ii)
\(v_{\max }=\frac{q \mathrm{BR}}{m}\) …………………………….. (vi)

The kinetic energy of the ion when it leaves the apparatus is,
\(\text { K.E. }=\frac{1}{2} m v^{2} \max =\frac{q^{2} B^{2} R^{2}}{2 m}\) ………………….. (vii)

When charged particle crosses the gap between dees it gains KE = qV

In one revolution, it crosses the gap twice, therefore if it completes ^-revolutions before emerging the dees, the kinetic energy gained = 2nqV ………………. (viii)
\(\text { Thus, K.E. }=\frac{g^{2} \mathrm{B}^{2} \mathrm{R}^{2}}{2 m}=2 n q \mathrm{V}\)

Question 22.
Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid.
Answer:
Magnetic field due to a toroidal solenoid : A long solenoid shaped in the form of closed ring is called toroidal solenoid (or endless solenoid). Let n be the number of turns per unit of toroid and I the current flowing through it. The current causes the magnetic field inside the turns of the solenoid. The magnetic lines of force inside the toroid are in the form of concentric circle. By symmetry of the magnetic field it has the same magnitude at each point of circle and is along the tangent at every point on the circle.

For points inside the core of toroid: Consider a circle of radius r in the region enclosed by turns of toroid. Now we apply Ampere’s circuital law to this circular path, i.e.,
\(\vec{f} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0} I\) ………………. (i)

\(\vec{f} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=f B d l \cos 0=B \cdot 2 \pi r\) Length of toroid = 2πr Number of turns in toroid = n (2πr) and current in one- turn = I.

∴ Current enclosed by circular path = (n 2πr) • 1
∴ Equation (i) gives
\(\mathrm{B} 2 \pi r=\mathrm{m}_{0}(n 2 \pi r l) \Rightarrow \mathrm{B}=\mu_{0} n \mathrm{l}\)

Derivation of formula for magnetic field due to a current carrying wire using Biot-Savart law : Consider a wire EF carrying current I in upward direction. The point of observation is P at a finite distance R from the wire. If PM is perpendicular dropped from P on wire; then PM = R. The wire may be supposed to be formed of a large number of small current elements. Consider a small element CD of length δf at a distance f from M.

\(\text { Let } \angle \mathrm{CPM}=\phi \text { and } \angle \mathrm{CPD}=\delta \phi, \angle \mathrm{PDM}=\theta\) The length δf is very small, so that ∠PCM may also be taken as equal to θ.

The perpendicular dropped from C on PD is CN.

The angle formed between element I \(\overrightarrow{\delta l}\) and \(\overrightarrow{r}\) (= cp is (π – θ), Therefore according to Biot-Savart law, the magnetic field due to current element I \(\overrightarrow{\delta l}\) at P is

If the wire is of finite length and its ends make angles a and p with line MP, then net magnetic field (B) at P is obtained by summing over magnetic fields due to all current elements, i.e.,

This is expression for magnetic field due to current carrying wire of finite length (or very long), then α = β ⇒ π/2.

Question 23.
Draw the labelled ray diagram for the formation of image by a compound microscope.
Derive an expression for its total magnification (or magnifying power), when the find image is formed at the near point.
Why both objective and eye piece of a compound microscope must have their short focal lengths?
Answer:
Compound Microscope :
It consists of a long cylindrical tube, containing at one end of a convex lens of small aperture and small focal length. This is called the objective lens (O). At the other end of the tube another co¬axial smaller and wide tube is fitted which carries a convex lens (E) at its outer end. This lens is towards the eye and is called the eye-piece. The focal length and aperture of eye piece are somewhat large than those of objective lens. Cross¬wires are mounted at a definite distance before the eye piece. The entire tube can be moved forward and backward by the rack and pinion arrangement.

Magnifying power of a microscope:
is defined as the ratio of angle ((3) subtended by final image on the eye to the angle (a) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e.,

Magnifying power \(M=\frac{\beta}{\alpha}\)

As object is very small, angles α and β are very small and so tan α = α and tan β = β. By  efinition the object AB is placed at the least distance of distinct vision

Question 24.
State the principle of working of p-n diode as a rectifier. Explain with the help of a circuit diagram, the use of p-n diode as full wave rectifier. Draw a sketch of the input and output wave forms. Or, Draw a circuit diagram of a full wave rectifier. Explain the working principle. Draw the input/output wave forms indicating clearly the functions of the two diodes used.
Answer:
Rectification :
Rectification means conversion of a.c. into d.c. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode conducts only when it is forward biased; it does not conduct when reverse biased.

In a full wave rectifier, if input frequency is / hertz, then output frequency will be 2/ hertz because for each cycle of input, two positive half cycles of output are obtained.

Working:
The AC input voltage across secondary s₁ and s₂ cli anges polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal s₁ is positive relative to centre tap O and s₂ is negative relative to O. Then diode D₁ is forward biased and diode D₂ is reverse biased. Therefore diode D₁ conducts while diode D₂ does not. The direction of current (i₁) due to diode D₁ in load rsistance R_L is directed from A to 13. In next half cycle, the terminal s is negative and s₂ is positive relative to centre tap O.

