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Students can Download Bio Botany Chapter 1 Living World Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 1 Living World

Samacheer Kalvi 11th Bio Botany Living World Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Which one of the following statement about virus is correct?
(a) Possess their own metabolic system
(b) They are facultative parasites
(c) They contain DNA or RNA
(d) Enzymes are present
Answer:
(c) They contain DNA or RNA

Question 2.
Identify the incorrect statement about the Gram positive bacteria.
(a) Teichoic acid absent
(b) High percentage of peptidoglycan is found in cell wall
(c) Cell wall is single layered
(d) Lipopolysaccharide is present in cell wall
Answer:
(a) Teichoic acid absent

Question 3.
Identify the Archaebacterium.
(a) Acetobacter
(b) Erwinia
(c) Treponema
(d) Methanobacterium
Answer:
(d) Methanobacterium

Question 4.
The correct statement regarding Blue green algae is
(a) lack of motile structures
(b) presence of cellulose in cell wall
(c) absence of mucilage around the thallus
(d) presence of floridean starch
Answer:
(a) lack of motile structures

Question 5.
Identify the correctly matched pair
(a) Actinomycete – 1. Late blight
(b) Mycoplasma – 2. Lumpy jaw
(c) Bacteria – 3. Crown gall
(d) Fungi – 4. Sandal spike
Answer:
(c) Bacteria – 3. Crown gall

Question 6.
Differentiate Homoiomerous and Heteromerous lichens.
Answer:
Homoiomerous and Heteromerous Lichens:

Homoiomerous	Heteromerous
1. Algal cells evenly distributed in the thallus	1. A distinct layer of algae and fungi present

Question 7.
Write the distinguishing features of Monera.
Answer:
Distinguishing Features of Monera:

1. This kingdom includes all prokaryotic organisms. Example: Mycoplasma, bacteria, actinomycetes and cyanobacteria.
2. These are microscopic. They do not have a true nucleus and membrane bound organelles.
   
3. Many other bacteria like Rhizobium, Azotobacter and Clostridium can fix atmospheric nitrogen into ammonia.
   
4. Some bacteria are parasites and others live as symbionts.

Question 8.
Why do farmers plant leguminous crops in crop rotations/mixed cropping?
Answer:
Rotations / Mixed Cropping:

1. Legumes have bacteria on nodules which are on the roots of the plants. The bacteria on the nodules takes nitrogen from the air and fixes it into the soil, so that other plants that require nitrogen can use it as well.
2. Rotation of crops improves the fertility of the soil and hence brings about an increase in the production of food grains.
3. Rotation of crops helps in saving on nitrogenous fertilizers, because leguminous plants grown during the rotation of crops can fix atmospheric nitrogen in the soil with the help of nitrogen fixing bacteria.
4. Crop rotation adds diversity to an operation.

Question 9.
Briefly discuss on five kingdom classification. Add a note on merits and demerits.
Answer:
R.H. Whittaker, an American taxonomist proposed five kingdom classification in the year 1969. The kingdoms include Monera, Protista, Fungi, Plantae and Animalia. The criteria adopted for the classification include cell structure, thallus organization, mode of nutrition, reproduction and phylogenetic relationship. A comparative account of the salient features of each kingdom is given in table.

Merits:

• The classification is based on the complexity of cell structure and organization of thallus.
• It is based on the mode of nutrition.
• Separation of fungi from plants.
• It shows the phylogeny of the organisms

Demerits:

• The kingdom monera and protista accommodate both autotrophic and heterotrophic organisms, cell wall lacking and cell wall bearing organisms thus making these two groups more heterogeneous.
• Viruses were not included in the system.

Question 10.
Give a general account on lichens.
Answer:
The symbiotic association between algae and fungi is called lichens. The algal partner is called Phycobiont and the fungal partner is called Mycobiont. Algae provide nutrition for fungal partner and also help in fixing the thallus to the substratum through rhizinae. Asexual reproduction takes place through fragmentation, Soredia and Isidia. Phycobionst reproduce by akinetes, hormogonia, aplanospore, etc. Mycobionts undergo sexual reproduction and produce ascocarps.
Classification:

1. Based on the habitat lichens are classified into following types: Corticolous (on bark) Lignicolous (on wood) Saxicolous (on rocks) Terricolous (on ground) Marine (on siliceous rocks of sea) and Fresh water (on siliceous rock of fresh water).
2. On the basis of morphology of the thallus they are divided into Leprose (a distinct fungal layer is absent) Crustose – crust like; Foliose – leaf like; Fruticose – branched pendulous shrub like.
3. The distribution of algal cells distinguishes lichens into two forms namely Homoiomerous (Algal cells evenly distributed in the thallus) and Fleteromerous (a distinct layer of algae and fungi present).
4. If the fungal partner of lichen belongs to ascomycetes, it is called Ascolichen and if it is basidiomycetes it is called Basidiolichen.

Textbook Activity Solved

Get a button mushroom. Draw diagram of the fruit body. Take a thin longitudinal section passing through the gill and observe the section under a microscope. Record your observations.

Samacheer Kalvi 11th Bio Botany Living World Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
Earth was formed around billion years ago …………… .
(a) 3.3
(b) 5.6
(c) 4.6
(d) 5.9
Answer:
(c) 4.6

Question 2.
According to Mora et al., in 2011, the number of estimated species on Earth is …………… .
(a) 8.7 million
(b) 9.7 million
(c) 7.7 million
(d) 9.7 billion
Answer:
(a) 8.7 million

Question 3.
Which of the following is NOT a prokaryote?
(a) Bacteria
(b) Blue green algae
(c) Oedogonium
(d) Nostoc
Answer:
(c) Oedogonium

Question 4.
Which of the following organism undergoes regeneration?
(a) Spirogyra
(b) Planaria
(c) Yeast
(d) Aspergillus
Answer:
(b) Planaria

Question 5.
Vaccination for small pox was discovered by …………… .
(a) W.M. Stanley
(b) Adolf Mayer
(c) Robert Koch
(d) Edward Jenner
Answer:
(d) Edward Jenner

Question 6.
Who coined the term ‘Bacteriophage’?
(a) F.W. Twort
(b) d’Herelle
(c) Ivanowsky
(d) Robert Gallo
Answer:
(b) d’Herelle

Question 7.
The size of TMV is …………… .
(a) 300 × 20 mm
(b) 300 × 200 µm
(c) 300 × 20 nm
(d) 300 × 20 Å
Answer:
(c) 300 × 20 nm

Question 8.
One nanometer equals to metres …………… .
(a) 10^-9
(b) 10^-6
(c) 10^-5
(d) 10^-12
Answer:
(a) 10^-9

Question 9.
Which is a non – living character of viruses?
(a) Undergoes mutation
(b) Host – specific
(c) Crystallized
(d) Irritability
Answer:
(c) Crystallized

Question 10.
According to David Baltimore, the viruses are classified …………… into classes.
(a) 6
(b) 5
(c) 7
(d) 8
Answer:
(c) 7

Question 11.
Identify the criteria not used in classifying viruses by Baltimore …………… .
(a) ss (or) ds
(b) use of RT
(c) capsid
(d) sense or antisense
Answer:
(c) capsid

Question 12.
Viruses with dsRNA is …………… .
(a) Toga viruses
(b) Retro viruses
(c) Reo viruses
(d) Rhabdo viruses
Answer:
(c) Reo viruses

Question 13.
Which of the plant virus contains DNA as genome?
(a) Tobacco mosaic virus
(b) Cauliflower mosaic virus
(c) Sugarcane mosaic virus
(d) Cucumber mosaic virus
Answer:
(b) Cauliflower mosaic virus

Question 14.
Parvo viruses have …………… .
(a) ssDNA
(b) dsDNA
(c) ssRNA
(d) dsRNA
Answer:
(a) ssDNA

Question 15.
Molecular weight of TMV is dalton …………… .
(a) 39 × 10⁶
(b) 39 × 10^-6
(c) 39 × 10⁹
(d) 39 × 10^-9
Answer:
(a) 39 × 10⁶

Question 16.
Approximate number of capsomeres in TMV is …………… .
(a) 3120
(b) 1203
(c) 2130
(d) 3021
Answer:
(c) 2130

Question 17.
The empty protein coat left outside after penetration is …………… .
(a) host
(b) ghost
(c) capsid
(d) capsomeres
Answer:
(b) ghost

Question 18.
The genome of viroid is …………… .
(a) linear ssRNA
(b) dumb – bell shaped ssRNA
(c) circular ssRNA
(d) linear dsRNA
Answer:
(c) circular ssRNA

Question 19.
Viroids were discovered by …………… .
(a) Ivanowsky
(b) Robert Gallo
(c) Diener
(d) d’Herelle
Answer:
(c) Diener

Question 20.
Mad cow disease is caused by …………… .
(a) viroids
(b) virusoids
(c) prions
(d) viruses
Answer:
(c) prions

Question 21.
Match the Following:

(a) 1- d, 2 – b, 3 – a, 4 – c
(b) 1 – c, 2 – d, 3 – a, 4 – b
(c) 1 – c, 2 – a, 3 – b, 4 – d
(d) 1 – d, 2 – a, 3 – b, 4 – c
Answer:
(c) 1- c, 2 – a, 3 – b, 4 – d

Question 22.
Identify the correct sequence regarding lytic cycle of viruses …………… .
(A) Penetration
(B) Adsorption
(C) Assembly
(D) Synthesis

(a) BADC
(b) CABD
(c) BDAC
(d) ADBC
Answer:
(a) BADC

Question 23.
Mycophages infect …………… .
(a) blue green algae
(b) bacteria
(c) fungi
(d) cyanobacteria
Answer:
(c) fungi

Question 24.
Rice tungro is caused by …………… .
(a) fungi
(b) bacteria
(c) mycoplasma
(d) viruses
Answer:
(d) viruses

Question 25.
Father of Botany …………… .
(a) Aristotle
(b) Theophrastus
(c) Lederberg
(d) Whittaker
Answer:
(b) Theophrastus

Question 26.
Three kingdom classification was proposed by …………… .
(a) Copeland
(b) Theophrastus
(c) Linnaeus
(d) Haeckel
Answer:
(d) Haeckel

Question 27.
Which Is not a part of five kindgom classification?
(a) Viruses
(b) Monera
(c) Protista
(d) Mycoplasma
Answer:
(a) Viruses

Question 28.
Six kingdom classification was proposed by …………… .
(a) Haeckel
(b) Copeland
(c) Woese
(d) Cavalier – Smith
Answer:
(d) Cavalier – Smith

Question 29.
Ruggerlo et al., in 2015 proposed …………… kingdom classification.
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 30.
…………… is a new kingdom in seven kingdom classification.
(a) Eubactena
(b) Plantae
(c) Chromista
(d) Archaebacteria
Answer:
(c) Chromista

Question 31.
Actinomycetes comes under …………… kindgom.
(a) fungi
(b) chromista
(c) monera
(d) protista
Answer:
(c) monera

Question 32.
The sourness of curd is due to …………… .
(a) acetic acid
(b) galactic acid
(c) lactic acid
(d) lactone
Answer:
(c) lactic acid

Question 33.
Who is the founder of Modern Bacteriology …………… ?
(a) Aristotle
(b) Robert Koch
(c) Pasteur
(d) Linnaeus
Answer:
(b) Robert Koch

Question 34.
The term bacterium was coined by …………… .
(a) Stanley
(b) Ehrenberg
(c) Gram
(d) Koch
Answer:
(b) Ehrenberg

Question 35.
Plasmids were discovered by …………… .
(a) Ehrenberg
(b) H.Bergy
(c) Joshua Lederberg
(d) Koch
Answer:
(c) Joshua Lederberg

Question 36.
Genophore is seen in …………… .
(a) Amoeba
(b) Cyanobacteria
(c) Chlamydomonas
(d) Euglena
Answer:
(b) Cyanobacteria

Question 37.
Number of domains of life are there according to Carl Woese …………… .
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(a) 3

Question 38.
Which is not a component of bacterial cell?
(a) Mesosomes
(b) Glycocalyx
(c) Polysomes
(d) Histones
Answer:
(d) Histones

Question 39.
The most abundant polypeptide in bacterial cell wall is …………… .
(a) chitin
(b) amylopectin
(c) porin
(d) pectin
Answer:
(c) porin

Question 40.
Extra chromosomal element in bacterial cells are …………… .
(a) plasmids
(b) mesosomes
(c) histones
(d) genophores
Answer:
(a) plasmids

Question 41.
Bacteriocins are found in …………… .
(a) genophore
(b) plasmids
(c) nucleoids
(d) mesosomes
Answer:
(b) plasmids

Question 42.
Colour revealed by Gram positive bacteria after Gram staining is …………… .
(a) red
(b) indigo
(c) dark violet
(d) blue
Answer:
(c) dark violet

Question 43.
How many number of basal body rings seen in the flagella of Gram negative bacteria?
(a) 2
(b) 9
(c) 4
(d) 1
Answer:
(c) 4

Question 44.
Capnophilic bacteria require for growth …………… .
(a) C₂
(b) CO
(c) CO₂
(d) O₃
Answer:
(c) CO₂

Question 45.
The pigment present in green sulphur bacteria is …………… .
(a) bacterioviridin
(b) bacteriochlorophyll
(c) chlorophyll a
(d) xanthophyll
Answer:
(b) bacteriochlorophyll

Question 46.
The hydrogen donor of purple sulphur bacteria is …………… .
(a) H₂S
(b) thiosulphate
(c) ethanol
(d) acetic acid
Answer:
(b) thiosulphate

Question 47.
Campylobacter is a …………… .
(a) obligate aerobe
(b) obligate anaerobe
(c) capnophilic
(d) aerobe
Answer:
(c) capnophilic

Question 48.
Mycobacterium is a …………… .
(a) parasite
(b) symbiont
(c) saprophyte
(d) free – living
Answer:
(a) parasite

Question 49.
Which is the most common mode of asexual reproduction in bacteria?
(a) Endospore formation
(b) Fission
(c) Budding
(d) Conidia
Answer:
(b) Fission

Question 50.
…………… are thick walled resting spores.
(a) Aplanospores
(b) Endospores
(c) Conidia
(d) Zoospores
Answer:
(b) Endospores

Question 51.
In which of the following method genetic recombination does not occur?
(a) Generalised transduction
(b) Conjugation
(c) Transformation
(d) Fission
Answer:
(d) Fission

Question 52.
During conjugation in bacteria, which of the following is transferred from donor to recipient cell?
(a) R factor
(b) F factor
(c) Ti factor
(d) Ri factor
Answer:
(b) F factor

Question 53.
Griffith used …………… for his experiment.
(a) rat
(b) rabbit
(c) mice
(d) monkey
Answer:
(c) mice

Question 54.
Transformation in bacteria was demonstrated by …………… .
(a) Lederberg
(b) Zinder
(c) Edward
(d) Griffith
Answer:
(d) Griffith

Question 55.
Lederberg studied transduction in bacterium …………… .
(a) Diplococcus pneumoniae
(b) Streptococcus
(c) Salmonella typhi
(d) Escherichia coil
Answer:
(c) Salmonella typhi

Question 56.
Bacteria used in the curing of tea is …………… .
(a) Mycococcus candisans
(b) Eseherichia coli
(c) Acetobacter aceti
(d) Streptococcus lactis
Answer:
(a) Mycococcus candisans

Question 57.
Syphilis k caused by …………… .
(a) Mycococcus candisans
(b) Treponema pallidum
(c) Yersinia pestis
(d) Mycohacterium leprae
Answer:
(b) Treponema pallidum

Question 58.
Methanobacterium is …………… .
(a) Cyanobacteria
(b) Malobacteria
(c) Eubacteria
(d) Archaebacteria
Answer:
(d) Archaebacteria

Question 59.
…………… is NOT a phycobiont in lichens.
(a) Gloeocapsa
(b) Dermacarpa
(c) Scytonema
(d) Nostoc
Answer:
(b) Dermacarpa

Question 60.
Red sea is red colour due to …………… .
(a) Dermacarpa sps.
(b) Trichodesmium sps.
(c) Scytonema sps.
(d) Gloeocapsa sps.
Answer:
(b) Trichodesmium sps.

Question 61.
Filamentous trichome is the plant body of …………… .
(a) Chroococcus
(b) Gloeocapsa
(c) Nostoc
(d) Oscillatoria
Answer:
(c) Nostoc

Question 62.
Stromatolites are the colonies of cyanobacteria bind with …………… .
(a) calcium carbonate
(b) calcium hydroxide
(c) magnesium sulphate
(d) calcium silicate
Answer:
(a) calcium carbonate

Question 63.
…………… sps. is an endophyte in coralloid roots of Cycas.
(a) Gioeocapsa
(b) Scytonerna
(c) Nostoc
(d) Azolla
Answer:
(c) Nosloc

Question 64.
Myxophyceae refers to …………… .
(a) Algae
(b) Fungi
(c) Archaebacteria
(d) Cyanobacteria
Answer:
(d) Cyanobacteria

Question 65.
…………… is used in single cell protein.
(a) Spirulina
(b) Azolla
(c) Dermacarpa
(d) Nostoc
Answer:
(a) Spirulina

Question 66.
…………… is a pleomorphic organism.
(a) Fungi
(b) Mycoplasma
(c) Bacteria
(d) Algae
Answer:
(b) Mycoplasma

Question 67.
Pleuropneumonia is caused by …………… .
(a) bacteria
(b) fungi
(c) mycoplasma
(d) viruses
Answer:
(c) mycoplasma

Question 68.
…………… is also called as Ray fungi.
(a) Basidiomycetes
(b) Ascomycetes
(c) Actinomycetes
(d) Deuteromycetes
Answer:
(c) Actinomycetes

Question 69.
Earthy odour of soil after rain is due to …………… .
(a) Basidiomycetes
(b) Ascomycetes
(c) Actinomycetes
(d) Deuteromycetes
Answer:
(c) Actinomycetes

Question 70.
Viruses that attack blue green algae are called as …………… .
(a) Mycophages
(b) Phycophages
(c) Cyanophages
(d) Bacteriophages
Answer:
(c) Cyanophages

Question 71.
Cell membrane of Archaebacteria has …………… .
(a) glycine and isopropyl ethers
(b) glycerol and isobutyl ethers
(c) glycerol and isopropyl ethers
(d) cellulose and isobutyl ethers
Answer:
(c) glycerol and isopropyl ethers

Question 72.
Which is a true bacteria?
(a) Halobacterium
(b) Thermoplasma
(c) Methanobacteriurn
(d) Azotobacter
Answer:
(d) Azolobacter

Question 73.
Study of fungus is called as …………… .
(a) phycology
(b) mycology
(c) algology
(d) biology
Answer:
(b) mycology

Question 74.
Who is considered as the founder of mycology?
(a) K.C.Mehta
(b) G.C.Ainsworth
(c) P.A.Micheli
(d) T.S.Sadasivan
Answer:
(c) P.A.Micheli

Question 75.
Asexual phase of fungi is called as …………… .
(a) telomorph
(b) holomorph
(c) metamorph
(d) anamorph
Answer:
(d) anamorph

Question 76.
In which mycelium, the hyphae are arranged loosely?
(a) Prosenchyma
(b) Plectenchyma
(c) Pseudoparenchyrna
(d) Arenchyma
Answer:
(a) Prosenchyma

Question 77.
Number of nuclei in coenocytic mycelium …………… .
(a) 2
(b) many
(c) nil
(d) 9
Answer:
(b) many

Question 78.
Thallospores are produced by …………… .
(a) Aspergillus
(b) Erysiphe
(c) Saccharomyces
(d) Fusarium
Answer:
(b) Erysiphe

Question 79.
In Agaricus, …………… type of sexual reproduction occurs.
(a) spermatization
(b) somatogamy
(c) oogamy
(d) isogamy
Answer:
(b) somatogamy

Question 80.
Albugo belongs to …………… .
(a) oomycetcs
(b) zygomycetes
(c) ascomycetes
(d) deuteromycetes
Answer:
(a) oomycetes

Question 81.
Fungi growing on dung is called as …………… .
(a) Mold fungus
(b) Saprophytes
(c) Capnophilous
(d) Coprophilous
Answer:
(d) Coprophilous

Question 82.
Coprophilous belongs to …………… group.
(a) basidiomycetes
(b) ascomycetes
(c) zygomycetes
(d) oomycetes
Answer:
(c) zygomycetes

Question 83.
Which of the following is a coprophilous fungi?
(a) Albugo
(b) Entomophthora
(c) Rhizopus
(d) Pilobolus
Answer:
(d) Pilobolus

Question 84.
Cup fungus belongs to …………… .
(a) zygomycetes
(b) oomycetes
(c) ascomycetes
(d) actinomycetes
Answer:
(c) ascomycetes

Question 85.
Which group of fungus is called as Sac fungi?
(a) Deuteromycetes
(b) Zygomycetes
(c) Ascomycetes
(d) Oomycetes
Answer:
(c) Ascomycetes

Question 86.
Number of ascospores in an asci is …………… .
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 87.
Shape of perithecium is …………… .
(a) cup shaped
(b) flask shaped
(c) completely closed
(d) open type
Answer:
(b) flask shaped

Question 88.
…………… are called as Club fungi.
(a) Ascomycetes
(b) Zygomycetes
(c) Basidiomycetes
(d) Deuteromycetes
Answer:
(c) Basidiomycetes

Question 89.
Parasexual cycle is observed in …………… .
(a) basidiomycetes
(b) zygomycetes
(c) deuteromycetes
(d) ascomycetes
Answer:
(c) deuteromycetes

Question 90.
Which is called as imperfect fungi?
(a) Basidiomycetes
(b) Zygomycetes
(c) Deuteromycetes
(d) Ascomycetes
Answer:
(c) Deuteromycetes

Question 91.
In basidiomycetes, clamp connections are formed to maintain …………… condition.
(a) monokaryotic
(b) coenocytic
(c) dikaryotic
(d) zygotic
Answer:
(c) dikaryotic

Question 92.
…………… is a single celled fungus used in dairy industry.
(a) Volvariella
(b) Agaricus
(c) Penicillin
(d) Yeast
Answer:
(d) Yeast

Question 93.
Ergot alkaloids are produced by …………… .
(a) Penicillium notatum
(b) Acremonium chrysogenum
(c) Claviceps purpurea
(d) Penicillium griseofulvum
Answer:
(c) Claviceps purpurea

Question 94.
Kojic acid is produced by …………… .
(a) Aspergillus terreus
(b) Aspergillus niger
(c) Aspeigillus oryzae
(d) Agaricus hisporus
Answer:
(c) Aspergillus oryzae

Question 95.
…………… infest dried foods and produce carcinogenic toxin.
(a) Aspergillus flavus
(b) Amanita verna
(c) Amanita phalloides
(d) Rhizopus
Answer:
(a) Aspergillus flavus

Question 96.
Rust of wheat is produced by …………… .
(a) Albugo candida
(b) Puccinia graminis tritici
(c) Candida albicans
(d) Colletotrichum sps
Answer:
(b) Puccinia graminis tritici

Question 97.
VAM is a type of …………… .
(a) Endomycorrhiza
(b) Ectomycorrhiza
(c) Ectendomycorrhiza
(d) Endectomycorrhiza
Answer:
(a) Endomycorrhiza

Question 98.
Algal partner of lichen is …………… .
(a) phycobiont
(b) phytobiont
(c) mycobiont
(d) both (a) & (c)
Answer:
(a) phycobiont

Question 99.
Asexual reproduction by Soredia is seen in …………… .
(a) fungi
(b) lichen
(c) mycorrhiza
(d) algae
Answer:
(b) lichen

Question 100.
Saxicolous lichen grow on …………… .
(a) ground
(b) bark
(c) wood
(d) rock
Answer:
(d) rock

Question 101.
In leprose form of lichen distinct …………… layer is absent.
(a) fungal
(b) algal
(c) both
(d) none
Answer:
(a) fungal

Question 102.
…………… are used as pollution indicators.
(a) Algae
(b) Lichen
(c) Fungi
(d) Mycorrhiza
Answer:
(b) Lichen

Question 103.
…………… acid is obtained from lichen acting as antibiotics.
(a) Alginic
(b) Acetic
(c) Oxalic
(d) Usnic
Answer:
(d) Usnic.

II. Very Short Answer Type Questions 

Question 1.
Differentiate plant growth from animal growth.
Answer:
Plant Growth From Animal Growth:

S. No.	Plant growth	Animal growth
1.	1. Growth is indefinite.	1. Growth is definite.
2.	2. It occurs throughout life.	2. It occurs for some period.

Question 2.
Define Growth.
Answer:
Growth is an intrinsic property of all living organisms through which they can increase cells both in number and mass.

Question 3.
Growth of living thing is an intrinsic property – Justify.
Answer:
Living cells grow by the addition of new protoplasm within the cells. Therefore, growth in living thing is intrinsic.

Question 4.
Define reproduction and Mention its types.
Answer:
Reproduction is the tendency of a living organism to perpetuate its own species. There are two types of reproduction namely asexual and sexual.

Question 5.
What is metabolism? Mention its types.
Answer:
The sum total of all the chemical reactions taking place in a cell of living organism is called metabolism. It is broadly divided into anabolism and catabolism.

Question 6.
What is consciousness and irritability?
Answer:
Animals sense their surroundings by sense organs. This is called consciousness. Respond of plants to the stimuli is called irritability.

Question 7.
List out few attributes of living organisms.
Answer:
The attributes of living organisms are growth, metabolism, movement, reproduction, nutrition, excretion, etc.

Question 8.
Define cyclosis.
Answer:
The movement of cytoplasm inside the cell is called cytoplasmic streaming or cyclosis.

Question 9.
How will you define viruses?
Answer:
Viruses are sub – microscopic, obligate intracellular parasites. They have nucleic acid core surrounded by protein coat.

Question 10.
Mention the size of Bacteriophage and tobacco mosaic virus (TMV).
Answer:
Bacteriophage measures about 10 – 100 nm in size. The size of TMV is 300 × 20 nm.

Question 11.
Classify viruses based on nature of nucleic acid with example.
Answer:
On the basis of nature of nucleic acid viruses are classified into four categories. They are viruses with ssDNA (Parvo viruses), dsDNA (Bacteriophages), ssRNA (TMV) and dsRNA (wound tumour virus).

Question 12.
Distinguish between deoxyviruses and riboviruses.
Answer:
Deoxyviruses and Riboviruses:

1. Deoxyviruses: Viruses having DNA are called deoxyviruses. E.g. Animal viruses except HIV
2. Riboviruses: Viruses having RNA are called riboviruses. E.g. Plant viruses except cauliflower mosaic virus (CMV)

Question 13.
Write the constituents of virions.
Answer:
The virion is made up of two constituents, a protein coat called capsid and a core called nucleic acid.

Question 14.
What are capsomeres?
Answer:
The protein coat of viruses is made up of approximately 2130 identical protein subunits called capsomeres.

Question 15.
Name the two types of phage multiplication.
Answer:
Phages multiply through two different types of life cycle:

1. Lytic or Virulent cycle
2. Lysogenic or Avirulent life cycle

Question 16.
What do you mean by a ‘ghost’ in virology?
Answer:
The empty protein coat left outside by the phage after penetrating the host cell is called as ghost.

Question 17.
What do you understand by “pinning” of phage?
Answer:
Once the contact is established between tail fibres of phase and bacterial cell, tail fibres bend to anchor the pins and base plate to the cell surface. This step is called pinning.

Question 18.
What is prophage?
Answer:
As soon as the phage injects its linear DNA into the host cell, it becomes circular and integrates into the bacterial chromosome by recombination. The integrated phage DNA is now called prophage.

Question 19.
When does a prophage enters lytic cycle?
Answer:
On exposure to UV radiation and chemicals the excision of phage DNA may occur and results in lytic cycle.

Question 20.
Define virion?
Answer:
Virion is an intact infective virus particle which is non – replicating outside a host cell.

Question 21.
What are viroids?
Answer:
Viroid is a circular molecule of ssRNA without a capsid. RNA is of low molecular weight.

Question 22.
Name any two disease caused by viroids.
Answer:
Two Disease:

1. Citrus exocortis
2. Potato spindle tuber disease

Question 23.
What are virusoids?
Answer:
Virusoids are the small circular RNAs which are similar to viroids but they are always linked with larger molecules of the viral RNA.

Question 24.
Who discovered viroids and virusolds?
Answer:
Viroids were discovered by T.O. Diener in 1971. Virusoids were discovered by J.W.Randles in 1981.

Question 25.
Name the causative organism for mad cow disease.
Answer:
Prions are the causative organisms for mad cow disease, Prions are the proteinaceous infectious particles.

Question 26.
What are cyanophages? Who reported it first?
Answer:
Viruses infecting blue green algae are called Cyanophages and are first reported by Safferman and Morris in the year 1963.

Question 27.
Name any two disease caused by Prions.
Answer:
Two Disease:

1. Bovine Spongiform Encephalopathy (BSE) (mad cow disease)
2. Creutzfeldt – Jakob Disease (CJD)

Question 28.
What are mycophages? Who first reported it?
Answer:
Viruses infecting fungi are called mycophages or mycoviruses. Mycophages were first reported by Hollings in 1962.

Question 29.
Expand the following acronyms: (a) SARS and (b) AIDS.
Answer:
Acronyms:
(a) SARS: Severe Acute Respiratory Syndrome
(b) AIDS: Acquired Immuno Deficiency Syndrome

Question 30.
Name the two groups of aninmals according to Aristotle.
Answer:
Two Groups of Aninmals:

1. Enaima – animals with red blood.
2. Anaima – animals without red blood.

Question 31.
Which are the major setbacks of Linnaeus classification?
Answer:
Linnaeus classification faced major setback because prokaryotes and eukaryotes were grouped together. Similarly fungi, heterotrophic organisms were placed along with the photosynthetic plants.

Question 32.
Name the viruses which are employed as potential insecticides?
Answer:
Cytoplasmic polyhedrosis Granulo viruses and Entomopox virus were employed as potential insecticides.

Question 33.
Who proposed five kingdom classification? Mention the five kingdoms.
Answer:
R.H. Whittaker proposed the five kingdom classification. It includes Monera, Protista, Fungi, Plantae and Animalia.

