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Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 3 Vegetative Morphology Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 3 Vegetative Morphology

11th Bio Botany Guide Vegetative Morphology Text Book Back Questions and Answers

Part-I

I. Choose the Right Answer:

Question 1.
Roots are
a. Descending-negatively geostrophic positively phototrophic
b. Descending-positively geostrophic negatively phototrophic
c. Ascending, positively geostrophic negatively phototrophic
d. Ascending, negatively geostrophic negatively phototrophic
Answer:
b. Descending-positively geostrophic negatively phototrophic

Question 2.
When the root is thick and fleshy but does not take a definite shape is said to be
a. Nodulose root
b.Tuberous root
c. Moniliform root
d. Fasciculated root
Answer:
b. Tuberous root

Question 3.
Example for negatively geotrophic roots
a. Ipomoea, Dahlia
b. Asparagus, Ruellia
c. Vitis, Portulaca
d. Avicennia, Rhizophora
Answer:
d. Avicenniarhizophora

Question 4.
Cureumaamada curcuma domestica Asparagus maranta are examples of
a. Tuberous root
b. Beaded root
c. Moniliform root
d. Nodulose root
Answer:
d. Nodulose root

Question 5.
Bryophyllum and Dioscorea are examples for
a. Foliar bud, apical bud
b.Foliar bud, cauline bud
c. Cauline bud, apical bud
d.Cauline bud, fohar bud
Answer:
b. Foliar bud, cauline bud

Two Marks

Question 6.
Why lateral roots are endogenous?
Answer:
Lateral roots arise from the pericycle, part Eg. Inner part – so it is known as endogenous in origin.

Question 7.
Write the similarities and differences between

1. Avicennia & trapa
2. Banyan & silk cotton
3. Fusiform and Napiform root

Answer:
I. Avicennia & trapa

II. Banyan & silk cotton

III. Fusiform and Napiform root

Question 8.
How root climbers differ from stem climbers?
Answer:

Question 9.
Compare sympodial branching with monopodial branching.
Answer:

Question 10.
Compare pinnate unicostate and palmate multicoastate venation?
Answer:

Part – II

11th Bio Botany Guide Vegetative Morphology Additional Important Questions and Answers

Choose the Correct Answer:

Question 1.
The study about external features of an organism is known as …………… .
(a) morphology
(b) anatomy
(c) physiology
(d) taxonomy
Answer:
(a) morphology

Question 2.
Onion lettuce, fennel, radish, cabbage are examples of
a. perennial
b. annual.
c. centennial
d. biennial
Answer:
d.biennial

Question 3.
The branch of science that deals with the classification of organisms is called as …………… .
(a) taxonomy
(b) morphology
(c) physiology
(d) anatomy
Answer:
(a) taxonomy

Question 4.
Palmately reticulate, convergent venation is seen in
a. zizipus
b. mango
c. cucurbita
d. carica papaya
Answer:
a. zizipus

Question 5.
Rolling or folding of individual leaves may be as follows
a. pteryix
b. ptyxis
c. typxis
d. xyptes
Answer:
b. ptyxis

Question 6.
The general form of a plant is referred to as …………… .
(a) habitat
(b) structure
(c) habit
(d) shape and size
Answer:
(c) habit

Question 7.
These are examples for shrubs
a. coconut and Palmyra
b. mango and bamboo
c. hibiscus and castor
d. cotton and bougainvillea
Answer:
c. Hibiscus and Castor

Question 8.
Angiosperms are also known as
a. Bryophytes
b. pteridophytes
c. Magnoliophytes
d. Tracheophytes
Answer:
c.Magnoliophytes

Question 9.
Climbers are also called as …………… .
(a) herbs
(b) trees
(c) vines
(d) shrubs
Answer:
(c) vines

Question 10.
The phyllotaxy seen in Nerium is known as
a. whorled
b. opposite
c. alternate
d. ternate
Answer:
d.Ternate

Question 11.
…………… is an example for xerophytes.
(a) Lichens
(b) Euphorbia
(c) Ficus
(d) Ipomoea
Answer:
(b) Euphorbia

II. FILL UP THE BLANKS

Question 1.

Answer:
a. Heterophylly
b. isobilateral leaf
c. citrus
d. acalypha

III. Identify the Diagram

Question 1.
Identify The Diagram and Label ABCD
Answer:

Question 2.
IDENTIFY THE DIAGRAM and Label ABCD from the diagram
Answer:

Question 3.
Identify the Diagram and Label ABCD from the diagram
Answer:

IV. Read the following Assertion and Reason Find the correct answer

Question 1.
Assertion: Rootstock laek root cap and root hairs but they possess terminal but which is a characteristic of stem
Reason: Rootstocks also known as underground stem
a. Assertion and Reason are correct reason explaining stem
b. Assertion and Reason are correct but the reason is not explaining assertion
c. Assertion is true, but Reason wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct -Reason is explaining assertion

Question 2.
Assertion: Avieennia develop special kinds of the root (negatively-geotropic) known as respiratory roots
Reason: They are mangrove plants
a. Assertion and Reason are correct Reason is explaining assertion
b. Assertion and Reason are correct but, the reason is not explaining assertion
c. Assertion is true, but Reason is wrong
d. Assertion is true, but Reason is not explaining assertion
Answer:
Assertion and reason are correct, but the reason is not explaining assertion.

V. Find out the Wrong answer

Question 1.
Buttress roots are not traced in
a. Terminalia arjuna
b. Delonixregia
c. Bombax spp
d. Piper betel
Answer:
d. Piper betel

Question 2.
Among the given which one doesn’t have foliar roots
a. Bryophyllum
b. Begnonia
c. Zamiaculeas
d. Ranunculus
Answer:
d. Ranunculus

Question 3.
Among the given, Find out the odd man with reference to the fibrous root system.
a. Eleusineeoracana
b. Pennisetumamericanum
c. Zingiferaoffieinalis
d. Ficusbenahaliensis
Answer:
d. Ficusbenahaliensis

VI. Form the match and Find the Wrong Pair

Question 1.
(1) Tendril as stem modification – Passiflora
(2) Tendril as leaf modification – Lathyrus
(3) Tendril as stipule modification- Smilax
(4) Tendril as a modification of petiole of the leaf – Nepenthes
Answer:
(4) Tendril as a modification of petiole of the leaf – Nepenthes

Question 2.
(1) Zingifereffienalis – Rhizome
(2) Eolehicum – Eorm
(3) Allium cepia – Tunicatedbulle
(4) Tuber – Tulipa.spp
Answer:
(4) Tuber – Tulipa .spp

Question 3.
(1) Leaf base – Hypopodium
(2) petiole – Mesopodium
(3) Midrib – Endopodium
(4) Lamenia – Epipodium
Answer:
(3) Midrib – Endopodium

Question 4.
(1) Flattened – Cynodon dactyla
(2) Cylindrical cladode – Asparagus
(3) Flattened phylloclade – Opuntia
(4) Cylindrical – Euphorbia antiquorum
Answer:
(3) Flattened phylloclade- Opuntia

Question 5.
(1) Additional outgrowth between leafe base & lamina – Ligute
(2) Sheathing leaf base – Mesopodium
(3) Stiples occur in – Fabaceae
(4) Stiples absent in – Monocots
Answer:

VII. Match And Find The Correct Answer

Question 1.
(1) Unipennate – Eaesalpinia A
(2) Bipinnate – Eoriandrumsativum B
(3) Tripinnate – Neem C
(4) Decompound – Moringa D

Answer:
b)C-A-D-B

Question 6.
Tabulate the Aerial Stem modification
Answer:

Question 7.
Draw the structure of prop roots
Answer:

Question 8.
Differentiate between Excurrent and decurrent types of stem.
Answer:

Question 9.
Differentiate between Runner and Sucker:
Answer:

Question 10.
Differentiate between Ternate and Whorled Phyilotaxy
Answer:

Question 11.
What is plant morphology?
Answer:
Plant morphology is also known as external morphology deals with the study of shape, size, and structure of plants and their parts like (roots, stems, leaves, flowers, fruits, and seeds).

Question 12.
Name any 2 brace roots and write down their botanical name
Answer:

1. Sugarcane – Saccharum officinarum
2. Maize- Zea mays

Question 13.
Draw the Regions of the root tip and label the parts
Answer:

Question 14.
What is meant by ‘Eye’ of a potato?
Answer:
The axillary bud ensheathes by the scale appears as eye-like on the potato surface each and every eye can develop into a potato plant.
‘S’ Scale Leaf Auxiliary bud

Question 15.
Draw the structure of a typical leaf and label the parts
Answer:

Give Short Answer

Question 1.
The morphological study is important in Taxonomy. Why?
Answer:
Morphological features are important in determining the productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.

Question 2.
Classify leaves on the basis of duration
Answer:

1. Cauducuous (fagaceous)- falling off soon after formation – Opuntia
2. Deciduous – Falling at the end of the growing season (winter& summer-leaf less)- Erythrina indica
3. Evergreen- persistent throughout the year tree never remain leafless Mimusops
4. Marcescent- no falling-but withering on the plants – Fagaceae

Question 3.
Classify compound leaf types
Answer:

Question 4.
Give the diagrammatic representation of leaf modifications
Answer:

Question 5.
Give a brief account on the tap root system.
Answer:
Primary root is the direct prolongation of the radicle. When the primary root persists and continues to grow as in dicotyledons, it forms the main root of the plant and is called the tap root. Tap root produces lateral roots that further branch into finer roots. Lateral roots along with their branches together called secondary roots.

Question 6.
Classify venation
Answer:

Question 7.
Notes on Heterophylly.
Answer:
Definition:
Morphologically 2 different kinds of leaves in the same plant is called heterophylly.
Types-2

1. Structural
2. Developmental

1. Structural – In Limnophyllaheterophylla, aquatic plant half of its plant body is submerged and half is above water level. Here aerial leaves are normal & the submerged leaves are highly dissected.

2. Developmental – In Sterculiavillosa Varying structure during different developmental stages- Young leaves – lobed or dissected Mature leaves – entire

Question 8.
How the leaf hooks helps the Bignonia plant?
Answer:
In cat’s nail (Bignonia unguiscati) an elegant climber, the terminal leaflets become modified into three, very sharp, stiff, and curved hooks, very much like the nails of a cat. These hooks cling to the bark of a tree and act as organs of support for climbing.

Phyllode- It is a winged leaf petiole or stalk or rachis Eg. Nepenthus – modified to perform the function as leaf
Acacia auriculiformis Leaf- petiole modification to do photosynthesis

Essay Question – Five marks

Question 1.
Classify terrestrial habitats
Answer:

Question 2.
Classify aquatic habitat.
Answer:

Question 3.
Draw the structure of a typical plant and neatly label the parts
Answer:

Question 4.
What are the various types of root modification
Answer:

Question 5.
Give a clear-cut distinction of horks Spines & prickles.
Answer:

Question 6.
List out various types of Phyllotaxy.
Answer:

Question 7.
Compare & Contrast pitcher from bladderwort.
Answer:

Question 8.
Define Ptyxis or Vernation list out the various types
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 2 Plant Kingdom
Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 2 Plant Kingdom

11th Bio Botany Guide Plant Kingdom Text Book Back Questions and Answers

Part-I

Choose the Right Answer:

Question 1.
Which of the plant group has gametophyte as a dominant phase ?
a. Pteridophytes
b. Bryophytes
c. Gymnosperm
d. Angiosperm
Answer:
b. Bryophytes

Question 2.
Which of the following represent gametophytic generation in Pteridophytes?
a. Prothallus
b. Thallus
c. Cone
d. Rhizophore
Answer:
a. Prothallus

Question 3.
The haploid number of chromosome for an Angiosperm is 14, the number of chromosome in its endosperm would be
a. 7
b. 14
c. 42
d. 28
Answer:
c. 42

Question 4.
Endosperm is gymnosperm is formed
a. At the time of fertilization
b. Before fertilization
c. After fertilization
d. Along with the development of embryo
Answer:
b. Before fertilization

Question 5.
Differentiate Haplontic and Diplontic life cycle.
Answer:

Question 6.
What is plectostele – Give example.
Answer:
Plectostele: Xylem plates alternates with phloem plates. Example: Lycopodium clavatum.

Question 7.
What do we infer from the term Pucnoxylic?
Answer:

• Secondary growth is also traced in gymnosperms, E.g. Cycas and Pinus.
• The wood may be compact with narrow medullary ray this condition known as Pycnoxlic seen in Pinus.
• It is opposite to Manoxylic condition which is seen in Cycas.

Question 8.
Mention two characters shared by Gymnosperms and Angiosperms.
Answer:

Question 9.
Do you think shape of chloroplast is unique for algae. Justify your answer.
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup-shaped (chlamydomonas). Discoid ((Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 10.
Do you agree with the statement that Bryophytes need water for fertilization? Justify your answer.
Answer:
Being amphibians of plant kingdom they are simplest land inhabiting cryptogams and restricted to moist shady habitats. They need water for the male gametes to reach Archegonia to effect fertilization. So water is needed for completing their fertilization and their life cycle.

Part – II

11th Bio Botany Guide Plant Kingdom Additional Important Questions and Answers

I. Choose the correct option.

Question 1.
Gametophytic phase is …………… .
(a) triploid
(b) tetraploid
(c) haploid
(d) diploid
Answer:
(c) haploid

Question 2.
The numbers of known species of Angiosperms in the world is
a. 268600
b. 286600
c. 224400
d. 274832
Answer:
a. 268600

Question 3.
Which algae leads an endozoic life in Hydra?
(a) Chlorella
(b) Gracilaria
(c) Ulothrix
(d) Chlamydomonas
Answer:
(a) Chlorella

Question 4.
The protein bodies found in chromatophores & assist in the synthesis and storage of starch is
a. Leucoplasts
b. Floridean starch
c. Pyrenoids
d. Mannitol
Answer:
c. Pyrenoids

Question 5.
Postelia palmaeformis is commonly known as
a. Sea kelp
b. Sea shell
c. Sea palm
d. Sea worth
Answer:
c. Sea palm

Question 6.
In Chara, thallus is encrusted with …………… .
(a) calcium carbonate
(b) hydrogen sulphate
(c) silica
(d) ammonium carbonate
Answer:
(a) calcium carbonate

Question 7.
Gemmae formation is not traced in which three of the given four options
a. Marchanlia
b. Riella
c. Ricciocarpus
d. Anthoceros
(i) ab & c
(ii) be & d
(iii) ab & d
(iv) ac & d
Answer:
(ii) be & d

Question 8.
Find out the aquatic bryophytes of the following.
a. Riella
b. Ricciocarpus
c. Riccia
d. Bryopteris
(i) a&c
(ii) b&c
(iii) c&d
(iv) a&b
Answer:
(iv) a&b

Question 9.
…………… are thin-walled non – motile spores.
(a) Zoospores
(b) Akinetes
(c) Aplanospores
(d) Genunae
Answer:
(c) Aplanospores

Question 10.
which one of the following is a terrestrial chiorophycea
a. Chara
b. Zygnema
c. Trentipohlia
d. Ulva
Answer:
c. Trentipohlia

Question 11.
Thick walled spores meant for perrennation are known as
a. Aplanospores
b. Akinetes
c. Endospores
d. Zoospores
Answer:
b. Akinetes

Question 12.
The photosynthetic part of the Phaeophyceae thallus Is called as
(a) holdfast
(b) stipes
(c) lamina
(d) fronds
Answer:
(d) fronds

Question 13.
The female sex organ of red algae is known as
a. Archegonium
b. Spermatogonium
c. Carpogonium
d. Oogonium
Answer:
c. Carpogonium

Question 14.
Endosperm is triploid and haploid in
a. Pteridophyta & Gymnosperm
b. Angiosperm & Gymnosperm
c. Gymnosperm & monocot
d. Gymnosperm & dicot
Answer:
b. Angiosperm & Gymnosperm

Question 15.
Gelidium belongs to …………… members.
(a) Rhodophyccae
(b) Phaeophyceae
(c) Cyanophyccae
(d) Dinophyceae
Answer:
(a) Rhodophyceae

Question 16.
……………….. is known as leather leaf fern
a. Marsilea
b. Azolla
c. Selaginella
d. Rumohra adiantiformis
Answer:
d. Rumohra adiantiformis

Question 17.
Stele includes
a. Xylem, phloem & cambium
b. Xylem, phloem & medulla
c. Xylem, phloem & pericycle
d. Xylem, phloem, pericycle & medulla
Answer:
d. Xylem, phloem, pericycle & medulla

Question 18.
Marchantia vegetatively propagates by …………… .
(a) tubers
(b) gemmae
(c) buds
(d) brood bodies
Answer:
(b) gemmae

Question 19.
The organ in bryophytes that help to attach the thallus to the substratum is
a. Hold fast
b. Rhizoids
c. Rhizopore
d.Roots
Answer:
a. Holdfast

Question 20.
Coralloid roots of cycas have a symbiotic association with …………….
(a) Blue-green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(a) Blue-green algae

II. Match the following & find the correct answer.

Question 1.
(i) Spores look similar to parental cell – Zoospore (A)
(ii) Thick walled aplanospores – Autospore (B)
(iii) Thin walled non-motile spores – Hypnospore (C)
(iv) Thin walled motile spores – Aplanospore (D)

Answer:
(b) B-C-D-A

Question 2.
(i) Helianthusannum – Amphiboloicstele (A)
(ii) Lycopodium serratum – Eustele(B)
(iii) Zeamays – Actinostele (C)
(iv) Adiantumpedatum – Atactostele (D)

Answer:
(a) B-C-D-A

Question 3.
(i) Fossil bryophyte – Lepidodendron, Williamson
(ii) Fossil Algae – Calamites Baragwanthia
(iii) Fossil pteridophyte – Naiadita, Hepaticites
(iv) Fossil Gymnosperm – Palaeoporella,Dimorphosiphon

Answer:
(d) C-D-B-A

Question 4.
(i) Abies balsamea – Drug for cancer treatment (A)
(ii)Taxus brevifolia – Wood for making door, boat & railway sleepers (B)
(iii) Cedrus deodara – Treatment for asthama & bronchitis(C)
(iv) Ephedra gerardiana – Slide mounting medium(D)

Answer:
(c) D-A-B-C

Question 5.
Find the chromosome number of the following by choosing the correct option.
(i) Embryo ofblyophyta
(ii) Embryo ofAngiosperm
(iii) Endosperm ofAngiosperm
(iv) Sporophyte ofpteridophyta

Answer:
(a) (I) n (II) 2n (III) 3n (IV) 2n

III. Choose the wrong statement.

Question 1.
The following statement is not applicable to which one of the following options
The sporophyte is dominant, photosynthetic, and independent. The gametophytic phase is represented by a single to few celled gametophyte
a. Fucus
b. Mango
c. Pinus
d. Marchantia
Answer:
d. Marchantia

Question 2.
One of the following is not a Marine Algae.
a. Gracilaria
b. Sargassum
c. Oedogonium
d. Cladophora
Answer:
c. Oedogonium

Question 3.
Which one of the following is not a Vascular cryptogam
a. Lycopodium
b. Anthoceros
c. Equisetum
d. Selaginella
Answer:
b. Anthoceros

Question 4.
Which one among the given four doesn’t belong to Chlorophyceae
a. Chiorella
b. Volvox
c. Chara
d. Sargassum
Answer:
d. Sargassum

Question 5.
Pollination is not entomophilous in
a. Hibiscus
b. Mangifera
c. Chrysanthemum
d. Cycas
Answer :
d. Cycas

Question 6.
The following one is not a monoecious plant
a. Pinus
b. Cycas
c. Alnus
d. Ginkgo
Answer:
b. Cycas

Question 7.
Which one of the following is not the correct statement regarding Algae?
a. The study of Algae is known as Phycology
b. A wide range of thallus organization is found in Algae
c. Algae are eukaryotic except Blue Green Algae
d. They are the simplest plant group with root stem and leaves
Answer:
d. They are the simplest plant group with root stem and leaves

IV. Find out the true or false statements from the following and on that basis find the correct answer.

Question 1.
(i) Chara thallus is encrusted with calcium carbonate
(ii) Siliceous wall occurs in the cell wall of Diatom
(iii) Soil inhabiting algae – Fritshchiella
(iv) Cladophora crispate is growing now

Answer:
c. (I) True (II) True (III) False (IV) True

Question 2.
(i)  Prothallus develop into a sporophyte
(ii)  Algae growing on snow is known as cryptophytes
(iii)  The common name of Postelia Palmaeformis is known as sea palm.
(iv)  Endosperm is triploid in pinus.

Answer:
a. (I) False (II) True (III) True (IV) False

Question 3.
(i)  Apogamy and Apospory is common in pteridophytes
(ii) Spore bearing leaves in pteridophytes are known as a sorus
(iii) Branches of limited growth and branches of unlimited growth are seen in gymnosperm
(iv) Cambium occur in gymnosperm as in dicots

Answer:
d. (I) True (II) False (III) False (IV) True

Question 4.
(i) Fungi play important role in soil conservation
(ii) Vascular cryptogams were predominant in the paleozoic era
(iii) Gymnosperms were dominant in the early cretaceous period.
(iv) Angiosperms appeared during the Jurassic period

Answer:
b (I) False (II) True (III) False (IV) False

Question 5.
(i) Polyembryony is traced in Pteridophyta
(ii) Vessels are present in Gnetum and ephedra
(iii) Heterosporus condition is seen in Lycopodium
(iv) Corolloid root occur in Cycas

Answer:
a. (I) False (II) True (III) False (IV) True

Question 6.
Which one of the following is the correct statement regarding Phaeophyta
a.  It is commonly known as Red Algae
b. The plant body has fronds, stipe & hold fast
c. The reserve material is Floridian starch
d. Sexual reproduction is isogamous
Answer:
b. The plant body has fronds, stipe & holdfast

Question 7.
Choose the right statement regarding leaves of Pteridiumsp.
a. It is used as food
b. Green dye is derived from it.
c. It is used as bio-fertilizer
d. It is an ornamental foliage plant
Answer:
b.Green dye is derived from it

Question 8.
Which one of the following is a correct statement regarding Bryophyta
a. Mostly terrestrial plants so water is not essential for reproduction
b. The gametophyte is dominant but the sporophyte is independent
c. They have well-developed xylem and phloem tissues
d. They are the simplest land inhabiting cryptogams lacking vascular tissues.
Answer:
d.They are the simplest land inhabiting cryptogams lacking vascular tissues

Question 9.
Choose the correct statement regarding Gymnosperm
a. The spores are generally homosporous
b. The leaves are dimorphic, foliage and scale leaves are present
c. The stem is aerial erect and unbranched in conifers
d. Xylem mostly consists of only vessels.
Answer:
b. The leaves are dimorphic, foliage and scale leaves are present.

Question 10.
Choose the correct statement regarding the common characters of Gymnosperm and Angiosperm only
a. Pollen tube help in the transfer of male nucleus & fertilization is Siphonogamous
b. Heterospory is of common occurrence
c. Vessels are the chief water-conducting elements
d. Pollination is by Anemophilous method only
Answer:
a. Pollen tubes help in the transfer of the male nucleus & fertilization is Siphonogamous.

