Category: Class 11

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 3 Kingdom Plantae Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 3 Kingdom Plantae

1. Choose the correct option.

Question (A)
Which is the dominant phase in Pteridophytes?
(a) Capsule
(b) Gametophyte
(c) Sporophyte
(d) Embryo
Answer:
(c) Sporophyte

Question (B)
The tallest living gymnosperm among the following is
(a) Sequoia sempervirens
(b) Taxodium mucronatum
(c) Zamia pygmaea
(d) Ginkgo biloba
Answer:
(a) Sequoia sempervirens

Question (C)
In Bryophytes
(a) sporophyte and gametophyte generation are independent
(b) sporophyte is partially dependent upon gametophyte
(c) gametophyte is dependent upon sporophyte
(d) inconspicuous gametophyte
Answer:
(b) sporophyte is partially dependent upon gametophyte

Question (D)
A characteristic of Angiosperm is
(a) Collateral vascular bundles
(b) Radial vascular bundles
(c) Seed formation
(d) Double fertilization
Answer:
(d) Double fertilization

Question (E)
Angiosperms differ from gymnosperms in having
(a) Vessels in wood
(b) Mode of nutrition
(c) Siphonogamy
(d) Enclosed seed
Answer:
Both (a) Monocotyledons and (d) Enclosed seed

Question 2.
How you place the pea, jowar and fern at its proper systematic position? Draw a flow chart.
Answer:

Question 3.
Complete the following table.

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch
2. Cell wall	—	Cellulose and algin
3. Major pigments	—		Chl-a, d and phycoerythrin

Answer:

Groups of algae	Chlorophyceae	Phaeophyccac	Rhodophyceae
1. Stored food	Starch	Mannitol, laminarin	Floridean starch
2. Cell wall	Cellulose	Cellulose and algin	Cellulose, pectin
3. Major pigments	Chl-a, b	Chl-a, c, fucoxanthin	Chl-a, d and phycoerythrin

Question 4.
Differentiate between Dicotyledonae and Monocotyledonae based on the following characters:
a. Type of roots
b. Venation in the leaves
c. Symmetry of flower
Answer:

Characters	Dicotyledonae	Monocotyledonae
1. Type of roots	Taproots	Fibrous roots
2. Venation in the leaves	Reticulate venation	Parallel venation
3. Symmetry of flower	Tetramerous or Pentamerous symmetry	Trimerous symmetry

Characters Dicotyledonae Monocotyledonae
1. Type of roots Tap roots Fibrous roots
2. V enation in the leaves Reticulate venation Parallel venation
3. Symmetry of flower Tetramerous or Pentamerous symmetry Trimerous symmetry

5. Answer the following questions.

Question (A)
We observe that land becomes barren soon after monsoon. But in the next monsoon it flourishes again with varieties we observed in season earlier. How you think it takes place?
Answer:

1. After monsoon, plants like mosses (bryophytes), ferns (pteridophytes), small herbaceous plants, etc become dry, due to which land becomes barren.
2. However, spores of bryophytes, pteridophytes and seeds of herbaceous plants, grass remain in barren land.
3. During next monsoon, these spores and seeds germinate due to availability of water and other favourable conditions.
4. Bryophytes and pteridophytes require water for reproduction. Hence they flourish during monsoon season.
5. Along with bryophytes and pteridophytes varieties of higher plants like grasses, some seasonal herbs or shrubs grow on barren land during monsoon due to favourable conditions.

Question (B)
Fern is a vascular plant. Yet it is not considered a Phanerogams. Why?
Answer:

1. Fern belongs to sub-kingdom Cryptogamae.
2. Cryptogams produce spores but do not produce seeds.
3. Also, in cryptogams the sex organs are concealed.
4. Phanerogams are seed producing plants and their sex organs are visible.
5. Hence, fern is a vascular plant. Yet it is not considered a Phanerogams.

Question (C)
Chlamydomonas is microscopic whereas Sargassum is macroscopic; both are algae. Which characters of these plants includes them in one group?
Answer:

1.  Both Chlamydomonas and Sargassum belong to division Thallophyta.
2. Members of Thallophyta range from unicellular (e.g. Chlamydomonas) to multicellular (e.g. Sargassum).
3. Both are aquatic plants containing photosynthetic pigments.
4. In both Chlamydomonas and Sargassum plant body is not differentiated into root, stem and leaves.
5. The stored food is mainly in the form of starch and its other forms.
6. Cell wall is made up of cellulose and other components. Due to these characters, both Chlamydomonas and Sargassum are included in one group i.e. Thallophyta.

Question 6.
Girth of a maize plant does not increase over a period of time. Justify.
Answer:

1. Maize plant belongs to class monocotyledonae.
2. In monocotyledonous plants, vascular bundles are closed type.
3. Thus, cambium is absent between xylem and phloem, due to which secondary growth does not occur in these plants.
4. Increase in girth of a stem occurs by secondary growth. Thus, girth of a maize plant does not increase over a period of time.

Question 7.
Radha observed a plant in rainy season on the compound wall of her school. The plant did not have true roots but root like structures were present. Vascular tissue was absent. To which group the plant may belong?
Answer:
The plant observed by Radha belongs may belong to division Bryophyta, as it shows root like structures i.e. rhizoids and absence of vascular tissue.

8. Draw neat labelled diagrams

Question 1.
Draw neat and labelled diagram of:
(A) Spirogyra
(B) Chlamydomonas
Answer:

Question (C)
Draw neat and labelled diagram of Funaria.
Answer:

Question (D)
Draw neat and labelled diagram of Nephrolepis.
Answer:

[Note: Frond: Fern leaf, originating from rhizome. It consists of blade and petiole, Blade: Main part of the frond which is rich in chlorophyll]

Question (E)
Draw neat and labelled diagram of Haplontic and Haplo-diplontic life cycle.
Answer:

Question 9.
Identify the plant groups on the basis of following features:
A. Seed producing plants
B. Spore producing plants
C. Plant body undifferentiated into root, stem and leaves
D. Plant needs water for fertilization
E. First vascular plants
Answer:
1. Phanerogams (Angiospermae and Gymnospermae)
2. Cryptogams (Thallophyta, Bryophyta and Pteridophyta)
3. Thallophyta, Bryophyta
4. Thallophyta, Bryophyta, Pteridophyta
5. Pteridophytes

Practical/Project:

Question 1.
Study the Nephrolepis plant in detail.
Answer:

1. Nephrolepis belongs to division pteridophyta.
2. They grow abundantly in cool, shady, moist places.
3. Roots are adventitious (fibrous) growing from the underground stem.
4. Leaves are well developed on the stem (Rhizome).
5. They show presence of well-developed conducting system for transportation of water and food.
6. They reproduce asexually by spores produced within sporangia, which are present in sori. These sori are located along the posterior surface of leaflets.
7. These plants have neither fruits nor flowers.
8. Some ferms are used as food, medicine or as ornamental plants.

Question 2.
Study the coralloid roots, scale leaf and megasporophyll of Cycas in detail.
Answer:
1. Coralloid roots of Cycas:
Coralloid roots of Cycas show association with blue green algae for nitrogen fixation.
Coralloid roots are coral-like, dichotomously branched and fleshy. They grow upward toward the surface of the soil. These roots arise from the lateral branches of normal roots.
2. Scale leaf of Cycas:
In Cycas leaves are dimorphic i.e. foliage leaves and scale leaves. Scale leaves are minute, membranous and brown. These are non- photosynthetic and provide protection to the stem apex.
3. Megasporophyll of Cycas:
Megasporophylls are usually arranged in compact structures called female cones or female strobili. Megasporophyll contains megasporangia (ovule) which produce megaspores.
[Students are expected to collect more information about coralloid roots, scale leaf and megasporophyll of Cycas.]

Question 10.
Observe the following diagram. Correct it and write the information in your words.

Answer:

1. The given figure indicates alternation of generation.
2. The life cycle of a plant includes two generations, sporophytic (diploid = 2n) and gametophytic (haploid = n)
3. Some special diploid cells of sporophyte divide by meiosis to produce haploid cells.
4. These haploid cells divide mitotically to produce gametophyte.
5. On maturation, gametophyte produces male and female gametes which fuse during fertilization and produce diploid zygote.
6. Diploid zygote divides by mitosis and forms diploid sporophyte.

11th Biology Digest Chapter 3 Kingdom Plantae Intext Questions and Answers

Can you recall? (Textbook Page No. 19)

Why do we call plants as producers on land?
Answer:
Plants can prepare their own food by the process of photosynthesis. Hence, they are called as producers on land.

Can you recall? (Textbook Page No. 19)

What are differences between sub-kingdoms cryptogamae and Phanerogamae?
Answer:

Cryptogamae	Phanerogamae
1. Plants belonging to this sub-kingdom are non­flowering.	Plants belonging to this sub-kingdom are flowering.
2. Sex organs are concealed.	Sex organs are visible.
3. These plants do not produce seeds.	These plants produce fruits and seeds.
4. An ovule is not formed.	An ovule is formed.
5. It is further divided into three divisions, viz.	It is further divided into two divisions, viz.
6. Thallophyta, Bryophyta and Pteridophyta.	Gymnospermae and Angiospermae.

Observe and Discuss (Textbook Page No. 19)

Collect different water samples of fresh water. Mount them on a glass slide and observe under a compound microscope. Try to identify the organisms which are visible under it.
Answer:
Micro-organisms like Paramoecium, Amoeba, blue-green algae, unicellular algae, filamentous algae can be observed under compound microscope.
[Students are expected to observe different water samples of fresh water under compound microscope and identify the organisms.]

Can you tell? (Textbook Page No. 21)

Give salient features of algae.
Answer:
Algae belongs to division Thallophyta.
Salient features of algae:
1. Habitat: Algae are mostly aquatic, few grow on other plants as epiphytes and some grow symbiotically. Some algae are epizoic i.e. growing or living non-parasitically on the exterior of living organisms.
Aquatic algae grow in marine or fresh water. Most of them are free-living while some are symbiotic.

2. Structure: Plant body is thalloid i.e. undifferentiated into root, stem and leaves. They may be small, unicellular, microscopic like Cblorella (non-motile), Chlamydomonas (motile). They can be multicellular, unbranched, filamentous like Spirogyra or branched and filamentous like Chara. Sargassum is a huge macroscopic sea weed which measures more than 60 meters in length.

3. Cell wall: The algal cell wall contains either polysaccharides like cellulose / glucose or a variety of proteins or both.
Reserve food material: Reserve food is in the form of starch and its other forms.

4. Photosynthetic pigments: Photosynthetic pigments like chlorophyll – a, chlorophyll – b, chlorophyll – c, chlorophyll – d, carotenes, xanthophylls, phycobilins are found in algae.

5. Reproduction: Reproduction takes place by vegetative, asexual and sexual method.

6. Life cycle: The life cycle shows phenomenon of alternation of generation, dominant haploid and reduced diploid phases.

Internet my friend (Textbook Page No. 20)

Write different pigments found in algae.
Answer:
Various types of photosynthetic pigments are found in algae.
1. Chlorophyll-a (Essential photosynthetic pigment) is present in all groups of algae.
2. The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Name the accessory pigments of algae.
Answer:
The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Can you tell? (Textbook Page No. 21)

Differentiate between Chlorophyceae and Phaeophyceae.
Answer:

Chlorophyceae (Green algae)	Phaeophyceae (Brown algae)
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-b.	Photosynthetic pigments are chlorophyll-a, chlorophyll-c and fucoxanthin.
2. Reserve food is in the form of starch.	Reserve food is mannitol and laminarin.
3. e.g. Chlorella, Chlamydomonas, Spirogyra, Chara, I Volvox, Ulothrix	Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Can you tell? (Textbook Page No.21)

Enlist examples of Chlorophyceae and Rhodophvceae.
Answer:
1. Examples of Chlorophyceae:
Chlorella, Chlamydomonas, Spirogyra, Char a, Volvox, Ulothrix, etc.
2. Examples of Rhodophyceae:
Chondrus, Batrachospermum, Porphyra, Gelidium, Gracillaria, Polysiphonia, etc.