Question 34.
List out the criteria undertaken for Whittaker’s classification.
Answer:
The criteria adopted for the classification include cell structure, thallus organization, mode of nutrition, reproduction and phylogenetic relationship.

Question 35.
Point out the demerits of five kingdom classification.
Answer:
(a) The kingdom Monera and Protista accommodate both autotrophic and heterotrophic organisms, cell wall lacking and cell wall bearing organisms thus making these two groups more heterogeneous.
(b) Viruses were not included in the system.

Question 36.
Who proposed six kingdom classification? Mention the kingdoms.
Answer:
Thomas Cavalier – Smith proposed six kingdom classification.
The kingdom includes:

1. Archaebacteria
2. Eubacteria
3. Protista
4. Fungi
5. Plantae and
6. Animalia.

Question 37.
How milk is changed into curd, if a few drops of curd is added to it? What is the reason for its sourness?
Answer:
The change is brought by Lactobacillus lactis, a bacterium present in the curd. The sourness is due to the formation of lactic acid.

Question 38.
Define bacteria and bacteriology.
Answer:
Bacteria are prokaryotic, unicellular, ubiquitous, microscopic organisms. The study of bacteria is called bacteriology.

Question 39.
What is Porin? How it helps the bacteria?
Answer:
Porin is an abundant polypeptide present in bacterial cell walls. It helps in the diffusion of solutes.

Question 40.
List out the cytoplasmic inclusions of bacterial cell.
Answer:
Glycogen, poly-β-hydroxybutyrate granules, sulphur granules and gas vesicles.

Question 41.
Define Genophore.
Answer:
The bacterial chromosome is a single circular DNA molecule, tightly coiled and is not enclosed in a membrane as in Eukaryotes. This genetic material is called nucleoid or genophore.

Question 42.
Write the chemical composition of bacterial cell wall.
Answer:
The chemical composition of cell wall is rather complex and is made up of peptidoglycan or mucopeptide (N – acetyl glucosamine, N – acetyl muramic acid and peptide chain of 4 or 5 aminoacids).

Question 43.
What are polysomes?
Answer:
During protein synthesis, the ribosomes are held together by mRNA and form the polysomes.

Question 44.
What are Pili?
Answer:
Pili or fimbriae are hair like appendages found on surface of cell wall of gram – negative bacteria.

Question 45.
What are capnophilic bacteria? Give an example.
Answer:
Bacteria which require CO₂ for their growth are called as capnophilic bacteria.
Example: Campylobacter.

Question 46.
Distinguish between Photolithotrophs and Photoorganotrophs.
Answer:
Between Photolithotrophs and Photoorganotrophs:

1. Photolithotrophs: In photolithotrophs, the hydrogen donor is an inorganic substance. E.g. Chlorobium
2. Photoorganotrophs: In Photoorganotrophs, the hydrogen donor is an organic acid or alcohol. E.g. Rhodospirillum

Question 47.
Name the hydrogen donor of green sulphur bacteria and purple sulphur bacteria.
Answer:
Hydrogen donor of green sulphur bacteria is H₂S. Hydrogen donor of purple sulphur bacteria is thiosulphate.

Question 48.
Name the bacterial pigment of green sulphur bacteria and purple sulphur bacteria.
Answer:
Bacteria’s:

1. Green sulphur bacteria – Bacterioviridin
2. Purple sulphur bacteria – Bacteriochlorophyll

Question 49.
What are endospores?
Answer:
Endospores are thick walled resting spores developed by bacteria during unfavourable condition.
E.g. Clostridium tetani produces endospores.

Question 50.
Mention the various ways by which genetic recombination occurs.
Answer:
Genetic recombination in bacteria occurs by conjugation, transduction and transformation.

Question 51.
Name the eminent persons who demonstrated the conjugation process.
Answer:
J. Lederberg and Edward L. Tatum.

Question 52.
What is transformation? Name the bacteriologist who described it.
Answer:
The Bacteriologist Who Described it:

1. Transfer of DNA from one bacterium to another is called transformation.
2. Fredrick Griffith demonstrated the transformation process.

Question 53.
Which organism and bacterial species was used in Griffith’s transformation experiment?
Answer:
Mice and Diplococcus pneumoniae.

Question 54.
List out the asexual modes of reproduction of bacteria.
Answer:
Asexual reproduction in bacteria includes binary fission, conidia formation and endospore formation.

Question 55.
Who discovered transduction? Define it.
Answer:
Zinder and Lederberg (1952) discovered transduction in Salmonella typhimurum. Phage mediated DNA transfer is called transduction.

Question 56.
Name any two bacterial species and the antibiotic produced by them.
Answer:

Bacteria	Antibiotic
1. Streptomyces griseus	1. Streptomycin
2. Bacillus polymyxa	2. Polymyxin

Question 57.
How bacteria helps in vinegar production?
Answer:
Acetobacter aceti bacteria oxidises ethanol obtained from molasses by fermentation to form vinegar.

Question 58.
Name any two ammonifying bacteria.
Answer:
Two Ammonifying Bacteria:

1. Bacillus ramosus and
2. Bacillus mycoides.

Question 59.
What do you mean by retting of fibres?
Answer:
The fibres from the fibre yielding plants are separated by the action of Closiridium is called retting of fibres.

Question 60.
Name any two plant disease caused by the bacteria and mention the host.
Answer:

Question 61.
Name any four animal disease caused by bacteria.
Answer:
Four Animal Disease:

1. Anthrax
2. Brucellosis
3. Bovine tuberculosis and
4. black leg.

Question 62.
Name any four human disease caused by bacteria.
Answer:
Four Human Disease:

1. Cholera
2. Typhoid
3. Tuberculosis and
4. Leprosy.

Question 63.
What are Archaebacteria?
Answer:
Archaebacteria are primitive prokaryotes and are adapted to thrive in extreme environments like hot springs, high salinity and low pH.
E.g., Thermoplasma.

Question 64.
How stromatolites are formed?
Answer:
Stromatolites are deposits formed when colonies of cyanobacteria bind with calcium carbonate.

Question 65.
What is the reason for the colour of Red Sea?
Answer:
A cyanobacteria called Trichodesmium erythraeum imparts red colour to sea.

Question 66.
Mention the cyanobacteria leading endophytic relation with Cycas roots.
Answer:
Nostoc and Anabaena.

Question 67.
Define Cyanobacteria.
Answer:
Cyanobacteria are popularly called as ‘Blue green algae’ or ‘Cyanophyceae’. They are photosynthetic, prokaryotic organisms. Cyanobacteria are primitive forms and are found in different habitats.

Question 68.
Blue green algae can also be called as Myxophyceae. How?
Answer:
The presence of mucilage around the thallus is characteristic feature of cyanobacteria group. Therefore, this group is also called Myxophyceae.

Question 69.
Define mycoplasma?
Answer:
The mycoplasma are very small (0.1 – 0.5 μm), pleomorphic gram negative microorganisms.

Question 70.
Name few plant disease caused by mycoplasma.
Answer:
Little leaf of brinjal, witches broom of legumes, phyllody of cloves and sandal spike are some plant diseases caused by mycoplasma.

Question 71.
Draw and label the structure of mycoplasma.
Answer:
The Structure of Mycoplasma:

Question 72.
What is the reason behind the earthy odour after raining?
Answer:
Streptomyces is a mycelial forming Actinobacteria which lives in soil, they impart “earthy odour” to soil after rain which is due to the presence of geosmines (volatile organic compound).

Question 73.
When and by whom the penicillin was discovered?
Answer:
Penicillin was discovered by Alexander Flemming in 1928.

Question 74.
Define Fungi.
Answer:
Fungi are ubiquitous, eukaryotic, achlorophyllous heterotrophic organisms. They exist in unicellular or multicellular forms.

Question 75.
Define mycology. Who is the founder of mycology?
Answer:
Study of fungi is called mycology. P.A. Micheli is considered as the founder of mycology.

Question 76.
Name few eminent Mycologists.
Answer:
John Webster, G.C. Ainsworth, T.S. Sadasivan and C.V. Subramanian.

Question 77.
With example define coenocytic mycelium.
Answer:
In lower fungi the hypha is aseptate, multinucleate and is known as coenocytic mycelium (Example: Albugo).

Question 78.
What is plectenchyma? Mention its types.
Answer:
The mycelium is organised into loosely or compactly interwoven fungal tissues called plectenchyma. It is further divided into two types: prosenchyma and pseudoparenchyma.

Question 79.
Distinguish between Anamorph and Telomorph.
Answer:
Between Anamorph and Telomorph:

Anamorph	Telomorph
The asexual phase of fungi is called anamorph.	The sexual phase of fungi is called telomorph.

Question 80.
What is holomorph?
Answer:
Fungi showing both sexual and asexual phases are called holomorph.

Question 81.
What is planogametic copulation? Mention its types.
Answer:
Fusion of motile gamete is called planogametic copulation.
Types:

1. Isogamy
2. Anisogamy and
3. Oogamy.

Question 82.
List out the asexual spores produced by fungus.
Answer:
Zoospores, conidia, oidia and chlamydospores.

Question 83.
What are coprophilous fungi? Give an example.
Answer:
Fungi growing on dung are called coprophilous fungi.
Example: Pilobolus.

Question 84.
Ascomycetes are called sac fungi. Give reason.
Answer:
In ascomycetes the ascospores are found inside a bag like structure called ascus. Due to the presence of ascus, this group is popularly called “Sac fungi”.

Question 85.
Name the four types of ascocarps produced by ascomycetes.
Answer:
Four Types Of Ascocarps Produced By Ascomycetes:

1. Cleistothecium
2. Perithecium
3. Apothecium and
4. Pseudothecium.

Question 86.
Basidiomycetes are called club fungi. Give reason.
Answer:
In basidiomycetes the basidium is club shaped with four basidiospores, thus this group of fungi is popularly called “Club fungi”. The fruit body formed is called Basidiocarp.

Question 87.
Name the special structures in deuteromycetes that produces conidia.
Answer:
Pycnidium, acervulus, sporodochium and synnemata.

Question 88.
Deuteromycetes are imperfect fungi – Justify.
Answer:
The fungi belonging to deuteromycetes lack sexual reproduction and are called imperfect fungi.

Question 89.
List out the antibiotics produced by fungi.
Answer:
Penicillin, cephalosporins and griseofiilvin.

Question 90.
Name some toxins produced by Fungus.
Answer:
Alfatoxin, Patulin and Ochratoxin – A.

Question 91.
Name 2 fungal species employed as Biopesticides.
Answer:
Two Fungal Species Employed As Biopesticides:

1. Beauveria bassiana and
2. Metarhizium anisopliae.

Question 92.
Name few fungal diseases in plants.
Answer:
Blast of paddy, rust of wheat, red rot of sugarcane and white rust of crucifers.

Question 93.
Name few fungal diseases in Humans?
Answer:
Fungal Diseases in Humans:

S. No.	Human Diseases	Causative Fungi
1.	Athlete’s foot	Epidermophyton floccosum
2.	Candidiasis	Candida albicans
3.	Coccidioidomycosis	Coccidioides immitis
4.	Aspergillosis	Aspergillus fumigatus

Question 94.
What is mycorrhiza?
Answer:
The symbiotic association between fungal mycelium and roots of plants is called as mycorrhiza.

Question 95.
What are 3 types of mycorrhiza?

Answer:
3 types of mycorrhiza:

1. Ectomycorrhiza
2. Endomycorrhiza and
3. Ectendomycorrhiza

Question 96.
Define lichen.
Answer:
Lichen is a symbiotic association between algae and fungi.

Question 97.
What is a phycobiont and mycobiont?
Answer:
Fungal partner of lichen is called as mycobiont. Algal partner of lichen is called as phycobiont.

Question 98.
How symbiosis workout in lichen?
Answer:
In lichens, algae provide nutrition for fungal partner in turn fungi provide protection and also help to fix the thallus to the substratum through rhizinae.

Question 99.
Classify lichens based on morphology.
Answer:
Morphology:

1. Leprose – Absence of distinct fungal layer
2. Crustose – Crust-like
3. Foliose – Leaf-like
4. Fruticose – Branched pendulous shrub-like

Question 100.
What is homoiomerous and heteromerous?
Answer:
In homoiomerous (algal cells evenly distributed in the thallus) and heteromerous (a distinct layer of algae and fungi present).

Question 101.
Define ascolichen and basidiolichen.
Answer:
If the fungal partner of lichen belongs to ascomycetes, it is called ascolichen and if it is basidiomycetes it is called basidiolichen.

Question 102.
Lichens are pollution indicators. How?
Answer:
Lichens are sensitive to air pollutants especially to sulphur-di-oxide. Therefore, they are considered as pollution indicators.

Question 103.
Classify lichens based on habitat.
Answer:

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Prokaryotes and Eukaryotes.
Answer:
Between Prokaryotes and Eukaryotes:

Question 2.
List out various types of asexual reproduction in living organisms.
Answer:
Reproduction in Living Organisms:

Question 3.
Define Homeostasis. Why it is essential?
Answer:
Property of self – regulation and tendency to maintain a steady state within an external environment which is liable to change is called homeostasis. It is essential for the living organism to maintain internal condition to survive in the environment.

Question 4.
Viruses are considered as biological puzzle – Justify.
Answer:
Viruses are considered as biological puzzle since they exhibit both living and non – living characters.

Question 5.
State the 3 types of viral symmetry.
Answer:
3 Types Of Viral Symmetry:

1. Cuboid symmetry – Example: Adenovirus and Herpes virus.
2. Helical symmetry – Example: Influenza virus and TMV.
3. Complex or Atypical – Example: Bacteriophage and Vaccinia virus.

Question 6.
List out the non – living characters of viruses.
Answer:
The non – living characters of viruses:

• Can be crystallized.
• Absence of metabolism.
• Inactive outside the host.
• Do not show functional autonomy.
• Energy producing enzyme system is absent.

Question 7.
Name any one RNA animal virus and DNA plant virus.
Answer:
RNA animal virus and DNA plant virus:L

1. RNA animal virus is HIV.
2. DNA plant virus is Cauliflower mosaic virus.

Question 8.
What are the prominent symptoms of TMV affected tobacco plants?
Answer:
The first visible symptom of TMV is discoloration of leaf colour along the veins and show typical yellow and green mottling which is the mosaic symptom. The downward curling and distortion of young apical leaves occurs, plant becomes stunted and yield is affected.

Question 9.
What are bacteriophages? Where can we find it?
Answer:
Viruses infecting bacteria are called bacteriophages. It literally means ‘eaters of bacteria’. Phages are abundant in soil, sewage water, fruits, vegetables and milk.

Question 10.
Sequencely mention the types of lytic cycle.
Answer:
Types of Lytic Cycle:

1. Adsorption
2. penetration
3. synthesis
4. assembly
5. maturation and
6. release.

Question 11.
Why classification of organisms is important?
Answer:
Classification is essential to achieve following needs:

1. To relate things based on common characteristic features.
2. To define organisms based on the salient features.
3. Helps in knowing the relationship amongst different groups of organisms.
4. It helps in understanding the evolutionary relationship between organisms.

Question 12.
List out the merits of five kingdom classification.
Answer:
Five kingdom Classification:

1. The classification is based on the complexity of cell structure and organization of thallus.
2. It is based on the mode of nutrition.
3. Separation of fungi from plants.
4. It shows the phylogeny of the organisms.

Question 13.
Who is called as founder of modern bacteriology? Mention his contribution?
Answer:
Robert Heinrich Hermann Koch is considered as the founder of modern bacteriology. He identified the causal organism for Anthrax. Cholera and Tuberculosis. He proved experimental evidence for the concept of infection (Koch’s postulates).

Question 14.
Draw Amphitrichous, Lophotrichous and Peritrichous flagellation in bacteria.
Answer:
Amphitrichous, Lophotrichous and Peritrichous flagellation in bacteria:

Question 15.
What are the three layers of bacterial cell?
Answer:
The bacterial cell reveals three layers:

1. capsule / glycocalyx
2. cell wall and
3. cytoplasm.

Question 16.
What is a capsule? Mention its role.
Answer:
Some bacteria are surrounded by a gelatinous substance which is composed of polysaccharides or polypeptide or both. A thick layer of glycocalyx bound tightly to the cell wall is called capsule. It protects cell from desiccation and antibiotics. The sticky nature helps them to attach to substrates like plant root surfaces, Human teeth and tissues. It helps to retain the nutrients in bacterial cell.

Question 17.
What are plasmids? How it helps the bacteria?
Answer:
Plasmids are extrachromosomal double stranded, circular, self-replicating, autonomous elements. They contain genes for fertility, antibiotic resistant and heavy metals. It also help in the production of bacteriocins and toxins which are not found in bacterial chromosome.

Question 18.
Classfiy plasmids based on function.
Answer:
Plasmids are classified into different types based on the function. Some of them are F (Fertility) factor, R (Resistance) plasmids, Col (Colicin) plasmids, Ri (Root inducing) plasmids and Ti (Tumor inducing) plasmids.

Question 19.
Give a brief note on Mesosomes.
Answer:
Mesosomes are localized infoldings of plasma membrane produced into the cell in the form of vesicles, tubules and lamellae. They are clumped and folded together to maximize their surface area and helps in respiration and in binary fission.

Question 20.
How Gram positive and Gram negative bacteria react on Gram staining process?
Answer:
The Gram positive bacteria retain crystal violet and appear dark violet, whereas Gram negative type loose the crystal violet and when counterstained by safranin appear red under a microscope.

Question 21.
Name the three components of gram negative cell wall.
Answer:
Three Components Of Gram Negative Cell Wall:

1. Lipoprotein
2. outermembrane and
3. lipopolysaccharide.

Question 22.
What are Magnetosomes?
Answer:
Intracellular chains of 40 – 50 magnetite (Fe₃O₄) particles are found in bacterium Aquaspirillum magnetotacticum and it helps the bacterium to locate nutrient rich sediments.

Question 23.
Write a note on binary fission.
Answer:
Under favourable conditions the cell divides into two daughter cells. The nuclear material divides first and it is followed by the formation of a simple median constriction which finally results in the separation of two cells.

Question 24.
How do archaebacteria thrive at extreme temperatures and against lytic agents?
Answer:
The unique feature of archaebacteria is the presence of lipids like glycerol and isopropyl ethers in their cell membrane. Due to the unique chemical composition the cell membrane show resistance against cell wall antibiotics and lytic agents.
E.g. Methanobacterium.

Question 25.
Name few members of cyanobacteria which act as phycobiont in lichen thalli.
Answer:
Gloeocapsa, Nostoc and Scytonema.

Question 26.
Describe in brief about Actinomycetes.
Answer:
Actinomycetes are also called ‘Ray fungi’ due to their mycelia like growth. They are anaerobic or facultative anaerobic microorganisms and are Gram positive. They do not produce an aerial mycelium. Their DNA contain high guanine and cytosine content (E.g., Streptomyces).

Question 27.
Mention few antibiotics produced by Streptomyces group of fungi.
Answer:
Streptomycin, Chloramphenicol and Tetracycline.

Question 28.
Explain in brief about the plant body of fungi?
Answer:
Majority of fungi are made up of thin, filamentous branched structures called hyphae. A number of hyphae get interwoven to form mycelium. The cell wall of fungi is made up of a polysaccharide called chitin (polymer of N – acetyl glucosamine).

Question 29.
Point out the steps involved in sexual reproduction of fungi.
Answer:
Sexual reproduction in fungi includes three steps:

1. Fusion of two protoplasts (plasmogamy)
2. Fusion of nuclei (karyogamy) and
3. Production of haploid spores through meiosis.

Question 30.
Differentiate between Anisogamy and Oogamy with an example of fungus.
Answer:
Between Anisogamy and Oogamy with an example of fungus:

Anisogamy	Oogamy
Fusion of morphologically or physiologically dissimilar gametes. E.g. Allomyces	Fusion of both morphologically and physiologically dissimilar gametes. E.g. Monoblepharis

Question 31.
Define Spermatization.
Answer:
In spermatization method a uninucleate pycniospore / microconidium is transferred to receptive hyphal cell (Example: PuccinialNeurospora).

Question 32.
Draw a simple diagram showing the budding of yeast.
Answer:
Diagram showing the budding of yeast:

Question 33.
Write a simple note on Oomycetes.
Answer:
Coenocytic mycelium is present. The cell wall is made up of glucan and cellulose. Zoospore with one whiplash and one tinsel flagellum is present. Sexual reproduction is Oogamous.
Example: Albugo.

Question 34.
Give a brief account on fungal food.
Answer:
Mushrooms like Lentinus edodes. Agaricus bisporus and Volvariella volvaceae are consumed for their high nutritive value. Yeasts provide vitamin B and Eremothecium ashbyii is a rich source of vitamin B₁₂

Question 35.
List out the importance of mycorrhiza.
Answer:
The Importance Of Mycorrhiza:

1. Helps to derive nutrition in Monotropa, a saprophytic angiosperm.
2. Improves the availability of minerals and water to the plants.
3. Provides drought resistance to the plants.
4. Protects roots of higher plants from the attack of plant pathogens.

Question 36.
How symbiotic relationship is executed in mycorrhiza?
Answer:
In mycorrihiza, relationship fungi absorbs nutrition from the root and in turn the hyphal network of mycorrhiza forming fungi helps the plant to absorb water and mineral nutrients from the soil.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Define metabolism. List out the differences between the types of metabolism.
Answer:
Metabolism: The sum total of all the chemical reactions taking place in a cell of living organism is called metabolism. It is broadly divided into anabolism and catabolism. The difference between anabolism and catabolism is given in table.

Question 2.
Draw a flow chart representing the various levels of organisation and integration in living organisms.
Answer:
A flow chart representing the various levels of organisation and integration in living organisms:

Question 3.
Enumerate the living and non – living characters of viruses.
Answer:
Living characters:

1. Presence of nucleic acid and protein.
2. Capable of mutation.
3. Ability to multiply within living cells.
4. Able to infect and cause diseases in living beings.
5. Show irritability.
6. Host – specific.

Non – living characters:

1. Can be crystallized.
2. Absence of metabolism.
3. Inactive outside the host.
4. Do not show functional autonomy,
5. Energy producing enzyme system is absent.

Question 4.
Draw and describe the structure of tobacco mosaic virus.
Answer:

Electron microscopic studies have revealed that TMV is a rod shaped helical virus measuring about 280 × 150 pm with a molecular weight of 39 × 10⁶ Daltons. The virion is made up of two constituents, a protein coat called capsid and a core called nucleic acid. The protein coat is made up of approximately 2130 identical protein subunits called capsomeres which are present around a central single stranded RNA molecule. The genetic information necessary for the formation of a complete TMV particle is contained in its RNA. The RNA consists of 6,500 nucleotides.

Question 5.
Write a note on David Baltimore’s classification of viruses.
Answer:
David Baltimore (1971) classification is based on mechanism of RNA production, the nature of the genome (single stranded -ss or double stranded – ds ), RNA or DNA, the use of reverse transcriptase (RT), ss RNA may be (+) sense or (-) antisense. Viruses are classified into seven classes.

Question 6.
Give an account of viral genome.
Answer:
Each virus possesses only one type of nucleic acid either DNA or RNA. The nucleic acid may be in a linear or circular form. Generally nucleic acid is present as a single unit but in wound tumour virus and in influenza virus it is found in segments. The viruses possessing DNA are called ‘Deoxyviruses’ whereas those possessing RNA are called ‘Riboviruses’.

Majority of animal and bacterial viruses are DNA viruses (HIV is the animal virus which possess RNA). Plant viruses generally contain RNA (Cauliflower Mosaic virus possess DNA). The nucleic acids may be single stranded or double stranded. On the basis of nature of nucleic acid viruses are classified into four categories. They are viruses with ssDNA (Parvo viruses), dsDNA (Bacteriophages), ssRNA (TMV) and dsRNA (wound tumour virus).

Question 7.
Explain the structure of T₄ bacteriophage with a labelled diagram.
Answer:

The T₄ phage is tadpole shaped and consists of head, collar, tail, base plate and fibres (figure). The head is hexagonal which consists of about 2000 identical protein subunits. The long helical tail consists of an inner tubular core which is connected to the head by a collar. There is a base plate attached to the end of tail The base plate contains six spikes and tail fibres. These fibres are used to attach the phage on the cell wall of bacterial host during replication. A dsDNA molecule of about 50 μm is tightly packed inside the head. The DNA is about 1000 times longer than the phage itself.

Question 8.
Describe in detail about the lytic cycle of phages with diagram.
Answer:
Lytic cycle: During lytic cycle of phage, dis integration of host bacterial cell occurs and the progeny virions are released.

The steps involved in the lytic cycle are as follows:
(i) Adsorption: Phage (T₄) particles interact with cell wall of host (E. coli). The phage tail makes contact between the two, and tail fibres recognize the specific receptor sites present on bacterial cell surface. The lipopolysaccharides of tail fibres act as receptor in phages. The process involving the recognition of phage to bacterium is called landing. Once the contact is established between tail fibres of phage and bacterial cell, tail fibres bend to anchor the pins and base plate to the cell surface. This step is called pinning.

(ii) Penetration: The penetration process involves mechanical and enzymatic digestion of the cell wall of the host. At the recognition site phage digests certain cell wall structure by viral enzyme (Jysozyrne). After pinning the tail sheath contracts (using ATP) and appears shorter and thicker. After contraction of the base plate enlarges through which DNA is injected into the cell wall without using metabolic energy. The step involving injection of DNA particle alone into the bacterial cell is called Transfection. The empty protein coat leaving outside the cell is known as ‘ghost’.

(iii) Synthesis: This step involves the degradation of bacterial chromosome, protein synthesis and DNA replication. The phage nucleic acid takes over the host biosynthetic machinery. Host DNA gets inactivated and breaks down. Phage DNA suppresses the synthesis of bacterial protein and directs the metabolism of the cell to synthesis the proteins of the phage particles and simultaneously replication of phage DNA also takes place.

(iv) Assembly and Maturation: The DNA of the phage and protein coat are synthesized separately and are assembled to form phage particles. The process of assembling the phage particles is known as maturation. After 20 minutes of infection about 300 new phages are assembled

(v) Release: The phage particle gets accumulated inside the host cell and are released by the lysis of the host cell wall

.

Question 9.
Explain the lysogenic multiplication of phages.
Answer:
In the lysogenic cycle the phage DNA gets integrated into host DNA and gets multiplied along with nucleic acid of the host. No independent viral particle is formed.

As soon as the phage injects its linear DNA into the host cell, it becomes circular and integrates into the bacterial chromosome by recombination. The integrated phage DNA is now called prophage. The activity of the prophage gene is repressed by two repressor proteins which are synthesized by phage genes. This checks the synthesis of new phages within the host cell. However, each time the bacterial cell divides, the prophage multiplies along with the bacterial chromosome. On exposure to UV radiation and chemicals the excision of phage DNA may occur and results in lytic cycle.

Question 10.
Draw a tabular column and compare the characters of five kingdoms.
Answer:

Question 11.
List out the general characters of bacteria.
Answer:
The General Characters of Bacteria:

1. They are prokaryotic organisms and lack nuclear membrane and membrane bound organelles.
2. The genetic material is called nucleoid or genophore or incipient nucleus.
3. The cell wall is made up of polysaccharides and proteins.
4. Most of them lack chlorophyll, hence they are heterotrophic (Vibrio cholerae) but some are autotrophic and possess Bacteriochlorophyll (Chromatium).
5. They reproduce vegetatively by binary fission and endospore formation.
6. They exhibit variations which are due to genetic recombination and is achieved through conjugation, transformation and transduction.

Question 12.
Draw and label the ultra structure of a bacterium.
Answer:
The ultra structure of a bacterium:

Question 13.
Write in detail about Plasmids.
Answer:
Plasmids are extrachromosomal double stranded, circular, self – replicating, autonomous elements. They contain genes for fertility, antibiotic resistant and heavy metals. It also help 1 in the production of bacteriocins and toxins which are not found in bacterial chromosome.

The size of a plasmid varies from 1 to 500 kb usually plasmids contribute to about 0.5 to 5% of the total DNA of bacteria. The number of plasmids per cell varies. Plasmids are classified into different types based on the function. Some of them are F (Fertility) factor, R (Resistance) plasmids, Col (Colicin) plasmids, Ri (Root inducing) plasmids and Ti (Tumour inducing) plasmids.

Question 14.
Describe the structure of Gram positive and Gram negative bacterial cell wall using diagram.
Answer:
Most of the gram positive cell wall contain considerable amount of teichoic acid and teichuronic acid. In addition, they may contain polysaccharide molecules. The gram negative cell wall contains three components that lie outside the peptidoglycan layer:

1. Lipoprotein
2. Outer membrane and
3. Lipopolysaccharide.

Thus the different results in the gram stain are due to differences in the structure and composition of the cell wall.

Question 15.
Tabulate the differences between Gram positive and Gram negative bacteria.
Answer:

Question 16.
Give an account on respiration types of bacteria.
Answer:
Two types of respiration is found in bacteria.
They are:

1. Aerobic respiration
2. Anaerobic respiration

1. Aerobic respiration: These bacteria require oxygen as terminal acceptor and will not grow under anaerobic conditions (i.e. in the absence of O₂) Example: Streptococcus.

(i) Obligate aerobes: Some Micrococcus species are obligate aerobes (i.e. they must have 1 oxygen to survive).

2. Anaerobic respiration:
These bacteria do not use oxygen for growth and metabolism but obtain their energy from fermentation reactions. Example: Clostridium.

(i) Facultative anaerobes: There are bacteria that can grow either using oxygen as a terminal electron acceptor or anaerobically using fermentation reaction to obtain energy. When a facultative anaerobe such as E. coli is present at a site of infection like an abdominal abscess, it can rapidly consume all available O₂ and change to anaerobic metabolism producing an anaerobic environment and thus allow the anaerobic bacteria that are present to grow and cause disease.
Example: Escherichia coli and Salmonella.

(ii) Capnophilic bacteria: Bacteria which require CO₂ for their growth are called as capnophilic bacteria.
Example: Campylobacter.