Question 11.
Look at the picture and find out the correct answer.
a. Gemmae of Marchantia
b. Thallus of Riccia
c. The gametophyte of Anthoceos
d. The tubers of Anthoceos

Answer:
d.The tubers of Anthoceos

Question 12.
Look at the picture and find out the correct answer
a. The Asexual reproduction in Ultrix by the formation of zoospores
b. Akinetes formation in Pithophora
c. Scalariform conjugation in Zygnema
d. Zygospore formation in spirogyra

Answer:
c. Scalariform conjugation in zygoma.

V.

Question 1.
Which one of the following is a wrong pair?
a. National wood fossil park – Thuruvakkarai
b. Shiwalik fossil park – Arunachala Pradesh
c. Mandal fossil park – Madhya Pradesh
d. Raj mahal hill – Jharkhand
Answer:
b. Shiwalik fossil park – Arunachala Pradesh

Question 2.
Which one of the following is a wrong pair?
a. Halophytic Algae – Dunaliella salina
b. Epiphytic Algae – Rhodymenia
c. Endophytic algae – Cladophora crisp ala
d. Endozoic Algae – Chlorella
Answer:
c. Endophytic Algae – Cladophora crisp ala

Question 3.
Which one of the following is a wrong pair?
a. Father of Indian Bryology- Prof. Shiv. Ram Kashyap
b. Father of Indian Phycology – F.E. Fritsch
c. The classification of gymnosperm – Sporn
d. The father of Indian Paleobotany – Prof. Birbal Sahni
Answer:
b.Father of Indian Phycology F.E.Fritsch

VI. Read the following assertion & reason. Find the correct answer.

Question 1.
Assertion: Chlorophyceae is known as green algae.
Reason: These plants have chlorophyll & chlorophyll as their major photosynthetic pigments.
(a) Assertion and Reason are correct. The reason is explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and Reason are correct. The reason is explaining Assertion.

Question 2.
Assertion: In Bryophytes haploid gametophyte, alternate with diploid sporophyte phase.
Reason: Bryophytes lack vascular tissue and hence called non-vascular cryptogams.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion

Question 3.
Assertion: Gnetum has flowers and it also has vessels as conducting elements like angiosperm.
Reason: Gnetum is a primitive gymnosperm.
(a) Assertion and Reason are correct, Reason explaining assertion.
(b) Assertion and Reason are correct, but Reason not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(c) Assertion is true, but Reason is wrong.

Question 4.
Assertion: The endosperm is haploid and develops before fertilization in a gymnosperm.
Reason: The endosperm is triploid and develops after fertilization in angiosperm.
(a) Assertion and Reason are correct. The reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.

Question 5.
Assertion: The embryogeny is endoscopic in Bryophytes.
Reason: The first division of the zygote is transverse & the apex of the embryo develops from the outer cell.
(a) Assertion and Reason are correct, Reason explaining Assertion.
(b) Assertion and Reason are correct, but Reason is not explaining Assertion.
(c) Assertion is true, but Reason is wrong.
(d) Assertion is true, but Reason is not explaining Assertion.
Answer:
(a) Assertion and reason are correct, Reason explaining Assertion.

VII. In the following diagram what are the parts (A) (B) (C) (D) representing?

Question 1.
(a) (A) Epidermis (B) Xylem
(C) Phloem (D) Cambium
(b) (A) Epidermis (B) Phloem
(C) Cambium (D) Xylem
(c) (A) Epidermis (B) Cambium
(C) Xylem (D) Phloem
(d) (A) Phloem (B) Xylem
(C) Cambium (D) Epidermis

Ans: (b) (A) Epidermis-(B) Phloem-(C) Cambium-(D) Xylem

Additional Questions – 2 Marks

Question 1.
Define alternation of generation.
Answer:
Alternation of the haploid gametophytic phase (n) with diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 2.
Chlorella – structure label the parts.
Answer:

Label (1) nucleus
(2) cup-shaped chloroplast

Question 3.
Name any two freshwater algae.
Answer:
Two freshwater algae:

1. Oedogonium and
2. Ulothrix

Question 4.
What is the use of Diatomaceous earth?
Answer:
Diatomaceous earth is got from the siliceous frustules (Diatom). It belongs to bacillario phyta. The transparent cell walls of a diatom are made up of hydrated silica. Generally known as Diatomaceous earth.
Use:

• It is used in water filters as an insultation material.
• Reinforcing agent in concrete and rubber.

Question 5.
What is the use of chlorella in sewage treatment?
Answer:

• Chlorella, Scenedesmus, Chlamydomonas are used in the sewage treatment plants. For their photosynthetic activity, they utilize the carbon dioxide from sewage and release oxygen.
• The aerobic bacteria, by utilising this oxygen degrade & decompose organic materials in the sewage. Thus play a vital role in sewage treatment plants. (STPs)

Question 6.
Identify and label the given diagram.

Answer:
The given figure represents Marchantia

1. Apical notch
2. Sex organs
3. Gametophyte of thallus
4. Rhizoids
5. Gemma

Question 7.
Differentiate between Sorus, Sporangia, Sporophyll.
Answer:

Question 8.
Bryophytes are amphibians of plant kingdom – Justify.
Answer:
Bryophytes are called ‘amphibians of plant kingdom’ because they need water for completing their life cycle.

Question 9.
Label the given diagram.

Answer:
The given diagram is the sporophyte of Funaria parts

1. Calyptra
2. Capsule
3. Leaves
4. Rhizoids

Question 10.
What are the aspects that helped Pteridophtes to evolve into terrestrial habitats?
Answer:

• Pteridophytes are the first to acquire vascular tissues, xylem and phloem, and a well-developed root system. So known as vascular cryptogam.
• This aspect had helped pteridophytes to evolve into terrestrial habitats.

Question 11.
Assign few plants in the group Pteridophytes.
Answer:
Type Botanical name
Club moss – Lycopodium
Horsetails – Equisetum
Quill worts – Isoetes
Water ferns – Salviniales
Tree ferns – Dryopteris

Question 12.
Write down Reimer’s classification of Pteridophytes.
Answer:
5 subdivisions – 19 orders – 48 families

1. Psilophytopsida
2. Psilotopsida
3. Lycopsida
4. Sphenopsida
5. Pteropsida

Question 13.
What Is amber? Which group of plants produce amber?
Answer:
Amber is a plant secretion that is an efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Question 14.
What is meant by Siphonostele?
Answer:
In siphonostele, the xylem is surrounded by phloem with pith in the centre- eg. Marsilea. It includes ectophloic, siphonoslele, Amphiphloic-siphonoslele, soleonslete, eustele, Atactostele, and polycyclic stele thus altogether 6 types.

Question 15.
What is Amber? Where have you noticed it?
Answer:

• In Spielberg’s Jurassic park-movie (A mosquito embedded in a transparent substance called amber is
  mentioned)
• Amber is thus a plant secretion, used as an efficient preservative, that doesn’t get degraded.
• Pinites succinifera a Gymnosperm plant produces Amber.

Question 16.
What is the word Jurassic – denote?
Answer:

• ‘Jurassic’ is a specific period of the dinosaurs. It comes under the Mesozoic era.
• Gymnosperms were also dominant in that period.

3 Marks

Question 1.
Give the widely accepted outline classification for plants.
Answer:

Question 2.
Give the total number of plant groups in the world and India.
Answer:

Question 3.
Name the 3 types of life cycles seen in plants?
Answer:
The 3 types of life cycles seen in the plant:

1. Haplontic life cycle
2. Diplontic life cycle
3. Haplodiplontic life cycle

Question 4.
More than half of the total productivity of the world is done by Marine Algae -Justify.
Answer:

• Yes two-third of earths surface is covered by oceans and seas.
• Of this the photosynthetic plants called algae are the major primary producers Nearly 1/2 of total productivity of the world is done by marine Algae. All other marine organisms depend upon them for the very existence.

Question 5.
Classify Algae according to F.E.Fritsch.
Answer:
F.E. Fritsch in his the structure and reproduction of Algae (1935)-classified Algae into 11 classes.

1. Chlorophyceae
2. Xanthophyceae
3. Chrysophyceae
4.  Bacillariophyceae
5. Cryptophyceae
6. Dinophyceae
7. Chloromondineae
8. Euglenophyceae
9. Phaeophyceae
10. Rhodophyceae
11. Cyanophyceae

Question 6.
Write about Reproduction in Chlorophyceae.
Answer:

Question 7.
Write down the economic importance of Bryophyte?
Answer:

Question 8.
List down the economic importance of Pteridophytes.
Answer:

Question 9.
Name the three classes of Bryophytes, according to Proskauer.
Answer:
Three Classes of Bryophytes, According to Proskauer:

1. Hepaticopsida
2. Anthocerotopsida and
3. Bryopsida.

Question 10.
List down the salient features of Angiosperm.
Answer:

• Vascular system – Xylem and Phloem well developed
• Flowers are produced (instead of cones)
• Ovules (embryosac and seeds) – remain enclosed in ovary/fruit Pollen tube -Help in fertilization water no necessary
• Double fertilization & triple fusion present is one of the unique features.

Question 11.
Differentiate between Dicotyledons & Monocotyledons.
Answer:

5 Marks

Question 1.
Classify Algae on the basis of their habitats.
Answer:

Question 2.
Give an account of General Characteristics of Algae?
Answer:
Criteria I

Criteria II

Criteria III

Question 3.
Give an account of Phaeophyceae.
Answer:

Question 4.
Give an account of Rhodophyceae (red algae) criteria.
Answer:

Question 5.
Economic importance of Algae.
Answer:

Question 6.
General characteristic features of Bryophytes.
Answer:

Question 7.
Write down differences between Gymnosperm & Angiosperm.
Answer:

Question 8.
Write down the Economic Importance of Gymnosperm.
Answer:

Question 9.
Give an account of Fossil plants.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Physics Guide Pdf Chapter 8 Heat and Thermodynamics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 8 Heat and Thermodynamics

11th Physics Guide Heat and Thermodynamics Book Back Questions and Answers

Part – I:
I. Multiple choice questions:

Question 1.
In hot summer after a bath, the body’s:
(a) internal energy decreases
(b) internal energy increases
(c) heat decreases
(d) no change in internal energy and heat
Answer:
(a) internal energy decreases

Question 2.
The graph between volume and temperature in Charle’s law is:
(a) an ellipse
(b) a circle
(c) a straight line
(d) a parabola
Answer:
(c) a straight line

Question 3.
When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:
(a) isothermal
(b) adiabatic
(c) isobaric
(d) isochoric
Answer:
(b) adiabatic

Question 4.
An ideal gas passes from one equilibrium state (P₁, V₁, T₁, N) to another equilibrium state (2P₁, 3V₁, T₂, N). Then:
(a) T₁ = T₂
(b) T₁ = \(\frac{\mathrm{T}_{2}}{6}\)
(c) T₁ = 6T₂
(d) T₁ = 3T₂
Answer:
(b) T₁ = \(\frac{\mathrm{T}_{2}}{6}\)

Question 5.
When a uniform rod is heated, which of the following quantity of the rod will increase:
(a) mass
(b) weight
(c) centre of mass
(d) moment of inertia
Answer:
(d) moment of inertia

Hint:
M.I. = MK²
K – radius of gyration.
During heating K would be increased. Hence moment of inertia will increase.

Question 6.
When food is cooked in a vessel by keeping the lid closed, after some time the steam pushes the lid outward. By considering the steam as a thermodynamic system, then in the cooking process:
(a) Q > 0, W > 0
(b) Q < 0, W > 0
(c) Q > 0, W < 0
(d) Q < 0, W < 0
Answer:
(a) Q > 0, W > 0

Hint:
Q > 0; W > 0
∆Q = ∆U₁ + ∆W.
∴ ∆Q ∝ ∆W
When ∆Q > 0
∆W > 0.

Question 7.
When you exercise in the morning, by considering your body as a thermodynamic system, which of the following is true?
(a) ∆U > 0, W > 0
(b) ∆U < 0, W > 0
(c) ∆U< 0, W < 0
(d) ∆U = 0, W > 0
Answer:
(b) ∆U < 0, W > 0

Hint:
∆Q = ∆U + ∆W
When exercise is performed, internal energy is utilised (∆U < 0). So it will decrease.
But as internal energy is utilised, the exercise (∆W) will be more. So work will be more.
∴ ∆W > 0
∆U < 0 W > 0

Question 8.
A hot cup of coffee is kept on the table. After some time it attains a thermal equilibrium with the surroundings. By considering the air molecules in the room as a thermodynamic system, which of the following is true?
(a) ∆U > 0, Q = 0
(b) ∆U > 0, W < 0
(c) ∆U > 0, Q > 0
(d) ∆U = 0, Q > 0
Answer:
(c) ∆U > 0, Q > 0

Hint: During the thermal equilibrium with surrounding as heat energy (Q) is increased, internal energy will be increased.
∴ ∆U > 0 Q > 0

Question 9.
An ideal gas is taken from (P_i, V_i) to (P_f, V_f) in three different ways. Identify the process in which the work done on the gas the most.

(a) Process A
(b) Process B
(c) Process C
(d) Equal work is done in Process A, B & C
Answer:
(b) Process B

Hint:
When work is done on the gas, compression takes place. Final pressure P_f will be increased and remains constant. It is shown by process B.

Question 10.
The V-T diagram of an ideal gas which goes through a reversible cycle A → B → is shown below. (Processes D → A and B → C are adiabatic)

The corresponding PV diagram for the process is (all figures are schematic)

Answer:

Question 11.
A distant star emits radiation with maximum intensity at 350 nm. The temperature of the star is:
(a) 8280 K
(b) 5000 K
(c) 7260 K
(d) 9044 K
Answer:
(a) 8280 K

Hint:

Question 12.
Identify the state variables given here.
(a) Q, T, W
(b) P,T,U
(c) Q, W
(d) P,T,Q
Answer:
(b) P,T,U

Question 13.
In an isochoric process, we have:
(a) W = 0
(b) Q = 0
(c) ∆U = 0
(d) ∆T = 0
Answer:
(a) W = 0

Hint:
Work done in an isochoric process is zero.

Question 14.
The efficiency of a heat engine working between the freezing point and boiling point of water is: (NEET2018)
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Answer:
(b) 20%

Hint:
T₂ = 0° C = 0 + 273 = 273 K
T₁ = 100° C = 100 + 273 = 373 K

Question 15.
An ideal refrigerator has a freezer at a temperature of -12°C. The coefficient of performance of the engine is 5. The temperature of the air (to which the heat ejected) is:
(a) 50°C
(b) 45.2°C
(c) 40.2°C
(d) 37.5°C
Answer:
(c) 40.2°C

Hint:

I. Short Answer Questions:

Question 1.
‘An object contains more heat’- is it a right statement? If not why?
Answer:
When heated, an object receives heat from the agency. Now object has more internal energy than before. Heat is the energy in transit and which flows from an object at a higher temperature to an object at lower temperature. Heat is not a quantity. So the statement I would prefer “an object contains more thermal energy”.

Question 2.
Obtain an ideal gas law from Boyle’s and Charles’law.
Answer:
According to Boyle’s law, Pressure ∝ \(\frac { 1 }{ 2 }\)
i.e., P ∝ \(\frac { 1 }{ V }\)
According to Charle’s law,
Volume ∝ Temperature, i.e., V ∝ T
By combining these two laws, we get
PV= CT … (1)
Where C is a positive constant
But C = k x Number of particles (N)
C = kN
Where k – Boltzmann’s constant.
∴ Equation (1) becomes
PV = NkT

Question 3.
Define one mole.
Answer:
One mole of any substance is the amount of that substance which contains Avogadro number (NA) of particles (such as atoms or molecules).

Question 4.
Define specific heat capacity and give its unit.
Answer:
Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or 1 °C.
The SI unit for specific heat capacity is J kg^-1 K^-1.

Question 5.
Define molar specific heat capacity.
Answer:
Heat energy required to increase the temperature of one mole of substance by IK or 1°C.

Question 6.
What is a thermal expansion?
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Question 7.
Give the expressions for linear, area and volume thermal expansions.
Answer:
(i) Linear expansion:
\(\frac { ∆L }{ L }\) = α ∆T ⇒ α_L = \(\frac { ∆L }{ L∆T }\)

(ii) Area expansion:
\(\frac { ∆A }{ A }\) = 2α ∆T ⇒ α_A = \(\frac { ∆A }{ A∆T }\)

(iii) Volume expansion:
\(\frac { ∆V }{ V }\) = 3α ∆T ⇒ α_v = \(\frac { ∆V }{ V∆T }\)

Question 8.
Define latent heat capacity. Give its unit.
Answer:
Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material. The SI unit for latent heat capacity is J kg^-1.

Question 9.
State Stefan-Boltzmann law.
Answer:
The total amount of heat energy radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature.

Question 10.
What is Wien’s law?
Answer:
The wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.

Question 11.
Define thermal conductivity. Give its unit.
Answer:
The quantity of heat transferred through a unit length of a material in a direction normal to Unit surface area due to a unit temperature difference under steady-state conditions is known as the thermal conductivity of a material.

Question 12.
What is a black body?
Answer:
A black body is one which neither reflects nor transmits but absorbs whole of the radiation incident on it.

Question 13.
What is a thermodynamic system? Give examples.
Answer:
Thermodynamic system: A thermodynamic system is a finite part of the universe. It is a collection of a large number of particles (atoms and molecules) specified by certain parameters called pressure (P), Volume (V), and Temperature (T). The remaining part of the universe is called the surrounding. Both are separated by a boundary.
Examples: A thermodynamic system can be liquid, solid, gas, and radiation.

Question 14.
What are the different types of thermodynamic systems?
Answer:
(i) Open system can exchange both matter and energy with the environment.

(ii) Closed system exchange energy but not matter with the environment.

(iii) Isolated system can exchange neither energy nor matter with the environment.

Question 15.
What is meant by ‘thermal equilibrium’?
Answer:
Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

Question 16.
What is mean by state variable? Give example.
Answer:
The quantities that are used to describe the equilibrium states of a Thermodynamic system. .
Example: Pressure, Volume, Temperature.

Question 17.
What are intensive and extensive variables? Give examples.
Answer:
Extensive variable depends on the size or mass of the system.
Example: Volume, total mass, entropy, internal energy, heat capacity etc.
Intensive variables do not depend on the size or mass of the system.
Example: Temperature, pressure, specific heat capacity, density etc.

Question 18.
What is an equation of state? Give an example.
Answer:
The equation that relates the state variables in a specific manner is called equation of state.
Examples:
(i) Gas equation
PV = NkT

(ii) Vander Waal’s equation, .
(p +\(\frac{a}{v^{2}}\))(v – b) = RT

Question 19.
State Zeroth law of thermodynamics,
Answer:
The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Question 20.
Define the internal energy of the system.
Answer:
The sum of kinetic and potential energies of all the molecules of the system with respect to the centre of mass of the system.

Question 21.
Are internal energy and heat energy the same? Explain.
Answer:
Internal energy and thermal energy do not mean the same thing, but they are related. Internal energy is the energy stored in a body. It increases when the temperature of the body rises, or when the body changes from solid to liquid or from liquid to gas.
“Heat is the energy transferred from one body to another as a result of a temperature difference.”

Question 22.
Define one calorie.
Answer:
To raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat energy was measured in calorie.
1 cal = 4.186J

Question 23.
Did joule converted mechanical energy to heat energy? Explain.
Answer:
In Joule’s experiment, when the masses fall, the paddle wheel turns. The frictional force between paddle wheel and water causes a rise in temperature of water. It is due to the work done by the masses. This infers that gravitational potential energy is transformed into internal energy of water. Thus Joule converted mechanical energy into heat energy.

Question 24.
State the first law of thermodynamics.
Answer:
‘Change in internal energy (AU) of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surroundings’.

Question 25.
Can we measure the temperature of the object by touching it?
Answer:
No, we can’t measure the temperature of the object touching it. Because the temperature is the degree of hotness or coolness of a body. Only we can sense the hotness or coolness of the object.

Question 26.
Give the sign convention for Q and W.
Answer:

Question 27.
Define the quasi-static process.
Answer:
A quasi-static process is an infinitely slow process in which the system changes its variables (P, V, T) so slowly such that it remains in thermal, mechanical, and chemical equilibrium with its surroundings throughout.

Question 28.
Give the expression for work done by the gas.
Answer:
In general, the work done by the gas by increasing the volume from V_i to V_f is given by
W = \(\int_{V_{i}}^{V_{f}} \mathrm{P} d \mathrm{~V}\)

Question 29.
What is PV diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 30.
Explain why the specific heat capacity at constant pressure is greater than the specific heat capacity at constant volume.
Answer:
When increasing the temperature of a gas at constant pressure, gas expands and consume some heat to do work. It is not happened, while increasing the temperature of a gas at constant volume. Hence less heat energy is required to increase the temperature of the gas at constant volume. Hence CP is always greater than Cv.

Question 31.
State the equation of state for an isothermal process.
Answer:
The equation of state for isothermal process is given by,
PV = constant

Question 32.
Give an expression for work done in an isothermal process.
Answer:
W = μRTln \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\)

Question 33.
Express the change in internal energy in terms of molar specific heat capacity.
Answer:
When the gas is heated at the constant volume the temperature increases by dT. As no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU. dU = µωdT

Question 34.
Apply first law for (i) an isothermal (ii) adiabatic (Hi) isobaric processes.
Answer:
(i) For an isothermal process temperature is maintained as constant. Hence,
∆V= 0
∴ W = P∆V = 0
According to first law of thermodynamics
∆U = Q – W
= Q – 0
= Q
∴ ∆U = Q

(ii) For adiabatic process, Q = 0
By applying first law of thermodynamics
∆U = Q – W
∆U = |- W| = W
∆U = W = \(\frac{\mu R}{\gamma-1}\) = (T_i – T_f)

(iii) (a) For isobaric expansion Q > O
According to first law of thermodynamics
∆U = – Q – W
∆U = Q – P∆V

(b) For isobaric compression Q < O.
According to first law of thermodynamics
∆U = – Q – W
= – Q – P∆V

Question 35.
Give the equation of state for an adiabatic process.
Answer:
The equation of state for an adiabatic proc ess is given by,
PV_γ = Constant

Question 36.
State an equation state for an isochoric process.
Answer:
The equation of state for an isochoric process is given by,
P = (\(\frac { μR }{ V }\))T

Question 37.
if the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? ff not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 38.
Draw the PV diagram for,
(a) isothermal process
(b) Adiabatic process
(c) lsobaric process
(d) lsochoric process
Answer:
(a) isothermal process:

(b) Adiabatic process:

(c) lsobaric process:

(d) lsochoric process:

Question 39.
What is a cyclic process?
Answer:
It is a process in which the thermodynamic system returns to its initial state after undergoing a series of changes.