Internet my friend (Textbook Page No. 21)

Different forms of green, red, brown and blue green algae.
Answer:
1. Forms of green algae:
Unicellular motile: e.g. Chlamydomonas Unicellular non-motile: E.g. Chlorella Colonial forms: e.g. Volvox Filamentous branched: e.g. Cladophora, Chara Filamentous unbranched: e.g. Ulothrix, Spirogyra

2. Forms of red algae:
The red thalli of most of the red algae are multicellular, macroscopic, e.g. Gracilaria, Gelidium, Porphyra, Polysiphonia, etc. .

3. Forms of brown algae:
Simple, branched and filamentous: Sargassum, Fucus, Ectocarpus Profusely branched: Laminaria, Dictyota, Kelps (Seaweed)

4. Forms of blue-green algae:
Unicellular, colonial or filamentous, freshwater or marine water or terrestrial algae.
[Note: Blue-green algae are cyanobacteria which are photosynthetic autotrophs.]
[Students are expected to collect more information from internet.]

Internet my friend (Textbook Page No. 20)
Enlist the forms of filamentous algae.
Answer: The forms of filamentous algae:
1. Filamentous branched: e.g. Cladophora, Chara, Ectocarpus, Dictyota, etc.
2. Filamentous unbranched: e.g. Ulothrix, Spirogyra, etc.

Internet my friend (Textbook Page No. 21)

Economic importance of algae.
Answer:
(a) Many species of algae are used as food. For e.g. Chlorella (rich in cell proteins hence used as food supplement, even by space travelers), Sargassum, Laminaria, Porphyra, etc.
(b) Alginic acid is produced commercially from Kelps.
(c) Hydrocolloids like algin and carrageen are obtained from brown algae and red algae respectively.
(d) ‘Agar’ which is used as solidifying agent in tissue culture is obtained from red algae like Gelidium and Gracilaria.
(e) Brown algae like sea weeds are used a fodder for sheep, goat, etc.
[Students are expected to collect more information about the economic importance of algae.]
(f) Role of algae in environment.
Answer:
(a) Being photosynthetic, algae help in increasing the level of dissolved oxygen in their immediate environment.
(b) Algae are primary producers of energy rich compounds which forms the basis of food cycles in aquatic animals.
[Students are expected to find out more information about the role of algae in environment on internet.]

Can you recall? (Textbook Page No. 19)

Differentiate between Thallophytes and Bryophytes.
Answer:

Thallophytes	Bryophytes
1. Mostly aquatic in habitat.	Mostly terrestrial, occurs on moist and shady places.
2. Thallus may be unicellular or multicellular.	Thallus is multicellular.
3. Motile and non-motile forms are present.	Non-motile forms present, except male gametes.
4. Rhizoids are absent.	Rhizoids are present.

Can you tell? (Textbook Page No. 23)

Why Bryophyta are called amphibians of Plant Kingdom?
Answer:
Members of Bryophyta are mostly terrestrial plants which depend on water for fertilization and completion of their life cycle. Hence, they are called ‘amphibians of Plant Kingdom’.

Observe and Discuss (Textbook Page No. 21)

You may have seen Funaria plant in rainy season. Why is it called amphibious plant?
Answer:
Funaria belongs to division Bryophyta.
It is a terrestrial plant but requires water for fertilization and completion of its life cycle. Hence, it is called as an amphibious plant.

Observe and Discuss (Textbook Page No. 23)

You may have seen the various plants which do not bear flowers, fruits and seeds but they have well developed root, stem and leaves. Discuss.
Answer:
1. The plants which do not bear flowers, fruits and seeds, but have true roots, stem and leaves belong to division Pteridophyta.
2. These plants are cryptogams as they do not produce seeds and flowers.
3. They have primitive conducting system.

Can you tell? (Textbook Page No. 23)

Pteridophytes are also known as vascular Cryptogams – Justify.
Answer:
1. The reproductive organs of pteridophytes are hidden.
2. Pteridophytes do not produce flowers, fruits and seeds. They reproduce asexually by forming spores and sexually by forming gametes, hence they belong to Cryptogamae.
3. These plants possess a primitive conducting system. Thus, conduction of water and food occurs through vascular tissue.
Hence, Pteridophytes are also known as vascular Cryptogams.

Can you tell? (Textbook Page No. 23)

Give one example of aquatic and xerophytic Pteridophytes.
Answer:
Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).

Can you recall? (Textbook Page No. 19)

Give any two examples of Pteridophyta.
Answer:
Nephrolepis, Selaginella, Azolla, Marsilea, Equisetum, Lycopodium, Psilotum, Dryopteris, Pteris, Adiantum.

Can you tell? (Textbook Page No. 25)

Give general characters of Gymnosperms and Angiosperms.
Answer:
1. General characters of Gymnosperms:
(a) Types: Most of the gymnosperms are evergreen, shrubs or woody trees.
(b) Vascular tissues: They are vascular plants having xylem with tracheids and phloem with sieve cells.
(c) Flower: These are primitive group of flowering plants producing naked seeds.
(d) Body: The plant body is sporophyte. It is differentiated into root, stem and leaves.
(e) Roots: The root system is tap root type. In some gymnosperms, the roots form symbiotic association with other life forms. Coralloid roots of Cycas show association with blue green algae and roots of Pinus show association with endophytic fungi called mycorrhizae.
(f) Stem: In gymnosperms, stem is mostly erect, aerial, solid and cylindrical. Secondary growth is seen in Gymnosperms due to the presence of cambium. In Cycas it is usually unbranched, while in conifers it is branched, (e.g. Pinus, Cedrus).
(g) Leaves: The leaves are dimorphic. The foliage leaves are green, simple needle like or pinnately compound, whereas scale leaves are small, membranous and brown.
(h) Spores: Spores are produced by microsporophyll (Male) and megasporophyll (Female).

(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Observe and Discuss (Textbook Page No. 23)

Observe all garden plants like Cycas, Thuja, Pinus, Sunflower, Canna and compare them. Note similarities and dissimilarities among them.
Answer:
1. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following similarities can be observed:
Plant body is divided into root, stem and leaves.
2. When we observe garden plants like Cycas, Thuja, Pinus, Sunflower, Canna, following dissimilarities can be observed:
(a) In Cycas, Thuja and Pinus seeds are not enclosed within a fruit, whereas in Sunflower and Canna seeds are enclosed within a fruit.
(b) Plants like Cycas, Thuja, Pinus show cones bearing microsporophylls and megasporophylls, whereas sunflower and Canna plant bear flowers.
(c) In Cycas, Thuja and Pinus green, simple needle like or pinnately compound foliage leaves and brown, membranous scaly leaves can be observed, whereas in Sunflower, Canna green foliage leaves can be observed.

Can you recall? (Textbook Page No. 24)

What are the salient features of angiosperms?
Answer:
(ii) General characters of angiosperms:
(a) Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.
(b) Alternation of generations: Angiosperms show heteromorphic alternation of generation in which the sporophyte is diploid, dominant, autotrophic and independent. The gametophytes (male or female) are haploid, reduced and concealed.
(c) Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.
(d) Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Can you recall? (Textbook Page No. 24)

What is double fertilization?
Answer:
(a) Double fertilization is a characteristic feature of angiosperms.
(b) In this process one male gamete fuses with egg cell and another male gamete fuses with secondary nucleus, to form an embryo and endosperm respectively.

Can you recall? (Textbook Page No. 24)

Explain in brief the two classes of Angiosperms? Draw and label one example of each class.
Answer:
Two classes of Angiosperms are Dicotyledonae and Monocotyledonae.
а. Dicotyledonae:

1. These plants have two cotyledons in their embryo.
2. They have a tap root system and the stem is branched.
3. Leaves show reticulate venation.
4. Flowers show tetramerous or pentamerous symmetry.
5. Vascular bundles are conjoint, collateral and open type.
6. Cambium is present between xylem and phloem for secondary growth.
7. In dicots, secondary growth is commonly found. e. g. Helianthus annuus (Sunflower)

b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Try This (Textbook Page No. 24)

Study the leaves of Hibiscus, Peepal, Canna, Grass and Tulsi. Classify them as Monocot and Dicot.
Answer:

Monocot leaves	Dicot leaves
Canna. Grass (Parallel venation)	Hibiscus, Peepal, Tulsi (Reticulate venation)

Can you tell? (Textbook Page No. 25)

(i) Distinguish between Dicotyledonae and Monocotyledonae.
Answer:
Spores and Sporophylls: Angiosperms are heterosporous. Microspores (commonly called pollens) are formed in microsporangia (or anthers). They develop in highly specialized microsporophyll or stamens while megaspores are formed in megasporangia (or ovules) borne on highly specialized megasporophyll called carpel.

(ii) Why do Dicots show secondary growth while Monocots don’t?
Answer:
(a) In dicots, vascular bundles are conjoint, collateral and open type. Cambium is present between xylem and phloem for secondary growth.
(b) Whereas in monocots, vascular bundles are conjoint, collateral and closed type. Thus, due to absence of cambium, secondary growth does not occur in majority of monocots.

Observe and Discuss (Textbook Page No. 23)

Which differences did you notice between Gymnosperms and Angiosperms?
Answer:

Gymnosperms	Angiosperms
1. In gymnosperms, the seeds arc naked.	In angiosperms, the seeds are enclosed within the fruit.
2. Plants are evergreen, shrubs or woody trees.	Plants are annual, biennial or perennial herbs, shrubs or trees, either woody or herbaceous.
3. Xylem is made up of tracheids only.	Xylem is made up of vessels and tracheids.
4. Phloem is with sieve cells only.	Phloem is with sieve tubes and companion cells.
5. Usually two types of leaves are present, i.e. green foliage leaves and scale leaves.	Leaves are of usually one type only, such as green foliage leaves.
6. Double fertilization absent.	Double fertilization occurs.

Can you tell? (Textbook Page No. 26)

What is alternation of generations?
Answer:
The sporophytic and gametophytic generations generally occur alternately in the life cycle of a plant. This phenomenon is called alternation of generations.

Can you tell? (Textbook Page No. 26)

Which phase is dominant in the life cycle of Bryophyta and Pteridophyta?
Answer:
In the life cycle of Bryophyta, gametophyte is the dominant phase whereas in the life cycle of Pteridophyta, sporophyte is the dominant phase.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 8 Plant Tissues and Anatomy Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 8 Plant Tissues and Anatomy

1. Choose the correct option.

Question (A)
Location or position of meristematic regions is divided into _______ types.
(A) one
(B) two
(C) three
(D) none of the above
Answer:
(C) three

Question (B)
Cambium is also called
(A) apical meristem
(B) intercalary meristem
(C) lateral meristem
(D) none of the above
Answer:
(C) lateral meristem

Question (C)
Collenchyma is a type of ________ tissue.
(A) living
(B) dead
(C) living and dead
(D) none of the above
Answer:
(A) living

Question (D)
_______ is a complex permanent tissue.
(A) Parenchyma
(B) Sclerenchyma
(C) Chlorenchyma
(D) Xylem
Answer:
(D) Xylem

Question (E)
Mesophyll tissue is present in ________ .
(A) root
(B) stem
(C) leaf
(D) flower
Answer:
(C) leaf

2. Answer the following questions

Question (A)
A fresh section was taken by a student but he was very disappointed because there were only few green and most colourless cells. Teacher provided a pink colour solution. The section was immersed in this solution and when observed it was much clearer. What is the magic?
Answer:
1. The pink coloured solution given by teacher must be a saffanin stain.
2. Saffanin is used to stain plant tissues, especially lignified tissues such as cell wall and xylem.

Question (B)
While observing a section, many scattered vascular bundles could be seen. Teacher said, in spite of this large number the stem cannot grow in girth. Why?
Answer:

1. Students must have observed monocot stems.
2. It is because, monocot stem shows scattered vascular bundles.
3. In monocot stem, vascular bundles are closed i.e. without cambium.
4. Thus, secondary growth does not occur which is required for increase in girth. Hence, in spite of having large number of scattered vascular bundles, monocot stems do not grow in girth.