The diode D is reverse biased and diode D₂ is forward biased. Therefore, diode D₂ conducts while D₁ does not. The direction of current (i₂) due to diode D₂ in load resistance R_L is still from A to B. Thus the current in load resistance R_L is in the same direction for both half cycles of input AC voltage. Thus for input AC signal the output current is a continuous series of unidirectional pulses.

In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 1.
Single-celled animals are said to be immortal because
(a) they grow indefinitely in size
(b) they can tolerate any degree of change in temperature
(c) they can reproduce throughout their life span
(d) they continue to live as their daughter cells.
Answer:
(d) they continue to live as their daughter cells.

Question 2.
Which of the following has the longest life span?
(a) Banyan tree
(b) tortoise
(c) parrot
(d) Elephant
Answer:
(a) Banyan tree

Question 3.
Select the option which arranges the given organisms in ascending order of their life span.
(a) Parrot < Crow < Butterfly < Banyan tree
(b) Butterfly < Crow < Parrot < Crocodile
(c) Fruit fly < Crocodile < Parrot < Banyan tree
(d) Parrot < Tortoise < Dog < Crow
Answer:
(c) Fruit fly < Crocodile < Parrot < Banyan tree

Question 4.
________ is a life process that is not essential for an individual’s survival but for survival of the species.
(a) Growth
(b) Reproduction
(c) Respiration
(d) Nutrition
Answer:
(b) Reproduction

Question 5.
‘Clones’ are individuals that have exactly the same
(a) Lifespan
(b) physiology
(c) growth rate
(d) genetic makeup.
Answer:
(d) genetic makeup.

Question 6.
Which one of the following processes results in the formation of clone of bacteria?
(a) Regeneration
(b) Budding
(c) Binary fission
(d) Fragmentation
Answer:
(c) Binary fission

Question 7.
Asexual reproduction is seen in members of Kingdom
(a) Monera
(b) Plantae
(c) Animalia
(d) All of these.
Answer:
(d) All of these.

Question 8.
During binary fission in Amoeba which of the following organelles is duplicated?
(a) Plasma membrane
(b) Nucleus
(c) Contractile
(d) All of these
Answer:
(b) Nucleus

Question 9.
Vegetative propagation is the term used for
(a) sexual reproduction in animals
(b) sexual reproduction in plants
(c) asexual reproduction in animals
(d) asexual reproduction in plants.
Answer:
(d) asexual reproduction in plants.

Question 10.
Which of the following is not used for vegetative propagation?
(a) Bud
(b) Bulbil
(c) Turion
(d) Antherozoid
Answer:
(d) Antherozoid

Question 11.
Identify the given organism and find its maximum life span.

(a) Sparrow, 25 years
(b) Crow, 30 years
(c) Crow, 15 years
(d) Eagle, 40 years
Answer:
(c) Crow, 15 years

Question 12.
Which of the following options shows two plants in which new plantlets arise from the same organ?
(a) Dahlia and ginger
(b) Potato and sweet potato
(c) Dahlia and rose
(d) Potato and sugarcane
Answer:
(d) Potato and sugarcane

Question 13.
Refer to the given figure and identify X in it.

(a) Offset
(b) Eyes
(c) Runner
(d) Bulb
Answer:
(b) Eyes

Question 14.
Fleshy buds produced in the axil of leaves, which grow to form new plants when shed and fall on ground, are called
(a) bulbs
(b) bulbils
(c) tubers
(d) offsets.
Answer:
(b) bulbils

Question 15.
In which one pair, both the plants can be vegetatively propagated by leaf pieces?
(a) Bryophyllum and Kalanchoe
(b) Chrysanthemum and Agave
(c) Agave and Dioscorea
(d) Bryophyllium and Asparagus
Answer:
(a) Bryophyllum and Kalanchoe

Question 16.
Identify the given vegetative propagule.

(a) Bulb
(b) Runner
(c) Rhizome
(d) Bulbil
Answer:
(d) Bulbil

Question 17.
If a leaf cell of Agave has x chromosomes then what will
be the number of chromosomes in a cell of its bulbil?
(a) 2 x
(b) x/2
(c) x/4
(d) x
Answer:
(d) x

Question 18.
Which of the following cannot serve as a vegetative propagule?
(a) A piece of potato tuber with eyes
(b) A middle piece of sugarcane internode
(c) A piece of ginger rhizome
(d) A marginal piece of Bryophyllum leaf
Answer:
(b) A middle piece of sugarcane internode

Question 19.
Which of the following options correctly identifies artificial and natural methods of vegetative propagation?
Artificial methods – Natural methods
(a) Grafting – Cutting
(b) Layering – Bulbils
(c) Offset – Tissue culture
(d) Tubers – Rhizomes
Answer:
(b) Layering – Bulbils

Question 20.
Sexual reproduction is considered more beneficial than asexual reproduction because
(a) it is not affected by adverse environmental conditions
(b) fertilization is a chance factor
(c) it rapidly multiplies the population
(d) it assists in evolution by producing variations.
Answer:
(d) it assists in evolution by producing variations.