Question 40.
What is meant by a reversible and irreversible processes?
Answer:
Reversible processes: A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
Irreversible processes: All natural processes are irreversible. Irreversible process cannot be plotted in a PV diagram, because these processes cannot have unique values of pressure, temperature at every stage of the process.

Question 41.
State Clausius form of the second law of thermodynamics.
Answer:
“Heat energy always flows from hotter object to colder object spontaneously”. This is known as the Clausius form of second law of thermodynamics.

Question 42.
State Kelvin-Planck statement of second law of thermodynamics.
Answer:
It is impossible to construct a heat engine that operates in a cycle, whose sole effect is to convert the heat completely into work.

Question 43.
Define heat engine.
Answer:
Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

Question 44.
What are processes involves in a Carnot engine?
Answer:

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression.

Question 45.
Can the given heat energy be completely converted to work in a cyclic process? If not, when can the heat can completely converted to work?
Answer:
No.
According to second law it can’t be completely converted. It is not possible to convert heat completely converted into work, in a cyclic process. But, in Non-cyclic process, heat can be completely converted into work.

Question 46.
State the second law of thermodynamics in terms of entropy.
Answer:
“For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural processes should occur.

Question 47.
Why does heat flow from a hot object to a cold object?
Answer:
Because entropy increases when heat flows from hot object to cold object. If heat were to flow from a cold to a hot object, entropy will be decreased. So second law thermodynamics is violated.

Question 48.
Define the coefficient of performance.
Answer:
The ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = ß = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{W}}\).

III. Long Answer Questions:

Question 1.
Explain the meaning of heat and work with suitable examples.
Answer:
The spontaneous flow of energy from the object at higher temperature to the one at lower temperature. When they are contact. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating.

Example: ‘A hot cup of coffee has more heat’. This statement is correct or wrong?

When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and that flows from a body at higher temperature to an other body at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat’ is wrong, instead ‘coffee is hot’ will be appropriate.

Meaning of Work: If a body undergoes a displacement then only work is said to be done by the body. Work is also the transfer of energy by other means such as moving a piston of a cylinder containing gas.

Example: When you rub your hands against each other the temperature of the hand is increased. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now when you place your hands on the chin, the temperature of the chin increases.

This is because the hands are at higher temperature than the chin. In the above example, the temperature of hands is increased due to work and temperature of the chin is increased due to heat transfer from the hands to the chin.

The temperature in the system will increase and sometimes may not by doing work on the system. Like heat, work is also not a quantity and through the work energy is transferred to the system.

Question 2.
Discuss the ideal gas laws.
Answer:
(i) Boyles law: When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume.
P ∝ \(\frac { 1 }{ v }\)

(ii) Charles law: When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. V ∝ T

(iii) By combining these two equations we have PV= CT … (1)

(iv) Let the constant C as k times the number of particles N. Here k is the Boltzmann constant and it is found to be a universal constant. So the ideal gas law can be stated as follows,
PV = NkT … (2)
If the gas consists of p mole of particles then total number of particles is
N = μN_A … (3)
Where N_A is Avogadro’s number.
Substituting the value of N in the equation (2) we get
PV = μN_AkT
Here N_Ak = R
So, the ideal gas law can be written for μ mole of gas is
PV= pRT
It is called equation of state for an ideal gas.

Question 3.
Explain in detail the thermal expansion.
Answer:
Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature. When a solid is heated, its atoms vibrate with higher amplitude about their fixed points. At that time, relative change in the size of solids is small.

Liquids expand more than solids because, they have less intermolecular forces than solids. In the case of gas molecules, the intermolecular forces are almost negligible and so they expand much more than solids.

Linear Expansion: In solids, for a small change in temperature ∆T, the fractional change in length \(\frac { ∆L }{ L }\) is directly proportional to ∆T.
\(\frac { ∆L }{ L }\) = α_L∆T
Therefore, α_L = \(\frac { ∆L }{ L∆T }\)

Area Expansion: For an infinitesimal change in temperature ∆T the fractional change in area of a substance is directly proportional to ∆T and it can be written as
\(\frac { ∆A }{ A }\) = α_A∆T
Therefore, α_A = \(\frac { ∆A }{ A∆T }\)
Volume Expansion: For a very small change in temperature ∆T the fractional change in volume \(\frac { ∆V }{ V }\) of a substance is directly proportional to AT.
\(\frac { ∆V }{ V }\) = α_V∆T
Therefore, α_V = \(\frac { ∆L }{ V∆T }\)

Question 4.
Describe the anomalous expansion of water. How is it helpful in our lives?
Answer:
On heating liquids expand and contract on cooling at moderate temperatures. But water exhibits an anomalous behavior. It contracts on heating between 0°C and 4°C. The volume of the given amount of water decreases as . it is cooled from room temperature, until it reach 4°C . Below 4°C the volume increases and so the density decreases. It means that the water has a maximum density at 4°C. This behaviour of water is called anomalous expansion of water.

During winter in cold countries, the surface of the lakes would be at lower temperature than the bottom. Since the solid water (ice) has lower density than its liquid form,’below 4°C, the frozen water will be on the top surface above the liquid water (ice floats). It is due to the anomalous expansion of water. The water in lakes and ponds freeze only at the top. So, the species living in the lakes will be safe at the bottom.

Question 5.
Explain Calorimetry and derive an expression for final temperature when two thermodynamic systems are mixed.
Answer:
Calorimetry means the measurement of the amount of heat released or absorbed by thermodynamic system during the heating process. When a body at higher temperature is brought in contact with another body at lower temperature, due to transformation of heat. The heat lost by the hot body is equal to the heat gained by the cold body. No heat is allowed to escape to the surroundings. It can be mathematically expressed as
Q_gain = – Q_lost
Q_gain + Q_lost = 0
Heat gained or lost can be measured with a calorimeter.
When a sample is heated at high temperature (T₁) and immersed into water at room temperature (T₂) in the calorimeter. After some time both sample and water reach a final equilibrium temperature T_f. Since the calorimeter is insulated, heat given by the hot sample is equal to heat gained by the water.
Q_gain = – Q_lost
As per the sign convention, the heat energy lost is denoted by negative sign and heat gained is denoted as positive.
From the definition of specific heat capacity,
Q_gain= m₂s₂(T_f – T₂)
Q_lost = m₁s₁(T_f – T₁)
Here s₁ and s₂ specific heat capacity of hot sample and water respectively.
So we can write,

Question 6.
Discuss various modes of heat transfer.
Answer:
There are three modes of heat transfer: Conduction, Convection and Radiation.

Conduction: Conduction is the process of direct transfer of heat through matter due to temperature difference. If two objects are placed in direct contact with one another, then heat will be transferred from the hotter object to the colder one.

Convection: Convection is the process in which heat energy transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.
Example: Boiling of water in a pot.

Radiation: Radiation is a form of energy transfer from one body to another by electromagnetic waves.
Example: Solar energy from the Sun.

Question 7.
Explain in detail Newton’s law of cooling.
Answer:
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac { dQ }{ dt }\) ∝ – (T – T_s) … (1)
The negative sign implies that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T = Temperature of the surrounding

From the above graph, it is clear that the rate of cooling is high initially and decreases with falling temperature.

Consider a body of mass m and specific heat capacity s at temperature T. Let T be the temperature of the surroundings. If the temperature falls by a small amount dl in time dt, then the amount of heat lost is,
dQ = msdT
Dividing both sides of equation (2) by dt
\(\frac { dQ }{ dt }\) ∝ \(\frac { msdT }{ dt }\) … (3)
From Newton’s law of cooling
\(\frac { dQ }{ dt }\) ∝ (T – T_s)
\(\frac { dQ }{ dt }\) ∝ (T – T_s) … (4)
Where a is some positive constant.
From equation (3) and (4)
– a(T – T_s) = ms\(\frac { dT }{ dt }\)
\(\frac{d \mathrm{~T}}{\mathrm{~T}-\mathrm{T}_{s}}\) = – \(\frac { a }{ ms }\)dt … (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides, we get
T = T_s + b₂ \(e^{-\frac{a}{m s} t}\)
here b₂ = \(e^{b_{1}}\) = constant.

Question 8.
Explain Wien’s law and why our eyes are sensitive only to visible rays?
Answer:
Wien’s law states that, ‘the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body’.
λ_m ∝ \(\frac { 1 }{ T }\)
(or)
λ_m = \(\frac { b }{ T }\) … (1)
It is implied that if temperature of the body increases, maximal intensity wavelength (λ_m) shifts towards lower wavelength (higher frequency) of electromagnetic spectrum.

Wien’s law and Vision: Why our eye is sensitive to only visible wavelength (in the range 400 nm to 700nm)?

The Sun is approximately considered as a black body. Since any object above 0 K will emit radiation, Sun also emits radiation. Its surface temperature is about 5700 K. By substituting this value in the equation (1),

It is the wavelength at which maximum intensity is 508nm. Since the Sun’s temperature is around 5700K, the spectrum of radiations emitted by Sun lie between 400 nm to 700 nm which is the visible part of the spectrum.

The humans evolved under the Sun by receiving its radiations. The human eye is sensitive only in the visible.

Question 9.
Discuss the,
(i) Thermal equilibrium
(ii) Mechanical equilibrium
(iii) Chemical equilibrium
(iv) Thermodynamic equilibrium.
Answer:
(i) Thermal equilibrium: Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, which will not change with time.

(ii) Mechanical equilibrium: A system is said to be in mechanical equilibrium if no unbalanced force acts on the thermodynamic system or on the surrounding by thermodynamic system.

(iii) Chemical equilibrium: If there is no net chemical reaction between two thermodynamic systems in contact with each other then it is said to be in chemical equilibrium.

(iv) Thermodynamic equilibrium: If two systems are set to be in thermodynamic equilibrium, then the systems are at thermal, mechanical and chemical equilibrium with each other. In a state of thermodynamic equilibrium the macroscopic variables such as pressure, volume and temperature will have fixed values and do not change with time.

Question 10.
Explain Joule’s Experiment of the mechanical equivalent of heat.
Answer:
Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were tied with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy which is equal to 2mgh. When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel.

This causes a rise in temperature of the water. This confirms that gravitational potential energy is converted to internal energy of water. The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat energy. He found that to raise lg of an object by 1°C , 4.186J of energy is required. In earlier days the heat was measured in calorie.
1 cal = 4.186 J

This is called Joule’s mechanical equivalent of heat.

Question 11.
Derive the expression for the work done in a volume change in a thermodynamic system.
Answer:
Let us consider a gas contained in the cylinder fitted with a movable piston. When the gas is expanded quasi-statically by pushing the piston by a small distance dx as shown in Figure. Because the expansion occurs quasi- statically. The pressure, temperature and internal energy will have unique values at every instant.
The small work done by the gas on the piston
dW = F dx … (1)

The force exerted by the gas on the piston F – PA.
A – area of the piston and
P – pressure exerted by the gas on the piston.
Equation (1) can be rewritten as
dW = PA dx … (2)
But A dx = dV change in volume during this expansion process.
Hence the small work done by the gas during the expansion is given by
dW = dV … (3)
It is noted that is positive since the volume is increased. In general the work done by the gas by increasing the volume from V_i to V_f is given by
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathbf{P} d \mathrm{~V}\)
Suppose if the work is done on the system, then V_i > V_f.
Then, W is negative.

Question 12.
Derive Meyer’s relation for an ideal gas.
Answer:
Let us consider μ mole of an ideal gas in a container with volume V, pressure P and temperature T. When the gas is heated at constant volume the temperature increases by dT. Since no work is done by the gas, the heat that flows into the system will increase only the internal energy. Let the change in internal energy be dU.
If C_v is the molar specific heat capacity at constant volume,.
dU= μC_vdT … (1)
When the gas is heated at constant pressure so that the temperature increases by dT. If ‘Q’ is the heat supplied in this process and ‘dV’ the change in volume of the gas.
Q = μC_pdT … (2)
If W is the workdone by the gas in this process, then
W = P dV … (3)
But from the first law of thermodynamics,
Q = dU + W … (4)
Substituting equations (1), (2) and (3) in (4), we get,
μC_pdT = μC_vdT + P dV … (5)
For mole of ideal gas, the equation of state is given by,
PV = μRT
⇒ PdV + VdP = μRdT … (6)
Since the pressure is constant, dP = 0
∴ C_pdT = C_vdT + RdT
∴ C_p = C_v + R
(or) C_p – C_v = R … (7)
This relation is called Meyer’s relation.

Question 13.
Explain in detail the isothermal process.
Answer:
It is a process in which both the pressure and volume of a thermodynamic system will change at constant temperature. The ideal gas equation is
PV = μRT
Here, T is constant for this process.
So the equation of state for isothermal process is given by,
PV= constant … (1)
It is implied that if the gas goes from one equilibrium state (P₁,V₁) to another equilibrium state (P₂,V₂) the following relation holds for this process
P₁,V₁ = P₂,V₂ … (2)
Since PV = constant, P is inversely proportional to V(P ∝ \(\frac { 1 }{ V }\)). This implies that PV graph is a hyperbola. The pressure-volume graph for constant temperature is also called isotherm.
The following figure shows the PV diagram for quasi-static isothermal expansion and quasi-static isothermal compression. For an isothermal process since temperature is constant, the internal energy is also constant. It implies that dU or ∆U – 0.

For an isothermal process, the first law of thermodynamics can be written as follows,
Q = W … (3)
From equation (3), it implies that the heat supplied to a gas is used to do only external work. It is a common misconception that when there is flow of heat energy to the system, the temperature will increase. For isothermal process it is not true. When the piston of the cylinder is pushed, the isothermal compression takes place. This will increase the internal energy which will flow out of the system through thermal contact.

Question 14.
Derive the work done in an isothermal process.
Answer:
Let us consider an ideal gas which is allowed to expand quasi-statically at constant temperature from initial state (P_i,V_i) to the final state (P_f, V_f. Let us calculate the work done by the gas during this process.
gas, W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
As the process occurs quasi-statically, at every stage the gas is at equilibrium with the surroundings. Since it is in equilibrium at every stage the ideal gas law is valid.
P = \(\frac { μRT }{ V }\) … (2)
Substituting equation (2) and (1) we get,

In equation (3), μRT is taken out of the integral, since it is constant throughout the isothermal process.
By performing the integration in equation (3), we get
W = μRTln\(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) … (4)
Since we have an isothermal expansion,
\(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) > 1, so \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) > 0. As a result the work done by the gas during an isothermal expansion is positive.
Equation (4) is true for isothermal compression also. But in an isothermal compression \(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\) < 1. So \(\left(\frac{\mathrm{V}_{f}}{\mathrm{~V}_{i}}\right)\) < 0.
Hence the work done on the gas in an isothermal compression is negative. In the PV diagram the work done during the isothermal expansion is equal to the area under the graph as shown in figure.

Similarly for an isothermal compression, the area under the PV graph is equal to the work done on the gas that turns out to be the area with a negative sign.

Question 15.
Explain in detail an adiabatic process.
Answer:
It is a process in which no heat energy flows into or out of the system (Q = 0). But the gas can expand by spending its internal energy or gas can be compressed through some external work. Hence the pressure, volume and temperature of the system may change in an adiabatic process.

From the first law for an adiabatic process, ∆U = W. This implies that the work is done by the gas at the expense of internal energy or work is done on the system that increases its internal energy. The adiabatic process can be achieved by the following methods

(i) Thermally insulating the system from surroundings. So that no heat energy flows into or out of the system. For instance, when thermally insulated cylinder of gas is compressed (adiabatic compression) or expanded (adiabatic expansion) as shown in the Figure.

(ii) If the process occurs so quickly so that there is no time to exchange heat with surroundings even though there is no thermal insulation

The equation (1). Implies that if the gas goes from an equilibrium state (P_i, V_i) to another equilibrium state (P_f,V_f) adiabatically then it satisfies the relation,
PV^γ = constant … (1)
The equation (1) Here γ is called adiabatic exponent ( γ = \(\frac{C_{\mathrm{P}}}{C_{\mathrm{V}}}\)) which depends on the nature of the gas.
The equation (1). Implies that if the gas goes from an equilibrium state (P_i,V_i) to another equilibrium state (P_f, V_f) adiabatically then it satisfies the relation,
P_iV_i^γ = P_fV_f^γ … (2)
The PV diagram of an adiabatic expansion and adiabatic compression process.

The PV diagram for an adiabatic process is also called adiabatic. It is noted that the PV diagram for isothermal and adiabatic processes look similar. But actually the adiabatic curve is steeper than isothermal curse.
Let us rewrite the equation (1) in terms of T and V. From ideal gas equation, the pressure P = \(\frac { μRT }{ V }\). Substituting this equation in the equation (1), we have
\(\frac { μRT }{ V }\)V^γ = constant
(or) \(\frac { T }{ V }\)V^γ = \(\frac { constant }{ μR }\)
Note here that is another constant. So it can
be written as
TV^γ-1 = constant … (3)
From the equation (3) it is known that if the gas goes from an initial equilibrium state (T_i,V_i) to final equilibrium state (T_f, V_f) adiabatically then it satisfies the relation
\(\mathrm{T}_{i} \mathrm{~V}_{i}^{\gamma-1}=\mathrm{T}_{f} \mathrm{~V}_{f}^{\gamma-1}\) … (4)
The equation of state for adiabatic process can also be written in terms of T arid P as
\(\mathrm{T}^{\gamma} \mathrm{P}^{1-\gamma}\) = constant

Question 16.
Derive the work done in an adiabatic process.
Answer:
Let us consider p moles of an ideal gas enclosed in a cylinder having perfectly non conducting walls and base. A frictionless and insulating piston A is fitted in the cylinder as shown in Figure. It’s area of cross-section be A.

Let W be the work done when the system goes from the initial state (P_iV_iT_i) to the final state (P_fV_fT_f) adiabatically.
W = \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\) … (1)
It is assumed that the adiabatic process occurs quasi-statically, at every stage the ideal gas law is valid. Under this condition, the adiabatic equation of state is PV^γ = constant (or) P = \(\frac{\text { constant }}{\mathrm{V}^{\gamma}}\) can be substituted in the equation (1), we get

From ideal gas law,

In adiabatic expansion, work is done by the gas. i.e., W_adia is positive. As T_i > T_f the gas cools during adiabatic expansion.

In adiabatic compression, work is done on the gas. i.e., W_adia is negative. As T_i < T_f hence the temperature of the gas increases during adiabatic compression.

Question 17.
Explain the isobaric process and derive the work done in this process.
Answer:
Isobaric process is a thermodynamic process that occurs at constant pressure. Inspite of the pressure consistency of this process, temperature, volume and internal energy are not constant. From the ideal gas equation, we have
V = (\(\frac { μR }{ P }\))T … (1)
Here, \(\frac { μR }{ P }\) = constant
In an isobaric process the temperature is directly proportional to volume.
V ∝ T(Isobaric process) … (2)
It is implied that for a isobaric process, the V-T graph is a straight line passing through the origin.
Work done by the gas
W= \(\int_{V_{i}}^{\mathrm{V}_{f}} \mathrm{P} d \mathrm{~V}\)
In an isobaric process, the pressure is constant, So P comes out of the integral,

∆V – change in the volume. If AV is negative, W is also negative. It is implied that the work is done on the gas. If ∆V is positive, W is also positive, implying that work is done by the gas. From the ideal gas equation.
equation (4) can be rewritten as
PV = μRT and V = \(\frac { μRT }{ P }\)
Substituting this in equation (4) we get,
W = μRT_f(1 – \(\frac{\mathrm{T}_{i}}{\mathrm{~T}_{f}}\)) … (5)
In the PV diagram, area under the isobaric curve is equal to the work done in isobaric process. The shaded area in the following Figure is equal to the work done by the gas.

The first law of thermodynamics for isobaric process is given by,
∆U = Q – P∆V

Question 18.
Explain in detail the isochoric process.
Answer:
Isochoric process is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables. The pressure-volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.

The equation of state for an isochoric process is given by,
P = \(\frac { μR }{ V }\) … (1)
Here \(\frac { μR }{ V }\) = constant
∴ P ∝ T
It is inferred that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin. If a gas goes from state (P_i, T_i) to (P_f, T_f) at constant volume, then the system satisfies the following equation,
\(\frac{P_{i}}{T_{i}}=\frac{P_{f}}{T_{f}}\)
For an isochoric processes, AV=0 and W-0. Then the first law of thermodynamics becomes
∆U = Q … (3)
It is implied that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.

Question 19.
What are the limitations of the first law of thermodynamics?
Answer:
The first law of thermodynamics explains well the inter convertibility of heat and work. But is not indicated the direction of change.
Example:
(i) When a hot object is in contact with a cold object, heat always flows from the hot object to cold object but not in the reverse direction. According to first law of thermodynamics, it is possible for the energy to flow from hot object to cold object and viceversa. But in nature the direction of heat flow is always from higher temperature to lower temperature.

(ii) When brakes are applied, a car stops due to friction and the work done against friction is converted into heat. But this heat is not reconverted into the kinetic energy of the car. Hence the first law is not sufficient to explain many of natural phenomena.

Question 20.
Explain the heat engine and obtain its efficiency.
Answer:
Heat engine is a device which takes heat as input and converts this heat into work by undergoing a cyclic process.
A heat engine has three parts:
(i) Hot reservoir
(ii) Working substance
(iii) Cold reservoir
A Schematic diagram for heat engine is given below in the figure.

• Hot reservoir (or) Source: It supplies heat to the engine. Which is always maintained at a high temperature T_H.
• Working substance: It is a substance like gas or water, that converts the heat supplied into work.
• Cold reservoir (or) Sink: The heat engine ejects some amount of heat (Q_L) into cold reservoir after it doing work. It is always maintained at a low temperature T_L.

The heat engine works in a cyclic process. After a cyclic process it returns to the same state. Since the heat engine returns to the same state after it ejects heat, the change in the internal energy of the heat engine is zero.

The efficiency of the heat engine is defined as the ratio of the work done (output) to the heat absorbed (input) in one cyclic process. Let the working substance absorb heat Q_H units from the source and reject Q_L units to the sink after doing work W units.
We can write,
Input heat = Work done + ejected heat
Q_H = W + Q_L
W = Q_H – Q_L
Then the efficiency of heat engine

Question 21.
Explain in detail Carnot heat engine.
Answer:
A reversible heat engine operating in a cycle between two temperatures in a particular way is called a Carnot Engine. The carnot engine consists of four parts that are given below.

1. Source: It is the source of heat maintained at constant high temperature TH. Without changing its temperature any amount of heat can be extracted from it.
2. Sink: It is a cold body maintained at a constant low temperature TL. It can absorb any amount of heat.
3. Insulating stand: It is made of perfectly non-conducting material. Heat is not conducted through this stand.
4. Working substance: It is an ideal gas enclosed in a cylinder with perfectly non-conducting walls and perfectly conducting bottom. A non-conducting and frictionless piston is fitted in it.

The four parts are shown in the following Figure.