Question (C)
A section of the stem had vascular bundles, where one tissue was wrapped around the other. How will you technically describe it?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question (D)
There were two cut logs of wood lying in the campus. One had growth rings and other didn’t. Teacher said it is due to differences in their pattern of grow th which is dependent on season. How?
Answer:
1. It is possible that one of the cut logs was of a tropical tree, whereas the other was of a temperate tree. Since tropical trees grow in a similar manner all year, growth rings are not apparent. Another explanation for this could be that the log which had growth rings must be of an old tree which has experience many seasons, whereas the log without growth rings must be of younger tree, that has not been subjected to seasonal changes and hence not developed prominent growth rings.

2. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.

3. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

4. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

5. These growth rings can be used to estimate the age of the tree. These are found more in older trees as compare to younger tree.

Question (E)
While on the trip to Kashmir, Pintoo observed that cut portions of large trees show distinct rings, which he never found in Maharashtra. Why is so?
Answer:
1. Cut portions of large tress show distinct rings which are annual rings formed due to activity of cambium during favourable and non-favourable climatic conditions.
2. Kashmir falls under temperate region where the climatic conditions are not uniform through the year. In the spring season, conditions are favourable due to which cambium is active, whereas in autumn season, conditions are unfavourable due to which cambium is less active. This leads to formation of spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.
3. Maharashtra falls under tropical region where climatic conditions are favourable throughout the year. In tropical areas, continuous growth of secondary xylem occurs. Thus, trees growing in tropical regions show less or no annual rings as compared to trees in temperate region.

Question (F)
A student was observing a slide with no label under microscope. The section had some vascular bundles scattered in the ground tissue. It is section of a monocot stem! He exclaimed. No! it is section of fern rachis, said the teacher. Teacher told to observe vascular bundle again. Student agreed, Why?
Answer:

1. In fern rachis, the number of vascular bundles is less as compared to number of vascular bundles in monocot stem. In monocot stem, vascular bundles are numerous.
2. In fern rachis, xylem consists of only tracheids whereas in monocot stem, xylem consists of vessels (protoxylem and metaxylem) as well as tracheids. Monocot stem shows presence of lysigenous cavity just below protoxylem.
3. In fern rachis, phloem consists of only sieve cells whereas in monocot stem, phloem consists of sieve tubes and companion cells. Thus, a student must hav e observed these differences in the given section and agreed to teacher’s statement that the given section is of fern rachis and not of monocot stem.

Question (G)
Student found a wooden stopper in lab. He was told by an old lab attendant that it is there for many years. He kept thinking how it did not rot?
Answer:
1. Wooden stopper or cork is obtained from the phellem (cork) part of a bark.
2. Phellem (cork) is impervious in nature and does not allow entry of water due to suberized walls.
3. Due to this it does not rot and remains as it is for many years.

Question (H)
Student while observing a slide of leaf section observed many stomata on the upper surface. He thought he has placed slide upside down. Teacher confirmed it is rightly placed. Explain.
Answer:
1. In a dicot leaf, stomata are generally absent on upper epidermis but are present on lower epidermis. Thus, the student must have thought that he has placed slide upside down.
2. According to teacher, the section was placed rightly, thus the given section must be of monocot leaf.
3. It is because, in monocot leaf stomata are present on both upper and lower epidermis.

3. Write short notes on the following points.

Question (A)
Structure of stomata.
Answer:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question (B)
Write a short note on secondary growth.
OR
With the help of neat and labelled diagram explain the secondary growth in dicot stem.
Answer:
Secondary growth:

1. Dicotyledonous plants and gymnosperms exhibit increase in girth of root and stem.
2. In dicot stem, secondary growth begins with the formation of a continuous cambium ring.
3. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
4. The cells of medullary rays adjoining these intrafascicular cambium strips become meristematic (regain the capacity to divide) and form the interfascicular cambium.
5. Thus, a complete and continuous ring of vascular cambium is formed.
6. The cambium ring cuts off new cells, towards both inner and outer sides.
7. The cells that are cut-off towards pith (inner side) mature into secondary xylem and cells that are cut-off towards periphery mature into secondary phloem.
8. Generally, amount of secondary xylem is more than the secondary phloem.

Question (C)
Write a short note on peculiarity of a sclerenchyma cell wall.
Answer:
Peculiarity of a sclerenchyma cell wall:
1. Cell wall of sclerenchyma is evenly thickened due to uniform deposition of lignin.
2. Cell wall of sclereids is extremely thick and strongly lignified.

4. Differentiate

Question (A)
Differentiate between vascular bundles of monocot and dicot.
Answer:

1. Vascular bundle of monocot and dicot root.
2. Vascular bundle of monocot and dicot stem.
3. Vascular bundle of monocot and dicot leaf.

Question (B)
Differentiate between xylem and phloem.
Answer:

Xylem	Phloem
1. It is a dead complex tissue.	It is a living complex tissue.
2. It is composed of xylem, tracheids, vessels, xylem fibres and xylem parenchyma.	It is composed of sieve tubes, sieve cells, companion cells, phloem parenchyma and phloem fibres.
3. It is also known as wood.	It is also known as bast.
4. The cell walls are thick due to lignin.	The cell walls are thin.
5. Xylem conducts water and minerals from roots to the stem and leaves. It also provides mechanical strength to the plant parts.	It is the chief food conducting tissue of vascular plants responsible for translocation of food from leaves to other plant parts.

5. Draw neat labelled diagrams

Question (A)
T.S. of dicot leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question (B)
T.S. of Monocot root.
Answer:

Question (C)
Draw neat labelled diagrams of T.S. of dicot stem.
Answer:

Question 6.
Write the information related to diagram given below.

Answer:


[Note: The labelled part can be considered as the ‘region of maturation ’ of root apical however, the region of maturation does not contain meristematic tissue ]
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 7.
Identify the following diagrams, label it and prepare a chart of characteristics.
Answer:
1. Figure ‘c’

Question 8.
Distinguish between dicot and monocot leaf on the basis of following characters.
Answer:

Characters	Dicot leaf	Monocot leaf
Stomata	Stomata are restricted to lower epidermis. Guard cells of stoma are kidney shaped.	Stomata occur on both epidermis. Guard cells of stoma are dumbbell shaped.
Intercellular space	More intercellular spaces due to presence of spongy parenchyma.	Less intercellular spaces as mesophyll is not differentiated into spongy and palisade tissue.
Venation	Reticulate venation	Parallel venation
Vascular bundle	Vascular bundles of varying size. The size of the vascular bundles is dependent on the size of the veins which vary in thickness in dicot leaf.	Vascular bundles are nearly of similar size (Except in main veins).
Mesophyll cells	Mesophyll tissue is differentiated into palisade parenchyma and spongy parenchyma.	Mesophyll tissue is not differentiated into palisade parenchyma and spongy parenchyma.

Practical/ Project:

Question 1.
Prepare detail anatomical charts with diagrammatic representation of dicot and monocot plants.
Answer:
Anatomy of dicot root: The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Anatomy of monocot stem: A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single-layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Anatomy of dicot leaf:

1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Anatomy of monocot leaf:

1.It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 2.
Observe different slides related to anatomy of flowering plants under the guidance of teacher.
[Students are expected to perform this practical own their own.]

11th Biology Digest Chapter 8 Plant Tissues and Anatomy Intext Questions and Answers

Can you recall? (Textbook Page No. 85)

(i) Which component brings about important processes in the living organisms?
Answer:
Cell is the component that brings about important processes in the living organisms.

(ii) What is tissue?
Answer:
A group of cells having essentially a common function and origin is called as tissue.

(iii) Explain simple and complex tissue.
Answer:
a. Simple tissue:
1. They are made up of only one type of cells.
2. They are found in all the plant parts.
3. They perform many functions.
4. Simple tissues in plants are Parenchyma, Collenchyma, Sclerenchyma.

b. Complex tissue:
1. They are made up of many types of cells.
2. They are found only in the vascular regions of the plant.
3. They mainly perform the function of conduction of food and water.
4. Complex tissues in plants are Xylem and Phloem.

(iv) Complete the flow chart.
Organisms → Organs → Cells
Answer:
Organism → Organ system → Organs → Tissue system → Tissue → Cells

Can you tell? (Textbook Page No. 86)

Enlist the characteristics of meristematic tissue.
Answer:
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Can you tell? (Textbook Page No. 86)

Classify meristematic tissue on the basis of origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Can you tell? (Textbook Page No. 89)

Write a note on parenchyma.
Answer:
Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Can you tell? (Textbook Page No. 89)

Describe sclerenchyma fibres.
Answer:
Sclerenchyma fibres:
1. Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls.
2. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique.
They provide mechanical strength.

Can you tell? (Textbook Page No. 89)

Sketch and label T.S. of phloem tissue.
Answer:
T.S. of phloem tissue: Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
2. It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
3. On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
4. It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Can you tell? (Textbook Page No. 92)

Concentric vascular bundles are always closed. Describe.
Answer:

1. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
2. When cambium is not present between xylem and phloem, it is known as closed vascular bundle.
3. Due to absence of cambium between xylem and phloem, concentric vascular bundles are always closed.

Can you tell? (Textbook Page No. 92)

How is the structure of vascular bundles of the root?
Answer:

1. Vascular bundles of the root are radial.
2. In radial vascular bundles, complex tissues are situated separately on separate radius as separate bundle.
3. The xylem and phloem bundles are arranged alternating with each other.

Can you tell? (Textbook Page No. 92)

Why vascular bundles of dicot stem are described as conjoint collateral and open?
Answer:
Vascular bundles of dicot stem are described as conjoint collateral and open because;
1. In dicot stem, the complex tissue is collectively present as neighbours of each other on the same radius in the form of xylem inside and phloem outside. Such type of vascular bundles are called as conjoint and collateral.
2. In dicot stem, a strip of cambium is present between xylem and phloem. Hence, it is called as open vascular bundle.

Can you tell? (Textbook Page No. 92)

How is the arrangement of vascular bundles in dicot and monocot stem?
Answer:
1. Vascular bundle in dicot stem: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
2. Vascular bundle in monocot stem: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and cloused (without cambium). Xylem is endarch and shows lysigenous cavity.

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 7 Cell Division Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 7 Cell Division

1. Choose the correct option

Question (A)
The connecting link between Meiosis – I and Meiosis – II is …………
(a) interphase – I
(b) interphase – II
(c) interkinesis – III
(d) anaphase – IV
Answer:
(c) interkinesis – III

Question (B)
Synapsis is pairing of ………………. .
(a) any two chromosomes
(b) non – homologous chromosomes
(c) sister chromatids
(d) homologous chromosomes
Answer:
(d) homologous chromosomes

Question (C)
Spindle apparatus is formed during which stage of mitosis?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) S-phase

Question (D)
Chromosome number of a cell is almost doubled up during _______ .
(a) G₁ – phase
(b) S – phase
(c) G₂-phase
(d) G₀-phase
[Note: Due to DNA replication the DNA content of cell doubles during S-phase. But the number of chromosomes remain the same.]
Answer:
(b) S – phase

Question (E)
How many meiotic divisions are necessary for formation of 80 sperms?
(a) 80
(b) 40
(c) 20
(d) 10
Answer:
(c) 20

Question (F)
How many chromatids are present in anaphase – I of meiosis – I of a diploid cell having 20 chromosomes?
(a) 4
(b) 6
(c) 20
(d) 40
Answer:
(d) 40

Question (G)
In which of the following phase of mitosis chromosomes are arranged at equatorial plane?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) Metaphase

Question (H)
Find incorrect statement.
(a) Condensation of chromatin material occurs in prophase.
(b) Daughter chromatids are formed in anaphase.
(c) Daughter nuclei are formed at metaphase.
(d) Nuclear membrane reappears in telophase.
Answer:
(c) Daughter nuclei are formed at metaphase.

Question (I)
Histone proteins are synthesized during
(a) G₁ phase
(b) S – phase
(c) G₂ – phase
(d) Interphase
Answer:
(b) S – phase

2. Answer the following questions

Question (A)
While observing a slide, student observed many cells with nuclei. But some of the nuclei were bigger as compared to others but their nuclear membrane was not so clear. Teacher inferred it as one of the phase in the cell division. Which phase may be inferred by teacher?
Answer:
Prophase.