Question 17.
Explain the mode of nutrition in bacteria.
Answer:
On the basis of their mode of nutrition bacteria are classified into two types namely autotrophs and heterotrophs.
I. Autotrophic bacteria:
Bacteria which can synthesis their own food are called autotrophic bacteria. They may be further subdivided as

A. Photoautotrophic bacteria – Bacteria use sunlight as their source of energy to synthesize food. They may be:

1. Photolithotrophs: In Photolithotrophs the hydrogen donor is an inorganic substance.
   • Green sulphur bacteria: In this type of bacteria the hydrogen donor is H₂S and possess pigment called Bacterioviridin. Example: Chlorobium.
   • Purple sulphur bacteria: For bacteria belong to this group the hydrogen donor is thiosulphate, Bacteriochlorophyll is present. Chlorophyll containing chlorosomes are present. Example: Chromatium.
2. Photoorganotrophs: They utilize organic acid or alcohol ‘as hydrogen donor. Example: Purple non-sulphur bacteria – Rhodospirillum.

B. Chemoautotrophic bacteria – They do not have photosynthetic pigment hence they cannot use sunlight energy. These type of bacteria obtain energy from organic or inorganic substance.

1. Chemolithotrophs: This type of bacteria oxidize inorganic compound to release energy.
Examples:

• Sulphur bacteria – Thiobacillus thiooxidans
• Iron bacteria – Ferrobacillus ferrooxidans
• Hydrogen bacteria – Hydrogenomonas and
• Nitrifying bacteria – Nitrosomonas and Nitrobacter.

2. Chemoorganotrophs: This type of bacteria oxidize organic compounds to release energy.
Examples:

• Methane bacteria – Methanococcus
• Acetic acid bacteria – Acetobacter and
• Lactic acid bacteria – Lactobacillus

II. Heterotrophic bacteria:
They are parasites (Clostridium and Mycobacterium), Saprophytes (Bacillus mycoides) or symbiotic (Rhizobium in root nodules of leguminous crops).

Question 18.
Describe the process of transformation.
Answer:
The process of transformation:

Transfer of DNA from one bacterium to another is called transformation. In 1928, the bacteriologist Fredrick Griffith demonstrated transformation in mice using Diplococcus pneumoniae. Two strains of this bacterium are present. One strain produces smooth colonies and are virulent in nature (S type), in addition another strain produced rough colonies and are avirulent (R type). When S-type of cells were injected into the mouse, the mouse died. When R-type of cells were injected, the mouse survived. He injected heat killed S-type cells into the mouse the mouse did not die.

When the mixture of heat killed S-type cells and R-type cells were injected into the mouse. The mouse died. The avirulent rough strain of Diplococcus had been transformed into S-type cells. The hereditary material of heat killed S-type cells had transformed R-type cell into virulent smooth strains. Thus the phenomenon of changing the character of one strain by transferring the DNA of another strain into the former is called transformation.

Question 19.
Define conjugation and its mechanism.
Answer:
J. Lederberg and Edward L Tatum demonstrated conjugation in E. coil in the year 1946. In this method of gene transfer the donor cell gets attached to the recipient cell with the help of pili.

The pilus grows in size and forms the conjugation tube. The plasmid of donor cell which has the F+ (fertility factor) undergoes replication. Only one strand of DNA is transferred to the recipient cell through conjugation tube. The recipient completes the structure of double stranded DNA by synthesizing the strand that complements the strand acquired from the donor.

Question 20.
Write in detail about transduction and its types.
Answer:
Zinder and Lederhcrg (1952) discovered transduction in Salmonella typhimurum. Phage mediated DNA transfer is called transduction.

Transduction is of two types:

1. Generalized transduction and
2. Specialized or restricted transduction.

1. Generalized transduction: The ability of a bacteriophage to carryr genetic material of any region of bacterial DNA is called generalised transduction.

2. Specialized or restricted transduction: The ability of the bacteriophage to carry only a specific region of the bacterial DNA is called specialized or restricted transduction.

Question 21.
List out the uses of bacteria in industries.
Answer:
The uses of bacteria in industries:

Question 22.
Explain the role of bacteria in antibiotic production and medicines.
Answer:
The role of bacteria in antibiotic production and medicines:

Question 23.
Explain the role of bacteria in soil fertility.
Answer:
The role of bacteria in soil fertility:

Question 24.
List out the salient features of cyanobacteria.
Answer:
The Salient Features Of Cyanobacteria:

• The members of this group are prokaryotes and lack motile reproductive structures.
• The thallus is unicellular in Chroococcus, colonial in Gloeocapsa and filamentous trichome in Nostoc.
• Gliding movement is noticed in some species (Oscillatoria).
• The protoplasm is differentiated into central region called centroplasm and peripheral region bearing chromatophore called chromoplasm.
• The photosynthetic pigments include c-phyocyanin and c-phycoerythrin along with myxoxanthin and myxoxanthophyll.
• The reserve food material is cyanophycean starch.
• In some forms a large colourless cell is found in the terminal or intercalary position called heterocysts. They are involved in nitrogen fixation.
• They reproduce only through vegetative methods and produce akinetes (thick wall dormant cell formed from vegetative cell), hormogonia (a portion of filament get detached and reproduce by cell division), fission and endospores.
• The presence of mucilage around the thallus is characteristic feature of this group. Therefore, this group is also called Myxophyceae.
• Sexual reproduction is absent.
• Microcystis aeruginosa and Anabaena flos-aquae cause water blooms and release toxins and affect the aquatic organism. Most of them fix atmospheric nitrogen and are used as biofertilizers (Example: Nostoc and Anabaena). Spirulina is rich in protein hence it is used as single cell protein.

Question 25.
Explain the various types of asexual reproduction in fungi.
Answer:
The Various Types Of Asexual Reproduction In Fungi:

1. Zoospores: They are flagellate structures produced in zoosporangia (Example: Chytrids)
2. Conidia: The spores produced on condiophores (Example: Aspergillus)
3. Oidia / Thallospores / Arthrospores: The hypha divide and develop into spores called oidia (Example: Erysiphe).
4. Fission: The vegetative cell divide into 2 daughter cells. (Example: Schizosaccharomyces – yeast).
5. Budding: A small outgrowth is developed on parent cell, which gets detached and become independent. (Example: Saccharomyces-yeast).
6. Chlamydospore: Thick walled resting spores are called chlamydospores. (Example: Fusarium).

Question 26.
Explain the various types of sexual reproduction in fungi.
Answer:
The Various Types Of Sexual Reproduction In Fungi:

1. Planogametic copulation: Fusion of motile gamete is called planogametic copulation.
   • Isogamy – Fusion of morphologically and physiologically similar gametes. (Example: Synchytrium).
   • Anisogamy – Fusion of morphologically or physiologically dissimilar gametes (Example: Allomyces).
   • Oogamy – Fusion of both morphologically and physiologically dissimilar gametes. (Example: Monoblepharis).
2. Gametangial contact: During sexual reproduction a contact is established between antheridium and oogonium (Example: Albugo).
3. Gametangial copulation: Fusion of gametangia to form zygospore. (Example: Mucor and Rhizopus).
4. Spermatization: Auninucleate pycniospore/microconidium is transferred to receptive hyphal cell (Example: Puccinia / Neurospora)
5. Somatogamy: Fusion of two somatic cells of the hyphae (Example: Agaricus).

Question 27.
List out the salient features of zygomycetes.
Answer:
The salient features of zygomycetes:

1. Most of the species are saprophytic and live on decaying plant and animal matter in the soil. Some lead parasitic life (Example: Entomophthora on housefly)
2. Bread mold fungi (Example: Mucor and Rhizopus) and Coprophilous fungi (Fungi growing on dung Example: Pilobolus) belong to this group.
3. The mycelium is branched and coenocytic.
4. Asexual reproduction by means of spores produced in sporangia.
5. Sexual reproduction is by the fusion of the gametangia which results in thick walled zygospore. It remains dormant for long periods. The zygospore undergoes meiosis and produce spores.

Question 28.
List out the salient features of ascomycetes.
Answer:
Features of Ascomycetes:

1. Ascomycetes include a wide range of fungi such as yeasts, powdery mildews, cup fungi, morels and so on.
2. Although majority of the species live in terrestrial environment, some live in aquatic environments both fresh water and marine.
3. The mycelium is well developed, branched with simple septum.
4. Majority of them are saprophytes but few parasites are also known. (Powdery mildew – Erysiphe).
5. Asexual reproduction takes place by fission, budding, oidia, conidia and chlamydospore.
6. Sexual reproduction takes place by the fusion of two compatible nuclei.
7. Plasmogamy is not immediately followed by karyogamy, instead a dikaryotic condition is prolonged for several generations.
8. A special hyphae called ascogenous hyphae is formed.
9. A crozier is formed when the tip of the ascogenous hyphae recurves forming a hooked cell. The two nuclei in the penultimate cell of the hypha fuse to form a diploid nucleus. This cell form young ascus.
10. The diploid nucleus undergo meiotic division to produce four haploid nuclei, which further divide mitotically to form eight nuclei. The nucleus gets organised into 8 ascospores.
11. The ascospores are found inside a bag-like structure called ascus. Due to the presence of ascus, this group is popularly called “Sac fungi”.
12. Asci gets surrounded by sterile hyphae forming fruit body called ascocarp.
13. There are 4 types of ascocarps namely Cleistothecium (Completely closed), Perithecium (Flask shaped with ostiole), Apothecium (Cup shaped and open type) and Pseudothecium.

Question 29.
List out the salient features of Basidiomycetes.
Answer:
Features of Basidiomycetes:

1. Basidiomycetes include puffballs, toad stools, Bird nest’s fungi, Bracket fungi, stink horns, rusts and smuts.
2. The members are terrestrial and lead a saprophytic and parasitic mode of life.
3. The mycelium is well developed, septate with dolipore septum (bracket like). Three types of mycelium namely primary (Monokaryotic), secondary (Dikaryotic) and tertiary are found.
4. Clamp connections are formed to maintain dikaryotic condition.
5. Asexual reproduction is by means of conidia, oidia or budding.
6. Sexual reproduction is present but sex organs are absent. Somatogamy or spermatisation results in plasmogamy. Karyogamy is delayed and dikaryotic phase is prolonged. Karyogamy takes place in basidium and it is immediately followed by meiotic division.
7. The four nuclei thus formed are transformed into basidiospores which are borne on sterigmata outside the basidium (Exogenous). The basidium is club shaped with four basidiospores, thus this group of fungi is popularly called “Club fungi”. The fruit body formed is called Basidiocarp.

Question 30.
Compare the characters of different types of Mycorrhiza.
Answer:
The characters of different types of Mycorrhiza:

Question 31.
List out the beneficial aspects of lichens.
Answer:
The Beneficial Aspects of Lichens:

1. Lichens secrete organic acids like oxalic acids which corrodes the rock surface and helps in weathering of rocks, thus acting as pioneers in Xerosere.
2. Usnic acid produced from lichens show antibiotic properties.
3. Lichens are sensitive to air pollutants especially to sulphur-di-oxide. Therefore, they are considered as pollution indicators.
4. The dye present in litmus paper used as acid base indicator in the laboratories is obtained from Roccella montagnei.
5. Cladonia rangiferina (Reindeer moss) is used as food for animals living in Tundra regions.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Viruses and viroids are infectious particles. How do you differentiate one from other?
Answer:
Viruses and viroids are infectious particles:

S.No.	Viruses	Viroids
1.	Viruses may have DNA or RNA as genetic material.	Viroids has ssRNA as genetic material
2.	Capsid is present	Capsid is absent

Question 2.
In R.H. Whittaker’s classification, how many kingdoms come under prokaryotes and how many kingdoms come under eukaryotes.
Answer:
Monera is the only prokaryotic kingdom in Whittaker’s classification, whereas protista, fungi, plantae and animalia comes under eukaryotes.

Question 3.
Arrange the following in a proper sequence with respect to fungal sexual cycle. Karyogamy, protoplasmic fusion, meiosis and spores production.
Answer:
Protoplasmic fusion → Karyogamy → Meiosis → Spores production.

Question 4.
List out major attributes and features that a cell must possess to call it as a living one.
Answer:
Growth, reproduction, metabolism, nutrition, movement and irritability, etc.

Question 5.
In five kingdom classification, actinomycetes and mycoplasma belongs to same kingdom.
(a) Name the kingdom.
(b) Which level of body organization does they exhibit.
Answer:
Actinomycetes and Mycoplasma belongs to kingdom Monera. Moneras are unicellular organisms.

Question 6.
Viruses are useful to us – Justify. Majority of viruses are harmful, causing wide range of diseases among organisms. But certain viruses of Baculoviridae group are commercially used as insecticides.
Answer:
E.g. Cytoplasmic polyhedrosis granulo viruses and Entomopox viruses.

Question 7.
Write the appropriate term for each of the following:
(a) Complex sugar that makes fungal cell wall.
(b) Blue green algae
Answer:
(a) Chitin
(b) Cyanobacteria.

Question 8.
Why fungi is not placed under kingdom plantae, though it has cell wall?
Answer:
Though fungi has cell wall, it differs from plants in their mode of nourishment. Fungi shows heterotropic mode of nutrition, whereas plants are autotrophs.

Question 9.
Which organisms is more complex and highly evolved among blue green algae, mushroom and maize? Give reason.
Answer:
Maize plant is more complex and highly evolved as it is a eukaryotic and autotrophic organism, showing tissue grade organisation. The blue green algae is a unicellular prokaryote and mushroom is a fungus, which is heterotrophic with no tissue grade of organisation.

Question 10.
Why viruses are not included in the category of microorganisms?
Answer:
Viruses possess characters of both living and non – living and also they does not have well defined body organisation. Hence they are not included in the category of microorganism.

Question 11.
Generally nucleic acid in viruses is present as single unit. Name any two viruses that possess segmented nucleic acid.
Answer:

1. Wound tumour virus
2. Influenza virus

Question 12.
Name any two recent viral diseases that threaten human life.
Answer:

1. Ebola
2. Nipah

Question 13.
When does a prophage enter lytic cycle?
Answer:
On exposure to UV radiation and chemicals, the prophage enters the lytic cycle.

Question 14.
Capsule layer helps the bacterium. How?
Answer:
Capsule in bacteria protects the cell from desiccation and antibiotics.

Question 15.
Imagine yourself as Carl Woese and explain your colleagues about your classification.
Answer:
I (Carl Woese) classified the living organisms into three domains based on the differences in rRNA nucleotide sequence and lipid structure of the cell membrane. The three domains are Bacteria, Archae and Eukarya.

Question 16.
Name the strains used in Gram staining procedure.
Answer:
Crystal violet and safranin.

Students can Download Tamil Nadu 12th English Model Question Paper 4 Pdf, Tamil Nadu 12th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th English Model Question Paper 4

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1= 20]

Choose the correct synonyms for the underlined words from the options given:

Question 1.
They had tangled hair.
(a) shiny (b) matted (c) straight (d) smooth
Answer:
(b) matted

Question 2.
The best manner of making it is the subject of violent disputes.
(a) agreements (b) applauses (c) conflicts (d) discussions
Answer:
(c) conflicts

Question 3.
Suffering seems cruelly prevalent in the world today.
(a) unique (b) uncommon (c) common (d) fair
Answer:
(c) common

Choose the correct antonyms for the underlined words from the options given:

Question 4.
They wear over-sized high altitude boots.
(a) height (b) light (c) depth (d) sound
Answer:
(c) depth

Question 5.
I felt a sense of freedom.
(a) liberty (b) slavery (c) equality (d) fraternity
Answer:
(b) slavery

Question 6.
They were now frozen solid.
(a) thick (b) liquid (c) concrete (d) firm
Answer:
(b) liquid

Question 7.
Choose the correct combination for the compound word “bleaching Powder”.
(a) Noun+ Verb (b) Verb + Noun (c) Gerund + Noun (d) Preposition + Noun
Answer:
(c) Gerund + Noun

Question 8.
Choose the correct expansion of CMO.
(a) Chief Marketing Officer (b) Chief Managing Officer (c) Chief Mediocre Officer (d) Chief Maritime Organisation
Answer:
(a) Chief Marketing Officer

Question 9.
Choose the meaning of the foreign word in the sentence:
‘Tina really loves to make croissants because they taste better than other types of bread’.
(a) sweater (b) pastry (c) beverage (d) soup
Answer:
(b) pastry

Question 10.
Choose the correct combination for the blended word “Brunch”.
(a) Break + lunch (b) Post + lunch (c) breaking + lunch (d) Breakfast + lunch
Answer:
(d) Breakfast + lunch

Question 11.
Choose the clipped form of the word diskette.
(a) disk (b) disc (c) dike (d) kette
Answer:
(b) disc

Question 12.
A collection of selected literary passages is called
(a) Eulogy (b) Arthrology (c) Analogy (d) Anthology
Answer:
(d) Anthology

Question 13.
Form a derivative by adding the right prefix to the word experienced.
(a) re (b) in (c) un (d) dis
Answer:
(b) in

Question 14.
Fill in the blanks with a suitable relative pronoun.
Most of the girl students Mrs. Jeyaseela had informed came for the special class.
(a) which (b) who (c) whom (d) that
Answer:
(c) whom

Question 15.
Fill in the blanks with a suitable preposition.
Shankar has been absent Tuesday.
(a) from (b) for (c) since (d) till
Answer:
(c) since

Question 16.
Choose the correct question tag for the following statement.
Amanda would like to get a scholarship,?
(a) wouldn’t she (b) was she (c) can’t she (d) has she
Answer:
(a) wouldn’t she

Question 17.
Choose the suitable meaning or idiom found in the following sentence.
I have given my bike to a new mechanic for service. Hope he delivers the goods!
(a) produce bike (b) to create a bike (c) give the bike (d) do what is expected or promised
Answer:
(d) do what is expected or promised

Question 18.
Substitute the underlined word with the appropriate polite alternative. Sam is buying a used car.
(a) pre-loved (b) tried (c) different (d) challenged
Answer:
(a) pre-loved

Question 19.
Choose the correct sentence pattern for the following sentence.
They were selling wild strawberries.
(a) SVOA (b) SVCO (c) SVIODO (d) SVIODOA
Answer:
(b) SVCO

Question 20.
Fill in the blank with a suitable phrasal verb.
Jacopo to the door handle of the car.
(a) looked at (b) brought forth (c) sent out (d) came up
Answer:
(d) came up

Part II
Section -1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“All through the summer at ease we lay,
And daily from the turret wall
We watched the mowers in the hay”
(a) How did the soldiers spend the summer days?
(b) What could they watch from the turret wall?
Answer:
(a) They spent the summer days gazing out of the castle. They were ready to shoot the enemy at sight who were at half-a kilometer distance. But none came near. So, they were relaxed.
(b) They could watch the farmers mowing at a distance from inside their turret walls.

Question 22.
“But not because of its magnificence Dear is the Casuarina to my soul:
Beneath it we have played; though years may roll,”
(a) What is not the cause for Toru Dutt’s love for the Casuarina tree?
(b) What makes the tree dear to the poet?
Answer:
(a) The magnificence or the impressive appearance of the tree is not the cause for Toru Dutt’s love for the Casuarina tree.
(b) As children, the poet and her friends had played under the tree. This experience has made the tree dear to the poet.

Question 23.
“And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances; ”
(а) Whom does justice refer to?
(b) Describe his appearance.
Answer:
(a) Justice refers to a man in his fifth stage when he becomes critical of everyone else’s opinion in life.
(b) He has a pot belly and is fond of eating delicacies.

Question 24.
“…I mete and dole
Unequal laws unto a savage race,
That hoard, and sleep, and feed, and know not me. ”
(a) What does Ulysses do?
(b) Did he enjoy what he was doing? Give reasons.
Answer:
(a) Ulysses, like a grocery shop owner, measures and delivers rewards and punishments unto a large number of uncivilized citizens.
(b) No, he did not enjoy his work. He does not like the idea of ministering variable justice to people who like “drones” or animals just eat, sleep and multiply their kind. He wants to leave such work to his son.

Question 25.
“Life is hard; be steel; be a rock.”
(а) How should one face life?
(b) Identify the figure of speech in the above line.
Answer:
(a) One should face life like a steel.
(b) Metaphor

Question 26.
“Legs wide, arms locked behind,
As if to balance the prone brow
Oppressive with its mind.”
(а) Whose action is described here?
(b) What is meant by prone brow?
Answer:
(a) Napoleon Bonaparte’s action is described here.
(b) “Prone brow” means brow accustomed or inclined down to contemplate on serious matters.

SECTION – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the following dialogue:
Answer:
Michael: “What are you doing here, Shil? I haven’t seen you since June.”
Shil : “I’ve just come back from my holiday in Ireland.” .
Michael asked Shil what she was doing there and that he hadn’t seen her since June. Shil said that she had just come back from her holiday in Ireland.

Question 28.
Sheena must consult a dermatologist. Otherwise she will not know what her problem is. (Combine using ‘if’)
Answer:
If Sheena does not consult a dermatologist, she will not know what her problem is.

Question 29.
Rewrite the sentence making an inversion in the conditional clause.
Answer:
If you should require any further information, please contact us at our office in Madley Street.
Should you require any further information, please contact us at our office in Madley street.

Question 30.
Leonardo da Vinci, who was a renowned painter and sculptor, was also an inventor and scientist. (Change the following into a simple sentence)
Answer:
Leonardo da Vinci, a renowned painter and sculptor, was also an inventor and scientist.

Part-III
Section -1

Explain any two of the following with reference to the context [2 x 3 = 6]

Question 31.
How can this shameful tale be told?
Answer:
Reference: This line is from Edwin Muir’s poem “The Castle”.

Context: The narrator feels disgraced that their strong castle was overtaken without a groan.

Explanation: There was no fight. They were helpless because under the cover of darkness, they were sold for gold by the aged warder. It was a shameful and treacherous act of betrayal. The castle had fallen due to the greed of an aged warder. The narrator is hesitant to disclose the shameful betrayal to outsiders.

Question 32.
When first my casement is wide open thrown At dawn, my eyes delighted on it rest;
Answer:
Reference: These lines are from the poem ‘Our Casuarina Tree” written by Torn Dutt.

Context: The poet says these words while talking about the early influences of the Casuarina tree in her life.

Explanation: The young poet’s day started with seeing the tree with delight. The very first thing she used to see was the Casuarina tree through her casement (i.e.) door like window. The emotional bonding between herself and the giant Casuarina tree is beyond reason.

Question 33.
I cannot rest from travel: I will drink Life to the lees:
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.

Context: Ulysses says these words after settling down in his kingdom. He feels restless as he has stopped travelling and having adventures.

Explanation: Ulysses, after spending many years in the seas returns to Ithaca and starts ruling his country. However, his heart is not in the administration of his kingdom. He wants to sail again. At this context he says these words. He wishes to enjoy life to the fullest and so he can’t afford to idle away his remaining life as a king.

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
Describe the appearance of Nicola and Jacopo.
Answer:
One boy had worn Jersey and cut-off khaki pants. The other had a shortened army tunic gathered in loose folds about his skinny frame. They had tangled hair and dark earnest eyes.

Question 35.
What thoughts troubled Dr. Christiaan Barnard as he neared the end of his career as a heart surgeon?
Answer:
Towards the end of his career, Dr. Christiaan Barnard was troubled by the suffering of people and especially of young children. He could not accept the fact that twelve million children are unlikely to reach the age of one and about six million children die annually before reaching the age of five.

Question 36.
How did the firm snow at the higher regions fill them with hope?
Answer:
They were a little perturbed by slippery soft snow. As they reached higher up, they felt better. As one bottle of oxygen got exhausted, their load was now less. As Hillary’s axe bit into the first steep slope of the ridge, his high hopes were realized. The snow was crystalline and film. They were able to make comfortable belays to haul themselves up slowly.

Section – 3

Answer any three of the follow ing questions in about 30 words each.

Question 37.
Study the pie chart given and answer the questions that follow.

(a) Name the drug which has the least sale record in the store.
(b) How many types of drugs are being sold at this store?
(c) Which drug is twice the sale of analgesics?
Answer:
(a) Sedatives have the least sale record in the store.
(b) Five types of drugs are sold in the store.
(c) Antibiotics is sold twice as much as analgesics.

Question 38.
Write a dialogue of minimum 3 exchanges between a tourist and a guide.
Answer:
Tourist: Could you please tell me which are the important locations in Kanchipuram to be seen?
Guide: Of course yes, you must visit Kailasanathar Temple and Thennangur Panduranga Temple.
Tourist: Could I cycle to those locations?
Guide: Yes of course, let me bring my bicycle too. I will show you important things just for a price of Rs. 1000/
Tourist: Don’t you think your price is a bit high?
Guide: Okay. Last price Rs. 750/-

Question 39.
Describe the process of making a cup of tea.
Answer:
Boil water in a bowel. Add tea leaves. Wait till the water turns red or black. Now turn off the stove. Use a filter to remove the tea leaves from the essence. Now add hot milk and sugar to your taste. Those who prefer black tea need not add milk.

Question 40.
Complete the proverbs using the word given below.
Answer:
(a) A soft answer turns away ……………….. (villain, wrath, friend)
(b) A picture is worth a ……………….. words, (beautiful, thousand, describe)
(c) A bird in hand is worth ……………….. (hundreds, the freedom, two in the bush)
Answer:
(a) wrath (b) thousand (c) two in the bush

PART – IV

Answer the following questions: [7 x 5 = 35]
Answer in a paragraph in about 150 words.

Question 41.
What message is conveyed through the story, ‘Two Gentlemen of Verona’?
Answer:
Adversity is a touchstone of virtue. Both the little boys resemble tea leaves. Their best essence comes to limelight when they find themselves in hot waters. But they don’t complain. One does not give up on family relations when misfortune strikes. The bond becomes strong as the bond demands selfless sacrifice. The nobility of human life emerges from the precious lessons one learns from the supreme sacrifices of Nicola and Jacopo. One who shoulders the responsibility of taking care of the loved ones, irrespective of age, is an exemplary gentleman worthy of emulation. The primary motivational force of an individual is to find meaning in life. Both the little boys had a specific purpose for their life. They had made up their minds to do what ever work possible to earn money to save their sister and restore her singing career.

Or

How did Hillary and Tenzing prepare themselves before they set off to the summit?
Answer:
They started up their cookers and drank lots of lemon juice and sugar. Then they took sardines and biscuits. Hillary cleaned the ice off the oxygen sets. He rechecked and tested them. He had removed his boots which had become wet the day before. They were now frozen solid. It would be very challenging to start climbing ice-cold Himalayas with such wet and chilling boots.

So, he cooked them over the fierce flame of Primus and managed to soften them up. They were also conscious of the probability of braving snow storms during the ascent. They fortified their clothing with wind proof and also pulled three pairs of gloves silk, woollen, and windproof on to their hands. At 6.30 a.m. they crawled out of their tent into the snow. They hoisted their 30 lb. of oxygen gear on their backs. Connecting their oxygen masks they turned on the valves to bring life-giving oxygen into their lungs. Taking a few deep breaths, they got ready to go.

Question 42.
Describe the reminiscences of the poet, when she sees the casuarina tree.
Answer:
The poet remembers how her days started with the sight of the Casuarina tree from her casement. She remembers how her loving companions played under the giant Casuarina tree. The memory of her beloved companions bring hot tears because they had succumbed to cruel tuberculosis. She remembers how well the tree accommodated birds to sing songs during days and nights. The tree had allowed the creeper to embrace it like a lady love. Though it sapped its vitality, like a gallant lover, allowed the creeper to stay around its neck like a scarf. She remembers how a baboon seated at the crest of the tree had watched beautiful sunrise while her young ones were leaping and playing in the lower branches of the giant tree.

Or

Explain how the poet guides his son who is at the threshold of manhood, to face the challenges of life.

The poet shares his wisdom with his son who is at the threshold of manhood. He persuades his son to be hard like steel or rock to withstand challenges and unforeseen betrayals in life. A person with soft heart will crumble before a breach of trust. Similarly he wants his son to be discerning enough to be soft when needed to grow like a frail flower plant splitting a rock. Occasionally one has to go with the current because life is at times fertile with a lot of opportunities to grow even among the harshest circumstances. ‘Rich soft wanting’ can help a person to win against all odds. He reiterates this idea by explaining how gentleness can reform a hardened criminal when lashes would, in contrast, harden them further.

Question 43.
Write a paragraph of about 150 words by developing the following hints:
Martha and John – grudge – underpaid two decades – punishment – condemned the bribe – abusive language – one hundred thousand dollars – changed attitude – testify – three words – corrupt practices.
Answer:
Martha and John cherished a grudge against Gresham for having underpaid Baldwin for about two decades. They openly said that he deserved punishment. Initially Gresham condemned the bribe claimed as a difference in salary paid. Had he been made a similar offer he claimed he would have asked him to go to the devil. Baldwin said that having been a friend he couldn’t use such an abusive language. Casually John asked how much he offered.

Baldwin said that it was one hundred thousand dollars. Suddenly Martha and John changed their attitude towards Gresham. John viewed a shame if Gresham got indicted because he shared his name. Martha also desperately tried to convince him not to testify against Gresham. John even suggested that he could say those three words “I don’t remember” as the depositor would not lose a cent. Baldwin felt miserable because the family members who he wanted to feel proud of his uprightness wanted him to crossover to the side of corrupt practices because of the generous offer made by Gresham.

Or

“Remember Caesar” – light hearted comedy – tragic hero – rose tree planting-great loss to England – great impartial judge – Mrs. Weston’s entry – eating a game pie – immediate worry – French and Dutch – assassinate – pepper – bread – judge and his assistant – pet dishes – handful of candles – stay underground – infernal machine – doctor disapproves of excitement – nonsense and whimsical behavior – 15th March.
Answer:
Remember Caesar is a light hearted comedy. The name Caesar is the name of a. tragic hero Julius Caesar in one of the plays of William Shakespeare. Weston gives an appointment to Mr. Caesar to discuss rose tree planting work in his garden. Just to remind him of the proffered appointment, he scribbles two words “Remember Caesar” and keeps that scrap of paper in his coat pocket. Roger, the assistant of the judge elevates him to the level of a tragic hero. He flatters the judge that his death could be a great loss to England as he was a great impartial judge whose integrity could not be bargained or bribed. Mrs. Weston’s entry lightens up the whole drama and the tragedy transitions into a dark comedy.