Carnot’s cycle: In camot’s cycle, the working substance is subjected to four successive reversible processes.
Let the initial pressure, volume of the working substance be P₁, V₁.
When the cylinder is placed on the source, the heat (Q_H) flows from source to the working substance (ideal gas) through the bottom of the cylinder.
The input heat increases the volume of the gas.So the piston is allowed to move out very’ slowly.
W₁ is the work done by the gas it expands from volume V₁ to volume V₂ with a decrease of pressure from P₁ to P₂.
Then the work done by the gas (working substance) is given by
∴ Q_H = Q_A→B \(\int_{V_{1}}^{\mathrm{V}_{2}} \mathrm{P} d \mathrm{~V}\)
When the cylinder is placed on the insulating stand and the piston is allowed to move out, the gas expands adiabatically from volume V₂ to volume V₃ the pressure falls from P₂ to P₃. The temperature falls to TLK.
The work done by the gas in an adiabatic expansion is given by,

Area under the curve BC.
Isothermal compression from (P₃, V₃, T_L) to (P₄, V₄, T_L).

Let the cylinder is placed on the sink and the gas is isothermally compressed until the pressure and volume become P₄ and V₄ respectively.

Let the cylinder is placed on the insulating stand again and the gas is compressed adiabatically till it attains the initial pressure P₁, volume V₂ and temperature T_H, Which is shown by the curve DA in the P – V diagram.

= – Area under the curve DA
Let ‘W’ be the net work done by the working substance in one cycle

The net work done by the Carnot engine in one cycle is given by,
W = W_A→B – W_C→D
Net work done by the working substance in one cycle is equal to the area (enclosed by ABCD) of the P-V diagram.

Question 22.
Derive the expression for Carnot engine efficiency.
Answer:
Efficiency is defined as the ratio of work done by the working substance in one cycle to the amount of heat extracted from the source.

Let us omit the negative sign. Since we are interested in only the amount of heat (Q_L) ejected into the sink, we have

By applying adiabatic conditions, we get,
T_H V₂^γ-1 = T_L V₃^γ-1
T_H V₁^γ-1 = T_L V₄^γ-1
By dividing the above two equations, we get
\(\left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right)\)^γ-1 = \(\left(\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\right)\)^γ-1
Which implies that
\(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}=\frac{\mathrm{V}_{3}}{\mathrm{~V}_{4}}\) … (5)
Substituting equation (5) and (4), we get
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}=\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
∴ The efficiency
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)

Question 23.
Explain the second law of thermodynamics in terms of entropy.
Answer:
\(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{Q}_{\mathrm{H}}}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\) … (1)
According to efficiency of Carnot’s engine equation, quantity \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) = \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\).
The quantity \(\frac { Q }{ T }\) is called entropy. It is a very important thermodynamic property of a System, which is also a state variable.
\(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\) is the entropy received by the Carnot engine from hot reservoir and \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\) is entropy given out by the Carnot engineLto the cold reservoir. For reversible engines (Carnot Engine) both entropies should be same, so that the change in entropy of the Carnot engine in one cycle is zero. It is proved in equation (1).
But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation \(\frac{\mathrm{Q}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{L}}}\) > \(\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{T}_{\mathrm{H}}}\).
In fact the second law of thermodynamics can be reformulated as follows “For all the processes that occur in nature(irreversible process), the entropy always increases. For reversible process entropy will not change”. Entropy determines the direction in which natural process must occur.

Question 24.
Explain in detail the working of a refrigerator.
Answer:
A refrigerator is a Carnot’s engine working in the reverse order which is shown in the figure.

8.17 Schematic diagram of a refrigerator The working substance (gas) absorbs a quantity of heat Q_L from the cold body (sink) at a lower temperature T_L. A certain amount of work W is done on the working substance by the compressor and a quantity of heat Q_H is rejected to the hot body (source) i.e., the atmosphere at T_H. When we stand beneath of the refrigerator, we can feel warmth air.
From the first law of thermodynamics, we have Q_L + W = Q_H
Hence the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.

IV. Numerical problems:

Question 1.
Calculate the number of moles of air is in the inflated balloon at room temperature as shown in the figure.
Answer:

The radius of the balloon is 10 cm, and pressure inside the balloon is 180 kPa.
The pressure inside the balloon P = 1.8 x 10⁵ P
Room temperature
T = 273 + 30 = 303K.
Let the radius of the balloon be R = 10 x 10^-2m.
Volume of the balloon

Question 2.
In the planet Mars, the average temperature is around – 53°C and atmospheric pressure is 0.9 kPa. Calculate the number of moles of
the molecules in unit volume in the planet Mars? Is this greater than that in earth?
Answer:

= 0.9 x 10³Pa
PV = NkT
Since volume is unity V = 1
Equation (1) becomes
P = NkT
Where k is Boltzmann constant

Question 3.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V₁ and contains ideal gas at pressure P₁ and temperature T₁. The other chamber has volume V₁ and contains ideal gas at pressure P₂ and temperature T₂. If the partition is removed without doing any work on the gases, calculate the final equilibrium temperature of the container.
Answer:
In first chamber let the pressure be P₁
In second chamber let the pressure be P₂
In first chamber let the volume be V₁
In second chamber let the volume be V₂
In first chamber let the temperature be T₁
In second chamber let the temperature be T₂
According to conservation of energy,

Question 4.
The temperature of a uniform rod of length L having a coefficient of linear expansion α_L is changed by ∆T. Calculate the new moment of inertia of the uniform rod about axis passing through its centre and perpendicular to an axis of the rod.
Answer:
Moment of inertia of uniform rod of mass M, and length L about axis passing through its centre and perpendicular to an axis of rod is given by
I = \(\frac{\mathrm{ML}^{2}}{12}\)
The increase in length of the rod
∆l = Lα_L∆T
α_L – Coefficient of linear expansion.
∆T – change in temperature
After the rod is heated,
Moment of inertia,

Question 5.
Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for
(a) Isochoric process
(b) Isothermal process
(c) isobaric process.
Answer:
(a) For isochoric process:
V = V₀ = constant

The points a and b are represented as a = (P₁V₀T₀) b = (P₂, V₀, T₂)

(b) For isothermal process:
T = T₀ = constant
PV = nRT₀

P(T) = multivalue
P(T) = \(\frac{n \mathrm{RT}_{0}}{\mathrm{~V}}\)
The points a and b are represented as
a = (P₁V₁,T) b = (P₂V₀T₀)

(c) For isobaric process:
P = P₀ = constant
T(V) = \(\frac{\mathrm{P}_{0} \mathrm{~V}}{n^{2}}\)
P₀V = nRT
P(T) = P₀

Question 6.
A man starts bicycling in the morning at a temperature around 25°C, he checked the pressure of tire which is equal to be 500 kPa. Afternoon he found that the absolute pressure in the tyre is increased to 520 kPa. By assuming the expansion of tyre is negligible, what is the temperature of tyre at afternoon?
Answer:

Question 7.
The temperature of a normal human body is 98.6°F. During high fever if the temperature increases to 104°F, what is the change in peak wavelength that emitted by our body? (Assume human body is a black body)
Answer:
Normal temperature of human body = 98.6°F
Conversion of F° to C°
C = \(\frac { (F-32) }{ 1.8 }\)
= \(\frac { (98.6-32 }{ 1.8 }\)
= \(\frac { 66.6 }{ 1.8 }\)
= 37°C
Conversion of C° to Kelvin
T = 37 + 273 = 310
Peak wavelength at 98.6°F is
λ_m = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{-3}}{310}\)
= 0.009348 x 10^-3 m
= 9348 x 10^-9
= 9348 nm
Final temperature of the human body = 104°F
Conversion of F° to C°
C = \(\frac { F-32 }{ 1.8 }\)
= \(\frac { 104-32 }{ 1.8 }\)
= \(\frac { 72) }{ 1.8 }\)
Conversion of 40° to Kelvin
T = 40+ 273 = 313 K
Peak wavelength at 104°F is
λ_m = \(\frac { b }{ T }\)
= \(\frac{2.898 \times 10^{3}}{313}\)
= 0.009258 x 10^-3 m
= 9258 x 10^-9 = 9258 nm
λ_max = 9348 nm at 98.6°F
λ_max = 9258 nm at 104°F

Question 8.
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure? (For air γ = 1.4)
Answer:
Percentage of increased volume = 4%
\(\frac { ∆V }{ V }\) x 100 = 4%
γ = 1.4
In adiabatic process
PV^γ = Constant

Question 9.
In a petrol engine, (internal combustion engine) air at atmospheric pressure and temperature of 20°C is compressed in the cylinder by the piston to 1/8 of its original volume. Calculate the temperature of the compressed air. (For air γ = 1.4)
Answer:
P₁ = 1 atmospheric pressure, V₁ = V, V₂ = \(\frac { V }{ 8 }\)
T₁ = 20 + 273 – 293K, T₂ – ?, γ = 1.4.

∴ The temperature of the compressed air = 400°C

Question 10.
Consider the following cyclic process consist of isotherm, isochoric and isobar which is given in the figure.

Answer:
Draw the same cyclic process qualitatively in the V-T diagram where T is taken along x direction and V is taken along y-direction. Analyze the nature of heat exchange in each process.

Process 1 to 2: increase in volume. So heat must be added.

Process 2 to 3: Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.

Process 3 to 1: Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.

Explanation: In the graph, during the process (1 to 2), the gas undergoes isothermal expansion. It receives certain amount of heat from the surroundings. It uses this heat in doing the work. Hence internal energy of the gas remains unchanged.

During the process represented by (2 to 3) the gas is heated at constant volume. Since no work is done and volume does not change, the process is isochoric process.

Since heat is transferred to the gas from the surroundings, the internal energy of the gas is increased. During the process represented by (3 to 1) the gas is compressed isobarically. Work is done on the gas. Since temperature drops internal energy is reached. Hence the gas gives up heat to the surroundings.

Question 11.
An ideal gas is taken in a cyclic process as shown in the figure. Calculate
(a) work done by the gas.
(b) work done on the gas.
(c) Net work done in the process

Answer:
(a) Let the work done by the gas along AB be W. From the fig pressure,
P = 600 N/m²
From the fig change in volume,
∆V = 3
∴Work done,
W = PAV
= 600 x 3 = 1800J
= 1.8 kJ

(b) Isobaric compression take place and work is done on the gas along BC.
W = – PAV
∆V = (6 – 3) = 3
P = +400N/m²
∴ W= – 400 x 3
= – 1200
= – 1.2 kJ

(c)

Question 12.
For a given ideal gas 6 x 10⁵ J heat energy is supplied and the volume of gas is increased from 4 m³ to 6 m² at atmospheric pressure. Calculate
(a) the work done by the gas
(b) change in internal energy of the gas
(c) graph this process in PV and TV diagram.
Answer:
Heat energy supplied to the gas
Q = 6 x 10⁵J

(b) Change in internal energy
∆U = ∆Q – AW
= ∆Q – P∆V
= 6 x 10⁵ – 2.026 x 10⁵
= 3,974 x 10³
∆U = 3.974 kJ

(c)

Question 13.
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency.
(a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant
(b) By increasing the temperature of the hot reservoir from 300°C tc 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Answer:
Temperature of cold reservoir
T₁ = 100°C
= 100 + 273 – 373 K
Temperature of hot reservoir
T_H = 300 + 273 = 573 K
η = 1 – \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}}\)
η = 1 – \(\frac { 373 }{ 573 }\)
η = 1 – \(\frac { 573 – 373 }{ 573 }\)
= \(\frac { 200 }{ 753 }\)
= 0.349 x 100
Initial efficiency = 34.9%

(a) By decreasing the temperature of the cold reservoir from 100°C to 50PC.

(b) By increasing the temperature of the hot reservoir from 300°C to 350°C,
T₁ = 350 + 273 = 623K
T₁ = 100 + 273 =373K
∴ Efficiency,
η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
η = 1 – \(\frac { 373 }{ 623 }\)
η = \(\frac { 623 – 373 }{ 623 }\)
η = \(\frac { 250 }{ 623 }\)
η = 0.4012
η = 0.4012 x 100 = 40.12%
∴ Method (a) is more efficient than method (b).

Question 14.
A Carnot engine whose efficiency is 45% takes heat from a source maintained atatemperature of 327°C. To have an engine of efficiency 60% what must be the intake temperature for the same exhaust (sink) temperature?
Answer:
Temperature of a source,
T₁ = 327°C = 327 + 273 = 600 k
η = \(\frac { 45 }{ 100 }\) = 0.45
Efficiency η = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\)
∴ 0.45 = 1 – \(\frac{\mathrm{T}_{2}}{600}\)
\(\frac{\mathrm{T}_{2}}{600}\) = 1 – 0.45 = 0.55
∴ T₂ = 0.55 x 600 = 330 K
η = 60% = \(\frac { 60 }{ 100 }\) = 0.6
0.6 = 1 – \(\frac{\mathrm{T}_{\mathrm{2}}}{\mathrm{T}_{\mathrm{1}}}\) = 1 – \(\frac{330}{\mathrm{~T}_{1}}\)
\(\frac{330}{\mathrm{~T}_{1}}\) = 1 – 0.6 = 0.4
In take temperature
T₁ = \(\frac { 330 }{ 0.4 }\) = 825K
= 825 – 273
= 552°C
The intake temperature = 552°C

Question 15.
An ideal refrigerator keeps its content at 0°C while the room temperature is 27°C. Calculate its coefficient of performance.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 1 Living World Text Book Back Questions and Answers, Notes.

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

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

Part-I

Choose the Right Answer: 

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

Question 2.
Identify the Archaebacterium
a. Acetobacteria
b. Erwinia
c. Treponema
d. Methanobacterium
Answer:
d. Methanobacterium

Question 3.
Identify the correctly matched pair
a. Actinomycete – a) Late blight
b. Mycoplasma  – b) lumpy jaw
c. Bacteria – c) crown gall
d. Fungi – d) sandal spike
Answer:
a. Actinomycete – Lumpy jaw
b. Mycoplasma – sandal spike
c. Bacteria – crown gall
d. Fungi – late blight

Question 4.
Identify the incorrect statement about the gram-positive bacteria.
a. Teichoic acid absent
b. A high percentage of peptidoglycan is found in the cell wall.
c. Cell wall is single-layered
d. Lipopolysaccharide is present in the cell wall.
Answer:
a. Teichoic acid absent

Question 5.
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 Floridian starch
Answer:
a. lack of motile structure

Question 6.
Differentiate Homoiomerous and Heteromerous lichens.
Answer:

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:
Rhizobium- Nitrogen-fixing bacteria,
Living in the root modules of leguminous plants has a symbiotic association with it, fix atmospheric nitrogen, and convert it into nitrates, thereby increases the fertility of the soil. Growing legumes alternatively with paddy can help paddy to give high yield- This method of growing paddy, alternatively with leguminous plants is known as crop rotation.

Mixed cropping:
Amidst, other crops. The leguminous crop is also raised as a mixed crop – so that it enriches the soil and increases the yield by fixing atmospheric nitrogen.

Question 9.
Briefly discuss the 5 kingdom system of classification. Add a note on their merits and demerits.
Answer:
a. Proposed by R.H. Whittaker (American taxonomist)
b. Criteria considered – cell structure, Thallus Organization, Mode of Nutrition, Reproduction, and Phytogenitic Relations.
5 kingdom classifications include: –
a. Monera b. Protista c. Fungi d. Plantae e. Animalia

Question 10.
Give a general account of lichens
Answer:
a. Definition: A symbolic association of algae and Fungi helping each other & living together known as lichens.
b. Partners: Algal partner known as Phycobiont & Fungal partner known as Mycobiont
c. Role of Algal partner – Autotrophic prepare food – give nutrition to fungal partner also
d. Role of fungal partner – gives protection- helps in fixing to the substratum by rhizines.
Classification:

Economic importance:

II. a. Pollution Indicators – Lichens sensitive to air pollutants- (pollution indicators)
b. Rocella Montagne – Produces a dye used in litmus paper (acid-base indicator)
c. Cladonia rangiferina – Food for animals in tundra regions

Part – II.

11th Bio Botany Guide Living World Additional Important Questions and Answers

Choose The Right Answer:

Question 1.
Earth has formed around billion years ago ……………
(a) 3.3
(b) 5.6
(c) 4.6
(d) 5.9
Answer:
(c) 4.6

Question 2.
The organism that is reproductively sterile is
a. Wasp
b. Worker bees
c. Housefly
d. Drosophila
Answer:
c. Worker bees

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.
Recombination is the result of
a. Binary fission
b. Asexual reproduction
c. Sexual reproduction.
d. Vegetative propagation
Answer:
c. sexual reproduction

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

Question 6.
Blister-like pustules occur due to
a. Chickenpox
b. Rust
c. Smut
d. Mumps
Answer:
a. Chickenpox

Question 7.
Expand Bt-toxin
a. Biotechnology
b. Biotoxin
c. Beta-toxin
d. Bacillus thuringiensis
Answer:
d. Bacillus thuringiensis

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.
Saprophytic angiosperm with mycorrhiza
a. Clostridium
b. Azolla
c. Monotropa
d. Viscum
Answer:
c. Monotropa

Question 10.
The famous roqueforti cheese is produced by employing
a. Aspergillus roquefortic
b. Penicillium camemberti
c. Penicillium notatum
d. Aspergillus terreus
Answer:
b.Penicillium camémberti

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.
Both viruses and bacteria contain
a. Plasma membrane
b. Protein wat
c. Peptidoglycan
d. Nucleic acids
Answer:
d. Nucleic acids

Question 13.
Lactobacillus bulgaricus is responsible for the formation of
a. Lactic acid
b. Cheese
c. Yogurt
d. Curd
Answer:
C. Yoghurt

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

Question 15.
Bacterial chlorophyll is also known as
a. Chlorophyll
b. Bilirubin
c. Chromatium
d. Chioridin .
Answer:
Chromatium

Question 16.
This drug is also known as wonder drug.
a. Streptomycin
b. Aureomycin
c. Bacitracin
d. Pencillin
Answer:
Pencillin

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

Question 18.
Among the given 4 – one is not viral diseases- find it out.
a. Cucumber mosaic
b. Citrus canker
c. Ricetungro
d. Potato leaf roll
Answer:
b. citrus canker

Question 19.
Bacillus thuringiensis is an
a. Biofertilizer
b. Bio-fuel
c. Bio-pesticide
d. Bio-medicine
Answer:
c. Bio-pesticide

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

Question 21.
Bacteria that grow in high salinity condition is known as
a. Methane bacterium
b. Halobacterium
c. Thermos aquaticus
d. Agrobacterium
Answer:
b. Halobacterium

Question 22.
Budding is an unique feature of
a. Schizo saccharomyces
b. Aspergillus
c. Penicillium
d. Neurospora
Answer:
Schizo saccharomyces

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

Question 24.
The colourless cell in Nostoc in the intercalary position is responsible for nitrogen fixation is
a. Holoblast
b. Heterozygote
c. Homocyst
d. Heterocyst
Answer:
d. Heterocyst

Question 25.
Genetic trait carried in the bacterial
a. Cell wall
b. Chromosome
c. Plasmid
d. Cell membrane
Answer:
c. Plasmid

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

Question 27.
Developing a vaccine for SARS is difficult because
a. It spreads through nucleic acid
b. It is an enveloped virus
c. It has RNA
d. It constantly changes its form
Answer:
d. It constantly changes its form

Question 28.
Arrange correctly the following viruses according to the given shape and symmetry, Cuboidal, spherical, helical and complex respectively
I. a. Vaccinovirus b. Influenza, c. HIV, d. Herpes
II. a. Influenza b. HIV c, Herpes d. Vaccinovirus
III. a. Herpes b. HIV, c. Influenza d. Vaccinovirus
IV. a. HIV b. Herpes c. Vaccinio Virus d. Influenza
Answer:
III. a. Herpes, b, HIV, c, Influenza, d, Vaccinovirus

II.Match the following and find the correct answer.

Question 1.
I. Five kingdom system of classification
II. Three kingdom system of classification
III. Four kingdom system of classification
IV. Two kingdom system of classification

Answer:
a. C-D-B-A

Question 2.
I. TMV Discovered by world
II. Bacterium word coined by
III. Father of Mycology
IV. Classification virus given by

Answer:
b. B-D-A-C

Question 3.
I. Genophore
II. Bacteria
III. Extra Chromosomal DNA
IV. Fimbrial

Answer:
b. D-A-B-C

Question 4.
I. Plasmid
II. Heterocyst
III. Glycocalyx
IV. Mesosome

Answer:
c. B-A-D-C

III.

Question 1.
Which one of the following is a false statement regarding Prions
a. Prions were discovered by B. Prusiner in 1982
b. They are infectious particles of lipo protein
c. They cause about a dozen fatal degenerative disorders of CNS
d. Creutzfeldt-Jakob disease(cjD) and bovine spongiform encephalopathy (BSE) are some commonly known diseases
Answer:
b. They are infectious particles of lipoprotein

Question 2.
Which one of the following is a false statement regarding Ribosomes
a. Ribosomes are the sites of protein synthesis
b. The number of ribosomes per cell varies from 1000 to 1500.
c. The ribosomes are 70s type and consists of a 2 subunits (50s and 30s)
d. The nbosomes are held together by mRNA and form polyribosomes or polysomes.
Answer:
b. The number of ribosomes per cell varies from 1000 to 1500.

IV. Find out the True and False statements from the following and on that basis find the correct answer:

Question 1.
(i) Poly-B hydroxybutysate is a microbial plastic which is biodegradable
(ii) Transfer of DNA from one bacterium to another is known as transduction
(iii) Micrococcus must have oxygen to survive-known as an obligate aerobe
(iv) Spirulina is rich in carbohydrates so treated as an alternative food.

Answer:
c. True False True False

Question 2.
(i) Toad stools are known as an edible mushroom
(ii) Volvariella volvaceae and Agaricus bisporous are known for their high poisonous nature
(iii) Claviceps purpurea produces ergot-used as vasoconstrictor
(iv) Aspergillus flavus infest dried foods and produce carcinogenic toxin called -aflatoxin

Answer:
B. False False True True

Question 3.
Which one of the following is a correct statement regarding TMV
a. David Baltimore in 1971 discovered TMV
b.First visible symptom of TMV is visible discoloration of leaves along the veins-but with typical yellow and green symptom molting-(mosaic symptom)
c. The plant grow abnormally at the nodal point
d. Infection spread by house flies and mosquitoes
Answer:
b. First visible symptom of TMV is discoloration of leaves along with the veins-but visible with typical yellow and green symptom molting-(mosaic symptom)

V.

Question 1.
Which one of the following is a correct statement regarding bacterial antibiotic
a. Chloromycetin got from Streptomyces venezuelae cure T.B
b. Bactracin is got from bacillus mycoides- used to treat UTI
c. Aurecomycin got from Streptomyces aureofaciens is used to treat whooping cough and eye infections
d. Streptomycin got from Streptmyces griseus cure typhoid fever
Answer:
c. Aurecomycin got from streptomyces aureofaciens is used to treat whooping cough and eye infections

Question 2.
Which one of the following is a correct statement regarding mycoplasma
a.  Mycoplasm are very small (0.1-0.5mm) pleomorphic gram negative micro organisms
b.  The have whole body appear like boiled egg-like structure in culture
c.  Little leaf of tomato Witches broom of solanum.
d.  Mycoplasma is also known as mollicutes
Answer:
Mycoplasm arevery small (0.1- 0.5mm) pleomorphic gram negative micro organisms

VI.