Question 21.
The growth phase of an organism before attaining sexual maturity is referred to as
(a) juvenile phase
(b) vegetative phase
(c) both (a) and (b)
(d) none of these.
Answer:
(c) both (a) and (b)

Question 22.
Select the monocarpic plant out of the following.
(a) Bamboo
(b) Lite hi
(c) Mango
(d) All of these
Answer:
(a) Bamboo

Question 23.
Clear cut vegetative, reproductive and senescent phases cannot be observed in
(a) annual plants
(b) perennial plants
(c) biennial plants
(d) ephemeral plants.
Answer:
(b) perennial plants

Question 24.
Strobilanthus kunthiana flowers once in
(a) 5 years
(b) 12 years
(c) 20 years
(d) 50 years.
Answer:
(b) 12 years

Question 25.
Strobilanthus kunthiana differs from bamboo in
(a) being monocarpic
(b) length of juvenile phase
(c) being polycarpic
(d) none of these.
Answer:
(b) length of juvenile phase

Question 26.
Oestrous cycle is reported in
(a) cows and sheep
(b) humans and monkeys
(c) chimpanzees and gorillas
(d) none of these.
Answer:
(a) cows and sheep

Question 27.
Which of the following animals show menstrual cycle ?
(a) Gorillas and chimpanzees
(b) Monkeys and humans
(c) Orangutans and monkeys
(d) All of these
Answer:
(d) All of these

Question 28.
Senescent phase of an organism’s life span can be recognised by
(a) slow metabolism
(b) cessation of reproduction
(c) decreased immunity
(d) all of these
Answer:
(d) all of these

Question 29.
If a fungal thallus has both male and female reproductive structures, it will be called
(a) heterothallic
(b) homothallic
(c) dioecious
(d) monoecious
Answer:
(b) homothallic

Question 30.
Staminate flowers produce
(a) eggs
(b) antherozoids
(c) fruits
(d) all of these
Answer:
(b) antherozoids

Question 31.
Which of the following is a unisexual organisam?

Answer:

Question 32.
Which of the following groups is formed only of the hermaphrodite organisms?
(a) Earthworm, tapeworm, housefly, frog
(b) Earthworm, tapeworm, sea horse, housefly
(c) Earthworm, leech, sponge, roundworm
(d) Earthworm, tapeworm, leech, sponge
Answer:
(d) Earthworm, tapeworm, leech, sponge

Question 33.
Which of the following options shows bisexual animals only?
(a) Amoeba, sponge, leech
(b) Sponge, cockroach, Amoeba
(c) Earthworm, sponge, leech
(d) Tapeworm, earthworm, honeybee
Answer:
(c) Earthworm, sponge, leech

Question 34.
Read the following statements and select the incorrect one.
(a) Cucurbits and coconuts are monoecious plants.
(b) Papayas and date palms are dioecious plants.
(c) Leeches and tapeworms are bisexual animals.
(d) Sponges and coelenterates are unisexual animals.
Answer:
(d) Sponges and coelenterates are unisexual animals.

Question 35.
Meiosis does not occur in
(a) asexually reproducing diploid individuals
(b) sexually reproducing haploid individuals
(c) sexually reproducing diploid individuals
(d) all of these.
Answer:
(a) asexually reproducing diploid individuals

Question 36.
A diploid parent plant body produces ________ gametes and a haploid parent plant body produces ________ gametes.
(a) diploid, haploid
(b) haploid, diploid
(c) diploid, diploid
(d) haploid, haploid
Answer:
(d) haploid, haploid

Question 37.
Which of the following organisms has the highest number of chromosomes?
(a) Housefly
(b) Butterfly
(c) Ophioglossum
(d) Onion
Answer:
(c) Ophioglossum

Question 38.
In maize, a meiocyte has 20 chromosomes. What will be the number of chromosomes in its somatic cell?
(a) 40
(b) 30
(c) 20
(d) 10
Answer:
(c) 20

Question 39.
If a butterfly has chromosome number 360 in its meiocyte (2n). What will be the chromosome number in its gametes?
(a) 380
(b) 190
(c) 95
(d) 760
Answer:
(b) 190

Question 40.
In flowering plants, both male and female gametes are non-motile. The method to bring them together for fertilisation is
(a) water
(b) air
(c) pollination
(d) apomixis
Answer:
(c) pollination

Question 41.
Development of new individual from female gamete without fertilisation is termed as
(a) syngamy
(b) embryogenesis
(c) oogamy
(d) parthenogenesis.
Answer:
(d) parthenogenesis.

Question 42.
Fertilisation cannot occur in absence of surface water in
(a) Fucus
(b) Funaria
(c) Marsilea
(d) all of these.
Answer:
(d) all of these.