Lady Weston keeps hinting about earlier attempts when he was eating a game pie. She becomes seriously playful when she says, “you’ve always wanted to be a great man and now you’ve got your wish. They don’t assassinate nobodies.” She cool-headedly tells her husband that he can stay indoors and have drinks. Lady Weston asks her husband if someone was planning to murder him. He says, “obviously”. She gives a curt retort, “I wonder someone hasn’t done it long ago. A great many people must hate judges. And you are a strict judge, they say.”

When Weston says he has instructed Roger to barricade all doors, her immediate worry is about the grocery awaited. She asks if he was expecting both French and Dutch together in the attempt to assassinate. When Weston asks if a little pepper was more important to her than her husband’s life, she says that he would be the first person to complain if the bread was short and the gravy thin. The cook disobeys the order. She has never been “behind the bars”. She will handle any one who enters by the kitchen. A judge and his assistant want the door locked but a cook is not afraid. Lady Weston doesn’t want to send away the cook. She is practical. She wants her to stay back to cook his pet dishes.

Lady Weston brings a handful of candles to keep the room lit if Mr. Weston has to stay underground for a while. A velvet coat is construed as an infernal machine that could blow up the whole place is smothered with books inside a pail of water to the great annoyance of Lady Weston. She casually reminds her husband that doctor disapproves of excitement. It was only on this account, Lady Weston bears the nonsense and whimsical behaviour of Mr. Weston. The arrival of Mr. Caesar only sorts out the knot. The judge remembers to meet Mr. Caesar on 15th March.

Question 44.
Write a summary or Make notes of the following passage.
Answer:
The greatness of a country depends upon its people. India is fortunate to have vast human resources. Our countrymen are second to none in intelligence and in doing hard work. From the ancient period, Indians excelled in art architecture, knowledge of metals, medicines, literature etc. After our independence in 1947, the government took steps to improve our country in all spheres. By the five-years plans, provisions are made for the development of the country.

The first five-year plan was specially designed to improve irrigational methods. By the green revolution, we attained self-sufficiency in the field of agricultural production. The present age is the atomic age. India too established an Atomic Energy Commission under the guidance of Dr. Bhabha. India made the first successful nuclear explosion on 18th May, 1974. This made India the sixth member of the world nuclear club. It was purely the efforts of the Indian Scientists alone. We have sent our Indian Cosmonaut to space also on 3rd April, 1984. Shri Rakesh Sharma, the best pilot of the India Air Force travelled.into space with two Soviet Spaceman in Soyuz I Spaceship.

Summary

No. of words given in the original passage: 185
No. of words to be written in the summary: 185/3 = 62 ± 5

Rough draft
Indians have been great human resources. Since ancient times, the have excelled in various fields such as metallugry nedicineand architecture in ancient period itself. After 1947, the Government through five year plan attend selfi-sufficiency in agricultural production. Gradually the Government focused on developing atoimic energy andspace technology. Atomic implosion in 1974 and conquest of space in 1984 are best examples of Indians success in joining space club and energy development.

Fair draft True Indians
Ancient Indians proved their expertise in metallurgy, medicine and architecture. After 1947, the country focused on agricultural self-sufficiency through five year plans. She attained it in, agricultural production. Then India joined the small band of atomic energy producing countries and by atomic explosion in 1974 proved her power. She also joined space club, in 1984 by sending Rakesh Sharma to space.

No. of words in the summary: 61

Or

Notes
Title: True Indians

Indians
best human resources
excelled in architecture, metallurgy and medicine

Five year plans
initial agriculture development
Energy development focus

Drive future
atomic energy nuclear implosion on 18th may 1974.

Space
Indian Cosmonaut on 3rd April 1984 – Rakesh Sharma I Cosmonaut.

Question 45.
As Cultural Secretary of your school, write a letter to the Cultural Program Coordinator, DAV Public School requesting him/her for details regarding the inter¬school competitions to be held on 20.3.2020.
Answer:
05 Jan, 2019
From
B. Amrita
Chettinad School
Ramapuram
Chennai

To
Cultural Program
Co-ordinator
DAV Public School
Perambur
Chennai
Sir,

Sub: Information regarding inter-school competitions

We have received the notification about the inter-school competitions to be held in your school. Kindly confirm the details that is mentioned in the poster.

• Participants from school: 30 No.’s
• Registration fees : Rs. 500/- (DD attached)
• Time of entry: 8.30 a.m.
• No. of competitions a participant can take part in: 3
• No. of teachers to accompany students: 3
• Last date for registration: 10.02.2020

The specific conditions and eligibility criteria may please be intimated to us.
I hope you will provide the necessary information at the earliest.

Yours faithfully
B. Amrita

Sports Secretary

Address on the envelope
To :
The Cultural Programme Co-ordinator
DAV Public School, Perambur
Chennai

[OR]

Write a paragraph in about 150 words on “The Importance Of Education That Gives Power To An Individual.”

Education Gives Power

Education provides us knowledge. It trains our mind and sharpens our skills and abilities. Education refines our tastes and temperaments and builds our thought process. Vocational courses help young boys to earn and learn together. They provide means of earning livelihood and open the route to employment. Professional courses, as is evident from the name itself, equip us for adopting various professions. Some of these highly skilled professionals seek placements or jobs in esteemed companies and business concerns. Thus education is important for our survival.

Decent living is impossible without good income or high salary. Education improves the quality of our life and frees us of superstition, foolish, meaningless mind-blocks and rituals. If women are educated the whole family benefits as the food is hygienically prepared, children are healthy, well-mannered and disciplined. Education gives us power over our environment. We can control the situation and shape our destiny. Education spreads awareness among people and gives them freedom from socialills. It makes people independent by providing them means to earn their living. They become responsible citizens and realise their rights and duties. In short, education gives one power.

Question 46.
Spot the errors and rewrite the sentences correctly.
(a) The collection ‘Fragrant flowers’ are quite interesting.
(b) It is an great honour to welcome you.
(c) Beside tennis, I play Cricket.
(d) The Chief Manager along with his deputies are visiting the factory tomorrow.
(e) Many have arrested at the meeting.
Answer:
(a) The collection ‘Fragrant flowers’ is quite interesting.
(b) It is a great honour to welcome you.
(c) Besides tennis, I play Cricket.
(d) The Chief Manager along with his deputies is visiting the factory tomorrow.
(e) Many have been arrested at the meeting.

[OR]

Fill in the blanks correctly.
(а) This ………………. is made out of plaster of Paris ………………. and painted after it solidifies. (flower/flour)
(b) I ………………. speak two languages. (Use a modal in the given blank.)
(c) I am so tired! I ………………. take a nap. (use a semi-modal)
(d) I ………………. (sing) since I was four years old. (use a proper tense)
Answer:
(a) flower/flour (b) can (c) need to (d) have been singing

Question 47.
Identify each of the following sentences with the fields given below
(a) The flight was cancelled due to fog.
(b) Spicy food can cause acidity in the stomach.
(c) Meena stumbled upon a chance to practice running a race.
(d) The company has recommended a dividend of 75 percent.
(e) The monitor is not working. Get it repaired.
[Sports, Weather, Commerce, Nutrition and Dietetics, Computer]
Answer:
(a) Weather (b) Nutrition and Dietetics (c) Sports (d) Commerce (e) Computer

[OR]

Read the following passage and answer the questions in your own words.
During the holidays, many people hang mistletoes over doorways. People share kisses under this evergreen plant, which is a popular Christmas tradition. However, one should never let the image of a romantic plant used during the happy times of the holidays fool. In the forests where they’re from, mistletoes can do some real damage. Let’s see the different ways in which they do harm others.

The mistletoe plant is evergreen. This means it has leaves that remain green throughout the year. It is poisonous and has white berries and small, yellow flowers. The mistletoe lives on other plants, taking water and nutrients from these plants. For this reason, mistletoes are considered parasites.

The white berries of the mistletoes contain seeds. Some birds and mammals like to feed on these berries. When they do, the seeds may attach to the animal eating the berries. The animal may carry the seeds to another part of the tree or shrub. They may also carry the seeds to another plant altogether. The seeds start to grow roots that dig through the bark of the tree or shrub. The roots grow into the tissues of the plant they’ve taken over.

That’s how mistletoes take nutrients and water away from the host plants. Mistletoe can be hard to remove once it infects a plant. The best way to fight off a mistletoe infestation is to cut off the infected branch completely. If the mistletoe takes over more parts of the plant, it can start to weaken the plant and make it harder for it to grow.

As mistletoes grow in the trees, they become a thick mix of branches and stems. This big mass is sometimes called a “witch’s broom.” Some animals nest in these witches’ brooms. These animals include chickadees, house wrens, and most Cooper’s hawks.

Questions:
(a) What is a popular Christmas tradition?
(b) Why is mistletoes considered a parasite?
(c) What is the best way to ward off the mistletoe infestation? id) What is called a ‘witch’s broom’?
(ie) Which animals nest in the witches’ brooms?
Answers:
(a) During Christmas holidays, many people hang mistletoes over doorways. People share kisses under mistletoes, an evergreen plant hung over doorways. This is a popular Christmas tradition.
(b) The mistletoe lives on other plants, taking water and nutrients from these plants. Hence, mistletoes are considered parasites.
(c) Mistletoe can be hard to remove once it infects a plant. The best way to fight off a mistletoe infestation is to cut off the infected branch completely.
(d) Mistletoes that grow in the trees become a thick mix of branches and stems. This big mass is called a “witch’s broom.”
(e) Some animals like chickadees, house wrens, and most Cooper’s hawks nest in the witches’ brooms.

Students can Download Accountancy Chapter 5 Trial Balance Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 5 Trial Balance

Samacheer Kalvi 11th Accountancy Trial Balance Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Trial balance is a ………………
(a) Statement
(b) Account
(c) Ledger
(d) Journal
Answer:
(a) Statement

Question 2.
After the preparation of ledger, the next step is the preparation of ………………
(a) Trading account
(b) Trial balance
(c) Journal
(d) Profit and loss account
Answer:
(b) Trial balance

Question 3.
The trial balance contains the balances of ………………
(a) Only personal accounts
(b) Only real accounts
(c) Only nominal accounts
(d) All accounts
Answer:
(d) All accounts

Question 4.
Which of the following is/are the objective(s) of preparing trial balance?
(a) Serving as the summary of all the ledger accounts
(b) Helping in the preparation of final accounts
(c) Examining arithmetical accuracy of accounts
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Question 5.
While preparing the trial balance, the accountant finds that the total of the credit column is short by ₹ 200. This difference will be ………………
(a) Debited to suspense account
(b) Credited to suspense account
(c) Adjusted to any of the debit balance
(d) Adjusted to any of the credit balance
Answer:
(b) Credited to suspense account

Question 6.
A list which contains balances of accounts to know whether the debit and credit balances are matched is ………………
(a) Journal
(b) Day book
(c) Trial balance
(d) Balance sheet
Answer:
(c) Trial balance

Question 7.
Which of the following method(s) can be used for preparing trial balance?
(a) Balance method
(b) Total method
(c) Total and Balance method
(d) (a), (b) and (c)
Answer:
(d) (a), (b) and (c)

Question 8.
The account which has a debit balance and is shown in the debit column of the trial balance is ………………
(a) Sundry creditors account
(b) Bills payable account
(c) Drawings account
(d) Capital account
Answer:
(c) Drawings account

Question 9.
The difference of totals of both debit and credit side of trial balance is transferred to ………………
(a) Trading account
(b) Difference account
(c) Suspense account
(d) Miscellaneous account
Answer:
(c) Suspense account

Question 10.
Trial balance is prepared ………………
(a) At the end of the year
(b) On a particular date
(c) For a year
(d) None of the above
Answer:
(b) On a particular date

II. Very Short Answer Questions

Question 1.
What is trial balance?
Answer:
Trial balance is a statement containing the debit and credit balances of all ledger accounts on a particular date. It is arranged in the form of debit and credit columns placed side by side and prepared with the object of checking the arithmetical accuracy of entries made in the books of accounts and to facilitate preparation of financial statements.

Question 2.
Give the format of trial balance.
Answer:
Trial balance is prepared in the following format under the balance method:

Question 3.
What are the methods of preparation of trial balance?
Answer:

1. Balanced method
2. Total method
3. Total and balance method

Question 4.
State whether the balance of the following accounts should be placed in the debit or the credit column of the trial balance:

1. Carriage outwards
2. Carriage inwards
3. Sales
4. Purchases
5. Bad debts vi. Interest paid
6. Interest received
7. Discount received
8. Capital
9. Drawings
10. Sales returns
11. Purchase returns

Answer:
Trial balance as on 31^st March, 2017

III. Short Answer Questions

Question 1.
What are the objectives of preparing trial balance?
Answer:
Trial balance is prepared with the following objectives:
1. Test of arithmetical accuracy:
Trial balance is the means by which the arithmetical accuracy of the book – keeping work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that posting from subsidiary books, balancing of ledger accounts, etc. are arithmetically correct. However, there may be some errors which are not disclosed by trial balance.

2. Basis for preparing final accounts:
Financial statements, namely, trading and profit and loss account and balance sheet are prepared on the basis of summary of ledger balances obtained from the trial balance.

3. Location of errors:
When the trial balance does not tally, it is an indication that certain errors have occurred. The errors may have occurred at one or more of the stages of accounting process, namely, journalising or recording in subsidiary books, totalling subsidiary books, posting in ledger accounts.

Balancing the ledger accounts, carrying ledger account balances to the trial balance and totalling the trial balance columns, etc. Hence, the errors should be located and rectified before preparing the financial statements.

4. Summarised information of ledger accounts:
The summary of ledger accounts is shown in the trial balance. Ledger accounts have to be seen only when details are required in respect of an account.

Question 2.
What are the limitations of trial balance?
Answer:
The following are the limitations of trial balance:

1. It is possible to prepare trial balance of an organisation, only if the double entry system is followed.
2. Even if some transactions are omitted, the trial balance will tally.
3. Trial balance may tally even though errors are committed in the books of account,
4. If trial balance is not prepared in a systematic way, the final accounts prepared on the basis of trial balance may not depict the actual state of affairs of the concern.
5. Agreement of trial balance is not a conclusive proof of arithmetical accuracy of entries made in the accounting records. This is because there are certain errors which are not : disclosed by trial balance such as complete omission of a transaction, compensating errors and error of principle.

Question 3.
‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records’. Do you agree with this statement? Give reasons.
Answer:
Yes, I agree with the statement. ‘A trial balance is only a prima facie evidence of the arithmetical accuracy of records.

Reasons:
Trial balance is the means by which the arithmetical accuracy of the book – keeping. work is checked. When the totals of debit column and credit column in the trial balance are equal, it is assumed that porting from subsidiary books, balancing of ledger accounts, etc.

IV. Exercises

Question 1.
Prepare a trial balance with the following information: (3 Marks)

Answer:
Trial balance

Question 2.
Prepare the trial balance from the following information: (3 Marks)

Answer:
Trial Balance

Question 3.
Prepare the trial balance from the following balances of Chandramohan as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Chandramohan as on 31^st March, 2017

Question 4.
Prepare the trial balance from the following balances of Babu as on 31^st March, 2016. (3 Marks)

Answer:
Trial balance of Babu as on 31^st March, 2016

Question 5.
From the following balances of Aijun, prepare the trial balance as on 31^st March, 2018. (3 Marks)

Answer:
Trial balance of Arjun as on 31^st March, 2018

Question 6.
Prepare the trial balance from the following balances of Rajesh as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Rajesh as on 31^st March, 2017

Question 7.
Prepare the trial balance from the following balances of Karthik as on 31^st March, 2017. (3 Marks)

Answer:
Trial balance of Karthik as on 31^st March, 2017

Question 8.
From the following balance of Rohini, Prepare the trial balance as on 31^st March, 2016. (3 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 9.
Balan who has a car driving school gives you the following ledger balances. Prepare trial balance as on 31^st December, 2016. (3 Marks)

Answer:
Trial balance of Balan as on 31^st March, 2017

Question 10.
The following balances are extracted from the books of Ravichandran on 31^st December, 2016. Prepare the trial balance. (5 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 11.
From the following balances, Prepare trial balanceof Baskar as on 31^st March 2017. Transfer the difference, if any, to suspense account. (5 Marks)

Answer:
Trial balance of Rohini as on 31^st March, 2017

Question 12.
From the following balances extracted from the books of Rajeshwari as on 31^st March, 2017, prepare the trial balance.

Answer:
Trial balance of Rajeshwari as on 31^st March, 2017

Question 13.
Correct the following trail balance:

Answer:
The Corrected Trial balance

Textbook Case Study Solved

Question 1.
Mary runs a textile store. She has prepared the following trial balance from her ledger balances. Her trial balance does not tally. She needs your help to check whether what she has done is correct.

Solution:
There are some errors in Mary’s Trial balance. The corrected trial balance has been written below:
The Corrected Trial balance of Mary

Students can Download Computer Applications Chapter 2 An Introduction to Adobe Pagemaker Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Applications Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 2 An Introduction to Adobe Pagemaker

Samacheer Kalvi 12th Computer Applications An Introduction to Adobe Pagemaker Text Book Back Questions and Answers

I. Choose The Correct Answer:

Question 1.
DTP stands for ………………………..
(a) Desktop Publishing
(b) Desktop Publication
(c) Doctor To Patient
(d) Desktop Printer
Answer:
(a) Desktop Publishing

Question 2.
……………………….. is a DTP software.
(a) Lotus 1-2-3
(b) PageMaker
(c) Doctor To Patient
(d) Lotus 1-2-3
Answer:
(b) PageMaker

Question 3.
Which menu contains the New option?
Answer:
(a) File menu
(b) Edit menu
(c) Layout menu
(d) Type menu
Answer:
(a) File menu

Question 4.
In PageMaker Window, the area outside of the dark border is referred to as ……………………………
(a) page
(b) pasteboard
(c) blackboard
(d) dashboard
Answer:
(b) pasteboard

Question 5.
Shortcut to close a document in PageMaker is ……………………………
(a) Ctrl + A
(b) Ctrl + B
(c) Ctrl + C
(d) Ctrl + W
Answer:
(d) Ctrl + W

Question 6.
A …………………………… tool is used for magnifying the particular portion of the area.
(a) Text tool
(b) Line tool
(c) Zoom tool
(d) Hand tool
Answer:
(c) Zoom tool

Question 7.
…………………………… tool is used for drawing boxes.
(a) Line
(b) Ellipse
(c) Rectangle
(d) Text
Answer:
(c) Rectangle

Question 8.
Place option is present in …………………………… menu.
(a) File
(b) Edit
(c) Layout
(d) Window
Answer:
(a) File

Question 9.
To select an entire document using the keyboard, press ……………………………
(a) Ctrl + A
(b) Ctrl + B
(c) Ctrl + C
(d) Ctrl + D
Answer:
(a) Ctrl + A

Question 10.
Character formatting consists of which of the following text properties?
(a) Bold
(b) Italic
(c) Underline
(d) All of these
Answer:
(d) All of these

Question 11.
Which tool lets you edit text?
(a) Text tool
(b) Type tool
(c) Crop tool
(d) Hand tool
Answer:
(a) Text tool

Question 12.
Shortcut to print a document in Pagemaker is ……………………………
(a) Ctrl + A
(b) Ctrl + P
(c) Ctrl + C
(d) Ctrl + V
Answer:
(b) Ctrl + P

Question 13.
Adobe PageMaker is a …………………………… software.
Answer:
Page layout

Question 14.
…………………………… Bar is the topmost part of the PageMaker window.
Answer:
TItle

Question 15.
…………………………… is the process of moving up and down or left: and right through the document window.
Answer:
Scrolling

Question 16.
…………………………… tool is used to draw a circle.
Answer:
Ellipse

Question 17.
The Insert pages option is available on clicking the …………………………… menu.
Answer:
Layout

Question 18.
Match the following.
(a) Cut – (i) Ctrl + Z
(b) Copy – (ii) Ctrl + V
(c) Paste – (iii) Ctrl + X
(d) Undo – (iv) Ctrl + C
Answer:
(iii), (iv), (ii), (i)

Question 19.
Choose the odd man out.
(i) Adobe PageMaker, QuarkXPress, Adobe InDesign, Audacity
(ii) File, Edit, Layout, Type, Zip
(iii) Pointer Tool, Line tool, Hide Tool, Hand Tool
(iv) Bold, Italic, Portrait, Underline
Answer:
(i) Audacity, (ii) Zip, (iii) Hide Tool, (iv) Potrait

Question 20.
Choose the correct statement.
(i) (a) Text can be selected using mouse only.
(b) Text can be selected using mouse or the keyboard.
(ii) (a) DTP is an abbreviation for Desktop publishing.
(b) DTP is an abbreviation for Desktop publication.
Answer:
(i) (b) & (ii) (a)

Question 21.
Choose the correct pair.
Answer:
(a) Edit and Cut
(b) Edit and New
(c) Undo and Copy
(d) Undo and Redo
Answer:
(d) Undo and Redo

PART – II
II. Short Answer:

Question 1.
What is desktop publishing?
Answer:
Desktop publishing (abbreviated DTP) is the creation of page layouts for documents using DTP software.

Question 2.
Give some examples of DTP software?
Answer:
Some of the popular DTP software are Adobe PageMaker, Adobe InDesign, QuarkXPress, etc.

Question 3.
Write the steps to open PageMaker?
Answer:
In the Windows 7 operating system, we can open Adobe PageMaker using the command sequence Start →All Programs → Adobe → Pagemaker 7.0 → Adobe PageMaker 7.0.

Question 4.
How do you create a New document in PageMaker?
Answer:
To create a new document

1. Choose File > New in the menu bar. (or) Press Ctrl + N in the keyboard. Now Document Setup dialog box appears.
2. Enter the appropriate settings for your new document in the Document Setup dialog box.
3. Click on OK.

Question 5.
What is a Pasteboard in PageMaker?
Answer:
A document page is displayed within a dark border. The area outside of the dark border is referred to as the pasteboard. Anything that is placed completely in the pasteboard is not visible when you print the document.

Question 6.
Write about the Menu bar of PageMaker?
Answer:
Below the title bar is the Menu bar. It contains the following menus File, Edit, Layout, Type, Element, Utilities, View, Window, Help. When you click on a menu item, a pulldown menu appears.

Question 7.
Differentiate Ellipse tool from Ellipse frame tool?
Answer:
Ellipse tool:
Used to draw circles and ellipses.
Ellipse frame tool:
Used to create elliptical placeholders for text and graphics.

Question 8.
What is text editing?
Answer:
Editing meAnswer:Answer:Answer:Answer:making changes to the text. Editing encompasses many tasks, such as inserting and deleting words and phrases, correcting errors, and moving and copying text to different pfaces in the document.

Question 9.
What is text block?
Answer:
A text block contains text you type, paste, or import. You can’t see the borders of a text block until you select it with the pointer tool.

Question 10.
What is threading text blocks?
Answer:
All text in PageMaker resides inside containers called text blocks. A Text block can be connected to other text block so that the text in one text block can flow into another text block. Text blocks that are connected in this way are threaded.

Question 11.
What is threading text?
Answer:
A Text block can be connected to other text block so that the text in one text block can flow into another text block. Text blocks that are connected in this way are threaded. The process of connecting text among Text blocks is called threading text.

Question 12.
How do you insert a page in PageMaker?
Answer:

1. Go to the page immediately before the page you want to insert.
2. Choose Layout > Insert Pages in the menu bar. The Insert Pages dialog box appears.
3. Type the number of pages you want to insert.
4. To insert pages after the current page, choose ‘after’ from the pop-up menu.
5. Click on Insert.
6. The new pages are inserted in your publication.

PART – III
III. Explain in Brief Answer:

Question 1.
What is PageMaker? Explain its uses?
Answer:
Adobe PageMaker is a page layout software. It is used to design and produce documents that can be printed. You can create anything from a simple business card to a large book.
Page layout software includes tools that allow you to easily position text and graphics on document pages. For example, using PageMaker, you could create a newsletter that includes articles and pictures on each page. You can place pictures and text next to each other, on top of each other, or beside each other wherever you want them to go.

Question 2.
Mention three tools in PageMaker and write their keyboard shortcuts?
Answer:
Tools:

1. Printer Tool
2. Rotating Tool
3. Line Tool

Keyboard Short Cut:

1. F9
2. Shift + F2
3. Shift + F3

Question 3.
Write the use of any three tools in PageMaker along with symbols?
Answer:

Question 4.
How do you rejoin split blocks?
Answer:
Rbjoining split blocks
To rejoin the two text blocks

1. Place the cursor on the bottom handle of the second text.block, click and drag the bottom handle up to the top.
2. Then place the cursor on the bottom handle of the first text block, and click and drag the bottom handle down if necessary.

Question 5.
How do you link frames containing text?
Answer:
Linking Frames containing Text
A single frame may not be large enough to hold an entire story when you are using a large amount of text, you can link frames together so that an entire story is visible.
To link Frames containing text

1. Draw a second frame With the Frame tool of your choice.
2. Click the first frame to select it.
3. Click on the red triangle to load the text icon.
4. Click the second frame. PageMaker flows the text into the second frame.

Question 6.
What is the use of Master Page?
Answer:
Any text or object that you place on the master page will appear on the entire document pages to which the master is applied. It shortens the amount of time because you don’t have to create the same objects repeatedly on subsequent pages. Master Pages commonly contain repeating logos, page numbers, headers, and footers. They also contain non printing layout guides, such as column guides, ruler guides, and margin guides.

Question 7.
How to you insert page numbers in Master pages?
Answer:
Inserting Page Numbers in Master Pages
To make page numbers appear on every page

1. Click on Master Pages icon.
2. Then click on Text Tool. Now the cursor changes to I – beam.
3. Then Click on the left Master page where you want to put the page number.
4. Press Ctrl + Alt + P.
5. The page number displays as ‘LM’ on the left master page.
6. Similarly click on the right Master page where you want to put the page number.
7. O Press Ctrl + Alt + P.
8. The page number displays as ‘RM’ on the right master page, but will appear correctly on the actual pages.

PART – IV
IV. Explain in detail.

Question 1.
Explain the tools in PageMaker toolbox?
Answer:

Question 2.
Write the steps to place the text in a frame?
Answer:
Placing Text in a Frame
You can also use frames to hold text in place of using text blocks.
To place text in a Frame

1. Click on one of a Frame tool from the Toolbox.
2. Draw a frame with one of PageMaker’s Frame tools (Rectangle frame tool or Ellipse Frame Tool or Polygon frame Tool). Make sure the object remains selected.
3. Click on File. The File menu will appear.
4. Click on Place. The Place dialog box will appear.
5. Locate the document that contains the text you want to place, select it.
6. Click on Open.
7. Click in a frame to place the text in it. The text will be placed in the frame.

Question 3.
How can you convert text in a text block to a frame?
Answer:
Converting text in a Text block to a Frame
After created text in a text block, if you want to convert it to a frame. You can do this by using these steps.

1. Draw the frame of your choice using one of the PageMaker’s Frame tool.
2. Select the text block you want to insert in the frame.
3. Click the frame while pressing the Shift key. Now both elements will be selected.
4. Choose Element > Frame > Attach Content on the Menu bar.
5. Now the text appears in the frame.

Question 4.
Write the steps to draw a star using polygon tool?
Answer:
Drawing a Star using Polygon tool
To draw a Star
(i) Click on the Polygon tool from the toolbox.
The cursor changes to a crosshair.

(ii) Click and drag anywhere on the screen. As you drag, a Polygon appears.

(iii) Release the mouse button when the Polygon is of the desired size.

(iv) Choose Element > Polygon Settings in the menu bar.
Now Polygon Settings dialogue box appears.

(v) Type 5 in the Number of sides text box.

(vi) Type 50% in Star inset textbox.

(vii) Click OK. Now the required star appears on the screen.

Drawing a star with given number of sides and required inset
1. The value of ‘Star inset’ is 50%. The number of sides is 15.

2. The value of ‘Star inset’ is 25%. The number of sides is 25.

3. The value of ‘Star inset’ is 35%. The number of sides is 70.

Samacheer Kalvi 12th Computer Applications An Introduction to Adobe Pagemaker Additional Question and Answers

I. Choose the Best Answer:

Question 1.
How will you get Document setup dialog box?
(a) Document → setup
(b) File → New
(c) File → Document
(d) File → setup
Answer:
(b) File → New

Question 2.
…………………………. is the shortcut key to create a new document.
Answer:
Ctrl + N

Question 3.
………………………… is used to temporarily hold elements while designing.
Answer:
Pasteboard

Question 4.
How many control buttons are present in the title bar?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 5.
The default name of the new document created is …………………………..
(a) untitled 1
(b) doc 1
(c) default 1
(d) untitled
Answer:
(a) untitled 1

Question 6.
When you move the mouse pointer on a button in the toolbar, a short text that appears is called ……………………………..
(a) Tool tip
(b) Text
(c) Show text
(d) Short text
Answer:
(a) Tool tip

Question 7.
Match the following.
(i) polygon – (a) Shift + alt + F4
(ii) cropping – (b) Shift + F6
(iii) constrained line – (c) Shift + alt + F2
(iv) Rectangle Frame – (d) Shift + alt + F4
(a) (i)-(a) (ii)-(b) (iii)-(c) (iv)-(d)
(b) (i)-(d) (;ii)-(b) (iii)-(c) (z’v)-(a)
(c) (z’)-(c) (ii)-(b) (iii)-(d) (iv)-(a)
(d) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
Answer:
(d) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)

Question 8.
……………………. is the shortcut for the pointer tool.
Answer:
F9

Question 9.
Choose the correct pair.
(a) Zoom tool

(b) Polygon

(c) Ellipse

(d) line

Answer:
(a) Zoom tool

Question 10.
……………………….. tool is used to trim imported graphics.
(a) Text
(b) Rotating
(c) Trim
(d) Crop
Answer:
(d) Crop

Question 11.
How many scroll bars are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 12.
How many ruler bars are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 13.
The rulers are present at the ………………….. and …………………
(a) top, right
(b) right, left
(c) top, left
(d) bottom, top
Answer:
(c) top, left

Question 14.
…………………………. tool is used for creating text blocks.
(a) Block
(b) Text
(c) Footer
(d) Insertion point
Answer:
(b) Text

Question 15.
The flashing verticle bar is called ……………………………
(a) sroll bar
(b) ruler
(c) Footer
(d) Insertion point
Answer:
(d) Insertion point

Question 16.
Which key should be pressed at the end of the paragraph or when a blank line is to be inserted?
(a) Enter
(b) Tab
(c) Spacebar
(d) esc
Answer:
(a) Enter

Question 17.
Choose the correct statement.
(i) double click with I-beam to select a word.
(ii) double click with I beam to select a paragraph
Answer:
(i) double click with I-beam to select a word.