Question 1.
Sac fungi & club fungi are common names of
a.  Ascomycetes
b.  Basidiomycetes
c.  Deuteromycetes
d.  Phycomycetes
(i) a & b
(ii) b & c
(iii) a & c
(iv) c & d
Answer:
(i) a & b

Question 2.
Recurrence of fungal skin disease is due to
a. Resistance to antibiotics
b. Dormant spores become active at the onset of favourable condition
c. Non-availability of specific drugs.
d. Moisture favor fungal mycelium to spread,
(i) a & C
(ii) b & c
(iii) b & d
(iv) c & d
Answer:
(i) b & d

Question 3.
Find out the symbiotic associations from the given options.
a. Nitrogen fixing bacteria on the leguminous plants.
b. Rhizoids of Neprolepis
c. Lichens on rocks
d. Mycorrhizal roots.
(I) ac & d
(II) ab & c
(III) ab & d
(IV) a & b
Answer:

VII. Find out the wrong statement

Question 1.
Which one of the following is a wrong statement regarding ehemo lithotrophs
a. Sulphur bacteria
b. Iron bacteria
c. Methane bacteria
d. Hydrogen bacteria
Answer:
c. Methane bacteria

Question 2.
Among the following, which one is not viral?
a. Cucumber mosaic
b. Citrus canker
c. Rice tungro
d. Potato leaf roll
Answer:
b. Citrus canker

Question 3.
Which one of the following is not a Ribovirus?
a. Tobacco mosaic virus
b. Cauliflower mosaic virus
c. Human immune deficiency virus.
d. Wound tumour virus d. Pilobolus
Answer:
d. Wound tumour virus

Question 4.
Find out from the given, which one is not a Zygomycetes fungi.
a. Mucor
b. Rhizopus
c. Yeast
d. Pilobolus
Answer:
c. yeast

VIII. Read the following Assertion A and Reason R. Find the correct Answer

Question 1.
Assertion ‘A’: Viruses have genetic material but cannot divided on its own. They also don’t have in built metabolic machinery Reason ‘R’: Virus kept between living and non living
(a) A & R correct. R is explaining A
(b) A & R correct R is not explaining A
(c) A is true but R is wrong
(d) A is true but R is not explaining A
Answer:
a. A & R correct R is explaining A.

Question 2.
Assertion ‘A : Some bacteria have the capacity to retain gramstain after treatment with acid alcohol.
Reason ‘R’: Known as gram +ve as attracted towards positive pole under the influence of electric current.
Answer:
c. A is true but R is wrong.

Question 3.
Assertion’A’: Aflatoxin produced by Aspergillus flavus.
Reason ’R’: These toxin are useful to mankind to cure few disease
Answer:
c. A is true but R is wrong

Question 4.
Assertion ’A’: In septal mycelium the septa complete the partition walls between cells Reason
‘R’: There is no cytoplasmic connection between adjacent cells.
Answer:
d. A is true but R is not explaining A.

I. Additional 2 Marks

Question 1.
Define Growth.
Answer:
Growth is an intrinsic property of all living organisms through which they can increase cells both in number and mass.

Question 2.
Tabulate Milestones in Virology
Answer:

Question 3.
Draw the structure of TMV and label the parts and explain in a word or two
Answer:

Answer:
a. Nucleic acid – It is ss RNA(Single Standed)
b. Capsomere – Protein Units
c. Capsid – Protein Coat.

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.
Bacteria is a indeed friend- discuss
Answer:
Even though Bacteria cause many diseases to plants animals and human beings they are beneficial to day-to-day life also. Ex: Milk (lactobacillus acidophobus)/(lactobaci llus lacti)
a. Curd b. Butter c. Cheese d. Yogurt These are a few of the beneficial activities.

Question 6.
Draw the ultra structure of bacterial cell.
Answer:

Question 7.
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 8.
Why is it essential to do classification?
Answer:

• To relate on the basis of common features
• To define, on the basis of salient features
• To know the relationship among different groups.
• To understand evolutionary relationship.

Question 9.
Ruggerio etal’s recent classification-Explain.
Answer:

• Ruggerio etal in 2015-published-7-kingdom system of classification
• It is an extension of Cavalier’s 6 – Kingdom Scheme’Include 2 superkingdom
  

Question 10.
What is Prophage?
Answer:
a. In the lysogenic cycle, the injected phage DNA become circular and integrates into bacterial chromosome by recombination.
b. The integrated DNA of phage and bacteria is known as Prophase.

Question 11.
Distinguish between Cyanophage and Mycophage
Answer:

Question 12.
Differentiate between flagella of Prokaryotes and Eukaryotes
Answer:

Question 13.
Differentiate between Photolithotrophs and Photo organotrophs
Answer:

Question 14.
If you think endospore formation not a reproduction method then justify you answer
Answer:
Yes Endospores are formed not during reproduction, but during unfavorable season the thick walled endospores are resting spores when favorable condition comes, they germinate and form bacteria.

Question 15.
What are the 3 different methods by which gene recombination occur in bacteria ? Or write about Sexual reproduction in bacteria?
Answer:
Sexual reproduction is so simple formation and fusion of gametes is absent. However by  Three different methods gene recombination can occur

1. Conjugation
2. Transduction
3. Transformation

2 Marks

Question 16.
Define chemolithotrophs – give examples
Answer:

• The type of bacteria oxidise in organic compound to release energy
• Eg Sulphur bacteria – Thiobacillus thio oxidants
• Iron bacteria – Ferrobacillus ferro oxidants
• Hydrogen bacteria – Hydrogenomonas

Question 17.
Label the given diagram properly
Answer:

Answer:
1. F- plasmid
2. Conjugation pilus
3. Chromosome
4. F+ cell

Question 18.
Write any 2 vitamin yielding bacteria
Answer:

Question 19.
Name any 2 bacteria diseases affecting Potato
Answer:

Question 20.
What is meant by Probiotics
Answer:

• Microorganism such as lactobacillus bifidobacterium when consume as a dietary supplement help to maintain or restores beneficial bacteria to the digestive tract. They are called friendly or good bacteria. They keep our gut healthy
• They help to increase the immunity of the body
• Eg: Probiotic Yoghurt
• Probiotie tooth paste.

Question 21.
What is the meant by Ray fungi? Give example
Answer:

• Actinomycetes are called as ray fungi due to their mycelia like growth
• They are anaerobic or facultative anaerobic
• They are gram + ve
• Don’t produce aerial mycelium
• DNA contain high guanine and cytosine content Eg strepotmyces

Question 22.
What is this structure?
Answer:
The figure is the structure of mycoplasma

Answer:
1. Cell membrane
2. Ribosome
3. DNA strain
4. Cytoplasm

Question 23.
Name few Renowed mycologists?
Answer:
A. Arthur H.R. Buller, John Webster D.L. Hawksworth, G.C Ainsworth
B. B mundkur, K.C. Meta, C.V. Subramanian and T.S. Sadasivam & Father of Indian Mycology -E.J. Butler-

Question 24.
Identify the diagram and label any three parts.
Answer:

Conidia formation-Penicillium

1. Conidiophores
2. Ramus
3. Metula
4. Sterigma
5. Conidium or Conidiospore

Question 25.
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 conditions to survive in the environment.

Question 26.
Name 4 fungi, from which we derive organic acids.
Answer:

1. Citric acid& Gluconic acid – Aspergillus niger
2. Itaconicacid – Aspergillus terreus
3. Kojicacid – Aspergillus oryzae

Question 27.
Name 4 common basidiomycetes
Answer:

1. Puffballs,
2. Toadstools,
3. Birds nest’s fungi,
4. Bracket fungi,
5. Smuts
6. Rusts,
7. Smuts.

Question 28.
Define aflatoxin.
Answer:
Aspergillus, Polyporus, Mucor and Penicillium are involved in spoilage of food material Aspergillus flavus infest dried food & produce caranogenic toxin known as aflatoxin.

Question 29.
Name 3 Dermatophytes
Answer:

1. Trichophyton
2. Tinea
3. Microsporum
4. Epidermophyton are some fungi causing skin problems

3 Marks Additional Questions

Question 1.
State the living and Non-living character of the Virus.
Answer:
(i) Living characters:

• Presence of nucleic acid & protein
• Capable of mutation
• Ability to multiply with living cells
• Able to infect and cause diseases
• Show irritability and host-specific.

(ii) Non-living characters:

• Can be crystallized
• Don’t have metabolic machinery or functional autonomy
• In active outside the host
• Energy producing enzyme system is absent

Question 2.
Draw the ultra structure of bacterial cell. Image Parts:
Answer:

Question 3.
What are the economic importance of Cyanophyceae
Answer:

Question 4.
Explain any 3 asexual method of reproduction in fungi
Answer:

Question 5.
Tabulate animal diseases caused by bacteria
Answer:

Question 6.
Distinguish between Ammonification & Nitrification.
Answer:

Question 7.
By drawing diagrams, classify bacteria on the basis of flagellatin
Answer:

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.
Define Transduction and explain
Answer:
The three types of Transduction

1. Phage mediated DNAtransfer is Transduction
2. Zinder and Lederberg (1952) discovered it in Salmonella typhimurum
3. 2 types – Generalised and Specalised

I. Generalised Transduction:
a. Ability of bacteriophage to carry the genetic material of any region of bacteria DNA is called generalized transduction.

II. Specialised Transduction:
a. Ability of bacteriophage to carry only a specific region of the bacterial DNA is called Specialized or Restricted Transduction.

Question 10.
Identify from the diagram – & table correctly
Answer:
The given diagram is Basidiocarp of Agaricus

Answer:
1. Rhizoids
2. Stipe
3. Pileus
4. Gills

Question 11.
Identify from the diagram & label correctly.
Answer:

The given diagrams sporangium of mucor
1. Rhizoids
2. Sporangiophore
3. Zygosporangium
4. Zygospores

Question 12.
Define biopesticides and give 2 examples from fungi.
Answer:

1. The substances derived from microbes and plants can be used to kill or eradicate pests weeds and diseases causing germs of crops.
2. This is known Bio-pesticide. They are ecofriendly, non hazardous, non phytotoxic, e.g. Beauveria bassiana Metarhizium anisopliae

Question 13.
Write down any 4 uses of Mycorrhiza.
Answer:

• Nutrition – Saprophytic Angiosperm cant prepare food-due to absence of green leaves & it get nutrition via mycorrhiza e.g. Monotropa
• Availability of water and minerals: improve availability of water & minerals
• Protection-help plant to resist drought & attack of plant pathogen

Question 14.
Progametangium is formed at the tip.
Answer:
The fusion occurs & Zygosporangium is formed then undergo meiosis & zygospores These zygospores germinate are formed, during favourable condition into fungal hyphae.

Question 15.
Explain briefly the characteristics of Oomycetes
Answer:

Question 16.
Explain briefly the characteristics of zygomycetes
Answer:

Question 17.
Planogametic copulation in fungi has 3 types Explain.
Answer:
Planogametic copulation means fusion of motile gametes it has 3 types

1. Isogamy
2. Anisogamy
3. Oogamy
   Isogamy — Morphologically smilar gametes Anisogamy Morphologically dissimilar gametes
   Oogamy – But in Oogamy it is highly advanced anisogamy

Question 18.
Explain gametangial copulation in Rhizopus with the help of diagrams.
Answer:

• In rhizopus & in Mucor there occur heterothallism- there are 2 strains of the hyphae.
• In a sexual copulation only the 2 opposite strains +ve and -ve strains come together.
  

Question 19.
The tea or tobacco is not a result of mere drying process. Explain the value addition in these things?
Answer:
Yes tea & tobacco they are not mere dried leaves of Camellia sinensis (tea) & Nicotiana tobaccum so tea and tobacco to reach the utility stage, they have to be subjected to biological process known as curing, where specific bacteria are added & curing occur by a process of fermentation specific flavor and aroma of tea and tobacco are due to this fermentation process.

Tea – Mycococcus candisans
Tobacco – Bacillus megatherium

Question 20.
Write about the harmful activities of fungi.
Answer:

Question 21.
Mycorrhiza is known as bio fertilizer. Explain
Answer:

• Symbolic association between fungal mycelium and roots of higher plants is known as mycorrhiza.
• Fungi absorbs nutrition from root of higher plant, and intum fungal hyphae helps the paint to absorb water and minerals nutrients from soil.
• Any nutrient of biological orgin is biofertilizer & This is Bio fertilizer it is non hazardous, Non phytotoxic and ecofriendly.

Question 22.
Name the Antibiotics derived from fungi.
Answer:

5 Marks

Question 1.
Compare the five-kingdom system of classification
Answer:
Kingdom

Question 2.
Ultrastructure of a typical bacterial cell.
Answer:

Question 3.
Explain industrial uses of bacteria.
Answer:
Industrial Uses

Question 4.
List out bacterial diseases caused to plants, animals & human brings.
Answer:
Plant diseases caused by bacteria

Animal diseases caused by bacteria

Human diseases caused by bacteria

Question 5.
Tabulate the salient features of Cyanophyceae?
Answer:

Question 6.
Give an account of Mycoplasma as Mollicutes.
Answer:

• Small-(0.1-0.5gm)
• Pieomorphic- gram-negative
• 1 st isolation by Nocard & co in 1898-from pleural fluid of cattle affected with bovine pleuropneumonia.
• Cell wall-absent
• In culture appear as fried egg.
• DNA contains low guanine and cytosine than true bacteria
• Cause disease in Animals / Plants
• Little leaf of brinjal – Witches broom of legumes Phyllody of cloves Sandal spike, -plant diseases
• Pleuropneumonia – animal disease caused by Mycoplasma mycoide.

Question 7.
Differentiate between Gram-positive and Gram-Negative bacteria.
Answer:

Question 8.
Explain transformation in bacteria.
Answer:
Definition: Transfer of DNA from one bacteria to another is called transformation.

• 1928 Frederick Griffth demonstrated it in mice using Diplococcus pneumonia.
• 2 strains
  • Smooth colonies – virulent type (S)
  • Rough colonies – Avirulent type (R)
• S type injected – mouse died because it is virulent
• R type injected – mouse lived because R type is Avirulent
• Heat killed S type injected-mouse lived
• Heat killed S type + R type called when injected mixedly} mouse died

Question 9.
Classify the nitrogen-fixing biological systems.
Answer:

I. Symbiolic Angiosperm:
Leguminous Rhizobium root nodules of Ground nut Peas etc. (soil bacterium) Non-Leguminous (Alnus casuarinas) Frankia actinomycetes (root nodules)
II. Symbiolic gymnosperm Cycas unidentified spices of Blule green algae (corolloid roots)
III. Symbiotic ferns Azolla- Anbaena azollae – leaf pockets
IV. Symbiotic (fungi & Algae) Lichen (phycobiont mycobiont mutually benefited).

Question 10.
Tabulate the human disease caused by bacteria.
Answer:

Question 11.
Explain General characteristics of fungi
Answer:

• Plant body- the plant body is filamentous & branched hyphae No of hyphae interwoven to form – my celium
• Cell wall – made up of chitin (polymer of N acetyl glucose)
• Mycelium 2 types septate cross wall between cells Aseptate hyphae (multi nucleus coenocytic mycelium) E.g Albugo
• Mycelium of higher fungi. Septum present between cells if the hyphae Eg. Fusarium.
  

Types of mycelium
Mycelium organized to compactly interwoven tissue Pletenchyma – 2 Types

• Prosenchyma-hyphae loosely arranged
• Pseudoparenchyma hyphae compactly arranged Holocarpic form- Entire thallus- become reproductive structure.
• In Eucarpic form Some region of thallus reproductive & some other region vegetative Reproduction there are 3 types
  • Anamorph-Asexual
  • Teleomorph -Sexual
  • Holomorph-both sexual & Asexual.

Question 12.
Tabulate various types of Mycorrhizae
Answer:
Mycorrhizae – 3 Types

Question 13.
List out diseases caused by fungi
Answer:
Diseases caused by fungi

Human diseases

Question 14.
Fruit bodies or Ascocarps 4types
Answer:

• Cleistothecium
• Ascothecium(flask shape)
• Apothecium(cup shape)
• Psedothecium
  1. Image structure of cleistothecium
  2. Structure of PeritheciumV.S
  3. Structure of Apothecium

Question 15.
Give an account of Basidiomycetes (club fungi).
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 10 Communication Systems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems

12th Physics Guide Communication Systems Text Book Back Questions and Answers

Part – I:

Textbook Evaluation:

I. Multiple Choice questions:

Question 1.
The output transducer of the communication system converts the radio signal into
a) Sound
b) Mechanical energy
c) Kinetic energy
d) None of the above
Answer:
a) Sound

Question 2.
The signal is affected by noise in a communication system
a) At the transmitter
b) At the modulator
c) In the channel
d) At the receiver
Answer:
c) In the channel

Question 3.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called
a) Amplitude modulation
b) Frequency modulation
c) Phase modulation
d) Pulse width modulation
Answer:
b) Frequency modulation

Question 4.
The interntionally accepted frequency deviation for the purpose of FM broadcasts.
a) 75 kHz
b) 68 kHz
c) 80 kHz
d) 70 kHz
Answer:
a) 75 kHz

Question 5.
The frequency range of 3 MHz to 30 MHz is used for
a) Ground wave propagation
b) Space wave propagation
c) Sky wave propagation
d) Satellite communication
Answer:
c) Sky wave propagation

II. Short Answers Questions:

Question 1.
Give the factors that are responsible for transmission impairments.
Answer:
Attenuation:
The strength of signal decreases with increasing distance which causes loss of energy.
Distortion:
Change in the shape of the signal
Noise:
Random or unwanted signal mixes up with the original sound.

Question 2.
Distinguish between wireline and wireless communication? Specify the range of electromagnetic waves in which it is used.
Answer:

Question 3.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called center frequency or resting frequency.

Question 4.
What does RADAR stand for?
Answer:

• RADAR stands for Radio Detection and Ranging
• It is used to sense, detect and locate distant objects like aircraft, ships, spacecraft, etc.

Question 5.
What do you mean by the Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

III. Long Answers Questions:

Question 1.
What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation:
For long-distance transmission, the low-frequency baseband signal (input signal) is superimposed onto a high-frequency radio signal by a process called modulation.

i) Amplitude modulation:
1. If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal.
2. The frequency and the phase of the carrier signal remains constant.
3. It is used in radio and TV broadcasting.

Advantages of AM:
1. Easy transmission and reception
2. Lesser bandwith
3. Low cost

Disadvantages of AM:
1. High noise level
2. Low efficiency
3. Small operating range

ii) Frequency modulation:
1. Frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal.
2. The amplitude and the phase of the carrier signal remain constant.
3. Increase in the amplitude of the baseband signal increases the carrier signal frequency.
4. Louder signals form compressions and weaker signals form rarefactions.
5. When the frequency of the baseband signal is zero, there is no change in the frequency of the carrier wave. This is known as centre frequency or resting frequency.

Advantages:
1. Decrease in noise
2. Efficiency is high
3. Operating range is quite large
4. Better quality

Disadvantages:
1. Requires wider channel
2. Less area is covered.
3. Transmitters and receivers are complex and costly.

Phase Modulation:

1. The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant.
2. The frequency shift depends on

• the amplitude of the modulating signal and
• the frequency of the signal.

3. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal.
4. The carrier signal is compressed or its frequency is increased.
5. The negative half cycle of the baseband signal produces lag in the carrier signal.
6. Phase modulated wave also comprises of compressions and rarefactions.
7. When the signal voltage is zero, (A, C and E) the carrier frequency is unchanged.

Advantages:
1. FM Signal produced from PM signal is very stable.
2. The center frequency called resting frequency is stable.

Question 2.
Elaborate on the basic elements of a communication system with the necessary block diagram.
Answer:
1. Information (Baseband or input signal):
i. Information can be in the form of a sound signal like speech, music, pictures or computer data.
ii. These can be given as input to the input transducer.

2. Input transducer:
i. It converts variation in a physical quantity such as pressure, temperature, sound into equivalent electrical signals or vice versa.
ii. It converts the information into corresponding electrical signals.
iii. The electrical equivalent of the original information is called baseband signal.

3. Transmitter:
It is located at the broadcasting station.

a) Amplifier:
Very weak transducer output is amplified

b) Oscillator:
i. For long-distance, the high-frequency carrier wave is transmitted into space.
ii. Energy of the wave is proportional to its frequency, so the carrier wave has very high energy.

c) Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

d) Power amplifier:
To cover a large distance, the power level of the electrical signal is increased.

4. Transmitting antenna:
i. It travels in the form of electromagnetic waves with the velocity of light
ii. Used for long-distance transmission
iii. Ex: Mobile, radio or TV broadcasting, satellite communication.

5. Communication channel:
To carry the electrical signal from transmitter to receiver with less noise or distortion.

Wireline Communication:
i. Medium – Wires, Cables, and optical fibers
ii. Cannot be used for long-distance transmission
iii. Ex: Telephone, intercom and Cable TV

Wireless communication:
i. Medium – free space
ii. Signals are transmitted in the form of electromagnetic waves.

6. Noise:
i. Undesirable electrical signal
ii. Reduces the quality of the transmitted signal.
iii. It may be man-made or natural.
iv. It cannot be completely eliminated but it can be reduced.

7. Receiver:
i. The transmitted signals are fed into the receiver.
ii. Receiver consists of the demodulator, amplifier, detector etc.
iii. Demodulator extracts the baseband signal from the carrier.
iv. Then it is amplified and detected, then fed to the output transducer.

8. Repeaters:
i. Combination of transmitter and receiver
ii. Increase the range or distance through which the signals are sent.
iii. Signals are received, amplified, and retransmitted with a different frequency to the destination.
iv. Ex. Communication satellite in space.

9. Output transducer:
i. Converts the electrical signal back to its original form such as sound, music, pictures or data.
ii. Ex : Loudspeakers, Picture tubes, Computer monitor etc.

10. Attenuation:
The loss of strength of a signal while propagating through a medium.

11. Range:
i. Maximum distance between the source and the destination.
ii. The signal is received with sufficient strength.

Question 3.
Explain the three inodes of propagation of electromagnetic waves through space.
Answer:
According to the frequency range, the electromagnetic wave transmitted by the transmitter travels in three different modes.