Question (B)
Students prepared a slide of onion root tip. There were many cells seen under microscope. There was a cell seen under microscope. There was a cell with two groups of chromosomes at opposite ends of the cell. This cell is in which phase of mitosis?
Answer:
Anaphase.

Question (C)
Students were shown some slides of cancerous cells. Teacher made a comment as if there would have been a control at one of its cell cycle phase, there wouldn’t have been a condition like this. Which phase the teacher was referring to?
Answer:
The phase teacher was referring would be G_i phase.

Question (D)
Some Mendelian crossing experimental results were shown to the students. Teacher informed that there are two genes located on the same chromosome. He enquired if they will be ever separated from each other?
Answer:

1. Genes are located on chromosomes at specific distance and position.
2. The greater this distance, the greater the chance that a crossover can occur between the genes and the greater the chances of recombination.
3. The chances of recombination are less between the genes that are placed closed to each other on the chromosome.
4. Therefore, due to recombination the two genes located on the same chromosome have possibility of separating from each other.

Question (E)
Students were observing a film on Paramoecium. It underwent a process of reproduction. Teacher said it is due to cell division. But students objected and said that there was no disappearance of nuclear membrane and no spindle formation, how can it be cell division? Can you clarify?
Answer:

1. Paramoecium is a unicellular organism. The division in Paramoecium occurs by amitosis.
2. It is the simplest mode of cell division.
3. In amitosis, nucleus elongates and a constriction appears. This constriction deepens and divides the nucleus in two daughter nuclei followed by the division of cytoplasm.

Question (F)
Is the meiosis responsible for evolution? Justify your answer.
Answer:

1. Meiosis ensures that organisms produced by sexual reproduction contain correct number of chromosomes.
2. Meiosis exhibits genetic variation by the process of recombination.
3. Variations increase further after union of gametes during fertilization creating offspring with unique characteristics. Thus, it creates diversity of life and is responsible for evolution.

Question (G)
Why mitosis and meiosis – II are called as homotypic division?
Answer:
1. In mitosis, the chromosome number and genetic material of daughter cells remain same as that of the parent cell.
2. In meiosis – II, two haploid cells formed during first meiotic division divide further into four haploid cells. This division is identical to mitosis. The daughter cells formed in second meiotic division are similar to their parent cells with respect to the chromosome number formed in meiosis -1. Hence mitosis and meiosis – II are called homotypic division.

Question (H)
Write the significance of mitosis.
Answer:

1. As mitosis is equational division, the chromosome number is maintained constant.
2. It ensures equal distribution of the nuclear and the cytoplasmic content between the daughter cells, both quantitatively and qualitatively. Therefore, the process of mitosis also maintains the nucleo-cytoplasmic ratio.
3. The DNA is also equally distributed.
4. It helps in growth and development of organisms.
5. Old and worn-out cells are replaced through mitosis.
6. It helps in the asexual reproduction of organisms and vegetative propagation in plants.

Question (I)
Enlist the different stages of prophase – I.
Answer:
1. Prophase -I:
It is the most complicated and longest phase of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:
The volume of the nucleus increases.
The chromosomes become long distinct and coiled.
They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies. j Lep
The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during Gj phase of interphase.]

b. Zygotene:
Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:
Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points. These points are called chiasmata (Appear like a cross-X).
Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:
The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
The terminal chiasmata exist till the metaphase.
The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

3. Draw labelled diagrams and write explanation

Question (A)
With the help of suitable diagram, describe the cell cycle.
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):
Metabolic activities essential for cell division occur during this phase.
Various proteins which are necessary for the cell division are also synthesized in this phase.
Apart from this, RNA synthesis also occurs during this phase.
In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Question (B)
Distinguish between mitosis and meiosis.
Answer:

Mitosis	Meiosis
(a) It occurs in somatic cells and stem cells.	It occurs in germ cells.
(b) In this nucleus divides only once.	In this nucleus divides twice (Meiosis I and Meiosis II)
(c) In these two daughter cells are formed.	In these four daughter cells are formed.
(d) Daughter cells formed by mitotic division are diploid (2n).	Daughter cells formed by meiotic division are haploid (n)•
(e) In mitosis, crossing over does not take place.	In meiosis, crossing over takes place.
(f) Mitosis plays an important role in growth, repair, healing and development.	Meiosis is important for formation of haploid gametes and spores.

Question (C)
Draw labelled diagrams and write explanation Draw the diagram of metaphase.
Answer:
Metaphase:
a. Chromosomes are completely condensed and appear short.
b. Centromere and sister chromatids become very prominent.
c. All the chromosomes are arranged at equatorial plane of cell. This is called metaphase plate.
d. Mitotic spindle is fully formed in this phase.
e. Centromere of each chromosome divides horizontally into two, each being associated with a chromatid. [Note: The centromeres divide at the beginning of anaphase so that the two chromatids of each chromosome become separated from each other.
Source: Cell Division, Donald B. McMillan, Richard J. Harris, in An Atlas of Comparative Vertebrate Histology, 2018.]

Question 4.
Match the following column – A with column – B

Column I (Phases)	Column II (Their events)
1. Leptotene	(a) Crossing over
2. Zygotene	(b) Desynapsis
3. Pachytene	(c) Synapsis
4. Diplotene	(d) Bouquet stage

Answer:

Column I (Phases)	Column II (Their events)
1. Leptotene	(d) Bouquet stage
2. Zygotene	(c) Synapsis
3. Pachytene	(a) Crossing over
4. Diplotene	(b) Desynapsis

Question 5.
Is the given figure correct? Why?

Answer:
1. The given figure is incorrect as the spindle fibres are not attached to centromere of the chromosomes.
2. During metaphase, chromosomes are attached to spindle fibres with the help of centromeres.

Question 6.
If an onion has 16 chromosomes in its leaf cell, how many chromosomes will be there in its root cell and pollen grain.
Answer:
1. The chromosomes in root cell will be 16 as root cell is a diploid cell.
2. The chromosomes in pollen grain will be 8 as pollen grain is a haploid cell.

7. Identify the following phases of mitosis and label the ‘A’ and ‘B’ given in diagrams.

Question (i)

Answer:
The diagram shown is of Metaphase.
A: Chromosomes arranged on metaphase plate

Question (ii)

Answer:
The diagram shown is of Anaphase.
B: Chromatids moving to opposite poles.

Practical / Project:

Question 1.
Fix the onion root tips at different durations of the day starting from 6am up to 9am at the intervals of half an hour. Prepare the slide of each fixed root tip and analyse the relation between time and phase of mitosis.
Answer:
Mitotic division is an equational division in which one parent cell give rise to two daughter cells with equal number of chromosomes in daughter cells and mother cell. It has four sub phases: prophase, metaphase, anaphase, telophase.

Mitosis is affected by temperature and time. Mitotic index is high in morning so the mitosis is observed clearly in the morning. (Mitotic index is defined as the ratio between the number of cells in a population undergoing mitosis to the total number of cells in a population. )
[Note: Students catt use above information for reference and perform this activity on their own.]

11th Biology Digest Chapter 7 Cell Division Intext Questions and Answers

Can you recall? (Textbook Page No. 76)

How do your wounds heal?
Answer:
a. A wound is an injury to living tissue.
b. Healing of wound take place by mitosis.
c. Repetitive mitotic divisions near the site of injury results in healing of wound.

Can you tell? (Textbook Page No. 79)

What is cell cycle?
Answer:

1. Sequential events occurring in the life of a cell is called cell cycle.
2. Interphase and M – phase are the two phases of cell cycle.
3. Cell undergoes growth or rest during interphase and divides during M – phase.

Discuss with teacher (Textbook Page No. 76)

Some cells do not have gap phase in their cell cycle whereas some cells spend maximum part of their life in gap phase. Search for such cells. Some cells are said to be in G₀ phase. What is this G₀ phase?
Answer:

1. G₀ is the phase of the cell cycle in eukaryotes in which many cell types stop dividing. It is also called a quiescent stage.
2. If cells are deprived of appropriate growth factors, they stop at the Gi checkpoint of the cell cycle. Their growth and division are arrested and they remain in G₀ phase.
3. Mature neurons and muscle cells remain in G₀ phase.

Question 5.
Can you tell? (Textbook Page No. 79)
Answer:
1. Series of events occurring in the life of a cell is called cell cycle. Interphase and M – phase are the two phases of cell cycle.
2. Interphase: It is the stage between two successive cell divisions. It is the longest phase of a cell cycle during which the cell is highly active and prepares itself for cell division.
The interphase is subdivided into three sub-phases as G₁ – phase, S-phase and G₂-phase.
a. G₁ – phase (First gap period/First Gap Phase):
It begins immediately after cell division.
RNA (mRNA, rRNA and tRNA) synthesis, protein synthesis and synthesis of membranes take place during this phase.
b. S – phase (Synthesis phase):
In this phase DNA is synthesized (replicated), so that amount of DNA per cell doubles.
Synthesis of histone proteins takes place in this phase.
c. G₂ – phase (Second growth phase/Second Gap Phase):

1. Metabolic activities essential for cell division occur during this phase.
2. Various proteins which are necessary for the cell division are also synthesized in this phase.
3. Apart from this, RNA synthesis also occurs during this phase.
4. In animal cells, a daughter pair of centrioles appears near the pre-existing pair.

Internet my friend (Textbook Page No. 77)

What is Karyogram or Karyotype?
Answer:
1. A karyotype is a representation of condensed chromosomes arranged in pairs.
2. Analysis of the karyotype of a particular individual indicates whether the individual has a normal set of chromosomes or whether there are abnormalities in number or appearance of individual chromosomes.

Can you tell? (Textbook Page No. 79)

Which are the steps of mitosis?
Answer:
Steps in mitosis are Karyokinesis and Cytokinesis. Karyokinesis includes four stages – Prophase, Metaphase, Anaphase and Telophase.

Internet my friend (Textbook Page No. 79)

How the life span of a cell is decided?
Answer:

1. Life span of different cells vary greatly.
2. Life span of a cell is decided by its growth rate, metabolic activities and cell size.
3. The life span of a cell can be analysed in laboratory by applying carbon-14 technique to DNA.
4. This method is commonly used in archaeology and paleontology to find the age of fossils. Same can be applied to determine the life span of a cell.

Do yourself (Textbook Page No. 80)

Write down the explanation of prophase I in your own words.
Answer:
1. Prophase -I:
It is the most complicated and longest phas0e of meiotic division.
It is further divided into five sub-phases viz. leptotene, zygotene, pachytene, diplotene and diakinesis.

a. Leptotene:

1. The volume of the nucleus increases.
2. The chromosomes become long distinct and coiled.
3. They orient themselves in a specific fonn known as bouquet stage. This is characterized with the ends of chromosomes converged towards the side of nucleus where the centrosome lies.
4. The centriole duplicates into two and migrates to opposite poles. [Note: Centrioles divide during G_j phase of interphase.]

b. Zygotene:

1. Pairing of non-sister chromatids of homologous chromosomes takes place by formation of synaptonemal complex. This pairing is called synapsis.
2. Each pair consists of a maternal chromosome and a paternal chromosome. Chromosomal pairs are called bivalents or tetrads.

c. Pachytene:

1. Each individual chromosome begins to split longitudinally into two similar chromatids. Therefore, each bivalent now appears as a tetrad consisting of four chromatids.
2. The homologous chromosomes begin to separate but they do not separate completely and remain attached to one or more points.
3. These points are called chiasmata (Appear like a cross-X).
4. Chromatids break at these points and broken segments are exchanged between non-sister chromatids of homologous chromosomes resulting in recombination.

d. Diplotene:
The chiasma becomes clearly visible in diplotene due to beginning of repulsion between synapsed homologous chromosomes. This is known as desynapsis. Synaptonemal complex also starts to disappear.

e. Diakinesis:

1. The chiasmata begin to move along the length of chromosomes from the centromere towards the ends of chromosomes. The displacement of chiasmata is termed as terminalization.
2. The terminal chiasmata exist till the metaphase.
3. The nucleolus and nuclear membrane completely disappear and spindle fibres begin to appear.

Curiosity Box: (Textbook Page No. 81)

(i) What is exact structure of synaptonemal complex?
Answer:
Synaptonemal complexes are zipper like structures assembled between homologous chromosomes during the prophase of first meiotic division.
[Source: ncbi.nlm. nih.gov/pubmed/8743892]

(ii) What is structure of chiasma?
Answer:
Chiasma is a X-shaped point of attachment between two non-sister chromatids of a homologous chromosomes.

(iii) Which type of proteins are involved in formation of spindle fibres?
Answer:
Spindle fibres are formed from microtubules with many accessory proteins.

(iv) Why and how spindle fibres elongate and some contract?
Answer:
a. Spindle fibres elongate for assembly of chromosomes at equatorial plane of the cell during metaphase and spindle fibres contract for pulling chromosomes towards opposite poles during anaphase.
b. The spindle fibres elongate (polymerize) by incorporating subunits of the protein tubulin and contract

(v) What is the role of centrioles in formation of spindle apparatus?
Answer:
Centriole plays an important role in cell division. Centrioles help organize microtubule assembly and forms spindle apparatus that separate the chromosomes during cell division.

Curiosity box (Textbook Page No. 81)

What would have happened in absence of meiosis?
Answer:

1.  Gametes are produced by the process of meiosis which are essential for sexual reproduction.
2. Diploid organisms have two set of chromosomes (one paternal and one maternal).
3. For a diploid organism to undergo sexual reproduction it needs to create gametes that contain only one set of chromosomes so the number of chromosomes remains same in the next generation.
4. In absence of meiosis, the chromosome number of parents and their offsprings will differ in every generation; hence no species will hold its characters.
5. Also, there will be no crossing over of homologous chromosomes. Thus, there will be no variations with respect to the changing environment in progeny to maintain their existence, which may lead to extinction of species.

Can you tell? (Textbook Page No. 82)

(i) What is the difference between mitosis and meiosis?
Answer:

Mitosis	Meiosis
(a) It occurs in somatic cells and stem cells.	It occurs in germ cells.
(b) In this nucleus divides only once.	In this nucleus divides twice (Meiosis I and Meiosis II)
(c) In these two daughter cells are formed.	In these four daughter cells are formed.
(d) Daughter cells formed by mitotic division are diploid (2n).	Daughter cells formed by meiotic division are haploid (n) •
(e) In mitosis, crossing over does not take place.	In meiosis, crossing over takes place.
(f) Mitosis plays an important role in growth, repair, healing and development.	Meiosis is important for formation of haploid gametes and spores.

(ii) What is difference between meiosis – I and meiosis – II?
Answer:

Meiosis I	Meiosis II
(a) Diploid cell is divided into two haploid cells.	Two haploid cells formed in meiosis I divides further into four haploid cells.
(b) This division is called heterotypic division.	This division is called homotypic (equational) division.
(c) It consists of prophase – I, metaphase – I, anaphase -1, telophase -1 and cytokinesis.	It consists of prophase – II, metaphase – II, anaphase – II, telophase – II and cytokinesis.
(d) Number of chromosomes is reduced to half, i.e. from diploid to haploid state.	In meiosis II number of chromosomes remain the same.
(e) It is complicated and long duration division.	It is simple and short duration division.
(f) Telophase I results into 2 daughter cells.	Telophase II results in 4 daughter cells.

(iii) Elaborate the process of recombination.
Answer:
a. Recombination is exchange of genetic material between paternal and maternal chromosomes during gamete formation.
b. The points where crossing over takes place is known as chiasmata.
c. Chromatids acquire new combinations of alleles by physically exchanging segments in crossing-over.
d. The exchange of genetic material between homologous chromosomes involves accurate breakage and joining of DNA molecules through a complex mechanism.
e. It is catalyzed by enzymes.

Do Yourself (Textbook Page No. 82)

Prepare a concept map on cell division in following box.
Answer:
Refer Quick Review

Internet My Friend (Textbook Page No. 82)

Different types of proteins like cyclins, maturation promoting factor (MPF), cyclosomes, enzymes like cyclin dependent kinases (CDK) play important role in control of cell cycle. Collect more information about these proteins and enzymes from internet, prepare a power-point presentation and present it in the class.
Answer:

1. The regulation of the cell cycle involves an internal control system consisting of proteins called cyclins and enzymes called cyclin-dependent kinases.
2. A Cdk is a protein kinase. When the kinase of the Cdk is activated upon binding to a cyclin, it phosphorylates target proteins in the cell, regulating their activities.
3. Those proteins play important roles in initiating or regulating significant events of the cell cycle, such as DNA replication, mitosis, and cytokinesis.
4. Maturation Promoting Factor (MPF) triggers the cell’s passage into the mitotic phase.
   [Note: Students are expected to perform the above activity by their own with the help of information provided in the answer.]

Balbharti Yuvakbharati English 11th Digest Chapter 3.1 Expansion of Ideas Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 11 English Yuvakbharati Solutions Chapter 3.1 Expansion of Ideas

11th English Digest Chapter 3.1 Expansion of Ideas Textbook Questions and Answers

Question 1.
Discuss the different ideas connoted by the word ‘season’.
Answer:

1. A season is a division of the year based on weather, ecology etc.
2. India experiences six seasons round the year, namely, summer (grisha), rainy season (varsha), early autumn (sharad), late autumn (hemanta), winter (sheet), spring (vasanta).
3. The change of season allows many people to plan their activities (not shifting of house during rainy season), food, recreation, celebrations, etc.

Question 2.
Select a season of your choice and give the following details:

• Time of the year –
• Characteristics- crops, festivals etc.
• Features/changes – climate/weather/temperature etc.
• Advantages/Disadvantages-

Answer:

Winter
Duration	October to January
Climate	Cold
Crops	Wheat, Onion, Grapes, Sunflower
Work	Sowing seeds
Festivals	Diwali, Dashera, Christmas, Makarsankranti
Advantages	Cool weather useful for the growth of crop, rebirth of natural scenery

Question 3.
Mention some proverbs associated with the word season, guess their meanings and write them along with a sentence of your own.
Answer:
(a) Make hay while the sun shines.
1. Meaning: Make the most benefit out of an opportunity
2. Sentence: Having got admission in a good college you should make hay while the sun shines.

(b) For everything there is a season
1. Meaning: For everything there is appropriate time.
2. Sentence: This is not the time to waste it talking to your friend, you must know, for everything ‘
there is a season

(c) No winter lasts for ever, no spring skips its turn.
1. Meaning: bad days do not stay for ever, as spring always replaces winter in the natural course of
nature.
2. Sentence: Cheer up, my friend, as no winter lasts for ever, no spring skips its turn.

Express your views and opinions in favor and against the topic.

Question (i)
Are sports essential in Jr. Colleges?
Answer:
Favour: (a) Sports help in increasing the physical and mental ability of students.
(b) Sports boost confidence and keep sportspersons fit for everyday living.
(c) Sports teach students team work.
Against: (a) Concentrating on sports is a waste of time. Students should focus on their career.
(b) Focus on entrance exams to persue better career opportunities.

Question (ii)
Are college council elections essential in Jr. Colleges?
Answer:
Favour: (a) The college council elections train the students to take up leadership roles and help them develop decisiveness.
Against: (a) Through elections in colleges children will follow the wrong path.
(b) They will be affected by national politics.
(c) Might get into bad company.

Question (iii)
Is cell-phone the need of the times?
Answer:
Favour: (a) Cell-phone has become the main and the fastest source of communication.
(b) Without a cell-phone one may feel crippled as it is repository of essential data and also is a source of knowledge.
Against: (a) Regular use of cell-phones affects our health conditions.
(b) Increases crimes.

Question (iv)
Should the ‘Earn and Learn’ concept be mandatory for students?
Answer:
Favour:
(a) It can make students economically independent.
(b) It can develop perseverance among students.
(c) Students will be aware of the value of money and hard work.
Against:
(a) This is an age for enjoyment.
(b) Students should concentrate on their health and family time.

(A1)

Question 1.
Expand the idea inherent in the following proverbs:
Answer:
1. A Bad Workman Blames his Tools
This proverb is an useful guide in practical life. It has been generally found that an incompetent person always grumbler. If a student does not do well in examination, he/she sometimes takes an easy resort to blame game that the question paper is tough. The person never admits that the preparation has not been up to the mark. On the other side, a sincere and good workman never tries to find excuses for his mediocre or bad performance. He probably thinks grumbling is a confession of his personal weakness.

Difficulties are parts and parcels of our lives and we have to learn to overcome by putting our best foot forward rather than finding fault with others or may be unfavourable situations. It is better to find out the reason for the performance which is not up to the mark. One has to find out the remedy for the shortcomings and need to have the faith on one’s ownself to raise the bar of his execution.

If the tools are bad, they can be mended, not by grumbling but by removing the defects. A good workman does exactly that and does not waste time complaining. “Patience and perseverance can fetch definite rewards. Everything is possible to achieve for a sincere worker whereas all things are impossible for a lazy person who is always on the look out for a scapegoat to bear the blame of his own failure.

2. One Should Eat to Live, Not Live to Eat – (Franklin)

We all have heard the proverb “health is wealth”. A sound mind can only exist in a sound body. If we are healthy, we can handle any situation in life. Eating plays a major role in maintaining a person’s health. The eating habits depend on own discretion and if we are conscious about the decision where to stop, “this far and no farther”, we can avoid many critical conditions of life with a healthy body. That is the biggest wealth any person can have.

But one needs a strong willpower not to live just to eat. The temptations are spread all around us. Some people have the habit of eating to their heart’s content and consoling themselves saying that it is just one day only. But they are greedy enough to forget their promise easily at the sight of mouth-watering dishes and continue their theory of “living to eat”.

Apparently, to some people, the eating habits do not need to be given so much importance, because they feel that the modern technologies used in the gyms will compensate for the loss of over-eating. Work hard in the gym and you can eat anything, is their motto of life. So, after coming out of the gym, they consume a big mug of cold coffee with cream from a reputed coffee shop and do not feel guilty about it. Not only home-cooked delicious foods, but all sorts of junk foods are included in their list of foods. Food is essential for one’s survival but excess of anything is bad. It is not about restrictions only, it is about how one can balance and enjoy the food.

3. If Winter Comes, Can Spring be Far Behind? – (Shelley)

Think positive and live happy, celebrate life ideally this should be the motto of a person’s life. But how many of us honestly follow this motto? Life, indeed, is beautiful but it does not move in a straight line. There has to be ups and downs and both these ups as well as downs have something to teach us, as P.B. Shelley said, “If winter comes, can spring be far behind?”. These are the natural courses of life. One has to follow the other as one season follows the other.

Life has problems and every problem is bound to have some solutions if we can try to see the brighter side. “No one makes a lock without a key. That is why God won’t give you problems without solutions.” In God’s own world also, if severe winter creates difficulties, He has kept the spring ready to bring smile on the faces of those who faced the difficult situations bravely. Winter, being the symbol of destruction here, spring will bring with it abundance everywhere.

Our duty is to maintain our patience and wait with hope for welcoming the good days because “At the end of every tunnel, there is light”. Night follows Day, that is also God’s design. If we crumble with the pressure of frustration and make ourselves fatigued, how will we enjoy the brightness of the day or the charming weather of the spring?

4. Beauty is Truth, Truth is Beauty – (Keats)

“Ode on a Grecian Urn”, the immortal poem of the poet John Keats, brings out a fact of life, that has its own beauty where joys and sorrows live together. He shows in the poem, the pictures on the urn has paintings of a combination of happiness and sadness to depict the truth of human life. A work of art has the power to express this truth of life so explicity.

“Beauty lies in the eyes of the beholder”. The person who is seeing the beauty can interpret it his or her own way, but the truth will remain the same. Truth is the permanent and ultimate beauty in the world and no one has the power to destroy it.

So, to ignore truth will be a futile attempt and only the inward beauty has the power to be the ever-lasting truth and the outward appearances are momentary. But our thoughts want to find out the truth and our feelings are inspired by the beauty. Thus, thought and feeling, truth and beauty need to go hand in hand.

5. Fools Rush in Where Angels Fear to Tread – (Alexander Pope)

“Experience is the best teacher”. The inexperienced people do not judge the pros and cons of the situation and take a step without much thinking. The experienced people become mature enough to take. a cautious step before taking a hasty decision. Their experience has taught them to “wait and watch” and then decide whether to avoid or get involved.

“A little learning is a dangerous thing”. The prudent and intelligent person think twice before opening their mouth and are also good observers. They are actually “afraid” but they are also careful that their actions should be safe and so they stay away from unnecessary risks. Even if they take risk believing “no risk, no gain”, they are capable of measuring the extent of the risk to reach their final decision. But a so called “fool” or rather an unwise person does not bother to think and can be easily influenced to do stupid things to invite problems in life.

“Ignorance” is not always “bliss”, since ignorance can lead to a irreversible damage. But, if a wise person is ignorant about certain things, he knows how to keep a distance from the unknown territory. The proverb actually tries to create an awareness against quick decisions or may be judgements, because every step of life is important. One wrong step, taken in a hurry, can lead to a major set back which probably will bring the disaster. History stands proof for that.

(A2)

Question 1.
Complete the tabular columns to specify Dos and Don’ts associated with ‘Expansion of Ideas’.
Answer:

Expansion of ideas
Dos	Don’ts
1. Begin impressively	1. Don’t go off-track
2. Use clear symbolism	2. Do not remove topic sentence
3. Focus on words and expression	3. Do not add irrelevant points
4. Should be unity and clarity of thoughts	4. Don’t use too many ideas

Yuvakbharati English 11th Digest Chapter 3.1 Expansion of Ideas Additional Important Questions and Answers

Question 1.
Expand the idea inherent in the following proverbs:
Answer:
(i) Rome was not Built in a Day

Big cities cannot be built very quickly. This actually refers to jobs that we undertake, our careers, our life and our achievements.

Just as Rome was not built in a day, so also our career, our life and our ambitions cannot be achieved in a short period of time. We must bear in mind, before we take up a project, that hard work has to go into making any endeavor a success. Success is 99% perspiration and 1% inspiration. We are aware of this, but do we honestly put in hard work or do we keep putting off our hard work for another day?

Rome took years to be built and once it was built it turned out to be a city beyond compare. Its beauty and its allure was incomparable. This was and is the result of dedication and hard work. We will achieve our goal and rise to the top only, and only if we understand the meaning of dedication, perseverance and hard work.

The three D’s must always be in our mind when we undertake a task to be done. The three D’s are- Dedication, Determination and Devotion. We will be able to touch the stars if we work hard and devote ourselves to the task ahead of us. We should not think of finishing our job in haste because haste makes waste. Only our determination and hard work will help us to achieve our aim in life.

(ii) Cut Your Coat According to Your Cloth

This proverb is something that all of us must bear in mind and abide by. The message that is given is that we must not spend beyond our means but be very careful with our expenditure.

The explanation is that if we give a piece of cloth to a tailor to stitch a coat, he will first measure the cloth and tell us whether it is possible, for him to make the coat we have asked for or not. If the cloth is insufficient, he will not be able to stitch the coat.

The same is the case with our income and expenses. We must always stay within the limits and not spend more than we can afford to or we will end up repenting. I know of a young man who wanted everything he saw advertised and he kept on buying the articles on installment-basis. Finally, he realized that the total amount he had to pay by way of the installments exceeded his income. The young man borrowed the money to make up the deficit.

How long could he go on in this way? His loans increased. People refused to give him more loans, since he could not pay back what he had already borrowed. Finally no one could wait to get the money they had given him on loan. He lost everything he had bought on installments and he landed in jail. His entire life had become one big mess – No money, no friends.

Could we too, end up like this young man? Yes, we could end up like this if we do not keep a track of the money we have and if we do not spend according to our means. This proverb teaches us to economies and to be frugal. We must learn how to manage our resources and live within our means. This will surely keep us out of trouble.

(iii) Empty Vessels Make the Most Sound

We have experienced this fact a number of times at home or even in school. When we strike on an empty vessel, we get a sonorous deep sound and if we strike on a vessel that is full of some liquid, we get only a dull thud. The above adage is metaphorically and literary correct.

We see around us people with no knowledge or very little knowledge making themselves heard above the rest and when they are questioned we realise that all is empty talk, they are just ignorant people who are trying to impress the crowd. Those who really have the knowledge are the ones who are not making a loud noise. They check out on the situation and open their mouths. What they say is the correct thing. They have knowledge and they use it wisely and correctly.

When we are in company, we must not try to prove to all present there, that we are the best, there is no one as knowledgeable as we are. We must use our etiquette and let the others have their say. We must realise that there are many who have more knowledge than we do.

It is very important to learn that when in company, we must give others the chance to have their say and not monopolies all the time such people are respected by others.

(iv) As you Sow, So shall you Reap

If we want to earn good things in life, we must do good things. If a farmer wants to cultivate rice, he will sow rice and not wheat. There was once a person who was not well to do and was really downtrodden and poor. This man had a very large heart. Whatever little food he got, he shared with others. There were times when he went without food, but he saw to it, that the beggars around had food to eat.

The neighbours observed this and decided to help this poor, warm-hearted person. They gave him an education. The man studied very hard and did well. The neighbours were very pleased. They gave him a job and he did very well. This man however never forgot the beggars and always bought them food and clothing.

Soon the poor man rose higher and higher and became the manager of the factory. He helped all the poor and even started a special “Society for the poor”. The man helped others and in return, he was helped. All of us must do well if we expect others to do good to us.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.3

Question 1.
What would be the simple interest on an amount of ₹ 9,600 at the rate of 6% per annum after 3 years?
Solution:
Given Principal P = ₹ 9600
Rate of interest R = 6% p.a.
Number of years = T = 3
Simple Interest I = \(\frac{\text { PRT }}{100}\)
= \(\frac{9600 \times 3 \times 6}{100}\)
= 96 × 18
= 1728
∴ Simple interest after 3 years would be ₹ 1728

Question 2.
What would be the simple interest at the rate of 9\(\frac{1}{2}\)% per annum on ₹ 6,000 for 2\(\frac{1}{2}\) years?
Solution:
Rate of interest per annum R = 9\(\frac{1}{2}\)% = \(\frac{19}{2}\)%
Principal P = ₹ 6000
Duration T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
∴ Simple Interest, I = \(\frac{\text { PRT }}{100}\)
= 6000 × \(\frac{19}{2} \times \frac{5}{2} \times \frac{1}{100}\)
= 15 × 19 × 5
= 1425
∴ Simple interest would be ₹ 1425.

Question 3.
What would be the simple interest on ₹ 8,400 in 9 months at the rate of 8.25 percent per annum?
Solution:
Principal P = ₹ 8400
Rate of interest R = 8.25%
Duration T = 9 months = \(\frac{3}{4}\) years

∴ Simple interest would be ₹ 519.75.

Question 4.
What would be the compound interest on ₹ 4200 for 18 months at 10% per annum compounded half yearly?
Solution:
Principal P = ₹ 4200
Rate of interest R = 10%
Duration T = 18 months = 1.5 years
compounding is done half yearly

= \(\frac{4200 \times 9261}{2000}\)
= 4862.025
I = A – P
= 4862.025 – 4200
= 662.025
∴ Compound interest would be ₹ 662.025.

Question 5.
Find compound interest on ₹ 10,000 for 2 years at 8% per annum compounded half yearly.
Solution:
Principal P = ₹ 10,000
Rate of interest R = 8% p.a. compounded half yearly
Duration T = 2 years

I = A – P
= 11648.58 – 10000
= 1698.58
∴ Compound interest is ₹ 1698.58.

Question 6.
In how many years ₹ 1,00,000 will become ₹ 1,33,100 at compound interest rate of 10% per annum?
Solution:
Principal P = ₹ 1,00,000
Amount A = ₹ 1,33,100
Rate of interest R = 10% p.a.

∴ ₹ 1,00,000 will become ₹ 1,33,100 after 3 years.

Question 7.
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many does it become double of itself at the same rate of simple interest?
Solution:
Given that, sum of money triples itself in 20 years
∴ P + I = 3P
∴ I = 2P
and T = 20 years
Now simple interest I = \(\frac{\text { PRT }}{100}\)
∴ 2P = \(\frac{\mathrm{P} \times \mathrm{R} \times 20}{100}\)
∴ R = 10
∴ Rate of interest = 10% per annum
The time period is to be calculated for the condition that the sum doubles itself i.e. for the condition
P + I = 2P
i.e. I = P
\(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = P
∴ \(\frac{10 \times T}{100}\) = 1
∴ T = 10
∴ The sum will become double of itself in 10 years.

Question 8.
A person borrows 10,000 for 2 year at 4% p.a. simple interest he immediately lends it to another person at 6.5% p.a. for 2 years. Find his total gain in the transaction.
Solution:
Person borrows money at 4% per annum and lends it at 6.5% per annum.
∴ His gain is (6.5 – 4) = 2.5% on ₹ 10000 for 2 years
i.e. gain = \(\frac{1000 \times 2.5 \times 2}{100}\)
= 100 × 5
= ₹ 500
∴ The person will gain ₹ 500 in this transaction.

Question 9.
A man deposits X 200 at the end of each year in recurring account at 5% compound interest. How much will it become at the end of 3 years?
Solution:
At end of 1st year, 2nd year and 3rd year ₹ 200 were deposited.
Rate of interest R = 5% p.a.
At end of 3 years, amount
= 200 + \(200\left[1+\frac{5}{100}\right]+200\left[1+\frac{5}{100}\right]^{2}\)
= 200 [1 + 1.05 + (1.05)2]
= 200 [2.05 + 1.1025]
= 200 [3.1525]
= 630.5
At end of 3 years, the account will have a balance of ₹ 630.5.

Question 10.
A man gets a simple interest of ₹ 2,000 on a certain principal at the rate of 5% p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate.
Solution:
Let Principal amount = P
Simple Interest I = ₹ 2000
Rate of interest R = 5% p.a.
Time duration T = 4 years
I = \(\frac{\text { PRT }}{100}\)
∴ 2000 = \(\frac{\mathrm{P} \times 5 \times 4}{100}\)
∴ P = 10000
Twice the principal was invested for compound interest with the same rate of interest for 2 years.
Here, P = 2 × 10,000 = ₹ 20,000
∴ Amount received,

I = A – P = 22050 – 20000 = 2050
The man will receive ₹ 2050 as compound interest.

Question 11.
The difference between simple interest and compound interest on a certain sum of money is ₹ 32 at 8% per annum for 2 years. Find the amount.
Solution:
Compound Interest = A – P = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-\mathrm{P}\)
Simple interest = \(\frac{\text { PRT }}{100}\)
Given R = 8%, T = 2 years and
compound interest – simple interest = ₹ 32

∴ The man will receive a compound interest of ₹ 5000.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

I. Integrate the following functions w.r.t. x:

(i) x³ + x² – x + 1
Solution:

(ii) \(x^{2}\left(1-\frac{2}{x}\right)^{2}\)
Solution:

(iii) \(3 \sec ^{2} x-\frac{4}{x}+\frac{1}{x \sqrt{x}}-7\)
Solution:

(iv) \(2 x^{3}-5 x+\frac{3}{x}+\frac{4}{x^{5}}\)
Solution:

(v) \(\frac{3 x^{3}-2 x+5}{x \sqrt{x}}\)
Solution:

II. Evaluate:

(i) ∫tan² x . dx
Solution:

(ii) \(\int \frac{\sin 2 x}{\cos x} \cdot d x\)
Solution:

(iii) \(\int \frac{\sin x}{\cos ^{2} x} \cdot d x\)
Solution:

(iv) \(\int \frac{\cos 2 x}{\sin ^{2} x} \cdot d x\)
Solution:

(v) \(\int \frac{\cos 2 x}{\sin ^{2} x \cdot \cos ^{2} x} \cdot d x\)
Solution:

= -cot x – tan x + c

(vi) \(\int \frac{\sin x}{1+\sin x} \cdot d x\)
Solution:

(vii) \(\int \frac{\tan x}{\sec x+\tan x} \cdot d x\)
Solution:

(viii) \(\int \sqrt{1+\sin 2 x} \cdot d x\)
Solution:

(ix) \(\int \sqrt{1-\cos 2 x} \cdot d x\)
Solution:

(x) ∫sin 4x cos 3x dx
Solution:

III. Evaluate:

(i) \(\int \frac{x}{x+2} \cdot d x\)
Solution:

(ii) \(\int \frac{4 x+3}{2 x+1} \cdot d x\)
Solution:

(iii) \(\int \frac{5 x+2}{3 x-4} \cdot d x\)
Solution:

(iv) \(\int \frac{x-2}{\sqrt{x+5}} \cdot d x\)
Solution:

(v) \(\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x\)
Solution:

(vi) \(\int \frac{\sin 4 x}{\cos 2 x} \cdot d x\)
Solution:

(vii) \(\int \sqrt{1+\sin 5 x} \cdot d x\)
Solution:

(viii) ∫cos² x . dx
Solution:

(ix) \(\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \cdot d x\)
Solution:

(x) \(\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \cdot d x\)
Solution:

IV.

Question 1.
If f'(x) = x – \(\frac{3}{x^{3}}\), f(1) = \(\frac{11}{2}\), find f(x).
Solution:
By the definition of integral,

Balbharti Maharashtra State Board Class 11 History Solutions Chapter 5 Janapadas and Republics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 History Solutions Chapter 5 Janapadas and Republics

1A. Choose the correct alternative and write the complete sentences.

Question 1.
A region occupied by the ‘janas’ was called as __________
(a) Ganarajya
(b) Janapada
(c) Mahajanapada
(d) Gotra
Answer:
(b) Janapada

Question 2.
The principal functionary of a Ganasangha was known as __________
(a) Senapati
(b) Bhandagarika
(c) Raja
(d) Uparaja
Answer:
(c) Raja

Question 3.
The author of the ‘Ashtadhyayi’ which mentions ‘Janapadin’ was __________
(a) Kautilya
(b) Panini
(c) Chanakya
(d) Vyas
Answer:
(b) Panini

Question 4.
The sense of territoriality and the ensuing awareness __________ were the main factors responsible for the formation of ancient Janapadas in India.
(a) of unity
(b) of authority
(c) of autonomy
(d) of public authority
Answer:
(c) of autonomy

1B. Find the incorrect pairs from set ‘B’ and write the correct ones.

Question 1.

Set ‘A’	Set ‘B’
(a) Praachya	of the east
(b) Praatichya	of the west
(c) Udichya	of the north
(d) Aparanta	the region to the north of Vindhya ranges

Answer:
(d) Aparanta – the region to the south of Vindhya ranges

2. Choose the correct reason and complete the sentence.

Question 1.
The Ganasangha of the Youdhey, Malav, Kshudraka were mentioned as Ayudhajivi. Because-
(a) it was in the northeast region of the Indian Subcontinent.
(b) these people were skilled warriors and warfare was the means of their livelihood.
(c) they were skilled in trade and commerce.
(d) these were the ganasanghas dependent on agriculture and animal husbandry.
Answer:
(b) these people were skilled warriors and warfare was the means of their livelihood.

3. Complete the concept map.

Question 1.

Answer:

4. Explain the concepts with examples.

Question 1.
Ganarajya and Sangharajya
Answer:

• ‘Gana’ means the ruling class comprising members of equal social status.
• Similarly, ‘sangha’ means a state formed by many kulas or janapadas by coming together.
• By the 6th century B.C.E. many sangharajyas had come into existence.
• There were three main types of the ancient federation of states in India:
• Ganarajya of the members of the same kula. For example, Malava and Shibi.
• Ganarajya was created by more than one kulas coming together. For example, Vajji Ganasangha. It included eight kulas. Vajji, Lichchhavi, Dnyatruk, and Videha were the important ganas among them.
• More than one ganrajyas coming together to create a sangharajya. For example, Yaudheya- Kshudrak Sangh.

Question 2.
Vartashastarpajivi Ganasnagh
Answer:

• Ancient Indian literature mentions two more types of ganasanghas.
• ‘Ayudhjivi’ sangh and ‘Varta-Shastropajivi’ sangh.
• ‘Varta’ means trade and commerce.
• The people in the Varta-Shastropajivi ganasanghas lived by trade and commerce, agriculture and animal husbandry, as well as their skills in warfare.
• People in the Kamboj and Saurashtra ganasanghas earned their livelihood by these means.

Question 3.
‘Jana’ and ‘Janapada’
Answer:

• Vedic people used the term Jana to designate a group of people, united under a common bond of singular kinship structure.
• Their settlement was known as ‘Grama’.
• A cluster of gramas consisting of the same Jana was known by the name of that particular Jana.
• A region occupied by a Janas was called Janapada.
• Gradually the Janapadas had more formal administrative structures transforming them into independent states.
• These were the first well-established states of ancient India.
• However, this does not necessarily mean that every Janapada evolved into an independent state.

5. Answer the following questions in detail.

Question 1.
Describe the democratic and oligarchic states in ancient India.
Answer:
Democratic States:

• Some of the ganasanghas were divided into regional zones called ‘Khanda’.
• They functioned through a group of elected individuals, who were found capable.
• Each of the elected members represented his respective khanda.
• These elected members were installed with collective authority for the smooth running of the ganasangha.
• This was a democratic system. Ganasanghas which functioned in this democratic way existed in Punjab and Sindh at the time of Alexander’s invasion.
• Each elective representative of the respective regional zone was designated as ‘Ganamukhya’.
• Every ganamukhya was a member of the assembly known as ‘ganaparishada’.
• The decisions made by the ganaparishada were implemented by designated functionaries of various cadres.
• He was known as the ‘Adhyaksha’ or ‘Raja’.

Oligarchic States:

• In this type the elite class in the society held all the powers of decision-making and administration,
• Panini and Kautilya mention them as ‘Rajshabdopajivi’ Sangh.
• Panini includes Vajji, Andhaka, Vrishni, Yaudheya in the Rajashabdopjivi type.
• Kautilya includes the Vrijji or Vajji, Madrak, Kuru, Panchala, etc. in this type.
• This type of ganasanghas was more prevalent in the eastern region of Uttar Pradesh and Bihar.

Activity

Present an act in the class based on the simulation of the administrative system of an oligarchic state.
Answer:
Students have to make the presentation in class.

Balbharti Maharashtra State Board Class 11 History Solutions Chapter 4 Vedic Period Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 History Solutions Chapter 4 Vedic Period

1A. Choose the correct alternative and write the complete sentences.

Question 1.
The importance of agriculture is clearly emphasised in the ______________ mandala of the Rigveda.
(a) 4th
(b) 10th
(c) 8th
(d) 6th
Answer:
(b) 10th

Question 2.
A potter is mentioned as an artisan in the ______________ for the first time.
(a) Rigveda
(b) Yajurveda
(c) Samaveda
(d) Atharvaveda
Answer:
(b) Yajurveda

Question 3.
The god who protected the animals was known as ______________
(a) Indra
(b) Pushan
(c) Ashwin
(d) Varun
Answer:
(b) Pushan

1B. Find the incorrect pair from set ‘B’ and write the correct ones.

Question 1.

Set ‘A’	Set ‘B’
(a) Nishka	Gold ornament
(b) Barley	Maincrop of the Vedic people
(c) Krishtya	The apparatus of drawing water from the well
(d) Navya	River transport

Answer:
(c) Krishtya – Tribal settlements

1C. Write the names.

Question 1.
A branch of linguistics.
Answer:
Philology

Question 2.
A battle was fought among the ten tribal chiefs.
Answer:
Dasharajnya Yuddha

Question 3.
Cultivable land prepared by ploughing.
Answer:
Urvara

2. Complete the concept map given below:

Question 1.

Answer:

3. Choose the correct reason and complete the sentence.

Question 1.
Panis were looked upon as enemies by the Vedic people because-
(a) they belonged to a different tribe.
(b) their language was impure.
(c) Panis used to steal the cattle of the Vedic people.
(d) Panis did not obey the orders of the Vedic people.
Answer:
(c) Panis used to steal the cattle of the Vedic people.

4. State your opinion.

Question 1.
There is a debate about the original home of the Aryans.
Answer:

• The debate began in the 16th Century. Till then the concept of the ‘Aryans’ was not known.
• The European academics became aware of the similarities between Sanskrit and Latin-Greek languages.
• This resulted in the notion of the Indo-European language family which gave momentum to the search for a Mother language from which developed the Indo-European languages.

Question 2.
The Rigvedic people subsisted on agriculture.
Answer:

• The Rigvedic tribal settlements have been mentioned as ‘Krishtya’ in the Rigveda. ‘Krish’ means ploughing.
• Hence the people who ploughed and also their settlements were mentioned as ‘Krishtya’.
• The importance of agriculture is clearly emphasised in the 10th mandala of Rigveda.
• It states that for the farmer his ploughshare is the means of obtaining.

5. Explain the following concepts.

Question 1.
Origin of Aryan people.
Answer:

• There are multiple unanswered questions, such as, who were the Aryans, did they arrive in India from a distant place or were they natives of India, which are the archaeological sites where the remains of their culture are found, how to identify those remains, etc.
• Most of the information about their culture is derived from Vedic literature. There are various v opinions about the chronology of the Vedic culture.
• However, there is a general agreement that the Vedic people composed Rigveda in India around 1500 B.C.E.
• However, Lokmanya Tilak calculated this date as 6000 B.C.E. on the basis of astronomical events.
• He was also of the opinion that the original home of the Aryans was in the Arctic region.
• This debate began in the 16th century. Till then the concept of the ‘Aryans’ was not known.

Question 2.
Indo-European family of languages.
Answer:

• In the year 1583, an Italian merchant by the name of Filippo Sassetti came to Kochi (Cochin) the port city in Kerala.
• He never returned to his motherland. He stayed in Kerala and Goa. During his stay, he wrote detailed letters to his family members about Indian life, language and culture. He happens to be the first European who wrote about his observations of Indian society.
• He also studied Sanskrit. He was the first one who observed the similarities between Sanskrit and Latin.
• His observations could be said to be the first, to give impetus to the notion of a family of Indo-European languages.

Balbharti Maharashtra State Board Class 11 History Solutions Chapter 3 Chalcolithic Villages in India Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 History Solutions Chapter 3 Chalcolithic Villages in India

1A. Choose the correct alternative and write the complete sentences.

Question 1.
On one of the cemetery H pots, dead humans are seen carried by ____________ in their stomach.
(a) deer
(b) peacock
(c) fish
(d) bull
Answer:
(b) peacock

Question 2.
The archaeological evidence shows that Balathal was a center of mass production of ____________
(a) stone pots
(b) copper pots
(c) earthen pots
(d) glassware
Answer:
(c) earthen pots

Question 3.
Permanent villages of farmers were first established in Maharashtra by ____________ people.
(a) Savalda
(b) Malwa
(c) Harappa
(d) Kayatha
Answer:
(b) Malwa

1B. Find the incorrect pair from set ‘B’ and write the correct ones.

Question 1.

Set ‘A’	Set ‘A’
1. Savalda Culture	Daimadabad
2. Malwa Culture	Navadatoli
3. Ahar Culture	Sonpur
4. Jorwe Culture	Inamgaon

Answer:
3. Ahar Culture – Balathal

2A. Explain the statements with reasons.

Question 1.
Harappan people had to migrate.
Answer:

• When the Mature (urban) Harappan civilisation collapsed completely, the people of Late Harappan cultures who had settled on the ruins of Mature Harappan cities had to migrate elsewhere.
• The urban Harappans and the Late Harappans dispersed. Wherever these people reached, new rural cultures came into being.
• Thus, as the Harappan civilisation collapsed, Harappan people had to migrate.

Question 2.
People of ‘Malwa’ culture were the first farmers of Maharashtra.
Answer:

• The people of Malwa culture reached Maharashtra around 1600 B.C.E. Permanent villages of farmers were first established in Maharashtra by the Malwa people.
• After arriving in Maharashtra, they came into contact with the neolithic people in Karnataka.
• It resulted in a few changes in the pot-making technology of Malwa people as far as shapes of the pots and designs are concerned.
• Thus, they were considered the first farmers of Maharashtra as they were the first to establish in Maharashtra.

3. State your opinion.

Question 1.
It seems that the Harappan people had gone as far as Bihar.
Answer:

• Chalcolithic sites have been discovered in Bihar, Bengal, Odisha, and Madhya Pradesh.
• The excavations at sites like Chirand, Sonpur, etc. yielded earthen pots of black-and-red ware.
• The shapes of these pots are similar to those of Harappan culture.
• Thus, it seems that the Harappan people had gone as far as Bihar, and the local cultures were influenced by them.

4. Write short notes.

Question 1.
Banas Culture
Answer:

• The chalcolithic culture in India generally belongs to, the Post-Harappan period.
• However, the ‘Ahar’ or ‘Banas’ culture in the Mewad region of Rajasthan was contemporary to the Harappan civilisation.
• Balathal and Gilund are important sites of Ahar culture.
• It was first discovered at Ahar near Udaipur, so it was named as ‘Ahar’ culture.
• Ahar is a tributary of the river Banas, so it is also known as ‘Banas culture’.

Question 2.
Malwa Culture
Answer:

• The name ‘Malwa’ tells us that this culture originated and spread first in the Malwa region.
• It flourished in Madhya Pradesh during 1800 B.C.E. – 1200 B.C.E. Navadatoli, situated on the bank of Narmada is an important site of Malwa culture.
• The people of Malwa culture reached Maharashtra around 1600 B.C.E.
• Permanent villages of farmers were first established in Maharashtra by the Malwa people. They were the first farmers of Maharashtra.

Question 3.
Kayatha Culture
Answer:

• Kayatha is a chalcolithic site situated on the banks of the river known as Chhoti Kali Sindh at a distance of 25 km from Ujjain in Madhya Pradesh.
• Kayatha culture was contemporary to the Harappan civilisation.
• The Kayatha people followed agriculture and animal husbandry.
• They mainly used handmade pots and microliths.

5. Write about the chalcolithic cultures in Gujarat with the help of the given points.

Question 1.
(a) Period
(b) Means of livelihood
(c) Geographical spread
(d) Evidence of cultural contact with other people.
Answer:
(a) Period: The chalcolithic settlements in Gujarat coincide with the following phases of the Harappan culture:

• Early Harappan phase (3950-2600 B.C.E.)
• Mature (urban) phase (2600-1900 B.C.E.)
• Post-Harappan phase (1900-900 B.C.E.)

(b) Means of livelihood: There are ample sources of semi-precious stones in Gujarat. Making beads of these stones was a big industry during Harappan times. The Neolithic settlements in Gujarat played a major role in procuring these stones. People residing in the neolithic settlements of Gujarat were mainly pastoral, that is people whose primary occupation was animal husbandry. Thus, making beads, animal husbandry, pottery making were some of the means of livelihood.

(c) Geographical spread: There are regional variations in the characteristics of the chalcolithic cultures of Gujarat. The chalcolithic pottery of Kutch – Saurashtra and Northern Gujarat are distinct from each other. The chalcolithic villages in Kutch-Saurashtra were abandoned by 1900 B.C.E.

(d) Evidence of cultural contact with other people: In the post-Harappan period there were two chalcolithic cultures in Gujarat. The culture in south Gujarat was known as ‘Prabhas’ culture and the one in northeastern Gujarat was known as ‘Rangpur’ culture. These cultures existed till 1800-1200 B.C.E.

Activity

With the help of the Internet, reference books, field trips, newspaper articles, etc. obtain pictures of excavated artifacts and architectural remains and arrange an exhibition under the guidance of your teachers.
Answer:
To be done by students.

Balbharti Maharashtra State Board Class 11 History Solutions Chapter 2 First Cities of India Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 History Solutions Chapter 2 First Cities of India

1. Choose the correct alternative and write the complete sentences.

Question 1.
In the early phase of agriculture, making __________ pots and farming were the jobs of women.
(a) copper
(b) bronze
(c) earthen
(d) stone
Answer:
(c) earthen

Question 2.
Lothal is known for its ancient __________
(a) agriculture
(b) dock
(c) textiles
(d) tools
Answer:
(b) dock

Question 3.
A general impression prevailed that the Harappan seals had some connection with the __________ civilisation.
(a) Chinese
(b) Greek
(c) Mesopotamian
(d) Egyptian
Answer:
(c) Mesopotamian

Question 4.
The bodies (mummies) of dead royalties in Egypt were wrapped in __________ coloured cloth.
(a) white
(b) black
(c) red
(d) blue
Answer:
(d) blue

2A. Choose the correct reason and complete the sentence.

Question 1.
The major reason for the decline of the Mesopotamian civilisation was __________
(a) foreign invasion
(b) deteriorating environment
(c) loss in trade
(d) migration
Answer:
(b) deteriorating environment

2B. Find the incorrect pair from set ‘B’ and write the correct ones.

Question 1.

Set ‘A’	Set ‘B’
(a) Dilmun	Bahrain
(b) Makan	Oman-Iran-Baluchistan Coastal region
(c) Shortugai	Mesopotamia
(d) Meluhha	Region of Harappa civilisation

Answer:
(c) Shortugai – Badakshan province of Afghanistan

3. Explain the statements with reasons.

Question 1.
The remains found in cities like Harappa, Mohenjodaro, Kalibangan, Lothal, Dholavira, Rakhigarhi, etc. are evidence of the past glory of this civilization.
Answer:

• A well-developed and rich civilisation flourished in the Indian subcontinent in the period from 3500-3000 B.C.E.
• This period is characterised by systematic town planning, with houses of baked brick, granaries, excellent drainage systems, public baths, and impressive public movements.
• Good administrative control over the distribution of water and other resources.
• Remains also prove a good social organisation, a proper social hierarchy based on the position of power.
• Flourishing internal and distant trade, mass production of goods, and craft specialisation developed writing system of script on Harappan seals give evidence of the past glory of the civilisation.

Question 2.
Lapis lazuli had a very important place in the Harappan trade.
Answer:

• A network of small towns, big and small villages, and campsites of semi-nomadic people functioned to cater to the needs of major urban centres of the Harappan civilisation.
• The Harappan settlement of ‘Shortugai’ in the Badakshan province of Afghanistan, is rich with mines of lapis lazuli, a semi-precious stone in Mesopotamia.
• The Mesopotamian epics describe Goddess Inanna’s palace, the walls of which were embedded with this stone, lapis lazuli.
• This stone was a very important commodity in the Harappan trade with Mesopotamia.

Question 3.
Harappan civilisation declined.
Answer:

• Sir Mortimer Wheeler in his study has concluded that the Harappan civilisation was destroyed by Indra, who destroyed the fortified cities.
• Vedic Aryans destroyed the Harappan cities under the leadership of Indra.
• The civilisation also declined due to the cumulative effects of factors such as the decline in trade, climatic changes, and the weakening of the economy.

4. State your opinion.

Question 1.
The Harappan cities and villages in the vicinity were interdependent.
Answer:

• The interrelationship between Harappan cities and villages was dependent on the mechanism of making available food grains and raw materials.
• A network of small towns, big and small villages, and campsites of semi-nomadic people, functioned to cater to the needs of major urban centres of Harappan civilisation.
• The city people were dependent on natural sources and villages in their vicinity, to meet the needs of urban lifestyle and urban administration.

Question 2.
The Harappan cities seem to have a well-organised administrative system.
Answer:

• The Harappan cities had a well organised administrative system to manage industrial production, import-export, the interrelationship between cities managing trading operations and villages around them.
• The town planning, standardization of bricks, weights, seals, shapes, and ornamentation of various objects, confirm the presence of an efficient administrative system.
• Cities like Harappa and Mohenjodaro were perhaps regional capitals.
• Lothal and Kalibangan were important religious centres.
• However, the nature of Harappan polity, whether it was a single state or a federation of small states, is not yet known.

5. Answer the following questions with the help of given points.

Question 1.
Write about the characteristics of Harappan cities with the help of the given points:
(a) Town planning
(b) Social organisation
(c) Administration
(d) Economy
Answer:
(a) Town planning:

• The town planning of Harappan cities was very systematic.
• Houses were of baked bricks, which included bathrooms, toilets, wells.
• Granaries existed, with impressive public monuments.
• The excellent drainage systems, public baths, and independent fortification walls are highlights.
• The grid pattern was used, where streets crossed each other in right angles, and the resulted blocks were used for building houses.
• The English bond masonry method was used with two headers and two stretchers to build a wall, which was especially useful for earthquake-prone areas.

(b) Social organisation:

• Social hierarchy was based on the position of power.
• Classes of skilled artisans and individuals were based on craft specialisation.
• Belief systems existed, with evidence of burials indicating rituals after death.
• Artifacts and architecture also indicate belief systems.

(c) Administration:

• Administrative control existed over the distribution of water and other resources.
• The size of bricks indicates the use of standardisation and ratio.
• Weights, set style of shapes and painted designs of pottery, majestic and non-residential buildings for public administrative offices are also seen.

(d) Economy:

• Harappan civilisation practiced mass production of goods for trade purposes.
• The concentration of factories and residences of artisans in a particular area of the city indicates the purpose of convenience of production, flourishing internal and distant trade, and administrative control over trade transactions.
• Well-shaped, beautiful earthen pots, statues, metal objects of gold, silver, copper, and bronze were made.
• Various types of beads were prepared, indicating a sound economy.

Activity

Collect information and illustrations with the help of the internet about the town planning of the Harappan cities and Chandigarh. Compare them.
Answer:
The town planning of the Harappan Cities:

• The Harappans were the first to build planned cities with a scientific drainage system.
• Their cities were built on a uniform plan.
• The people of Indus valley lived a highly civilized and developed life.
• This highly developed and scientific plan can be seen in the following areas.

Streets:

• The streets were straight and cut each other at right angles
• They were 13 to 34 feet wide and were well lined.
• The streets and roads divided the city into rectangular blocks.
• Lamp posts were provided at regular intervals.
• Dust bins were also provided on the streets which proves the presence of a good municipal administration.

Drainage System:

• The city was provided with an excellent closed drainage system.
• Each house had its own drainage and soak pit which was connected to the public drainage.
• Brick-laid channels were found through every street.
• The drains were covered and had manholes at regular intervals for cleaning and clearing.
• Large brick culverts were constructed on the outskirts of the city to carry excess water.
• The Indus valley civilization had a perfect underground drainage system.

The Great Bath:

• The most striking feature of Mohenjo Daro is the Great Bath.
• It consists of a large quadrangle. In the center, there is a large swimming pool approx. 39 ft long, 23 ft wide, and 8 ft deep.
• This swimming pool had rooms and galleries on all four sides.
• It had a flight of steps at either end and a well in one of the adjoining rooms. The water was discharged by a huge drain.
• The Great Bath had 8ft thick outer walls.

Granaries:

• The largest building in Mohenjo Daro is the granary which is 45.71 m long and 15.23 m wide.
• Granaries have also been found in Harappa and the southern parts of Kalibangan.
• These granaries were used to store grains which were probably collected as revenue or storehouses to be used in emergencies.

Buildings:

• People of the Indus valley civilization built houses and other buildings on the side streets.
• Built terraced houses of burnt bricks.
• Every house had two or more rooms, there were also more than one-storied houses.

The town planning of Chandigarh City:

• Chandigarh, the capital of the northern Indian states of Punjab and Haryana was designed by the Swiss-French modernist architect, Le Corbusier.
• Buildings include the Capitol Complex with its High Court, Secretariat, and Legislative Assembly.