Question 43.
Spirogyra is a sexually reproducing alga in which vegetative thallus is haploid. In Spirogyra, meiosis
(a) never occurs
(b) occurs at time of gamete production
(c) occurs after fertilisation
(d) occurs during vegetative growth.
Answer:
(c) occurs after fertilisation

Question 44.
Life begin in all sexually reproducing organisms as a
(a) single-celled zygote
(b) double-celled zygote
(c) haploid zygote
(d) haploid gametes.
Answer:
(a) single-celled zygote

Question 45.
Which of the following is not correct regarding sexual reproduction ?
(a) It is usually biparental.
(b) Gametes are always formed.
(c) It is a slow process
(d) It involves only mitosis.
Answer:
(a) It is usually biparental.

Question 46.
Offsprings of oviparous animals are at greater risk of survival as compared to those of viviparous animals because
(a) proper embryonic care and protection is absent
(b) embryo does not develop completely
(c) progenies are of smaller size
(d) genetic variations do not occur.
Answer:
(a) proper embryonic care and protection is absent

Question 47.
Deposition of calcareous shell around zygote occurs in
(a) birds and reptiles
(b) birds and mammals
(c) mammals and reptiles
(d) all of these.
Answer:
(a) birds and reptiles

Question 48.
Select the option which shows viviparous animals only,
(a) Lizard, Turtile
(b) Platypus, Crocodile
(c) Cow, Crocodile
(d) Whale, Mouse
Answer:
(d) Whale, Mouse

Question 49.
Which of the following animals give birth to young ones?
(a) Ornithorhynchus and Echidna
(b) Macropus and Pteropus
(c) Balaenoptera and Homo sapiens
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

Question 50.
Viviparity is found in
(a) Sharks
(b) lizards
(c) frogs
(d) birds
Answer:
(a) Sharks

Question 51.
In which of the following plants, sepals do not fall off after fertilisation and remain attached to the fruit?
(a) Brinjal
(b) Cucumber
(c) Papaya
(d) Bitter gourd
Answer:
(a) Brinjal

Question 52.
Which of the labelled parts in the transverse section of tomato fruit, is/are diploid?

(a) X
(b) Y
(c) Both X and Y
(d) None of these
Answer:
(c) Both X and Y

Question 53.
The wall of the ovary forms
(a) pericarp
(b) fruit wall
(c) fruit
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 54.
The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because
(a) offspring do not possess exact copies of parental DNA
(b) DNA of only one parent is copied and passed on to the offspring
(c) offspring are formed at different times
(d) DNA of parent and offspring are completely different
Answer:
(a) offspring do not possess exact copies of parental DNA

Question 55.
The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively
(a) 12,24,12
(b) 24,12,12
(c) 12, 24, 24
(d) 24, 12, 24.
Answer:
(c) 12, 24, 24

Question 56.
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) nodes are shorter than intemodes
(b) nodes have meristematic cells
(c) nodes are located near the soil
(d) nodes have non-photosynthetic cells.
Answer:
(b) nodes have meristematic cells

Question 57.
There is no natural death in single celled organisms like Amoeba and bacteria because
(a) they cannot reproduce sexually
(b) they reproduce be binary fission
(c) parental body is distributed among the offspring
(d) they are microscopic.
Answer:
(c) parental body is distributed among the offspring

Question 58.
There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) the habitat and morphology of the organism
(b) morphology of the organism
(c) morphology and physiology of the organism
(d) the organisms habitat, physiology and genetic makeup.
Answer:
(d) the organisms habitat, physiology and genetic makeup.

Question 59.
Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer:
(b) Embryo development

Question 60.
The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the microspore mother cells of the same plant shall be
(a) 20
(b) 10
(c) 40
(d) 15
Answer:
(a) 20

Bihar Board 12th Biology Model Papers

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Time : 3 Hours 15 Min
Total Marks : 70

Instructions for the candidates:

1. Candidates are required to give answers in their own words as far as practicable.
2. Figures in the right-hand margin indicate full marks.
3. While answering the questions, the candidate should adhere to the word limit as for as practicable.
4. 15 Minutes of extra time has been allotted for the candidates to read the questions carefully.
5. This question paper is divided into two sections: Section – A and Section – B
6. In Section – A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/ Blade/Nail on OMR Sheet otherwise result will be treated as invalid.
7. In Section – B, there are Non-objective type questions. There are 18 Short answer type questions, out of which any 10 questions are to be answered. Each question carries 2 marks. Apart from this, there are 6 long answer type questions, out of which any 3 of them are to be answered. Each question carries 5 marks
8. Use of any electronic device is prohibited.