1. Ground Wave propagation (or) Surface Wave propagation:
i. The electromagnetic waves are transmitted by the transmitter glide over the surface of the earth to reach the receiver.
ii. Both transmitting and receiving antennas must be close to the earth.
iii. The size of the antenna decides the efficiency of the radiation of signals.
iv. Some reasons for attenuation are

a) Increasing distance:
Attenuation of the signal depends on
(i) Power of the transmitter
(ii) frequency of the transmitter
(iii) Condition of the earth surface.

b) Absorption of energy by the Earth:
i. Transmitted signal is in contact with the Earth, it induces charges in the Earth and constitutes a current.
ii. The Earth behaves like a leaky capacitor which leads to the attenuation of the wave.

c) Tilting of the wave:
i. The wavefront starts gradually tilting according to the curvature of the Earth.
ii. When the tilt increases, the electric field strength decreases.
iii. At some distance, the surface wave dies out due to energy loss.
iv. The frequency of the ground waves is mostly less than 2 MHz
v. It is used in local broadcasting, radio navigation, for ship – to ship, ship – to shore communication and mobile communication.

2. Skywave propagation (or) Ionospheric propagation:
i. The electromagnetic waves are radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth.
ii. Due to the absorption of ultraviolet rays, cosmic ray and other high energy radiations like α, β rays from sun get ionized.
iii. These charged ions provide a reflecting medium for the reflection of radio waves.
iv. The phenomenon of bending the radio waves back to earth is due to total internal reflection.
v. The shortest distance between the transmitter and the point of reception of the skywave along the surface is called as the skip distance.
vi. When the angle of emission increases, the reception of ground waves decreases.
vii. A zone where there is no reception of electromagnetic waves neither ground nor sky is called skip zone or skip area.
viii. In this mode of propagation, the frequency range of EM waves is 3 to 30 MHz.
ix. It is used for short wave broadcast services.
x. Extremely long-distance communication is possible.
xi. A single reflection helps to travel a distance of approximately 4000 km.

3. Space Wave propagation:
i. The process of sending and receiving information signal through space.
ii. EM waves of very high frequencies above 30 MHz are used.
iii. Waves travel in a straight line from the transmitter to the receiver.
iv. The transmission towers must high.
v. For high frequencies, the signals will not encounter the curvature of the earth.
vi. It travels with less attenuation and loss of signal strength.
vii. TV broadcast, satellite communication and RADAR are based on this propagation
viii. It has some advantages, larger bandwith, high data rates, better directivity, small antenna size, low power consumption etc.
ix. Range depends on the height of the antenna (h).
x. Range or distance (d) = \(\sqrt{2 \mathrm{Rh}}\)
R – Radius of the earth
R = 6400 Km.

Question 4.
What do you know about GPS? Write a few applications of GPS.
Answer:
GPS:

1. GPS stands for Global Positioning System.
2. It offers geolocation and time information anywhere on or near the earth.
3. It works with the assistance of a satellite network.
4. It broadcasts a precise signal like an ordinary radio signal.
5. It convey the location data which is translated by the GPS software.
6. This software recognize the satellite, its location ahd time taken to travel from each satellite.
7. Then the data from each satellite is processed to estimate the location of the receiver.

Applications:

1. It is used in fleet vehicle management for tracking cars, trucks and buses.
2. In wildlife management for counting of wild animals.
3. In engineering for making tunnels, bridges etc.

Question 5.
Give the applications of ICT is mining and agriculture sectors.
Answer:
ICT in mining:

1. Improves operational efficiency, remote monitoring and disaster locating system.
2. Provides audio – visual warning to the trapped underground miners.
3. Helps to connect remote sites.

ICT in agriculture:

1. Increases food productivity and farm management.
2. Optimize the use of water, seeds and fertilizers etc.
3. It is used in farming to decides the suitable place for the species to be planted.
4. It can be used in sophisticated technologies like robots, temperature, moisture sensors, aerial images and GPS technology.

Question 6.
Modulation helps to reduce the antenna size in wireless communication Explain.
Answer:

1. Antenna is used at both transmitter and receiver.
2. Antenna height is important in wireless communication.
3. The height of the antenna must be a muiltiple of \(\frac{\lambda}{4}\).
   i.e h = \(\frac{\lambda}{4}\)
   Where, λ = \(\frac{C}{\gamma}\), C = Velocity of light
   γ – frequency λ – wavelength
4. Consider two base band signals.
5. One is modulated and the other is not modulated
6. Frequency of the original baseband signal, γ = 10 KHz
7. Frequency of the modulated signal, γ = 1 MHz
8. Height required to transmit the original baseband signal is,
   h₁ = \(\frac{\lambda}{4}=\frac{C}{4 \gamma}\)
   = \(\frac{3 \times 10^{8}}{4 \times 10 \times 10^{3}}\) = 7.5 Km
   Height required to transmit the modulated signal is
   h₂ = \(\frac{\lambda}{4}=\frac{C}{4 \gamma}\)
   = \(\frac{3 \times 10^{8}}{4 \times 10 \times 10^{6}}\) = 75 m
9. From the above example, it is clear that modulated signals reduce the height of the antenna.

Question 7.
Fiber-optic communication is gaining popularity among the various transmission media – justify.
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems. Light has very high frequency (400 THz – 790 THz) than microwave radio systems.  The fibers are made up of silica glass or silicon dioxide which is highly abundant on Earth.

Now it has been replaced with materials such as chalcogenide glasses, fluoro aluminate crystalline materials because they provide larger infrared wavelength and better transmission capability. As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and isolation is required from electrical and electromagnetic interference.

Applications:
Optical fiber system has a number of applications namely, international communication, inter-city communication, data links, plant and traffic control and defense applications.

Merits:

1. Fiber cables are very thin and weight lesser than copper cables.
2. This system has much larger bandwidth. This means that its information-carrying capacity is larger.
3. Fiber optic system is immune to electrical interferences.
4. Fiber optic cables are cheaper than copper cables.

Demerits:

1. Fiber optic cables are more fragile when compared to copper wires.
2. It is an expensive technology.

PART – II:

12th Physics Guide Communication Systems Additional Questions and Answers

I. Matching Type Questions:

Question 1.

Answer:

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

Question 2.

Answer:

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

Question 3.

Answer:

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

II. Fill in the blanks:

Question 1.
_______ gives the difference between the upper and lower frequency limits of the signal.
Answer:
Bandwidth

Question 2.
For a frequencv less than the critical frequencv, skip distance is _______.
Answer:
Zero

Question 3.
Multimode fibers operate at the speed of _______.
Answer:
10 Mbps

Question 4.
Radar uses _______ for communication.
Answer:
electromagnetic waves.

III. Choose the Odd man out:

Question 1.
a) Input transducer
b) Amplifier
c) Oscillator
d) Demodulator
Answer:
d) Demodulator

Question 2.
a) Telephone
b) mobile
c) intercom
d) Cable TV
Answer:
b) Mobile

Question 3.
a) Weather satellites
b) Communication
c) satellites
d) Navigation satellites
e) RADAR
Answer:
e) RADAR

Question 4.
a) Total internal reflection
b) skip distance
c) amplitude modulation
d) skip zone
Answer:
c) Amplitude modulation

IV. Choose the incorrect pair:

Question 1.
i) Space wave propagation – LOS
ii) Tracking cars – ICT
iii) Counting of wild animals – GPS
iv) Home automation using
a mobile phone – IoT
Answer:
ii) Tracking cars – ICT

Question 2.
i) Input transducer – Microphone
ii) Output transducer – Loudspeakers
iii) Transmitter – Broadcasting station
iv) Oscillator – Low-frequency carrier wave
Answer:
iv) Oscillator – Low-frequency carrier wave

Question 3.
i) Bandwidth, BW – γ₁ – γ₂
ii) Amplitude modulation system – 10 KHz
iii) Single side – band system – 5 KHz
iv) Height of the Antenna – \(\frac{\lambda}{4}\)
Answer:
i) Bandwidth, BW – γ₁ – γ₂

Question 4.
i) Amplitude – High-efficiency modulation
ii) Baseband signal – input signal
iii) Carrier signal – a radio signal
iv) Resting frequency – centre frequency
Answer:
i) Amplitude modulation – High efficiency

V. Choose the correct pair:

Question 1.
i) Attenuation – Amplitude of the transmitter
ii) High frequency – High skip distance
iii) Skip distance – The longest distance
iv) RADAR – Ground wave propagation
Answer:
ii) High frequency – High skip distance

Question 2.
i) Audio frequency – 200 to 2000 Hz
ii) Carrier wave – Cosine wave
iii) Louder signal – Compressions
iv) Frequency shift – Rarefaction
Answer:
iii) Louder signal- Compressions

Question 3.
i) Velocity of light – 3 × 10⁶ m/ s
ii) Radius of the earth – 6800 Km
iii) Date speed for homes – 2 Gbps
iv) Antenna height – d = \(\sqrt{2 \mathrm{Rh}}\)
Answer:
iv) Antenna height – d = \(\sqrt{2 \mathrm{Rh}}\)

Question 4.
i) Skywave propagation – 3 MHz to 30 MHz
ii) Space wave propagation – 30 MHz to 40 MHz
iii) Ground wave propagation – 20 KHz to 20 MHz
iv) Wireless communication – 1 KHz to 2 KHz
Answer:
i) Skywave propagation – 3 MHz to 30 MHz

VI. Assertion and Reason:

Question 1.
Assertion:
Short wave bands are used for the transmission of radio waves to a large distance.
Reason:
Short waves are reflected by the ionosphere.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion:
Television broadcasting becomes weaker with increasing distance.
Reason:
The power transmitted from the TV transmitter varies inversely as the distance of the receiver.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
c) Assertion is true but reason is false

Question 3.
Assertion:
Optical fibre communication has immunity to cross talk.
Reason:
Optical interference between fibres is zero.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 4.
Assertion:
A dish antenna is highly directionals
Reason:
This is because a dipole antenna is omnidirectional.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.

VII. Choose the correct statement:

Question 1.
i) A transducer is a device which converts an electrical signal into variations in a physical quantity.
ii) Transmitting antenna radiates the radio signal into space in all directions.
iii) The power amplifier decreases the power level of the signal
iv) The minimum distance between the source and the destination is called a range
Answer:
Statement (ii) is correct.

Question 2.
i) Attenuation depends on the frequency of the transmitter.
ii) Increase in the tilt increases the electric field strength.
iii) Transmitting and receiving antenna must be close to the earth.
iv) The frequency of the ground waves is most less than 1 MHz.
Answer:
Statements (i) and (iii) are correct

Question 3.
i) The higher the frequency, lower is the skip distance.
ii) Bending of radio waves back to earth is called refraction.
iii)The frequency range of EM wave in skywave propagation is 30 to 300 MHz.
iv) When the angle of emission increases, the reception of ground waves decreases.
Answer:
Statement (iv) is correct.

Question 4.
i) The transducer output is very weak
ii)The electrical equivalent of the original information is called the carrier signal.
iii) Microphone converts electrical energy into sound energy.
iv) Wireline communication uses free space as a communication medium.
Answer:
Statement (i) is correct.

VIII. Choose the incorrect statement:

Question 1.
i) Noise is the undesirable electrical signal.
ii) It attenautes or reduces the quality of the signal
iii) It may be man – made or natural
iv) Noise can be completely eliminated.
Answer:
Statement (iv) is correct.

Question 2.
i) Operating range is quite large in FM
ii) In FM, there is a large increase in noise
iii) Am radio has better quality compared to FM radio.
iv) FM requires a much under channel
Answer:
Statement (ii) & (iii) are incorrect

Question 3.
i) Noise level is low in AM
ii) There is a lesser landwidth in AM
iii) In A.M the efficiency is low
iv) Small operating range in AM
Answer:
Statement (i) is incorrect

Question 4.
i) ICT in mining helps to connect remote sites.
ii) ICT is widely used in decreasing food productivity
iii) ICT improves operational efficiency
iv) ICT are also used in fisheries
Answer:
Statement (ii) is incorrect

IX. Choose the best answer:

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz
Hint:
A frequency of 10 kHz will require a very large radiating antenna while frequencies 1GHz and 1000 GHz will penetrate the ionosphere and cannot be reflected by it.

Question 2.
Cellular phones use radio waves in ______ band.
a) long-wave
b) short wave
c) medium wave
d) ultra-high frequency
Answer:
d) ultra-high frequency

Question 3.
Antenna is
(a) inductive
(b) capacitive
(c) resistive above its resonant frequency
(d) resistive at the resonant frequency
Answer:
(d) resistive at the resonant frequency.
Hint:
An antenna is a tuned circuit consisting of an inductance and a capacitance. At the resonant frequency, it is resistive.

Question 4.
In space wave propagation, the range of the propagation depends on the height (h) of the antenna given by the equation.
a) \(\sqrt{2 \mathrm{Rh}}\)
b) \(\sqrt{\frac{R h}{2}}\)
c) \(\sqrt{2 R^{2} h}\)
d) \(\frac{h}{2 R}\)
Answer:
a) \(\sqrt{2 \mathrm{Rh}}\)

Question 5.
_______ modulation is used in radio and TV broadcasting.
a) Frequency
b) Phase
c) Amplitude
d) Carrier
Answer:
c) Amplitude

Question 6.
If a ________ wave is used as the baseband signal, then phase reversal takes place in the modulated signal.
a) Square
b) Sine
c) Cosine
d) triangular
Answer:
a) Square

Question 7.
An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulation index?
(a) 0.67
(b) 5.00
(c) 0.20
(d) 1.5
Answer:
(b) 5.00
Hint:
m_f = \(\frac { ∆f }{ f }\) = \(\frac { 10kHz }{ 2kHz }\) = 5

Question 8.
The best example of transducer is________
a) Loudspeaker
b) picture tubes
c) computer monitor
d) microphone
Answer:
d) microphone

Question 9.
________ generates high – frequency carrier wave.
a) Modulator
b) Oscillator
c) Amplifier
d) Power amplifier
Answer:
b) Oscillator

Question 10.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(a) h^1/2
(b) h
(c) ^3/2
(d) h²
Answer:
(a) h^1/2
Hint:
d = \(\sqrt { 2Rh } \) ; d ∝ h^1/2

Question 11.
The height of the antenna must be a multiple of _______.
a) \(\frac{\lambda}{4}\)
b) \(\frac{\lambda}{8}\)
c) 4λ
d) 2λ
Answer:
a) \(\frac{\lambda}{4}\)

Question 12.
The relation between the frequency, the velocity of light, and wavelength is given by ________.
a) C = γλ
b) C = \(\frac{\lambda}{\gamma}\)
c) λ = \(\frac{C}{\gamma}\)
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 13.
In AM, the channel bandwidth is ______ the signal frequency
a) thrice
b) twice
c) equal to
d) half
Answer:
b) twice

Question 14.
The height of the transmitting and receiving antenna must be a multiple of
a) 2λ
b) λ/4
c) λ/2
d) 4λ
Answer:
b) λ/4

Question 15.
A laser beam is used for carrying out surgery because it
(a) is highly monochromatic
(b) is highly coherent
(e) is highly directional
(d) can be sharply focused
Answer:
(d) can be sharply focused
Hint:
A laser beam is highly monochromatic, directional, and coherent and hence it can be sharply focused for carrying out surgery.

Question 16.
The _______ of the antenna plays a major role in deciding the efficiency of the radiation of the signals.
a) shape
b) distance
c) direction
d) size
Answer:
d) size

Question 17.
A single reflection helps the radio waves to travel a distance of approximately
a) 40 km
b) 400 km
c) 4000 km
d) 4 km
Answer:
c) 4000 km

Question 18.
When the angle of emission increases, the reception of ground waves _______.
a) increases
b) decreases
c) remains constant
d) none of the above
Answer:
b) decreases

Question 19.
The principle used in the transmission of signals through an optical fiber is
(a) total internal reflection
(b) refraction
(c) dispersion
(d) interference
Answer:
(a) total internal reflection
Hint:
Signals propagate through an optical fiber by suffering repeated total internal reflections.

Question 20.
The fibres are made up of _______.
a) glass
b) silicon
c) silica glass
d) both a & c
Answer:
d) both a & c

Question 21.
_______ provide the fastest transmission rate compared to any other form of transmission
a) Cable wires
b) Fibers
c) Fiber optic cables
d) Copper wire
Answer:
c) Fiber optic cables

Question 22.
________ is a fast-growing technology in the field communication system.
a) Mobile
b) Satellite
c) Fiber optics
d) Internet
Answer:
d) Internet

Question 23.
The skywave propagation is suitable for radio waves of frequency
(a) from 2 MHz to 50
(b) upto 2MHz
(c) from 2 MHz to 30 MHz
(d) from 2MHz to 20 MHz
Answer:
(c) from 2 MHz to 30 MHz

Question 24.
To store all the information available on the internet, we need over
a) 2 billion DVDs
b) 1 billion DVDs
c) 3 billion DVDs
d) 4 billion DVDs
Answer:
b) 1 billion DVDs

Question 25.
_______ is used to search for information on the World Wide Web
a) Communication
b) Satellite
c) E-Commerce
d) Search engine
Answer:
d) Search engine

Question 26.
_______ is used for making tunnels, bridges etc.
a) GSM
b) TDMA
c) GPS
d) GPRS
Answer:
c) GPS

Question 27.
Fiber optic cables provide data speed of ____ for homes and businesses.
a) 1 Gbps
b) 2 Gbps
c) 10 Mbps
d) 5 Gbps
Answer:
a) 1 Gbps

Question 28.
A modem is a device which performs
(a) modulation
(b) demodulation
(c) rectification
(d) modulation and demodulation
Answer:
(d) modulation and demodulation

X. Two Marks Questions:

Question 1.
What is a communication system?
Answer:
The setup used to transmit information from one point to another is called a communication system.

Question 2.
What is called a baseband signal?
Answer:
The electrical equivalent of the original information.

Question 3.
What is a transducer?
Answer:
A device that converts variations in a physical quantity into an equivalent electrical signal.

Question 4.
Define bandwidth.
Answer:
The frequency range over which the baseband signals or the information signals such as voice, music, the picture is transmitted.

Question 5.
Define bandwidth?
Answer:
The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc. is transmitted is known as bandwidth.

Question 6.
What is skip distance?
Answer:
The shortest distance between the transmission and the point of reception of the skywave along the surface.

Question 7.
What is skip zone or skip area?
Answer:

• The zone in which there is no reception of electromagnetic waves.
• It is also called as skip area.

Question 8.
What is called fibre optic communication?
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber.

Question 9.
What is mean by fibre optic communication?
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication.

Question 10.
Expand GPS and write a note on it.
Answer:

• GPS stands for Global Positioning System.
• It is a global navigation satellite system that offers geolocation and time anywhere on or near the earth.

Question 11.
Why do we need carrier waves of very high frequency in the modulation of signals?
Answer:
High-frequency carrier waves are used

1. to increase operating range
2. to reduce antenna length
3. to convert the wideband signal into the narrowband signal.

Question 12.
Why is modulation needed at all?
Answer:
Modulation is needed.

1. to transmit a low-frequency signal to a distant place.
2. for protecting the waveform of the signal
3. to keep the height of antenna small.

Question 13.
Which is better for high fidelity reception FM or AM?
Answer:
FM transmission gives high fidelity due to the presence of a large number of sidebands.

Question 14.
Why is the transmission of signals through a coaxial cable not possible for frequencies greater than 20 MHz?
Answer:
For frequencies greater than 20 MHz, dielectric loss becomes quite high.

Question 15.
Why are short wave bands used for long-distance transmission of signals?
Answer:
Radio waves of short wave bands can be easily reflected bv the ionosphere. So they are used in long-distance transmission.

Question 16.
Why microphone is used as an input transducer?
Answer:
A transducer is a device that converts sound into an equivalent electrical signal So, a microphone which converts sound into electrical energy is used as a transducer.

Question 17.
What are the elements in the transmitter?
Answer:

1. Amplifier
2. Modulator
3. Oscillator
4. Power amplifier

Question 18.
Draw the block diagram of Transmission.
Answer:

Question 19.
Define range.
Answer:
The maximum distance between the source and the destination upto which the signal is received with sufficient strength.

Question 20.
Mention the application of ICT is fisheries.
Answer:

• Satellite vessel monitoring system helps to identify fishing zones.
• Use of barcodes helps to identify time and date of catch, species name, quality of fish.

Question 21.
What is satellite communication?
Answer:
A mode of communication of signal between transmitter and receiver via satellite.

Question 22.
What are the basic elements required for the transmission and reception of a signal?
Answer:
Transducer, amplifier, carrier signal, modulator, power amplifier, medium of transmission, tansmitting and receiving antenna, demodulator and detector.

Question 23.
Give the reasons for attenuation in the Ground wave propagation.
Answer:

1. Increasing distance
2. Absorption of energy by the earth
3. Tilting of the wave.

XI. Three Marks questions:

Question 1.
What are the advantages of FM?
Answer:

1. Large decrease in noise. An increase in signal noise ratio
2. The operating range is quite large
3. The transmission efficiency is very high
4. FM bandwidth covers the entire frequency range.
5. FM radio has better quality compared to AM radio.

Question 2.
What are the disadvantages of FM?
Answer:

1. FM requires a much wider channel
2. FM transmitters and receivers are more complex and costly.
3. In FM reception, less area is covered compared to AM.

Question 3.
Give the reason why transmission of TV signals via sky wave is not possible.
Answer:

1. Television frequencies lie in the range 100 – 220 MHz which cannot be reflected by the ionosphere.
2. So, sky wave propgation is not used in TV transmission.

Question 4.
Show in diagram, the skip distance and the skip zone.
Answer:

Question 5.
It is necessary to use satellites for long distance TV transmission. Why?
Answer:

1. TV signals being of high frequency are not reflected by the ionosphere.
2. Ground wave transmission is possible only up to a limited range.
3. So, satellites are used for long distance TV transmission.

Question 6.
Which is more efficient mode of transmission FM or AM?
Answer:

1. FM transmission is more efficient because all the transmitted power is useful.
2. In AM transmission, most of the power goes waste in transmitting the carrier alone.

Question 7.
Give an account of E-commerce and search engine.
Answer:
E-Commerce:
Buying and selling of goods and services, transfer of funds are done by an electronic network.
Search engine:
A web-based service tool used to search for information on World Wide Web.

Question 8.
Expand GSM. Give a note on it.
Answer:

• GSM – Global System for Mobile Communication.
• Increases the utilization of bandwidth of the network, sharing of the networks, error detections etc.

Question 9.
Mention some of the applications of optical fiber system.
Answer:

1. International Communication
2. Intercity communication
3. data links
4. plant and traffic control
5. defence applications.

Question 10.
What are repeaters in an electronic communication system?
Answer:

1. Repeaters are a combination of transmitter and receivers.
2. Used to increase the range or distance through which the signals are sent.
3. The received signals are transmitted with a carrier signal of different frequencies to the destination.

Question 11.
State two factors by which the range of transmission of T.V. signals can be increased.
Answer:
The range of TV transmission can be increased by using

• tall antenna and
• geostationary satellites.

Question 12.
Compare the difference between PM and FM.
Answer:

Question 13.
Why should transmitters broadcasting programmes use different carrier frequencies?
Answer:

1. Different audio signals fall in the same spectral range.
2. Different transmitting stations are allowed different slots in radio frequency range.
3. A single receiver can tune into these frequencies without any confusion or overlap.