Question 18.
Which key is used to select the text with keyboard?
(a) Ctrl
(b) alt
(c) shift
(d) ctrl+tab
Answer:
(c) shift

Question 19.
Identify the statement which is wrong?
(a) One line down – alt + ↓
(b) Beginning of current line – shift + home
(c) One character to the left – shift + ←
Answer:
(a) One line down – alt + ↓

Question 20.
Which key deletes the character to the left?
(a) Delete
(b) Backspace
(c) Remove
(d) Escape
Answer:
(a) Delete

Question 21.
To delete the characters to the right of the insertion point, key is used.
(a) delete
(b) backspace
(c) alt
(d) ctrl
Answer:
(b) backspace

Question 22.
To delete a block of text, press
(a) delete
(b) Backspace
(c) Edit → clear
(d) All of these
Answer:
(d) All of these

Question 23.
Undo option is present in ………………………….. menu.
(a) File
(b) Edit
(c) Help
(d) Insert
Answer:
(b) Edit

Question 24.
Which command is not present in the Edit menu?
(a) Cut
(b) Paste
(c) Copy
(d) place
Answer:
(d) place

Question 25.
How many methods of creating text blocks are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 26.
While selecting a text block with the pointer tool, there will be a dark ……………………….. on both ends of the handle.
(a) rectangle
(b) circle
(c) square
(d) polygon
Answer:
(c) square

Question 27.
A red triangle in the bottom window shade meAnswer:Answer:Answer:Answer:…………………………..
(a) there is more text in the text block
(b) there is no text in the text block
(c) text block is deleted
(d) new text block appears
Answer:
(a) there is more text in the text block

Question 28.
Rearrange the sentences
(i) It has squiggly arrow
(ii) place the loaded text icon, click then page will automatically flow
(iii) Select Layout → Autoflow
(iv) Import the text
(a) (iii), (iv), (i), (ii)
(b) (i), (ii), (iii), (iv)
(c) (iv), (iii), (ii), (i)
(d) (iv), (i), (ii), (iii)
Answer:
(a) (iii), (iv), (i), (ii)

Question 29.
Text that flows through one or more threaded blocks is
(a) blocks
(b) story
(c) thread
(d) unthread
Answer:
(b) story

Question 30.
A threaded block is identified by …………………….. sign in the handles.
(a) +
(b)
(c)
(d)
Answer:
(a) +

Question 31.
In page maker, text and graphics that you draw or import are called ……………………………
(a) event
(b) handle
(c) tool
(d) objects
Answer:
(d) objects

Question 32.
Choose the wrong statement.
(a) An object can be on a page
(b) An object can be on the pasteboard
(c) An object cannot be on the pasteboard
(d) An object can be on a page or pasteboard
Answer:
(c) An object cannot be on the pasteboard

Question 33.
Which command is used to convert text into frame?
(a) Element → Frame → Attach Content
(b) Element → Frame → add content
(c) Add → Frame
(d) Add → content
Answer:
(a) Element → Frame → Attach Content

Question 34.
Which dialog box appears when File → Save command is given?
(a) Save
(b) Save as
(c) Save page
(d) Save publication
Answer:
(d) Save publication

Question 35.
Instead of File → Save as, we can use the shortcut key
(a) Ctrl + S
(b) alt + S
(c) shift + alt + S
(d) Ctrl + shift + S
Answer:
(d) Ctrl + shift + S

Question 36.
How will you get open publication dialog box?
(a) Ctrl + O
(b) Alt + O
(c) Shift + 0
(d) Shift + O + P
Answer:
(a) Ctrl + O

Question 37.
Find the wrong pair?
(a) Beginning of line – Home
(b) End of line – end
(c) One character to left-left arrow
(d) Up one line – Ctrl + up arrow
Answer:
(d) Up one line – Ctrl + up arrow

Question 38.
To toggle between magnification and reduction, press …………………………………. key.
(a) alt
(b) Ctrl
(c) shift
(d) esc
Answer:
(b) Ctrl

Question 39.
Find the wrong statement.
(a) Press Ctrl + spacebar to zoom in
(b) Press alt + spacebar to zoom out
(c) Press alt + Ctrl + spacebar to zoom out
Answer:
(b) Press alt + spacebar to zoom out

Question 40.
………………………… is the process of changing the general arrangement of text
(a) Text
(b) Alignment
(c) Formatting
(d) Editing
Answer:
(c) Formatting

Question 41.
To apply character formatting
(i) choose type → character
(ii) choose format → character
(iii) press Ctrl → T
(iv) Open character specification dialog box
(a) (i), (iv) are wrong
(b) (i), (iii) are wrong
(c) (ii) is wrong
(d) All are correct
Answer:
(c) (ii) is wrong

Question 42.
How will you get control palette?
(a) alt + ’
(b) shift + ’
(c) esc
(d) Ctrl + ’
Answer:
(d) Ctrl + ’

Question 43.
How many line tools are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 44.
When a line tool is selected to draw a line, cursor becomes …………………………..
(a) +
(b) –
(c) double arrow
(d) cross hair
Answer:
(d) cross hair

Question 45.
Find the false statement.
(a) Custom Stroke dialog box is used to draw dotted line.
(b) Rounded comers dialog box is used to draw rounded comer rectangle.
(c) Both are correct.
Answer:
(c) Both are correct.

Question 46.
The constrained line tool draws line only at the increments of ………………………. degrees.
(a) 30
(b) 45
(c) 60
(d) 90
Answer:
(b) 45

Question 47.
Press the ………………………….. key while you are drawing to constrain the shape to a square or circle.
(a) alt
(b) shift
(c) del
(d) Ctrl
Answer:
(b) shift

Question 48.
Which menu has polygon settings options?
(a) Tools
(b) Layout
(c) Element
(d) Place
Answer:
(c) Element

Question 49.
To get color palette, press
(a) shift + J
(b) alt + J
(c) del + J
(d) Ctrl + J
Answer:
(d) Ctrl + J

Question 50.
Press alt + Ctrl + G to open go to page dialog box.
(a) true
(b) false
Answer:
(a) true

Question 51.
What are the two letters associated with master page option?
(a) L and R
(b) S R
(c) RT
(d) TB
Answer:
(a) L and R

Question 52.
By default, all the page maker document have a master page already created
(a) Master page
(b) Document Master
(c) Document Page
(d) Page Master
Answer:
(b) Document Master

Question 53.
Find the correct statement. .
(а) A master item can be selected on a document page.
(b) A master item cannot be selected on a document page.
Answer:
(b) A master item cannot be selected on a document page.

Question 54.
A sign or special character that can be inserted in the page maker document is on ………………………
(a) Symbol
(b) Menu
(c) Formula
(d) Sign
Answer:
(a) Symbol

Question 55.
…………………………. is the text that is repeated at the top of each page.
Answer:
Header

Question 56.
A ……………………………. is a set of letters, numbers or symbols in a certain style.
Answer:
Font

Question 57.
When the publication has reached its maximum magnification or reduction level, center becomes …………………………………
Answer:
Blank

Question 58.
To Scroll a relative distance …………………………… is used.
Answer:
Scroll box

Question 59.
To down one paragraph, …………………………. is used.
Answer:
ctrl+down arrow

Question 60.
……………………….. can be contained in text blocks or text frames.
Answer:
Text

Question 61.
Press ………………………….. for rotating tool.
Answer:
Shift+F2

II. Short Answer:

Question 1.
When will you get the cursor like this Find the name of the tool & write its uses.?
Answer:
It is hand tool which is used to scroll the page (an alternative to the scroll bar).

Question 2.
Write note on scroll bars?
Answer:
Scrolling is the process of moving up and down or left and right through the document window. There are two scroll bars namely Vertical and Horizontal scroll bars for scrolling the document vertically or horizontally.

Question 3.
What is meant by wrapping the text?
Answer:
When the text being typed reaches the end of the text block, PageMaker will automatically wrap the text to the next line.

Question 4.
What is the use of enter key?
Answer:
The Enter key should be pressed only at the end of a paragraph or when a blank line is to be inserted.

Question 5.
How will you select the word and a paragraph?
Answer:
To select:

1. A word
2. A paragraph

Press:

1. Double click with I – beam
2. Triple click with I – beam

Question 6.
Name the tools where you get cursor with *+’ symbol?
Answer:
Line tool, Constrained line tool, Rectangle tool, Rectangle frame tool, Ellipse tool, Ellipse frame tool, Polygon tool, and Polygon frame tool.
A Paragraph Triple-click with I-beam

Question 7.
What are the two ways of creating text blocks?
Answer:

1. Click or drag the text tool on the page or pasteboard, and then type
2. Click a loaded text icon in an empty column or page.

Question 8.
Define Windowshades?
Answer:
When you select a text block with the Pointer tool, the block’s boundaries become visible. Two handles are seen above and below the text block. These handles are called Windowshades.

Question 9.
How will you separate the text from the frame?
Answer:
To separate text from a frame

1. Click the frame with the Pointer tool.
2. Choose Element > Frame > Delete Content in the menu bar. The text will not appear in the frame.

Question 10.
Define font?
Answer:
A font is a set of letters, numbers or symbols in a certain style. Each font looks different from other fonts.

Question 11.
What are the two line tools in page maker?
Answer:
PageMaker has two Line tools. The first one creates a straight line at any orientation. The second is a constrained Line tool that draws only at increments of 45 degrees.

Question 12.
How will you change text colors in page maker?
Answer:
To color characters

1. Select the text you want to colour.
2. Choose Window > Show Colors in Menu bar. The Colors palette appears. Click the color you want to apply to the selected text.

Question 13.
Name any 4 options from the layout menu?
Answer:

1. Go to pages
2. Insert pages
3. Remove pages
4. Sort pages

Question 3.
How wOl you Show / hide the rulers?
Answer:
To show the ruler

1. Click on View. The View menu will appear.
2. Click on Show Rulers. Rulers appear along the top and left sides of the document window. To hide the ruler
3. Click on View. The View menu will appear.
4. Click on Hide Rulers to hide the rulers.

Question 4.
Differentiate moving and copying text?
Answer:
The Copy and Paste commands of PageMaker can be used to copy text from one location in a document and paste it at another location. The Copy command creates a duplicate of the selected text, leaving the original text unchanged. The Paste command pastes the copied text at the position where the insertion point is placed. The Cut and Paste commands can be used to move text from one position in a document to the other. The Cut command deletes the selected text from its original position. The Paste command then places this text at the position where the insertion point is placed.

Question 5.
How will you create a text block with text tool?
Answer:
To create a text block with the text tool:
(i) Select the text tool (T) from the toolbox. The pointer turns into an I-beam.

(ii) On an empty area of the page or pasteboard, do one of the following:
Click the I-beam where you want to insert text. This creates a text block the width of the column or page. By default, the insertion point jumps to the left side of the text block.

(iii) Type the text you want.
Unlike with a text frame, you do not see the borders of a text block until you click the text with the pointer tool.

Question 6.
How will you Split a text block into two?
Answer:
To split a text block into two

1. Place the cursor on the bottom handle, click and drag upwards. When you release the bottom handle will contain a red triangle.
2. Click once on this, and the cursor changes to a loaded text icon.

Question 7.
How will you save a document with a new name or in a different location?
Answer:
To save a document with a new name or in a different location:

1. Choose File > Save As in the menu bar.
2. Press Shift + Ctrl + S in the keyboard.

Now Save Publication dialog box will appear:

1. Type a new name or specify a new location.
2. Click the Save button.

Question 8.
Give the functions of those keys give below?
Answer:

1. Left arrow
2. right arrow
3. Ctrl + left arrow
4. Ctrl + right arrow
5. up arrow
6. down arrow

Press:

1. Left Arrow
2. Right Arrow
3. Ctrl + Left Arrow
4. Ctrl + Right Arrow
5. Up Arrow
6. Down Arrow

Move:

1. One character to the left
2. One character to the right
3. One word to the left
4. One word to the right
5. Up one line
6. Down one line

Question 9.
How will you go to a specific page? (or) what are the 3 methods to go to a particular page?
Answer:
Pagemaker provides several methods for navigating the pages in your publication.
Method 1:
You can move from one page to another by using the Page up and Page down keys on your keyboard.

Method 2:
You can move from one page to another by using the page icons at the left bottom of the screen.

Method 3:
Using the Go to Page dialog box.

IV. Answer in Detail.

Question 1.
What are the different ways of selecting the text?
Answer:
Selecting Text:
Text can be selected using the mouse or the keyboard.
Selecting Text using the mouse:
To select text using a mouse, follow these steps :

1. Place the Insertion point to the left of the first character to be selected.
2. Press the left mouse button and drag the mouse to a position where you want to stop selecting.
3. Release the mouse button.
4. The selected text gets highlighted.

To Select Press:

1. A Word Double-click with I-beam
2. A Paragraph Triple-click with I-beam

Selecting Text using the Keyboard:
To select text using a keyboard, follow these steps :

1. Place the Insertion point to the left of the first character you wish to select.
2. The Shift key is pressed down and the movement keys are used to highlight the required text.
3. When the Shift key is released, the text is selected.

To Select – Press
One character to the left – Shift + ←
One character to the right – Shift + →
One line up – Shift + ↑
One line down – Shift + ↓
To the end of the current line – Shift +End
To the beginning of the current line – Shift + Home,
Entire Document – Ctrl + A

Question 2.
How will import (place) the text from other software programs in the page maker?
Answer:
You can insert text from other software program like MS-Word in the PageMaker documents.

1. Choose File > Place. The Place dialog box will appear.
2. Locate the document that contains the text you want to place and select it.
3. Click on Open in the Place dialog box. The pointer changes to the loaded text icon ( )
4. Make a text block to place the text. (Or) Click in the page to place the text.

The text will be placed in the page.
If the text to be placed is too big to fit on one page, PageMaker allows you to place it on several pages. This can be done manually or automatically.

Manual text flow:

1. Position the loaded text icon at a comer of the area where you want to place text, hold down the mouse button, and drag to define the text block. Release the mouse button.

2. Text flows into the defined area. If there is more text than fits in the text block you defined, a red triangle appears in the bottom windowshade handle.

3. Click once on this and the loaded text icon reappears. Now generate a new text block and # click. Repeat this process until there is no more text to place.

Question 3.
Explain how will you draw a rounded corner rectangle?
Answer:
Drawing a Rounded Corner Rectangle
To draw a rounded-comer rectangle:

1. Double-click the Rectangle tool in the toolbox.
2. The Rounded Comers dialog box appears.
3. Choose a comer setting from the preset shapes.
4. Click on OK. The cursor changes to a cross hair.
5. Click and drag anywhere on the screen.
6. Release the mouse button when the rectangle is the desired size.
7. Press the Shift key as you draw to constrain the shape to a rounded-comer square.

Question 4.
Explain how will you draw a dotted line?
Answer:
To draw a Dotted line:

1. Double click the Line tool from the toolbox. A Custom Stroke dialogue box appears.
2. Select the required Stroke style in the drop-down list box.
3. Then click OK button. Now the cursor changes to a cross hair.
4. Click and drag on the screen to draw your dotted line. As you drag, the line appears.
5. Release the mouse button and the line will be drawn and selected, with sizing handles on either end.
   Resize the line by clicking and dragging the handles, if necessary.

Question 5.
Explain how will you print a document in the page maker?
Answer:
Print a document:

(i) Choose File > Print in the menu bar (or) Press Ctrl + P in the keyboard. The Print Document dialog box appears
(ii) Choose the settings in the Print Document dialog box as

• Select the printer from the Printer drop-down list box.
• Choose the pages to be printed in the Pages group box by selecting one of the following available options:

All:
This option prints the whole document.

Ranges:
This option prints individual pages by the page number or a range of pages.
You can use commas to separate the page numbers (e.g., 5,7,19).
Use a hyphen to print page ranges(e.g., 10-17; this will print all pages from page numbers 10 . to 17).
To print from a particular page to the end of the document, enter the starting page number followed by a hyphen (e.g., 5 -).
You may also combine individual page numbers and a range of pages (e.g., 5, 9, 15-26).
Print: You can also print only odd-numbered or even-numbered pages. Select the Odd pages or Even pages Option from the Print drop-down list box.

• Type the number of copies you want in the Copies text box.
• You can choose whether to collate the pages or not. Suppose you want to print 4 copies of a 5 pages document

If the Collate option is not selected. PageMaker will first print 4 copies of page 1, then 4 copies of page 2, and so on.
If the Collate option is selected, PageMaker will print a complete set of pages 1 to 5, then a second set, and so on.

(iii) After choosing from the options in the Print Document dialog box, click Print button to print the document. Make sure the printer is switched on.

Students can Download Tamil Nadu 12th English Model Question Paper 1 Pdf, Tamil Nadu 12th English Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th English Model Question Paper 1

Time: 2 1/2 Hours
Maximum Marks: 90

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the sections in each part. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by writing the correct answer along with the corresponding option code.
• Part II has got two sections. The questions are of two marks each. Question numbers 21 to 26 in Section I and Question numbers 27 to 30 in Section II are to be answered in about one or two sentences each.
• Question numbers 31 to 40 in Part III are of three marks each and have been divided in three sections. These are to be answered as directed.
• Question numbers 41 and 47 in Part IV are of five marks each. These are to be answered as directed.

Part -1

I. Answer all the questions. [20 x 1= 20]
Choose the correct synonyms for the underlined words from the options given:

Question 1.
There was little hope for his recovery.
(a) discovery
(b) loss
(c) damage
(d) recuperation
Answer:
(d) recuperation

Question 2.
If used sparingly, the newly found cylinders could get us down to south.
(a) extravagantly
(b) lavishly
(c) economically
(d) sufficiently
Answer:
(c) economically

Question 3.
Individual liberty would become social anarchy.
(a) vanity
(b) civility
(c) lawlessness
(d) cooperation
Answer:
(c) lawlessness

Choose the correct antonyms for the underlined words from the options given:

Question 4.
The driver disapproved of their shabby appearance.
(a) dirty
(b) unkempt
(c) smart
(d) untidy
Answer:
(c) smart

Question 5.
which are not to be despised.
(a) liked
(b) defeated
(c) hated
(d) dare
Answer:
(a) liked

Question 6.
“Don’t buy,” warned Luigi, our cautious driver.
(a) sloppy
(b) repulsive
(c) reckless
(d) proud
Answer:
(c) reckless

Question 7.
Choose the correct combination for the compound word “outlaw”.
(a) Noun + Verb
(b) Verb + Noun
(c) Gerund + Noun
(d) Preposition + Noun
Answer:
(d) Preposition + Noun

Question 8.
Choose the correct expansion of “DSLR” camera.
(a) Digital Single Lens Refraction
(b) Digital Single Light Reflex
(c) Digital Service Lens Reflex
(d) Digital Single Lens Reflex
Answer:
(d) Digital Single Lens Reflex

Question 9.
Choose the meaning of the foreign word in the sentence:
Alan is going to travel to Europe this summer, but he’s planning on only taking one rucksack.
(a) rug
(b) quilt
(c) back pack
(d) suitcase
Answer:
(c) back pack

Question 10.
Choose the correct combination for the blended word “edutainment”.
(a) education + entertainment
(b) edn + ment
(c) edu + tent
(d) ed + tenant
Answer:
(a) education + entertainment

Question 11.
Choose the clipped form for “necktie”.
(a) entie
(b) tie
(c) neckie
(d) necty
Answer:
(b) tie

Question 12.
A passion for writing is known as …………………… .
(a) genocide
(b) philosophy
(c) graphomania
(d) cardiologist
Answer:
(c) graphomania

Question 13.
Form a derivative by adding the right suffix to the word ‘devote’.
(a) -tion
(b) -ly
(c) -ness
(d) -ment
Answer:
(a) -tion

Question 14.
Fill in the blanks with a suitable relative pronoun.
Nicola …………………… was glaring at his young brother was in vexation.
(a) whose
(b) who
(c) when
(d) which
Answer:
(b) who

Question 15.
Fill in the blanks with a suitable preposition.
This material is different …………………… that.
(a) to
(b) with
(c) of
(d) from
Answer:
(d) from

Question 16.
Choose the correct question tag for the following statement.
Everybody was very happy ……………………?
(a) wasn’t they
(b) haven’t they
(c) weren’t they
(d) isn’t they
Answer:
(c) weren’t they

Question 17.
Choose the suitable meaning or idiom found in the following sentence.
My class teacher examined my paper with a fine tooth comb before handing it to the external examiner.
(a) examining in great detail
(b) casually seeing
(c) carelessly reviewing
(d) examining with indifference
Answer:
(a) examining in great detail

Question 18.
Substitute the underlined word with the appropriate polite alternative.
Vincent Parker is a reputed burier in this locality for Christians.
(a) disposer
(b) mortician
(c) expeller
(d) end maker
Answer:
(b) mortician

Question 19.
Choose the correct sentence pattern for the following sentence.
They put on quite a show that day.
(a) SVAOA
(b) SVOC
(c) SVO
(d) SVCA
Answer:
(a) SVAOA

Question 20.
Fill in the blank with a suitable phrasal verb.
The guest lecturer asked the students to …………………… and pay attention.
(a) pipe over
(b) be serious
(c) be playful
(d) pipe down
Answer:
(d) pipe down

Part – II
Section – 1

Read the following sets of poetic lines and answer any four from it. [4 x 2 = 8]

Question 21.
“We could do nothing, being sold.”
(a) Why couldn’t they do anything?
(b) Why did they feel helpless?
Answer:
(a) The enemies had entered through the wicket gate stealthily and easily occupied their castle.
(b) They felt helpless because they had no weapon to fight “Gold” their invisible enemy. Their castle fell due to the unnoticed greed of their warder, an aged man.

Question 22.
“And oft at nights the garden overflows
With one sweet song that seems to have no close,
Sung darkling from our tree, while men repose.”
(a) What overflows the garden at night?
(b) What is called the darkling’?
Answer:
(a) One sweet song overflows in the garden at night.
(b) Nightingale is called the darkling.

Question 23.
‘‘‘‘And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress ’ eyebrow. ”
(a) What does the lover do for his mistress?
(b) What is the poetic device used in the second line?
Answer:
(a) The lover is always sighing and longing for his beloved. He writes a sad ballad describing the eyebrow of his mistress.
(b) ‘Simile’ is used as a poetic device in the second line.

Question 24.
“This is my son, mine own Telemachus,
To whom I leave the sceptre and the isle Well-loved of me,”
(а) Who does Ulysses entrust his kingdom to, in his absence?
(b) Bring out the significance of the ‘sceptre’.
Answer:
(a) Ulysses entrusts his kingdom to his beloved son Telemachus in his absence.
(b) Sceptre is an ornamental staff carried by a King on ceremonial occasions as a symbol of sovereignty. It symbolizes the power of a king.

Question 25.
“Tell him time as a stuff can be wasted.
Tell him to be a fool ever so often”
(a) Why does the poet suggest that time can be wasted?
(b) Identify the figure of speech in the above line.
Answer:
(a) Unless one wastes one’s time, one may not commit mistakes and learn from them. Besides learning not to repeat those mistakes would naturally make him wiser.
(b) Simile

Question 26.
“A film the mother-eagle’s eye When her bruised eaglet breathes”
(а) Who is compared to the mother eagle in the above lines?
(b) Explain the comparison.
Answer:
(a) Napoleon Bonaparte is compared to mother eagle.
(b) A mother eagle will be proud when the eaglet takes on a stronger predator. When the eaglet is hurt, the mother eagle will be naturally sad as no mother will want the young one to perish in a combat. The king, like a mother, is sad about the impending death of a valiant boy-soldier.

Section – 2

Answer any three of the following questions. [3 x 2 = 6]

Question 27.
Report the dialogue.
John : What were you doing, Mary?
Mary : I was making these paper flowers.
Answer:
John asked Mary what she had been / was doing. Mary replied that she had been/was making those paper flowers.

Question 28.
He did not invite her. She didn’t come, (combine using if)
Answer:
She would have come if he had invited her.

Question 29.
Rewrite the sentence making an inversion in the conditional clause. If the people had not been informed, the situation would have been much worse.
Answer:
Had the people not been informed, the situation would have been much worse. Or Had the people been informed, the situation would not have been much worse.

Question 30.
We saw the flash of lightning, and seconds later we heard a rumble of thunder. (Change the following into a complex sentence)
Answer:
When we saw the flash of lightning, seconds later we heard a rumble of thunder.

Part-III
Section -1

Explain any two of the following with reference to the context. [2 x 3 = 6]

Question 31.
Dear is the Casuarina to my soul;
Answer:
Reference: These words are from the poem, ‘Our Casuarina Tree” written by Torn Dutt.
Context: The poet says this while highlighting the importance of Casuarina tree in her life.
Explanation: The poet explains that she, her friends and siblings have spent long hours playing and enjoying themselves under the tree’s shade. The tree has been her friend since childhood. There is a special bonding between the poet and the tree. So, it is dear to her soul.

Question 32.
“……And then the justice,
In fair round belly with good capon lined,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;
Answer:
Reference: These lines are from the poem, ‘All the world’s a stage” written by William Shakespeare.
Context: The poet says these words while describing the fifth stage of life.
Explanation: At this stage, he behaves like a judge pronouncing his decisive opinions with the modem instances. He quotes wise maxims from his own life experiences to influence other people. He is fond of eating delicacies unmindful of the protruding belly size.

Question 33.
To follow knowledge like a sinking star,
Beyond the utmost bound of human thought.
Answer:
Reference: These lines are from the poem, ‘Ulysses” written by Alfred Tennyson.
Context: The poet says these words while describing the quest Ulysses has for adventure and fulfillment.
Explanation: Similar to a sinking star, Ulysses wants to pursue in his failing old age to pursue knowledge like the goal of Goethe’s Faust, his quest is defined by the pursuit of new and unique knowledge “beyond the utmost bound of human thought”.

Section – 2

Answer any two of the following questions in about 30 words. [2 x 3 = 6]

Question 34.
Who took the author to the cubicle?
Answer:
A trained nurse who was very familiar with Nicola and Jacopo took the narrator in. She led him through a cool, tiled vestibule into the hospital, the villa had become. She left him at the door of a little cubicle from where he can watch unseen Nicola and Jacopa chatting with their ailing sister.

Question 35.
Whom does the author call ‘misguided people’? What is his advice to them?
Answer:
Some misguided people admit that they drink tea for stimulation and warmth. They are not interested in tea. So, they add sugar to take the taste away. He advises them to drink tea without adding sugar for a fortnight. They would never again ruin the real taste of tea by adding sugar.

Question 36.
The soft snow was difficult and dangerous. Why?
Answer:
The soft snow made a route on top of the ridge both dangerous and difficult. Sometimes it held Hillary’s weight. But often it gave way suddenly. Thus.it was dangerous for the climbers. But both persisted and trudged ahead for 400 feet and reached the southern summit.

Section – 3

Answer any three of the following.

Question 37.
Study the graph given and answer the questions, that follow. [3 x 3 = 9]

(а) Which type of tourist centre is preferred by the third largest number of tourists?
(b) Which type of tourist centre is preferred more than Beach resorts?
(c) Name the tourist centres preferred the least.
Answer:
(a) Pilgrim centres is preferred by the third largest number of tourists.
(b) Hill stations are preferred more than beach resorts. ‘
(c) Historical monuments are preferred the least.

Question 38.
Write a dialogue between of minimum three exchanges an abandoned sick man and a student-volunteer.
Answer:
Student: Grandpa, how is your health?
Old man: (Tears trickling down the cheek) Did you call me grandpa?
Student : Yes, you are of the same age as my own grandpa.
Old man: I too have a grandson of your age who was separated from me years ago.
Student : Oh, I am so sorry. Here, please take some idlis that I have brought for you.
Old man: God bless you my child.

Question 39.
How will you fix a fuse?
Answer:
I will switch off the main line. Pull out the carrier.
I shall verify if the filament of wire is “intact” or burnt out. If it is burnt out, I shall tie the ends of the carrier with a thin wire and carefully insert it in the carrier plug point.
Then I shall switch on the main line. The lights will come back to life.

Question 40.
Complete the proverbs using the words given below:.
(a) A penny saved is a penny …………. (borrowed, earned, invested)
(b) Darkest hour is just before the …………. (dusk, midnight, dawn)
(e) Don’t change horses …………. (mid stream, always, again)
Answer:
(a) earned
(b) dawn
(c) mid stream

Part – IV

Answer the following questions:
Answer in a paragraph in about 150 words.

Question 41.
Write a character sketch of Nicola and Jacopo.
Answer:
Nicola was thirteen and Jacopo only twelve. They were brothers. They were tanned, had tangled hair and dark earnest eyes. Though they were just kids, they were serious about their work. They did hundreds of errands for the tourists. They were found doing brisk business shining shoes or selling wild berries. They had the skill to find seats in theater for the tourists and also guide the tourists through many important sites of the city of Verona such as Juliet’s tomb. Jacopo was lively as a squirrel. Nicola’s smile was steady and engaging yet in their innocent faces one could find seriousness far beyond their years. What struck one was the extreme willingness of both the boys to work. Under the scorching Sun, they hawked newspapers. When the narrator enquired what they did with the money they earned as they were not spending it on clothes or food or saving it for emigrating to America, Nicola coloured but he did not reveal the secret family adversity. Both were gentlemen because they did not want any one’s sympathy.

[OR]

Civilization can only exist when the public collectively accepts constraints on its freedom of action – Explain.

Human beings, by nature are quick to find fault with the imperfections of others. They don’t usually realize the truth that they have to accommodate their liberty to the liberty of others. A reasonable consideration for the rights and feelings of others is the foundation of social conduct. It is in small matters of conduct, in the observance of the rules of road, that we pass judgement on ourselves. We assess ourselves as civilized or uncivilized. We are civilized if we enjoy our rights without violating the rights of others or eroding into the privacy of others. It is the little habits of common place interaction that make up the great sum of life and sweeten or make bitter the journey. In a civilized society, public collectively accepts the fact that freedom is an accommodation of interests of others. It means curtailing a part of one’s own liberty to the liberty of others. So, it is true that civilization can only exist and prosper when the public collectively accepts constraints to its freedom of action.

Question 42.
Describe the various stages of a man’s life picturised in the poem “All the World’s a stage.”
Answer:
Shakespeare has beautifully portrayed this world as a huge open theatre where in all humans play seven acts/ages. In the first act, he is a helpless infant puking on the nurse’s arms mewling like a kitten. In the second stage, he is the grumbling/whining school student. He moves to school like a snail/unwillingly with his slate and bag. In the third Act, he is a lover sighing and yearning for the attention of his lady love. He composes romantic ballads complaining his love that he needs a better deal. In the fourth Act, he becomes a quick-tempered soldier, aggressive and ambitious, ready to stake his life for the sake of bubble reputation. As he matures, he becomes a wise judge of contemporary life quoting wise maxims to endorse his opinion. He is firm and serious. In the sixth act, his stout legs become thin making his trousers of youth unsuitable. Thin and lean legs easily travel through them but are unable to stay due to a slimmed waist. His bass voice has become treble like that of a child. In the last act, he is sans teeth, sans eyes, sans taste and sans everything (i.e.) loses all senses. He departs the world.

[OR]

In the poem ‘Incident of the French Camp’ the young soldier matched his emperor in courage and patriotism. Elucidate your answer.

Emperor Napoleon was an astute planner planning the moves of the battle observing each step the French army made. Emperor Napoleon being a bold and wise warrior always had two plans, one to advance forward if the battle brings victory and the next as to what to be done in case the battle is lost. He was not resting at a tent during the battle. He was very close to the place of battle planning the strategic steps. Similarly, the boy-soldier was also equally brave. Unlike the emperor, the boy soldier flung himself in the midst of battle and risked his life. He did not bother about his death. He doggedly carried out the mission of hoisting French national flag. Instead of being carried away for first aid, he hurried on horse back to communicate the news of conquest of Ratisbon despite his chest being split into two. So, it is obvious the boy soldier’s patriotism and gallantry are equal to that of Napoleon.

Question 43.
Write a paragraph of about 150 words by developing the following hints into a paragraph:

Liquid life – precious than gold – ship wreck – exquisite nectar – Pi’s forehead – expressing joy – heart beat – blood flow – Strength and suppleness – return – world of dead – ecstatic basked in bliss and plentitude – experiences of Pi-indisputable – water, elixir of life.
Answer:
The author calls it ‘liquid life’. For a thirsty man, a cup of water is more precious than gold. He had not drunk water for two and a half days after the ship wreck. Pi drank two litres of the most exquisite nectar (i.e.) water. Suddenly Pi’s forehead was wet with fresh perspiration. Everything in him right down to the pores of his skin was expressing joy. A sense of well¬being quickly overcame him. His skin relaxed, his joints moved with greater ease. His heart began to beat like a merry drum. Blood started flowing through his veins like cars from a wedding party honking their way through the town. Strength and suppleness came back to his muscles. His head became clearer. Truly he was returning to life from the world of dead. After being thirsty for a while, to be drunk on water is noble and ecstatic basked in bliss and plentitude for several minutes. These experiences of Pi highlight the indisputable fact that water is the elixir of life.

[OR]

Margot – vivid memory – Sun – five years old – planet earth – Venus planet – Sunny day – tantrums – thousands of dollars – Sun shape and warmth – students hatred -convictions – conflict with Margot and other children.
Answer:
Margot had a vivid memory of having seen the Sun till she was five years old on the planet earth. Though she had come to Venus planet school, her heart longed for the ‘Sunny day’. Preparations were going on to send her back to earth because one day she threw tantrums refusing to take a shower. Her intended visit could cost thousands of dollars to her parents. She drew paintings of the Sun and clarified the shape of the Sun and the warmth it generated. All the other students hated her superior understanding of the Sun and her possible return to the earth. She was not their “kind”. She kept her convictions. She refused to mix with them. This was the conflict with Margot and other children in the story.

Question 44.
Write a summary or Make notes of the following passage.
Answer:
Compared to a motorbike or car, the bicycle is a slow moving vehicle but its popularity has been on the increase in recent years. There are many reasons for this. Firstly, the bicycle does not require petrol or diesel as a car or motorbike does. In these days of fuel shortage this is definite advantage; it draws power from the rider. The bicycle, moreover, is small, light and convenient. There are no parking problems. Another reason for its popularity is that it takes one, to one’s destination with more certainty than cars or large motorbikes which are often held up in traffic jams. Then, there is the fact that the bicycle causes absolutely no pollution. It produces no noise and no smoke. Finally, it is now known that riding a bicycle is good exercise which tones up one’s muscles. No wonder, more and more people prefer a bicycle to other modes of transport.

Summary

No. of words given in the original passage: 153
No. of words to be written in the summary: 153/3 = 51 ± 5
Rough Draft
Cycle’s popularity is on the high. It doesn’t need petrol or diesel. Due to fuel shortage, cycle is a definite advantage. It is a smoll light convenient. No parking problems and traffic jams. Moreover cycle causes no pollution, noise or smoke. Cycle is a good exercise that tones up muscles.

Fair Draft Cycle – the best mode of transport
Cycle has many advantages over other modes of transport. Its popularity is on the increase. It doesn’t need petrol or diesel. Cycle is a definite advantage. It is small, light, convenient. Parking problems and traffic jams do not affect the cyclist’s progress. Moreover, cycle causes no pollution such. Cycling keeps one fit, tones up one’s muscles.

No. of words in the summary: 56
Or
Notes
Title: Cycle – the best mode of transport
Cycles: advantageous over cars and bikes
Popularity: no petrol or diesel
Size: small, no parking issue, traffic jam
Environment friendly: no pollution, a good exercise.

Question 45.
You are A. Shanti, in-charge of the Physical Education Department of your school. You need various items of sports for the department. Write a letter to Mr. Rahmat Singh & Sons, a well known supplier of sports goods, Dharmapuri enquiring him about the rates of each item and the maximum discount that they can offer to you.
12 Feb 2020
From
A.V. Sharmila
10th Main Road,
Valadi
Dharmapuri

To
M/S Rahmat Singh & Sons
Shop No. 59
Dharmapuri
Dear Sir,

Sub: Enquiring about rates & discount
We are old customers of yours and need no introduction. As we are placing a bulk order with you, we want to enquire at what rates you can supply each item of different sports. The details of the items are furnished below.

Equipment	Numbers
Goal posts	16
Rods and tackle	14
Wickets and bases	30
Nets	20
Flying discs	35
Hockey stick	40

Kindly study them thoroughly and give the lowest quotations.
Let us remind you that you have been giving us 25% flat discount on various items. As we are an educational institution, we would expect a.special discount that dealers generally give to such institutions. Kindly do the needful and furnish all the details regarding the inquiries at the earliest.

Yours faithfully,
A. Shanti

To
M/S Rahmat Singh & Sons
Shop No. 59
Dharmapuri

[OR]

Write a paragraph on the topic “How will you protect the earth from the evilof deforestation”.

Environmental Pollution and need for Conservation

Deforestation leaves wild animals and birds homeless. We, humans face a lot of undesirable environmental changes. The most threatening one is Global Warming. Deforestation affects seasonal cycles. We experience droughts or floods, long summer or winter, thus, deforestation affects seasonal cycles. One more harmful effects of deforestation is soil erosion. Thus, the harmful consequences of deforestation are manifold. Tropical rainforests are being felled substantially for the purpose of farming and human habitation. Afforestation should be done in a large scale. This should be initiated by government and we all should actively participate and involve and most importantly act to protect mother nature.

Question 46.
Spot the errors and rewrite the sentences correctly
(a) A million rupees are a big sum.
(b) Beside Tennis, I play cricket.
(c) Both Giri as well as Raghu have scored centum.
(d) Brittney spears are a popular singer.
(e) Every one of us were given a prize.
Answer:
(a) A million rupees is a big sum.
(b) Besides Tennis, I play cricket.
(c) Both Giri and Raghu have scored centum.
(d) Brittney spears is a popular singer.
(e) Every one of us was given a prize.

[OR]

Fill in the blanks correctly.
(a) I need to get a new ……………. (sole/soul) and restore my favourite ……………. (pear/pair) of walking shoes.
(b) You ……………. have a cookie after dinner. (Use a modal in the given blank)
(c) My sister ……………. read a book at night(use a semi-modal)
(d) Martha ……………. (walk) three miles a day before she broke her leg. (use a proper tense)
Answer:
(a) sole/pair
(b) may
(c) used to
(d) had been walking

Question 47.
Identify each of the following sentences with the fields given below:
(a) A normal bulb uses almost 80% energy to create heat and only 20% for production of light.
(b) Dropouts who come for mid-day meal are returning to class.
(c) In the world’s largest open-air theatre, stories of Krishna and Kansa are magically retold.
(d) Sports was started for the differently abled children.
(e) Technology develops machines that can substitute for humans and replicate human actions.
[Education; Paralympics; Science; Robotics; Media]
Answer:
(a) Science
(b) Education
(c) Media
(d) Paralympics
(e) Robotics

[OR]

Read the following passage and answer the questions in your own words.
Three days before John Wilkes Booth fatally shot the President, Lincoln relayed a dream he had to his wife in which he was wandering through the rooms of the White House hearing sobs and crying as he went. When he reached the East Room, he noticed a casket. He asked a soldier who was in the casket and the soldier replied that it was the President – killed by an assassin. Lincoln woke up from that dream but failed to sleep for the rest of the night.

Little did the President know that as he dreamed of his own assassination, the actual plot was being formulated by John Wilkes Booth, an actor and Confederate sympathizer. Booth’s Original plan was simply to kidnap the President, but as his anger grew over Lincoln’s support for former slaves, he resolved to kill the president. On the night of April 14, 1865, Lincoln was to attend the performance of Our American Cousin at Ford’s Theater. It was a perfect opportunity for Booth, who knew every line of the play, and knew every comer and corridor of the theater. Lincoln’s presidential box was supposed to be manned by a police officer named John Frederick Parker.

Parker, however, left his post to visit a tavern, and may or may not have returned. He might have fallen asleep on the job. Booth, with easy access to the President’s box, waited for the right moment during the play, rushed into the box, and shot the President in the back of the head. As the President fell over, Mary Todd Lincoln caught him then began screaming. Soon, chaos broke out as the audience attempted to flee the theater. Booth vaulted from the box to the stage below but caught his boot spur in a treasury flag and broke his leg. Before he escaped from the theater, a visibly limping Booth was said to have yelled “Sic Semper Tyrannis,” which means “Thus Always to Tyrants” in Latin. It is also the motto of the state of Virginia.

Questions:
a. What disturbed Abraham Lincoln just three days before his assassin?
b. What did he notice in the East room?
c. Who formulated the plot of killing Abraham Lincoln?
d. What was the original plan made by Booth? Why did he change?
e. What happened to Booth when he attempted to flee the theatre?
Answers:
(a) Abraham Lincoln had a dream where he heard sobs and crying in the rooms of the white House. On enquiry he was also told that the president was killed by an assassin. This disturbed Lincoln.
(b) Lincoln noticed a casket in the East room.
(c) John Wilkes Booth, an actor and Confederate sympathizer formulated the plot of killing Abraham Lincoln.
(d) The original plan was to just kidnap the President but since Lincoln’s support for former slaves angered Booth, he resolved to kill the president.
(e) When Booth vaulted from the box to the stage below, he trampled over the treasury flag and broke his leg.

Students can Download Chemistry Chapter 13 Hydrocarbons Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 13 Hydrocarbons

Samacheer Kalvi 11th Chemistry Hydrocarbons Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Hydrocarbons Multiple Choice Questions.

Question 1.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ……….. [NEET]
(a) the eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) the staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) the staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 2.
The above reaction is an example of which of the following?
(a) Reirner Tiemann reaction
(b) Wurtz reaction
(c) Aldol condensation
(d) Hoffmann reaction
Answer:
(b) Wurtz reaction

Question 3.
An alkyl bromide (A) reacts with sodium in ether to form 4, 5-diethyloctane, the compound (A) is ……….
(a) CH₃(CH₂)₃Br
(b) CH₃(CH₂)₅Br
(c) CH₃(CH₂)₃CH(Br)CH₃



Answer:

Question 4.
The C-H bond and C-C bond in ethane are formed by which of the following types of overlap ………..
(a) sp³ – s and sp³ – sp³
(b) sp² – s and sp³ – sp³
(c) sp – sp and sp – sp
(d) p – s and p – p
Answer:
(a) sp³ – s and sp³ – sp³

Question 5.
In the following reaction
the major product obtained is ………

Answer:

Question 6.
Which of the following is optically active?
(a) 2 – Methylpentane
(b) Citric acid
(c) Glycerol
(d) none of these
Answer:
(a) 2 – Methylpentane

Question 7.
The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate are ……….
(a) CH₄ and H₂
(b) CH₄ and CO₂
(c) C₂ H₆ and CO₂
(d) C₂ H₆ and Cl₂
Answer:
(c) C₂ H₆ and CO₂

Question 8.
The general formula for cycloalkanes is …………
(a) C_nH_n
(b) C_nH_2n
(c) C_nH_2n-2
(d) C_nH_2n+2
Answer:
(b) C_nH_2n

Question 9.
The compound that will react most readily with gaseous bromine has the formula ………….[NEET]
(a) C₃H₆
(b) C₂H₂
(c) C₄H₁₀
(d) C₂H₄
Answer:
(a) C₃H₆

Question 10.
Which of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction? [NEET]

(b) CH₃ – CH₂ – CH₂ – OH
(c) H₂C – C = O
(d) CH₃ – CH₂ – CH₂Br
Answer:
(c) H₂C = C = O

Question 11.
Which among the following alkenes on reductive ozonolysis produces only propanone?
(a) 2 – Methylpropene
(b) 2 – Methylbut – 2 – ene
(c) 2, 3 – Dimethylbut – 1 – ene
(d) 2, 3 – Dimethylbut – 2 – ene
Answer:
(d) 2, 3 – Dimethylbut – 2 – ene

Question 12.
The major product formed when 2 bromo – 2 – methylbutane is refluxed with ethanolic KOH is ……..
(a) 2 – methylbut – 2 – ene
(b) 2 – methylbutan – 1 – ol
(c) 2 – methyl but – 1 – ene
(d) 2 – methylbutan -2- ol
Answer:
(a) 2 – methylbut – 2 – ene

Question 13.
Major product of the below mentioned reaction is ……….

(a) 2 – chloro – 1 – iodo – 2 – methylpropane
(b) 1 – chloro – 2 – iodo – 2 – methylpropane
(c) 1 ,2 – dichioro – 2 – methylpropane
(d) 1, 2 diiodo 2 – methylpropane
Answer:
(a) 2 – chioro – 1 – lodo – 2 – methylpropane

Question 14.
The IUPAC name of the following compound is ………

(a) trans – 2 – chloro-3 iodo – 2 – pentane
(b) cis – 3 – iodo – 4 – chloro – 3 – pentane
(c) trans – 3 – iodo – 4 – chloro – 3 – pentene
(d) cis – 2 chloro – 3 – lodo -2 – pdntene
Answer:
(a) trans -2 – chloro -3 – iodo – 2 – pentane

Question 15.
cis – 2 – butene and trans – 2 – butene are ……….
(a) conformational isomers
(b) structural isomers
(c) configurational isomers
(d) optical isomers
Answer:
(c) configurational isomers

Question 16.
Identify the compound (A) in the following reaction.


Answer:

Question 17.
where a is ………
(a) Zn
(b) Conc. H₂SO₄
(c) Alc. KOH
(d) Dil. H₂SO₄
Answer:
(c) Alc. KOH

Question 18.
Consider the nitration of benzene using mixed conc. FeSO₄ and HNO₃, if a large quantity of KHSO₄ is added to the mixture, the rate of nitration will be ………
(a) unchanged
(b) doubled
(c) faster
(d) slower
Answer:
(d) slower

Question 19.
In which of the following molecules, all atoms are co-planar?

Answer:
(d) both (a) and (b)

Question 20.
Propyne on passing through red hot iron tube gives ………

Answer:

Question 21.


Answer:

Question 22.
Which one of the following is non-aromatic?

Answer:

Question 23.
Which of the following compounds will not undergo Friedal – crafts reaction easily? [NEET]
(a) Nitrobenzene
(b) Toluene
(c) Cumene
(d) Xyiene
Answer:
(a) Nitrobenzene

Question 24.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) – COOH
(b) – NO₂
(c) – C N
(d) – SO₃H
Answer:
(b) – NO₂

Question 25.
Which of the following can be used as the halide component for friedal – crafts reaction?
(a) Chiorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride
Answer:
(d) Isopropyl chloride

Question 26.
An alkane is obtained by decarboxylation of sodium propionate. Same alkane can be prepared by ……..
(a) Catalytic hydrogenation of propene
(b) action of sodium metal on iodomethane
(c) reduction of 1 – chloropropane
(d) reduction of bromomethane
Answer:
(b) action of sodium metal on iodomethane

Question 27.
Which of the following is aliphatic saturated hydrocarbon?
(a) C₈H₁₈
(b) C₉H₁₈
(c) C₈H₁₄
(d) All of these
Answer:
(a) C₈H₁₈

Question 28.
Identify the compound ‘Z’ in the following reaction.

(a) Formaldehyde
(b) Acetaldehyde
(c) Formic acid
(d) None of these
Answer:
(a) Formaldehyde

Question 29.
Peroxide effect (Kharasch effect) can be studied in case of ………
(a) Oct – 4 – ene
(b) Hex – 3 – ene
(c) Pent – 1 – ene
(d) But – 2 – ene
Answer:
(a) Pent – 1 – ene

Question 30.
2 – butyne on chlorination gives ………
(a) 1 – chlorobutane
(b) 1, 2 – dichlorobutane
(c) 1, 1, 2, 2 – tetrachlorobutane
(d) 2, 2, 3, 3 – tetrachlorobutane
Answer:
(d) 2, 2, 3, 3 tetra chiorobutane

Samacheer Kalvi 11th Chemistry Hydrocarbons Short Answer Questions

Question 31.
Give IUPAC names for the fllowing compounds …………
(i) CH₃ CH = CH – CH= CH – C ≡C – CH₃
(ii)

(iii) (CH₃)₃ C-C≡C-CH (CH₃)₂
(iv) Ethyl – isopropyl – acetytene
(v) CH≡C – C = C – C≡CH
Answer:

Question 32.
Identify the compound A. B, C and D in the following series of reactions.

Answer:

Question 33.
Write a short note on ortho – para directors aromat&c e1ropiJic substitution reactions.
Answer:
The group which increases the eleiron deity at oìo and para positions of the ring are known as ortho-para directors.
Example:
-OH, -NH₂ -NHR -CH₃, -OCH₃ etc.
Let us consider the directive influences of phenolic( -OH) group. Phenol is the resonace hybrid of following structure.

In these resonance structures the negative charge residue is present on ortho and para posrtions of the ring structure. Therefore the electron density at ortho and para positions increases as compared to the metci position, thus phenolic group activities the benzene ring for electrophilic attack at ortho and para positions and hetice – OH group is an ortho-para direction or and an activator.

Question 34.
How is propyne prepared from an alkyene dihalide?
Answer:

Question 35.
An alkyl halide with molecular formula C₆H₁₃Br on dehydrohalogenation gave two isomeric alkenes X and Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃. CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide.
Answer:
1. C₆H₁₃Br is 3 – Bromo – 4 methylpentanc.

2. 3 – Bromo -4 methylpentane on dehydrogenation give two isomers X and Y as follows:

There fore C₆H₁₃ Br is 3 – Bromo – 4 – methy ipentane.

Question 36.
Describe the mechanism of Nitration of benzene.
Answer:
Step-1 :
Generation of \(\overset { \oplus }{ { NO }_{ 2 } } \) electrophile.
HNO₃ + H₂SO₄ → \(\overset { \oplus }{ { NO }_{ 2 } } \) + H\(\overset { \oplus }{ { SO }_{ 4 } } \) + H₂O

Step-2:
Attack of the electrophile on benzene ring to form arenium ion.

Step-3:
Rearomatisation of arenium ion.

Overall Reaction:

Question 37.
How does Huckel rule help to decide the aromatic character of a compound?
Answer:
A compound is said to be aromatic, if it obeys the following rules:

• The molecule must be cyclic.
• The molecule must be co-planar.
• Complete delocalisation of it-electrons in the ring.
• Presence of (4n + 2) π electrons in the ring where n is an integer (n = 0,1,2 …)

This is known as Huckel’s rule.
Example:

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised ir electrons.
4. 4n + 2 = 6
4n = 6 – 2
4n = 4
n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

Question 38.
Suggest the route for the preparation of the following from benzene.
1. 3 – chioro – nitrobenzene
2. 4- chlorotoluene
3. Bromobenzene
4. in – dinitrobenzene
Answer:
1. Preparation of 3 – chloronitro – benzene from benzene:
Benzene undergoes nitration and followed by chlorination and it leads to the formation of 3- chloronitrobenzene.

2. Preparation 4-chiorotoluene from benzene:
Benzene undergoes Fnedel crafi’s alkylation followed by chlorination and it leads to the formation of 4-chiorotoluene.

3. Prepar2tion of Bromobenzene from benzene:
Bezene undergo bromination to give bromobenzene.

4. Preparation of m-dinitrobenzene from benzene:
Benzene undergo twice the time nitration to give m-dinitrobenzene.

Question 39.
Suggest a simple chemical test to distinguish propane and propene.
Answer:
Chemical test to distinguish between propane and propene:
1. Bromine water test:
Propene contains double bond, therefore when we pour the bromine water to propene sample, it decolounses the bromine water whereas propane which is a saturated hydrocarbon does not decolourise the bromine water.

2. Baeyer’s test:
When propene reacts with Bayer’s reagent it gives 1,2 dihydroxypropene. Propane does not react with Baeyer’s reagent.

Question 40.
What happens when isobutylene is treated with acidified potassium permanganate?
Answer:
Isobutylene is treated with acidified KMnO₄ to give acetone.

Question 41.
how will you convert ethyl chloride in to –
1. ethane
2. n – butane
Answer:
1. Conversion of ethyl chloride into ethane:

2. Conversion of ethyl chloride into n-butane:
Wurtz reaction:

Question 42.
Describe the conformers of n-butane.
Answer:
n-butane may be considered as a derivative of ethane as one hydrogen on each carbon atom is replaced by a methyl group.

Edipsed conformation:
In this conformation, the distance between the two methyl groups is minimum so there is maximum repulsion between them and it is the least stable conformer.

Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. It is the most stable conformer. The following potentially energy diagram shows the relative stability of various conformers of n-butane.

Question 43.
Write the chemical equations for combustion of propane.
Answer:
Chemical equations for combustion of propane:
The general combustion reaction for any alkane is:

Question 44.
Explain Markovnikoffs rule with suitable example.
Answer:
Markovnikoff’s rule: When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon atom that has more number of hydrogen and halogen adds to the carbon atom having fewer hydrogen atoms.
Example:
Addition of HBr to Propene:

Question 45.
What happens when ethylene is passed through cold dilute alkaline potassium permanganate.
Answer:
Ethylene reacts with cold dilute alkaline KMnO₄ solution to give ethylene glycol:

Question 46.
Write the structures of following alkanes.
1. 2, 3 – Dimethyl – 6 – (2 – methylpropyl) decane
2. 5 – (2 – Ethylbutyl) – 3, 3 – dimethyldecane
3. 5 (1,2 – Dimethyipropyl) – 2 – methylnonane
Answer:
1. 2, 3 – Dimethyl – 6 – (2 – rnethylpropyl) decane

2. 5 – (2 – Ethylbutyl) – 3, 3- dimethyldecane

3. 5 – (1,2 – Dimethylpropyl) – 2- methylnonane

Question 47.
How will you prepare propane from a sodium salt of fatty acid?
Answer:

Sodium salt of butyric acid on heating with sodalime gives propane.

Question 48.

Answer:

Question 49.
How will you distinguish 1 – butyne and 2 – butyne?
Answer:

Answer:

Question 50.
How will you distinguish 1 – butyene and 2 – butyne?
Answer:

In 1-butyne, terminal carbon atom contains atom one acidic hydrogen, therefore it will react with silver nitrate in the presence of ammonium hydroxide to give silver butynide. Whereas 2-butyne does not undergo such type of the reaction, because of the absence of acidic hydrogen.

Samacheer Kalvi 11th Chemistry Hydrocarbons Evaluate Your self

Question 1.
Write the structural formula and carbon skeleton formula for all possible chain isomers of C₆H₁₄ (Hexane).
Answer:
C₆H₁₄ has five possible isomeric structures:

Question 2.
Give the IUPAC name for the following alkane.


Answer:

Question 3.
Draw the structural formula for 4,5 -diethyl -3,4,5-trimethyloiane
Answer:

Question 4.
Water destroys Grignard reagents. Why?
Answer:
CH₃MgX + HOH → CH₄ – Mg(OH)X.
Water would protonate the grignard reagent and destroy the gngnard reagent, because the grignard carbon atom is highly nucleophilic. This would form a hydrocarbon. Therefore to make a grignard solution, only ether is the best solvent and water or alcohol are not used for that purpose.

Question 5.
Is it possible to prepare methane by Kolbe’s electrolytic method.
Answer:
Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.

Question 6.
Write down the combustion reaction of propane whose AH° = -2220 kJ
Answer:

Question 7.
Why ethane is produced in chlorination of methane?
Answer:
Chlorination of methane involves free radical mechanism. During the propagation step methyl free radical is produced, which is involved in the termination step, the two methyl free radical then combines to form ethane.

Question 8.
How toluene can be prepared by this method?
(i) From n-heptane,
(ii) From 2-methyihexane
Answer:

(ii)

Question 9.
Write the IUPAC names for the following alkenes.

Answer:

Question 10.
Draw the structures for the following alkenes.

1. 6 – Bromo – 2,3 – dimethyl 2 – hexene
2. 5 – Bromo – 4 – chloro 1 – heptene
3. 2,5 – Dimethyl 4 – octene
4. 4 – Methyl – 2 pentene

Answer:
1. 6 – Bromo – 2,3 – dirnethyl – 2 – hexene:

2. 5 Bromo -4 – Chloro – 1 – heptene:

3. 2.5 – dimethyl – 4 – Octene:

4. 4 – methyl 2 pentene:

Question 11.
Draw the structure and write down the IUPAC name for the isomerism exhibited by the molecular formulae:
1. C₅H₁₀ – Pentene (3 isomers)
2. C₆H₁₂ – Hexene (5 isomers)
Answer:
1. C₅H₁₀ – Pentene (3 isomers)
(a) CH₃ – CH₂ – CH₂ – CH = CH₂ → 1- penlene (or) pent- 1-ene
(b) CH₃ – CH₂ – CH = CH – CH₃ → 2- pentene (or) pent-2-ene

2. C₆H₁₂ – Hexene (5 isomers)
(a) CH₃ – CH₂ – CH₂ – CH₂ – CH = CH₂ → Hex-I-ene
(b) CH₃ – CH₂ – CH₂ – CH = CH – CH₃ → Hex – 2 – ene
(c) CH₃ – CH₂ – CH = CH – CH₂ – CH₃ → Hex – 3 – ene


These two compounds exhibits constitutional isomerism.

Question 12.
Determine whether each of the following alkenes can exist as cis-trans isomers?
(a) 1 – Chloropropene
(b) 2 – Chloropropene
Answer:
(a) 1 – Chloropropene:
CH₃ – CH = CHCl

Therefore – 1 – Chloropropene has cis-trans isomers.

(b) 2 – Chloropropene:

In 2-Chloropropene, it deviates from the rule, that is “At least one same group is present on the two doubly bonded carbon atom”. Therefore-2-chloropropenc cannot exist as cis – trans isomers.

Question 13.
Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃ – CH= CH – CH₂ – CH₃
Answer:
(a) 2-Chloro-2-butene:

(b) CH₃ – CH = CH – CH – CH₃

Question 14.
How propene is prepared form 1, 2-dichloropropane?
Answer:

Question 15.
How ozone reacts with 2-methyl propene?
Answer:

Question 16.
An organic compound (A) on ozonolysis gives only acetaldehyde. (A) reacts with Br₂ /CCl₄ to give compound (B). Identify the compounds (A) and (B). Write the IUPAC name of (A) and (B). Give the geometrical isomers of (A).
Answer:
2-Rutene undergo ozonolysis to give acetaldehyde only.


Geometrical isomers of 2 – Butene (A):

Question 17.
An organic compund (A) C₂H₄ decolourises bromine water. (A) on reaction with chlorine gives (Br). A reacts with HBr to give (C). Identify (A),(B),(C). Explain the reactions.
Answer:
(i) C₂H₄ (A), decolourises the bromine water. Therefore it contains double bond. Hence (A) is ethylene.

Question 18.
Prepare propyne from its corresponding alkene.
Answer:
Preparation of propyne from propene:

Question 19.
Write the products A & B for the following reaction.

Answer:

Question 20.

Answer:

Question 21.
Calculate the number of rings present in C₁₈H₁₂.
Answer:
Double bond equivalent formula = ( C – \(\frac {1}{2}\) + \(\frac {1}{2}\))
Where C = no. of carbon atoms. H = no. of hydrogen and halogen atoms and N = no. of nitrogen atoms.
So. in C₁₈H₁₂ Double bond equkalent = 18 – \(\frac {12}{2}\) + 0 + 1 = 18 – 6 + 1 = 13

One ring is equal to one double bond equivalent.
∴ here four rings are there, four double bond equivalent arc used. So remaining, 13 – 4 = 9.
nine double bonds are present in the ring. Hence, C₁₈H₁₂ contain four aromatic rings.

Question 22.
write all possible isomers for an aromatic benzenoid compound having the molecular formula C₈H₁₀.
Answer:
Possible isomers for C₈H₁₀:

Question 23.
Write all possible isomers Iòr a monosubstituted aromatic benzenoid compound having the molecular formula C₉H₁₂
Answer:
Possible isomers for C₉H₁₂:

Question 24.
how benzene can be prepared by Grignard Reagent?
Answer:
Phenyl magnesium bromide reagent reacts with a water molecule to give benzene:

Question 25.
Why benzene undergoes an electrophilic substitution reaction whereas alkenes undergo an addition reaction?
Answer:

• Benzene possesses an unhybridised p-orbital containing one electron. The lateral overlap of their p-orbitals produces 3 it bond.
• The six electron of the p-orbitais cover all the six carbon atoms and arc said to be delocalised.
• Due to delocalisation, strong it-bond is formed which makes the molecule stable. Therefore benzene undergoes electrophilic substitution reaction, whereas alkenes undergo addition reaction.

Question 26.
Convert Ethyne to Benzene and name the process.
Answer:
Conversion of Ethyne into Benzene:

This process is one of the cyclic polymerisation process.

Question 27.
Toluene undergoes nitration easily than benzene.Why?
Answer:
1. Toluene has a methyl group on the benzene ring which is electron releasing group and hence activate the benzene ring by pushing the electrons on the benzene ring.
2. CH₃ group is ortho – para director and ring activator. Therefore in toluene, ortho and para positions are the most reactive towards an electrophile, thus promoting electrophilic substitution reaction.
3. The methyl group hence makes it around 25 times more reactive than benzene. Therefore it undergoes nitration easily than benzene.

Samacheer Kalvi 11th Chemistry Hydrocarbons Additional Questions Solved

Choose the correct statement.

Question 1.
The IUPAC name of neopentane is
a) 2 – methyl butane
b) 2, 2 – dimethyl propane
c) 2 – methyl propane
d) 2, 2- dimethyl butane
Answer:
b) 2, 2 – dimethyl propane

Question 2.
Statement – I : n-butane and iso – butane are isomers.
Statement – II : Because they are having same molecular formula but differs only in the structural formula.
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I
(h) Statement – I and II are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but statement – II s wrong.
(d) Statement – I is wrong but statement – II is correct.
Answer:
(a) Statement -I and II are correct and statement – II is correct explanation of statement-I

Question 3.
The compressed gas available in cooking gas cylinders is a mixture of:
a) C₆H₆ + C₆H₅CH₃
b) C₂H₄ + C₂H₂
c) C₂H₄ + CH₄
d) C₄H₁₀ + C₃H₈
Answer:
d) C₄H₁₀ + C₃H₈

Question 4.
Find out the branched hydrocarbon from the following compounds.
(a) 1 – propane
(b) n – propane
(c) iso – butane
(d) n – butane –
Answer:
(c) iso – butane

Question 5.
Adam’s catalyst is:
a) platinum metal
b) palladium
c) nickel metal
d) PtO₂
Answer:
d) PtO₂

Question 6.
Statement – I : Boiling point of methane is lower than that of butane.
Statement – II : The boiling point of continuous chain alkanes increases with increase in length of carbon chain.
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I
(b) Statement – I and II are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but a statement – II is wrong.
(d) Statement – I is wrong but a statement – I is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement -I.

Question 7.
Methyl bromide is converted into ethane by heating it in an ether medium with
a) Al
b) Mg
c) Na
d) Cu
Answer:
c) Na

Question 8.
Consider the following statements.
(i) The process of reduction using sodium in liquid ammonia is called Birch reduction.
(ii) Birch reduction is stereospecific in reaction.
(iii) Aleynes can be reduced to cis – alkenes using Birch reduction. Which of the above statement is/are correct’?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) and (ii)

Question 9.
Statement – I : Alkenes are more reactive than alkanes.
Statement – II : Because of the presence of a double bond.
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I
(b) Statement – I and Il are correct but statement – II is not correct explanation of statement -I
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – I is wrong but statement – II is correct.
Answer:
(a) Statement – I and II are correct and statement – II is correct explanation of statement – I

Question 10.
Select the correct statement about alkanes
a) they are polar in nature
b) they are soluble in water
c) they are non – combustible
d) their dipole moment is zero
Answer:
d) their dipole moment is zero

Question 11.
Which one of the following has garlie odour?
(a) Ethane
(b) Ethene
(c) Ethyne
(d) Ethanol
Answer:
(c) Ethyne

Question 12.
Identify the X.
(a) Propane
(b) Acetone
(c) Acetaldehyde
(d) Formaldehyde
Answer:
(b) Acetone

Question 13.
Which one of the following is not a monocyclic aromatic hydrocarbon?
(a) Benzene
(b) Phenol
(c) Toluene
(d) Naphthalene
Answer:
(d) Naphthalene

Question 14.
Which one of the following is a polynuclear aromatic hydrocarbon?
(a) Anthracene
(b) Phenol
(c) Benzene
(d) Toluene
Answer:
(a) Anthracene

Question 15.
Which one of the following is an aromatic compounds?

Answer:

Question 16.
Conformational isomers are due to
a) Free rotation about C – C single bond
b) Frozen rotation about C – C single bond
e) Frozen rotation about C – C double bond
d) Restricted rotation about C – C single bond
Answer:
a) Free rotation about C – C single bond

Question 17.
Statements-I: Unlike alkenes and alkynes benzene undergoes substitution reactions rather than addition reactions under normal conditions.
Statements-II: Because of the delocalisation of electrons a strong t bond is formed which makes the molecule stable.
(a) Statement-I and II are correct and statement-II is the correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not the correct explanation of statement-I
(c) Statement-I ¡s correct but statement-II is wrong.
(d) Statement-I is wrong but statement-II is correct.
Answer:
(a) Statement-I and II are correct and statement-II is the correct explanation of statemen-I.

Question 18.
A → B + H₂O. Identify A and B.

Answer:

Question 19.
Benzene undergoes Birch reduction to form .

Answer:

Question 20.
How many types of carbon atoms are present in 2, 2, 3 – trimethyl pentane
a) one
b) Two
c) Three
d) Four
Answer:
d) Four

Question 21.
Which one of the following is not a meta director’?
(a) – NH₂
(b) – NO₂
(c) – COOR
(d) – SO₃H
Answer:
(a) – NH₂

Question 22.
Which one of the following benzene ring deactivator?
(a) – CHO
(b) – OH
(c) – CH₃
(d) – OCH₃
Answer:
(a) – CHO

Question 23.
Fine the odd one out:
(a) Benzene
(b) Ethane
(c) Ethene
(d) Propyne
Answer:
(a) Benzene

Question 24.
Which among these is not associated with aliphatic compounds?
(a) They contain (4n+2) π electrons
(b) They contain straight chain compounds.
(c) They contain branched-chain compounds.
(d) They have an appropriate number of H-atoms and functional groups.
Answer:
(a) They contain (4n+2) π electron.

Question 25.
Which of the following compounds will exhibit cis-trans isomerism
(a) 2 – Butene
(b) 2 – Butyne
(c) 1 – Butene
(d) 2 – Butanol
Answer:
(a) 2 – Butene

Question 26.
Which conformation of ethane has the lowest potential energy?
(a) Eclipsed
(b) Staggered
(c) Skew
(d) All will have equal potential energy.
Answer:
(b) Staggered

Question 27.
Arrange the following in the decreasing order of their boiling points
i) n – butane
ii) 2 – methylbutane
iii) n – pentane
iv) 2 – methylbutane
a) i > ii > iii > iv
b) ii > iii > iv > i
c) iv > iii > ii > i
d) iii > ii > iv > i
Answer:
d) iii > ii > iv > i

II. Match the following
Question 1.


Answer:

Question 2.

Answer:

Question 3.


Answer:

Question 4.


Answer:

Question 5.


Answer:

Question 6.


Answer:

Question 7.

Answer:

Question 8.


Answer:

III. Fill in the blanks.

Question 1.
Liquefied petroleum gas consisting of a mixture of ………..
Answer:
Propane + butane

Question 2.
Mangoes contain ………….
Answer:
Cyclo hexane

Question 3.
Methane gas is also called as ………….
Answer:
Marsh gas

Question 4.
The IUPAC name of the following compound is:

Answer:
3-Ethyl, 2-methylpentane

Question 5.
Sodalime is the mixture of ……………
Answer;
NaOH + CaO

Question 6.
Wurtz reaction used in the preparation of …………
Answer:
higher alkanes

Question 7.
The major reagent present in Corey-House reaction is ………..
Answer:
Lithium dimethyl cuparate

Question 8.
General formula for grignard reagents is ………..
Answer:
R – MgX

Question 9.
The rotation of C – C single bond leads to different isomenc structure called as …………
Answer:
conformers

Question 10.
The least stable conformer of ethane is formed……….
Answer:
eclipsed

Question 11.
The potential energy difference between the staggered and eclipsed conformation of ethane is …………
Answer:
12.5 kJ /mol

Question 12.
The most stable conformer of butane is ………..
Answer:
Staggered

Question 13.
Paraffin is the older name for the group family of compounds.
Answer:
alkane

Question 14.
Paraffin means ……….
Answer:
Little activity

Question 15.
The preparation of methyl chloride is followed by a mechanism.
Answer:
Free radical

Question 16.
n-hexane passed over chromic oxide supported on alumina at 873 K will give ……….
Answer:
Benzene

Question 17.
The number of possible isomers of C₆H₁₂ is …………..
Answer:
5

Question 18.
When ethanol is heated at 440 K with excess of concentrated H2SO4, it will give …………
Answer:
Ethene

Question 19.
Alkynes undergo reduction using Lindlar’s catalyst to give ………….
Answer:
cix – alkenes

Question 20.
Alkynes undergo reduction using sodium in liquid ammonia to give …………
Answer:
trans – alkenes

Question 21.
An aqueous solution of potassium succinate is electrolysed to give ………….
Answer:
Ethane

Question 22.
The order of reactivity of different hydrogen halides (HCl, HI , HBr) is ………..
Answer:
HI >HBr >HCl

Question 23.
Addition of hydrohalides to alkene is an example for …………
Answer:
Electrophilic additton

Question 24.
Ethane reacts with HBr to form …………
Answer:
Bromo ethane

Question 25.
Homolytic fission of benzoyl peroxide will give ………..
Answer:
C₆H₅

Question 26.
Propene reacts with HBr in the presence of peroxide to form …………
Answer:
1 – Bromopropane

Question 27.
Baeyer’s reagent is …………
Answer:
alkaline KMNO₄

Question 28.
Ozonolysis of ethene produces ………….
Answer:
HCHO

Question 29.
Electrolysis of potassium maleate yields ………
Answer:
Ethyne

Question 30.
Ozonolysis of acetylene gives ………….
Answer:
HCOOH

Question 31.
Three molecules of acetylene undergoes polymerisation to give …………
Answer:
benzene

Question 32.
Benzene reacts with bromine in the presence of AlCl₃ to form bromobenzene and it is an example of reaction
Answer:
Electrophilic substitution

Question 33.
The six carbon atoms of benzene are hybridised.
Answer:
sp²

Question 34.
Bond angle in benzene is …………
Answer:
120º

Question 35.
Benzene contains a bonds and it bonds …………
Answer:
12,3

Question 36.
Wurtz-fittig reaction helps to prepare compounds …………
Answer:
aromatic

Question 37.
When phenol reacts with Zn-dust under dry distillation conditions it gives ………..
Answer:
Benzene

Question 38.
Benzene is insoluble in …………
Answer:
water

Question 39.
Benzene reacts with hydrogen in the presence of Pt to yield ……….
Answer:
Cyclohexane

Question 40.
Benzene reacts with Cl₂ in the presence of sunlight to give ………..
Answer:
BHC (Ben.zene hexachioride)

Question 41.
The step in which Cl-Cl bond homolysis occurs is called ……….
Answer:
Initiation step

Question 42.
Dienes are the name given to compounds with ……….
Answer:
Exactly two double bonds

Question 43.
The hybridisation state of a carbocation is …………
Answer:
sp²

Question 44.
The peroxide effect in anti-markovnikoff addition involves a ………. mechanism.
Answer:
free radical.

IV. Choose the odd one out.
Question 1.
(a) Ethane
(b) Benzene
(c) Ethene
(d) Ethyne
Answer:
(b) Benzene. it is an aromatic hydrocarbon whereas others are aliphatic hydrocarbons.

Question 2.
(a) Zn + HCl
(b) Zn + CH₃COOH
(c) LiAlH₄
(d) Acidified K₂Cr₂O₇
Answer:
Acidified K₂Cr₂O₇. It is an oxidising agent whereas the other three reagents are reducing

Question 3.
(a) Soft drink bottle
(b) Jars
(c) Vegetable oil bottle
(d) Straws
Answer:
(d) Straws. It is made up of polypropylene whereas others are made of PET (Polyethylene terephthalate).

Question 4.
(a) Straws
(b) Foam cups
(c) Diapers
(d) Toys
Answer:
(b) Foam cups. It is made up of polystyrene whereas others are made up of polypropylene.

Question 5.
(a) Orlon
(b) Neoprene rubber
(c) PVC
(d) PET
Answer:
(d) PET. It is prepared by the polymerization of glycol and terephthalic acid, whereas others are prepared from acetylene.

Choose the correct pair.

Question 1.
(a) Propene + O₃ : HCHO + CH₃ CH0
(b) Ethene + O₃ : CH₃ CHO + H₂O₂
(c) But-2-ene + O₃ : HCHO + H₂ O₂

Answer:
(a) Propene + O₃ : HCHO + CH3CHO

Question 2.
(a) PET : shampoo bottles
(b) PS : disposable utensils
(c) PP : grocery bags
(d) HDPE : plastic pipes
Answer:
(b) PS : disposable utensils

Question 3.

Answer:

Question 4.

Answer:

VI. Choose the incorrect pair.

Question 1.

Answer:

Question 2.
(a) CH₂ = CH₂ + O₃ : 2HCOH
(b) CH₃ – CH = CH₂ + O₃ : CH₃CHO + HCHO
(c) CH₃ – CH = CH – CH₃ + O₃ : 2CH₃COCH₃
(d) CH₃ – CH = CH – CH₃ + O₃ : 2CH₃CHO
Answer:
(c) CH₃ – CH = CH – CH₃ + O₃ : 2CH₃COCH₃

Question 3.
(a) PET: Jars
(b) HDPE: Juice containers
(c) PS: Disposable utensils
(d) PVC: Grocery bags
Answer:
(d) PVC: Grocery bags

VII. Assertion & Reason.
Question 1.
Assertion (A): Methane is called marsh gas.
Reason (R): Decomposition of plant and animal matter in an oxygen-deficient environment like swamps, marshes and bogs produce methane gas.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A) : Water destroys grignard reagent and so it is not used as solvent for RMgX.
Reason (R) : Water decomposes grignard reagent (RMgX) to give alkane.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): The boiling point of straight-chain isomers have higher boiling point as compared to branched-chain isomers.
Reason (R): The boiling point decreases with increase in branching as the molecule becomes compact and the area of contact decreases.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A)is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A) : The eclipsed conformation of ethane is less stable than staggered conformation of ethane.
Reason (R) : In eclipsed conformation, the distance between the two methyl group is minimum and so there is maximum repulsion between them and it is the least stable conformer.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A)and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Hydrocarbons 2 Mark Questions and Answers

Question 1.
What are unsaturated hydrocarbons?
Answer:
hydrocarbons having localised carbon-carbon multiple bonds are called unsaturated hydrocarbons.
Example :
alkenes and alkynes.

Question 2.
What is marsh gas?
Answer:
Answer:

• Methane is the major component of the atmosphere of jupiter, Saturn, Uranus and Neptune but only minor component of earth atmosphere.
• Decomposition of plant and animal matter in an oxygen deficient environment like swamps, marshes bogs and the sediments of lakes produces methane gas. It is otherwise known as marsh gas.

Question 3.
Write a note on methane clathrates.
Answer:

• A frozen mixture of water and methane gas is chemically known as methane clathrate.
• The methane molecule which is produced by biological process under the deep -ocean (at 4°C and 50 atm) does not simply reach the surface instead each molecule is trapped inside the clusters of 6 to 18 water molecules forming methane clathrates.

Question 4.
What are alkenes? Give example.
Answer:
Alkenes are unsaturated hydrocarbons that contain carbon-carbon double bond. They are represented by the general formulae C_nH_2n where ‘n’ stands for a number of carbon atoms in the molecule. Alkenes are also known as olefins (in Latin – oil maker) because the first member ethene combines with chlorine gas to form an oily liquid as a product.
Example:
Ethylene (C₂H₄)

Question 5.
Draw and name the possible structural formula for C₄H₁₀
Answer:
C₄H₁₀ has two possible structural formula, they are,
(i) CH₃-CH₂-CH₂-CH₃ n-Butane

Question 6.
Give the IUPAC name of the following compounds.

Answer:

Question 7.
What is Sabatier- Sendersens reaction?
Answer:
The process of addition of H₂ to unsaturated compounds (alkenes or alkynes) in known as hydrogenation. The above process can be catalysed by nickel at 298 K. This reaction is known as Sabalier-Sendersens reaction.
For example:

Question 8.
What are decarboxylation reactions? Given an example.
Answer:
When a mixture of sodium salt of carboxylic acid and sodalime is heated an alkane is formed. During this process CO₂ molecule is eliminated process and this process is known as decarboxylation reaction.

Question 9.
Write a note on Kolbe’s electrolytic method?
Answer:
When potassium or sodium salt of carboxylic acid is electrolysed, a higher alkane is formed.

Question 10.
How will you prepare propane from Chloropropane?
Answer:
Nascent hydrogen reacts with chloropropane to give propane.

Question 11.
What is Wurtz reaction?
Answer:
When a solution of haloalkanc in dry ether is treated with sodium metal, higher alkanes are produced. This reaction is known as Wurtz reaction. For example:

Question 12.
Write shnrt flotes on Corey-House reaction?
Answer:
An alkyl halide and lithium dialkyl cupratc are reacted to give higher alkanes. This reaction is known as Corey-House reaction.

Question 13.
How will you prepare methane from grignard reagent’?
Answer:
Methyl magnesium chloride reacts with water to give methane.

Question 14.
What are conformers?
Answer:
The rotation about C-C single bond axis yielding several arrangements of a hydrocarbon called conformers.

Question 15.
Draw the conformations of ethane using Newman projection formula method?
Answer:

Question 16.
What are combustion reactions?
Answer:
A combustion reaction is a chemical reaction between a substance and oxygen with evolution of heat and light. In the presence of sufficient oxygen alkanes undergoes combustion when ignited and produces carbon dioxide and water.

Question 17.
What is arornatisation?
Answer:
Alkanes with six to ten carbon atoms are converted into homologous of benzene at higher temperature and in the presence of a catalyst. This process is known as aromatisation. For example, n-Hexane passed over Cr₂O₃ supported on alumina at 873 K gives benzene.

Question 18.
Write notes on isomerisation.
Answer:
1. Isomerisation is a chemical process by which a compound is transformed into any of its isomeric form.
2. Normal alkanes can be converted into branched alkanes in the presence ofAlCl₃ and HCl at 298 K.

3.  This process is of great industrial importance, the quantity of gasoline is improved by isomerising its components.

Question 19.
Mention the uses of alkanes.
Answer:

• Alkanes are extensively used as fuels.
• Methane present in natural gas is used in home heating.
• A mixture of propane and butane is known as LPG gas which is used for domestic cooking purpose.
• Gasoline is a complex mixture of many hydrocarbons used as a fuel for internal combustion engines.

Question 20.
Draw and name the structural formula for C4H8’?
Answer:
(i) CH₃ – CH = CH – CH₃ -w 1 – Butene
(ii) CH₂ = CH – CH₂ – CH₂ – 2 Butene

Question 21.
cis-isomers are less stable than rans-isomers?
Answer:
Consider:

Among cis and trans isomers, cis isomer is less stable than trans isomer. In the cis isomer, similar groups arc very near to each other. vander Waal’s repulsion and steric hindrance make the molecule much more unstable. But in trans isomer, similar groups are diagonally opposite to each other and there is no such steric hindrance. Due to more steric interaction cis isomers is less stable than Irons isomer.

Question 22.
How will you prepare ethene from ethanol?
Answrer:
When ethanol is heated at 430 K – 440 K with excess cone. H₂SO₄. a molecule of water from alcohol is removed and ethene or ethylene is formed.

Question 23.
How will you convert 1-brornopropane into popene?
Answer:

1-bromopropane reacts with alcoholic KOH and eliminate hydrogen bromide resulting in the formation of propene.

Question 24.
How will you prepare ethene by Kolbe’s electrolytic method?
Answer:
When an aqueous solution of potassium succinate is electrolysed between two platinum electrodes, ethene is produced at the anode.

Question 25.
Write any two test for alkenes.
Answer:

• Rapid decolourisation of bromine in CCl₄ without evolution of hydrogen bromide.
• Decolourisation of cold dilute aqueous solution of KMnO₄

Question 26.
State Markovnikoff’s rule.
Answer:
When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen atom adds to the carbon atom that has more number of hydrogen atoms and halogen add to the carbon atom having fewer hydrogen atoms.

Question 27.
What is peroxide effect?
Answer:
The addition of HBr to an alkene in the presence of organic peroxide gives the antì-Markovnikofl’s product. This effect is called as peroxide effect.

Question 28.
Identify the products A and B from the following reaction.

Answer:

Question 29.
What happens when propcne reacts with concentrated H,S04?
Answer:
Propene reacts rjth cone. H2S04 to form 2 – propyl hydrogen sulphate in accordance with Markovnikoffs rule Further hydrolysis yields 2-propanol.

Question 30.
Complete the following reaction and identify A, B and C.

Answer:

Question 31.
Mention the uses of alkenes.
Answer:

• Alkenes are used as starting material in the synthesis of alcohols, plastics, detergents and fuels.
• Ethenc is the most important organic fcedstock in the polymer industry. Examples are, PVC, Sarans and Polythene. These polymers are used in the manufacture of floor tiles, shoe-soles, synthetic fibres, raincoats, pipes etc.

Question 32.
What are gem dihalides? How will you prepare propyne from gem dihalides?
Answer:
Compounds containing two halogen atoms on the same carbon atom are called gem dihalides. On heating 1,1 -dichloropropane with alcoholic KOH, it will give propyne.

Question 33.
How will you prepare acetylene from potassium maleate?
Answer:
Electrolysis of potassium maleate yields acetylene. This is one of Kolbe’s electrolytic reaction.

Question 34.
How will you prepare acetylene from calcium carbide?
Answer:
Acetylene can be manufactured in large scale by action of calcium carbide with water.

Question 35.
How will you convert ethyne into ethanol?
Answer:
Ethyne undergo hydration on warming with mercuric sulphate and dil H₂SO₄ at 333 K to form ethanol (Acetaldehyde).

Question 36.
Mention the uses of Acetylene.
Answer:

• Acetylene is used in oxy acetylene torch used for welding and cutting metals.
• It is used for manufacture of PVC, polyvinyl acetate, Polyvinyl ether, orlon and neoprene rubbers.

Question 37.
What are all the conditions for aromaticity?
Answer:
Huckel proposed that aromaticity is a function of electronic structure of an organic compound.
A compound may be aromatic, if it obey the following rules:

• The molecule must have a cyclic structure.
• The molecule must be co-planar.
• Complete delocalisation of it electrons in the ring.
• Presence of (4n + 2) it electrons in the ring where n is an integer (n = 0,1,2….), this is known as Huckel’s rule.

Question 38.
Classify the following compounds by using aromaticity concepts:

Answer:

• It is co-planar molecule.
•  It has six delocalised ir electrons.
•  4n + 26
  4n = 6 – 2
  4,n = 4
  n = 1, obeys Huckel’s rule with n = 1

(b) Cyclo-octatetraene.

1. Molecule is non-planar.
Hence it is a non-aromatic compound.

(c) Cyclopropyl cation.

• It has a co-planar structure.
• It has two delocalised it electrons.
  4n + 2 = 2
  4n = 0
  n = 0

Hence is an aromatic compound.

Question 39.
Write notes on Resonance of benzene?
Answer:
1. The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance.
2. The actual structure of the molecule is said to be a resonance hybrid of the various possible alternative structures.

3.  In benzene, Kekule’s structures I and II represented the resonance structures, and structure III is the resonance hybrid of structures I and II.
4. The structures I and II exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

Question 40.
How will you convert phenol into benzene?
Answer:
When phenol vapours are passed over zinc dust then it is reduced to benzene:

Question 41.
What is Wurtz-fitting reaction?
Answer:
When a solution of bromobenzene and iodomethane in dry ether is treated with metallic sodium, toluene is formed.

Question 42.
What are activating and deactivating groups?
Answer:

• When mono substituted benzene undergoes an electrophilic substitution reaction, the rate of the reaction and the site of attack of the incoming electrophile depends on the functional group already attached to it.
•  Some groups increases the reactivity of benzene ring and are known as activating groups.
• Some groups decreases the reactivity of benzene ring and are known as de-activating groups.
• Example:
  Activating groups: – NH₂ – OH. – CH₃ etc.
  Deactivating groups: – NO₃. CN, – CHO etc,

Question 43.
Why does benzcnc undergo clectrophilic substitution reactions easily and nucleophilic substitution with difficulty?
Answer:
Due to the presence of an electron cloud containing 6π – electrons above and below the plane of the ring, benzene is a rich source of electrons, Consequently, it attracts the electrophilic reagents towards it and repels the nucleophilic reagents. As a result, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitution with difficulty.

Question 44.
Out of benzene, m – dinitrobenzene and toluene which will undergo nitration most easily and why?
Answer:
CH₃ group is an electron-donating group, while -NO₂ group is electron withdrawing group. Therefore, maximum electron density will be there in toluene, followed by benzene and it is least in pn-dinitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene > benzene > m – dinitrobenzene.

Question 45.
Why are alkanes called paraffins?
Answer:
Paraffins means little affinity. Alkanes due to strong C-C and C-H bonds are relatively chemically inert. They are thus called as paraflins.

Samacheer Kalvi 11th Chemistry Hydrocarbons 3 Mark Questions and Answers

Question 1.
Explain the classification of hydrocarbons.
Answer:

Question 2.
How to write the possible isomers of C₅H₁₂?
Answer:
1. To begin draw the carbon backbone of the straight chain isomer.
C-C-C-C-C

2. To determine the carbon backbone structure of the other isomers, arrange the carbon atoms in the other way:

3. Fill in ail the hydrogen atoms so that each carbon forms four bonds,

Question 3.
Explain how to draw the structural formula for 3-ethyl, 2, 3-dimethylpentane.
Answer:
3-ethyl, 2, 3-dimethylpentane.
Step 1:
The parent hydrocarbon is pentane. Draw the chain of five carbon atoms and number

Step 2:
Complete the carbon skeleton by attaching the alkyl group as they are specified in the name. An ethyl group is attached to carbon atom 3 and two methyl groups are attached to carbon atoms 2 and 3.

Step 3:
Add hydrogen atoms to the carbon skeleton so that each carbon atoms have four bonds.

Question 4.
In alkane compounds with same number of carbon atoms, straight chain isomers have higher boiling point as compared to branched chain isomers-justify statement.
Answer:

• The boiling point of continuous chain alkanes increases with increase in the length of carbon chain roughly about 30°C for every added carbon atom to the chain.
• Alkanes are non-polar compounds and having weak Vander Wal’s force which depends upon molecular surface area and hence increases with increase in molecular size.
• The boiling point decreases with increase in branching on the molecule i.e. as it becomes compact and the area of the contract decreases. Hence in alkanes with same number of carbon atoms, straight chain isomers have higher boiling point as compared to branched chain isomers.

Question 5.
Why oil spills in aqueous environment spread so quickly?
Answer:

• Water molecules are polar and akanes are non-polar. The insolubility of alkanes in water makes them a good water repellent for metals which protects the metal surface from corrosion.
• Because of their lower density than water they tòrm two layers and occupies the top layer. The density difference between alkanes and water explains why oil spills in aqueous environment spread so quickly.

Question 6.
Explain pyrolysis method.
Answer:
1. Pyrolysis is defined as the thermal decomposition of an organic compound into smaller fragments in the absence of air through the application of heat. Pyrolysis of alkanes is also named as cracking.

2. In the absence of air, whai alkane are vapours passed through red-hot metal it breaks into simpler hydrocarbons.

3. The product depends upon the nature of alkane, temperature, pressure and presence or absence of the catalyst.

Question 7.
Write notes on Geometrical isomerism or cis-trans isomerism.
Answer:
1. It is a type of stereoisomerism and it is also called cis-trans isomerism. Such type of isomerism results due to the restricted rotation of doubly bounded carbon atoms.

2. If the similar groups lie on the same side then the geometrical isomers are called as cis-is omers.

3. If the similar groups lie on the opposite side then the geometrical isomers are called as trans-isomers.

Question 8.
Explain how 2-butyne reacts with (a) Lindlar’s catalyst and (b) Sodium in liquid ammonia?
Answer:
(a) 2-butyne reacts with Lindlar’s catalyst:
2-butyne can be reduced to cis-2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis-2-butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans-2-butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans-2-butene.

Question 9.
What are vicinal dihalides? How will you prepane alkene from vicinal dihaldes?
Answer:
1. The compounds in which two halogen atoms are attached to adjacent carbon-atoms are called as vicinal dihalides.

2. When vicinal dihalides are warmed with granulated zinc in methanol they lose a molecule

Question 10.
Why alkenes are more reactive than alkanes?
Answer:
1. Alkenes are more reactive than alkanes due to the presence of a double bond.

2. The a-bond is strong but the it-bond is weak. The typical reactions of alkenes involve the addition of an electrophile across the double bond proceeding through ionic mechanism. However addition reactions proceed through free-radical mechanism also. Ozonolysis and nolvmerisation are some of the characteristic reaction of alkenes.

Question 11.
Explain the mechanism of addition of HBr to propenc.
Answer:
Step 1:
Formation of electrophile:
In HBr, br is more electronegative than H. When bonded electron move towards Br, polarity is deve[oped and it creates electrophile H⁺ which attacks to the double bond to form a carbocation.

Step 2 :
Secondary carbocation is more stable than primary carhocation and it predominates over the primary carbocation.

Step 3 :
The Br^-1 ion attack the 2°- carbocation to form 2-Bromo propane as the major product.

Question 12.
Explain the mechanism of addition of FlBr to 3-methyl-1-butene.
Answer:
Consider the addition of HBr to 3-methyl-1-butene. Here the expected product according to Markovnikoff’s rule is 2-bromo-3-methylbutane but the actual major product is 2-bromo- 2-methylbutane. This is because, the secondary carbocation formed during the reaction is rearranged to give the more stable tertiary carbocation. Attack of Br^-1 on this tertiary carbocation gives the major product 2-bromo-2-methylbutane

Question 13.
Why peroxide effect is not observed in HCl and HI?
Answer:
The H-Cl bond is stronger (430.5 k.J mol^-1) than H-Hr bond (363.7 kJ mol^-1), thus H-Cl is not cleaved by the free radical. The H-I bond is further weaker (296.8 kJ mol^-3) than H-Cl bond. Thus H-I bond breaks easily hut iodine free radicals combine to form iodine molecules instead of adding to the double bond and hence peroxide effect is not observed in case of HCl and HI.

Question 14.
Explain the ozonolysis of (a) Ethene and (b) propene
Answer:
Ozonolysis is a method ofoxidative cleavage of alkenes using ozone and it form two carbonyl compounds. Alkenes react with ozone to form ozonide and it is cleaved by Zn/H₂O to form smaller molecules.
This reaction is often used to identify the structure of unknown alkene by detecting the position of double or triple bond.

Question 15.
What is polymerisation? Explain with suitable example.
Answer:
A polymer is a large molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation. A few examples are:

Question 16.
Complete the following reactions:

Answer:

Question 17.
Explain the ozonolysis of (a) Acetylene (b) Propyne
Answer:
Ozone adds to carbon-carbon triple bond of alkynes to form ozonides. The ozonides are hydrolysed by water to form carbonyl compounds. The hydrogen peroxide formed in the reaction may oxidise the carbonyl compound to carboxylic acid.
(a) Acetylene:

(b) propyne:

Question 18.
Explain the polyrnerisation of acetylene molecules.
Answer:
Acetylene undergoes two types of polymerisation reactions, they are:

1. Linear polymerisation
2. Cyclic polymerisation

1. Linear polymerisation:
Acetylene forms linear polymer, when passed into a solution of cuprous chloride and ammonium chloride.

2. Cyclic polymerisation:
Acetylene undergoes cyclic polymerisation on passing through red hot iron tube. Three molecules of acetylene polymerises to form benzene.

Question 19.
Discuss the Kekule’s structure of benzene.
Answer:
Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
1. Bnzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two ortho disubstituted products as shown:

2. Kekule’s structure failed to explain why benzene with three double bonds did not give addition reaction like alkenes. To overcome this objection, Kekule suggested that henzene was a resonance hybrid of two forms(1 and 2) which are in rapid equilibrium.

Question 20.
Why benzene undergoes substitution reaction rather than addition reactions under normal conditions?
Answer:
1. Each carbon atom in henzene possess an unhybridised p-orbital containing one electron. The lateral overlap of their p-orbitais produces 3π- bond, six electrons of the p-orbitais over all the six carbon atoms and are said lo be delocalised.

2. Due to delocalisation, strong it-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes, benzene undergoes substitution reactions rather than addition reactions under normal conditions.

Question 21.
Explain the industrial preparation of benzene from coal tar.
Answer:
Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds, namely, benzene, toluene, xylene in the temperature range of 350 K to 443 K. These vapours are collected at the upper part of the fractionating column.

Question 22.
Explain the suiphonation of benzene.
Answer:
Benzcne reacts with fuming sulphuric acid and give benzene suiphonic acid. Although SO₃ molecule it does not have positive charge, yet it is a strong electrophile. This is because the octet of electrons around the sulphur atom is not reacted. This reached is reversible nd desulphonation occus readily in aqueous medium.

Question 23.
What is BHC? How will you prepare BHC? Mention its uses.
Answer:
1. BHC is I3enzene hexachioride.
2. Benzene reacts with three molecule of Cl₂in the presence of sunlight or UV light to yield BHC.
This is also called as gammaxane or Lindane.

3. BHC is a powerful insecticide.

Question 24.
In aryl halides, halogen group is a oriho -para director and a deactivator towards electrophilic substitution reactions, why?
Answer:
1.  In aryl halides, the strong -I effect of the halogens decreases the electron density of benzene ring, thereby deactivating it for electrophilic attack.
2. The presence of lone pair on halogens is involved in resonance with π-electrons of the benzene ring and it increases the electron density at ortho and para position. Hence the halogen group is an ortho-para director and a deactivator.

Question 25.
Explain the carcinogenity and toxicity of aromatic hydrocarbons.
Answer:
Benzene and polycyclic aromatic hydrocarbons (PAH) are ubiquitous environmental pollutants generated during incomplete combustion of coal, oil, petrol and wood. Some (PAH) originate from open burning, natural seepage of petroleum and coal deposits and volcanic activities. They are toxic, mutagenic and carcinogenic. It has hematological, immunological and neutrological effect on humans They are radioactive and prolonged exposure leads to genetic damage. Some of the examples of PAH arc “L” shaped polynuclear hydrocarbons, which are much more toxic and carcinogenic.

It is found in cigarette smoke, in tobacco and and charcoal boiled food.

It is found in gasoline exhaust and barbecued food.

Question 26.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behavior. Also give reason for this behavior?
Answer:

Since s-electrons arc closer to the nucleus, therefore as the s-character of the orbital making the C-H bond increases the electrons of C-H bond lies closer and closer to the carbon atom. In other words, the partial +ve charge on the H-atom and hence the acidic character increases as the s-character of the orbital increases. Thus, the acidic character decreases in the order
Ethyne > Benzene > Hexane

Samacheer Kalvi 11th Chemistry Hydrocarbons 5 Mark Questions

Question 1.
Explain the conformational analysis of ethane.
Answer:
The two tetrahedral methyl groups can rotate about the carbon-carbon bond axis yielding several arrangements called conformers. The extreme conformations are staggered and eclipsed conformations. There can be number of other arrangements between staggered and eclipsed forms and their arrangements are known as skew forms.

Eclipsed conformation:

In this conformation, the hydrogen of one carbon atom is directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.

Staggered conformation:

In this conformation, the hydrogen’s of both the carbon atoms are tir apart from each other. The repulsion between the atoms is minimum and it is the most stable conformer.

Skew formation:
The infinite numbers of possible intermediate conformations between the two extreme conformations are referred as skew conformations. The stabilities of various conformations of ethane are
Staggered > Skew> Eclipsed
The potential energy difference between the staggered and eclipsed conformation of ethanc is around 12.5 kJ mol^-1HBr The various conformations can be represented by Newman projection formula.

Newman Projection formula for Ethane:

Question 2.
Explain the structure of benzene.
Answer:
1. Molecular formula:
Elemental analysis and molecular weight determination have proved that the molecular formula of benzene is C₆H₆. This indicates that benzene is a highly unsaturated compound.

2. Straight chain structure is not possible:
Benzene could be constructed as a straight chain but it not feasible since it does not show the properties of alkenes or alkynes. For example, it does not decolorise the bromine water in CCl₄.

3. Evidence of cyclic structure:
(1) In the presence of Nickel, benzene reacts with hydrogen to give cyclohexane, a six membered ring. This proves that benzene is a hexagonal
molecule with three double bonds.

(2) Benzene reacts with bromine in the presence of iron to give substituted C₆H₅Br. No isomers of C₆H₅Br was identified. On further reaction with bromine three isomeric disubstituted products C₆H₄Br₂ are formed. On this basis Kekule proposed that benzene consists of ring of carbon atoms with alternate single and double bonds.

4. Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position ofatomsis called resonance. The actual structure of the molecule is said to be a resonance hybrid of various possible alternative structures. In benzene. Kekule’s structure (I) and (II) represented the resonance structures and structure (III) is the resonance hybrid of structure I and II.

5. Spectroscopic measurements:
X-ray and electron dîflìaction studies indicated that all carbon-carbon bonds are of equal length which is in between that of a single bond (1 .45Å ) and that of a double bond (1.34Å ).

6. Molecular orbital structure :
(1). Benzene is a flat hexagonal molecule with all carbons and hydrogen lying in the same plane with a bond angle of 120º. Each carbon atom has sp2 hybrid orbitais of carbon, overlap with each other and with s-orbitals of six hydrogen atoms forming six sigma (σ) C-H bonds and six sigma (σ) C – C bonds.

(2). All the σ-bonds in benzene lies in one plane with bond angle of 120º. Each C-atom in benzene possess an unhybridised p-orbital containing one electron. The lateral overlap of their p-orbitais produces 3π-bond, the six electrons of the p-orbitais cover all the six C-atoms and are said to be delocalised. Due to this delocalisation, strong π-bond is formed which makes the molecule stable.

7. Representation of benzene:
Hence, there are three ways is which benzene can be represented.

Question 3.
Explain the mechanism of the reaction between methane and chlorine.
Answer:
Methane reacts with chlorine in the presence of light which proceeds through the free radical chain mechanism. This mechanism is characterised by three steps: initiation, propagation and termination.

1. Chain initiation:
The chain is initiated by UV light, leading to homolytic fission of chlorine molecules into free radicals (chlorine atoms).

Here we choose Cl-Cl bond for fission because C- C and C-H bonds are stronger than Cl-Cl bond.

2. Chain propagation:
It proceeds as follows –
(a) Chlorine free radiais attack the methane molecule and breaks the C-H bond resulting in the generation of methyl free radicals.

(b) The methyl free radicals thus obtained attacks the second molecule of chlorine to give chloromethane (CH₃Cl) and a chlorine free radical as follows –

(c) This chlorine free radical then cycles back to step (a) and both steps (a) and (b) are repeated many times and thus a chain of reaction is set up.

3. Chain termination:
Aller sometime, the reaction stops due to the consumption of reactant and the chain is terminated by the combination of free radicals.

Question 4.
Write the mechanism for following reaction.

Answer:

The reaction proceeds through freee radical mechanism:

Step 1.
The weak O-O single bond linkage of peroxide undergoes homolytic cleavage to generate free radical.

Step 2.
The radicals abstracts a hydrogen atom from HBr thus generating bromine free radical.

Step 3.
The Bromine free radical adds lo the double bond in the way so as to form the more stable alkyl free radical,

Step 4.
Addition of HBr to secondary free radical.

Question 5.
Explain the acidic nature of alkynes.
Answer:
An alkyiic shows acidic nature only if it contains terminal hydrogen atom. This can be explained by considering the Sp hybrid orbitais of carbon atom in alkynes. The percentage of s-character of sp hybrid orbital (50%) is more than sp² hybrid orbital ofalkenes (33%) and sp³ hybrid orbital of alkanes (25%). Because of this, carbon atom becomes more electronegative, thus facilitating donation of H ions to bases. So hydrogen attached to triply bonded carbon atoms is acidic in nature.

Question 6.
Write the mechanism of chlorination of benzene.
Answer:
Step 1:
Generation of Cl^⊕ electrophile.
AlCl + Cl – Cl^⊕ + AlCl^Θ

Step 2:
Attack of the electrophile on the benzene ring to form arenium ion:

Step 3:
Rearomatisation of arenium ion:

Question 7.
Describe the mechanism of suiphonation of benzene.
Answer:
Step 1:
Generation of SO3 electrophile:
2H₂SO₄ → H₃O^⊕ + SO₃ + HSO₄^Θ

Step 2:
Attack of the electrophile on benzene ring to form arenium ion:

Step 3:
Rearomatisation of Areniurn ion:

Question 8.
Describe the mechanism of Freidel craft’s alkylation.
Answer:
Step 1:
Generation of ^⊕CH₃ electrophile:
AlCl₃ + CH₃Cl → ^⊕CH₃ + AlCl^Θ

Step 2 :
Attack of the electrophile on benzene ring to form areniurn ion:

Step 3:
Rearornatisation of arenium ion:

Question 9.
Write the mechanism of Freidel craft’s acylation.
Answer:
Step 1:
Generation of CH₃CO electrophile:

Step 2 :
Attack of the electrophile on benzene ring to form arenium ion.

Step 3 :
Rearomatisation of arenium ion:

Question 10.
An organic compound (A) of a molecular formula C₆H₆ which is a simple aromatic hydrocarbon. A undergoes hydrogenation to give a cyclic compound (B). A reacts with chlorine in the presence of UV-light to give C which is used as insecticide. Identify A, B and C. Explain the reactions with equation.
Answer:
1. Simple aromatic hydrocarbon, C₂H₆ is benzene.
2. Benzene (A) reacts with H₂ in the presence of Pt to give cyclohexane (B).

3. Benzene (A) reacts with Cl₂ in presence of UV-light to give benzene hexachioride (C).

Question 11.
What are ortho-para directors? Explain why -OH group is an ortho-para director and activator.
Answer;
The group which increases the electron density at ortho and para positions are called as ortho-para directors,
Example:
-OH. -NH₂ -NHR. -OCH₃, -CH₃, -C₂H₅ etc.

Let us consider the directive influence of phenolic group. Phenol is the resonance hybrid of the following structures:

In these resonance structures, the negative charge residue in the present at orilio and para positions of the ring structure. h is quite evident that the lone pair of electrons on the atom which is attached lo the ring is involved in resonance and it makes the ring more electron rich than benzene.

The electron density at oriho and para position increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at oriho and para positions. Hence 01-1 group is an orilio para director and activator.

Question 12.
An organic compound (A) of a molecular formula C₆H₆ is a simple aromatic hydrocarbon. A reacts with O₂ in the presence of VO₅ at 773 K to give B. A is further treated with sodium and liquid ammonia to give C which is a dienc compound. Identify A, B, and C and explain the reactions.
Answer:
1. A is benzene (C₆H₆), a simple aromatic hydrocarbon

2. Benzene (A) reacts with O₂ is the presence of V₂O₅ at 773K to give maleic anhydride (B).

3. Benzene (A) is treated with sodium and liquid ammonia to give 1.4 – cyclohexadiene (C)

Question 13.
What are inea drectors’? Explain with suitable example.
Answer:
The group which increases the electron density at mcta position are called as rneta directors.
For example:
-NO₂. -CN, -CHO, -COOH. -SO₃H etc.
Let us consider the directive influence of aldehydic (-CHO) group. Benzaldehyde is a resonence hybrid of the following structures.

In these resonance structures, the positive charge residue is present on the ring structure. It is quite evident that resonance deLocalises the positive charge on the atoms of the ring, making the ring less electron rich than henzene. Here overall electron density of benzene ring decreases due to -I effect of’ -CHO group, thereby deactivating the henzene for electrophilic atlack. However resonating structures shows that electron density is more at meto position as compared to ortho and para positions. Flence -CHO group is a ineta -director and a deactivator.

Question 14.
An organic compound (A) of molecular formula C₂H₄ which is a simple alkene reacts with Baeyer’s reagent to give B of molecular formula C₂H₆O₂ A again reacts with ozone followed by hydrolysis in the presence of zinc to form C of molecular formula CH₂O. Identify A, B and C. Explain with reactions.
Answer:
(i). A is CH₂ = CH₂ (Ethylene)

(ii). A (ethylene) reacts with Baeyer’s reagent (cold alkaline KMnO₄) to give ethylene glycol (ethane 1 ,2 diol).

Question 15.
An organic compound 1,1-dichioropropane reacts with aicholic KOH to give A of molecular formula C₃H₄. A reacts with mercuric sulphate and dil. H₂SO₄ at 333 K to give B. A on passing through red hot iron tube at 873 K will give C which is a cyclic compound. Identify A, B and C. Explain the reactions.
Answer:
1. 1,1-dichioropropane reacts with aicholic KOI-I to give propyne (A).

2. Propyne (A) reacts with mercuric sulphate and dil. H₂SO₄ at 333 K to give acetone (B).

3. Propyne (A) on passing through red hot iron tube at 873 K will give mesitylene (C).

Students can Download Accountancy Chapter 4 Ledger Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Accountancy Solutions Chapter 4 Ledger

Samacheer Kalvi 11th Accountancy Ledger Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the Correct Answer

Question 1.
Main objective of preparing ledger account is to ……………..
(a) Ascertain the financial position
(b) Ascertain the profit or loss
(c) Ascertain the profit or loss and the financial position
(d) Know the balance of each ledger account
Answer:
(d) Know the balance of each ledger account

Question 2.
The process of transferring the debit and credit items from journal to ledger accounts is called ……………..
(a) Casting
(b) Posting
(c) Journalising
(d) Balancing
Answer:
(b) Posting

Question 3.
J.F means ……………..
(a) Ledger page number
(b) Journal page number
(c) Voucher number
(d) Order number
Answer:
(b) Journal page number

Question 4.
The process of finding the net amount from the totals of debit and credit columns in a ledger is known as ……………..
(a) Casting
(b) Posting
(c) Journalising
(d) Balancing
Answer:
(d) Balancing

Question 5.
If the total of the debit side of an account exceeds the total of its credit side, it means
(a) Credit balance
(b) Debit balance
(c) Nil balance
(d) Debit and credit balance
Answer:
(b) Debit balance

Question 6.
The amount brought into the business by the proprietor should be credited to ……………..
(a) Cash account
(b) Drawings account
(c) Capital account
(d) Suspense account
Answer:
(c) Capital account

II. Very Short Answer Questions

Question 1.
What is a ledger?
Answer:
Ledger account is a summary statement of all the transactions relating to a person, asset, liability, expense or income which has taken place during a given period of time and it shows their net effect. From the transactions recorded in the journal, the ledger account is prepared. Ledger is known as principal book of accounts.

Question 2.
What is meant by posting?
Answer:
The process of transferring the debit and credit items from the journal to the ledger accounts is called posting.

Question 3.
What is debit balance?
Answer:
When the total of the debit side is more than the total of credit side the difference is debit balance and is placed on the credit side as ‘By Balance c/d’.

Question 4.
What is credit balance?
Answer:
If the credit side total is more than the total of debit side, the difference is credit balance and is placed on the debit side as ‘To Balance c/d’.

Question 5.
What is balancing of an account?
Answer:
Balancing means that the debit side and credit side amounts are totalled and the difference between the total of the two sides is placed in the amount column as ‘Balance c/d’ on the side having lesser total, so that the total of both debit and credit columns are equal.

III. Short Answer Questions

Question 1.
Distinguish between journal ad ledger.
Answer:
Following are the differences between journal and ledger:

Question 2.
What is ledger? Explain its utilities.
Answer:
1. Quick information about a particular account: Ledger account helps to get all information about a particular account like sales, purchases, machinery, etc., at a glance. For example, where there are several transactions with a debtor, the net amount due from a debtor can be known from the ledger account.

2. Control over business transactions: From the ledger balances extracted, a thorough analysis of account balances can be made which helps to have control over the business transactions.

3. Trial balance can be prepared: With the balances of ledger accounts, trial balance can be prepared to check the arithmetical accuracy of entries made in the journal and ledger.

4. Helps to prepare financial statements: From the ledger balances extracted, financial statements can be prepared for ascertaining net profit or loss and the financial position.

Question 3.
How is posting made from the journal to the ledger?
Answer:
The process of transferring the debit and credit items from the journal to the ledger accounts is called posting. The procedure of posting from journal to ledger is as follows:
1. Locate the ledger account that is debited in the journal entry. Open the respective account in the ledger, if already not opened. Write the name of the account in the top middle. If already opened, locate the account from the ledger index. Now entries are to be made on the debit side of the account.

2. Record the date of the transaction in the date column on the debit side of that account.

3. Record the name of the account credited in the journal with the prefix ‘To’ in particulars column.

4. Record the amount of the debit in the ‘amount column’.

5. Locate the ledger account that is credited in the journal entry. Open the respective account in the ledger, if already not opened. Write the name of the account in the top middle. If already opened, locate the account from the ledger index. Now entries are to be made on the credit side of the account. Record the date of the transaction in the date column. Record the name of the account debited in the journal entry in the particulars column with the prefix ‘By’ and write the amount in the amount column.

Question 4.
Explain the procedure for balancing a ledger account.
Answer:
Following is the procedure for balancing an account:
1. The debit and credit columns of an account are to be totalled separately.

2. The difference between the two totals is to be ascertained.

3. The difference is to be placed in the amount column of the side having lesser total. ‘Balance c/d’ is to be entered in the particulars column against the difference and in the date column the last day of the accounting period is entered.

4. Now both the debit and credit columns are to be totalled and the totals will be equal. The totals of both sides are to be recorded in the same line horizontally. The total is to be distinguished from other figures by drawing lines above and below the amount.

5. The difference has to be brought down to the opposite side below the total. ‘Balance b/d’ is to be entered in the particulars column against the difference brought down and in the date column, the first day of the next accounting period is entered.

6. If the total on the debit side of an account is higher, the balancing figure is debit balance and if the credit side of an account has higher total, the balancing figure is credit balance. If the two sides are equal, that account will show nil balance.

IV. Exercises

Question 1.
Journalise the following transactions and post them to ledger. (3 Marks)
Answer:

Journal

Ledger Cash Account

Capital Account

Bank Account

Purchase Account

Question 2.
Give journal entries for the following transactions and post them to ledger. (3 Marks)

Journal Entries

Ledger Cash Account

Somu Account

Sales Account

Furniture Account

Interest Received Account

Question 3.
Pass journal entries for the following transactions and post them to ledger. (3 Marks)

Answer:
Journal Entries

Ledger in the Books of Dharma
Cash Account

Capital Account

Ganesan Account

Rent Account

Received Commission Account

Question 4.
Record the following transactions in the journal of Banu and post them to the ledger.

Answer:
Journal Entries in the Book of Banu

Cash Account

Capital Account

Rent Received Account

Furniture Account

Question 5.
The following balances appeared in the books of Vinoth on Jan 1, 2018
Assets: Cash ₹ 40,000; Stock ₹ 50,000; Amount due from Ram ₹ 20,000;
Machinery ₹ 40,000 Liabilities: Amount due to Vijay ₹ 10,000
Pass the opening journal entry and post them to Vinoth’s Capital account. (2 Marks)
Answer:
Opening Entry in the Books of Vinoth

Vinoth’s Capital Account

Question 6.
Prepare Furniture A/c from the following transactions (2 Marks)

Answer:

Question 7.
The following balances appeared in the books of Kumaran on April 1, 2017. (5 Marks)
Assets: Cash ₹ 1,00,000; Stock ₹ 40,000; Amount due from Rohit ₹ 10,000,
Furniture ₹ 10,000; Liabilities: Amount due to Anush ₹ 40,000;
Kumaran’s capital ₹ 1,20,000
Find the capital and show the ledger posting for the above opening balances.
Answer:
Opening Entry

Cash Account

Stock Account

Rohit (Debtors) Account

Furniture Account

Creditors Account

Kumaran’s Capital Account

Question 8.
Give journal entries and post them to cash account. (3 Marks)

Answer:
Journal Entry

Cash Account

Question 9.
Give journal entries from the following transactions of Mohit, dealing in Textiles and post them to ledger: (3 Marks)

Answer:
Journal Entries in the Book of Mohit

Ledger Cash Account

Mohit’s Capital Account

SBI Account

Furniture Account

Question 10.
Give journal entries from the following transactions and post them to ledger: (3 Marks)

Answer:
Journal Entries

Ledger Cash Account

Purchase Account

Capital Account

Hema Account

Purchase Returns Account

Question 11.
Give journal entries from the following transactions and post them to Cash A/c and Sales A/c: (3 Marks)

Answer:
Journal Entries

Cash Account

Sales Account

Question 12.
Journalise the transactions given below and post them to ledger. (5 Marks)

Answer:
Journal Entries

Trade Expenses Account

Stationary (Postage Stamps) Account

Commission Received Account

Rent Account

Cash Account

Question 13.
Journalise the transactions given below and post them to ledger. (5 Marks)

Answer:
Journal Entries

Cash Account

Sales Account

Purchase Account

Kumar Account

Prabu Account

Salary Account

Question 14.
Enter the following transactions in the books of Ganesan and post them into ledger. (5 Marks)

Answer:
Journal Entries in the Books of Ganesan

Cash Account

Ganesan’s Capital Account

Bank Account

Furniture Account

Purchase Account

Sales Account

Vasu Account

Drawing Account

Question 15.
Journalise the following transactions in the books of Aran and post them to ledger accounts. (5 Marks)

Answer:
Journal Entries in the Books of Arun

Cash Account

Arun’s Capital Account

Purchase Account

Sales Account

Krishna Account

Govind Account

Question 16.
Journalise the following transactions and post them to ledger in the books of Raja. (5 Marks)

Answer:
Journal Entries in the Books of Raja

Cash Account

Sales Account

Murali Account

Purchase Account

Mani Account

Discount Allowed Account

Question 17.
Journalise the following transactions and post them to ledger. (5 Marks)

Answer:
Journal Entries in the Books of Raja

Cash Account

Capital Account

Purchase Account

Insurance Account

Machinery Account

Interest Received Account

Sales Account

Question 18.
Journalise the following transactions in the books of Vasu and post them to ledger accounts. (5 Marks)
2017 November:

1 – Cash in hand ₹ 1,00,000; Cash at bank: ₹ 30,000.
2 – Vasu sold goods to Jothi for ₹ 25,000 against a cheque and deposited the same in the bank.
4 – Received as commission ₹ 5,000.
8 – Bank paid ₹ 15,000 directly for insurance premium of Vasu.
15 – Cash deposited into bank ₹ 30,000.
18 – Cash withdrawn from bank for personal use ₹ 45,000.

Answer:
Journal entries in the Books of Vasu

Cash Account

Bank Account

Sales Account

Commission Received Account

Insurance Account

Drawings Account

Question 19.
Prepare Anand’s account from the following details. (3 Marks)

Answer:
Anand’s A/c

Question 20.
Prepare a Sales account from the following transactions

Sales Account
Answer:

Question 21.
Show the direct ledger postings for the following transactions: (5 Marks)
2017 June:

1 Raja commenced business with cash ₹ 50,000.
6 Sold goods for cash ₹ 8,000.
8 Sold goods to Devi on credit ₹ 9,000.
15 Goods purchased for cash ₹ 4,000.
20 Goods purchased from Shanthi on credit ₹ 5,000.

Answer:
Cash Account

Capital Account

Sales Account

Devi Account

Purchase Account

Shanthi Account

Question 22.
Show the direct ledger postings for the following transactions: (5 Marks)
2017 July:

1 – Shankar commenced business with cash ₹ 1,00,000.
6 – Sold goods for cash ₹ 10,000.
9 – Wages paid ₹ 6,000.
19 – Salaries paid ₹ 8,000.
20 – Advertisement expenses paid ₹ 4,000.

Answer:
Cash Account

Shankar’s Capital Account

Sales Account

Wages Account

Salaries Account

Advertisement Account

Textbook Case Study Solved

Imagine you have been called for an interview with an auditor. The auditor shows you the following ledger account of Mr. Raheem, a dealer in food products.
Sita A/c

The auditor wants you to explain each posting in the above account and also to state where will the double entry for each posting be found.
Here Sita is a debtor of Mr. Reheem. So in Raheem ledger Sita A/c shows the balances.

1. April 1. The opening balance of Sita A/c ₹ 7,000
2. April 12. She deposited ₹ 5,000
3. May 2 Cash received from Sita ₹ 2,000
4. Sept. 25 Credit sales for Sita ₹ 10,000
5. Oct. 3 Goods returned by Sita ₹ 1,000
6. Nov. 17 Cash paid by Sita ₹ 3,000
7. Dec. 21 Sita deposited ₹ 4,000
8. Dec. 29 Credit sales to Sita ₹ 10,000
9. Dec. 31 Closing debtors ₹ 12,000

So from the Sita ledger the owner of the business Raheem can find out the closing balance of debtor (Sita) ₹ 12,000.

Samacheer Kalvi 11th Accountancy Ledger Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
………………. is known as principal book of accounts.
(a) Journal
(b) Ledger
(c) Trial balance
(d) Transaction
Answer:
(b) Ledger

Question 2.
………………. accounts show the values of assets.
(a) Real
(b) Personal
(c) Nominal
(d) Journal
Answer:
(a) Real

Question 3.
………………. accounts give the net amount due to creditor and the net amount due from debtors.
(a) Real
(b) Personal
(c) Nominal
(d) Ledger
Answer:
(b) Personal

Question 4.
Net position of an account cannot be ascertained from ……………….
(a) Journal
(b) Ledger
(c) Trial balance
(d) Balance sheet
Answer:
(a) Journal

Question 5.
Net position of an account can be ascertained from ……………….
(a) Journal
(b) Ledger
(c) Trial balance
(d) Balance sheet
Answer:
(b) Ledger

Question 6.
The term balance brought down is used in the name of ……………….
(a) balance b/d
(b) balanced c/d
(c) debit balance
(d) credit balance
Answer:
(a) balance b/d

Question 7.
When a journal entry has more than one debit or more than one credit or both, it is called ……………….
(a) Compound entry
(b) Single entry
(c) Journal entry
(d) Ledger entry
Answer:
(a) Compound entry

Question 8.
Total of credit > Total of debit =
(a) Debit balance
(b) Credit balance
(c) Nil balance
(d) Trial balance
Answer:
(b) Credit balance

Question 9.
Total of debit > Total of credit =
(a) Debit balance
(b) Credit balance
(c) Nil balance
(d) Trial balance
Answer:
(a) Debit balance