Objective Type Questions

There are 1 to 35 objective type questions with 4 options. Choose the correct option which is to be answered on OMR Sheet. (35 x 1 = 35)

Question 1.
Pollination by birds is called
(a) Entomophily
(b) Cheiropterophily
(c) Ornithophily
(d) All of the above
Answer:
(c) Ornithophily

Question 2.
Which one is the hotspot of biodiversity?
(a) Eastern Ghat
(b) Western Ghat
(c) Aravalli Hills
(d) None of the above
Answer:
(b) Western Ghat

Question 3.
Who is considered as Father of Genetics?
(a) Morgan
(b) Darwin
(e) Hugo-de-Vries
(d) Mendel
Answer:
(d) Mendel

Question 4.
Noise pollution is measured in
(a) Heartz
(b) Fathoms
(c) Decibels
(d) Nanometer
Answer:
(c) Decibels

Question 5.
Indicator of water pollution is
(a) Entamoeba histolytica
(b) E. Coli
(c) Vibrio cholera
(d) All of the above
Answer:
(b) E. Coli

Question 6.
mRNA formation takes place in
(a) Ribosome
(b) Nucleus
(c) Cytoplasm
(d) Mitochondria
Answer:
(b) Nucleus

Question 7.
Cancer affecting connective tissue is called
(a) Lymphoma
(b) Sarcoma
(c) Carcinoma
(d) None of the above
Answer:
(b) Sarcoma

Question 8.
IgM antibody is found in
(a) Serum
(b) Saliva
(c) Blood
(d) All of the above
Answer:
(c) Blood

Question 9.
How many Peptide chains are found in each antibody molecule?
(a) 4
(b) 3
(c) 2
(d) None of the above
Answer:
(a) 4

Question 10.
Green plants are known as producers, which convert
(a) Chemical energy into light energy.
(b) Light energy into chemical energy
(c) All of the above
(d) None of the above
Answer:
(b) Light energy into chemical energy

Question 11.
Which of the following cry gene prevents crop from borer
(a) Cry AC
(b) Cry / Ab
(c) Cry // Ab
(d) All of the above
Answer:
(d) All of the above

Question 12.
Human blood groups are
(a) A, B, AB, O
(b) A, B, C, O
(c) B, C, O
(d) None of the above
Answer:
(a) A, B, AB, O

Question 13.
Khorana got Nobel prize for
(a) Discovery of RNA
(b) Discovery of DNA
(c) Chemical synthesis of gene
(d) None ‘
Answer:
(c) Chemical synthesis of gene

Question 14.
Secondary pollutant is
(a) CO
(b) C02
(c) PAN
(d) None of these
Answer:
(c) PAN

Question 15.
Chipko movement is launched for protection of
(a) Grass land
(b) Forest
(c) Animal Community
(d) All of the above
Answer:
(b) Forest

Question 16.
TTxttis
(a) Test cross
(b) Back cross
(c) Hybridization
(d) None of the above
Answer:
(a) Test cross

Question 17.
How many types of DNA polymerases are associated with eukaryotic cell ?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 18.
Example of In Situ conservation is
(a) Zoological garden
(b) Botanical garden
(c) Biological safari park
(d) All of the above
Answer:
(d) All of the above

Question 19.
Which of the following is diploid ?
(a) Male gamete
(b) Ovum
(c) Zygote
(d) All of the above
Answer:
(c) Zygote

Question 20.
Pattern of any population depends on
(a) Density
(b) Distribution
(c) Specific structure
(d) All of the above
Answer:
(d) All of the above

Question 21.
Montreal Protocol is related to control the deleterious effect of
(a) Acid rain
(b) Ozone depletion
(c) Deforestation
(d) Global warming
Answer:
(b) Ozone depletion

Question 22.
Fertilization takes place in
(a) Ovary
(b) Uterus
(c) Fallopian tube
(d) All of the above
Answer:
(c) Fallopian tube

Question 23.
Hugo de-Vries worked on which plant ?
(a) Pea
(b) Evening Primrose
(c) Both
(d) None
Answer:
(b) Evening Primrose

Question 24.
Salmonella is associated with
(a) Polio
(b) T.B,
(c) Typhoid
(d) All of the above
Answer:
(c) Typhoid

Question 25.
AIDS is identified first of all in
(a) USA
(b) France
(c) Russia
(d) All of the above
Answer:
(a) USA

Question 26.
Bones of fore limbs of Bat, and humans are
(a) Homologous
(b) Analogous
(c) Both
(d) None
Answer:
(a) Homologous

Question 27.
Which enzyme is used to join cut ends of DNA?
(a) Ligase
(b) Cellulose
(c) Pectinase
(d) None of the above
Answer:
(a) Ligase

Question 28.
Which of the following serve as biofertilizer in paddy fields?
(a) Yeast
(b) Bacteria
(c) Cyano bacteria
(d) Fungi
Answer:
(c) Cyano bacteria

Question 29.
Which of the following is used in preparing Cheese?
(a) Viruses
(b) Algae
(c) Fungi
(d) Micorobes
Answer:
(d) Micorobes

Question 30.
Statins used for lowering blood cholesterol level
are extracted from
(a) Yeast
(b) Algae
(c) Virus
(d) All of the above
Answer:
(a) Yeast

Question 31.
Microbes are found in
(a) air
(b) water
(c) soil
(d) All of the above
Answer:
(d) All of the above

Question 32.
Ranikhet disease is connected with
(a) Pigs
(b) Cow
(c) Fish
(d) Hens
Answer:
(d) Hens

Question 33.
Induced breeding is done in case of
(a) Apiculture
(b) Pisci culture
(c) Lac culture
(d) Sericulture
Answer:
(b) Pisci culture

Question 34.
Taichung Native-1 is a variety of
(a) Rice
(b) Wheat
(c) Maize
(d) None
Answer:
(a) Rice

Question 35.
Zeba cattle is
(a) Goat
(b) Cat
(c) Cow
(d) Sheep
Answer:
(c) Cow

Non-Objective Type Questions

Short Answer Type Questions

In this section, there are 18 short answer type question (each carryies 2 marks) out of which answers any ten (10) questions. (10 x 2 = 20)

Question 1.
What is co-dominance? Explain with example.
Answer:
In co-dominance both dominant alleles of a gene express themselves equally in F, hybrids. The phenotype ratio matches with genotypic ratio i.e, 1: 2: 1 in F2 generation of offspring. Example of co-dominance can be seen in goat colour in

Question 2.
What is significance of double fertilization?
Answer:
Significance of double fertilization:
(i) Double fertilization is very important in angiospermic plants. If only syngamy occurs in plants and triple fusion does not occur, only 3 zygote will be formed. Endosperm will not be formed and this will result seed with undeveloped embryo or seed without embryo.

(ii) Endosperm is formed due to double fertilization. This provides nutrition to embryo and it contains maternal and paternal chromosomes. A body shows physiological aggressiveness due to hybrid vigour in inner endospermic cells.

Question 3.
What is radio-active pollution? What are the sources of radio-active pollution?
Answer:
Radio-active Pollution: Certain elements such as Radium, Thorium, Uranium emit protons (Alpha particles), electrons (Beta particles) and Gamma particles by disintegration of atomic nuclei. This phenomenon is called radio-activity and these elements are known as radio-active. When the radio-active radiations contaminate to water, air, soil and food materials it is called radio-active pollution.

Question 4.
Define Commensalism and Parasitism.
Answer:
Commensalism: An interaction between two species in which one is benefitted and the other one is neither harmed nor benefitted.
e.g. Orchids growing on trunk/branch of a tree.

Parasitism: An interaction between two species where parasite not only obtains food from its host but also completes its life cycle on the host. e.g. Bacteria, Fungi etc.

Question 5.
What are the ways for controlling amoebiosis?
Answer:
Amoebiosis is caused by a protozoan parasite Entamoeba histolytica, in which pain in stomach and stool with mucus and blood are seen as specific symptom.
Following are the control major of amoebiosis:

1. Patient should take rest and after some interval electrolytic fluid must be taken.
2. For control of disease, emetin, vioform, chinoform and diodoquin should be taken as drug and antibiotics like Terramycine, Erythromycine, Metronidazole and Aureomycin may be helpful in treatment and control of this disease.

Question 6.
What is Jumping Gene (Transposons)?
Answer:
These are capable of changing their locations from one place to another, so it is called jumping genes. It is a mobile fragment of DNA that can change position in a genome, a cause of sequence insertions and deletions in some organisms.

Question 7.
Name four living fossils.
Answer:
A living fossil is a living animal of ancient origin with many primitive characters. A living fossil has been living as such from the time of origin without many changes.
Examples are:

• Peripatus of phylum Arthropoda.
• Limulus of phylum Arthropoda.
• Latimeria-a bony fish.
• Sphenodon-a reptile.

Question 8.
Write any four benefits of micro-propagation.
Answer:
Benefits of micro-propagation:

• By such method crop production increases.
• By such method storage of germplasm can be taken place.
• By Meristem tip culture method disease resistant plants are produced.
• Micropropagation is also beneficial in conservation of biodiversity.

Question 9.
What is fishery? Name few common freshwater and marine edible fishes.
Answer:
Fishery: It is an industry concerned with catching, processing or selling of fish, shell fish or other aquatic animals.
Freshwater edible fishes:

• Catla
• Rohu
• Cirrhina mrigala.

Marine water edible fishes: (a) Hilsa (b) Sardines (c) Mackerel (d) Pomfrets.

Question 10.
Write short notes on gene cloning.
Answer:
The production of living structures genetically identical to their parent structure is known as gene cloning. It is therefore an exact copy of a single living parent. Gene cloning is the act of making copies of single gene. Once a gene is identified clones can be used.

Question 11.
Why is the Human Genome Project called a megaproject?
Answer:
Human Genome Project is called a mega project because it is aimed to sequence every base in human genome. It has developed the ways of mapping the human genome at increasing fine level of precision and to store this information in database and develops tools for data analysis. The magnitude and the requirements for this project are more for opening up new areas and avenues.

Question 12.
What are the functions of t-RNA in the protein synthesis?
Answer:

1. It carries specific amino acids from the cytoplasm to the ribosomal sites for the formation of polypeptide chain according with the sequence specified by m-RNA.
2. The t-RNA charged with the amino acid serves as a adaptor molecule to decode the information on the m- RNA.
3. It comes to elongate the polypeptide chain by the addition of several newly synthesised amino acids.

Question 13.
Define homologous orgAnswer: Give examples.
Answer:
These are the organs of similar structure and origin but dissimilar in functions and form.This is due to common ancestary.
Examples:
(i) The forelimbs of a frog, the wings of a bird, legs of a horse, the hands of a man and the flipper of a whale are homologous organs because all of them have similar pattern of basic plan (pentadacty) i.e. same number of bones, muscles, nerves and blood vessels etc. but they do the different functions such as hopping (frog), flying (bird), running (horse), graping (man) and swimming (whale).

(ii) Phylloclade of Opuntia and Cladode of Ruscus are homologous organs as both are modified stems. Similarly, a thorn of Bougainvillea and a tendril of Cucurbita are homologous as both arise in axillary position.

Question 14.
What is green revolution? List two factors that have led green revolution in India.
Answer: Green revolution is the enormous increase in the food production due to intensive cultivation by use of improved varieties of seeds and fertilizers.
The factors that contributed to the green revolution in India are:

• The development and introduction of high yielding varieties of crop and extension of the high yielding varieties over large area in the country.
• Transfer of technology of scientific farming from research farm to village farmers.

Question 15.
What do you understand by restriction enzyme?
Answer:
Restriction enzymes: These enzymes are nucleases which cut DNA into short pieces containing identifiable genes at specific sites. These pieces are then introduced into plastids, yeasts or plant cells.
Ex. ECORI, Hind-III.

Question 16.
Write short notes on plasmids.
Answer:
A plasmid is small circular double stranded DNA molecule that is distinct from a cell, chromosomal DNA.
Plasmids naturally exist in bacterial cells and they also occur in some eukaryotes. Often, the genes carried in plasmid provide bacteria with genetic advantage, such as antibiotic resistance.

Question 17.
Write short notes on organic evolution.
Answer:
The process by which changes in the genetic composition of populations of organism occur in response to environmental changes is known as organic evolution. Examples of organic evolution include organs that are different in structure but perform similar function. A Dolphin’s flipper’ and human arms are example of organic evolution.

Question 18.
Discuss the concept of IPM.
Answer:
Integrated Pest Management (IPM) aims at minimum use of pesticides to prevent agro-chemical pollution and to adopt natural methods of pest control as far as possible. The natural methods of pest control are the part of a larger agricultural strategy and these include:

1. Use of resistant varieties of crop to local pests.
2. To carry the practice of crop rotation and improved sanitation method.
3. To adopt biological control method and starvation method.
4. To grow planting or a preferred target crop to lure away insects.
5. Use of sterilization strategy, mechanical control, repellents, chemical attractants and natural insecticides etc.

Long Answer Type Questions

There are 6 long answers type question (each carrying 05 marks) out of which answers any three questions. (5 x 3 = 15)

Question 19.
Discuss the characters chosen by Mendel in garden pea. Mention the advantages of selecting this plant for experiment.
Answer:
Gregor Mendel, conducted hybridisation experiments on garden peas for seven years (1856-1863) and proposed the laws of inheritance. During Mendel’s investigations into inheritance patterns it was for the first time that statistical analysis and mathematical logic were applied to problems in biology. His experiments had a large sampling size, which have greater credibility to the data

that he collected. Also, the confirmation of his inferences from experiments on successive generations of his test plants, proved that his results pointed to general rules of inheritance rather than being unsubstantiated ideas. Mendel selected only pea plants for his experiment due to presence of many types of contrasting characters like:

• Round and wrinkled seeds
• Yellow and green seeds
• Violet and white flowers
• Inflated and constricted pods
• Green and yellow pods
• Axillary and terminal flowers
• Tall and dwarf plants

Mendel investigated characters in the garden pea plants that were manifested as two opposite traits which allowed him to set up a basic frame work of rules governing inheritance. Mendel conducted such artificial pollination/ cross pollination experiments using several true breeding pea lines.

Advantages of selecting pea plants for experiment by Mendel:

1. Due to short height of pea plants, it was easy to do experiment on that.
2. Pea plants had a short life cycle so that results could be had within a year.
3. Pea plants produced many seeds in one generation which helped in drawing correct conclusions.
4. Artificial cross-pollination could be easily achieved because androecium and gynoecium were completely enclosed.
5. Pea plants having each of the seven characters he selected were readily available.
6. Mendel took only one character at one time for his experiment.
7. Pea plants are self pollinated and pure.

Question 20.
Give an accout of Neo-Darwinism/Modern concept of evolution.
Answer:
Neo-Darwinism is the improvement of Darwin’s orginal theory of Natural Selection in order to remove its defects.
Important mechanisms for Neo-Darwinism are:
(a) Variations (b) Natural selection (c) Isolation

(a) Variations: Variations which influence evolution are the ones that develop in the genetic machinery of the germ cells. They are formed due to Mutations, Recombinations, Migrations, Hybridisation and Genetic Drift.

(b) Natural Selection: It operates through differential reproduction. Individuals with advantageous variations survives, reach maturity and leave there genes in the gene pool through differential reproduction. Repeated reproductive selection of individuals with specific traits increases the frequency of their genes in the gene pool. Individuals with harmful neutral variations are eliminated at various stages.

(c) Isolation: It is the separation of a segment of population from the rest due to apearance of a barrier. Gene pool of the isolated population changes due to genetic drift and accumulation of different types of reproduction.

Question 21.
Give a labelled diagram of the electron microscopic view of a human sperm and describe it briefly.
Answer:
The human sperm is a microscopic, long, flagellated and haploid motile cell and is formed of head, neck, middle piece and tail.

(i) Head: It constitutes the anterior region of sperm containing nucleus and acrosome. The nucleus is a narrow, flat and oval structure consisting of densely packed nuclear chromatin material. The latter is formed of DNA and nucleo protein. A small pointed sheathy acrosome is found at the anterior region of head and is derived from the golgi bodies. It contains hydrolytic enzyme-hyalourinidase which helps the sperm to penetrate into ovum during fertilization by dispersing the cells of corona radiata.

(ii) Neck: It is very short or indistinguishable region lying in between head and middle piece. It contains the proximal and distal centriole with 9 + 0 arrangement of micro-tubules. The proximal centriole forms the spindle in the first cleavage divison of the fertilised ovum. The distal centriole gives rise to many fine micro-tubules which run upto sperm tail by passing through middle piece.

(iii) Middle piece: It is a cylindrical region lying behind the neck. It contains number of mitochondrial spiral (nebenkern) which encircles the axial fdament arising from distal centriole. The mitochondrial spiral contains the oxidative enzymes which provide energy for movement of sperm by the process called oxidative phosphorylations.

(iv) Tail: It is the longest part of the sperm and is formed of main central axial filament and the outer protoplasmic sheath with small amount of cytoplasm. At the posterior end of tail, the axial filament is naked without any sheath and is called as end piece. The tail undulates rapidly and provides mobility to the sperm with the head forward in the fluid medium.

Question 22.
What is deforestation? Write its causes and effects.
Answer:
Deforestation is the conversion of forest areas to non-forest ones. It is a serious threat to the quality of life, national economy and future of environment. According to an estimate, almost 40>percent forests have been lost in the tropics, compared to only one percent in the temperate region. The present-scenario of deforestation is particularly grim in India.

The main cause of decrease in forest cover in India is deforestation due to expansion of agriculture, urbanization, industrialization, excessive commercial use of timber, fuel, wood etc. The causes of deforestation are population explosion, fires, pests, grazing and gnawing mammals, weather, jhum cultivation, etc.

Effects of deforestation are as follows:

1. Soil erosion has increased;
2. Flood and droughts have become more frequent;
3. Pattern of rain fall is changing;
4. Land slides and avalanches are on the increase;
5. Climate has become warmer;
6. Forest dwilling species are becoming extinct;
7. Consumption of CO₂ and production of O₂ are getting adversely affected.

Question 23.
Give an account of Apiculture.
Answer:
The maintenance of hives of honey bees for the production of honey is called bee keeping or apiculture.
Bee keeping is a small scale industry for the production of honey which have following economic importance:

• Bee keeping gives honey which is an important nutritive substance.
• Bee-hives gives bee-wax which is utilized in making polishes.
• Bee keeping helps in pollination of various crops useful for mankind

Question 24.
Write short notes on Homology and Analogy organ, vestigial organ, Atavism and Connecting links.
Answer:
(i) Homology and Analogy: The organs of similar structure and origin but dissimilar in functions and forms are called as homologous organs and this phenomenon is called homology. The presence of homologous organs implies a common evolutionary origin of amphibians, reptiles, birds and mammals from some ancient fish ancestor. The homologous structures seen in successive generations indicate actual relationship and the possessors are the diverse descendants of common ancestry (Divergent evolution).

The organs that perform the same function but differ in their origin and structure, are called as analogous organs and the phenomenon is called analogy. The wings of an insect are analogous to those of birds and bats because they perform the same function but have dissimilar structure and origin. The wings of an insect are modified outgrowth of the body wall whereas wings of birds and bats are forelimbs.

These organs have arisen in evolutionary process through adaptation of quite different organisms to a similar mode of life (Convergent evolution). On the same line, the similarities in proteins are genes performing a given function among diverse organisms give clues to common ancestry. These biochemical similarities point to the same shared ancestry as structural similiarities among diverse organisms.

(ii) Vestigial organs: These are the reduced and functionless organs which are of no use to the possessor but they still persists generation after generation in reduced form in an individual. They were complete and functional in the ancestors e.g. appendix in man is considered as the remnant of large intestine (caecum) but it is considered to be storage organ for cellulose digestion in herbivorous mammals. The vestigial organs which used to perform a normal function in the ancestor but during the course of evolution, they have been reduced to vestiges.

(iii) Atavism: It is reappearance of ancestral characters other than parents in the newly born offspring, which have either completely disappeared or reduced. The reappearance of short tail in some babies, multiple mammals in some individuals and dense hairy body etc. are the examples of atavism. The reappearance of such ancestral characters favour evolution.

(iv) Connecting links: The organisms which possess the characters of two different groups of organisms are known as connecting links e.g., duckbilled platypus and spiny ant-eater serve as a connecting link between the mammals and reptiles. Similarly lung fishes (Protopterus) is a connecting link between fishes and amphibians.