Question 14.
Give one example each of a system that uses the (i) sky – wave (ii) space wave mode of propagation.
Answer:

1. Short broadcast services use skywave propagation.
2. TV broadcast, microwaves links, satellite communication use space wave propagation.

Question 15.
A transmitting antenna has a height of 40m and the height of the receiving antenna is 30m. What is the maximum distance between them for line-of-sight communication?
[The radius of the earth is 6.4 × 10⁶ m]
Solution:
The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage,
d = d₁ + d₂
= \(\sqrt{2 R}\left(\sqrt{h_{1}}+\sqrt{h_{2}}\right)\)
= \(\sqrt{2 \times 6.4 \times 10^{6}}\)
= 16 × 10²√5 × (6.32+ 5.48)
= 42217m
d = 42.217 km

XII. Five Mark Questions:

Question 1.
Explain mobile communication? Write its applications.
Answer:
Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables. It allows the transmission over a wide range of area without the use of the physical link. It enables people to communicate with each other regardless of a particular location like office, house, etc. It also provides communication access to remote areas.

It provides the facility of roaming:
that is. the user may move from one place to another without the need of compromising on the communication. The maintenance and cost of installation of this communication network are also cheap.

Applications:

1. It is used for personal communication and cellular phones offer voice and data connectivity with high speed.
2. Transmission of news across the globe is done within a few seconds.
3. Using the Internet of Things (IoT), it is made possible to control various devices from a single device.
   Example: home automation using a mobile phone.
4. It enables smart classrooms, online availability of notes, monitoring student activities etc. in the field of education.

Question 2.
Give a brief account on RADAR and its applications.
Answer:

1. RADAR stands for Radio Detection and Ranging system.
2. The angle, range or velocity of the objects which are invisible to the human eye can be determined.
3. Electromagnetic waves are used for communication.
4. Signal is initially radiated into space by an antenna in all direction.
5. Then it gets reflected or reradiated in many directions as it strikes the object.
6. Radar antenna receives the reflected radio signal and it is delivered to the receiver.
7. These signals are processed and amplified to determine the geographical statistics of the object.
8. the range is determined by calculating the time taken by the signal to travel from RADAR to the target and back.

Applications:
RADAR is used

1. to sense, detect and distant objects like aircraft, ships, spacecraft etc.
2. in military for locating and detecting the targets.
3. in navigation systems such as ship-borne surface search, air search and weapons guidance system.
4. to measure precipitation rate and wind speed in meteorological observation
5. to locate and rescue people in emergency situations.

Question 3.
What do you know about mobile communication? Give its applications.
Answer:

1. Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables.
2. It allows the transmission over a wide range of areas without the use of the physical link.
3. It also provides communication access to remote areas.
4. It provides the facility of roaming.
5. User may move from one place to another without the need of compromising on the communication.
6. The maintenance and cost of installation are cheap.

Applications:

1. It is used to communicate with each other regardless of a particular location like office, house etc.
2. It is used for personal communication.
3. Cellular phones offer voice and data connectivity with high speed.
4. Transmission of news across the globe is done within a few seconds.
5. IoT is used to control various devices from a single device
   Ex. home automation using a mobile phone.
6. It enables smart classrooms, online availability of notes, monitoring student activities etc in the field of education.

Question 4.
What is bandwidth? Explain the bandwidth of the transmission system.
Answer:

1. The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc is transmitted.
2. Each of these signals has different frequencies. The communication system depends on the nature of the frequency band for a given signal.
3. Bandwidth gives the difference between the upper and lower frequency limits df the signal.
4. The portion of the electromagnetic spectrum occupied by the signal.
5. If γ₁ and γ₂ are the lower and upper-frequency limits of a signal.
6. Bandwidth, BW = γ₂ – γ₁

Bandwidth of Transmission system:

1. The range of frequencies required to transmit a piece of specified information in a particular channel.
2. An amplitude modulation system requires a channel bandwidth of 10 kHz to transmit a 5 kHz signal.
3. A single side-band system requires only 5 kHz channel bandwidth for a 5 kHz signal.
4. Because in amplitude modulation, the channel bandwidth is twice the signal frequency.
5. It is required to reduce the channel bandwidth to accommodate more channels in the available electromagnetic spectrum.

Question 5.
Explain the process of modulation.
Answer:

1. For long-distance transmission, the low-frequency baseband signal is superimposed onto a high-frequency radio signal.
2. The energy of the information signal is sufficient to send directly.
3. A very high-frequency signal called carrier signal is used to carry the baseband signal.
4. As the frequency of the carrier signal is very high, it can be transmitted to long distances with less attenuation.
5. The carrier wave is a sine wave signal.
6. The carrier signal will be more compatible.
7. It propagates in free space with greater efficiency.
8. A sine wave can be represented as
   e_c = E_c sin (2πγ_ct + φ)
   Where, E_c – amplitude γ_c – frequency
   φ – the initial phase of the carrier wave at time ‘t’
9. The characteristics of the carrier signal are modified in three types.
   • Amplitude modulation
   • Frequency modulation
   • Phase modulation.

Question 6.
What do you know about INTERNET? Write its few applications?
Answer:
Internet is a fast-growing technology in the field of a communication system with multifaceted tools. It provides new ways and means to interact and connect with people. Internet is the largest computer network recognized globally that connects millions of people through computers. It finds extensive applications in all walks of life.

Applications:

1. Search engine: The search engine is basically a web-based service tool used to search for information on World Wide Web.
2. Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services, and social networking tools.
3. E-Commerce: Buying and selling of goods and services, transfer of funds are done over an electronic network.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 9 Semiconductor Electronics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 9 Semiconductor Electronics

12th Physics Guide Semiconductor Electronics Text Book Back Questions and Answers

Part I:

Textbook Evaluation:

I. Multiple choice questions:

Question 1.
The barrier potential of a silicon diode is approximately
a) 0.7 V
b) 0.3V
c) 2.0 V
d) 2.2V
Answer:
a) 0.7 V

Question 2.
Doping a semiconductor results in
a) The decrease in mobile charge carriers
b) The change in chemical properties
c) The change in the crystal structure
d) The breaking of the covalent bond
Answer:
c) The change in the crystal structure

Question 3.
A forward-biased diode is treated as
a) An open switch with infinite resistance
b) A closed switch with a voltage drop of 0V
c) A closed switch in series with a battery voltage of 0.7V
d) A closed switch in series with a small resistance and a battery.
Answer:
d) A closed switch in series with a small resistance and a battery.

Question 4.
If a half -wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
a) 0° – 90°
b) 90° – 180°
c) 0° – 180°
d) 0° – 360°
Answer:
c) 0°-180°

Question 5.
The primary use of a zener diode is
a) Rectifier
b) Amplifier
c) Oscillator
d) Voltage regulator
Answer:
d) Voltage regulator

Question 6.
The principle in which a solar cell operates
a) Diffusion
b) Recombination
c) Photovoltaic action
d) Carrier flow
Answer:
c) Photovoltaic action

Question 7.
The light emitted in an LED is due to
a) Recombination of charge carriers
b) Reflection of light due to lens action
c) Amplification of light falling at the junction
d) Large current capacity.
Answer:
a) Recombination of charge carriers

Question 8.
When a transistor is fully switched on, it is said to be
a) Shorted
b) Saturated
c) Cut-off
d) Open
Answer:
b) Saturated

Question 9.
The specific characteristic of a common emitter amplifier is
a) High input resistance
b) Low power gain
c) Signal phase reversal
d) Low current gain
Answer:
c) Signal phase reversal

Question 10.
To obtain sustained oscillation in an oscillator,
a) Feedback should be positive
b) Feedback factor must be unity
c) Phase shift must be 0 or 2π
d) All the above
Answer:
d) All the above

Question 11.
If the input to the NOT gate is A = 1011, its output is
a) 0100
b) 1000
c) 1100
d) 0011
Answer:
a) 0100

Question 12.
The electrical series circuit in digital form is
a) AND
b) OR
c) NOR
d) NAND
Answer:
a) AND

Question 13.
Which one of the following represents a forward bias diode? (NEET)

Answer:
a) A node must have a high potential

Question 14.
The given electrical network is equivalent to

a) AND gate
b) OR gate
c) NOR gate
d) NOT gate
Answer:
c) NOR gate

Question 15.
The output of the following circuit is 1 when the input ABC is

a) 101
b) 100
c) 110
d) 010
Answer:
a) 101

II. Short Answer Questions:

Question 1.
Define electron motion in a semiconductor.
Answer:
To move the hole in a given direction, the valence electrons move in the opposite direction. Electron flow in an N-type semiconductor is similar to electrons moving in a metallic wire. The N-type dopant atoms will yield electrons available for conduction.

Question 2.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Question 3.
What do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

Question 4.
How electron-hole pairs are created in a semiconductor material?
Answer:
When external voltage is applied in semiconductor free electrons from valence band moves to conduction band and recombined with holes. So electron-hole pairs are created.

Question 5.
A diode is called a unidirectional device. Explain
Answer:
A diode is called a unidirectional device, i.e., current flows in only one direction (anode to cathode internally) when a forward voltage is applied, the diode conducts and when a reverse voltage is applied, there is no conduction. A mechanical analogy is a rat chat, which allows motion in one direction only.

Question 6.
What do you mean by leakage current in a diode?
Answer:

1. Under biasing very small current (μA) flows across the junction in a diode.
2. It is due to minority charge carriers.
3. This current is called leakage current or reverse saturation current.

Question 7.
Draw the output waveform of a full-wave rectifier.
Answer:

Input and output waveforms

Question 8.
Distinguish between avalanche and Zener breakdown.
Answer:

Question 9.
Discuss the biasing polarities in NPN and PNP transistors.
Answer:
In a PNP transistor, base and collector will be negative with respect to emitter indicated by the middle letter N whereas base and collector will be positive in an NPN transistor [indicated by the middle letter P.

Question 10.
Explain the current flow in an NPN transistor.
Answer:

1. In NPN transistor electron flow from Emitter to collector. So conventional current flow from collector to emitter.
2. Electrons from emitter region flow towards base region constitute emitter current (I_E). Electrons after reaching base region recombine with holes – Most of electrons reach collector region. This constitute collector current (I_c). After recombination of holes in base region by bias voltage V_EE constitute base current I_B.
   I_E = I_B + I_C [Since base current is very low I_c ≈ I_C]

Question 11.
What is the phase relationship between the AC input and output voltages in a common emitter amplifier? What is the reason for the phase reversal?
Answer:
In a common emitter amplifier, the input and output voltages are 180° out of phase or in, opposite phases. The reason for this can be seen from the fact that as the input voltage rises, so the current increases through the base circuit.

Question 12.
Explain the need for a feedback circuit in a transistor oscillator.
Answer:
If the portion of the output fed to the input is in phase with the input, then the magnitude of the input increases. It is necessary for sustained oscillations, so feed back circuit is needed for transistor oscillator.

Question 13.
Give circuit symbol, logical operation, truth table, and Boolean expression of AND, OR, NOT, NAND, NOR, and EX-OR gates.
Answer:

Question 14.
State De Morgan’s first and second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Second Theorem:
The complement of the product is equal to sum of its complements
\(\overline{A \cdot B}=\bar{A}+\bar{B}\)

III. Long Answer Questions:

Question 1.
Elucidate the formation of a N – type and P – type semiconductors.
Answer:
1. N – type semiconductor:
N – type semiconductor is obtained by doping a pure Germanium or silicon with dopant from group V pentavalent elements like phosphorous, Arsenic, Antimony.

n-type extrinsic semiconductor:
(a) Free electron which is loosely attached to the lattice
(b) Representation of donar energy level.

1. Dopants has 5 valence electrons, Germanium has 4 valence electrons.
2. During the process of doping few Ge atoms’ are replaced by group V dopants.
3. Four of the five valence electrons of impurities bound with 4 valence electrons of neighbouring Ge atoms.
4. The 5th valence electron of impurity atom is loosely attached, it has not formed covalent bond.
5. Energy level loosely attached electron is just below the conduction band edge which is called donor energy level.
6. At room temperature electrons easily moves to conduction band with absorption of thermal energy.
7. These thermally generated electrons leave holes in valance band.
8. Such a semiconductor doped with a pentavalent impurity is called n-type semiconductor.

2. P-type semiconductor.
1. Trivalent atom from group III elements such as Boron, Aluminium, Gallium, Indium is added with Germanium or silicon is a p-type semiconductor.
2. The dopants with 3 valence electrons bound with neighbouring Germanium atom one electron position of the dopant in Ge lattice will remain vacant.
3. The missing electron position in covalent band is denoted as a hole.
4. To make complete covalent bond with all 4 neighbouring atom, dopant needs one more electrons.
5. Dopants accept electron from neighbouring atoms. This impurity is called acceptor impurity.
6. Energy level of hole created by each impurity atom is just above valence band, which is called acceptor energy level.
7. In such extrinsic semiconductor holes are majority carriers and thermally generated electrons are minority carriers. This semiconductor is called a p-type semiconductor.

P-type extrinsic semiconductor
(a) Hole generated by the dopant
(b) Representation of acceptor energy level.

Question 2.
Explain the formation of PN junction diode. Discuss its V-I characteristics.
Answer:

p-n junction diode
(a) Schematic representation
(b) Circuit symbol

i. A p-n junction diode is formed when p-type semiconductor is fused with N-type semiconductor.
ii. Depending on polarity of external source to the p-n junction there are two types of biasing
1. Forward bias
2. Reverse bias

1. Forward bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side forward bias takes place.
ii. Electron moves to n-side holes move to p side Recombination takes place near junction and reduce depletion region.
iii. Electron from n-side accelerate towards p side it experience reduced potential barrier at junction.
iv. Applied voltage is increased, width of depletion region and barrier potential further reduced.
v. So large number of electrons pass through junction.

2. Reverse bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side reverse bias takes place.

ii. Depletion region is increased potential barrier is also increased.
iii. Majority charge carriers from both sides experience a great barrier to cross the junction. So diffusion current reduces.
iv. The current flows under reverse bias is called reverse saturation current I_S

V-I Characteristics:

1. Forward characteristics:
i. A graph is plotted by taking forward bias voltage in X axis and current i Y axis.
ii. Current flow is negligible when applied voltage is less than threshold voltage beyond that increase in current is significant even for small increase in voltage.
iii. Graph shows clearly current flow is non linear. It doesnot obey ohm’s law.
iv. Forward resistance r_f = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\)
v. Diode behaves as conductor when it is forward bias.

p-n junction diode
(a) diode under forward bias
(b) forward characteristics

2. Reverse characteristics:
i. In reverse bias p-region is connected to negative terminal and n-region to positive terminal, Graph is drawn for reverse bias characteristics.
ii. Very small current in pA flows across junction. This is due to leakage current or reverse saturation current.
iii. The reverse bias voltage can be increased only up to rated value otherwise diode will enter into breakdown region.

p-n junction diode
(a) diode under reverse bias
(b) reverse characteristics

Forward and reverse characteristics of a diode

Question 3.
Draw the circuit diagram of a half wave rectifier and explain its working.
Answer:
i. The circuit consist of transformer, p-n junction diode and resistor.
ii. In half wave rectifier either a positive half or the negative half of AC input is passed through it, other half is blocked.

(a) Input signal
(b) half wave rectifier circuit
(c) input and output waveforms

During positive half cycle:
1. When positive half cycle of input signal passes. A becomes positive with respect to B.
2. Diode is forward bias, current flows through R_L output voltage V₀ is developed.
3. Waveform is shown in figure.

During negative half cycle:
1. When negative half cycle of input signal passes A is negative with respect to B.
2. Diode is reverse bias, does not conduct No current is passed through R_L. No voltage drop across R_L.
3. Negative half cycle of ac supply is suppressed at output waveform is shown in figure.
4. Efficiency of half wave rectifier is 40.67%.

Question 4.
Explain the construction and working of a full wave rectifier.
Answer:
i. Full wave rectifier consist two p-n junction diodes a center tapped transformer and load resistor (R_L)
ii. Due to centre tap transformer output voltage rectified by each diode is only one half of total secondary voltage.

(a) Full wave rectifier circuit
(b) Input and output waveforms

During positive half cycle:
1. Input signal passes through the circuit M is positive, G is zero, N is negative potential.
2. D₁ forward bias, D₂ reverse bias.
3. Current flows through MD₁ AGC
4. Positive half cycle of voltage appears across R_L in direction G to C.

During negative half cycle:
1. Terminal N is positive, G is at zero, M is negative potential.
2. D₂ is forward bias, D₁ is reverse bias.
3. Current flows through ND₂BGC
4. Negative half cycle of voltage across R_L from G to C.
5. Hence in full wave rectifiers both postive and negative half cycle of input signals pass through the circuit in same direction.
6. Efficiency of full wave rectifier is 31.2%.

Question 5.
What is an LED? Give the principle of operation with a diagrajm?
Answer:
1. LED is a p-n junction diode which emit visible or invisible light when it is forward bias.
2. It converts electrical energy to light energy.
3. It consists of p-layer, n-layer and substrate. External resistance in series with biasing source is required to limit forward current through LED. It has anode and cathode.

(a) Circuit symbol of LED
(b) Inside view of LED
(c) Schematic diagram to explain recombination process

4. When p-n junction is forward bias, conduction band electron on n-side and valence band hole on p side diffuse across the junction.
5. When they cross junction, they become excess minortiy carriers. These excess minority carriers recombine with opposite charged majority carriers in respective region.
6. During recombination process energy is released is the form of light or heat.
7. The colour of the light is determined by energy band gap of the material.
8. LED’s with wide range colours such as blue (Sic), green (AlGaP), red(GaAsP), white(GaInN) is also available.

Question 6.
Write notes on Photodiode.
Answer:
                     
(a) Circuit symbol
(b) Schematic view of photodiode

1. Photodiode is a p-n junction diode which converts an optical signal into electric current.
2. It works in reverse bias.
3. It consists of F p-n junction semiconductor made of photosensitive material kept safely inside a plastic case.
4. It has small transparent window to allow light.
5. When photon (hυ) strikes depletion region of diode some valence band electrons elevated to conduction band. In turns holes are developed in valence band. This creates electron hole pair.
6. Amount of electron – hole pair generated depends on intensity of light.
7. These electrons and hole swept across the junction by electric field. Thus holes moves to -n side, electrons to p side.
8. When external circuit is made, electrons flow through external circuit and constitute the photo current.

Applications:
Alarm system, count items in conveyer belt, photo conductors, CD players, smoke detectors, detectors for computed tomography etc.

Question 7.
Explain the working principle of a solar cell. Mention its applications.
Answer:
11. Solar cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Cross-sectional view of a solar cell

2. It generates emf when radiations falls on p-n junction. A solar cell is of two type p-type and n-type.
3. Both types use a combinations of p-type and n-type silicon which together forms the p-n junction.
4. In solar cell, electron – hole pairs are generated due to absorption of light near junction.
5. Electrons move towards n-type silicon, holes moves towards p-type silicon layer.
6. Electrons reaching n-side are collected by front contact and holes reaching p-side are collected by back electrical contact.
7. Thus potential difference is developed across solar cell. When external load is connected, photocurrent flows through it.
8. Many solar cells are connected in series or parallel to form solar panel or module.

Applications:
1. Widely used in calculators, watches, toys, portable power supplies etc.
2. Used in satellites and space stations.
3. Solar panels are used to generate electricity.

Question 8.
Sketch the static characteristics of a common emitter transistor and bring out the essence of input and output characteristics.
Answer:

Static characteristics of a NPN transistor in common emitter configuration

i. The circuit to study the static characteristics of NPN transistor is given in figure
ii. Bias supply voltages V_BB and V_CC bias, base – emitter junction and collector – emitter junction
iii. Junction potentials are V_BE and Y_CE
iv. R₁ and R₂ are used to vary base and collector currents respectively.

1. Input Characteristics:
i. Input Characteristics curve gives relationship between I_B and V_BE at constant V_CE
ii. For constant collector emitter voltage V_CE, Base emitter voltage V_BE increases for corresponding Base current I_B which is recorded and graph is ploted.
iii. The curve looks like forward characteristics

Input characteristics of a NPN transistor in common emitter configuration

iv. Beyond knee voltage base current increases with increase in base emitter voltage for silicon 0.7 V & for Germanium 0.3 V
v. Increase is V_CE decreases I_B. This shift the curves outward.
vi. Input resistance R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C B}}\)

2. Output characteristics:

Output characteristics of a NPN transistor in common emitter configuration

i. The output characteristics gives the relation between ∆I_C with respect to ∆V_CE at constant I_B
ii. Initially base current I_B is set to particular value. Increasing collector emitter voltage V_CE corresponding collector current I_C increases. A graph is plotted.

iii. Saturation region:
When V_CE increased above 0V, I_C increases rapidly almost independent of I_B called knee voltage Transistor operated above this knee voltage

iv. Cut off region:
A small I_C exist even after I_B is reduced to zero. This current is due to presence of minority carriers across collector – base junction and the surface leakage current (I_CEO). This region is called cut-off region.

v. Active region:
In this region emitter – base junction is forward bias, collector – base junction is reverse bias. Transistor in this region can be used for voltage, current and power amplification.

vi. Breakdown region:
If V_CE is increased beyond rated value, given I_c increases enormously leading to junction breakdown of transistor. This avalanche breakdown can damages the transistor.
vii. Output resistance R_o = \(\left(\frac{\Delta V_{C E}}{\Delta I_{C}}\right)_{I_{B}}\)

Question 9.
Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.
Answer:
i. Amplification is the process of increasing the signal strength (increase in the amplitude)
ii. NPN transistor is connected in CE configuration
iii. To start with, Q point is fixed to get maximum signal swing at the output.
iv. R_C – to measure output voltage
C₁ allows only AC signals to pass,
Bypass capacitor C_E provides a low resistance path
Coupling capacitor C_C is used to couples next stage amplifier.
V_s input source signal
I_C = BI_B (∴ B = \(\frac{I_{C}}{I_{B}}\))
V_CE = V_CC – I_CR_C

                       
(a)Transistor as an amplifier
(b) Input and output waveform showing 180° phase reversal.

Working of the amplifier:

During positive half cycle:
1. Input signal V_s increases the forward voltage across emitter base. I_B increases. also increases I_C also increases β times.
2. This increase the voltage drop across R_c which decreaseV_CE.
3. Therefore input signal in the positive direction produces an amplified signal in negative direction at the output. Hence output signal is reversed by 180°

During negative half cycle:
1. Input signal V_s decreases the forwarded voltage across emitter – base. As a result I_B decreases, I_C increases.
2. Increase in I_C decreases potential drop across R_c and increases V_CE.
3. Input signal in negative direction produces amplified signal in the positive direction at the output.
4. Therefore 180° phase reversal is observed during negative half signal.

Question 10.
Transistor functions as a switch. Explain.
Answer:
1. In saturation and cut-off region transistor works as an electronic switch that helps to turn ON or OFF a circuit.
2. presence of dc source at the input (saturation region):

Transistor as a switch

3. When high input voltage (V_in = +5V) is applied, I_B increases and in turn increases I_C.
4. Transistor will move into saturation region (turned ON)
5. Increase in I_C increases voltage drop across R_c lowering the output voltage close to zero.
6. Transistor acts like a closed switch (ON condition)

Absence of dc source at the input (cut-off region) :
1. At low input voltage (V_in = 0V), decreases the I_B and in turn decreases I_C.
2. Transistor moves to cut off region (turned ON)
3. Decrease in I_C decreases voltage drop across R_c, increasing output voltage +5V transistor acts as open switch (OFF condition)
4. It is manifested that high input gives low output and low input gives high output.

Question 11.
State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
1. Complement law:
\(\overline{\mathrm{A}}\) = A
A Y = \(\overline{\mathrm{A}}\)
O Y = \(\overline{\mathrm{O}}\) = 1
I Y = \(\overline{\mathrm{I}}\) = \(\overline{\mathrm{O}}\)

2. OR laws:

3. AND laws:

4. Commutative laws:
A + B = B + A
A.B = B.A

5. Associative laws
A + (B + C) = (A + B) + C
A.(B.C) = (A.B).C

6. Distributive laws
A (B + C) = AB + AC
A + BC = (A + B) (A + C)

Question 12.
State and prove De Morgan’s First and Second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{A+B}=\bar{A} \cdot \bar{B}\)

Second Theorem:
The complement of the product of two inputs is equal to the sum of its complements.
\(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

IV. Numerical Problems:

Question 1.
The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R₁

Solution:
D₁ will act as Reverse bias so current will not pass through it
Total Resistance = 4Ω
V = 10 V
Current flow through R₁I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{10}{4}\) = 2.5 A

Question 2.
Four silicon diodes and a 10Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 18Ω resistor.

Solution:
Current flow through D₁ and D₃ = 2Ω
Current flow through D₂ and D₄ = 2Ω
These two are in parallel so
Net resistants = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1Ω
Total Resistance = 1Ω+ 18Ω = 19Ω
V = 2.5Ω
I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{2.5}{19}\)
I = 0.13 A

Question 3.
Assuming V_CESat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.

Solution:
V_CC = 3V
β = 50
I_B =?
Collector current
V_CC = 3 – 0.2 = 2.8 V
I_C = \(\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{c}}}\)
= \(\frac{2.8}{1 \times 10^{3}}\) = 2.8 × 10^-3 A

β = \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{B}}}\)
I_B = \(\frac{\mathrm{I}_{\mathrm{c}}}{\beta}\) = \(\frac{2.8 \times 10^{-3}}{50}\)
= 56 × 10^-6 A
I_B = 56 µA

Question 4.
A transistor having α = 0.99 and V_BE = 0.7 V, is given in the circuit. Find the value of the collector current.

Solution:
V_cc = 12 V
α = 0.99
V_BE = 0.7 V
I_C =?
Applying Kirchoff’s voltage law,
V_BE + (I_C + I_B) 1k + 10k.I_B + (I_C + I_B) 1k = 12 ………………(1)
I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{\beta}\)

β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.99}{1-0.99}\) = 0.99

I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)

I_C = β I_B
I_C = 99 I_B
I_C = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)
V_BE = 0.7 V
Substitute in equation (1)
\(\frac{0.7}{10^{3}}\) + (99 I_B + I_B) + 10 I_B + (99 I_B + I_B) = \(\frac{12}{10^{3}}\)
100 I_B + 10 I_B + 100 I_B = \(\frac{12-0.7}{10^{3}}\)
210 I_B = \(\frac{11.3}{10^{3}}\) ;
I_B = 0.054 × 10^-3 A
I_C = 99 I_B = 99 × 0.054 × 10^-3 A
I_C = 5.34 × 10^-3 A
I_C = 5.34 mA

Question 5.
In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter-base voltage V_EB = 600 mV, calculate the emitter-collector voltage V_EC (in volts)

Solution:
β = 50
V_EB = 600 × 10^-3V
R_B = 60 kΩ
R_C = 500 kΩ
V_B = V_E – V_EB
= 3 – 0.6 = 2.4 V
I_B = \(\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}\)
= \(\frac{2.4}{60 \times 10^{3}}\)
= 0.04 × 10^-3 A
I_B = 40 µA
I_C = β I_B
= 50 × 40 × 10^-6
I_C = 2mA
V_C = I_C R_C = 500I_C
= 500 × 2 × 10^-3
V_EC = V_E – V_C
= 3 – 1 = 2V

Part II:

12th Physics Guide Semiconductor Electronics Additional Questions and Answers

I. Choose the correct answer:

Question 1.
A transistor has α = 0.95, it has change in emitter current of 100 milliampere then the change in collector current is
a) 95 mA
b) 99.05 mA
c) 100.95 mA
d) 100 mA
Answer:
a) 95 mA
Solution:
α = \(\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{e}}}\)
∆I_C = ∆I_e × α
= 100 × 0.95
= 95 mA

Question 2.
Diode can works as
a) Demodulator
b) Modulator
c) Amplifier
d) Rectifier
Answer:
a) Demodulator

Question 3.
Three amplifiers of each with gain 10 volt are connected in series then the overall amplification is ______.
a) 10/3
b) 13
c) 1000
d) none of these
Answer:
c) 1000
Solution:
10 × 10 × 10 = 1000

Question 4.
A Transistor has an α = 0.95 then β is
a) 1/19
b) 19
c) 1.5
d) 0.95
Answer:
b) 19
Solution:
β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.95}{1-0.95}\)

Question 5.
In p – type semiconductor germanium is doped with _______.
a) aluminium
b) boron
c) gallium
d) all of these
Answer:
d) all of these

Question 6.
For an npn transistor, the collector current is 10 mA. If 90% of electron emitted reach the collector then
a) base current will be 1 mA
b) base current will be 10 mA
c) emitter current will be 11 mA
d) emitter current will be 9 mA
Answer:
a) & c)
Solution:
0.9 I_e10 = I_C
I_e = \(\frac{10}{0.9}\) = 11 mA
I_B = 11 – 10 = 1 mA

Question 7.
The depletion layer in p-n junction region is caused by _______.
a) drift of holes
b) diffusion of charge carriers
c) drift
d) drift of electrons
Answer:
b) diffusion of charge carriers

Question 8.
When – n – type semiconductor is heated ______.
a) number of electrons only increase
b) number of holes increase
c) number of electrons and holes remains same
d) number of electrons and holes increase equally
Answer:
d) number of electrons and holes increase equally

Question 9.
The device that can act as complete electronic circuit is _______.
a) junction diode
b) IC
c) transistor
d) diode
Answer:
b) IC

Question 10.
In following circuit the output Y for all possible input A and B is expressed by truth table

a)

b)

c)

d)

Answer:
c)

Y’ = \(\overline{A+B}\)
Y = \(\overline{Y^{\prime}}\) = \(\overline{\overline{A+B}}\)
Y = A + B

Question 11.
Which one is correct for reverse bias is applied in junction diode
a) Increase minority career
b) lower potential barrier
c) raise in potential barrier
d) increase in majority carrier
Answer:
c) raise in potential barrier

Question 12.
In common emitter configuration V_c = 1.5 V_c change in base current from 100 µA to 150 µA produce change in collector current from 5 mA to 10 mA. The current gain (β) is
a) 75
b) 100
c) 50
d) 67
Answer:
b) 100
Solution:
∆I_B = 50 µA
∆I_C = 5 mA
β = \(\frac{\Delta \mathrm{I}_{\mathrm{B}}}{\Delta \mathrm{I}_{C}}\)
= \(\frac{5 \mathrm{~mA}}{50 \mu \mathrm{A}}\)
= 100

Question 13.
To use a transistor as an amplifier
a) the emitter base junction is forward biased and base collector junction is reverse biased
b) no bias voltage is required
c) both junction are forward bias
d) both junctions are reverse bias
Answer:
a) the emitter base junction is forward biased and base collector junction is reverse biased

Question 14.
The circuit shown in figure contains two diodes each with forward resistance of 50Ω and with infinite backward resistance. If battery voltage is 6V the current through 100Ω resistance is

a) Zero
b) 0.02 mA
c) 0.03 mA
d) 0.36 mA
Answer:
b) 0.02 mA
Solution:
Current will not pass through D₂
Total resisatnce = 50 + 150 + 100 = 300Ω
Current = \(\frac{6}{300}\) = 0.02 V

Question 15.
Type of material which emits white light in LED
a) GaInN
b) Sic
c) AlGaP
d) GaInP
Answer:
a) GalnN

Question 16.
Blue colour LED is made up of
a) Sic
b) AIGaP
c) GaAsP
d) GaInP
Answer:
a) Sic

Question 17.
The energy band gap is maximum in
a) metals
b) super conductors
c) insulators
d) semiconductors
Answer:
c) insulators

Question 18.
The part of transistor which is mostly heavily doped to produce large majority carriers
a) emitter
b) base
c) collector
d) None of these
Answer:
a) emitter

Question 19.
In the middle of depletion layer of reverse – biased p-n junction
a) electric field is zero
b) potential is maximum
c) electric field is maximum
d) potential is zero
Answer:
c) electric field is maximum

Question 20.
What is the current flowing in the circuit?

a) 1.71 A
b) 2.0 A
c) 2.31 A
d) 1.33 A
Answer:
b) 2.0 A
Solution:
D₁ is reverse bias will not conduct = \(\frac{12}{6}\) = 2 A

Question 21.
A diode as a rectifier converts _______.
a) a.c to d.c
b) d.c to a.c
c) AC only
d) dc only
Answer:
a) ac to d.c

Question 22.
An oscillator is nothing but an amplifier with
a) positive feedback
b) large gain
c) no feed back
d) negative feedback
Answer:
a) positive feedback

Question 23.
In Intrinsic semiconductor at room temperature number of electrons and holes are
a) equal
b) zero
c) unequal
d) infinite
Answer:
a) equal

Question 24.
The level formed due to impurity atom in forbidden energy gap, very near to valence band is p-type semiconductor is called
a) An acceptor level
b) A donor level
c) Conduction level
d) A forbidden level
Answer:
a) An acceptor level

Question 25.
The symbol represents _______ gate.

a) NAND gate
b) OR gate y
c) AND gate
d) NOT gate
Answer:
d) NOT gate

Question 26.
In p-n junction, avalanche current flows is circuit when biasing
a) Forward
b) Reverse
c) Zero
d) Excess
Answer:
b) Reverse

Question 27.
For detecting light, which is correct?
a) Photodiode has to be forward biased
b) Photodiode has to be reversed biased
c) LED has to be connected in forwarded bias
d) LED in reverse bias
Answer:
b) Photodiode has to be reversed biased

Question 28.
What is the output Y in the above circuit, when all the three inputs A, B and C are 0?

a) 0
b) 1
c) 10
d) 11
Answer:
a) 0

Question 29.
Which type of semiconductor device does not need any bias voltage
a) Photodiode
b) Zero diode
c) Solar cell
d) Transistor
Answer:
c) Solar cell

Question 30.
The frequency of oscillator is_____ 2 1
a) f = \(\frac{1}{2 \pi \mathrm{LC}}\)

b) ω² = \(\frac{1}{\mathrm{LC}}\)

c) ω = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

d) f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
Answer:
d) \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

Question 31.
The logic circuit shown in figure represents ______ gate.

a) OR
b) AND
c) NOR
d) NAND
Answer:
d) NAND

Question 32.
Maximum efficiency of full wave rectifier
a) 100%
b) 81.2%
c) 40.6%
d) 95%
Answer:
b) 81.2%

Question 33.
The diffusion current in P-N junction is from
a) p side to n side
b) n side top side
c) Both (a) and (b)
d) None
Answer:
a) p side to n side

Question 34.
The current through an ideal PN junction shown in figure

a) 5 mA
b) 10 mA
c) 70 mA
d) 100 mA
Answer:
b) 10 mA
Solution:
V = 5-(-2)
V = 7V
I = \(\frac{7}{100}\)
I = 10 MA

Question 35.
Efficiency of half wave rectifier is
a) 81.2%
b) 100%
c) 40.6%
d) 95%
Answer:
c) 40.6%

Question 36.
In common base amplifier, phase difference between input voltage and output voltage
a) 0
b) π/4
c) π/2
d) π
Answer:
a) 0

Question 37.
What will be the input of A and B for Boolean expression \(\overline{(A+B)}\) \(\overline{(\mathrm{A} \cdot \mathrm{B})}\) = 1
a) 0,0
b) 0,1
c) 1,0
d) 1,1
Answer:
a) 0,0
Solution:
\(\overline{0} \cdot \overline{0}\) = 1
1.1 = 1

Question 38.
The value of current flowing through AB in this circuit

a) 10 mA
b) 20 mA
c) 15 mA
d) 11 mA
Solution:
PD = 3 -(-7) = 10V
I = \(\frac{10}{1000}\)
= 10^-2 A = 10 mA

Question 39.
To get output, 1 for following circuit. The correct choice for input is ________.

a) A = 0, B = 1, C = 0
b) A = 1, B = 0, C = 0
c) A = 1, B = 1, C = 0
d) A = 1, B = 0, C = 1
Answer:
c) A = 1, B = 1, C = 0
Solution:
Y’ = A+B
Y = (A + B) C = 1
(1 + 0).1 = 1

Question 40.
The given electrical network is equal to

a) AND gate
b) OR gate
c) NOR gate
d) Ex-OR gate
Answer:
c) NOR gate
Y₁ = \(\overline{A+B}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Y₂ = \(\overline{\mathrm{Y}}_{1}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) = A + B
Y = \(\overline{A+B}\)

Question 41.
Which gate will give high input when odd numbers of input are high
a) NAND gate
b) OR gate
c) NOR gate
d) EX-OR gate
Answer:
d) EX-OR gate

Question 42.
The current gain (β) of a transistor in common emitter mode is 40. To change the collector current by 160 mA at constant V_CE, The necessary change in the base current is
a) 0.2 mA
b) 4 µA
c) 4 mA
d) 40 mA
Answer:
c) 4 mA

Question 43.
Which one is Barkhausen condition for sustained Oscillation
a) phase shift 0°
b) loop gain is unity
c) \(|A B|\) = 0
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Question 44.
Tank circuit consists of ________.
a) Capacitor
b) Inductor
c) Both (a) and (b)
d) Diode
Answer:
c) Both (a) and (b)

Question 45.
The given electrical network is equivalent to:

(a) NAND gate
(b) OR gate
(c) NOT gate
(d) Ex-OR gate
Answer:
b) OR gate

Question 46.
The maximum reverse bias that can be applied before entering into zener region is referred as
a) PIV rating
b) PAV rating
c) RIM rating
d) None
Answer:
a) PIV rating

Question 47.
For reverse voltage between 4 and 6V ___________ are present
a) Zener effect
b) Avalanche effect
c) Both effects
d) None
Answer:
c) Both effects

Question 48.
Solar cell works on _______ effect.
a) Photo emissive
b) Photo electric
c) Photo conducting
d) Photo voltaic
Answer:
d) Photo voltaic

Question 49.
Which one is the symbol for photodiode
a)

b)

c)

d)
Answer:
c)

II. Choose the Correct Statement:

Question 1.
Which of the statement is not true?
a) The resistance of intrinsic semiconductor decreases with increase of temperature
b)Doping pure Si with trivalent impurities give p-type semiconductor.
c) The majority carriers in n type semiconductors are holes
d) A p-n type junction can act as a semiconductor diode
Answer:
c) The majority carriers in n type semiconductors are holes

Question 2.
Which one is correct statement A — p type semiconductor is
i) A silicon crystal is doped with arsenic impurity
ii) Si doped with aluminium impurity
iii) Ge doped with phosphorous impurity
iv) Ge doped with boron impurity
a) (i) and (ii) are correct
b) (ii) and (iii) are correct
c) (i) and (iv) are correct
d) (i) only correct
Answer:
b) (ii) and (iii) are correct

Question 3.
Read the following statement carefully
Y: The resistivity of semiconductor decreases with increase of temperature
Z: The resistivity value of semiconductor is 10^-5 to 10⁶ Ωm
a) Y is true but Z is false
b) Y is false but Z is true
c) Both Y and Z are true
d) Both Y and Z are false
Answer:
c) Both Y and Z are true

Question 4.
Which one of the following statement is correct?
a) The depletion region of P-N junction diode increase with forward bias
b) The depletion region of P-N junction diode decrease with reverse bias
c) The depletion region of PN junction diode does not change with biasing
d) The depletion region of PN junction diode decreases with forward biasing.
Answer:
d) The depletion region of PN junction diode decreases with forward biasing.

Question 5.
Assertion:
The size, capacity of chips are progressed enormously with advancement of technology
Reason:
Computers, mobile phones are possible in small size and low cost of ICs
a) Assertion and Reason are true and Reason is correct explanation of Assertion
b) Assertion and Reason are true but Reason is not correct explanation of Assertion
c) Assertion is true but Reason is false
d) Assertion is false but Reason is true
Answer:
a) Assertion and Reason are true and Reason is correct explanation of Assertion

III. Fill in the blanks:

Question 1.
The reverse saturation current of silicon diode doubles for every _______ rise in temperature.
Answer:
10°C

Question 2.
The external voltage applied to p-n junction is called _______.
Answer:
bias voltage

Question 3.
A semiconductor has ________ resistance coefficient.
Answer:
negative

Question 4.
The most commonly used semiconductor is ____________.
Answer:
silicon

Question 5.
Find the odd one out. Trivalent atoms used in doping
a) Boron
b) Aluminium
c) Indium
d) Arsenic
Answer:
d) Arsenic

Question 6.
Find the odd one out Pentavalent dopant used in n-type semiconductor
a) Phosphorus
b) Antimony
c) Galium
d) Arsenic
Answer:
c) Galium

IV. Match the following:

Question 1.

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

Question 2.

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

V. Two Mark Questions:

Question 1.
Define Energy band.
Answer:
Band of very large number of closely spaced energy levels in a very small energy range is known as energy band.

Question 2.
What is valence band?
Answer:
The energy band formed due to the valence orbits is called valence band.

Question 3.
What is called conduction band?
Answer:
The energy band formed by unoccupied orbits is called conduction band.

Question 4.
What is forbidden energy gap?
Answer:
The energy gap between valence band and conduction band is called forbidden energy gap.

Question 5.
What are passive and active components?
Answer:
Passive components:
Components that cannot generate power in a circuit.
Active components:
components that can generate power in a circuit.

Question 6.
Give example for Pentavalent elements.
Answer:
Phosphorus, Arsenic and Antimony

Question 7.
Write – the examples of trivalent impurities.
Answer:
Boron, Aluminum, Gallium and Indium

Question 8.
Differntiate Donor impurities, Acceptor impurities.
Answer:

Question 9.
What is barrier potential?
Answer:
The difference in potential across the depletion layer is called barrier potential.

Question 10.
What are extrinsic semiconductors?
Answer:
An extrinsic semiconductor is a semiconductor doped by a specific impurity which is able to deeply modify its electrical properties, making it suitable for electronic applications (diodes, transistors etc.) or optoelectronic applications (light emitters and detectors).

Question 11.
What is called bias voltage?
Answer:
The external voltage applied to the p-n junction is called bias voltage.

Question 12.
What is called Forward bias and reverse bias in p-n junction diode?
Answer:

1. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side is called forward bias.
2. If positive terminal of battery is connected to n-side and negative terminal to p-side. The junction is said
   to be reverse bias.

Question 13.
What is reverse saturation current?
Answer:
The current that flows under a reverse bias is called the reverse saturation current. It is represented as I_s.

Question 14.
Write down the applications of Zener diode.
Answer:
The Zener diode can be used as:

1. Voltage regulators
2. Peak clippers
3. Calibrating voltages
4. Provide fixed reference voltage in a network for biasing
5. Meter protection against damage from accidental application of excessive voltage.

Question 15.
In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B

Answer:
The output at the 1st AND gate : \(\mathrm{A} \overline{\mathrm{B}}\)
The output at the 2nd AND gate : \(\overline{\mathrm{A}} \mathrm{B}\)
The output at the OR gate : Y = \(\mathrm{A.} \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}\)

Question 16.
What are the characteristics of ideal diode?
Answer:

1. It acts like conductor when it is forward biased.
2. When it is reverse bias it acts like an insulator
3. The barrier potential is assumed to be zero and hence it behaves like a resistor.

Question 17.
What are Optoelectronic devices?
Answer:
Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.

Question 18.
Define efficiency of rectifier.
Answer:
Efficiency (η) is the ratio of the output DC power to the ac input power supplied to the circuit.

Question 19.
What is Zener diode. Draw its symbol
Answer:
Zener diode is a reverse biased heavily doped silicon diode. It is specially designed to operate in breakdown region

Question 20.
What is called optoelectronics?
Answer:

1. 1. Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.
2. Optoelectronic devices are LED, photodiode and solar cells.

Question 21.
What is light emitting diode?
Answer:
Light Emitting Diode (LED) is a p-n junction diode which emits visible or invisible light when it is forward biased. Symbol is

Question 22.
What is photodiode? Draw the circuit symbol.
Answer:
A p-n junction – diode which converts an optical signal into electric current is known as photodiode. Circuit symbol is

Question 23.
What is solar cell?
Answer:
A solar cell is photo voltaic cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Question 21.
Write down the applications of Oscillators.
Answer:
Applications of oscillators:

• to generate periodic sinusoidal or nonsinusoidal waveforms.
• to generate RF carriers.
• to generate audio tones
• to generate clock signals in digital circuits.
• as sweep circuits in TV sets and CRO.

Question 25.
What is Input resistance?
Answer:
The ratio of the change in base emitter voltage (∆V_BE) to change in base current (∆I_B) at constant collector emitter voltage (V_CE) is called input resistance (R_i)
R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C E}}\)

Question 26.
What is output resistance?
Answer:
The ratio of change in collector – emitter voltage (∆V_CE) to the corresponding change is collector current (∆I_C) at constant base current.
R₀ = \(\left(\frac{\Delta \mathrm{V}_{\mathrm{CE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right)_{\mathrm{I}_{\mathrm{B}}}\)

Question 27.
What is forward current gain in common emifter mode?
Answer:
The ratio of change in collector current (∆I_C) to change in base current (I_B) at constant collector – emitter voltage (V_CE) is called forward current gain (B)
B = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\)
The value is very high and range from 50 to 200.

Question 28.
What is amplification?
Answer:

1.  Amplification is the process of increasing the signal strength (increase the amplitude)
2. If large amplification is required, the transistors are cascaded.

Question 29.
Write the Barkhausen conditions for sustained oscillations?
Answer:

1. The loop phase shift is 0° or integral multiple of 2n.
2. The loop again must be unity \(|A B|\) = 1
   A – Voltage gain,
   B – Feedback ratio.

Question 30.
Draw the circuit diagrams of transistor in CB and CC modes.
Answer:

CB Mode	CC Mode

Question 31.
Differentiate damped and undamped Oscillations.
Answer: