Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 10 Secondary Growth Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 10 Secondary Growth

11th Bio Botany Guide Secondary Growth Text Book Back Questions and Answers

Part -I

I. Consider the following statements.

Question 1.
In spring vascular cambium
i) is less active
ii) Produces a large number of xylary elements
iii) forms vessels with wide cavities of these,
a) (i) is correct but (ii) and (iii) are not correct
b) (i) is not correct but (ii) and (iii) are correct
c) (i) and (ii) are correct but (iii) is not correct
d) (i) and (ii) are not correct but (iii) is correct
Answer:
b) (i) is not correct but (ii) and (iii) are correct

Question 2.
Usually, the monocotyledons do not increase their girth, because
a) They possess actively dividing cambium
b) They do not possess actively dividing cambium
c) Ceases activity of cambium
d) All are correct
Answer:
b) They do not possess actively dividing cambium

Question 3.
In the diagram of lenticel identify the parts marked as A, B, C, D

a) A. phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.
b) A. Complementary tissue, B.Phellem, C. Phellogcn,
c) A. Phellogen, B.Phellem, C. Phelloderm, D. Complementary tissue
d) A. Phelloderm, B. Phellem, C. Complementary tissue, D. Phellogen
Answer:
a) A. phellem, B. Complementary tissue, C. Phelloderm, D. Phellogen.

Question 4.
The common bottle cork is a product of
a) Dermatogen
b) Phellogen
c) Xylem
d) Vascular cambium
Answer:
b) Phellogen

Question 5.
What is the fate of primary xylem in a dicot root showing extensive secondary growth?
a) It is retained in the center of the axis
b) It gets crushed
c) May or may not get crushed
d) It gets surrounded by primary phloem
Answer:
b) It gets crushed

Question 6.
In a forest, if the bark of a tree is damaged by the horn of a deer, How will the plant overcome the damage?
Answer:
When the bark is damaged, the phellogen forms a complete cylinder around the stem and it gives rise to ring barks.

Question 7.
In which season the vessels of angiosperms are larger in size, why?
Ans:
In spring season the vessels are larger in size, because the cambium cells are very active during spring season.

Question 8.
Continuous state of dividing tissue is called meristem. In connection to this, what is the role is lateral meristem?
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

1. Vascular cambium and
2. Cork cambium

1. Vascular cambium:
The vascular cambium is the lateral meristem that produces the secondary vascular tissues, i.e.. secondary xylem and secondary phloem.

Origin and Formation of Vascular Cambium:

• A strip of vascular cambium originate from the procambium is present between xylem and phloem of the vascular bundle. This cambial strip is known as intrafascicular or fascicular cambium.
• In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of vascular cambium. It is called interfascicular cambium.

A. Organization of Vascular cambium:

• The active vascular cambium possesses cells with large central vacuole (or vacuoles) surrounded by a thin, layers of dense cytoplasm.
• The most important character of the vascular cambium is the presence of two kinds of initials, namely fusiform initials and ray initials.

Fusiform Initials:

• These are vertically elongated cells. They give rise to the longitudinal or axial system of the secondary xylem (tracheary elements, fibres, and Axia? parenchyma) and pholem (sieve, elements, fibres, and axial parenchyma).
• Based on the arrangement of the fusiform initials two types of vascular cambium are recognized.

Stoned (Stratified cambium) and Non – storied (Non – stratified cambium)

• If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in tangential longitudinal section (TLS) it is called storied (stratified) cambium. It is the characteristic of the plants with short fùsiform initials.
• In plants with long fusiform initials, they strongly overlap at the ends, and this type of cambium is called non – storied (non stratified) cambium.

Ray Initials:
These are horizontally elongated cells. They give rise to the ray cells and form the elements of the radial system of secondary xylem and pholem.

Activity of Vascular Cambium:

• The vascular cambial ring, when active, cuts off new cells both towards the inner and outer side. The cells which are produced outward form secondary phloem and inward secondary xylem.
• Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.

B. Phellogen (Cork Cambium)

• It is a secondary lateral meristem. It comprises homogenous meristematic cells unlike vascular cambium. It arises from epidermis, cortex, pholem or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells.
• The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phelloderm (secondary cortex).

Question 9.
A timer merchant bought 2 logs of wood from, a forest & named them A & B, The log A was 50 year old & B was 20 years old. Which log of wood will last longer for the merchant? Why?
Answer:

• In wood, the older it is, the stronger it becomes.
• Log A – Which was 50 years old is stronger and it will last longer.
• In a tree the central part of the wood will be darker in colour, dead in nature known as Heartwood or Duramen, and the outer sad wood is lighter in colour, living and conducting water.
• In the central Heartwood the conduction is blocked by the formation of tyloses from the nearby parenchyma cells, and dead.
• In the fully developed tyloses, starch crystals, resins, gums, oils tannins and coloured substances are found and it becomes very hard and durable.
• It is more resistant to the attack of microbes and insects like termites.
• Older woods have more heartwood than sapwood.
• Here log ‘A’ is older, has more heartwood and it is stronger and will last longer.

Question 10.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What are the significance of these rings?
Answer:
Growth (or) Annual Rings:
1. In the spring season cambium is very active and produces large number of xylary elements called Earlywood or Springwood. In the Winter season – cambium is less active and form few xylary elements – Latewood or Autumn Wood.

2. The springwood is lighter in color and has a lower density whereas the autumn wood is darker and has a higher density. The annual ring denotes the combination of earlywood and latewood and the ring becomes evident to our eye due to the high density of latewood. Sometimes annual rings are called growth rings

3. Pseudo – Annual Rings:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost defoliation, flood, mechanical, injury and biotic factors. Such rings arc called pseudo – or false – annual rings

4. Dendrochronology:
Each annual ring corresponds to one year’s growth and on the basis of these rings, the age of a particular plant can easily be calculated. The determination of the age of a tree by counting the annual rings is called dendrochronology.

Part – II.

11th Bio Botany Guide Secondary Growth Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
The roots and stems grow in length with the help of:
(a) cambium
(b) secondary growth
(c) apical meristem
(d) vascular parenchyma
Answer:
(c) apical meristem

Question 2.
The Gymnosperm in which vessel is present
a) Pinus
b) Cýcas
c) Ginkgo
d) Gnetum
Answer:
d. Gnetum

Question 3.
The secondary vascular tissues include:
(a) secondary xylem and secondary phloem
(b) secondary xylem, cambium strip and secondary phloem
(c) secondary phloem and fascicular cambium
(d) secondary xylem and primary phloem
Answer:
(a) secondary xylem and secondary phloem

Question 4.
In a dicotyledonous stem, the sequence of tissues from the outside to the inside of
a) Phellem, Pericycle, Endodermis, phloem
b) Phellem phloem, Endodermis, Pericycle
c) Phellem, Endodermis, Pericycle, phloem
d) Pericycle, Phellem, Endodermis, Phloem
Answer:
c. Phellem, Endodermis, Pericycle, Phloem

Question 5.
For a critical study of secondary growth in plants which one of the following pairs is suitable?
a) Sugarcane and sunflower
b) Teak and pine
c) Bamboo and Fem
d) Wheat and Fem
Answer:
b. Teak and pine

Question 6.
The axial system of the secondary xylem includes:
(a) treachery elements, sieve elements, fibers and axial parenchyma
(b) treachery elements, fibers and axial parenchyma
(c) treachery elements and fibers
(d) sieve elements and axial parenchyma
Answer:
(b) treachery elements, fibers and axial parenchyma

Question 7.
Tissues considered in an annual ring is/are
a) Secondary xylem and phloem
b) Primary xylem and phloem
c) Secondary xylem only
d) Primary phloem and secondary xylem
Answer:
c. Secondary xylem only

Question 8.
Ray cells are present between:
(a) primary xylem and phloem
(b) primary xylem and secondary xylem
(c) secondary xylem and phloem
(d) secondary phloem and cambium
Answer:
(c) secondary xylem and phloem

Question 9.
Interfascicular cambium is a
a) Primary meristematic tissue
b) Primordial meristem
c) Type of protoderm
d) Secondary meristematic tissue
Answer:
d. Secondary meristematic tissue

Question 10.
Interfascicular cambium develops from the cells of
a) Xylem parenchyma
b) endodermis
c) Pericycle
d) Medullary rays
Answer:
d. Medullary rays

Question 11.
Which of the statement is not correct?
(a) In temperate regions, the cambium is very active in the winter season.
(b) In temperate regions, the cambium is very active in the spring season.
(c) In temperate regions, cambium is less active in the winter season.
(d) In temperate regions earlywood is formed in the spring season.
Answer:
(a) In temperate regions, the cambium is very active in the winter season.

Question 12.
At maturity, the sieve plates become impregnated with
a) Cellulose
b) Pectin
c) Suberin
d) Callose
Answer:
d. Callose

Question 13.
You are given a fairly old piece of dicot stem and a dicot root, which of the following anatomical structures will you distinguish between the two?
a) Secondary xylem
b) Secondary phloem
c) Protoxylem
d) Cortical cells
Answer:
c. Protoxylem

Question 14.
determination of the age of a tree by counting the annual rings is called:
(a) chronology
(b) dendrochronology
(c) palaeology
(d) histology
Answer:
(c) palaeology

Question 15.
Which one of the following is dead and works efficiently?
a) Sieve tube
b) Companian cells
c) Vessels
d) Both (b) and (c)
Answer:
c. Vessels

Question 16.
Which one of the following pairs is an example for meristematic tissue
a) Phellogen and phelloderm
b) Phellogen and Fascicular cambium
c) Procambium and phelloderm
d) Interfascicular cambium and phellem
Answer:
b. Phellogen and Fascicular cambium

Question 17.
In fully developed tyloses:
(a) only starchy crystals are present
(b) resin and gums only are present
(c) oil and tannins are present
(d) starchy crystals, resins, gums, oils, tannins, or colored substances are present
Answer:
(d) starchy crystals, resins, gums, oils, tannins, or colored substances are present

Question 18.
Which wood is also known as Non-porous
a) Softwood
b) Heartwood
c) Hardwood
d) Sapwood
Answer:
a. Softwood

Question 19.
Which of the statement is not correct?
(a) Sapwood and heartwood can be distinguished in the secondary xylem
(b) Sapwood is paler in colour
(c) Heartwood is darker in colour
(d) The sapwood conducts minerals, while the heartwood conduct water
Answer:
(d) The sapwood conducts minerals, while the heartwood conduct water

Question 20.
Removal of a ring of wood tissue outside the vascular cambium from the tree trunk kills it because
a) Water cannot move up
b) Food does not travel down and root become starved
c) Shoot apex become starved
d) Annual rings are not produced
Answer:
food does not travel down root become starved

Question 21.
The trees growing in the desert will
a) Show alternate rings of xylem and sclerenchyma
b) Have only conjunctive tissue and phloem formed by the activity of cambium
c) do show distinct annual rings
d) do not show distinct annual rings
Answer:
d. do not show distinct annual rings.

Question 22.
Canada balsam is produced from:
(a) Pisum sativum
(b) the resin of Arjuna plant
(c) Abies balsamea
(d) the root of Vinca rosea
Answer:
(c) Abies balsamea

Question 23.
Choose the living cells from the given
I) Phellem
II) Phloem
III) Phellogen
IV) Xylem parenchyma
a) (I) (II) & (III)
b) (II) (III) & (IV)
c) (I) (III) & (IV)
d) (I) (II) & (IV)
Answer:
b) (II) (III) & (IV)

Question 24.
When we peel the skin of a potato tuber we remove
a) Periderm
b) Epidermis
c) Cuticle
d) Sapwood
Answer:
a) Periderm

Question 25.
Phelloderm is otherwise called as:
(a) primary cortex
(b) corkwood
(c) secondary cortex
(d) rhytidome
Answer:
(c) secondary cortex

Question 26.
The waxy substances associated with cell walls of cork cells are impervious to water because of the presence of ………………. which gets deposited on the cork cells
a) Cutin
b) Suberin
c) Eignin
d) Hemicellulose
Answer:
b) Suberin

Question 27.
The antimalarial compound quinine is, extracted from:
(a) seeds of cinchona
(b) bark of cinchona
(c) leaves of cinchona
(d) flowers of cinchona
Answer:
(b) bark of cinchona

Question 28.
Annual rings are distinct in plants growing in
a) Tropical regions
b) Arctic regions
c) Grasslands
d) Temperate region
Answer:
d) Temperate region

Question 29.
The bark does include three except one from the options.
I) Cortex
II) Periderm
III) Pith
IV) Secondary phloem
a) (I) (II) & (Ill)
b) (II) (III) & (IV)
c) (I) (III) & (IV)
d) (I) (II) & (IV)
Answer:
d) (I) (II) & (IV)

Question 30.
Rubber is obtained from:
(a) Bombax mori
(b) Hevea brasiliensis
(c) Quercus suber
(d) Morus rubra
Answer:
(b) Hevea brasiliensis

Question 31.
Wood actually means
a) Primary xylem
b) Secondary xylem
c) Primary phloem
d) Secondary phloem
Answer:
b) Secondary xylem

Question 32.
When a plant is wounded, the wound is healed by the formation of new cells, by the cavity of
a) Primary meristem
b) Apical meristem
c) Secondary meristem
d) Intercalary meristem
Answer:
c) Secondary meristem

Question 33.
Commercial cork is obtained from
a) Oak
b) Silver oak
c) Pine
d) Ficus
Answer:
a) Oak

II. Match Correctly And Choose The Right Answer

Question 1.
I) Spicy bark – A Turpentine
II) Ornamental Antique – B Quinine
III) Active drug – C Cinnamon
IV) Thinner and solvent – D Amber

Answer:
d) C-D- B-A

Question 2.
I) Phellogen – A. Cork
II) Phelloderm – B. Cork cambium
III) Phellem – C. Lack suberin
IV) Phelloids – D. Secondary cortex

Answer:
a) B-D- A-C

Question 3.
I) Sapwood – A. Softwood
II) Heartwood – B. Hardwood
III) Porous wood – C. Albumum
IV) Nonporous wood – D. Duramen

Answer:
c) C-D-B-A

III Identify True Or False And From The Given Option Choose The Right Answer:

Question 1.
A) In pinus wood is Non-porous
B) In Morns wood is porous
C) In Quercus the wood is Diffuse porous
D) In Acer the wood is Ring porous
a) A&DTrue B&C False
b) A & B True C & D False
c) A & C True B & D False
d) A&BFalseC &DTrue
Answer:
b) A & B True C & D False

Question 2.
Arrange the given plants in order from more distinct Annual rings to least distinct Annual rings.
a) Seashore plants, Desert plants, Tropical plants & Temperature plants
b) Temperature plants, Tropical plants, Desert plants, Seashore plants
c) Tropical plants, Desert plants, Temperature plants, Seashore plants
d) Temperature plants, Seashore plants, Tropical plants, Desert plants
Answer:
b) Temperature plants, Tropical plants, Desert plants, Seashore plants.

Question 3.
In wood, the Annual rings become clearly evident to our eyes due to
a) The high density and dark coloured latewood or Autumn wood
b) The low density and light coloured early wood or springwood
c) The high density and dark coloured early wood or springwood
d) The low density and light coloured latewood or Autumn wood
Answer:
a) The high density and dark coloured of latewood or Autumn wood

Question 4.
Column I and Column II – Match them correctly and Find out the right option.
Answer:

Column I	Column II
A. Springwood or earlywood B. Autumn wood or Latewood	1. Lighter in colour 2. Density high 3. Density low 4. Darker in colour 5. Larger number of xylem elements 6. Vessels with wider cavity 7. Lesser number of xylem elements 8. Vessels with a small cavity

Which of the following combination is correct?
a) A – 2, 4, 7, 8 B- 1,3, 5,6
b) A- 1,2, 7,8 B-3,4, 5, 6
c) A – 1,3, 5,6 B – 2,4, 7, 8
d) A- 1,3,7, 8 B – 2, 4, 5, 6
Answer:
b) A – 1, 2, 7,8 B-3,4,5, 6

IV. Assertion And Reason

Question 1.
ASSERTION: – A All tissues lying inside the Vascular Cambium are called as Bark
REASON -R: Bark is made up of Phellogen. Phellem and Phelloderm Cortex, primary and secondary phloem
a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true, but R is not the correct explanation of A
c) A is true but ‘R’ is false
d) A is false and ‘R’ is true
e) Both A and ‘R’ are false
Answer:
d) ‘A’ is false and ‘R’ is true

Question 2.
ASSERTION: -A In angiosperms, the conduction of water is more efficient because their xylem has vessels
REASON – R: Conduction of water by vessel element is an active process in which energy is supplied by xylem parenchyma with a large number of Mitochondria
Answer:
a) Both A and R – are true and ‘R’ is the correct explanation of A

Question 3.
ASSERTION: -A. All the endodermal cells of the root do not contain Casparian thickenings on their radial and transverse walls.
REASON-R: Passage cells are found in the root endodermis, which conducts water in to the xylem
Answer:
a) Both A and R are true and ‘R’ is the correct explanation of A

Question 4.
ASSERTION:-A Cambium is a lateral meristem and causes growth in width
REASON-R Cambium is made up of fusiform and ray initials in stem
Answer:
b) Both A and R are true, but R is not the correct explanation of A

Question 5.
ASSERTION: – A The lenticel is meant for gaseous exchange.
REASON-R Lenticel checks excessive evaporation of water.
Answer:
b) Both A and R are true but R is not the correct explanation of A

Question 6.
ASSERTION:-A Heartwood is more durable
REASON – R Heartwood contains organic compounds like tannins, resins, oil, gums, aromatic substances and essential oils help to resist microbial and termites attack.
Answer:
a) Both A and R are true and R is the correct explanation of A

V. 2 Marks Questions

Question 1.
Distinguish between Primary and Secondary growth.
Answer:
Primary growth

1. The roots and stem grow in length with the help of Apical meristem
2. It is known as longitudinal growth
3. Eg. Angiosperms & Gymnosperms

Secondary growth

1. The roots and stem show an increase in thickness or width with the help of Lateral meristem
2. It is also known as latitudinal growth or growth in girth
3. Eg. Most Angiosperms, including some Monocots and Gymnosperms

Question 2.
Mention the two Lateral meristems responsible for secondary growth.
Answer:
The secondary growth in dicots and gymnosperms is brought about by two lateral meristems.

1. Vascular Cambium and
2. Cork Cambium

Question 3.
Define interfascicular cambium?
Answer:
In between the vascular bundles, a few parenchymatous cells of the medullary rays that are in line with the fascicular cambium become meristematic and form strips of the vascular cambium. It is called interfascicular cambium.

Question 4.
Fill in the blanks

The botanical name of the plant	The common name of the product	Use
1. Abies balsamea	Canada balsam	………………………………..
2.  Acacia Senegal (meska)	…………………………………………	Natural gum used as a bind to water colour painting
3. …………………………………	Cork	Hydrophobic, impermeable, used as a bottle stopper.
4. ………………………………………………	Hematoxylin	Dye from the heartwood to stain plant materials view under a microscope

Answer:

1. The resin used as a mounting medium for microscopic slide preparation.
2. Gum Arabic
3. Quercus suber
4. Haematoxylon campechianum

Question 5.
Distinguish between the stratified cambium and Non-stratified cambium.
Answer:
Stratified cambium: Plants with short fusiform initials, produced, storied, cambium in horizontal tiers known as
stratified cambium

Non-stratified cambium: Plants with long fusiform initials produced non-storied cambium, strongly overlap at the ends, known as Non-stratified cambium

Question 6.
Why does porous wood be harder than non-porous wood?
Answer:

• Porous wood is wood with xylem vessels which appear as a pore in cross-section.
• When a tree stem become old, most of its vessels are blocked by tyloses with deposition of gum, resin, tannin, oils, etc. (Heartwood)
• So porous wood is harder and commercially important.

Question 7.
What is the source of turpentine? and What is its use?
Answer:

• Turpentine is a resin obtained from the bark of conifers Eg. pinus
• It is also used as a thinner for oil-based paints.
• It is also used as an organic solvent.
• It is also used as a balm to relieve muscular pain.

Question 8.
Distinguish between Periderm and Polydor
Answer:
Periderm:

1. The secondary growth replaces the epidermis and primary cortex and forms the Periderm
2. It consists of
   • Phellem
   • Phellogen
   • Phelloderm
3. Eg. Dicot stem & roots

Polyderm:

1. It is a special type of protective tissue consisting of a miserable suberized layer, alternating with multiseriate nonsuberized cells in periderm
2. Eg. Roots and underground stems of Rosaceae plants

Question 9.
Define Rhytidome?
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues ? formed during successive secondary growth, eg: Quercus.

Question 10.
What is the use of Canada balsam?
Answer:
From the resin ducts, the Abies balsamea plant produces an organic gum-like substance, used as a permanent mounting medium for microscopic slide preparation.
Eg. A slide of 60 years old holotype specimen of a flatworm is permanently mounted in Canada balsam.

VI. 3 Mark Questions

Question 1.
Distinguish between primary and secondary growth.
Answer:
1. Primary growth: The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is tailed primary growth or longitudinal growth.

2. Secondary growth: The gymnosperms and most angiosperms, including some monocots, show an increase in the thickness of stems and roots by means of secondary growth or latitudinal growth.

Question 2.
Notes on Lenticels.
Answer:

• The aerating pores are seen as raised opening on the surface of bark as scars on old stems and roots.
• It is fonned during secondary growth in stems.
• In this portion phellogen activity is more than elsewhere, a filling tissue known as complementary tissue (loosely arranged parenchyma) is formed.
• Lenticel is helpful in the exchange of gases and also facilitate the little amount of transpiration

Question 3.
Explain briefly about false annual rings.
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo – or false – annual rings.

Question 4.
Differences between Diffuse porous wood and Ring porous wood
Answer:
Diffuse porous wood:

1. This type of wood is formed where the climatic conditions are uniform
2. The vessels are more or less equal in diameter in any annual ring
3. The vessels are uniformly distributed throughout the wood

Ring porous wood:

1. This type of wood is formed where the climatic conditions are not uniform
2. The vessels are wide and narrow within an annual ring
3. The vessels are not uniformly distributed throughout the wood

Question 5.
Differences between Porous Wood and Non – porous wood
Answer:
Porous wood or Hardwood, Ex: Morus:

1. Common in Angiosperms
2. Porous because it contains vessels

Non-porous wood or softwood Ex: Pinus

1. Common in Gymnosperms
2. Non – porous because it does not contain vessels

Question 6.
Differences between Sapwood (alburnum) and Heart Wood (duramen)
Answer:

Sapwood (Alburnum)	Heartwood (Duramen)
1. Living part of the wood	1. Dead part of the wood
2. It is situated on the outer side of wood	2. It is situated in the centre part of wood
3. It is lighter in colour	3. It is dark coloured
4. Very soft in nature	4. Hard in nature
5. Tyloses are absent	5. Tyloses are present
6. It is not durable and not resistant to microorganisms	6. It is more durable and resists microorganisms insects and termites

Question 7.
Differences Between Phellem and Phelloderm
Answer:

Phellem (Cork)	Phelloderm (secondary cortex)
1. It is formed on the outer side of phellogen1	1.  It is formed on the inner side of the phellogen
2. Cells are compactly arranged in regular tires and rows without intercellular spaces.	2. Cells are loosely arranged with intercellular spaces.
3. Protective in function.	3. As it contains chloroplasts, it synthesizes and stores food
4. Consists of non-living cells with suberized walls	4. Consists of living cells, parenchymatous in nature and does not have suberin
5. Lenticels are present	5. Lenticels are absent

Question 8.
Explain the term lenticel.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems. When phellogen is more active in the region of lenticels, a mass of loosely arranged thin-walled parenchyma cells is formed. It is called complementary tissue or filling tissue. Lenticel is helpful in the exchange of gases and transpiration called lenticular transpiration.

Question 9.
Mention the benefits of bark in a tree.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation, and guards against variations of external temperature. It is insect repellent, decay proof, fireproof, and is used in obtaining drugs or spices. The phloem cells of the bark are involved in the conduction of food while secondary cortical cells involved in storage.

Question 10.
Annual rings are not clear and distinct in desert trees and seashore plants – Justify.
Answer:

• In the desert, as well seashore regions the climatic condition remain the same throughout the year.
• Secondary growth in plants is influenced by seasonal changes since in these areas seasonal changes are not significant enough to bring in distinct Annual rings with early and latewood formation alternatively.

Question 11.
A plant dies if the sapwood is damaged, but not with that of Heartwood – Give factual justification.
Answer:

• Sapwood is a living part of the wood, perform water conduction, that’s why it is known as Sapwood.
• Heartwood is a dead part of the wood, do not perform water conduction so if destroyed, no vital function of the plant is affected.
• If sapwood is damaged, or exposed conduction of water will be blocked, water loss is rapid leading to decay and decomposition of tissues and leads to death of the plant.

Question 12.
What is Dendrochronology? Add a note on the significance of studying the growth rings.
Answer:

• The annual ring of a tree corresponds to one year’s growth.
• If we count the rings we can determine the very age of the plant.
• This method of calculating the age of a tree by counting the annual ring is known as dendrochronology.

Significance of studying growth rings:

• Age of wood – calculated
• Age verified by Radioactive carbon dating
• Provides evidence in Forensic investigation.

Question 13.
If  ‘A’ is vascular cambium, then label other parts with reference to cambial activity.

Answer:
A – Vascular cambium
B – First formed phloem – (Primary phloem)
C – First formed xylem – (Primary xylem)
D – Second formed phloem – (Secondary phloem)
E – Second formed xylem – (Secondary xylem)

Question 14.

Answer:
Cross-section of wood showing Annual rings.
A-Bark
B-Sapwood
C – heartwood
D -Annual rings

Question 15.
Identify the diagram & Label the parts.

Answer:
Structure of Tyloses A – Parenchyma cell
B – Tyloses
C – Vessel wall
D – Vessel Lumen

Question 16.
In the given diagram, the parts labelled are A, B, C, D identify the part correctly with respect to its function.

a) A – Periderm for gaseous exchange
b) C – Secondary cortex for protection
c) B – Complementary tissues for gaseous exchange
d) D – Phellogen for xylem and phloem formation
Answer:
c) B – Complementary tissues for gaseous exchange

Question 17.

Answer:
Different stages of secondary growth in Dicot root
A – Cambial ring
B – Primary’ xylem
C – Secondary xylem
D – Primary phloem
E – Secondary phloem

VII. 5 Mark Questions

Question 1.
Distinguish between Phellem and Phelloderm.
Answer:
Phellem (Cork):

1. It is formed on the outer side of phellogen.
2. Cells are compactly arranged in regular tires and rows without intercellular spaces.
3. Protective in function.
4. Consists of nonliving cells with suberized walls.
5. Lenticels are present.

Phelloderm (Secondary cortex):

1. It is formed on the inner side of phellogen.
2. Cells are loosely arranged with intercellular spaces.
3. As it contains chloroplast, it synthesises and stores food.
4. Consists of living cells, parenchymatous in nature and does not have suberin.
5. Lenticels are absent.

Question 2.
Distinguish the significance of Cork and Bark
Answer:
Cork:

1. It includes only phellem layer of bark
2. It is composed of suberin a hydrophobic substance
3. It has impermeable buoyant, elastic and fibre retartant properties
4. Used in making bottle stoppers Eg. Bark of Quercus suber

Bark:

1. It includes all tissues outside vascular cambium (Periderm, Cortex, Primary and secondary phloem)
2. It has insect repellent, decay proof, fire proof properties.
3. Used as Drugs or spices.
4. Eg. Bark of Chichona – (AntimalariaJ drug), Bark of Cinnamomum (Used as spice)

Question 3.
Differentiate between, the vascular cambial components Fusiform initials and Ray initials.
Answer:
Fusiform initials:

1. Vertically elongated cells
2. Give rise to axial system of secondary tissues, xylem and phloem
3. Secondary xylem includes tracheary elements, fibres and axial parenchyma
4. Secondary phloem includes sieve elements
5. Based on arrangement of fusiform initials 2 types of vascular cambium recognised
   a – stratified cambium
   b – Nonstratified cambium

Ray initials

1. Horizontally elongated cells
2. Give rise to radial system of secondary xylem and phloem
3. Radial system consists of rows of parenchymatous cells oriented at right angles to the longitudinal axis of xylem elements
4. Secondary phloem include phloem rays fibres and axial parenchyma

Question 4.
Give an account of Primary & Secondary structure of Dicto Stern. (Flow chart)
Answer:

Question 5.
Give an account of Any 5 Commercial barks their, properties and uses
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 4 Hydrogen Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 4 Hydrogen

11th Chemistry Guide Hydrogen Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
Which of the following statements about hydrogen is incorrect?
(a) Hydrogen ion, H₃O⁺ exists freely in solution.
(b) Dihydrogen acts as a reducing agent.
(c) Hydrogen has three isotopes of which tritium is the most common.
(d) Hydrogen never acts as cation in ionic salts.
Answer:
(c) Hydrogen has three isotopes of which tritium is the most common.

Question 2.
Water gas is
(a) H₂O(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 3.
Which one of the following statements is incorrect with regard to ortho and para dihydrogen ?
(a) They are nuclear spin isomers
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
(c) The para isomer is favoured at low temperatures
(d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.
Answer:
(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin

Question 4.
Ionic hydrides are formed by
(a) halogens
(b) chalogens
(c) inert gases
(d) group one elements
Answer:
(d) group one elements

Question 5.
Tritium nucleus contains
(a) 1p + 0n
(b) 2p + 1n
(c) 1p + 2n
(d) none of these
Answer:
(c) 1p + 2n

Question 6.
Non-stoichiometric hydrides are formed by
(a) palladium, vanadium
(b) carbon, nickel
(c) manganese, lithium
(d) nitrogen, chlorine
Answer:
(b) carbon, nickel

Question 7.
Assertion:
Permanent hardness of water is removed by treatment with washing soda.
Reason:
Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 8.
If a body of a fish contains 1.2 g hydrogen in its total body mass, if all the hydrogen is replaced with deuterium then the increase in body weight of the fish will be
(a) 1.2 g
(b) 2.4 g
(c) 3.6 g
(d) \(\sqrt{4.8}\) g
Answer:
(a) 1.2 g

Question 9.
The hardness of water can be determined by volumetrically using the reagent
(a) sodium thio sulphate
(b) potassium permanganate
(c) hydrogen peroxide
(d) EDTA
Answer:
(d) EDTA

Question 10.
The cause of permanent hardness of water is due to
(a) Ca(HCO₃)²
(b) Mg(HCO₃)²
(c) CaCl₂
(d) MgCO₃
Answer:
(c) CaCl₂

Question 11.
Zeolite used to soften hardness of water is, hydrated
(a) Sodium aluminium silicate
(b) Calcium aluminium silicate
(c) Zinc aluminium borate
(d) Lithium aluminium hydride
Answer:
(a) Sodium aluminium silicate

Question 12.
A commercial sample of hydrogen peroxide marked as 100 volume H₂O₂, it means that
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP
(b) 1 L of H₂O₂ will give 100 ml O₂ at STP
(c) 1 L of H₂O₂ will give 22.4 L O₂
(d) 1 ml of H₂O2 will give 1 mole of O₂ at STP
Answer:
(a) 1 ml of H₂O₂ will give 100 ml O₂ at STP

Question 13.
When hydrogen peroxide is shaken with an acidified solution of potassium dichromate in presence of ether, the ethereal layer turns blue due to the formation of
(a) Cr₂O₃
(b) CrO₄^2-
(c) CrO(O₂)₂
(d) none of these
Answer:
(c) CrO(O₂)₂

Question 14.
For decolourisation of 1mole of acidified KMnO₄, the moles of H₂O₂ required is
(a) \(\frac{1}{2}\)

(b) \(\frac{3}{2}\)

(c) \(\frac{5}{2}\)

(d) \(\frac{7}{2}\)
Answer:
(c) \(\frac{5}{2}\)

Question 15.
Volume strength of 1.5 NH₂O₂ is
(a) 1.5
(b) 4.5
(c) 16.8
(d) 8.4
Answer:
(d) 8.4

Question 16.
The hybridisation of oxygen atom is H₂O and H₂O₂ are, respectively
(a) sp and sp³
(b) sp and sp
(c) sp and sp²
(d) sp³ and sp³
Answer:
(d) sp³ and sp³

Question 17.
The reaction H₃PO₂ + D₂O → H₂DPO₂ + HDO indicates that hypo-phosphorus acid is
(a) tribasic acid
(b) dibasic acid
(c) mono basic acid
(d) none of these
Answer:
(c) mono basic acid
Solution:
Hypophosphorus acid on reaction with D₂O, only one hydrogen is replaced by deuterium and hence it is mono basic.

Question 18.
In solid ice, oxygen atom is surrounded
(a) tetrahedrally by 4 hydrogen atoms
(b) octahedrally by 2 oxygen and 4 hydrogen atoms
(c) tetrahedrally by 2 hydrogen and 2 oxygen atoms
(d) octahedrally by 6 hydrogen atoms
Answer:
(a) tetrahedrally by 4 hydrogen atoms
Solution:

Question 19.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively
(a) inter molecular H-bonding and intra molecular f H-bonding
(b) intra molecular H-bonding and inter molecular H-bonding
(c) intra molecular H – bonding and no H – bonding
(d) intra molecular H -bonding and intra molecular H-bonding
Answer:
(b) intra molecular H-bonding and inter molecular H-bonding
                         

Question 20.
Heavy water is used as
(a) modulator in nuclear reactions
(b) coolant in nuclear reactions
(c) both (a) and (b)
(d) none of these
Answer:
(c) both (a) and (b)
Solution:
(c) Heavy water is used as moderator as well as coolant in nuclear reactions.

Question 21.
Water is a
(a) basic oxide
(b) acidic oxide
(c) amphoteric oxide
(d) none of these
Answer:
(c) amphoteric oxide
Solution:
Water is a amphoteric oxide.

II. Write brief answer to the following questions:

Question 22.
Explain why hydrogen is not placed with the halogen in the periodic table.
Answer:
The electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halide ions. In most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group -1 along with alkali metals and not placed with halogens.

Question 23.
A cube at the 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:
In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electron on oxygen atoms in each water molecule allows formation of a three-dimensional structure.

This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0°C. While in liquid water, unlike ice where hydrogen bonding occurs over a long-range, the strong hydrogen bonding prevails only in a short range and therefore the denser packing. Hence, ice cube sinks in water.

Question 24.
Discuss the three types of Covalent hydrides.
Answer:
Covalent hydrides are the compound in which hydrogen is attached to another element by sharing of electrons. The most common examples of covalent hydrides of non-metals are methane, ammonia, water and hydrogen chloride. Covalent hydrides are further divided into three categories, viz., electron
precise (CH₄ C₂H₆), electron deficient (B₂H₆) and electron-rich hydrides (NH₃H₂ O). Since most of the covalent hydrides consists of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Question 25.
Predict which of the following hydrides is a gas on a solid (a) HCl (b) NaH. Give your reason.
Answer:
Sodium hydride (NaH) is a gas on a soiid. NaH is prepared by direct reaction of hydrogen gas with liquid sodium and it has NaCl crystal structure. It is a tendency of deprotonating in many organic reactions. NaH is used in fuel cells.

Question 26.
Write the expected formulas for the hydrides of 4^th period elements. What is the trend in the formulas? In what way the first two numbers of the series different from the others?
Answer:
The expected formulas of the hydrides of 4^th period elements are MH or MH₂. However, except the first two members, many of the elements form non- stoichiometric interstitial hydrides with variable composition. The first two members of the period alkali metal (Potassium) and alkali earth metal (Calcium) forms ionic hydrides.

Question 27.
Write chemical equation for the following reactions.
Answer:
(i) reaction of hydrogen with tungsten (VI) oxide NO₃ on heating.
(ii) hydrogen gas and chlorine gas.
Answer:
(i) Hydrogen can be used to reduce metal oxide into metal. Hydrogen reduces tungsten(VI) oxide into tungsten.
WO₃ + 3H₂ → W + 3H₂O

(ii) Hydrogen reacts with chlorine at room temperature under light gives hydrogen chloride.
H_2(g) + Cl_2(g) → 2HCl_(g)

Question 28.
Complete the following chemical reactions and classify them into (a) hydrolysis (b) redox (c) hydration reactions.
(i) KMnO₄ + H₂O₂ →
(ii) CrCl₃ + H₂O →
(iii) CaO + H₂O →
Answer:
(i) KMnO₄ + H₂O₂ → 2KMnO₂ + 2KOH + 2H₂O + 3O₂
The reaction of potassium permanganate with hydrogen peroxide is a redox reaction.

(ii) CrCl₃ + H₂O → [Cr(H₂O)₆]Cl₃
It is a hydration reaction. Many salts crystallized from aqueous solutions form hydrated crystals. The water in the hydrated salt may form co-ordinate bonds.

(iii) CaO + H₂O → Ca(OH)₂
It is a hydrolysis reaction. Calcium oxide hydrolyses to calcium hydroxide.

Question 29.
Hydrogen peroxide can function as an oxidising agent as well as reducing agent. Substantiate this statement with suitable examples.
Answer:
Hydrogen peroxide can act both as an oxidizing agent and a reducing agent. Oxidation is usually performed in acidic medium while the reduction reactions are performed in basic medium.

In acidic conditions:
H₂O₂ + 2H⁺ + 2e^– → 2H₂O (E° = +1.77 V)
For example
2FeSO₄ + H₂SO₄ + H₂O₂ → Fe₂(SO₄)₃ + 2H₂O

In basic conditions:
HO₂^– + OH^– → O₂ + H₂O + 2e^– (E° = + 0.08V)

For Example,
2 KMnO_4(aq) + 3H₂O_2(aq) → 2MnO₂ + 2KOH + 2H₂O + 3O₂(g)

Question 30.
Do you think that heavy water can be used for drinking purposes?
Answer:
Heavy water cannot be used for drinking purposes because it does not form hydrogen bonding.

Question 31.
What is water-gas shift reaction?
Answer:
The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H₂O → CO₂ + H₂
The CO₂ formed in the above process is absorbed in a solution of potassium carbonate.
CO₂ + K₂CO₃ + H₂O → 2KHCO₃

Question 32.
Justify the position of hydrogen in the periodic table?
Answer:
The hydrogen has the electronic configuration of 1s¹ which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
1. It forms unipositive ion (H⁺) like alkali metals {Na⁺, K⁺, Cs⁺)
2. It forms halides (HX), oxides, (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, NaH₂OH₂, NaH₂S)
3. It also acts as a reducing agent.

However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1,314 kJ mol^-1 which is much higher than alkali metals.

Like the formation of halides (X^–) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion (H^–) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:
\(\frac{1}{2}\) H₂ + e^– → H^–                         ∆H = +36 kcalmol^-1
\(\frac{1}{2}\) Br₂ + e^– → Br^–                       ∆H = -55 kcalmol^-1

Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in +1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 33.
What are isotopes? Write the names of isotopes of hydrogen.
Answer:
Atoms of the same element having same atomic number and different mass number are called isotopes.
Hydrogen has three naturally occurring isotopes, viz., protium (₁H¹ or H), deuterium (₁H² or D) and tritium (₁H³ or T). Protium ₁H¹ is the predominant form (99.985 %) and it is the only isotope that does not contain a neutron. Deuterium, also known as heavy hydrogen, constitutes about 0.015 %. The third isotope, tritium is a radioactive isotope of hydrogen which occurs only in traces (~1 atom per 1018 hydrogen atoms). Due to the existence of these isotopes naturally occurring hydrogen exists as H₂, HD, D₂, HT, T₂, and DT.

Question 34.
Give the uses of heavy water.
Answer:
The uses of heavy water are as follows
(i) Heavy water is widely used as moderator in nuclear reactors as it can lower the energies of fast neutrons
(ii) It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
(iii) It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 35.
Explain the exchange reactions of deuterium.
Answer:
When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium. This reaction is known as exchange reaction of deuterium.
2NaOH + D₂O → 2NaOD + HOD

HCl + D₂O → DCl + HOD

NH₄Cl + 4D₂O → ND₄Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D₂O is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
H₃PO₂ + D₂O → H₂DPO₂ + HDO
It is also used to prepare some deuterium compounds:

Al₄C₃ + 12D₂O → 4Al(OD)₃ + 3CD₄
CaC₂ + 2D₂O → Ca(OD)₂ + C₂D₂
Mg₃N₂ + 6D₂O → 3Mg(OD)₂ + 2ND₃
Ca₃P₂ + 6D₂O → 3Ca(OD)₂ + 2PD₃

Question 36.
How do you convert parahydrogen into ortho hydrogen?
Answer:
At room temperature, normal hydrogen consists of about 75% ortho-form and 25% para-form. As the ortho-form is more stable than para-form, the conversion of one isomer into the other is a slow process. However, the equilibrium shifts in favour of para hydrogen when the temperature is lowered.

The para-form can be catalytically transformed into I ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric j discharge, heating above 800°C and mixing with paramagnetic molecules such as O₂, NO, NO₂ or with nascent/atomic hydrogen.

Question 37.
Mention the uses of deuterium.
Answer:
(i) Deuterium is used to prepare heavy water which is used as moderator in nuclear reactors.
(ii) Deuterium exchange reactions are useful in determining the number of ionic hydrogens present in a given compound.
(iii) It is also used to prepare some deuterium compounds.

Question 38.
Explain preparation of hydrogen using electrolysis.
Answer:
High purity hydrogen (> 99.9 %) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At anode:
2OH^– → H₂O + \(\frac{1}{2}\)O₂ + 2e^–
At cathode:
2H₂O + 2e^– → 2OH^– + H₂
Overall reaction:
H₂O → H₂ + \(\frac{1}{2}\)O₂

Question 39.
A group-1 metal (A) which is present ¡n common salt reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state. (B) on reaction with a gas to give universal solvent (D). The compound (D) on reacts with (A) to give (B), a strong base. Identify A, B, C, D and E. Explain the reactions.
Answer:
The metal belongs to Group-1 which is present in common salt is Sodium(A).
The metal reacts with (B) to give compound (C) in which hydrogen is present in -1 oxidation state.
2Na + H₂ → 2NaH
Hence, the (B) is hydrogen and (C) is sodium hydride. (B) on reaction with a gas to give universal solvent (D).
H₂ + O₂ → 2NaOH
The universal solvent (D) is water. The compound (D) on reaction with (A) to give (E) which is a strong base.
H₂O + 2Na → 2NaOH
The Compound (E) is Sodium hydroxide.
A – Sodium (Na)
B – Hydrogen (H₂)
C – Sodium Hydride (NaH)
D – Water (H₂O)
E – Sodium hydroxide (NaOH)

Question 40.
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderatorin nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D). Identify A, B, C and D.
Answer:
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderator in nuclear reaction.
2D₂ + O₂ → 2D₂O
The isotope of hydrogen is deuterium(A), diatomic element is oxygen and the compound (B) is heavy water.
(A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D).
CH₃ – CH = CH₂ + D₂ → CH₃ – CHD – CH₂D
(C)                                     (D)
The compound (C) is propene and (D) is 1, 2 deutropropane.
A – Deuterium (D₂)
B – Heavy water (D₂O)
C – Propene (CH₃ – CH = CH₂)
D – 1, 2 deutero propene (CH₃ – CHD – CH₂D)

Question 41.
NH₃ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15- Explain.
Answer:
When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity.

This interaction is called hydrogen bonding. Hence, NH₃ has exceptionally high melting point and boiling point as compared to those of the hydrides of the remaining element of group 15 due to intermolecular hydrogen bonding.

Question 42.
Why interstitial hydrides have a lower density than the parent metal?
Answer:
In interstitial hydrides, hydrogen occupies the interstitial sites. These hydrides show properties similar to parent metals. Most of these hydrides are non-stoichiometric with variable composition. Hence, interstitial hydrides have lower density than the parent metal.

Question 43.
How do you expect the metallic hydrides to be useful for hydrogen storage?
Answer:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Most of the hydrides are non-stoichiometric with variable composition (TiH _{1.5 – 1.8} and PdH_{0.6 – 0.8}), some are relatively light, inexpensive and thermally unstable which make them useful for hydrogen storage applications.

Question 44.
Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.
Answer:
When a hydrogen atom is covalently bonded to a highly electronegative atom such as nitrogen, the bond is polarized. Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity. This interaction is called hydrogen bonding. The magnitude of hydrogen bonding increases with the increase in electronegativity of the atom. Hence, the increasing magnitude of hydrogen bonding of NH₃, H₂O and HF follows the order NH₃ < H₂O < HF.

Question 45.
Compare the structures of HO and HO.
Answer:
Both in gas-phase and liquid-phase, the molecule adopts a skew conformation due to repulsive interaction of the OH bonds with lone-pairs of electrons on each oxygen atom. Indeed, it is the smallest molecule known to show hindered rotation about a single bond.

H₂O₂ has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in figure. Structurally, H₂O₂ is represented by the dihydroxyl formula in which the two OH-groups do not lie in the same plane.

One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spine. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°. Water has bent structure and the H-O-H bond angle is 104.5°.

11th Chemistry Guide Hydrogen Additional Questions and Answers

I. Choose the best Answer:

Question 1.
A simplest atom which contains one electron and one proton is
(a) Helium
(b) Deuterium
(c) Hydrogen
(d) Tritium
Answer:
(c) Hydrogen

Question 2.
Hydrogen has similarities with
(a) Alkali metals and halogens
(b) Alkaline earth metals and halogens
(c) Alkalimetals and noble gases
(d) Halogens and noble gases.
Answer:
(a) Alkali metals and halogens

Question 3.
Which of the following properties of hydrogen similar to alkali metals?
1. It forms uni negative ion.
2. It forms halides, oxides and sulphides similar to alkali metals.
3. It also acts as a reducing agent.
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1,2 and 3
Answer:
(b) 2 and 3

Question 4.
Ionisation energy of hydrogen is (in kJ mol^-1)
(a) 377
(b) 520
(c) 1413
(d) 1314
Answer:
(d) 1314

Question 5.
Hydrogen is placed in the ________ of the periodic table.
(a) group – 1
(b) group – 17
(c) group – 18
(d) group – 2
Answer:
(a) group – 1

Question 6.
The predominant isotope of hydrogen is
(a) deuterium
(b) protium
(c) tritium
(d) heavy hydrogen
Answer:
(b) protium

Question 7.
The radioactive isotope of hydrogen is
(a) protium
(b) deuterium
(c) tritium
(d) heavy hydrogen
Answer:
(c) tritium

Question 8.
The number of naturally occurring hydrogens due to existence of isotopes is
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(a) 6

Question 9.
The half life period of tritium is ______(in years).
(a) 13.2
(b) 10.5
(c) 12.3
(d) 15.8
Answer:
(c) 12.3

Question 10.
The correct order of melting point of isotopes of hydrogen is
(a) H < D < T
(b) T < D < H
(c) H < T < D
(d) D < H < T
Answer:
(a) H < D < T

Question 11.
The percentage of ortho and para forms of normal hydrogen at room temperature is
(a) 25% and 75%
(b) 40% and 60%
(c) 60% and 40%
(d) 75% and 25%
Answer:
(d) 75% and 25%

Question 12.
Which of the following statement is correct about ortho-para hydrogen?
(a) The magnetic moment of para hydrogen is twice that of a proton.
(b) Ortho form is more stable than para form
(c) Ortho form can be catalytically converted into para form using platinum.
(d) At room temperature, normal hydrogen consists of 75% para form.
Answer:
(b) Ortho form is more stable than para form

Question 13.
The composition of syngas is
(a) CO + N₂
(b) CO + H₂O
(c) CO + H₂
(d) CO₂ + H₂
Answer:
(c) CO + H₂

Question 14.
During electrolysis of water containing traces of acid hydrogen is liberated at
(a) cathode
(b) anode
(c) both anode and cathode
(d) none of the above
Answer:
(a) cathode

Question 15.
The conversion of carbon monoxide of the water gas into carbon dioxide is called water gas _______ reaction.
(a) displacement
(b) decomposition
(c) shift
(d) conversion
Answer:
(c) shift

Question 16.
The catalyst used in the water gas shift reaction is
(a) Copper
(b) Nickel
(c) Platinum
(d) Palladium
Answer:
(a) Copper

Question 17.
The CO₂ formed in the water gas shift reaction is absorbed in a solution of
(a) Potassium bicarbonate
(b) Sodium chloride
(c) Potassium sulphate
(d) Potassium carbonate
Answer:
(d) Potassium carbonate

Question 18.
The percentage of heavy water in normal water is
(a) 1.6 × 10⁴
(b) 1.6 × 10^-4
(c) 1.6 × 10^-3
(d) 1.6 × 10²
Answer:
(b) 1.6 × 10^-4

Question 19.
When water is completely electrolysed, the gas liberated is/are
(a) H₂
(b) D₂
(c) H₂ and D₂
(d) H₂ and T₂
Answer:
(c) H₂ and D₂

Question 20.
Lithium Aluminium Hydride is
(a) [LiAlH₃]
(b) Li[Al₂H₄]
(c) Li[AlH₂]
(d) Li[AlH₄]
Answer:
(d) Li[AlH₄]

Question 21.
Deuterium oxide is called
(a) hard water
(b) soft water
(c) heavy water
(d) heavy hydrogen
Answer:
(c) heavy water

Question 22.
Ammonia is synthesized by ______ process.
(a) Haber’s
(b) Bergius
(c) Decon’s
(d) Solvay
Answer:
(b) Bergius

Question 23.
Statement – I:
Tritium is a β-emitter.
Statement – II:
Radioactive decay of tritium gives \({ }_{2}^{3} \mathrm{He}\) and \({ }_{1}^{0} e\).
The correct statement/s is/are
(a) I alone
(b) II alone
(c) both I and II
(d) both are incorrect.
Answer:
(c) both I and II

Question 24.
The high melting and boiling points of water is due to
(a) Covalent bonding
(b) Hydrogen bonding
(c) Ionic bonding
(d) co-ordinate bonding
Answer:
(b) Hydrogen bonding

Question 25.
Chlorine reacts with water and forms _____ and _____ respectively.
(a) HCl and HOCl
(b) H₂ and HCl
(c) HOCl and H₂
(d) HCl and ClO₂
Answer:
(c) HOCl and H₂

Question 26.
Water is an ______ oxide.
(a) acidic
(b) basic
(c) amphoteric
(d) neutral
Answer:
(a) acidic

Question 27.
Hydrolysis of P₄O₁₀ gives
(a) HPO₂
(b) H₄P₂O₇
(c) H₃PO₃
(d) H₂PO₄
Answer:
(d) H₂PO₄

Question 28.
In CuSO₄.5H₂O, the number of water molecules form co-ordinate bonds is
(a) 4
(b) 5
(c) 3
(d) 1
Answer:
(a) 4

Question 29.
Flourine reacts with water and liberates
(a) hydrogen
(b) oxygen
(c) Fluorine dioxide
(d) HOF
Answer:
(b) oxygen

Question 30.
The number water molecules in hydrated crystal of Chromium chloride salt is
(a) 5
(b) 6
(c) 4
(d) 3
Answer:
(b) 6

Question 31.
The most common metal ions present in the hard water are
(a) Magnesium and Iron
(b) Calcium and Aluminium
(c) Magnesium and Calcium
(d) Manganese and Calcium
Answer:
(c) Magnesium and Calcium

Question 32.
Temporary hardness of water is removed by ______ method.
(a) Dewar
(b) Clark’s
(c) Leibeg
(d) Haber
Answer:
(b) Clark’s

Question 33.
Permanent hardness of water is due to the presence of soluble salts of _____ and ______ of magnesium and calcium.
(a) carbonates and bicarbonates
(b) chlorides and carbonates
(c) bicarbonates and sulphates
(d) chlorides and sulphates
Answer:
(d) chlorides and sulphates

Question 34.
The ion exchange bed used for the softening of hard water is
(a) Borates
(b) Zeolites
(c) Fluorides
(d) Phosphates
Answer:
(b) Zeolites

Question 35.
The general formula of zeolites is
(a) NaOAl₂O₃. ^xSiO₂. ^yH₂O
(b) Na₂O.Al₂O₃. ^ySiO₂. ^xH₂O
(c) NaOH.Al₂O₃. ^xSiO₂. ^yH₂O
(d) NaO.Al (OH)₃ .^xSiO₂. ^yH₂O
Answer:
(a) NaOAl₂O₃. ^xSiO₂. ^yH₂O

Question 36.
________ reactions are useful in determining the number of ionic hydrogens present in a given compound.
(a) Oxygen exchange
(b) Metal exchange
(c) Deuterium exchange
(d) Deuterium decomposition
Answer:
(c) Deuterium exchange

Question 37.
_______ is used as a moderator and coolant in nuclear reactors.
(a) Heavy hydrogen
(b) Ortho hydrogen
(c) Hydrogen peroxide
(d) Heavy water
Answer:
(d) Heavy water

Question 38.
Autoxidation of 2-alkyl anthraquinol gives
(a) Hydrogen peroxide
(b) Heavy water
(c) Hydrogen
(d) Water
Answer:
(a) Hydrogen peroxide

Question 39.
The percentage of hydrogen peroxide in ‘100 volume’ is
(a) 40
(b) 30
(c) 50
(d) 20
Answer:
(b) 30

Question 40.
Hydrogen peroxide solutions are stored in ______ container
(a) glass
(b) alkali metal
(c) plastic
(d) wooden
Answer:
(c) plastic

Question 41.
______ present in the glass catalyses the disproportionation reaction of hydrogen peroxide.
(a) Silica
(b) Alkali metals
(c) fluorine
(d) Oxygen
Answer:
(b) Alkali metals

Question 42.
Disproportionation of hydrogen peroxide gives
(a) oxygen and hydrogen
(b) hydrogen and water
(c) hydrogen and ozone
(d) oxygen and water
Answer:
(d) oxygen and water

Question 43.
Which of the following statement/s are true about hydrogen peroxide?
1. It can act both as an oxidizing agent and a reducing agent.
2. It is used in water treatment to oxidize pollutants.
3. It is used as mild analgesic.
4. It restores the white colour of the old paintings,
(a) 1, 2 and 3
(b) 1, 3 and 4
(c) 1, 2 and 4
(d) 2, 3 and 4
Answer:
(c) 1, 2 and 4

Question 44.
White pigment is
(a) Pb²(OH)₂(C0₃)₃
(b) Pb₃(OH)₂(C0₃)₂
(c) Pb₃(OH)(CO₃)₂
(d) Pb₂(OH)(CO₃)₃
Answer:
(b) Pb₃(OH)₂(C0₃)₂

Question 45.
The smallest molecule which shows hindered rotation about single bond is
(a) Hydrogen peroxide
(b) Water
(c) Deuterium oxide
(d) hydrogen
Answer:
(a) Hydrogen peroxide

Question 46.
Compounds in which hydrogen is attached to another element by sharing of electrons are called _________ hydrides.
(a) Interstitial
(b) Molecular
(c) Saline
(d) Metallic
Answer:
(b) Molecular

Question 47.
Which of the following molecule shows intramolecular hydrogen bond?
(a) Water
(b) Ammonia
(c) Salicylaldehyde
(d) Para-nitrophenol
Answer:
(c) Salicylaldehyde

Question 48.
Each water molecule is linked to ______ other molecules through hydrogen bonds.
(a) five
(b) four
(c) six
(d) two
Answer:
(b) four

Question 49.
Which one of the following is a covalent hydride?
(a) NH₃
(b) BeH₂
(c) NaH
(d) ZrH₂
Answer:
(a) NH₃

Question 50.
Hypo-phosphorus is a ______ acid.
(a) dibasic
(b) tribasic
(c) monobasic
(d) tetrabasic
Answer:
(c) monobasic

II. Very Short Question and Answers (2 Marks):

Question 1.
What are the isotopes of hydrogen?
Answer:
Hydrogen has three naturally occurring isotopes, viz., protium (₁H¹ or H), deuterium (₁H² or D) and tritium (₁H³ or T).

Question 2.
Write the physical properties of Hydrogen?
Answer:
Hydrogen is a colorless, odorless, tasteless, lightest and highly flammable gas. It is a non-polar diatomic molecule. It can be liquefied under low temperature and high pressure. Hydrogen is a good reducing agent

Question 3.
How is tritium prepared?
Answer:
Tritium is artificially prepared by bombarding lithium with slow neutrons in a nuclear fission reactor. The nuclear transmutation reaction for this process is as follows.
\({ }_{3}^{6} L i\) + \({ }_{0}^{1} n\) → \({ }_{2}^{4} \mathrm{He}\) + \({ }_{1}^{3} T\)

Question 4.
What are ortho and para hydrogens?
Answer:
In the hydrogen atom, the nucleus has a spin. When molecular hydrogen is formed, the spins of two hydrogen nuclei can be in the same direction or in the opposite direction as shown in the figure. These two forms of hydrogen molecules are called ortho and para hydrogens respectively.

Question 5.
Write the different forms of naturally occurring hydrogen?
Answer:
Due to the existence of three isotopes of hydrogen, Protium (H), Deuterium (D) and Tritium(T), naturally occurring hydrogen exists as H₂, HD, D₂, HT, T₂, and DT.

Question 6.
How will you convert para hydrogen into ortho hydrogen?
Answer:
The para-form can be catalytically transformed into ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric discharge, heating above 800°C and mixing with paramagnetic molecules such as O₂, NO, NO₂ or within ascent/atomic hydrogen.

Question 7.
How is pure hydrogen prepared?
Answer:
High purity hydrogen (> 99.9 %) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At anode:
2OH^– → H₂O + \(\frac{1}{2}\)O₂ + 2e^–

At cathode:
2H₂O + 2e^– → 2OH^– + H₂

Overall reaction:
H₂O → H₂ + \(\frac{1}{2}\)O₂

Question 8.
How is hydrogen prepared by steam reforming reaction?
Answer:
Hydrogen is produced in large scale by steam-reforming of hydrocarbons. In this method, hydrocarbon such as methane is mixed with steam and passed over nickel catalyst in the range 800- 900°C and 35 atm pressure.
CH₄ + H₂O → CO + 3H₂

Question 9.
What is water gas? How is it prepared?
Answer:
The mixture of carbon monoxide and hydrogen is called water gas. When steam is passed over a red- hot coke to produce carbon monoxide and hydrogen. The mixture of gases produced in this way is known as water gas (CO + H₂).
C + H₂O → (CO + H₂)Water gas /Syngas

Question 10.
What is syngas? Why is it called so?
Answer:
Water gas is also called as syngas as it is used in the synthesis of organic compounds such as methanol and simple hydrocarbons.

Question 11.
How is Deuterium prepared?
Answer:
Normal water contains 1.6 × 10^-4 percentage of heavy water. The dissociation of protium water (H₂O) is more than heavy water (D₂O). Therefore, when water is electrolysed, hydrogen is liberated much faster than D₂. The electrolysis is continued until the resulting solution becomes enriched in heavy water. Further electrolysis of the heavy water gives deuterium.

Question 12.
How is Tritium prepared?
Answer:
Tritium is present only in trace amounts. So it can be artificially prepared by bombarding lithium with slow neutrons in a nuclear fission reactor. The nuclear transmutation reaction for this process is as follows.
\({ }_{3}^{6} L i\) + \({ }_{0}^{1} n\) → \({ }_{2}^{4} \mathrm{He}\) + \({ }_{1}^{3} T\)

Question 13.
Write the physical properties of hydrogen.
Answer:
Hydrogen is a colorless, odorless, tasteless, lightest and highly flammable gas. It is a non-polar diatomic molecule. It can be liquefied under low temperature and high pressure. Hydrogen is a good reducing agent.

Question 14.
What is deuterium exchange reaction?
Answer:
Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium or heavy water.
CH₄ + 2D₂ → CD₄ + 2H₂
2NH₃ + 3D₂ → 2ND₃ + 3H₂

Question 15.
Write the physical properties of water.
Answer:
Water is a colourless and volatile liquid. The peculiar properties of water in the condensed phases are due to the presence of inter molecular hydrogen bonding between water molecules. Hydrogen bonding is responsible for the high melting and boiling points of water.

Question 16.
Water is an amphoteric oxide. Give reason.
Answer:
Water is an amphoteric oxide. It has the ability to accept as well as donate protons and hence it can act as an acid or a base. For example, in the reaction with HC1 it accepts proton where as in the reaction with weak base ammonia it donates proton.
NH₃ + H₂O → NH₄⁺ + OH^–
HCl + H₂O → H₃O⁺ + Cl^–

Question 17.
What is hard water? Give its types.
Answer:
Hard water contains high amounts of mineral ions. The most common ions found in hard water are the soluble metal cations such as magnesium & calcium, though iron, aluminium, and manganese may also be found in certain areas. Presence of these metal salts in the form of bicarbonate, chloride and sulphate in water makes water ‘hard’.

Question 18.
What is meant by temporary hardness of water?
Answer:
Temporary hardness is primarily due to the presence of soluble bicarbonates of magnesium and calcium.

Question 19.
What is permanent hardness of water?
Answer:
Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in it.

Question 20.
What are zeolites? Give its use.
Answer:
Zeolites are hydrated sodium alumino-silicates with a general formula, NaO.Al₂O₃ .xSiO₂ .yH₂O (x = 2 to 10, y = 2 to 6). Zeolites have porous structure in which the monovalent sodium ions are loosely held and can be exchanged with hardness producing metal ions (M = Ca or Mg) in water.

Question 21.
What is heavy water? How is it obtained?
Answer:
Heavy water (D₂O) is the oxide of heavy hydrogen. One part of heavy water is present in 5000 parts of ordinary water. It is mainly obtained as the product of electrolysis of water.

Question 22.
What is the effect of shielding on ionization energy?
Answer:
As we move down a group, the number of inner shell electron increases which in turn increases the repulsive force exerted by them on the valence electrons, i.e., the increased shielding effect caused by the inner electrons decreases the attractive force acting the valence electron by the nucleus. Therefore, the ionization energy decreases.

Question 23.
How does hard water produces less foam with detergents?
Answer:
The cleaning capacity of soap is reduced when used in hard water. Soaps are sodium or potassium salts of long chain fatty acids (e.g., coconut oil). When soap is added to hard water, the divalent magnesium and calcium ions present in hard water react with soap. The sodium salts present in soaps are converted to their corresponding magnesium and calcium salts which are precipitated as scum/precipitate.
M²⁺ + 2RCOONa → (RCOO)₂M + 2Na⁺
M = Ca or Mg, R = C₁₇H₃₅

Question 24.
How is hydrogen peroxide prepared?
Answer:
It can be prepared by treating metal peroxide with dilute acid.
BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂

Na₂O₂ + H₂SO₂ → Na₂SO₄ + H₂O₂

Question 25.
What are hydrides? How are they classified?
Answer:
Hydrogen forms binary compounds with many electropositive elements including metals and non¬metals are called hydrides. It also forms ternary hydrides with two metals. E.g., LiBH₄ and LiAlH₄. The hydrides are classified as ionic, covalent and metallic hydrides according to the nature of bonding.

Question 26.
What is a hydrogen bond?
Answer:
When a hydrogen atom (H) is covalently bonded to a highly electronegative atom such as fluorine (F) or oxygen (O) or nitrogen (N), the bond is polarized. Due to this effect, the polarized hydrogen atom is j able to form a weak electrostatic interaction with another electronegative atom present in the vicinity, f This interaction is called as a hydrogen bond.

Question 27.
What are intra and inter molecular hydrogen bonding?
Answer:
Hydrogen bonds can occur within a molecule is called intramolecular hydrogen bonding whereas between two molecules of the same type or different type is called intermolecular hydrogen bonding.

III. Short question and Answers (3 Marks):

Question 1.
Write notes on water-gas shift reaction?
Answer:
The carbon monoxide of the water gas can be converted to carbon dioxide by mixing the gas mixture with more steam at 400°C and passed over a shift converter containing iron/copper catalyst. This reaction is called as water-gas shift reaction.
CO + H₂O → CO₂ + H₂

The CO₄ formed in the above process is absorbed in a solution of potassium carbonate.
CO₂ + K₂CO₃ + H₂O → 2KHCO₃

Question 2.
Write the properties of hydrogen similar to alkali metals.
Answer:
The hydrogen has the electronic configuration of 1s¹ which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
(i) It forms unipositive ion (H⁺) like alkali metals (Na⁺, K⁺, Cs⁺)
(ii) It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
(iii) It also acts as a reducing agent.

Question 3.
Write notes on isotopes of hydrogen.
Answer:
Hydrogen has three naturally occurring isotopes, viz., protium (₁H¹ or H), deuterium (₁H² or D) and tritium (₁H³ or T). Protium (₁H¹) is the predominant form (99.985 %) and it is the only isotope that does not contain a neutron.
Deuterium, also known as heavy hydrogen, constitutes about 0.015%. The third isotope, tritium is a radioactive isotope of hydrogen which occurs only in traces (~1 atom per 1018 hydrogen atoms). Due to the existence of these isotopes naturally occurring hydrogen exists as H₂, HD, D₂, HT, T₂, and DT.

Question 4.
What are ortho and para hydrogen? How will you convert one form into another?
Answer:
In the hydrogen atom, the nucleus has a spin. When molecular hydrogen is formed, the spins of two hydrogen nuclei can be in the same direction or in the opposite direction as shown in the figure. These two forms of hydrogen molecules are called ortho and para hydrogens respectively.

At room temperature, normal hydrogen consists of about 75% ortho-form and 25% para-form. As the ortho-form is more stable than para-form, the conversion of one isomer into the other is a slow process. However, the equilibrium shifts in favour of para hydrogen when the temperature is lowered.

The para-form can be catalytically transformed into ortho-form using platinum or iron. Alternatively, it can also be converted by passing an electric discharge, heating above 800°C and mixing with paramagnetic molecules such as O₂, NO, NO₂ or with nascent/atomic hydrogen.

Question 5.
Discuss the methods of preparation of hydrogen.
Answer:
High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
At Anode:
2OH^– → H₂O + \(\frac{1}{2}\)O₂ + 2e^–

At Cathode:
2H₂O + 2e^– → 2OH^– + H₂

Overall Reaction:
H₂O → H₂ + \(\frac{1}{2}\)O₂
Hydrogen is conveniently prepared in laboratory by the reaction of metals, such as zinc, iron, tin with dilute acid.
Zn + 2HCl → ZnCl₂ + H₂↑

Laboratory preparation of Hydrogen

Question 6.
Write the chemical properties of deuterium.
Answer:
Like hydrogen, deuterium also reacts with oxygen to form deuterium oxide called heavy water. It also reacts with halogen to give corresponding halides.
2D₂ + O₂ → 2D₂O
D₂ + X₂ → 2DX (X = F, Cl, Br & I)

Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium or heavy water
CH₄ + 2D₂ → CD₄ + 2H₂
2NH₃ + 3D₂ → 2ND₃ + 3H₂

Question 7.
Write the reaction of halogens with water.
Answer:
The halogens react with water to give an acidic solution. For example, chlorine forms hydrochloric acid and hypo chlorous acid. It is responsible for the antibacterial action of chlorine water, and for its use as bleach.
Cl₂ + H₂O → HCl + HOCl
Fluorine reacts differently to liberate oxygen from water.
2F₂ + 2H₂O → 4HF + O₂

Question 8.
Discuss the nature of hydrated salts with suitable examples.
Answer:
Many salts crystallized from aqueous solutions form hydrated crystals. The water in the hydrated salts may form co-ordinate bond or just present in interstitial positions of crystals.
Examples: Cr(H₂O₆₎ Cl₃₎ – All six water molecules form co-ordinate bond. BaCl_2.2H₂O – Both the water molecules are present in interstitial positions.
CuSO₄ .5H₂O – In this compound four water molecules form co-ordinate bonds while the fifth water molecule, present outside the co-ordination, can form intermolecular hydrogen bond with another molecule. [Cu(H₂O)₄]SO₄. H₂O.

Question 9.
How is temporary hardness of water removed by boiling?
Answer:
Temporary hardness is primarily due to the presence of soluble bicarbonates of magnesium and calcium. This can be removed by boiling the hard water followed by filtration. Upon boiling, these salts decompose into insoluble carbonate which leads to their precipitation. The magnesium carbonate thus formed further hydrol used to give insoluble magnesium hydroxide.
Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂Mg(HCO₃) → MgCO₃ + H₂O
CO₂MgCO₃ + H₂O → Mg(OH)₂ + CO₂
The resulting precipitates can be removed by filtration.

Question 10.
How is temporary hardness of water removed by Clark’s method?
Answer:
In Clark’s method, calculated amount of lime is added to hard water containing the magnesium and calcium, and the resulting carbonates and hydroxides can be filtered-off.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O
Mg(HCO₃) + 2Ca(OH)₂ → 2CaCO₃+ Mg(OH)₂ + 2H₂O

Question 11.
Write the chemical properties of heavy water.
Answer:
When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium 2NaOH + D₂O → 2NaOD + HOD
HCl + D₂O → DCl + HOD
NH₄Cl + 4D₂O → ND₄Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D₂O is treated with of hypo-phosphorus acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
H₃PO₂ + D₂O → H₂DPO₂ + HDO
It is also used to prepare some deuterium compounds:
Al₄C₃ + 12D₂O → 4Al(OD)₃ + 3CD₄
CaC₂ + 2D₂O → Ca(OD)₂ + C₂D₂
MgN₂ + 6D₂O → 3Mg(OD)₂ + 2ND₃
Ca₃P₂ + 6D₂O → 3Ca(OD)₂ + 2PD₃

Question 12.
Write the uses of heavy water.
Answer:
(i) Heavy water is widely used as moderator in nuclear reactors as it can lower the energies of fast neutrons
(ii) It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
(iii) It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 13.
Write the uses of hydrogen peroxide.
Answer:
The oxidizing ability of hydrogen peroxide and the harmless nature of its products, i.e., water and oxygen, lead to its many applications. It is used in water treatment to oxidize pollutants, as a mild antiseptic, and as bleach in textile, paper and hair – care industry.

Hydrogen peroxide is used to restore the white colour of the old paintings which was lost due to the reaction of hydrogen sulphide in air with the white pigment Pb₃(OH)₂ (CO₃)₂ to form black colored lead sulphide. Hydrogen peroxide oxidises black coloured lead sulphide to white coloured lead sulphate, there by restoring the colour.
PbS + 4H₂O₂ → PbSO₄ + 4H₂O

IV. Long Question and Answers (5 Marks):

Question 1.
Justify the position of hydrogen in the periodic table.
Answer:
The hydrogen has the electronic configuration of 1s¹ which resembles with ns¹ general valence shell configuration of alkali metals and shows similarity with them as follows:
(i) It forms unipositive ion (H⁺) like alkali metals (Na⁺, K⁺, Cs⁺)
(ii) It forms halides (HX), oxides (H₂O), peroxides (H₂O₂) and sulphides (H₂S) like alkali metals (NaX, Na₂O, Na₂O₂, Na₂S)
(iii) It also acts as a reducing agent.

However, unlike alkali metals which have ionization energy ranging from 377 to 520 kJ mol^-1, the hydrogen has 1,314kJ mol^-1 which is much higher than alkali metals.

Like the formation of halides (X^–) from halogens, hydrogen also has a tendency to gain one electron to form hydride ion (H^–) whose electronic configuration is similar to the noble gas, helium. However, the electron affinity of hydrogen is much less than that of halogen atoms. Hence, the tendency of hydrogen to form hydride ion is low compared to that of halogens to form the halide ions as evident from the following reactions:

\(\frac{1}{2}\) H₂ + e^– → H^–                                 ∆H = + 36 kcalmol^-1
\(\frac{1}{2}\) Br₂ + e^– → Br^–                                 ∆H = -55 kcalmol^-1

Since, hydrogen has similarities with alkali metals as well as the halogens; it is difficult to find the right position in the periodic table. However, in most of its compounds hydrogen exists in+1 oxidation state. Therefore, it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

Question 2.
Discuss the reaction of hydrogen with (i) Oxygen (ii) Halogens (iii) Alkali metals.
Answer:
(i) Reaction of hydrogen with oxygen.
Hydrogen reacts with oxygen to give water. This is an explosive reaction and releases lot of energy. This is used in fuel cells to generate electricity.
2H₂ + O₂ → 2H₂O

(ii) Reaction of hydrogen with halogens.
Similarly, hydrogen also reacts with halogens to give corresponding halides. Reaction with fluorine takes place even in dark with explosive violence while with chlorine at room temperature under light. It combines with bromine on heating and reaction with iodine is a photochemical reaction.
2H₂ + O₂ → 2H₂O
In the above reactions the hydrogen has an oxidation state of +1.

(iii) Reaction of hydrogen with alkali metals:
It also has a tendency to react with reactive metals such as lithium, sodium and calcium to give corresponding hydrides in which the oxidation state of hydrogen is -1.
2Li + H₂ → 2 LiH
2Na + H₂ → 2NaH

These hydrides are used as reducing agents in synthetic organic chemistry. It is used to prepare other important hydrides such as lithium aluminium hydride and sodium boro hydride.
4 LiH + AlCl₃ → Li[AlH₄] + 3LiCl
4 NaH + B(OCH₃)₃ → Na[BH₄] + 3CH₃ONa

Question 3.
Explain the uses of hydrogen.
Answer:
1. Over 90 % hydrogen produced in industry is used for synthetic applications. One such process is Haber process which is used to synthesis ammonia in large scales. Ammonia is used for the manufacture of chemicals such as nitric acid, fertilizers and explosives.
N₂ + 3H₂  2NH
2. It can be used to manufacture the industrial solvent, methanol from carbon monoxide using copper as catalyst.
CO + 2H₂  CH₃OH
3. Unsaturated fatty oils can be converted into saturated fats called Vanaspati (margarine) by the reduction reaction with \(\frac{P_{t}}{H_{2}}\)
4. In metallurgy, hydrogen can be used to reduce many metal oxides to metals at high temperatures.
CuO + H₂ → Cu + H₂O

WO₃ + 3H₂ → W + 3H₂O
5. Atomic hydrogen and oxy-hydrogen torches are used for cutting and welding.
6. Liquid hydrogen is used as a rocket fuel.
7. Hydrogen is also used in fuel cells for generating electrical energy. The reversible uptake of hydrogen in metals is also attractive for rechargeable metal hydride battery.

Question 4.
What is permanent hardness of water? How is it removed?
Answer:
Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in it. It can be removed by adding washing soda, which reacts with these metal {M = Ca or Mg) chlorides and sulphates in hard water to form insoluble carbonates.

MCl₂ + Na₂CO₃ → MCO₃+ 2NaCl
MSO₄ + Na₂CO₃ → MCO₃ + Na₂SO₄

In another way to soften the hard water is by using a process called ion-exchange. That is, hardness can be removed by passing through an ion-exchange bed like zeolites or column containing ion-exchange resin. Zeolites are hydrated sodium alumino-silicates with a general formula, NaO . Al₂O₃ . ^xSiO₂. ^yH₂O (x = 2 to 10, y = 2 to 6).

Zeolites have porous structure in which the monovalent sodium ions are loosely held and can be exchanged with hardness producing metal ions (M= Ca or Mg) in water. The complex structure can conveniently be represented as Na₂ – Z with sodium as exchangeable cations.
Na₂ – Z + M²⁺ → M – Z + 2Na⁺

When exhausted, the materials can be regenerated by treating with aqueous sodium chloride. The metal ions (Ca² and Mg²⁺) caught in the zeolite (or resin) are released and they get replenished with sodium ions.
M – Z + 2NaCl → Na₂ – Z + MCl₂

Question 5.
Write the chemical properties and uses of heavy water.
Answer:
When compounds containing hydrogen are treated with D₂O, hydrogen undergoes an exchange for deuterium 2NaOH + D₂O → 2NaOD + HOD
HCl + D₂O → DCl + HOD
NH₄Cl + 4D₂O → ND₄Cl + 4HOD

These exchange reactions are useful in determining the number of ionic hydrogens present in a given compound. For example, when D₂O is treated with of hypo-phosphoric acid only one hydrogen atom is exchanged with deuterium. It indicates that, it is a monobasic acid.
H₃PO₂ +D₂O → H₂DPO₂ + HDO

It is also used to prepare some deuterium compounds:
Al₄Cl₃ + 12D₂O → 4Al(OD)₃ + 3CD₄
CaC₂ + 2D₂O → Ca(OD)₂ + C₂D₂
Mg₃N₂ + 6D₂O → 3Mg(OD)₂ + 2ND₃
Ca₃P₂ + 6D₂O → 3Ca(OD)₂ + 2PD₃

Uses of Heavy Water:
1. Heavy water is widely used as moderator innuclear reactors as it can lower the energies of fast neutrons
2. It is commonly used as a tracer to study organic reaction mechanisms and mechanism of metabolic reactions
3. It is also used as a coolant in nuclear reactors as it absorbs the heat generated.

Question 6.
Explain the structure of hydrogen peroxide.
Answer:
Both in gas-phase and liquid-phase, the molecule adopts a skew conformation due to repulsive interaction of the OH bonds with lone-pairs of electrons on each oxygen atom. Indeed, it is the smallest molecule known to show hindered rotation • about a single bond.

H₂O₂ has a non-polar structure. The molecular dimensions in the gas phase and solid phase differ as shown in figure 4.5. Structurally, H₂O₂ is represented by the dihydroxyl formula in which the two OH-groups do not lie in the same plane.

One way of explaining the shape of hydrogen peroxide is that the hydrogen atoms would lie on the pages of a partly opened book, and the oxygen atoms along the spine. In the solid phase of molecule, the dihedral angle reduces to 90.2° due to hydrogen bonding and the O-O-H angle expands from 94.8° to 101.9°.

Question 7.
What are hydrides? How are they classified?
Answer:
Hydrogen forms binary hydrides with many electropositive elements including metals and non-metals. It also forms ternary hydrides with two metals. E.g., LiBH₄ and LiAlH₄ The hydrides are classified as ionic, covalent and metallic hydrides according to the nature of bonding.

Hydrides formed with elements having lower electronegativity than hydrogen are often ionic, whereas with elements having higher electronegativity than hydrogen form covalent hydrides.

Ionic (Saline) hydrides:
These are hydrides composed of an electropositive metal, generally, an alkali or alkaline-earth metal, except beryllium and magnesium, formed by transfer of electrons from metal to hydrogen atoms. They can be prepared by the reaction of elements at about 400° These are salt-like, high-melting, white crystalline solids having hydride ions (H) and metal cations (Mn+).
2 Li + H₂ → 2LiH

Covalent (Molecular) hydrides:
They are compounds in which hydrogen is attached to another element by sharing of electrons. The most common examples of covalent hydrides of non-metals are methane, ammonia, water and hydrogen chloride. Covalent hydrides are further divided into three categories, viz., electron precise (CH₄, C₂H₆, SiH₄, GeH₄), electron-deficient(B₂H₆) and electron-rich hydrides (NH₃, H₂O). Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

Metallic (Interstitial) hydrides:
Metallic hydrides are usually obtained by hydrogenation of metals and alloys in which hydrogen occupies the interstitial sites (voids). Hence, they are called interstitial hydrides; the hydrides show properties similar to parent metals and hence they are also known as metallic hydrides.

Most of the hydrides are non-stoichiometric with variable composition (TiH_{1.5 – 1.8} and PdH_{0.6 – 0.8}), some are relatively light, inexpensive and thermally unstable which make them useful for hydrogen storage applications. Electropositive metals and some other metals form hydrides with the stoichiometry MH or sometimes MH₂ (M = Ti, Zr, Hf, V, Zn).

Question 8.
Write notes on intermolecular hydrogen bonding.
Answer:
Intermolecular hydrogen bonds occur between two separate molecules. They can occur between any numbers of like or unlike molecules as long as hydrogen donors and acceptors are present in positions which enable the hydrogen bonding interactions. For example, intermolecular hydrogen bonds can occur between ammonia molecule themselves or between water molecules themselves or between ammonia and water.

Water molecules form strong hydrogen bonds with one another. For example, each water molecule is linked to four others through hydrogen bonds. The shorter distances (100 pm) correspond to covalent bonds (solid lines), and the longer distances (180 pm) correspond to hydrogen bonds (dotted lines).

In ice, each atom is surrounded tetrahedrally by four water molecules through hydrogen bonds. That is, the presence of two hydrogen atoms and two lone pairs of electron on oxygen atoms in each water molecule allows formation of a three-dimensional structure. This arrangement creates an open structure, which accounts for the lower density of ice compared with water at 0°C. While in liquid water, unlike ice where hydrogen bonding occurs over a long-range, the strong hydrogen bonding prevails only in a short range and therefore the denser packing.

Question 9.
What is hydrogen bonding? Discuss its properties and applications.
Answer:
Hydrogen bonding is one of the most important natural phenomena occurring in chemical and biological sciences. These interactions play a major role in the structure of proteins and DNA. When a hydrogen atom (H) is covalently bonded to a highly electronegative atom such as fluorine (F) or oxygen (O) or nitrogen (N), the bond is polarized.

Due to this effect, the polarized hydrogen atom is able to form a weak electrostatic interaction with another electronegative atom present in the vicinity. This interaction is called as a hydrogen bond (20 – 50 kJ mol^-1) and is denoted by dotted lines (………).

It is weaker than covalent bond (> 100 kJ mol^-1) but stronger than the vander Waals interaction (< 20 kJ mol^-1). Hydrogen bond has profound effect on various physical properties including vapour pressure (H₂O and H₂S), boiling point, miscibility of liquids (H₂O and C₂H₅OH), surface
tension, densities, viscosity, heat of vaporization and fusion, etc. Hydrogen bonds can occur within a molecule (intramolecular hydrogen bonding) and between two molecules of the same type or different type (intermolecular hydrogen bonding).

Hydrogen bond occurs not only in simple molecules but also in complex biomolecules such as proteins, and they are crucial for biological processes. For example, hydrogen bonds play an important role in the structure of deoxyribonucleic acid (DNA), since they hold together the two helical nucleic acid chains (strands).

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 9 Tissue and Tissue System Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System

11th Bio Botany Guide Tissue and Tissue System Text Book Back Questions and Answers

Part-I.

Question 1.
Refer to the given figure and select the correct statement.

i) A, B, and C are histogen of shoot apex,
ii) A Gives rise to medullary rays.
iii) B Gives rise to cortex.
iv) C Gives rise to epidermis
a) I and ii only
b) ii and iii only
c) i and iii only
d) iii and iv only
Answer:
c) i and iii only

Question 2.
Read the following sentences and identify the correctly matched sentences.
i) In exarch condition, the protoxylem lies outside of the metaxylem.
ii) In endarch condition, the protoxylem lies towards the centre.
iii) In centrarach condition, metaxylem lies in the middle of the protoxylem
iv) In mesarch condition, protoxylem lies in the middle of the metaxylem
a) i, ii, and iii only
b) ii, iii, and iv only
c) i, ii, and iv only
d) All of these
Answer:
c) i, ii and iv only

Question 3.
In Gymnosperms, the activity of sieve tubes are controlled by.
a) Nearby sieve tube members
b) Pholem parenchyma cells
c) Nucleus of companion cell
d) Nucleus ofalbuminous cells
Answer:
d) Nucleus of albuminous cells

Question 4.
When a leaf trace extends from a vascular bundle in a dicot stem, what would be the arrangement of vascular in the veins of the leaf?
a) Xylem would be on top and the pholem on the bottom
b) Pholem would be on the top and the xylem on the bottom
c) Xylem would encircle the pholem
d) Pholem would encircle the xylem
Answer:
a) Xylem would be on top and the pholem on the bottom

Question 5.
Grafting is successful in dicots but not in monocots because the dicots have
a) Vascular bundles arranged ina ring
b) Cambium for secondary growth
c) Vessels with elements arranged end to end
d) Cork cambium
Answer:
b) Cambium for secondary growth

Question 6.
Why the cells of sclerenchyma and tracheids become dead?
Answer:
The cells of sclerenchyma and tracheids become dead because they lack protoplasm.

Question 7.
Explain sclereids with their types
Answer:
1. Sclereids – dead cells Isodiametric – but some elongated.
2. Cell wall is very thick due to lignification
3. Lumen – much reduced
4. Pits – may be simple or branched.


Question 8.
What are sieve tubes? Explain.
Answer:
Sieve tubes are long tube-like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube. The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls. They may possess simple or compound sieve plates.

The function of sieve tubes are believed to be controlled by campanion cells In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm. A special protein (P. Protein = Phloem Protein) called slime body is seen in it. In mature sieve tubes, the pores in the sieve plate are blocked by a substance called callose (callose plug). The conduction of food material takes lace through cytoplasmic strands. Sieve tubes occur only in Angiosperms.

Question 9.
Distinguish the anatomy of dicot root from monocot root
Answer:

Characters	Dicot root	Monocot root
1. Pericycle	Gives rise to lateral roots, phellogen and a part of vascular cambium	Gives rise to lateral roots only.
2. Vascular tissue	Usually limited number of xylem and phloem strips.	Usually more number of xylem and phloem strips,
3. Conjunctive tissue	Parenchymatous; Its cells are differentiated into vascular cambium.	Mostly sclerenchymatous but sometimes parenchymatous. It is never differentiated in to vascular cambium.
4. Cambium	It appears as a secondary meristem at the time of secondary growth.	It is altogether absent.
5. Xylem	Usually tetrach	Usually poly arch
6. Pith	Absent	Present at the centre

Question 10.
Distinguish the anatomy of dicot stem from monocot stem
Answer:

Characters	Dicot root	Monocot root
1. Hypodermis	collenchymatous	Sclerenchymatous
2. Ground tissue	Differentiated into cortex, endodermis and pericycle and pith	Not differentiated, but it is a continuous mass of parenchyma.
3. StarchSheath	Present	Absent
4. Medullary rays	Present	Absent
5. Vascular bundles	a) Collateral and open	a) Collateral and closed
	b) Arranged in a ring	b) Scattered in ground tissue
	c) Secondary growth occurs	c) Secondary growth usually does not occur.

Part-II

11th Bio Botany Guide Tissue and Tissue System Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
Who is the father of plant anatomy?
(a) David Muller
(b) Katherine Esau
(c) Nehemiah Grew
(d) Hofmeister
Answer:
(c) Nehemiah Grew

Question 2.
Father of Anatomy, as well as the scientist, who coined the term Meristem is
a) Hofmeister
b) Mettemius
c) Nehemiah Grew
d) Bloch
Answer:
c. Nehemia Grew

Question 3.
The book “Anatomy of seed plants” is written by:
(a) Hanstein
(b) Schmidt
(c) Nicholsen
(d) Katherine Esau
Answer:
(d) Katherine Esau

Question 4.
The fibres in which lignin is less and cellulose is more in the cell walls is known as
a) Gelatinous fibres,
b) Septate fibres
c) Libriform fibres
d) Hard fibres
Answer:
a. Gelatinous fibres

Question 5.
Which of the statement is not correct?
(a) Meristematic cells are self-perpetuating
(b) Meristematic cells are the most actively dividing cells
(c) Meristematic cells have large vacuoles
(d) Meristematic cells have dense cytoplasm with a prominent nucleus
Answer:
(c) Meristematic cells have large vacuoles

Question 6.
In mature sieve tubes, the pores in the sieve plates are blocked by a substance called
a) gum & nesins
b) Callose
c) Callus
d) Pectinose
Answer:
b. Callose

Question 7.
The tunica is:
(a) the peripheral zone of shoot apex, that forms cortex
(b) the inner zone of shoot apex, that forms stele
(c) the peripheral zone of shoot apex, that forms the epidermis
(d) the inner zone of shoot apex, that forms cortex and stele
Answer:
(c) the peripheral zone of shoot apex, that forms the epidermis

Question 8.
The tissue, that provide mechanical support and elasticity to the growing parts of the plant is
a) Sclerenchyma
b) Sclereids
c) Fibres
d) Collenchyma
Answer:
d. Collenchyma

Question 9.
The quiescent centre concept was proposed by:
(a) Lindall
(b) Clowes
(c) Holstein
(d) Sanio
Answer:
(b) Clowes

Question 10.
A meristem which divide in all planes is called
a) Lateral meristem
b) Apical meristem
c) Plate meristem
d) Mass meristem
Answer:
d. Mass meristem

Question 11.
Petioles of banana is composed of:
(a) storage parenchyma
(b) stellate parenchyma
(c) angular collenchyma
(d) prosenchyma
Answer:
(b) stellate parenchyma

Question 12.
The term ‘Hadrome’ for xylem and ‘Leptome’ for phloem were coined by
a) Sachs
b) Nageli
c) Hanstein
d) Haberlandt
Answer:
d. Haberlandt

Question 13.
The seed coat of groundnut is made up of:
(a) stone cells
(b) osteosclereids
(c) macrosclereids
(d) parenchyma cells
Answer:
(b) osteosclereids

Question 14.
The theory equivalent to Tunicia Corpus theory is
a) Histogen theory
b) Korperkappe theory
c) Apical cell theory
d) Quiescent center concept
Answer:
b. Korper Kappe theory

Question 15.
The term xylem was introduced by:
(a) Alexander
(b) Nageli
(c) Holstein
(d) Schmidt
Answer:
(b) Nageli

Question 16.
Trichoblasts are
a) Long cells seen in the root epidermis
b) the hair-like appendages seen on stem epidermis
c) the short cells seen in the piliferous layer of roots
d) the cells helping in the dispersal of seeds and fruits
Answer:
c. the short cells seen in the piliferous layer of roots

Question 17.
In cross-section, the tracheids are:
(a) hexagonal in shape
(b) rectangular in shape
(c) triangular in shape
(d) polygonal in shape
Answer:
(d) polygonal in shape

Question 18.
Stele include
a) Endodermis, pericycle, & Vascular bundle
b) Pericycle, Vascular bundle & pith
c) Cortex, endodermis, & Percycle
d) Xylem, phloem, cambium, & Pith
Answer:
b. Pericycle, Vascular bundle & Pith

Question 19.
Bulliform cells are present in:
(a) mango
(b) grasses
(c) groundnut
(d) potato
Answer:
(b) grasses

Question 20.
Water stomata occur in
a) Mangrove plants
b) Grass plants
c) Monocotyledon plants
d) Aquatic plants
Answer:
b. Grass Plants

Question 21.
In Ocimum the trichomes are:
(a) non – glandular
(b) fibrous
(c) glandular
(d) none of these
Answer:
(c) glandular

Question 22.
Sunken stomata is an adaptation seen in
a) Cycas
b) Neem
c) Ficus
d)Nerium
Answer:
d. Nerium

Question 23.
Casparian strips contain thickenings of:
(a) calcium carbonate and calcium oxalate
(b) carbohydrate, protein and lignin
(c) crystal of calcium oxalate
(d) lignin, suberin and some other carbohydrates
Answer:
(d) lignin, suberin and some other carbohydrates

Question 24.
The extension of pith cells that are involved in radial conduction of food and water is known as
a) Amphivasal vascular rays
b) Radial vascular parenchyma
c) Medullary ray
d) Inter fascicular parenchyma
Answer:
c. Medullary ray

Question 25.
Secondary phloem is derived from:
(a) apical meristem
(b) vascular cambium
(c) primary phloem
(d) none of the above
Answer:
(b) vascular cambium

Question 26.
Ground tissue includes all tissues except
a) Vascular bundles and pith
b) Epidermis and vascular strands
c) Cortex and vascular strands
d) Pith and conjunctive tissue
Answer:
b. Epidermis and vascular strands

Question 27.
In beans, the metaxylem vessels are generally:
(a) polygonal in shape
(b) circular in shape
(d) rectangular in shape
(d) triangular in shape
Answer:
(a) polygonal in shape

Question 28.
The thickening of which substance make endodermis impervious to water
a) Hemicellulose, cellulose, and pectin
b) Lignin, suberin, or cutin
c) Cellulose, Pectin, and Lignin
d) Pectin, Hemicellulose, and Suberin
Answer:
b. Lignin, Suberin, or Cutin

II. Match The Following & Find Out The Correct Order:

Question 1.
(I) Protoxylem lacuna – A. Liriodendron
(II) Multiple perforation plates – B. Gnetum
(III) Fibre like sclereids occur in – C. Zeamaysstem
(IV) Vessels occur in – D. Olea europaea

Answer:
a) C-A-D-B

Question 2.
(I) Apical Meristem – A. Cambium
(II) Lateral Meristem – B. Intemode
(III) Intercalary meristem – C. Root Apex
(IV) Secondary meristem – D. Cork cambium

Answer:
d) C-A-B-D

Question 3.
Name of the cell Occurence
(I) Bulliform cells or Motor cells – A. Rose&Ocimum
(II) Multilayered epidermis – B. Styrax & Hibiscus
(III) Glandular trichomes – C. Nerium & Ficus
(IV) Stellate hairs – D. Chloris & Grass

Answer:
a) D-C-A-B

Question 4.
Nature of vascular bundle Example
(I) Conjoint, Collateral & closed – A. Dicot root
(II) Conjoint, Collateral open Endarch – B. Monocot stem
(III) Radial, Tetrarch & Exarch – C. Dicot leaf & Monocot leaf
(IV) Conjoint, Collateral Close & Endarch- D. Dicot stem

Answer:
b) C-D-A-B

Question 5.
(i) Surface fibre – A. Jute
(II) Soft fibre – B. Agave
(III) Leaf fibre – C. Coconut
(IV) Septate fibre – D. Cofton
(V) Mesocarp fibre – E. Teak

Answer:
b) D – A – B – E

Question 6.
Lateral roots originate
(i) Endo genously
(ii) From pericycle cells
(iii) Exogenously
(iv) From endodermal cells
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
a. I & II

Question 7.
Monocot stem has
(I) Medulla or pith
(II) Atactostele
(III) Cambium – present
(IV) Scattered & skull-shaped bundles occur
a) I & II
b) II & III
c) II & IV
d) I & III
Answer:
c. II & IV

Question 8.
Which of the following statements are correct with reference to monocot stem
(I) Starch sheath is absent
(II) Pith is absent
(III) Pericycle absent
(IV) Phloem parenchyma is present
a) I, II, III
b) I and IV
c) II and IV
d) III & IV
Answer:
a. I, II,& III

III. State True Or False & On That Basis Choose The Right Answer

Question 1.
I) Lateral meristem – It occurs between the mature tissues, responsible for elongation of intemodes.
II) Inter calary meristem – It occurs along the longitudinal axis of stem and root, responisble for secondary growth
III) Protoderm – It gives rise to epiderminal tissue system, (i.e) epidermis, stomata & hairs
IV) Ground meristem – It gives rise to all tissues except Vascular strands and epidermis

Answer:
b) False – False – True – True

Question 2.
I) Phloem fibres and phloem parenchyma, are absence in primary phloem of monocot stem.
II) Phloem fibres are also known as Libriform fibres.
Ill Sieve cells are main food conducting elements of Angiosperms
IV) Phloem fibres are absent in primary phloem of Dicot stem

Answer:
a) True – False – False – True

Question 3.
I) The bundle cap of Dicto stem is known as Hard bast.
II) The bundle cap of Dicot stem is parenchymatous
III) The bundle sheath of Dicot leaf is sclerenchymatous walls of Endodermis in Endodermis is known as the outermost layer of stele.
IV) In Angiosperms pericycle gives rise to lateral roots

Answer:
b) False – True – False – True

Question 4.
I) Prickles are one type of epidermal emergences with vascular supply
II) Albuniinous cells! straburger cells – in conifers are analogous to companian cells of Angiosperm but
III) Piliferous layer, Epiblema are other names of Endodermis.
IV) Hypodermis of Dicot stem is living, whereas the Hypodermis of Moncot stem is dead.

Answer:
d) False – True – False – True

Question 5.
I) The inner most layer of cortex is known as pericycle
II) Suberin, lignin, and some other carbohydrates are present as strips in the radial and inner tangentious walls of the endodermis
III) Endodermis is known as the outer most layer of stele.
IV) InAngiosperms pericycle gives rise to lateral roots

Answer:
b) False – True – False – True

IV. With Reference To The Given Diagram, Identify The Incorrect Option Given Below:

Question 1.
a) Living cells with cell wall made up of more of hemicellulose and pectin besides cellulose
b) The type of tissue is of common occurrence in the hypodermis of Helianthus stem
c) Here cells are compactly arranged with thickening on the intercellular spaces
d) Here cells compactly arranged with thickening appear as successive tangential layers

Answer:
c) Here cells are compactly arranged with thickening on the intercellular spaces.

Question 2.
With reference to the given diagram/ figure of section of the plant organ, identify the in correct.
a) There is no epidermal growth, and hypodermis is sclerenchymatous
b) Cortex is absent but ground tissue is present
c) Endodermis, pericycle and pith are absent
d) Vascular bundles are scattered, skull shaped conjoint, collateral open and endarch

Answer:
d) Vascular bundles are scattered skull-shaped conjoint, collateral open and endarch

Question 3.
With reference to the given figure of the section of the plant part, identify the incorrect option given.
a) The vascular bundles are radial, tetrarch and endarch
b) The vascular bundles are radial tetrarch, and exarch
c) This is the cross-section of the primary structure root of Beam
d) Here xylem and phloem are arranged alternate to one another

Answer:
a) The vascular bundles are radial, tetrarch and endarchces.

Question 4.
With reference to the figure choose the right option.

	A	B	C	D
a	Lateral meristem	LeafPrimordia	Intercalary meristem	Apical meristem
b	Apical meristem	Lateral meristem	Intercalary meristem	Leaf primordia
c	Apical meristem	Intercalary meristem	Lateral meristem	Leaf primordia
d	Intercalary meristem	Lateral meristem	Leaf primordia	Apical meristem

V. Out of the given four options, find out the three relevant statements with reference to Quiescent centre.

Question 1.
a) It is the peripheral zone of shoot apex
b) This is the appearance in the active region of cells in root promeristem
c) It is located between calyptrate and other differentiating cells
d) It is the site of hormone synthesis.
i) a, b & C
ii) b, c & d
iii) a, c & d
iv) a, b & d
Answer:
ii) b, c & d

Question 2.
Out of the given four, find out the three relevant statements with reference to sclereids
a) These are dead cells, isodiametric, but some elongated to
b) The cell wall is very thick due to lignification
c) These are living, lignified cells with elongated tapering ends
d) These are only mechanical in function.
i) a,b& c
ii) a,c& d
iii) a,b,& d
iv) b,c, & d
Answer:
iii) a,b, & d

Question 3.
Generally, Ground tissue include
a) Cortex
b) Pericycle
c) Pith
d) Vascular bundle
i) a,b& d
ii) b,c,& d
iii) a,c,& d
iv) a,b,& c
Answer:
iv) a,b& c

VI. Find out the incorrect statement.

Question 1.
a) A stoma is surrounded by pa ir of guard cells
b) Each stoma opens into an air chamber
c) Guard cells contain no chloroplasts
d) The cuticle helps to check transpiration
Answer:
c. Guard cells contain no chloroplasts

Question 2.
a) Sieve cells occur in gymnosperms
b) Sieve tubes occur in Angiosperms
c) Sieve cells are absent in Angiosperms
d) Vessels are absent in Gnetum
Answer:
d. Vessels are absent in Gneturn

Question 3.
Read the following statements having two blank A and B
Collenchyma cell walls contain and find the correct option for A and B

Blank A	Blank B
a) Pectin	1. Lignin
b) cellulose	2. Aminosugar
e) Lignin	3. Cellulose
d) Pectin	4. Hemicellulose

Answer:
d. Pectin Hemicellulose

Question 4.
Read the following statements having two blank A and B Collenchyma cell walls contain A and B find the correct option for A and B

Answer:
C. C → E → A → D → B

VII. From the given choose the correct answer – Regarding Assertion & Reason

Question 1.
ASSERTION: – A The endodermis of root is homologous to starch sheath of dicot stem.
REASON -R The cells of endodermis are rich in starch grain and so-referred as starch sheath.
a. A & R correct and R is explaining A
b. A&R correct but R is not explaining A
c. A-correct but R is false
d. A – correct and R is not explaining ‘A’
Answer:
a. A&R correct and R is explaining A

Question 2.
ASSERTION: – A In Gymnosperm – plants show well developed vessels & fibres
REASON -R Companian cells are absent in Gymnosperm plants.
a. BothA&Rture, ‘R’is giving correct explanation of‘A’
b. Both A&R- true, but ‘ R’ is not correct explanation of ‘ A’
c. Both A & R are false
d. ‘A’ is false and ‘R’ is true.
Answer:
d. ‘A’ is false and ‘R’ is true.

Question 3.
ASSERTION:-A In grasses the bundle sheath is called kranz sheath
REASON -R It is involved in photsynthesis
a. ‘A’ and ‘R’ are right
b. A and R are wrong
c. R does not explain A
d. A is right and ‘R’ is wrong
Answer:
a. ‘A’ and ‘R’ are right

VIII. 2 Marks Questions

Question 1.
What is the Use of the study of Anatomy?
Answer:

• The organisation of cells and different kinds of tissues is understood.
• It is studied by means of dissection and microscopic examination.
• The organisation of cells and different kinds of tissues is understood by the study of anatomy
• The anatomical structure of different organs of plants can be compared
• The anatomical knowledge play an important role in taxonomical studies too.

Question 2.
What are the different types of plant tissue?
Answer:
The two types of principal groups are:

1. Meristematic tissues
2. Permanent tissues.

Question 3.
The pulp of pear is stony & gritty, whereas the seed coat of Pisum sativum seed coat is bony & shiny give reasons.
Answer:
The pulp of pear has Brachysclereids that make it stony and gritty, whereas the seed Coat of Peas seed coat is bony and shiny due to the presence of Osteosclereids

Question 4.
Mention the function of the apical meristem.
Answer:
Present in apices of root and shoot. It is responsible for the increase in the length of the plant, it is called primary growth.

Question 5.
Differentiate between Centrach and Mesearch xylem
Answer:

Centrach	Mesarch
Protoxylem lies in the centre, surrounded by metaxylem	Protoxylem lies in the centre, surrounded by metaxylem
Only one Vascular stand is developed Eg – Selaginella sp	Here unlike cent reach many vascular bundles are developed Eg – Ophioglossum sp.

Question 6.
Differentiate between Trichoblast and Trichomes
Answer:

Trichoblast	Trichomes
The root epidermis is made up of single layer of parenchyma, with big and small cells – The root hair are the extension of small cells known as trichoblast.	The epidermal layers of stems and leaves have unicellular or multicellular appendages that originate from the epidermal cells, known as trichomes, can be branched or unbranched, glandular or non glandular, helpful m dispersal of fruits & seeds. They are also protective infunction.

Question 7.
Differentiate between Exarch and Endarch condition.
Answer:

Exarch	Endarch
Protoxylem lies towards the periphery and metaxylem towards the centre is called Exarch condition. Eg. – Root Anatomy	Protoxylem lies towards the centre and metaxylem, towards the periphery is known as Endarch condition. Eg. Stem Anatomy

Question 8.
Explain briefly Branchysciereids or Stone cells.
Answer:
Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.

Question 9.
What is Protoxylent lacuna?
Answer:

• In Monoeoi stem, Xylem vessels occur in the form of letter ‘ Y’. The upper two arms of has two metaxylem vessels and at the base on or two protoxylem vessels occur.
• At maturity, the lowes, basal protoxylem disintegrates and form a cavity known as Protoxylem lacuna.

Question 10.
Distinguish between Eustele and Atactostele
Answer:

Eustele	Atactostele
Vascular bundles are arranged in the form of a ring around the pith is known as Eustele Eg. Dicot Stem (Sun flower)	Vascular bundles are simply scattered in the ground tissue. This condition is known as Atactostele Eg. Monocot Stem (Maize)

Question 11.
What are bast fibres?
Answer:
These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so-called pericyclic fibres are actually phloem fibres.

Question 12.
What is the significance of Quiescent centre?
Answer:

• The apparently inactive centre in the root anatomy, located between root cap and differentiating cells of the root.
• It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 13.
Differentiate between Meristematic Tissue and Permanent tissue.
Answer:

Meristematic tissue	Permanent tissue
Cells divide repeatedly	Donot divide but develop from meristematic tissue.
Cells are undifferentiated	Cells are differentiated
Produce other tissues	Perform specific functions.

Question 14.
Differentiate between xylary fibres and Extra xylary fibres (Phloem fibres)
Answer:

Xylary fibres	Bast fibres
Associated with sec xylem tissue	Present in phloem
Derived from the vascular cambium	Derived from phloem or outer bark
Many types Eg. Teak	Eg. Jute, Kenaf, Flax & hemp plant fibres

Question 15.
Explain bulliform cells in grasses.
Answer:
Some cells of the upper epidermis (eg: Grasses) are larger and thin-walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
What is meant by Sunken Stomata?
Answer:
In some Xerophytic plants (eg: Cycas, Nerium), stomata are sunken beneath the abaxial leaf surface within stomatal crypts. The sunken stomata reduce water loss by transpiration.

Question 17.
Distinguish, Protoxylem and Metaxylem from Protophloem and Metaphloem
Answer:

Proto & Metaxylem	Proto & Metaphloem
From the primary Xylem derived from procambium, the first formed elements are known as protoxylem and the later formed are known as metaxylem.	From the primary phloem derived from procambium, the first formed elements are known as proto phloem are known as meta phloem.

Question 18.
Distinguish the Bundle sheath of stem and leaf
Answer:

Bundle sheath of stem	Bundle sheath of leaf
1. Bundle sheath is the surrounding tissue of the vascular bundle	1. The sheath surrounding the dicot leaf and monocot leaf is known as bundle sheath
2. In monocot stem it is sclerenchymatous	2. It is parenchymatous both in Dicot and Monocot leaf
3. It is protective in function.	3.It is also known as border parenchyma, protective in function.

Question 19.
Distinguish Guard Cells and Subsidiary Cells
Answer:

Guard Cells	Subsidiary Cells
1. The two kidney-shaped cells in dicot leat and the two dumbbell-shaped cells in monocot leaf, which flank the stoma are called Guard Cells	1. These are specialised epidermal cells, distinct from other cells of the epidermis
2. Chloroplasts are present in the cells	2. Chloroplasts are absent in the cells
3.Help in opening and closing of stoma	3.She subsidiary ceils assist guard cells in the opening and closing of stoma

Question 20.
Differential between Radial and Collateral Vascular bundle.
Answer:

Radial	Collateral
1. Here, the xylem and phloem are arranged at different radius, (i.e) alternating with one another. Eg. Root Anatomy	1. this condition, the phloem and xylem lie in the same radius, one below another. 2. Here phloem is above and xylem lies below, this condition is known as conjoint, collateral Eg. Stem Anatomy

Question 21.
Describe briefly radial types of vascular Bundles.
Answer:
Xylem and phloem are present on different radii alternating with each other. The bundles are separated by parenchymatous tissue. (Monocot and Dicot roots).

Question 22.
What are Halophiles?
Answer:

• Plants adapted to grow in salty environmental conditions are known as Halophytes
• The secretion of ions by the salt glands, present in the leaves is the best mechanism to regulate the salt content of plant shoots.
• Eg. Mangrove Plants-Avicennia

Question 23.
Write down the function of Sclerenchyma.
Answer:

• Main function is to provide mechanical strength.
• Grittiness in the pulp of fruits like Guava, the presence of Pear, Pyrus etc is due to the presence of Sclerenchyma tissue.
• Provide rough and stiffness to seed coats nuts etc.
• Give various types of commercially useful fibres. Eg. Jute, hemp, cotton.

Question 24.
What are the special aspects of the trichomes on the leaves of insectivorous plants?
Answer:
The trichomes on the leaves of the insectivorous plants secrete mucopolysaccharides that help to trap bisects in the insectivorous plants living in marshy plants.

Question 25.
Define, hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth. Hydathodes occur in the leaves of submerged aquatic plants such as ranunculus fluitans as well as in many herbaceous land plants.

Question 26.
Notes on multilayered epidermis multiseriate epidermis.
Answer:

• In some leaves the upper and lower epidermis remain multilayered.
• The outer most layer has cuticle.
• In Nerium these multilayers and the culicle help to reduce the rate of transpiration.
• In Ficus the upper epidermal layer contain cystoliths made up of calcium carbonate crystals.
• These are plants that grow in dry climatic conditions and these are the Anatomical adaptations seen in xerophytic plants.

Question 27.
Notes on Medulla or Pith.
Answer:

• In the Dicot stem, Dicot root and Monocot root the central part is made up of ground tissue known as pith.
• Usually, starch, fatty substances, tannin, phenol, calcium oxalate crystals are stored in the pith.
• Function: storage

Question 28.
State Tunica corpus theory.
Answer:

• The theory was proposed by A. Schmidt (1924)
• There are two zones of tissues are found in apical meristem.
• Tunica-It is the peripheral zone of shoot apex that forms epidermis.
• Corpus – It is the inner zone of shoot apex that forms cortex and stele of the shoot.

Question 29.
State Koroperkappe theory.
Answer:

• The Korper Kappe theory was proposed by schuepp.
• This theory is equivalent to Tunica corpus theory of shoot apex.
• The two divisions are distinguished by the type of T division.
• Korper is characterised by inverted T divisions
• Kappe is characterised by straight T divisions.

Question 30.
Name the 4 types of xylary fibres.
Answer:

• Xylary fibres are associated with the secondary xylem tissue.
• These fibres are derived from the vascular cambium.

There are 4 types of xylary fibres.

• Libriform fibres – Long, narrow fibres with simple pits and lignified secondary walls
• Fibre tracheids – Shorter with moderate thickening pits-simple or bordered
• Septate fibres – Fibres have thin septa separating the lumen in to distinct chambers. Eg. Teak.
• Gelatinous fibres – Fibres with less lignin and more cellulose in the cell wall.

Question 31.
Distinguish single perforation plate from multiple perforation plate.
Answer:
Xylem vessels are perforated at the end walls.
If the entire cell wall is dissolved – and, give rise one pore then it is known as single perforation plate. Eg. Mangifera.
If the perforation plate has many pores it is called Multiple perforation plate. Eg. Liriodendron.

Question 32.
State Apical cell theory.
Answer:

• Proposed by Nagel
• Single apical cell-composes the Root meristem
• The apical initial is tetrahedral in shape and produces the root cap from one side.
• The remaining 3 sides produces epidermis cortex and vascular tissue.
• Fg. Vascular cryptogams.

IX. Identify the diagram & Label the parts.

Question 1.

Answer:
The given diagram is shoot apical meristem
A – Apical cell
B – Leaf Primordium

Question 2.

Answer :
Brachysciereid
A – Lumen Cell B – Thick cell wall

Question 3.
Name the tissue found ¡n these fruits Name the fruits a, b, c

Answer :
Sclerenchyma is the tissue found in these fruits
A – Pear fruit
B – Strawberry
C – Guava

Question 4.

Answer :
Asteroscie Reid
A – Thick cell wall
B – Lumen

X. 3 Mark Questions

Question 1.
Give an account of Prosenchyma and Chiorenchyma
Answer:
These are the two types of Parenchyma tissue
Prosenchyma:
Here the parenchyma cells become elongated, pointed and slightly thick-walled it provide mechanical support

Chiorenchyma:
Parenchyma cells with chiorephyll is known as chiorenchyma.
Eg. Mesophyll of leaves. it can be divided in Palisade tissue and spongy tissue in dicot leaf.

Question 2.
Distinguish libre and sclereids?
Answer:

Fibre	Sclereids
1. Long Cells	Short Cells
2. Narrow, Elongated pointed ends	Usually short and broad
3. Occurs in bundles	Occurs individually or in small groups
4. Commonly unbranched	Maybe branched
5. Derived directly from meristematic tissue	Develops from secondary sclerosis parenchyma cells

Question 3.
What is meant by the quiescent centre concept?
Answer:
Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. This centre is located between the root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 4.
Difference Between Meristernatic Tissue and Permanent Tissue.
Answer:

Meristematic tissue	Permanent tissue
1. Cells divide repeatedly	1. Do not divide
2. Cells are undifferentiated	2. Cells are fully differentiated
3. Cells are small and Isodiametric	3. Ceils are variable in shape and size
4. Intercellular spaces are absent	4. Intercellular spaces are present
5. Vacuoles are absent	5. Vacuoles are present
6.Cell walls are thin	6.Cell walls are may be thick or thin
7. Inorganic inclusions are absent	7. Inorganic inclusions are present

Question 5.
Draw and label three types of collenchyma
Answer:

Question 6.
Differentiate between Dicot leaf and Monocot leaf.
Answer:

Dicot Leaf	Monocot Leaf
1. Dorsiventral leaf	1. Isobilateral leaf
2. The mesophyll is differentiated into palisade and spongy parenchyma	2. Palisade parenchyma is present on both sides of the leaf and spongy parenchyma lies in the centre
3. Eg. Sunflower	3. Eg. Grass

Question 7.
Define tracheids & Draw the different types of cell wall thickening seen in tracheids & vessels
Answer:
Tracheids are dead, lignified and elongated cells with tapering ends. Its lumen is broader than that of fibres. In cross section, the tracheids are polygonal.

Question 8.
Draw the structure of stomata & label the parts.

Question 9.
Give a brief answer on subsidiary cells in plant leaves.
Answer:
Stomata are minute pores surrounded by two guard cells. The stomata occur mainly in the epidermis of leaves. In some plants addition to guard cells, specialised epidermal cells are present which are distinct from other epidermal cells. They are called Subsidiary cells. Based on the number and arrangement of subsidiary cells around the guard cells, the various types of stomata are recognized, The guard cells and subsidiary cells help in the opening and closing of stomata during gaseous exchange and transpiration.

Question 10.
Distinguish between Bulliform or motor cells, and silica cells
Answer:

Bulliform or motor cells	Silica cells
1. Some cells of upper epidermis in Grasses are larger and thin-walled, are known as bulliform or motor cells	1. Some of the epidermal cells of grass are filled with silica. They are called silica cells.
2. These cells are helpful for the rolling and unrolling of the leaf- according to the weather change in order to check transpiration	2. They provide mechanical stability and protection to the tissues

Question 11.
Explain the piliferous layer as epiblema.
Answer:
The outermost layer of the root is known as piliferous layer. It consists of single row of thin-walled parenchymatous cells without any intercellular space. Epidermal pores and cuticle are absent in the piliferous layer. Root hairs that are found in the piliferous layer are always unicellular. They absorb waer and mineral salt from the soil. Root hairs are generally short-lived. The main function of piliferous layer is protection of the inner tissues.

Question 12.
Differentiate between sieve tubes and vessels
Answer:

Sieve tube	Vessels
1. It is a component of phloem	1. It is a component of xylem
2. It is a syncyte (i.e) cell which is formed by fusion of cells is called syncyte	2. It is also a syncyte
3. Nucleus is absent but contain a lining layer of cytoplasm so known as living syncyte	3. Nucleus is absent but contain a lining layer of cytoplasm so known as living syncyte

Question 13.
Differentiate between Amphicribral (Halocentric) and Amphivasal (Leptocentric) vascular bundle.
Answer:
The above two come under concentric type of vascular bundle. Here xylem and phloem are present in concentric circles one around the other, in some stems.

Amphicribral – (Halocentric)
Here xylem lies in the centre and phloem surrounding it.
Eg. Ferns – (Polypodium) dicots – aquatic Amphivasal – (Leptocentric)
Here phloem lies in the centre and xylem surrounding it.
Eg. Dragon plant – Dracena & Yucca

Question 14.
Bring out the different between vascular bundles of Dicot and Monocot roots.
Answer:

	Dicot roots	Monocot root
1. Vascular tissue	Usually limited number of xylem and phloem strips	Usually more number of xylem and phloem strips.
2. Conjunctive tissue	Parenchymatous; Its cells are differentiated into vascular cambium	Mostly sclerenchymatous but sometimes parenchymatous. It is never differentiated in to vascular cambium
3. Cambium	It appears as a secondary meristem at the time of secondary growth	It is altogether absent
4. Xylem	Usually tetrarch	Usually poly arch

Question 15.
What is meant by kranz Anatomy? What is its importance.
Answer:

• In C₄ plants like maize, the tissue outside the vein, (vascular bundle), the bundle sheath is with large chloro- plasts where as its spongy tissue have few if any chloroplast.
• This anatomical uniqueness is known as Kranz anatomy. The border parenchyma has chloro plasts with out grana.
  This kranz sheath help in efficient CO₂, fixation in C₄ plants than C₃ plants.

Question 16.
Explain the nature of phloem in dicot stem.
Answer:
Primary phloem lies towards the periphery. It consists of protpphloem and metaphloem. Phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent in primary phloem. Phloem conduct organic foods material from the leaves to other parts of the plant body.

XI. 5 Marks Questions

Question 1.
A section enlarged – T.S. of Dicot leaf (Helianthus)
Answer:

Question 2.
Explain in detail about the vascular bundles of monocot stem.
Answer:
1. Vascular bundles: Vascular bundles are scattered (atactostele) in the parenchyma ground tissue. Each vascular bundle is surrounded by a sheath of sclerenchymatous fibres called bundle sheath. The vascular bundles are conjoint, collateral, endarch and closed. Vascular bundles are numerous, small and closely arranged in the peripheral portion. Towards the centre, the bundles are comparatively large in size and loosely arranged. Vascular bundles are skull or oval-shaped.

2. Phloem: The phloem in the monocot stem consists of sieve tubes and companion cells. Phloem parenchyma and phloem fibres are absent. It can be distinguished into an outer crushed protophloem and an inner metaphloem.

3. Xylem: Xylem vessels are arranged in the form of ‘Y’ the two metaxylem vessels at the base. In a mature bundle, the lowest protowylem disintegrates and forms a cavity known as protoxylem lacuna.

Question 3.
Korper Kappe theory:
Answer:

• Schuepp (1917)- proposed it
• According to it, Root system has 2 zones – Korper and Kappe
• Korper – zone forms body and Kappe forms the cap
• This theory is comparable to Tunica – corpus theory of shoot apex.

Question 4.
Compare and contrast simple and complex tissues by tabulation.
Answer:

Question 5.
Draw the different types of phloem elements and add a note on sieve tubes.
Answer:

• Sieve tubes are long tubes formed by a series of cells known as sieve tube elements.
• Arranged one above another to form vertical sieve tube.
• No. of pores occur on end walls – known as sieve plate.
• Sieve plates may be simple or compound.
• Sieve elements show nacreous thickening on their lateral walls.
• Mature sieve tube, nucleus is absent, only lining layer of cyto plasm, a special phloem protein (slimy body) is seen in it.
• Mature sieve tubes pores blocked by a substance known as callose (callose plug) sieve tube function as food conducting tissue Angiosperm.
  

Question 6.
Differentiate between collateral and Bicollateral vascular bundles.
Answer:

Question 7.
Tabulate the Anatomical differences between root and stem
Answer:

Characters	Root	Stem
1. Epidermis	Absence of Cuticle and epidermal pores.	Presence of cuticle and epidermal pores.
	Presence of unicellular root hairs.	Presence of unicellular and multicellular trichomes.
2. Outer cortical cells	Chlorenchyma absent	Chlorenchyma present
3. Endodermis	Well defined	ill-defined or absent
4. Vascular bundles	Radial arrangement	Conjoint arrangement
5. Xylem	Exarch	Endarch

Question 8.
Explain the various types of vascular bundles in a tabulation form.
Answer:

Question 9.
Tabulate the types and characteristics of tissue systems.
Answer:

Types/characters	Epidermal tissue system	Ground or fundamental tissue system	Vascular or conduction tissue system
1.Formation	Forms the outermost covering protoderm	Forms the ground meristem	Forms the procambial bundles
2. Components	epidermal cells, stomata and epidemic outgrowth	Simple permanent tissues – Parenchyma and Collenchyma	Xylem and Phloem
3.Functions	Protection of plant body; absorption of water in roots; gas exchange for photosynthesis and respiration; transpiration in shoots	Gives mechanical support to the organs; prepares and stores food in leaf and stem	Conducts is water and food: gives mechanical strength

Question 10.
Tabulate the Anatomical differences between stem and monocot stem
Answer:

Characters	Dicot stem	Monocot stem
1. Hypodermis	Collenchymatous	Sclerenchymatous
2. Ground tissue	Differentiated into cortex, endodermis, and peri cycle and pith	Not differentiated but it is a continuous mass of parenchyma
3. Starch sheath	Present	Absent
4. Medullary rays	Present	Absent
5. Vascular bundles	a. Collateral and open b. Arranged in a ring c. Secondary growth occurs	a. Collateral and closed b. Scattered in ground tissue c. Secondary growth usually does not occur

Question 11.
Explain the internal structure of Dicot root.
Answer:
The transverse section shows the following structure.
Piliferous layer or Epiblemma or Rhizodermis;

• Single-layer of parenchyma cells compactly arranged with out inter cellular space, devoid of circle and stomata (epidermal pores)
• Single called root hairs arise from the small cell known as trichoblast

Function: Protection & absorption
Cortex:

• Made of loosely arranged parenchyma cells with intercellular spaces.
• Starch grains are stored in, them leucoplasts occur in the cells.

Endodermis:

• Inner most layer of cortex – made up of single layer of barrel-shaped parenchyma cells.
• The radial and inner tangential walls have suberin and lignin thickening known as Casparian thickening.
• The cells opposite to protoxylem do not have Casparian thickening, known as Passage cells, which allow water to pass through but not the cells with Casparian thickening.

Stele:
All the tissue present inside endodermis comprise the stele, include pericycle & vascular bundle,
a) Pericycle:
Outer most layer of stele Single layer of parenchyma. The lateral roots originate from pericycle, so known to have endogenous origin.

Vascular bundle:
Made up of xylem and phloem.
Radial arrangement: In dicot root xylem and phloem are in different radii known as radial arrangement.

Exarch condition:
The protoxylem is pointing towards the periphery.

Tetrarch:
There are four protoxylem points, present this condition is known as tetrarch. Conjunctive tissue: The parenchyma tissue that separates xylem and phloem are known tissue.

Metaxylem – Vessels: are generally polygonal in cross-section.
Pith or Medulla: absent.

Question 12.
Explain the structure of the vascular bundle of Maize stem.
Answer:

• Vascular bundles are skull-shaped, numerous, bigger bundles towards the centre and numerous small bundles arranged in the periphery.
• Vascular bundles are scattered in the parenchymatous ground tissue. This condition is known as Atactostele.
• V – Bs Conjoint Collateral, Closed and Endarch in nature
• Pith or Medulla is absent.
  

Question 13.
Describe the Anatomy of Dicot stem.
Answer:
Epidermis:
A single layer of compactly arranged rectangular parenchymatous cells, with out intercellular space. Cuticle: on the outer walls check transpiration
Stomata: may be present here and their Chloroplasts: usually absent Multicellular hairs: occur in large numbers Function: Protective Cortex:

Lies below epidermis has 3 zones

1. Hypodermis:
Epidemial hay Made upof few layers of colknchyma cells liv- Cuticleing with thickenings at the successive tangential Epidenms avers giving mechanical inheiweeit

2. Chloresrchma:
A few layers below hypodermis with resin ducts in between

3. Parenchyma:
3rd zone, store food material.

Lndodermis or Starch Sheath:
inner most layer of cortex. barrel-shaped cells compactly managed without intercellular spaces.
Since starch grains are abundant in it. It is also a Medullary ray known as starch sheath, homologous to endodermis of root.

Stete: fonn a central ring inner to endodermis made up of Pericycle, VascuLar bundle & Pith.

Perlccle: A few layer of sclerenchyma outside the phloem, known as Hundk cap or Hard bast and also parenchynia cells between them constitute pericycic.

Vascular bundles:
Muscular bundles wedge-shaped arranged the form of a ring – (Eustelic)
Vascular bridle is made upon xylem. Phloem and cambium.
V – B is Conjoint, Collateral. Open and Endarch

Phloem: lies towards periphery.

Function : Conduction of organic food material

Cambium: brick-shaped thin-walled meristem responsible for secondary growth so. V – B is known as open V – B.

Function: Conduction of water and mineraLs from root to other parts.
Pith (Medulla): Central pith is present. It is parenchymatous.
Medullar ray: Pith extends between V – Bs as primary medullary ray.
FunctIon : Storage.

Question 14.
Explain the internal structure of monocot leaf. Epidermis:
Answer:

• A single layer of thin-walled cells with outer walls covered by thick cuticle.
• Stomata occur on both epidermis – stomata surrounded by dumbbell-shaped guard cells.
• Subsidiary cells: Surround guard cells.
• Bulliform cells : Occur on upper epidermis help for the rolling and un rolling of the leaf according to the weather change.
• Silica ceils: Some epidermal cells are filled with silica

Mesophyll:

• Grass, being isobilateral, mesophyll is not differentiated into palisade and spongy tissue, but compactly arranged cells with limited intercellular space.

Vascular Bundles:

• V.Bs differ in size – most of them are smaller, Large bundles occur at regular intervals
• Above and below large bundles sclerenchymatous patches occur – provide mechanical support, they are
• absent in small bundles.
• Bundle sheath – Each V.B is surrounded by a parenchymatous bundle sheath, generally contain starch grains.
• V.B has xylem upward and phloem towards the lower epidermis.
• V.Bs are Conjoint Collateral and Closed.
• In C₄ grasses the bundle sheath cells are called Kranz sheath, involve in C₄ cycle.
  

Question 15.
Draw the internal structure of Nerium leaf & Add a note on it’s special adaptive special features.
Answer:

• Multiseriate upper and lower Epidermis.
• Thick cuticle on the surface of upper epidermis
• Mesophyll is distinguished in to upper palisade and lower spongy parenchyma.
• Well developed vascular bundles with upper xylem and lower phloem (conjoint collateral closed V.B).
• Sunken stomata on the lower epidermis with trichomes, to reduce the rate of transpiration.
  

Question 16.
Difference between Stomata and Hydathodes.
Answer:

Stomata	Hydathodes
1. Occur in the epidermis of leaves, young stems	Occur at the tip or margin of leaves that are grown in moist shady place.
2. Stomatal aperture is guarded by two guard cells	The aperture of hydathodes are surrounded by a ring of cuticularized cells
3. The two guard cells are generally surrounded by subsidiary cell.	Subsidiary cells are absent
4. Opening and closing of the stomatal aperture is regulated by guard cells.	Hydathodes pores remain always open.
4. These are involved in transpiration and exchange of gases.	These are involved in guttation

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 6 Cell The Unit of Life Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life

11th Bio Botany Guide Cell The Unit of Life Text Book Back Questions and Answers

Choose The Right Answers:

Question 1.
The two subunits of ribosomes remain united at a critical level of
a) Magnesium
b) Calcium
c) Sodium
d) Ferrous
Answer:
a) Magnesium

Question 2.
Sequence of which of the following is used to know the phylogeny
a) mRNA
b) rRNA
c) tRNA
d) HnRNA
Answer:
d) HnRNA

Question 3.
Many cells function properly and divide mitotically even they do not have
a) Plasma membrane
b) Cyto skeleton
c) Mitochondria
d) Plastids
Answer:
d) Plastids

Question 4.
Keeping in view the Fluid mosaic model for the structure of cell membrane which one of the following statements is correct with respect to the movement of lipids & Proteins from one lipid mono layer to the other
a) Neither lipid nor protein can flip flop
b) Both lipid and protein can flip flop
c) While lipid can rarely flip flop proteins cannot
d) While proteins can flip flop but lipids cannot ,
Answer:
c) While lipids can rarely flip-flop proteins cannot

Question 5.
Match the columns and identify the correct option:
Answer:

Column I	Column II
a. Thylakoids	Disc shaped sacs in Golgi apparatus
b. Cristae	Condensed structure of DNA
c. Cistemae	Flat membrane sacs in stroma
d. Chromatin	In folding in Mitochondria

(a) (b) (c) (d)
(1) (iii) (iv) (ii) (i)
(2) (iv) (iii) (i) (ii)
(3) (iii) (iv) (j) (ii)
(4) (iii) (i) (iv) (ii)
Answer:
(3) (iii) (iv) (i) (ii)

Question 6.
Bring out the significance of Phase Contrast Microscope
Answer:
Phase-contrast microscope is used to observe living cells, tissues and the cells cultured invitro during mitosis.

Question 7.
State the Protoplasm theory
Answer:

• Fischer in 1894 & Hardy ( 1899 ) Proposed the Colloidal theory of Protoplasm (the physical basis of life)
• It is a colloidal system with water, many biological import things, glucose, fatty acids, amino acids minerals, vitamins hormones & enzymes are seen.
• Homogenous -These solutes are soluble
• Heterogenous – Solutes are not soluble – This Forms the basis for its colloidal nature.
• Protoplasm occur in 2 states but interconvertible

Question 8.
Distinguish between Prokaryotes & Eukaryotes.
Answer:

	Prokaryotes	Eukaryotes
Size of cell	1-5 cm	10 -100 cm True Nucleus
Nuclear character	Nucleoid or incipient nucleus only (No nuclear membrane or Nucleolus	Nucleolus & Nuclear membrane present
DNA	Usually Circular without histone protein	Usually linear with histone proteins
RNA/ Protein synthesis	Couples in Cytoplasm	RNA Synthesis inside Nucleus / Protein synthesis in the cytoplasm)
Ribosomes	50 s +30 s (70s)	60s + 40s ( 80s)
Organelles	Absent	Numerous
Cell Movement	Flagella	Flagella & Celia
Organisation	Usually unicellular	Single, Colonial and multicellular
Cell division	Binary Fission	Mitosis & Meiosis
Example	Bacteria & Archae Bacteria	Fungi, Plants, and Animals

Question 9.
Difference between plant and animal cell:
Answer:

Plant Cell	Animal Cell
1.Usually they are large than animal cell	Usually smaller than plant cell
2. Cell wall present in addition to plasma membrane and consists of middle lamellae. Primary and secondary walls	Cell wall absent
3. Plasmaodesmata present	Plasmodesmata absent
4. Chloroplast present	Chloroplast absent
5. Centrioles absent except motile cells of lower plants	Vacuole small and temporary
6. Vacuole larger and pennanent	Tonoplast absent
7. Tonoplast present around vacuole	Centrioles present
8. Nucleus present along the periphery of the cell	Nucleus at the centre of the cell
9. Lysosomes are rare	Lysosomes present

Question 10.
Draw the ultrastructure of a plant cell:
Answer:

Part-A

Choose The Right Answer:

Question 1.
Scientist who named the unicellular particles as ‘animalcules’ …………… .
(a) Aristotle
(b) Robert Brown
(c) Antonie von Leeuwenhoek
(d) Robert Hooke
Answer:
(c) Antonie van Leeuwenhoek

Question 2.
Compound microscope was invented by
a) Robert brown
b) Z. Sigmody
c) Z. Jansen
d) Zenike
Answer:
C) Z. Jansen

Question 3.
Micrometry is a technique of measurement of
a) Microtomy
b) Nanoparticles
c) Microscopic Objects
d) Moving objects
Answer:
c) Microscopic Objects

Question 4.
Which of the following electron opaque chemical is used in Electron microscope?
(a) Strontium
(b) Deuterium
(c) Palladium
(d) Uranium
Answer:
(c) Palladium

Question 5.
Who first observed Protoplasm
a) Corti
b) Felix Dujardin
c) Hugo Van Mohl
d) O. Hertwig
Answer:
a) Corti

Question 6.
Dinoflagellates and Protozoa are kept under
a) MegaKaryotes
b) Prokaryotes
c) Eukaryotes
d) Mesokaryola
Answer:
d) Mesokaryota

Question 7.
Which among the following is NOT an exception to cell theory?
(a) Viruses
(b) Viroids
(c) Prions
(d) Fungi
Answer:
(d) Fungi

Question 8.
Michondria was named by
a) A.kolliker
b) Altmann
c) Benda
d) Purkinje
Answer:
c) Benda

Question 9.
When Thylakoids are stacked together like a pile of coins known as
a) Grana
b) Cistemae
c) Quantosomes
d) Polysomes
Answer:
a) Grana

Question 10.
Dense particulars or granules observed by George Palade is known as
a) Cirtemae
b) Lamella
c) Locules
d) Ribosomes
Answer:
d) Ribosomes

Question 11.
Histone proteins are seen in the DNA of …………… .
(a) Pseudokaryotes
(b) Prokaryotes
(c) Mesokaryotes
(d) Eukaryotes
Answer:
(d) Eukaryotes

Question 12.
These are also known as Microbodies
a) Mitochondrial & Ribosomes
b) Ribosomes & Cistemao
c) Polysomes & Vacuoles
d) Peroxisomes & Glyoxysomes
Answer:
d) Peroxisomes & Glyoxysomes

Question 13.
The organelle made up of nine triplet peripheral fibrils are known as
a) Microbodies
b) Tululin
c) Centrosome
d) Centroles
Answer:
d) Centroles

Question 14.
Fungal cell wall is made of …………… .
(a) Cutin
(b) Chitin
(c) Hemicellulose
(d) Pectin
Answer:
(b) Chitin

Question 15.
‘Annule’ are circular structure seen around
a) Nuclear membrane
b) Nuclear Pore
c) Perinuclear Space
d) Annuli
Answer:
d) Annuli

Question 16.
The Chromosome that occur in the oocyte of Salamander and in Giant nucleus of Acetabularia is known as
a) Polytene Chromosome
b) Lamp brush chromosome
c) Mitochondrial chromosome
d) Chloroplast chromosome
Answer:
b) Lamp brush chromosome

Question 17.
Ordinary microscope can be made into Dark Field Microscope (DFM) by means of a special component is called
a) Patch stop carrier
b) Secondary Magnification lens
c) Stage
d) Phase plate
Answer:
a) Patch stop Carrier

Question 18.
In-plant cells, golgi bodies are found as small vesicles called …………… .
(a) Polysomes
(b) Cytosomes
(c) Cytosol
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 19.
Cisternae, tubule and Vesicles occur in which of the following:
a) Golgi apparatus
b) Lysosomes
c) Endoplasmic reticulum
d) Glyoxysomes
(i) a & b
(ii) b & c
(iii) c & d
(iv) a & c
Answer:
(iv) a & c

Question 20.
The Golgi apparatus in plant is known as
a) Dictyosomes
b) Glyoxysomes
c) Neo-particles
d) Microvesides
Answer:
a) Dictyosomes

Question 21.
Which of the three, come under the system of the membrane in Eukaryotic cell
a) Mitochondria
b) Nuclear Membrane
c) Golgi apparatus
d) Endoplasmic reticulum
(i) a, b & c
(ii) b, c & d
(iii) a, c & d
(iv) a, b & d
Answer:
(ii) b, c & d

Question 22.
DNA of mitochondrion is …………… .
(a) Helical
(b) Dumbbell
(c) Circular
(d) Spiral
Answer:
(c) Circular

Question 23.
Fluid droplets are engulfed by membrane, which form vesicles around them
a) Phagocytosis
b) Exocytosis
c) Endocytosis
d) Pinocytosis
Answer:
d) Pinocytosis

Question 24.
The 60 s large subunit of Eukaryotes contain
a) 23 s & 5 s – large subunit
b) 16 s r RNA in large subunit
c) 18 s r RNA in large subunit
d) 28 s, 5-8 sand 5 s in large subunit
Answer:
d) 28 s, 5-8 sand 5 s in large subunit

Question 25.
Elaioplasts store …………….
(a) Starch
(b) Lipid
(c) Protein
(d) Chlorophyll
Answer:
(b) Lipid

II. State whether the following statement True or False with reference to the origin of Eukaryotes.

1. A Prokaryote grow in size and develop infoldings in its cell membrane to increase surface area to volume ratio
2. Aerobic protea bacterium enter eukaryote as prey or parasite and become an endosymbiont
3. Proteobacteria eventually assimilated and became mitochondria
4.  Some Prokaryotes go on to acquire additional Exo symbionts the cyanobacteria evolve to become chloroplasts.

Answer:
b) True, True, True, False

Question 2.
Find out the true and false statements from the following and on that basis And the correct answer.
(i) In Prokaryotes the flagellar rotation, only proton movements are involved & not ATP.
(ii) In Eukaryotes to shift the adjacent microtubules to bend cilia or flagella, dynein use energy from ATP
(iii) Bacterial flagella are made up of helical polymers of protein known as Tubulin
(iv) In Eukaryotes the flagella are made up of microtubules and proteins known as dynein and nexin.

Answer:
b) True True False True

Question 3.
With reference to Eukaryotic flagellum Find out the true or false statements from the following and on that basis find the correct answer
(i) Flagellum is shorter than cilia as short as 200 µm
(ii) Flagella are microtubule projection of plasma membrane
(iii) Flagella composed of 8 pairs of microtubules with 2 pairs of microtubules in the center
(iv) Structure of Flagella has Axoneme made up of microtubules & protein tubules

Answer:
c) False True False True

Question 4.
(i) Cytoplasm is the physical basis of life
(ii) Cytoplasm inheritance occurs only through the plasma genes
(iii) Cytoplasm serves as a molecular soup where all the cellular organelles are suspended and bound together by a lipid bilayer plasma membrane
(iv) Cytoplasm is a very bad conductor of electricity.

Answer:
a) False True True False

Question 5.
Find out the true or false statements from the following and on that basis find the correct answer
(i) The contractibility of protoplasm is important for the absorption and removal of water, especially in stomatal operations
(ii) The viscosity of protoplasm is 2-20centipoises
(iii) The protoplasm is made of 10-20% of water
(iv) Brownian movement and Tyndall effect are colloidal properties, so not applicable to protoplasm

Answer:
d) True True False False

II. Choose The Wrong Match

Question 1.
(a) Cytoplith – Hypodermal leaf cells of ficus bengaliensis
(b) Raphides – Eichhomia leaves
(c) Sphaero raphides – Colocasia
(d) Silica – Oryza sativa
Answer:
d) Silica – Oryza sativa

Question 2.
Choose the wrong match with reference to mitochondria
(a) Protein – 73%
(b) Lipids – 25-30%
(c) DNA – 12%
(d) RNA – 5-7%
Answer:
c) DNA-12%

Question 3.
(i) Centrosome give rise to spindle fibers in Animal cell
(ii) Golgibodies play important role in packaging and secretion
(iii) Endoplasmic reticulum-SER is involved in protein synthesis
(iv) Vacuoles facilitate the transport of ions and materials in plant cell
Answer:
(iii) Endoplasmic reticulum SER is involved in protein synthesis

Question 4.
(i) The magnification of SEM & its resolving power is – 200000 &5-20nm
(ii) The magnification &resolution power of temis – 1 – 300000&2-10A
(iii) The magnification power of TEM is – 100000 then the light microscope
(iv) The magnification power of phase-contrast – 3 – 40000 & 8-10A microscope &its resolution power
Answer:
(iv) The magnification power contrast is microscope & its resolution power – 3-400000 & 8-10A

IV. Choose The Right Match From The Following

Question 1.
(i) Size of mycoplasma – 0.15-0.03 µm
(ii) Size of BGA – 60mm
(iii) Size of RBC – 0.25-0.06 µm
(iv) Size of chick egg – 7-811mm
Answer:
(i) Size of mycoplasma – 0.15-0.03µm

Question 2.
Choose the right match:
(i) Volutin granules occurin -Bacteria
(ii) Ttannin – Cassia auriculata
(iii) Calcium carbonate – Mimosa pudica
(iv) Heavy metals – Erchhornia
Answer:
(i) Volutin granules- Bacteria

Question 3.
Choose the right match:
(i) Cell theory – Cortix
(ii) Protoplasm theory – Max Schultze
(iii) Chromosomes physical carriers of genes – Strasburger
(iv) Endoplasmic reticulum word coined by – Benda
Answer:
(ii) Protoplasm theory – Max Schultze

V. Match The Following And Find The Correct Answer:

Question 1.
(i) Harry Beevers – (A) identified Lysosomes a Peroxisomes
(ii) Christian Do Duve – (B) Discovered Glycoxysome c
(iii) A-F-U- Schimper – (C) Coined the word Chromosome
(iv) Waldeyer – (D) Coined the word Plastids

Answer:
a) B A D C

Question 2.
(i) When the small pieces of golgibody pinches off from its tubules to form – A. Chioroplast
(ii) Fernandez moran particles occur in – B. Golgi apparatus
(iii) Zymogen granules occur in – C. Lysosome
(iv) Quantosomes are present in – D. Mitochondria

Answer:
b) C D B A

Question 1.
ASSERTION (A): A cell membrane shows fluid behavior
REASON (R): A membrane is- a mosaic or composite of diverse lipids and proteins
a) Assertion and Reason are correct ‘R’ explaining ‘A’
b) A and R-correct ‘R’ not explaining A
c) A is true, but R is wrong
d) A is true but R is not explaining A
Answer:
(a) Assertion A & Reason R are correct R is explaining A

Question 2.
Assertion (A): Chloroplast is an important cell organelle performing photosynthesis in plants
Reason (R): An organelle is a distinct part of a cell which has a particular structure and function.
a) A and R are correct R explaining A
b) A and R correct and R not explaining A
c) A is true, but R is wrong
d) A is true but R is not explaining A
Answer:
b) A and R correct R not explaining A

Question 3.
Assertion (A): The inheritance of Mitochondria is uniparental
Reason (R): Mitochondria of any one of the parenting divide and gets distributed to daughter cells.
a) A and R are correct R explaining A
b) A and R are correct R not explaining A
c) A is true but R is wrong
d) A is true but R is not explaining A
Answer:
(c) A is true but R is wrong

Question 4.
Assertion (A): The objective of optic lenses of a microscope are interchanged, then it can work as a telescope
Reason (R): The objective of the telescope has a small focal length
(a) A and R are correct R explaining A
(b) A and R are correct R not explaining A
(c) A is true but R is wrong
(d) A is true but R is not explaining A
Answer:
(d) A is true but R is not explaining A

Question 5.
Assertion (A): A polytene achieved by repeated replication of chromosomal DNA without nuclear division. The daughter chromatids aligned side by side called Endomitosis
Reason (R): Polytenes is observed in the salivary glands of Drosophila by C.G.Balbiani. 1881.
a) A and R are correct R explaining A
b) A and R are correct R not explaining A
c) A is true but R is wrong
d) A is true but R is not explaining A
Answer:
(b) A and R correct R not explaining A

Two Marks Questions

Question 1.
Name the scientist who proposed the cell theory.
Answer:
Matthias Schleiden and Theodor Schwann.

Question 2.
ER- can be referred to as the endoskeleton of the cell. Justify.
Answer:

• Yes. It connects plasma membrane & nuclear membrane, giving support to the Cytosol so we can call it the endoskeleton of the cell.
• It also helps in the exchange of substances in and out of the cell.

Question 3.
Why do we say that viruses are an exception to its cell theory?
Answer:
Viruses lack protoplasm, the essential parts of the cell, and are existing as obligate parasites (i.e)(subcellular particles).

Question 4.
Who said that different kinds of plastids can transform into one another?
Answer:

A-F-U Schimper said that the 3 different kinds of plastids can transform into one another according to the need or demand of the plant body.

Question 5.
In a Bright field microscope, where does the primary & secondary magnification occurs?
Answer:
Primary magnification is obtained through, objective lens, and secondary magnification is obtained through an eye piece lens.

Question 6.
State the functions of chloroplast
Answer:

• They are organs of Photosynthesis.
• Light reactions & dark reactions take place in the granum & stroma respectively.
• Chloroplast also play important role in the Photorespiration or C₂ cycle.

Question 7.
Distinguish between 70’s & 80’s Ribosomes.
Answer:

Question 8.
Name the types of cells based on nuclear characteristics.
Answer:
The types of cells based on nuclear characteristics:

1. Prokaryotes
2. Mesokaryotes and
3. Eukaryotes.

Question 9.
Distinguish between glyoxysomes, peroxysomes & sphaerosomes
Answer:

GLYOXYSOMES	PEROXYSOMES	SPHAEROSOMES
Single membrane-bound &sub cellular organelle	Single membrane-bound & subcellular organelle	Single membrane-bound & subcellular organelle
Contain enzymes of the glyoxylate pathway	Contain enzymes and play important role in C2 cycle or Photorespiration	They play important role in the storage of fats in the endoplasm cells of oilseeds
Beta oxidation of fatty acids occurs in the glyoxysomes of germinating seeds
Eg. Castor seeds	Eg. Occur in all green plants	Eg. Coconut Castor seeds

Question 10.
Distinguish between Resolution & Magnification:
Answer:

RESOLUTION	MAGNIFICATION
Ability of lenses to show the finest details between two points form Resolution R	It is the size of the image seen with eye, magnified by the microscope
Formula = \(R=\frac{0.61 \lambda}{(\mathrm{NA})}\) where,λ -wavelength of light NA-numerical aperture	Formula = Size of image seen with microscope Size of image seen with normal eyes

Question 11.
Differentiate 4 points of differences between Prokaryotes & Mesokaryotes
Answer:

PROKARYOTES	MESOKARYOTES
Nucleoid no true nucleus	Nucleus with nuclear membrane
1-5µm	5- 10µm
DNA usually circular without	DNA linear but without
histone proteins	histone proteins
Ribosomes 50S+30S	60S+40S
Organelles absent	Organelles present
Eg. bacteria & archaea	Eg. Dinoflagellate, Protozoa

Question 12.
Write down any 4 functions of cell wall
Answer:

NAME OF THE CELL WALL	FUNCTIONS OF THE CELL WALL
SHAPE	It offers definite Shape and Rigidity
BARRIER	It prevents the entry of several molecules into the cell
PROTECTION	Protects internal protoplasm against mechanical injury
Prevents cell from bursting	lit maintains osmotic pressure and prevent cell from bursting
DEFENSIVE DEVICE	It plays a major role by acting as a defensive device

Question 13.
Differentiate between TEM and SEM:
Answer:

TEM	SEM
It has a high resolving power	Resolving power Comparatively lower
Most commonly used	Occasionally used depending on the study
2-dimensional image is provided	3D image is provided
Magnification 1-3 lakhs times	Magnification 2 lakhs times
Resolving power 2-10A°	Resolving power 5-20 nm

Question 14.
Explain signal transduction:
Answer:
DEFINITION:

• It is a process by which a cell receives information from outside and respond to it is called signal transduction
• Nitric oxide → is the main signally molecule
• Cell membrane → site of chemical interaction of signal transduction

Question 15.
Draw the structure of the Golgi apparatus & label its parts.
Answer:

Question 16.
What is the cell wall composition of the following organism?
(a) Fungi
(b) Bacteria
(c) Algae
Answer:
(a) Fungi – Chitin and fungal cellulose.
(b) Bacteria – Peptidoglycan
(c) Algae – Cellulose, mannan and galactan.

Question 17.
What is meant by Holocentric chromosomes?
Answer:
If a chromosome has centromere activity distributed along the whole surface of the chromosome during mitosis (i.e) microtubules distributed all along the mitotic chromosome.
Eg. Caenorhabditis Elegans (transparent nematode) & many insects.

Question 18.
Differentiate between point centromere & Regional centromere.
Answer:

POINT CENTROMERE	REGION AL CENTROMERE
The kinetochore is assembled as a result of protein recognition of specific DNA sequences Kinetochores assembled on point centromere bind a single microtubule localized, Centromere Eg. Budding Yeasts	The kinetochore is assembled on a variable array of repeated DNA sequences Kinetochore assembled on regional centromeres, bind multiple microtubules Eg. Fission yeast cells, Human cells

Question 19.
Draw the structure of the polytene chromosome:
Answer:

Question 20.
Draw the structure of the lysosome.
Answer:

Three Mark Questions

Question 1.
Distinguish between autosomes & allosomes.
Answer:

AUTOSOMES	ALLOSOMES
In human diploid cells out of 46, only 44 chromosomes are Autosomes	Only 2 chromosomes are Allosomes or Sex chromosomes
They are controlling somatic characteristics of an organism	They are involved in Sex determination

Question 2.
Explain lampbrush chromosomes:
Answer:

• In 1882- observed by Flemming in Oocytes of animal Salamander &Giant nucleus of unicellular Algae Acetabularia
• The highly condensed chromosomes form a chromosomal axis, from which lateral loops of DNA formed as a result of intense RNA synthesis

Question 3.
Define cytoplasmic streaming.
Answer:
Cytoplasmic streaming refers to the movement of the cytoplasm along with the cellular materials inside the cell.

Question 4.
Draw the structure of the Eukaryotic flagellum.
Answer:

Question 5.
List out the functions of the Cell Wall.
Answer:
The cell wall plays a vital role in holding several important functions given below.

1. Offers definite shape and rigidity to the cell.
2. Serves as barrier for several molecules to enter the cells.
3. Provides protection to the internal protoplasm against mechanical injury.
4. Prevents the bursting of cells by maintaining the osmotic pressure.
5. Plays a major role by acting as a mechanism of defense for the cells.

Question 6.
Explain in detail about Fluid mosaic model.
Answer:
Jonathan Singer and Garth Nicolson (1972) proposed fluid model: It is made up of lipids and proteins together with a little amount of carbohydrate. The lipid membrane is made up of phospholipid. The phospholipid molecule has a hydrophobic tail and hydrophilic head. The hydrophobic tail repels water and water-loving polar molecule are called hydrophilic molecule. They have polar phosphate group responsible for attracting water. Water-hating non – polar molecule are called as a hydrophobic molecules. They have fatty acid which is non – polar which cannot attract water.

Hydrophilic head attracts water. The proteins of the membrane are globular proteins which are found intermingled between the lipid bilayer most of which are projecting beyond the lipid bilayer. These proteins are called as integral proteins. Few are superficially attached on either surface of the lipid bilayer which are called as peripheral proteins. The proteins are involved in the transport of molecules across the membranes and also act as enzymes, receptors or antigens.

Question 7.
Draw the structure of the chromosome & neatly label the parts:
Answer:

Question 8.
Based on the position of centromere classify the chromosomes with the help of diagrams.
Answer:

Eukaryotic chromosomes may be rod-shaped telo & acrocentric as well as meta & sub-meta-centric.

Question 9.
List out the functions of Golgi bodies.
Answer:
Functions of Golgi bodies:

1. Glycoproteins and glycolipids are produced.
2. Transporting and storing lipids.
3. Formation of lysosomes.
4. Production of digestive enzymes.
5. Cell plate and cell wall formation
6. Secretion of carbohydrates for the formation of plant cell walls and insect cuticles.
7. Zymogen granules (proenzyme / pre-cursor of all enzymes) are synthesized.

Question 10.
Explain the structure of Cilia.
Answer:

• Short cellular-numerous microtubule bound projections of plasma membrane.
• Each Cilium has membrane-bound structures, basal body,rootlets, basal plate shaft
• Shaft (axoneme) consists of nine pairs of microtubule doublets, arranged in a
• circle along the periphery with a two central tubules (9+2) arrangement of microtubules is present.
• Microtubules – made up of tubulin.
• Motor protein dynein – connects the outer microtubules pair & links them to the central pair.
• Nexin – links the peripheral doublets of microtubules.

Question 11.
Write in detail about the 3 types of centromere in eukaryotes.
Answer:
There are three types of centromere in Eukaryotes. They are as follows:

1. Point Centromere: The type of centromere in which the kinetochore is assembled as a result of protein recognition of specific DNA sequences. Kinetochores assembled on point centromere bind a single microtubule. It is also called a localized centromere. It occurs in budding yeasts.
2. Regional Centromere: In regional centromere where the kinetochore is assembled on a variable array of repeated DNA sequences. Kinetochore assembled on regional centromeres bind multiple microtubules. It occurs in fission yeast cell, humans and so on.
3. Holocentromere: The microtubules bind all along the mitotic chromosome. Example: Caenorhabditis Elegans (transparent nematode) and many insects.

Question 12.
Distinguish between primary wall & secondary wall of the plant cell wall.
Answer:

PRIMARY WALL	SECONDARY WALL
First formed	Formed later
Thin elastic, extensible	Thick inelastic
Matrix made up of Hemi cellulose-bind micro, fibrils with matrix Pectinase- filling material, Glycoprotein-control orientation of microfibrils Water	Here cellulose & pectin compactly arranged with different orients giving a laminated structure to give strength to the cell wall.
Only one layer	Has three sub-layers s1,s2,s3.
Does not determine shape of cell	Determine shape of cell

Question 13.
Describe the steps involved in cytological techniques.
Answer:
There are different types of mounting based on the portion of a specimen to be observed.

1. Whole-mount: The whole organism or smaller structure is mounted over a slide and observed.
2. Squash: This is a preparation where the material to be observed is crushed/squashed onto a slide so as to reveal its contents. Example: Pollen grains, mitosis, and meiosis in root tips and flower buds to observe chromosomes.
3. Smears: Here the specimen is in the fluid (blood and microbial cultures etc) are scraped, brushed, or aspirated from the surface of the organ. Example: Epithelial cells.
4. Sections: Freehand sections from a specimen and thin sections are selected, stained, and mounted on a slide. Example: Leaf and stem of plants.

Question 14.
List out any 3 stains used in histo- chemistry.
Answer:

S.No	Stain	Colour of staining	Affinity
1.	Eosin	Pink or red	Cytoplasm, Cellulose
2.	Methylene blue	Blue	Nucleus
3.	Saffranine	Red	Cell wall(lignin)
4.	Janus green	Greenish blue	Mitochondria

Question 15.
Identify the diagram and label the parts.

Answer:
This is a dark field microscope
A-objective lens
B-stage
C-condenser lens
D- patch stop
E-light source

Five Mark Questions

Question 1.
Differentiate between BFM & DFM.
Answer:

Question 2.
Differentiate between Light microscope & electron microscope.
Answer:

Light Microscope	Electron Microscope
Another name = compound microscope	1st introduced by Ernest Ruska & developed by   G.Binnin & H. Roher (1981)
Principle	Principle
The transmission of visible light from the source of eye through a sample	It uses a beam of accelerated electrons as source of illumination.
Resolving power – Lesser	Resolving power – Higher
Magnification – Less	Magnification-1,00,000 times than the light
Purpose – studying in schools & college	Purpose Microscope Research purpose -can be seen in scientific laboratories

Pattern of working: The microscope transmits visible light from eye through sample where interaction occur and magnified image is visible.	The specimen to be viewed under EM should be dehydrated and impregnated with election opaque chemicals like gold, palladium for withstanding electrons & also for contrast.
Types :1 Only one	Types: 2 types TEM, SEM

Question 3.
Write down the functions of the cell wall.
Answer:

NAME OF THE CELL WALL	FUNCTIONS OF THE CELL WALL
SHAPE	It gives definite Shape and Rigidity to the ceil
BARRIER	It prevents several molecules from entering the cells
PROTECTION	To the internal protoplasm against mechanical injury
MAINTAIN ANCE	It maintains osmotic pressure So, prevent bursting of cells
DEFENCE	They are acting as a source of defense for cells

Question 4.
Write down the functions of the Plasma Membrane or cell membrane.
Answer:

• Cell transport is the main function
• PM act as a channel of transport for molecules
• PM is selectively permeable to molecules

It transported by

1. Energy-dependent processes,
2. Energy independent processes Membrane proteins involved processes
3. Endocytosis & Exocytosis large quantity of solids and liquids are transported into a cell or out of cells.

I. Endocytosis 2 types
a) Phagocytosis particle is engulfed by membrance which fold around it forming vesicles, enzymes digest and products are absorbed.

b) Pinocytosis Fluid droplets are engulfed by forming vesicles.
II. Exocytosis -Vesicles fuse with the plasma membrane and eject contents.
-This may be a secretion in the case of digestive enzymes hormones or mucus.

Question 5.
Explain the fluid mosaic model of plasma membrane.
Answer:

• Jonathan Singer & Garth Nicolson (1972) proposed FM model.
• Plasma membrane made up of lipid (phospholipid), protein & little carbohydrate.

I. Phospholipid: Molecule has a hydrophobic tail(repel water) & hydrophilic head (water-loving)
II. Protein of membrane

• Globular in nature intermingles between lipid bipolar most perfect beyond Jt known
  as (integral proteins)

Few are superficially attached on either surface of lipid bilayer (peripheral proteins)

• They are involved in transport of molecules across the membrane
• They acts as enzymes
• They acts as receptors or antigens.

III Carbohydrate

• They are short chain of polysaccharides.
  (i.e) With protein glycoprotein With lipid glycolipids, glycocalyx

Flip Flapping:

• The movement of membrane lipids from one side of the membrane to the other side by vertical movement called flip-flap movement.

Answer:
A- hydrolipid tail,
B-hydrophilic head,} lipid
C-intrisic protein
D-extrinsic protein
This movement is very slow than lateral diffusion of lipid molecules.

• Phospholipids can flip flop due to smaller polar regions.
• Proteins cannot do so because of extensive polar regions.

Question 6.
Give an account of the structure and function of mitochondria.
Answer:

• 1st observed by A. Kolliker (1880)
• Altmann(1894) – named it as Bio-plasts
• Benda (1897) – named as Mitochondria

Structure

• Ovoid, rod-shaped, pleomorphic structures
• Double membrane
• Outer membrane smooth, & permeable- contain porins

2 compartments
1. outer chamber between 2 membranes
2. Inner chamber filled with matrix

Cristae – Infoldings of inner membrane:

• It contain enzymes for ETS(Electron Transport System)
• Inner membrane has FI particles or exosomes
• Each FI particle has a base, a stem & a rounded head
• Head has ATP synthetase to do oxidative phosphorylation content.
  • 73% protein
  • 25-30% lipids
  • 5-7% RNA, DNA & enzymes(about 60 circular DNA &70’s Ribosomes.
• All enzymes of Kreb’s cycle are found in the matrix except succinate dehydrogenase.
• Mitochondria is a semi-autonomous body
• It’s inheritance is uniparental (i.e) maternal
• It is used to track recent evolutionary time because it mutates 5-10 times faster than DNA in the nucleus.

Question 7.
Structure of chloroplast
Answer:

• A vital organ of green plants.
• Double membrane-bound organelle peripheral space in between the membrane
• Inner chloroplast is filled with gelatinous stroma
• Inside the stroma interconnected sacs called Thylakoids
• Inner space of the thylakoid is the thylakoid lumen
• Thylakoids stacked together like piles of coins known as grana.
• Light is absorbed and converted into chemical energy (carbohydrates) in the granum
  
• Chloroplast genome encodes for approximately 30 proteins involved in photosystem I & II – cytochrome, b, f, complex and ATP synthase & also one of the subunits of RUBISCO is enclosed by it.
• RUBISCO- is the major protein component of the stroma single most abundant protein on earth
• The thylakoid contain small, rounded photosynthetic units called Quantosomes
• The chloroplast is semi-autonomous, divided by fission.
  

Question 8.
Give an account of Ribosomes:
Answer:

• 1953 – 1 observed by George Palade
• Dense particles in the EM not membrane-bound

Electron microscope observation
1. Made up of 2 round subunits one large layer & one small unit to form a complete unit
2. Mg++ is required for complete cohesion.
Biogenesis – denova formation, auto replication and nucleolar origin
Function – Sites of protein synthesis.

Content – consists of

• RNA 60%,
• Protein 40%

Polysemes:
In protein synthesizing cells, many ribosomes attached to single m RNA – to form polysomes’ main role in the formation of several copies of particular.

Question 9.
Differentiate between chromoplast & leucoplast
Answer:

Chromoplast		Leucoplast
Nature	Coloured	Colourless
Types & occurence	Chloroplast: occur in green algae& higher plants. Pigment chlorophyll a & b Phaeoplast: occur in brown algae & dinoflagelletes Pigment-fucoanthin Rhodoplast: Occur in red algae Stores protein Pigment phycoerythrin	Amyloplast Stores starch occur in storage parts Eg. Tapioca rootElaioplast Stores- lipids Eg. Groundnut seeds Aleuroplast or proteoplast Eg. Moon dhal

Question 10.
State any 3 functions of Lysosomes
Answer:
polypeptide Intracellular digestion:
They digest carbohydrates, proteins & lipids present in the cytoplasm

Autophagy:
During the adverse condition, they digest their own organelles like mitochondria ER

Auto lysis:
Causes self-destruction of cell on the insight of disease

Aging:
Have autolytic enzymes that disrupts intracellular molecules.

Question 11.
Explain the structure of Centrioles
Answer:

• Central hub, surrounded by nine triplet peripheral fibrils (tubulin) connected to the tubules by radial spokes (9 + 0) pattern Cilia or Flagella Spindle fibres
• Centriole is the basal body of Flagella, Lilia or, Spindle fibers.
• It is a nonmembranous organelle

Question 12.
Differentiate between other inclusions of cells in Prokaryotes & Eukaryotes.
Answer:

	Prokaryotes	Eukaryotes
Reserse material	Phosphate granules & Cyanophycean granules	Starch grains Glycogen granules
Organic materials	Poly (3 hydroxyl granules sulphur granules, carboxysomes &Gas vacuoles	Aleurone grains, flat droplets
Other secretions	…………………………….	Essential oil, resins, gums, latex and tannin
Inorganic inclusions	metachromatic granules- such as polyphosphate granules (volutin granules) & sulfur granules	Calcium carbonate crystals, Calcium oxalate crystals, Silica crystals Eg.cystolith- hypodermal cells of Ficus bengalensis (calcium carbonate)
		Raphides- Eichhornia (calcium oxalate)
		Prismatic crystals – dry scales ofAlliumcepa (calcium oxalate)

Question 13.
Explain the structure of the Nucleus.

• It is important CPU of the cell, the largest part of it
• Control all activities of cell
• Hold the hereditary information

Nuclear envelope Nuclear space (nucleoplasm)

I Membrane:
Double membrane Nuclear envelope
a) Outer membrane

• Rough by the presence of ribosomes and with irregular intervals continues with ER
• It has nuclear pores that allow m RNA, ribosomal units, proteins & other macromolecules to pass in & out
• Nuclear pore enclosed by circular structure – annuli

b) Inner membrane:
Smooth without ribosomes in between the two membranes perinuclear space is present

II. Nucleoplasm:
A gelatinous matrix has 2 parts

• Nucleoli &
• Chromatin reticulum

a) Nucleoli:

• Small dense spherical structure occur in singly or in multiples.
• It possesses genes for r RNA &, tRNA

b) Chromatin network

• Uncoiled, indistinct , thread like structure(inter phase)
• Has little amount of RNA, DNA bound to histone proteins in Eukaryotes
• At the time of cell division – It get condensed to form Chromosome

Euchromatin With -2 parts
1. Euchromatin
2.Heterochromatin

• The portion that get transcribed into rn RNA – active genes that are not tightly condensed & stains lightly.
• Heterochromatin
• The portion of chromatin that does not get transcribed into m RNA – remain tightly condensed & stains intensively.

Question 14.
Explain the structure of Endoplasmic reticulum
Answer:

• The largest internal membrane (ER)
• Name given by K.R.Porter(1948)
  Consists of Vesicles &Tubules, Cisternae

Cisternae:

• Long broad, flat sac-like structures arranged in stacks to form lamella.
• In between membrane is filled with fluid

Vesicles:
Oval membrane-bound vascular structure

Tubules:
Irregular shaped, branched, smooth-walled structure enclosing a space

Function:

• It is associated with nuclear membrane and cell surface membrane
• When ribosomes present on ER- it is known as (RER) Rough Endoplasmic Reticulum
• When ribosomes absent on ER- it is known as Smooth Endoplasmic Retiöulum(SER).

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 3 Periodic Classification of Elements Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 3 Periodic Classification of Elements

11th Chemistry Guide Periodic Classification of Elements Text Book Back Questions and Answers

Textual Questions:

I. Choose the best Answer:

Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

Question 2.
The electronic configuration of the elements A and B are 1s², 2s², 2p⁶, 3s² and 1s², 2s², 2p⁵ respectively. The formula of the ionic compound that can be formed between these elements is
(a) AB
(b) AB₂
(c) A₂B
(d) none of the above
Answer:
(b) AB₂

Question 3.
The group of elements in which the differentiating electron enters the anti penultimate shell of atoms are called
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it?
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al³⁺ < Mg²⁺ < Na⁺ < F^– (increasing ionic size)
(d) B < C < O < N (increasing first ionisation enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

Question 5.
Which of the following elements will have the highest electronegativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Question 6.
Various successive ionisation enthalpies (in kjmol 1) of an element are given below.

The element is
(a) phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 7.
In the third period the first ionization potential is of the order.
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na < Al < Mg < P < Si
Answer:
(b) Na < Al < Mg < Si < P

Question 8.
Identify the wrong statement.
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Amongst isoelectric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius

Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy
(a) Al < O < C < Ca < F
(b) Al < Ca < O < C < F
(c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53 respectively is
(a) I > Br > Cl > F
(b) F > Cl > Br > I
(c) Cl > F > Br > I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Question 11.
Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen

Question 12.
The element with positive electron gain enthalpy is
(a) Hydrogen
(b) Sodium
(c) Argon
(d) Fluorine
Answer:
(c) Argon

Question 13.
The correct order of decreasing electronegativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y > A > Z
Answer:
(a) Y > Z > X > A

Question 14.
Assertion:
Helium has the highest value of ionisation energy among all the elements known
Reason:
Helium has the highest value of electron affinity among all the elements known
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Question 15.
The electronic configuration of the atom having | maximum difference in first and second ionisation j energies is
(a) 1s², 2s², 2p⁶, 3s¹
(b) 1s², 2s², 2p⁶, 3S²
(c) 1s², 2s², 2p⁶, 3s², 3s², 3p⁶, 4s¹
(d) 1s², 2s², 2p⁶, 3s², 3p¹
Answer:
(a) 1s², 2s², 2p⁶, 3s¹

Question 16.
Which of the following is second most electronegative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Question 17.
IE₁ and IE₂ of Mg are 179 and 348 kcal mol^-1 respectively. The energy required for the reaction Mg → Mg²⁺ + 2e^– is
(a) + 169 kcal mol^-1
(b) -169 kcal mol^-1
(c) +527 kcalmol^-1
(d) -527 kcal mol^-1
Answer:
(c) +527 kcalmol^-1

Question 18.
In a given shell the order of screening effect is
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Question 19.
Which of the following orders of ionic radii is correct?
(a) H > H⁺ > H
(b) Na⁺ > F^– > O^2-
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Question 20.
The First ionisation potential of Na, Mg and Si are 496, 737 and 786 kJ mol^-1 respectively. The ionisation potential of Al will be closer to
(a) 760 kJ mol^-1
(b) 575 kJ mol^-1
(c) 801 kJ mol^-1
(d) 419 kJ mol^-1
Answer:
(b) 575 kJ mol^-1

Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(b) Increases in a period and decreases in a group

Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases

Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

II. Write brief answer to the following questions:

Question 24.
Define modern periodic law.
Answer:
The modem periodic law states that “The physical and chemical properties of the elements are a periodic function of their atomic numbers.”

Question 25.
What are isoelectronic ions? Give examples.
Answer:
Two ions having the same number of electrons are called isoelectronic ions.
Example: Na⁺ (1s² 2s² 2p⁶) and F^– (1s² 2s² 2p⁶). Both these ions contain eight electrons.

Question 26.
What is an effective nuclear charge?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff = Z – S
Where Z = Atomic number
S = Screening constant calculated by using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
Answer:
“Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
The given statement is not correct. Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg⁺, Mg^2+, and Mg³ ions. Which step will have the highest ionization energy and why?
Answer:
Magnesium loses electrons successively in the following steps,
Step 1:
Mg_(g) + IE₁ → Mg⁺_(g) + 1e, Ionisation energy = I.E₁
Step 2:
Mg⁺_(g)+ IE₂ → Mg²⁺_(g) + 1e, Ionisation energy = I.E₂
Step 3:
Mg²⁺_(g) + IE₃ → Mg³⁺_(g) + 1e, Ionisation energy = I.E₃
The total number of electrons is less in the cation than the neutral atom while the nuclear charge remains the same. Therefore, the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus, the successive ionization energies, always increase in the following order I.E₁ < I.E₂ < I.E₃. Thus, Step-3 will have the ionization energy.

Question 29.
Define electronegativity.
Answer:
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionisation potential is always higher than the first ionisation potential?
Answer:

• The second ionization potential is always higher than the first ionization potential.
• Removal of one electron from the valence orbit of a neutral gaseous atom is easy so first ionization energy is less. But from a uni positive ion, removal of one more electron becomes difficult due to the more forces of attraction between the excess of protons and less number of electrons.
• Due to greater nuclear attraction, second ionization energy is higher than first ionization energy.

Question 31.
The energy of an electron in the ground state of the hydrogen atom is -2.8 × 10^-8 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol^-1
Answer:
Energy of an electron in the ground state of the hydrogen atom is – 2.8 × 10^-8.
The ionization energy of atomic hydrogen is 2.8 × 10^-18 × 6.023 × 10²³ J/mol
= 16.86 × 10⁵ J/mol
= 1686 kJ/mol.

Question 32.
The electronic configuration of an atom is one of the important factors which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:

• The electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half-filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s²  2s²  2p_x¹  2p_y¹ 2p_z¹ (half-filled electronic configuration) Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electrons is unfavorable and they have zero electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element Ununoctium (Oganesson, Z – 118) present in the 7^th period and 18^th group of the periodic table.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
The fifth period of the periodic table has 18 elements. 5^th period starts from Rb to Xe (18 elements).
5^th period starts with principal quantum number n = 5 and l = 0, 1, 2, 3 and 4.
When n = 5, the number of orbitals = 9.
1 for 5s
5 for 4d
3 for 5p
The total number of orbitals = 9.
Total number of electrons that can be accommodated in 9 orbitals = 9 x 2 = 18.
Hence the number of elements in the 5th period is 18.

Question 35.
Elements a, b, c and d have the following electronic configurations:
a: 1s², 2s², 2p⁶
b: 1s², 2s², 2p⁶, 3s², 3p¹
c: 1s², 2s², 2p⁶, 3s², 3p⁶
d: 1s², 2s², 2p¹
Which elements among these will belong to the same group of the periodic table?
Answer:
The elements ‘a’ [Ne (Z= 10), 1s², 2s², 2p⁶] and ‘c’ [Ar (Z = 18), 1s², 2s², 2p⁶, 3s², 3p⁶] have the same valence electronic configuration and hence, belongs to the same group, i.e., group 18 of the periodic table . Similarly, the elements ‘b’ [ Al (Z = 13), 1s², 2s², 2p⁶, 3s², 3p¹] and ‘d’ [B (Z= 5), 1s², 2s², 2p¹] have the same valence electronic configuration and hence, belongs to the same group, i.e., group 13 of the periodic table.

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-14 5d^0-16s².
• The electronic configuration of actinides is 5f^1-14 6d^0-1 7s².

Question 37.
Why do halogens act as oxidizing agents?
Answer:
Halogens are having the general electronic configuration of ns², np⁵ and readily accept an electron to get the stable noble gas electronic configuration. Therefore, halogens have high electron affinity. Hence, halogens act as oxidizing agents.

Question 38.
Mention any two anomalous properties of second-period elements.
Answer:

• In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
• In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

Question 39.
Explain the Pauling method for the determination of ionic radius.
Answer:
Ionic radius is defined as the distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud of the ion. The ionic radius of a uni-univalent crystal can be calculated using Pauling’s method from the interionic distance between the nuclei of the cation and anion. Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other and therefore,
d = r_c+ + r_A- …………..(1)
where ‘d’ is the distance between the centre of the nucleus of the cation C⁺ and A^–. r ^c+ and r^A- are the radius of the cation and anion respectively.

Pauling also assumed that the radius of the ion having noble gas electronic configuration (Na⁺ and Cl^– having 1s², 2s², 2p⁶ configuration) is inversely
proportional to the effective nuclear charge felt at the periphery of the ion.
i.e.,   r_c+ ∝ \(\frac{1}{\left(Z_{e f f}\right)^{C+}}\) ………(2)

and r_A- ∝ \(\frac{1}{\left(Z_{e f f}\right)^{A-}}\) ………….(3)

where Z_eff is the effective nuclear charge.
Z_eff = Z – S.
Dividing the equation (2) by (3)
\(\frac{r_{c^{+}}}{r_{A^{-}}}=\frac{\left(Z_{e f f}\right)^{A-}}{\left(Z_{e f f}\right)^{C+}}\)

On solving the equations (1) and (4), the ionic radius of cation and anion are calculated.

Question 40.
Explain the periodic trend of ionisation potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called ionization energy.
(b) Variation in a period:
Ionization energy is a periodic property. On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of the number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 41.
Explain the diagonal relationship.
Answer:
On moving diagonally across the periodic table, the second and third-period elements show certain similarities. Even though the similarity is not the same as we see in a group, it is quite pronounced in the following pair of elements.

The similarity in properties existing between the diagonally placed elements is called ‘diagonal relationship’.

Question 42.
Why the first Ionisation enthalpy of sodium is lower than that of magnesium while its second ionisation enthalpy Is higher than that of magnesium?
Answer:
The 1st ionization enthalpy of magnesium is higher than that of Na due to the higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of the first electron, Na⁺ formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to a very high second ionization enthalpy for sodium. On the other hand, Mg⁺ formed after losing the first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 43.
By using Pauling’s method calculate the ionic radii of K⁺ and Cl^– ions in the potassium chloride crystal. Given that d^K + – Cl^– = 3.14 Å.
Answer:
d = r_K+ + r_Cl- = 3.14 Å
\(\frac{r_{K+}}{r_{C l-}}=\frac{\left(Z_{e f f}\right)^{C l-}}{\left(Z_{e f f}\right)^{K+}}\)

(Z_eff)^Cl- = Z – S = 17 – 10.9 = 6.1
(Z_eff)^K+ = Z – S = 19 – 16.8 = 2.2
\(\frac{r_{K+}}{r_{C l-}}=\frac{6.1}{2.2}\) = 2.77

r_K+ = 2.77 r_Cl-
2.77 r_Cl- + r_Cl- = 3.17 Å
3.77 r_Cl- = 3.17 Å
r_Cl- = 0.83 Å
r_K+ = (3.14 – 0.83) Å = 2.31 Å
The ionic radius of the K⁺ ion is 2.31 Å and Cl^– ion is 0.83 Å.

Question 44.
Explain the following, give appropriate reasons.
(i) Ionisation potential of N is greater than that of O:
Answer:
Nitrogen with 1s², 2s², 2p³ electronic configuration has higher ionization energy than oxygen. Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from the 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

(ii) First ionisation potential of the C-atom is greater than that of the B atom, whereas the reverse is true is for the second ionisation potential.
Answer:
The first ionization potential of the C atom and B-atom are as follows:
C(1s², 2s², 2p²) + IE₁ → C⁺ (1s², 2s², 2p¹)

B(1s², 2s², 2p¹) + IE₁ → B⁺ (1s², 2s²)

The ionization energy usually increases along a period. Hence, the first ionization energy of carbon is greater than that of Boron.
The second ionization potential of the C atom and B atom is as follows:
C⁺ (1s², 2s², 2p¹) + IE₂ → C⁺(1s², 2s²)
B⁺ (1s², 2s²) + IE₂ → B⁺ (1s², 2s¹)

B⁺ has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of the C⁺ atom. Hence, the second ionization energy of Boron is greater than that of carbon.

(iii) The electron affinity values of Be, Mg, and noble gases are zero, and those of N (0.02 eV) and P (0.80 eV) are very low.
Answer:
Be, Mg and noble gases have completely filled stable configuration and the addition of further electron is unfavourable and requires energy. The addition of extra electrons will disturb their stable electronic configuration and hence, they have almost zero electron affinity.

Nitrogen and Phosphorus have a half-filled stable configurations and the addition of further electrons is unfavourable and requires energy. The addition of extra electrons will disturb their stable electronic configuration and hence, they have very low zero electron affinity.

(iv) The formation of from F^–_(g)  from F(g) is exothermic while that of O^2-_(g) from O(g) is endothermic.
Answer:
The sizes of oxygen and fluorine atoms are comparatively small and they have high electron density. The extra electron added to fluorine has to accommodate in the 2p orbital which is relatively compact. Hence, the formation of F- from F is exothermic. In the case of oxygen, the formation of O^2- from O is endothermic due to extra stability of the completely filled 2p orbital of O^2- formation.

Question 45.
What is the screening effect?
Answer:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect (or) screening effect.

Question 46.
Briefly give the basis for Pauling’s scale of electronegativity.
Answer:
Pauling’s scale:

• Electronegativity is the relative tendency of an element present in a covalently bonded molecule to attract the shared pair of electrons towards itself.
• Pauling assigned arbitrary values of electronegativities for hydrogen and fluorine as 2.2 and 4, respectively.
• Based on this the electronegativity values for other elements can be calculated using the following expression.
  (X_A-X_B) = 0.182 √E_AB – (E_AA E_BB)
  Where E_AB , E_AA, and E_BB are the bond dissociation energies of AB, A_2, and B₂ molecules respectively.
  X_A and X_B are electronegativity values of A and B.

Question 47.
State the trends in the variation of electronegativity in groups and periods.
Answer:
Variation of Electronegativity in a period:
The electronegativity generally increases across a period from left to right. As discussed earlier, the atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases. Hence, the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.

Variation of Electronegativity in a group:
The electronegativity generally decreases down a group. As we move down a group, the atomic radius increases, and the nuclear attractive force on the valence electron decreases. Hence, the electronegativity decreases. Noble gases are assigned zero electronegativity. The electronegativity values of the elements of 5-block show the expected decreasing order in a group. Except for 13^th and 14^th groups, all other p-block elements follow the expected decreasing trend in electronegativity.

11th Chemistry Guide Periodic Classification of Elements Additional Questions and Answers

I. Very Short Question and Answers (2 Marks):

Question 1.
State periodic law.
Answer:
Periodic law states that ‘‘the properties of the elements are the periodic functions of their atomic weights”.

Question 2.
Write a note about Chancourtois classification.
Answer:
In this system, elements that differed from each other in atomic weight by 16 or multiples of 16 fell very nearly on the same vertical line. Elements lying directly under each other showed a definite similarity. This was the first periodic law.

Question 3.
What is Lavoiser’s classification of elements?
Answer:
Lavoiser classified the substances into four groups of elements namely acid-making elements, gas-like elements, metallic elements, and earthy elements.

Question 4.
State Mendeleev’s periodic law.
Answer:
This law states that “The physical and chemical properties of elements are a periodic function of their atomic weights.”

Question 5.
What are groups and periods?
Answer:
All the elements are arranged in the modem periodic table which contains 18 vertical columns and 7 horizontal rows. In the periodic table, the vertical columns are called groups, and horizontal rows are called periods.

Question 6.
State modern periodic law.
Answer:
The modem periodic law states that, “The physical and chemical properties of the elements are periodic function of their atomic numbers.”

Question 7.
What are p-block elements? Give their general electronic configuration.
Answer:
The elements of groups 13 to 18 are called p-block elements or representative elements and have a general electronic configuration ns², np^{1 – 6}.

Question 8.
Mention the names of the elements with atomic numbers 101, 102, 109, and 110.
Answer:
Z = 101  IUPAC  name : Mendelevium
Z = 102  IUPAC  name : Nobelium
Z = 109  IUPAC  name : Meitnerium
Z = 110  IUPAC  name : Darmstadtium

Question 9.
What are f-block elements? Give their properties.
Answer:
The lanthanides and the actinides are called f-block elements. These elements are metallic in nature and have high melting points. Their compounds are mostly coloured. These elements also show variable oxidation states.

Question 10.
Give the name and electronic configuration of elements of group and 2 groups.
Answer:

• Elements of 1^st group are called alkali metals. Their electronic configuration is ns¹.
• Elements of 2^nd group are called alkaline earth metals. Their electronic configuration is ns².

Question 11.
Define Atomic radius.
Answer:
The atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Question 12.
Write any two characteristic properties of alkaline earth metals.
Answer:

• Alkaline earth metals readily lose their outermost electrons to form a +2 ion.
• As we go down the group. their metallic character and reactivity are increased.

Question 13.
Why is the covalent radius shorter than the actual atomic radius?
Answer:
The formation of a covalent bond involves the overlapping of atomic orbitals and it reduces the expected internuclear distance. Therefore, the covalent radius is always shorter than the actual atomic radius.

Question 14.
Define metallic radius.
Answer:
Metallic radius is defined as one-half of the distance between two adjacent metal atoms in the closely packed metallic crystal lattice.

Question 15.
Halogens and chalcogens have highly negative electron gain enthalpies. Why?
Answer:

• Group 16 (chalcogens) and Group 17 (halogens) are interested to add two or one electrons respectively to attain a stable noble gas configuration.
• Because of this interest, these elements have highly negative electron gain enthalpies.

Question 16.
What is the effective nuclear charge? How is it approximated?
Answer:
The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the below-mentioned equation Z_eff = Z – S.
where Z is the atomic number and S is the screening constant which can be calculated using Slater’s rules.

Question 17.
Elements Zn, Cd, and Hg with electronic configuration (n-1)d¹⁰ ns² do not show most of the transition elements properties. Give reason.
Answer:

• Zn, Cd, and Hg are having completely filled d-orbitais (d¹⁰ electronic configuration).
• They do not have partially filled d-orbitais Like other transition elements. So they do not show much of the transition elements properties.

Question 18.
Define Ionisation energy.
Answer:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
Why d-block elements are called transition elements?
Answer:
d-block elements form a bridge between the chemically active metals of s-block elements and the less active elements of groups of 13^th and 14^th and thus take their familiar name transition elements.

Question 20.
Beryllium has higher ionization energy than Boron. Give reason.
Answer:
Beryllium has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of Boron (2s² – 2p¹). Hence, Beryllium has higher ionization energy than Boron.

Question 21.
Write the electronic configuration of lanthanides and actinides?
Answer:

• The electronic configuration of lanthanides is 4f^1-144 5d^0-11 6s².
• The electronic configuration of actinides is 5f^1-14 6d^0-17s².

Question 22.
What is the effect of shielding on ionization energy?
Answer:
As we move down a group, the number of inner-shell electrons increases which in turn increases the repulsive force exerted by them on the valence electrons, i.e., the increased shielding effect caused by the inner electrons decreases the attractive force acting the valence electron by the nucleus. Therefore, the ionization energy decreases.

Question 23.
Define electron affinity.
Answer:
Electron affinity is defined as the amount of energy released when an electron is added to the valence shell of an isolated neutral gaseous atom in its ground state to form its anion.

Question 24.
What are periodic properties? Give example.
Answer:
The term periodicity of properties indicates that the elements with similar properties reappear at certain regular intervals of atomic number in the periodic table.
Example:

• Atomic radii
• Ionization energy
• Electron affinity
• Electronegativity.

Question 25.
Why is the electron affinity of Nitrogen almost zero?
Answer:
Nitrogen has half-filled stable (1s², 2s², 2p³) electronic configuration. The addition of extra electrons will disturb their stable electronic configuration and it has almost zero electron affinity.

Question 26.
The cationic radius is smaller than its corresponding neutral atom. Justify this statement.
Answer:

• When a neutral atom loses one or more electrons it forms cation.
  Na → Na⁺ + e^–
• The radius of this cation (r_Na⁺)is decreased than its parent atom (r_Na).
• When an atom is charged to cation, the number of nuclear charges becomes greater than the number of orbital electrons. Hence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

Question 27.
Define electronegativity.
Answer:
Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 28.
What are isoelectronic ions? Give example.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example: Na⁺ Mg²⁺, Al³⁺, F^–, O^2-, N^3-

Question 29.
What is the variation of electronegativity in a group?
Answer:
The electronegativity generally decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence, the electronegativity decreases.

Question 30.
The ionization energy of beryllium is greater than the ionization energy of boron. Why?
Answer:
Be (Z= 4) 1s² 2s². it has completely filled valence electrons, which requires high IE₁.
B (Z =5) 1s² 2s² 2p¹. It has incompletely filled valence electrons, which requires comparatively
less IE₁ Hence I.E₁ Be > I.E₁ B.

Question 31.
What is the diagonal relationship?
Answer:
On moving diagonally across the periodic table, the second and third-period elements show certain similarities. The similarity in properties existing between the diagonally placed elements is called the ‘diagonal relationship’.

Question 32.
Define electron gain enthalpy or electron affinity. Give its unit.
Answer:
The electron gain enthalpy of an element is the amount of energy released when an electron is added to the neutral gaseous atom.
A + electron → A^– + energy (E.A)
Unit of electron affinity is KJ mole^–.

II. Short Question and Answers (3 Marks):

Question 1.
How Moseley determined the atomic number of an element using X-rays?
Answer:

• Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
• He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding a clement with the high energy of electrons.
• Moseley correlated the frequency of the X-ray emitted by an equation as,
  \(\sqrt{v}\) = a (Z – b)
  Where υ = Frequency of the X-rays emitted by the elements.
  a and b = Constants.
• From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

Question 2.
Write notes on Newlands classification of elements.
Answer:
Newland made an attempt to classify the elements and proposed the law of octaves. On arranging the elements in the increasing order of atomic weights, he observed that the properties of every eighth element are similar to the properties of the first element. This law holds good for lighter elements up to calcium.

Question 3.
Describe Mendeleev’s periodic classification of elements.
Answer:
Dmitriev Mendeleev proposed that “the properties of the elements are the periodic functions of their atomic weights” and this is called periodic law. Mendeleev listed the 70 known elements at that time in several vertical columns in order of increasing atomic weights. Thus, Mendeleev constructed the first periodic table based on the periodic law.

In the periodic table, he left some blank spaces since there were no known elements with the appropriate properties at that time. He and others predicted the physical and chemical properties of the missing elements. Eventually, these missing elements were discovered and found to have the predicted properties. For example, Gallium of group-III and Germanium of group- IV were unknown at that time. But Mendeleev predicted their existence and properties. After the discovery of the actual elements, their properties were found to match closely to their predicted by Mendeleev.

Question 4.
Write notes on Moseley’s work.
Answer:
Henry Mosley studied the characteristic X-rays spectra of several elements by bombarding them with high energy electrons and observed a linear correlation between atomic number and the frequency of X-rays emitted which is given by the following expression, √υ = a(Z – b)
where, υ is the frequency of the X-rays emitted by the element with atomic number ‘Z’; ‘a’ and ‘b’ are constants and have same values for all the elements. The plot of √υ against Z gives a straight line. Using this relationship, we can determine the atomic number of an unknown element from the frequency of X-rays emitted.

Question 5.
What are the reasons behind Moseley’s attempt in finding an atomic number?
Answer:

• The number of electrons increases by the same number as the increase in the atomic number.
• As the number of electrons increases, the electronic structure of the atom changes.
• Electrons in the out can not shell of an atom (valence shell electrons) determine the chemical properties of the elements.

Question 6.
What are s-block elements? Give their properties.
Answer:
The elements of group-1 and group-2 are called s-block elements, since the last valence electron enters the ns orbital.

• The group-1 elements are called alkali metals while group-2 elements are called alkaline earth metals.
• These are soft metals and possess low melting and boiling points with low ionization enthalpies.
• They are highly reactive and form ionic compounds.
• They are highly electropositive in nature and most of the elements imparts colour to the flame.

Question 7.
Explain the classification of elements based on electronic configuration.
Answer:

• The distribution of electrons into orbitais, s, p, d, and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, l, m, and s. of these the principal quantum number (n) defines the main energy level known as shells.
• The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
• The electronic configuration of elements in the periodic table can be studied along with the periods and groups separately for the best classification of elements.
• Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
• A vertical column of the periodic table is called a group. A group consists of a series of elements having a similar configuration to the outermost shell. There are 18 groups in the periodic table.

Question 8.
What are d-block elements? Give their properties.
Answer:
The elements of the groups 3 to 12 are called d-block elements or transition elements with general valence shell electronic configuration ns^{1 – 2}, (n – 1 )d^{1 – 10}.

• These elements show more than one oxidation state and form ionic, covalent, and coordination compounds.
• They can form interstitial compounds and alloys which can also act as catalysts.
• These elements have high melting points and are good conductors of heat and electricity.

Question 9.
What is a covalent radius? Explain.
Answer:
Covalent radius is one-half of the internuclear distance between two identical atoms linked together by a single covalent bond. The internuclear distance can be determined using X-ray diffraction studies.
For example, the experimental intemuclear distance in Cl₂ molecule is 1.98 Å. The covalent radius of chlorine is calculated as follows,
d_{cl – cl} = r_cl + r_cl
r_cl = \(\frac{d_{c l-d t}}{2}=\frac{1.98}{2}\) = 0.99 Å .

The formation of a covalent bond involves the overlapping of atomic orbitals and it reduces the expected internuclear distance. Therefore, the covalent radius is always shorter than the actual atomic radius.

Question 10.
How is the covalent radius of an individual atom can be calculated?
Answer:
The covalent radius of an individual atom can be calculated using the internuclear distance (dA-B) between two different atoms A and B. The simplest method proposed by Schomaker and Stevenson is as follows.
d_{A – B} = r_A + r_B – 0.09(χ_A – χ_B)
where χ_A and χ_B are the electronegativities of A and B respectively in Pauling units. Here, χ_A > χ_B and radius is in Å.

Question 11.
Write about the electronic configuration of the 1^st and 2^nd periods.
Answer:
Electronic configuration of t period:
In 1 period only two elements are present. This period starts with the filling of electrons in the first energy level, n¹. This level has only one orbital as is. Therefore it can accommodate two electrons maximum.

Electronic configuration of 2nd period:
In the 2’ period 8 elements are present. This period starts with the filling of electrons in the second energy level, n = 2. In this level four orbitais (one 2s and three 2p) are present. Hence the second energy level can accommodate 8 electrons. Thus, the second period has eight elements.

Question 12.
Discuss the variation of electron affinity in the period.
Answer:
As we move from alkali metals to halogens in a period, generally electron affinity increases, i.e., the amount of energy released will be more. This is due to an increase in the nuclear charge and a decrease in the size of the atoms. However, in the case of elements such as beryllium and nitrogen the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have stable ns², np⁶ configurations, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns², np⁵ readily accept an electron to get the stable noble gas electronic configuration (ns², np⁶), and therefore, in each period the halogen has high electron affinity.

Question 13.
What are the two exceptions of block division in the periodic table?
Answer:
1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block but actually placed in s group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in the 18^th group. It also resembles 18^th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. it has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17th group elements). Because of these assumptions, the position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

III. Long Question and Answers (5 Marks):

Question 1.
Describe Mosley’s work and Modern Periodic Law.
Answer:
Henry Mosley studied the characteristic X-rays spectra of several elements by bombarding them with high energy electrons and observed a linear correlation between atomic number and the frequency of X-rays emitted which is given by the following expression,
√υ = a(Z – b)
where, υ is the frequency of the X-rays emitted by the element with atomic number ‘Z’; ‘a’ and ‘b’ are constants and have the same values for all the elements. The plot of √υ against Z gives a straight line. Using this relationship, we can determine the atomic number of an unknown element from the frequency of X-rays emitted.

Based on his work, the modern periodic law was developed which states that, “the physical and chemical properties of the elements are periodic functions of their atomic numbers”. Based on this law, the elements were arranged in order of their increasing atomic numbers. This mode of arrangement reveals an important truth that the elements with similar properties recur after regular intervals. The repetition of physical and chemical properties at regular intervals is called periodicity.

Question 2.
The first ionization enthalpy of magnesium is higher than that of sodium. On the other hand, the second ionization enthalpy of sodium is very much higher than that of magnesium. Explain.
Answer:
The 1^st ionization enthalpy of magnesium is higher than that of Na⁺ due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of the first electron, N& formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand. Mg⁺ formed after losing first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

Question 3.
By using Pauling’s method calculate the ionic radii of Na⁺ and F^– ions in the potassium chloride crystal. Given that dNa⁺ – F^– = 231 pm.
Answer:

Question 4.
Discuss the variation of electronic configuration along the periods.
Answer:
Each period starts with the element having general outer electronic configuration, ns¹ and ends with ns², np⁶ where n is the period number. The first period starts with the filling of valence electrons in is orbital, which can accommodate only two electrons. Hence, the first period has two elements, namely hydrogen and helium.

The second period starts with the filling of valence electrons in 2s orbital followed by three 2p orbitals with eight elements from lithium to neon. The third period starts with filling of valence electrons in the 3s orbital followed by 3p orbitals. The fourth period starts with filling of valence electrons from 4s orbital followed by 3d and 4p orbitals in accordance with Aufbau principle. Similarly, we can explain the electronic configuration of elements in the subsequent periods.

In the fourth period, the filling of 3d orbitals starts with scandium and ends with zinc. These 10 elements are called first transition series. Similarly, 4d, 5d and 6d orbitals are filled in successive periods and the corresponding series of elements are called second third and fourth transition series respectively.

In the sixth period the filling of valence electrons starts with 6s orbital followed by 4f, 5d and 6p orbitals. The filling up of 4f orbitals begins with Cerium (Z = 58) and ends at Lutetium (Z = 71). These 14 elements constitute the first inner-transition series called Lanthanides. Similarly, in the seventh period 5f orbitals are filled, and it’s -14 elements constitute the second inner-transition series called Actinides. These two series are placed separately at the bottom of the modem periodic table.

Question 5.
Describe the nomenclature of elements with Atomic Number greater than 100.
Answer:
Usually, when a new element is discovered, the discoverer suggests a name following IUPAC guidelines which will be approved after a public opinion. In the meantime, the new element will be called by a temporary name coined using the following IUPAC rules until the IUPAC recognizes the new name.
1. The name was derived directly from the atomic number of the new element using the following numerical
roots.

2. The numerical roots corresponding to the atomic number are put together and ‘ium’ is added as suffix.
3. The final ‘n’ of ‘enn is omitted when it is written before ‘nil’ similarly the final ‘i’ of ‘bi’ and ‘tri’ is omitted when it written before ‘ium’.
4. The symbol of the new element is derived from the first letter of the numerical roots.

Question 6.
Explain the merits of Moseley’s long form of the periodic table.
Answer:
Merits of Moseley’s long form of the periodic table:

•  As this classification is based on atomic number, it relates the position of an element to its electronic configuration.
• The elements having similar electronic configuration fall in a group. They also have similar physical and chemical properties.
• The completion of each periõd is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached.
• The position of zero group is also justified in the table as group 18.
• The table completely separates metals and non-metals.
• The table separates two subgroups. lanthanides and actinides, dissimilar elements do not fall together.
• The greatest advantage of this periodic table is that this can be divided into four blocks namely s, p. d, and f-block elements.
• This arrangement of elements is easier to remember, understand and reproduce.

Question 7.
Explain the variation of the atomic radius in periods and groups.
Answer:
Variation in a period:
Atomic radius tends to decrease in a period. As we move from left to right along a period, the valence electrons are added to the same shell. The simultaneous addition of protons to the nucleus increases the nuclear charge, as well as the electrostatic attractive force between the valence electrons and the nucleus. Therefore atomic radius decreases along a period.

Variation in a group:
In the periodic table, the atomic radius of elements increases down the group. As we move down a group, new shells are opened to accommodate the newly added valence electrons. As a result, the distance between the centre of the nucleus and the outermost shell containing the valence electron increases. Hence, the atomic radius increases.

Question 8.
What is an Effective nuclear charge? How is it calculated using slater’s rule?
Answer:
In addition to the electrostatic forces of attraction j between the nucleus and the electrons, there | exists repulsive forces among the electrons. The repulsive force between the inner shell electrons j and the valence electrons leads to a decrease in the j electrostatic attractive forces acting on the valence 1 electrons by the nucleus. Thus, the inner shell | electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective 1 nuclear charge. It is approximated by the below-mentioned equation.
Z_eff = Z – S
Where Z is the atomic number and ‘S’ is the screening constant which can be calculated using Slater’s rules as described below.
Step 1:
Write the electronic configuration of the atom and rearrange it by grouping ns and np orbitals together and others separately in the following form.
(1s)(2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) ……

Step 2:
Identify the group in which the electron of interest is present. The electron present right to this ! group does not contribute to the shielding effect. Each of the electrons within the identified group (denoted by ‘n’) shields to, an extent of 0.35 unit of nuclear charge. However, it is 0.30 unit for 1s electron.

Step 3:
Shielding of inner-shell electrons. If the electron of interest belongs to either s or p orbital,
(i) each electron within the (n – 1) group shields to an extent of 0.85 unit of nuclear charge, and
(ii) each electron within the (n – 2) group (or) even lesser group (n – 3), (n – 4) etc…
completely shields i.e. to an extent of 1.00 unit of nuclear charge.

If the electron of interest belongs to d or f orbital, then each of electron left of the group of electron of interest shields to an extent of 1.00 unit of nuclear charge.

Step 4:
Summation of the shielding effect of all the electrons gives the shielding constant ‘S’.

Question 9.
Explain the periodic variation of ionization energy in a period.
Answer:
The ionization energy usually increases along a period with few exceptions. When we move from left to right along a period, the valence electrons are added to the same shell, at the same time protons are added to the nucleus, This successive increase of nuclear charge increases the electrostatic attractive force on the valence electron and more energy is required to remove the valence electron resulting in high ionization energy. Consider the variation in ionization energy of second period elements. The plot of atomic number vs ionisation energy is given below.

In the graph, there are two deviations in the trends of ionisation energy. It is expected that boron has higher ionisation energy than beryllium since it has higher nuclear charge. However, the actual ionisation energies of beryllium and boron are 899 and 800kJ mol^-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s², 2p¹).

The electronic configuration of beryllium (Z = 4) in its ground state is 1s², 2s² and that of boron (Z = 5) 1s², 2s², 2p¹. Similarly, nitrogen with 1s², 2s², 2p¹ electronic configuration has higher ionisation energy (l402 kJmol^-1) than oxygen (1314 kJmol^-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

Question 10.
Explain the periodic variation of electron affinity in a group.
Answer:
Electron affinity is defined as the amount of energy released (required in the case of noble gases) when an electron is added to the valence shell of an isolated ^neutral gaseous atom in its ground state to form its anion. It is expressed in kJmol^-1
A + 1e^– → A^– + E_A

Variation of Electron Affinity in a period:
As we move from alkali metals to halogens in a period, generally electron affinity increases, i.e., the amount of energy released will be more. This is due to an increase in the nuclear charge and a decrease in size of the atoms. However, in case of elements such as beryllium (1s², 2s²), nitrogen (1s², 2s², 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have stable ns², np⁶ configurations, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns², np⁵ readily accept an electron to get the stable noble gas electronic configuration (ns², np⁶), and therefore in each period the halogen has high electron affinity, (high negative values)

Question 11.
Define electron affinity. How does it vary along with the group?
Answer:
Electron affinity is defined as the amount of energy released (required in the case of noble gases) when an electron is added to the valence shell of an isolated neutral gaseous atom in its ground state to form its anion. It is expressed in kJmoF1
A + 1e^– → A^– + E_A

Variation of Electron affinity in a group:
As we move down a group, generally the electron affinity decreases. It is due to an increase in atomic size and the shielding effect of inner-shell electrons. However, oxygen and fluorine have lower affinity than sulphur and chlorine respectively. The sizes of oxygen and fluorine atoms are comparatively, small and they have high electron density.

Moreover, the extra electron, added to oxygen and fluorine has to be accommodated in the 2p orbital which is relatively compact compared to the 3p orbital of sulphur and chlorine so, oxygen and fluorine have lower electron affinity than their respective group elements sulphur and chlorine.

Question 12.
Explain the salient features of groups.
Answer:
1. Number of electrons in the outermost shell:
The number of electrons present in the outermost shells does not change on moving down in a group, i.e remains the same. Hence, the valency also remains the same within a group.

2. Number of shells:
In going down a group the number of shells increases by one at each step and ultimately becomes equal to the period number to which the element belongs.

3. Valency:
The valencies of all the elements of the same group are the same. The valency of an element with respect to oxygen is same in a group.

4. Metallic character:
The metallic character of the elements increases in moving from top to bottom in a group.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 7 Cell Cycle Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

11th Bio Botany Guide Cell Cycle Text Book Back Questions and Answers

Choose The Correct Answer:

Question 1.
The correct sequence in cell cycle is
a) S M G1 G2
b) S G1 G2 M
c) G1 S G2 M
d) M G1 G2 S
Answer:
c) G1 S G2 M

Question 2.
If mitotic division is restricted in G, phase the cell cycle, then the condition is known as
a) S Phase
b) G₂ Phase
c) M Phase
d) G₀ phase
Answer:
d) G₀ phase

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cells. Which of the following is expected to occur?
a) Chromosomes will be fragmented
b) Chromosomes will not condense
c) Chromosomes will not regregate
d) Recombination of Chromosomes will occur
Answer:
c) Chromosomes will not segregate

Question 4.
In the S phase of cell cycle
a) Amount of DNA doubles in each cell
b) Amount of DNA remain same in each cell
c) Chromosome number is increased
d) Amount of DNA is reduced to half in each cell
Answer:
a) Amount of DNA doubles in each cell

Question 5.
The centromere is required for
a) Transcription
b) Crossing over
c) Cytoplasmic cleavage
d) movement of chromosome towards the pole
Answer:
d) movement of chromosome towards the pole

Question 6.
Synapsis occurs between
a) m RNA and ribosomes
b) spindle fibres and centromeres
c) two homologous chromosomes
d) a male and a female gemale
Answer:
c) two homologous chromosomes

Question 7.
In meiosis crossing over is initiated at
a) Diplotene
b) Pachytene
c) Leptotene
d) Zygotene
Answer:
b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage
a) Anaphase
b) Metaphase
c) Prophase
d) Interphase
Answer:
a) Anaphase

Question 9.
The pairing of homologous chromosomes on meiosis is known as
a) Bivalent
b) Synapsis
c) Disjunction
d) Synergids
Answer:
b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of
a) Lower animals
b) Higher animals
c) Higher plants
d) All living Organism
Answer:
c) Higher plants

Question 11.
Write any three significance of mitosis
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – daughter cells are genetically identical to parent cells.
2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
3. Regeneration – Arms of starfish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Differences between Mitosis and Meiosis

Mitosis	Meiosis
One division	Two division
The number of chromosomes remains the same	The number of chromosomes is halved
Homologous chromosomes line up separately on the metaphase plate	Homologous chromosomes line up in pairs at the metaphase plate
Homologous chromosome do not pair up	Homologous chromosome pair up to form bivalent
Chiasmata do not form and crossing over never occurs	Chiasmata form and crossing over occurs
Daughter cells are genetically identical	Daughter cells are genetically different from the parent cells
Two daughter cells are formed	Four daughter cells are formed

Question 13.
Differentiate Cytokinesis in plant cells & animal cells
Answer:

Question 14.
Given an account of the Go phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 15.
Write about the Pachytene and Diplotene stage of Prophase I.
Answer:

3. Pachytene stage

• Chromosome appear as bivalent or tetrads
• 4 chromatids & 2 centromeres are seen
• Synapsis of homologous chromosomes between non-sister chromatids completes except at chiasmata where crossing over occurs
• Recombination (exchange of chromosomal bits is completed by the end) – but chromosomes are linked at the sites of crossing over
• Enzyme – Recombinase mediates the process.

4. Diplotene

• Synaptonemal complex disassembled & dissolves
• Nonsister chromatids of homologous chromosomes get attached where x like shape occur at Crossing over known as chiasmata holding the homologous chromosomes together the homologous chromosomes tend to separate except at chiasmat
• The sub-stage last for days or years depending on the sex & the organism follows Pachytene
• synaptical complex disassembled & dissolves
• The chromosomes are actively transcribed in females as the eggs stores up materials for embryonic development
• Exception In Lamp brush chromosome prominent loops occur.
  

11th Bio Botany Guide Cell Cycle Additional Important Questions and Answers

I.Choose the correct answer: (1 Marks)

Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G₁
(b) S
(c) G₂
(d) G₀
Answer:
(d) G₀

Question 2.
Un differentiated cells include
a) Stem cells in animals
b) Meristematic cells in plants
c) RBC which carry out the transportation of oxygen
d) Mesophyll cells which carry out photosynthesis
(i) a & b
(ii) c & d
(iii) a & c
(iv) b & c
Answer:
(i) a & b

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilis
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
The proteins that activate the cell to perform cell division are
a) Actin and Myosin
b) Kinases and Cyclin
c) Histamine and Cyclin
d) Tubulin and Actin
Answer:
b) Kinases & Cyclin

Question 5.
The number of chromosomes in the onion cell is …………….
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Question 6.
Mitosis is called
a) Direct cell division
b) Indirect cell division
c) Mitotic Meiotic cell division
d) Reduction division
Answer:
a) Direct cell division

Question 7.
During Anaphase
(I) The daughter chromosome move to the opposite poles due to the shortening of the phragmoplast
(II) due to the thickening of chromosomes
(III) Shortening of microtubules
(IV) Shortening of asters
Answer:
(III) Shortening of microtubules

Question 8.
Cell cycle was discovered by …………….
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
The cells without nucleus are
a) RBC – platelets, tracheids & vessels
b) RBC – sievetube, companion cells thrombocytes
c) WBC – platelets, companion cells & vessels
d) RBC – platelets, companion cells & neurons
Answer:
a) RBC – platelets, tracheids & vessels

Question 10.
The stage between two meiotic division is called
a) Cytokinesis
b) Interphase I
c) Inter kinesis
d) Interphase II
Answer:
c) Inter kinesis

Question 11.
Replication of DNA occurs at …………… phase.
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(c) S

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

Question 1.
(I) Nucleolus disappear during the metaphase stage of mitosis
(II) The microtubules arrange to form asters in plant cells
(III) In plant cells phragmoplast is formed prior to the formation of cell plate
(IV) Mitosis is responsible for the Regeneration of lost arms of starfish

Answer:
b) False – False – True- True

Question 2.
(i) The word Protoplasm was coined by – Purkinje
(ii) Structure of bacteria was observed first – Edouard Van Beneden through a microscope by
(iii) Centrosome & Chromosome theory – Theodor Boveri Proposed by
(iv) Cell division in round Worm was – Walther Flemming Observed by

Answer:
b) True – False – True – False

Question 3.
(i) C Value is the amount in picograms of DNA contained within a haploid Nucleus
(ii) Nucleolan membrane disappear during Ana Phase stage of mitosis
(iii) The arrangement of microtubules is called to form Asters -, which is a unique feature of plant cells
(iv) One of the protein synthesis in G2 – phase is known as Maturation Promoting Factor

Answer:
a) True – False – False – True

III. Find out the correct match from the following.

Question 1.
Zygotene – Chromosomes appear as tetrads
Pachytene – Synapsis of homologous chromosomes occur
Diplotene – Condensation of chromosomes takes place
Diakinesis – Terminalisation of chiasmata occur & Nucleolus Disappear
Answer:
Diakinesis -Terminalisation of chiasmata occur & Nucleolus Disappear

Question 2.
Duration of different Phases of cell cycle given find out the correct match
(I) S Phase – 12 Hours
(II) G1 Phase – 11 Hours
(III) G2 Phase – 4 Hours
(IV) M Phase – 2 Hours
Answer:
(III) G2 Phase – 4 Hours

IV. Find out the Wrong match

Question 1.
(I) The chromosome does not divide as chromatids, for centromere does not divide – Anaphase I
(II) The chromatids move to the opposite poles by the splitting of the centromere – Anaphase
(III) Sister chromatids get separated by splitting of Centromere – Anaphase II
(IV) Homologous chromosomes appear as bivalent or tetrad – Metaphase II
Answer:
(IV) Homologous chromosomes appear as bivalent or tetrad – Metaphase II

Question 2.
Most nëurons remain in G-’o’ stage do not divide
(I) This technique can be applied to replace neurons in dementia patients
(II) neurons can be activated by giving electric shocks
(III) neurons can be replaced by surgical procedures
(IV) Dead or injured neurons can be replaced by stem cell therapy
Answer:
(IV) Dead or injured neurons can be replaced by stem cell therapy

V.

Question 1.
Match the following and find the
(I) Robert Brown – A) Coined the word cell
(II) Robert Hook – B) Coined the word Mitosis
(III) Schleiden & Schwann – C) Studied the presence of Nucleus in cells
(IV) Waither Flemming – D) Cell theory

Answer:
d) C-A- D- B

Question 2.
(I) Cell cycle – A) division that follows the nuclear division
(II) Restriction point – B) Longest part but not resting stage
(III) lnterphae – C) A series of events leading to the formation of a new Cell
(IV) Cytokinesis – D) The checkpoint at the end of Gi determine a cell’s fate

Answer:
c) C-D- B- A

VI. 

Question 1.
Read the following Assertion and Reason. Find the correct answer
Assertion A: The mitochondrial inheritance in higher animals is uniparental
Reason R: The mitochondria from the male partner either undergo degeneration or rejected and only mitochondria from egg or ova is accepted.
(a) Assertion and Reason are correct Reason is explaining assertion
(b) Assertion and Reason are correct, but the reason is not explaining assertion
(c) Assertion is true but Reason not explaining assertion
(d) Assertion is true but Reason is wrong
Answer:
a) Assertion and Reason are correct. Reason is explaining assertion

Question 2.
Assertion A: In Meiosis Prophase I is a longer but significant phase
Reason R: Chiasma formation and crossing over takes place and recombination takes place
(a) Assertion and Reason are correct Reason is explaining Assertion
(b) Assertion and Reason are correct but Reason is not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason not explaining Assertion
Answer:
b) Assertion and Reason are correct but Reason is not explaining Assertion

Question 3.
Assertion A: Interphase is the longest part of cell division and the cell actively involved protein synthesis & DNA synthesis
Reason: The Interphase is also known as the resting phase, & the cell takes rest between successive cell division
(a) Assertion and Reason are correct Reason is explaining Assertion
(b) Assertion and Reason are correct but Reason is not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason not explaining assertion
Answer:
c) Assertion is true but Reason is wrong

Question 4.
Assertion A: The sister chromatids of homologous chromosomes exchange chromosomal bits
Reason R: This process of exchange of chromosomal bits is known as crossing over
(a) Assertion and Reason are correct Reason is explaining Assertion
(b) Assertion and Reason are correct but Reason is not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason not explaining assertion
Answer:
d) Assertion is true but Reason not explaining assertion

VII. 

Question 1.
See the diagram & label the parts

Answer:
A – Leptotene
B – Chromosomes are visible under a light microscope
C – Paired sister chromatids begin to condense

Question 2.
See the diagram and write the correct answer.

A	B	C	D
a) Cell growth	interphase	Mitotic phase	Cyto kinesis
b) Interphase	Cell growth	Mitotic phase	Cytokinesis
c) Cell growth	Inter Phase	Cytokinesis	Mitotic phase
d) Interphase	Cell growth	Cytokineses	Mitotic phase

Answer:
a) Cell growth – Interphase – Mitotic phase – Cyto kinesis

Question 3.
See the diagram & label Find out the correct labelling.

	A	B	C	D
a)	Chromatin	Kineto chore	See constriction	Centromere
b)	Centromere	Chromatin	See constriction	Kineto chore
c)	Centromere	Kineto chore	Chromatin	Centomere
d)	Chromatin	See Constriction	Kineto chore	Centomere

Answer:
b) Centromere – Chromatin – Secondary constriction – Kinetochore

Question 4.
Label the diagram properly.

Answer:
A – Non-sister Chromatids
B – Centromere
C – Chiasma
D – Bivalent or Tetrad

Question 5.

Answer:
A – Outer kinetochore
B – Fibrous corona
C – Inner Kinetochore
D – Microtubules

2 Mark Questions.

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

1. Mitosis and
2. Meiosis.

Question 2.
What are the reasons for the arresting growth of cell during G1 Phase?
Answer:

• Deprivation of nutrition
• Lack of growth factors or density-dependent inhibition occur
• Some metabolic changes leads to Go – stage

Question 3.
Point out the reasons responsible for the arresting of the cell in the G₁ phase?
Answer:
Cells are arrested in G₁ due to:

• Nutrient deprivation
• Lack of growth factors or density dependant inhibition
• Undergo metabolic changes and enter into G₀ state.

Question 4.
Differentiate between karyokinesis and Cytokinesis of Amitosis.
Answer:

Karyokinesis	Cytokinesis
Division of Nucleus occur	Division of Cytoplasm occur
Nucleus develop contraction at the centre become dumbbell shaped contriction deepen divide nucleus into two	Plasma membrane develop constriction along with nuclear contraction, which deepen centripetally and the cell divides into two

Question 5.
Define amitosis & write about its drawbacks.
Answer:
Amitosis is known as Direct cell division or Incipient cell division.
No spindle formation, no condensation of chromatin material occur.

It has one 2 steps

1. Karyokinesis
2. Cytokinesis

Drawbacks: Causes unequal distribution of chromosomes Can lead to abnormalities in

• metabolism
• reproduction

Question 6.
Distinguish between Closed and Open mitosis.
Answer:

Closed mitosis	Open mitosis
Nuclear envelope remain intact & chromosomes migrate to opposite poles of a spindle with in the Nucleus Eg – Unicellular Eukaryotes – Yeast, Slime molds	Nuclear envelope breaks down and then reforms around the 2 sets of separated Chromosome Eg – Most higher Plants & Animals

Question 7.
Differentiate between Anastral and Amphiastral cell division
Answer:

Anastral	Amphiastral
Occur in plant cell No asters or centrioles are formed, only spindle fibres are formed Eg. Plants	Occur in animal cells Asters and centrioles formed at each pole of the spindle during cell division Eg. Animals

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. A cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally, the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Question 9.
Differentiate meiosis in Plants & Animals.
Answer:

Plants	Animals
In flowering plants meiosis occur during Microsporogenesis (anther) & in Mega sporogenesis (i.e) (ovule) development	It take place in reproductive organs at the time of production of gametes Spermatogenesis – produces haploid sperms Oogenesis – produces haploid eeas

Question 10.
Define Mitogens.
Answer:

• The biochemical substances or factors which promote cell cycle acceleration & proliferation is called Mitogen.
• Eg. Gibberellin, Ethylene, Indole Acelic Acid, Kinetin.
• They are also known as Growth promotors.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

3 Mark Questions

Question 1.
Write down the significance of Meiosis.
Answer:

• Maintain chromosome number constant
• Crossing over (exchange of genetic meterial) leads to variations
• Variation – the raw material for Evolution
• Finally, meiosis produces genetic variability by partitioning different combinations of genes into gametes through an independent assortment.
• Responsible for Adaptations of organisms to various environmental stress.

Question 2.
Differentiate between Mitosis in Plants & Animals.
Answer:

Plants	Animals
Centrioles are absent	Centrioles are present
Asters are not formed	Asters are formed
Cell division involves the formation of a cell	Cell division involves furrowing and cleavage of cytoplasm
Occurs mainly at the meristem	Occurs in tissues throughout the body

Question 3.
Explain Endomitosis.
Answer:
Sometimes, the replication of chromosomes occur in the absence of karyokinesis & cytokinesis resulting in numerous copies within each cell condition known as Endomitosis.

• Chromonema donot separate to form Chromosomes
• Nuclear membrance does not repture
• No spindle formation occur
• Each chromosome consisting of over thousands of synapsed Chromatids
  Eg – Salvary gland chromosome of Drosophila (Polytene or Chromosome).

Question 4.
Differentiate between Anaphase I & Anaphase II of Meiosis.
Answer:

5 Mark Questions

Question 1.
Explain in detail the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 2.
Explain karyokinesis in mitosis of plant cell
Answer:

Prophase – Longest Phase

• Chromosomes visible thread like condenses into thick chromosomes
• Initiation of spindle fibres occur
• Nucleolus disappear Nuclear envelope breaks down
• Golgi apparatus & ER not seen

Metaphase

• Sister chromatids attached to spindle fibres by kinetochore of centromere
• Chromosome align on the equatorial plane (metaphase plate)
  (spindle assembly check point decide the fate of the cell)

Anaphase

• Centromere split daughter chromatids move to opposite poles
• Shortening of spindle create pull divide centromere & divide chromosome into a chromatids
  (APC /C leads to degradation of protein lead to separation of chromatids

Telophase

• Genetic material division completed Nucleolus & Nuclear membrane reform.
• Sister chromatids become thick chromosomes with its own centromere.

Question 3.
Explain S phase & G2 phase, til S Phase
Answer:
(i) S Phase:

• Known as synthetic phase of interphase of mitosis
• Growth of cell continues
• Replication of DNA occur
• Histones synthesised and attach to DNA
• Duplication of centrioles occur
• DNA content doubles from 2C to 4C

(ii) G2 phase

• 4 C amount of DNA
• Cell growth continues
• Synthesis of
  • organelle,
  • mitochondria & chloroplast
  • tubulin synthesised, microtubules formed
• Spindle begin to occur
• Nuclear division follows
• MPF (Maturation Promoting factor) formed brings out condensation of interphase chromosomes into mitotic form.

Question 4.
Draw the Cell cycle.
Answer:

Question 5.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – Daughter cells are genetically identical to parent cells.
2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
6. Regeneration – Arms of starfish

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

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 11 Waves

11th Physics Guide Waves Book Back Questions and Answers

I. Multiple choice questions:

Question 1.
A student tunes his guitar by striking a 120 Hertz with a tuning fork, and simultaneously plays the 4^th string on his guitar. By keen observation, he hears the amplitude of the combined sound oscillating thrice per second. Which of the following frequencies is the most likely the frequency of the 4th string on his guitar?
(a) 130
(b) 117
(c) 110
(d) 120
Answer:
(b) 117

Hint:
Frequency of the fourth string can be derived from harmonic series generated by the strings.
Frequencies of first 4 strings are 120, 119, 118, 117 Hz.
f₄ = 117Hz

Question 2.
A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms^-1 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms^-1, respectively are:
(a) 120 Hz and 5 m
(b) 100 Hz and 5 m
(c) 120 Hz and 6 m
(d) 100 Hz and 6 m
Answer:
(d) 100 Hz and 6 m

Hint:
v_A = 500 ms^-1; λ_A = 5 m
∴ Frequency in medium B
f_B = \(\frac{v_{\mathrm{A}}}{\lambda_{\mathrm{A}}}\) = \(\frac { 500 }{ 5 }\) = 100 Hz
Wavelength in medium B is
λ_B = \(\frac{v_{\mathrm{B}}}{f_{\mathrm{B}}}\) = \(\frac { 600 }{ 100 }\) = 6m

Question 3.
For a particular tube, among six harmonic frequencies below 1000 Hz, only four harmonic frequencies are given: 300 Hz, 600 Hz, 750 Hz and 900 Hz. What are the two other frequencies missing from this list?
(a) 100 Hz, 150 Hz
(b) 150 Hz, 450 Hz
(c) 450 Hz, 700 Hz
(d) 700 Hz, 800 Hz
Answer:
(b) 150 Hz, 450 Hz

Hint: Fundamental frequency f₀ = 150 Hz
From the series of frequencies we get
f₁ = 2f₀
f₂ = 3f₀
= 3 x 150
= 450 Hz

Question 4.
Which of the following options is correct?

Options for (1), (2) and (3), respectively are:
(a) (b), (c) and (a)
(b) (c), (a) and (b)
(c) (a), (b) and (c)
(d) (b), (a) and (c)
Answer:
(a) (b), (c) and (a)

Question 5.
Compare the velocities of the wave forms given below, and choose the correct option.

where, v_A, v_B, v_C and v_D are velocities given in (a), (b), (c) and (d), respectively.
(a) v_A > v_B > v_D > v_C
(b) v_A < v_B < v_D < v_C
(c) v_A = v_B = v_D = v_C
(d) v_A > v_B = v_D > v_C
Answer:
(c) v_A = v_B = v_D = v_C

Hint:
The amplitude of the velocity waves is same in all four cases.

Question 6.
A sound wave whose frequency is 5000 Hz travels in air and then hits the water surface. The ratio of its wavelengths in water and air is:
(a) 4.30
(b) 0.23
(c) 5.30
(d) 1.23
Answer:
(a) 4.30

Hint:
Frequency = 5000 Hz
Speed of sound in air v_air = 332 m/s
Speed of sound in water v_water = 1450 m/s

Question 7.
A person standing between two parallel hills fires a gun and hears the first echo after t₁ sec and the second echo after t₂ sec. The distance between the two hills is:
(a) \(\frac{v\left(t_{1}-t_{2}\right)}{2}\)
(b) \(\frac{v\left(t_{1} t_{2}\right)}{2\left(t_{1}+t_{2}\right)}\)
(c) v(t₁ + t₂)
(d) \(\frac{v\left(t_{1}+t_{2}\right)}{2}\)
Answer:
(d) \(\frac{v\left(t_{1}+t_{2}\right)}{2}\)

Hint:
For first echo 2d₁ = vt₁
For second echo 2d₂ = vt₂
∴ d = d₁ + d₂ = \(\frac{v\left(t_{1}+t_{2}\right)}{2}\)

Question 8.
An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency 83 Hz. Then the length of the air column is:
(a) 1.5 m
(b) 0.5 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m

Hint:
l = \(\frac { λ }{ 4 }\); f = 83 Hz
v = 332 m/s
λ = \(\frac { v }{ f }\)
= \(\frac { 332 }{ 83 }\)
= 4 m
l = \(\frac { λ }{ 4 }\) = \(\frac { 4m }{ 4 }\) = 1m = 1.0 m

Question 9.
The displacement y of a wave travelling in the x direction is given by y = (2 x 10^-3) sin (300t – 2x + \(\frac { π }{ 4 }\)), where x and y are measured in metres and t in second. The speed of the wave is:
(a) 150 ms^-1
(b) 300 ms^-1
(c) 450 ms^-1
(d) 600 ms^-1
Answer:
(a) 150 ms^-1

Hint:
The given equation is similar to

Question 10.
Consider two uniform wires vibrating simultaneously in their fundamental notes. The tensions, densities, lengths and diameter of the two wires ,are in the ratio 8:1, 1:2, (x : y) and 4 : 1 respectively. If the note of the higher pitch has a frequency of 360 Hz and the number of beats produced per second is 10, then the value of (x : y) is:
(a) 36 : 35
(b) 35 : 36
(c) 1 : 1
(d) 1 : 2
Answer:
(a) 36 : 35

Hint:
No. of beats = 10
Frequency of higher pitch f₁ = 360 Hz
Frequency of lower pitch f₂ = 360 – 10 = 350 Hz
x : y = Ratio of length
l ∝ f
l₁ : l₂ = x : y = f₁ : f₂
= 360 : 350
= 36 : 35

Question 11.
Which of the following represents a wave:
(a) (x – vt)³
(b) x(x + vt)
(c) \(\frac { 1 }{ (x +vt) }\)
(d) sin(x + vt)
Answer:
(d) sin(x + vt)

Hint:
Wave must have a wave function.
Since y = a sin ωt

Question 12.
A man sitting on a swing which is moving to an angle of 60° from the vertical is blowing . a whistle which has a frequency of 2.0 k Hz. The whistle is 2.0 m from the fixed support point of the swing. A sound detector which detects the whistle sound is kept in front of the swing. The maximum frequency the sound detector detected is:
(a) 2.027 kHz
(b) 1.914 kHz
(c) 9.14 kHz
(d) 1.011 kHz
Answer:
(a) 2.027 kHz

Hint:

Question 13.
Let y = \(\frac{1}{1+x^{2}}\) at t = 0 be the amplitude of the wave propagating in the positive x-direction. At t = 2s, the amplitude of the wave propagating becomes y = \(\frac{1}{1+(x-2)^{2}}\). Assume that the shape of the wave does not
change during propagation. The velocity of the wave is:
(a) 0.5 ms^-1
(b) 1.0 ms^-1
(c) 1.5 ms^-1
(d) 2.0 ms^-1
Answer:
(b) 1.0 ms^-1

Hint:

Question 14.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?

Answer:

Hint:
Speed varies exponentially with height h.

Question 15.
An organ pipe A closed at one end is allowed to vibrate in its first harmonic and another pipe B open at both ends is allowed to vibrate in its third harmonic. Both A and B are in resonance with a given tuning fork. The ratio of the length of A and B is:
(a) \(\frac { 8 }{ 3 }\)
(b) \(\frac { 3 }{ 8 }\)
(c) \(\frac { 1 }{ 6 }\)
(d) \(\frac { 1 }{ 2 }\)
Answer:
(d) \(\frac { 1 }{ 2 }\)

Hint:
First harmonic of a closed organ pipe
L_C = \(\frac { 3 λ }{ 4 }\)
Third harmonic of an open organ pipe
L₀ = \(\frac { 3 λ }{ 2 }\)
\(\frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{O}}}\) = \(\frac { 3 π }{ 4 }\) x \(\frac { 3 }{ 3λ }\)
= \(\frac { 2 }{ 4 }\)
= \(\frac { 1 }{ 2 }\)
∴ \(\frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{O}}}\) = \(\frac { 1 }{ 2 }\)

II. Short Answer Questions:

Question 1.
What is meant by waves?
Answer:
The disturbance which carries energy and momentum from one point in space to another point in space without the transfer of the medium is known as a wave.

Question 2.
Write down the types of waves.
Answer:
Waves can be classified into two types –

• Transverse waves
• Longitudinal waves

Question 3.
What are transverse waves? Give one example.
Answer:
In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)

Question 4.
What are longitudinal waves? Give one example.
Answer:
The direction of vibration of particles in a medium is parallel to the direction of propagation of the wave.
Example: Sound waves traveling in air.

Question 5.
Define wavelength.
Answer:
For transverse waves, the distance between two neighbouring crests or troughs is known as the wavelength. For longitudinal waves, the distance between two neighbouring compressions or rarefactions is known as the wavelength. The SI unit of wavelength is meter.

Question 6.
Write down the relation between frequency, wavelength and velocity of a wave.
Answer:
Velocity of the wave is v = λf.

Question 7.
What is meant by the interference of waves?
Answer:
Interference is a phenomenon in which two waves moving in the same direction superimpose to form a resultant wave of greater, lower or the same amplitude.

Question 8.
Explain the beat phenomenon.
Answer:
When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two. sources, then their difference in frequency gives the beat frequency. Number of beats per second n = |f₁ – f₂| per second

Question 9.
Define intensity of sound and loudness of sound.
Answer:
“Loudness is defined as the degree of the sensation of sound produced in the ear or the perception of sound by the listener”. The intensity of sound is defined as “the sound power transmitted per unit area placed normal to the propagation of sound wave”.

Question 10.
Explain Doppler Effect.
Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound propagates, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

Question 11.
Explain red. shift and blue shift in Doppler Effect.
Answer:
If the spectral lines of the star are found to shift towards red end of the spectrum (called redshift) then the star is receding away from the Earth. If the spectral lines of the star are found to shift towards the blue end of the spectrum (called blue shift) then the star is approaching Earth.

Question 12.
What is meant by end correction in resonance air column apparatus?
Answer:
Since the antinodes are not exactly formed at the open end, we have to include a correction, called end correction e. It is calculated, by assuming that antinode is formed at a small distance above the open end e = 0.3 d.

Question 13.
Sketch the function y = x + a. Explain your sketch.
Answer:

Question 14.
Write down the factors affecting velocity of sound in gases.
Answer:
(i) Effect of pressure: Speed of sound is independent of pressure for a fixed temperature.

(ii) Effect of temperature: Speed of sound varies directly to the square root of temperature in kelvin. Speed of sound in air increases by 0.61 ms^-1 per degree celcius rise in temperature.

(iii) Effect of density: Velocity of sound in a gas is inversely proportional to the square root of the density of the gas.

(iv) Effect of moisture (humidity): Speed of sound increases with rise in humidity.

(v) Effect of wind: In the direction along the wind blowing, the speed of sound to wind blowing, the speed of sound decreases.

Question 15.
What is meant by an echo? Explain.
Answer:
An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces. The minimum distance from a sound-reflecting wall to hear an echo at 20°C is 17.2 meter.

III. Long Answer Questions:

Question 1.
Discuss how ripples are formed in still water.
Answer:
Suppose we drop a stone in a trough of still water, we can see a disturbance produced at the place where the stone strikes the water surface. We find that this disturbance spreads out (diverges out) in the form of concentric circles of ever-increasing radii (ripples) and strikes the boundary of the trough. This is because some of the kinetic energy of the stone is transmitted to the water molecules on the surface.

Actually, the particles of the water (medium) themselves do not move outward with the disturbance. This can be observed by keeping a paper strip on the water surface. The strip moves up and down when the disturbance (wave) passes on the water surface. This shows that the water molecules only undergo vibratory motion about their mean positions.

Question 2.
Briefly explain the difference between traveling waves and standing waves.
Answer:
Progressive waves:

1. Crests and troughs ate formed in transverse progressive waves. Compression and rarefaction are formed in longitudinal progressive waves. These waves move forward or backward in a medium i.e., they will advance in a medium with a definite velocity.
2. All the particles in the medium vibrate in such a way that the amplitude of the vibration for all particles is the same.
3. These waves carry energy while propagating.

Stationary waves:

1. Crests and troughs are formed in transverse stationary waves. Compression and rarefaction are formed in longitudinal stationary waves. These waves neither move forward nor backward in a medium i.e., they will not advance in a medium.
2. Except at nodes, all other particles of the medium vibrate such that the amplitude of vibration is different for different particles. The amplitude is minimum or zero at nodes and maximum at antinodes.
3. These waves do not transport energy.

Question 3.
Show that the velocity of a traveling wave produced in a string is v = \(\sqrt{\frac{T}{\mu}}\)
Answer:
Let us consider an elemental segment in the string as shown in the Figure. Let A and B be two points on the string at an instant of time. Let dl and dm be the length and mass of the elemental string, respectively. By definition, linear mass density, μ is

μ = \(\frac { dm }{ dl }\) … (1)
dm = μ dl … (2)
Consider an elemental string AB having a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O as shown in Figure. The angle 0 can be written in terms of arc length and radius as \(\frac { dl }{ R }\) = θ. The centripetal acceleration supplied by the tension in the string is
a_cp = \(\frac{v^{2}}{\mathrm{R}}\) … (3)
Then, centripetal force can be obtained when mass of the string (dm) is included in equation (3)
F_cp = \(\frac{(dm)v^{2}}{\mathrm{R}}\) … (4)
Experienced by elemental string can be calculated by substituting equation (2) in equation (4) we get the centripetal force as
\(\frac{(dm)v^{2}}{\mathrm{R}}\) = \(\frac{\mu v^{2} d l}{R}\) … (5)
The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. T can be resolved into horizontal component T cos (\(\frac { θ }{ 2 }\)) and vertical component T sin (\(\frac { θ }{ 2 }\)).

The horizontal components at A and B are equal in magnitude but opposite in direction. Hence, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial force F_r is
F_r = 2Tsin(\(\frac { θ }{ 2 }\)) … (6)
Since the amplitude of the wave is very small when it is compared with the length of the string, the sine of small angle is approximated
as sin (\(\frac { θ }{ 2 }\)) ≈ \(\frac { θ }{ 2 }\). Hence, equation (6) can be written as,
F_r = 2T x \(\frac { θ }{ 2 }\) = Tθ … (7)
But θ = \(\frac { dl }{ R }\), therefore substituting in equation (7), we get
F_r = T \(\frac { dl }{ R }\) … (8)
Applying Newton’s second law to the elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. By equating equation (5) and equation (8), we get,
T \(\frac { dl }{ R }\) = μv²dl
v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) measured in ms^-1

Question 4.
Describe Newton’s formula for velocity of sound waves in air and also discuss the Laplace’s correction.
Answer:
Newton assumed that when sound propagates in air, the formation of compression and rarefaction takes place in a very slow manner so that the process is isothermal in nature. It is found that, the heat produced during compression (pressure increases, volume decreases), and heat lost during rarefaction (pressure decreases, volume increases) occur over a period of time such in a way that the temperature of the medium remains constant. Hence, by treating the air molecules to form an ideal gas, the changes in pressure and volume obey Boyle’s law, Mathematically
PV= Constant …(1)
Differentiating equation (1), we get
PdV + Vdp = 0
(or) P = – V\(\frac { dp }{ dv }\) = B_T … (2)
where, B_T is an isothermal bulk modulus of air.
v = \(\sqrt{\frac{\mathrm{B}}{\rho}}\) … (3)
Substituting equation (2) in equation (3), the speed of sound in air is
v_T = \(\sqrt{\frac{\bar{B}_{\mathrm{T}}}{\rho}}\) = \(\sqrt{\frac{\mathrm{P}}{\rho}}\)
Since P is the pressure of air whose value at NTP (Normal Temperature and Pressure) is 76 cm of mercury, we have
P = (0.76 x 13.6 x 10³ x 9.8)Nm^-2
ρ = 1.293 kg m^-3.
Here p is density of air. Then the speed of sound in air at normal temperature and pressure (NTP) is

But the speed of sound in air at 0°C is experimentally observed as 332 ms^-1 that is close up to 16% more than theoretical value.

Laplace correction: Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a poor conductor of heat. Since, temperature is no longer considered as a constant here, propagation of sound is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), that is
PV’ = Constant … (4)
where, γ = \(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}\), that is the ratio between specific heat at constant pressure and specific heat at constant volume.
Differentiating equation (4) on both the sides, we get
V^γ dP + P (γV^γ-1 dV) = 0
or, γP = – V\(\frac { dp }{ dv }\) = B_A … (5)
where, B_A is the adiabatic bulk modulus of air.
v = \(\sqrt{\frac{\mathrm{B}}{\rho}}\)
Now, substituting equation (5) in equation (6), the speed of sound in air is

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_A = (\(\sqrt{1.4}\))(280ms^-1) = 331.30 ms^-1, which is very much closer to experimental data.

Question 5.
Write short notes on reflection of sound waves from plane and curved surfaces.
Answer:
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. When a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts. Hence, the reflected wave front on the plane surface is also spherical, such that its centre of curvature can be treated as the image of the sound source (virtual or imaginary loud speaker) that can be assumed to be at a position behind the plane surface. These are shown in figures.

The behaviour of sound is different when it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) that is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones.

We know that any surface (smooth or rough) can absorb sound. For instance, the sound produced in a big hall or auditorium or theatre is absorbed by the walls, ceilings, floor, seats etc. To avoid such losses, a curved sound board (concave board) is kept in front of the speaker, in such a way that the board reflects the sound waves of the speaker towards the audience. This method will minimize the spreading of sound waves in all possible direction in that hall it also enhances the uniform distribution of sound throughout the hall. Hence a person sitting at any position in that hall can hear the sound without any disturbance.

Question 6.
Briefly explain the concept of the superposition principle.
Answer:
When two or more waves in a medium overlap, their total displacement is the vector sum of the individual displacements.
Let us consider two functions which characterize the displacement of the waves, for example,
y₁ = A₁ sin(kx – ωt)
and y₂ = A₂ sin(kx – ωt)
Since both y₁ and y₂ satisfy the wave equation (solutions of wave equation) then their algebraic sum
y = y₁ + y₂
also satisfies the wave equation. It is meant that the displacements are additive.

Question 7.
Explain how the interference of waves is formed.
Answer:
Interference is a phenomenon in which two waves superimpose to form a resultant wave of greater, lower, or the same amplitude.

Let us consider two harmonic waves having identical frequencies, constant phase difference φ, and same waveform (can be treated as coherent source), but having amplitudes A₁ and A₂, then
y₁ = A₁ sin(kx – ωt) … (1)
y₂ = A₂ sin(kx – ωt) … (2)
Suppose they move simultaneously in a particular direction, then interference occurs (i.e., the overlap of these two waves). Mathematically y = y₁ + y₂ … (3)
Hence by substituting equation (1) and equation (2) in equation (3), we get

By squaring and adding equation (5) and (6), we get,
A² = A₁² + A₂² + 2A₁A₂cosφ … (8)
Since, intensity is square of the amplitude (I – A²), we get,
I = I₁ + I₂ + 2\(\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \varphi\) … (9)
This means the resultant intensity at any point depends on the phase difference at that point.

Question 8.
Describe the formation of beats.
Answer:
When two or more waves superimpose each other with slightly different frequencies, then a sound having periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency.
Number-of beats per second
n = \(\left|f_{1}-f_{2}\right|\) Per second

Question 9.
What are stationary waves? Explain the formation of stationary waves and also write down the characteristics of stationary waves.
Answer:
When the wave hits the rigid boundary it bounces back to the original medium and can interfere with the original waves. A pattern is formed, that is known as standing waves or waves stationary. Let us consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is
y₁ = A sin(kx – ωt) … (1)
(waves move toward right)
The displacement of the second wave (reflected wave) is
y₂ = A sin(kx + ωt) … (2)
(waves move toward left)
both will interfere with each other by the principle of superposition, the net displacement is
y = y₁ + y₂ … (3)
By substituting equation (1) and equation (2) in equation (3), we get
y = \(\left\{\begin{array}{l}
\mathrm{A} \sin (k x-\omega t) \\
+\mathrm{A} \sin (k x+\omega t)
\end{array}\right.\) … (4)
Using trigonometric identity, we rewrite equation (4) as
y(x, t) = 2A cos(ωt) sin(kx) … (5)
This represents a stationary wave or standing wave, It is meant that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. In addition, the displacement of the particle in equation (5) can be written in more compact form,
y(x, t) = A’cos(cot)
where, A’ = 2A sin(Ax). It is implied implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which
sin(kx) = 1
⇒ kx = \(\frac { π }{ 2 }\), \(\frac { 3π }{ 2 }\), \(\frac { 5π }{ 2 }\), … = nπ
where m takes half-integer or half-integral values. The position of maximum amplitude is known as an antinode. Expressing wave number in terms of wavelength, let us represent the anti-nodal positions as
x_m = (\(\frac { 2m+1 }{ 2 }\)) \(\frac { λ }{ 2 }\) … (6)
where, m = 0, 1, 2 ….
For m = 0 we have maximum at
x₀ = \(\frac { λ }{ 2 }\)
For m = 1 we have maximum at
x₀ = \(\frac { 3λ }{ 4 }\)
For m = 2 we have maximum at
x₂ = \(\frac { 5λ }{ 4 }\)
and so on.
The distance between two successive antinodes can be computed by,

Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting
sin(kx) = 0
⇒ kx = 0, π, 2π, 3π, … = nπ
where n takes integer or integral values. It is noted that the elements at these points do not vibrate (not move), and the points are called nodes. The n^th nodal positions is given by,
x_n = n\(\frac { λ }{ 2 }\) … (7)
where, n = 0, 1, 2, …
For n = 0 we have minimum at
x_o = 0
For n = 1 we have minimum at
x₁ = \(\frac { λ }{ 2 }\)
For n = 2 we have maximum at
x₂ = λ
and so on.
The distance between any two successive nodes can be calculated as
x_n – x_n-1 – n\(\frac { λ }{ 2 }\) – (n-1)\(\frac { λ }{ 2 }\) = \(\frac { λ }{ 2 }\)

Characteristics of stationary waves:

• Stationary waves are characterized by the confinement of a wave disturbance between two rigid boundaries. It is meant that the wave does not move forward or backward in a medium (does not advance), it remains steady at its place. Hence, they are called “stationary waves or standing waves”.
• Certain points in the region in which the wave exists have maximum amplitude, called anti-nodes. At certain points, the amplitude is minimum or zero, called nodes.
• The distance between two consecutive nodes (or) anti-nodes is \(\frac { λ }{ 2 }\).
• The distance between a node and its neighboring anti-node is \(\frac { λ }{ 4 }\).
• The transfer of energy along the standing wave is zero.

Question 10.
Discuss the law of transverse vibrations in stretched strings.
Answer:
There are three laws of transverse vibrations of stretched strings that are given as follows:
(i) The law of length: For a given wire with tension T (which is fixed) and mass per unit length µ. (fixed) the frequency varies inversely with the vibrating length. Therefore,
f ∝ \(\frac { 1 }{ f }\) ⇒ f = \(\frac { C }{ l }\)
⇒ l x f = C, when C is a constant.

(ii) The law of tension: For a given vibrating length l (fixed) and mass per unit length µ (fixed) the frequency varies directly with the square root of the tension T,
f ∝ \(\sqrt{T}\)
⇒ f = A\(\sqrt{T}\), when A is a constant.

(iii) The law of mass: For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length μ,
f ∝ \(\frac{1}{\sqrt{\mu}}\) when B is a constant.

Question 11.
Explain the concepts of frequency, harmonics and detail.
Answer:
(i) The fundamental frequency is the lowest natural frequency.

(ii) If natural frequencies are written as integral multiples of the fundamental frequency, then the frequencies are said to be in harmonics. So, the first harmonic is v₁ = v₁, (the fundamental frequency is called first harmonics), the second harmonics is v₂ = 2v₁ the third harmonics is v₃ = 3v_1, and so on.

(iii) The frequency higher than the fundamental frequency can be produced by blowing air strongly at the open end. Such frequencies are called overtones.

First overtone, (f₂ = 3f₁) since here, the frequency is three times the fundamental frequency it is called third harmonic. Second overtone, (f₃ = 5f₁) and since n = 5 here, this is called fifth harmonic.

Question 12.
What is a sonometer? Give its construction and working. Explain how to determine the frequency of tuning fork using a sonometer.
Answer:
Sonometer is used for sound-related measurements. Using this device, The following quantities can be determined.
(i) The frequency of the tuning fork or frequency of the alternating current.
(ii) The tension in the string.
(iii) The unknown hanging mass.

Construction: The sonometer is made up of a hollow box that is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in Figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increase the tension of the wire. Two adjustable wooden knives are put over the board, and their positions are adjusted to change the vibrating length of the stretched wire.

Procedure: A transverse stationary or standing wave is produced. So, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed.
If the length of the vibrating element is l then
l = \(\frac { λ }{ 2 }\) ⇒ λ = 2l
Let f be the frequency of the vibrating element, T the tension in the string, and p the mass per unit length of the string.
Then using the the equation v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) measured in ms^-1, we get,
f = \(\frac { v }{ λ }\) = \(\frac { 1 }{ 2l }\)\(\sqrt{\frac{\mathrm{T}}{\mu}}\) in Hertz.
Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length μ,

Question 13.
Write short notes on intensity and loudness.
Answer:
The intensity of sound is defined as “the sound power transmitted per unit area taken normal to the propagation of the sound wave”. The intensity of sound is inversely proportional to the square of the distance from the source.

This is known as the inverse square law of sound intensity. Two sounds with the same intensities need not have the same loudness. For example, the sound heard during the explosion of balloons in a silent closed room is very loud when compared to the same explosion happening in a noisy market.

If the intensity of sound is increased then loudness also increases. Loudness depends on both intensities of the sound wave and sensitivity of the ear.

Question 14.
Explain how overtones are produced in a,
(a) closed organ pipe
(b) Open organ pipe
Answer:
(a) Closed organ pipes: If one end of a pipe is closed, the wave reflected at this closed-end is 180° out of phase with the incoming wave. So, there is no displacement of the particles at the closed end. Hence, nodes are formed at the closed end and anti-nodes are formed at the open end.

Consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the open end and node at the closed end. From the above figure, let L be the length of the tube and the wavelength of the wave produced. For the fundamental mode of vibration, we have,
– L = \(\frac{\lambda_{1}}{4}\) (or)
Wave length λ₁ = 4L
The frequency of the note emitted is
f₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac { v }{ 4L }\)
which is called the fundamental note.
The frequencies higher than the fundamental frequency can be produced by blowing air strongly at the open end. Such frequencies are called overtones.
The second mode of vibration in open pipes is shown in the figure.

is called the first overtone, since here, the frequency is three times the fundamental frequency it is called third harmonic. The Figure shows the third mode of vibration having three nodes and three anti-nodes.

is called the second overtone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and the frequency of the n^th harmonic is f_n = (2n+l)f₁. Hence, the frequencies of harmonics are in the ratio,
f₁ : f₂ : f₃ : f₄ = 1 : 3 : 5 : 7 : …

(b) Open organ pipes:
The flute is an example of an open organ pipe. It is a pipe with both ends are opened. At both open ends, anti-nodes are formed. Consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe.

From above Figure, if L be the length of the tube, the wavelength of the wave produced is given by
L=\(\frac{\lambda_{1}}{2}\) (or) λ₁ = 2L
The frequency of the note emitted is
f₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac { v }{ 2L }\)
That is called the fundamental note.
The frequencies higher than the fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.

The second mode of vibration in open pipes is shown in figure. It has two nodes and three anti-nodes,
∴ L = λ₂ (or) λ₂ = L
The frequency,
f₂ = \(\frac{v}{\lambda_{2}}\) = \(\frac { v }{ L }\) = 2 x \(\frac { v }{ 2L }\) = 2f₁
is called first over tone. Since n = 2 here, it is called the second harmonic.

The Figure above shows the third mode of vibration having three nodes and four anti-nodes
∴ L = \(\frac { 3 }{ 2 }\)λ₃ (or) λ₃ = \(\frac { 2L }{ 3 }\)
The frequency,
f₃ = \(\frac{v}{\lambda_{3}}\) = \(\frac { 3v }{ 2L }\) = 3₁
is called second over tone. Since n = 3 here, it is called the third harmonic.
Hence, the open organ pipe has all the harmonics, and the frequency of the n^th harmonic is f_n = nf₁.
Hence, the frequencies of harmonics are in the ratio f₁ : f₂ : f₃ : f₄ = 1 : 2 : 3 : 4 : …

Question 15.
How will you determine the velocity of sound using the resonance air column apparatus?
Answer:

The resonance air column apparatus is used to measure the speed of sound in air at room temperature.

Construction:
(i) It consists of a cylindrical glass tube of one-meter length whose one end A is open and another end B is connected to the water reservoir R through a rubber tube as shown in Figure.

(ii) This cylindrical glass tube is mounted on a vertical stand with a scale attached to it. The tube is partially filled with water and the water level can be adjusted by raising or lowering the water in the reservoir R.

(iii) The surface of the water will act as a closed-end and others as the open end. Hence, it behaves like a closed organ pipe, forming nodes at the surface of water and antinodes at the closed end.

(iv) When a vibrating tuning fork is brought near the open end of the tube, longitudinal waves are formed inside the air column.

(v) These waves move downward as shown in Figure, and reach the surfaces of water and get reflected and produce standing waves.

(vi) The length of the air column is varied by changing the water level until a loud sound is produced in the air column.

(vii) At this particular length the frequency of waves in the air column resonates with the frequency of the tuning fork (natural frequency of the tuning fork).

(viii) At resonance, the frequency of sound waves produced is equal to the frequency of the tuning fork. This will occur only when the length of the air column is proportional to (\(\frac { 1 }{ 4 }\))^th of the wavelength of the sound waves produced. \(\frac { 1 }{ 4 }\)λ = L₁ … (1)
But since the antinodes are not exactly formed at the open end, a Correction is to include. It is called end correction e, by assuming that the antinode is formed at some small distance above the open end. Including this end correction, the first resonance is \(\frac { 1 }{ 4 }\)λ = L₁ + e … (2)
Now the length of the air column is increased to get the second resonance. Let L₂ be the length at which the second resonance occurs. Again taking end correction into account, we have
\(\frac { 3 }{ 4 }\)λ = L₂ + e … (3)
In order to avoid end correction, let us make the difference of equation (3) and equation (2), we get
\(\frac { 3 }{ 4 }\)λ – \(\frac { 1 }{ 4 }\)λ = (L₂ + e) – (L₁ + e)
⇒ \(\frac { 1 }{ 2 }\)λ = L₂ – L₁ = ∆L
⇒ λ = 2∆L
The speed of the sound in air at room temperature can be calculated by using the formula
v = fλ = 2f∆L.

Question 16.
What is meant by the Doppler effect? Discuss the following cases.
(i) Source in motion and Observer at rest

• Source moves towards the observer
• Source moves away from the observer

(ii) Observer in motion and Source at rest.

• The observer moves towards Source
• Observer resides away from the Source

(iii) Both are in motion

• Source and Observer approach each other
• Source and Observer resides from each other
• Source chases Observer
• Observer chases Source

Answer:
When the source and the observer are in relative motion with respect to each other and to the medium in which sound is propagated, the frequency of the sound wave observed is different from the frequency of the source. This phenomenon is called Doppler Effect.

(i) Source in motion and the observer at rest:
(a) Source moves towards die observer:

Suppose a source S moves to the right (as shown in Figure) with a velocity v_s and let the frequency of the sound waves produced by the source be f_s. It is assumed that the velocity of sound in a medium is v.

The compression (sound wavefront) produced by the source S at three successive instants of time are shown in the Figure. When S is at position x₁ the compression is at C₁. When S is at position x₂, the compression is at C₂ and similarly for x₃ and C₃.

It is assumed that if C₁ reaches the observer’s position A then at that instant C₂ reaches point B and C₃ reaches point C as shown in the Figure. Obviously it is seen that the distance between compressions C₂ and C₃ is shorter than the distance between C₁ and C₂.

It is meant that the wavelength decreases when the source S moves towards the observer O. But frequency is inversely related to wavelength and therefore, frequency increases.
Calculation:
Let λ be the wavelength of the source S as measured by the observer when S is at position x₁ and λ’ be the wavelength of the source observed by the observer when S moves to position x₂.

Then the change in wavelength is ∆λ = λ – λ’ = v_st, where t is the time taken by the source to travel between x₁ and x₂ Therefore,
λ’ = λ – v_st … (1)
But t = \(\frac { λ }{ v }\) … (2)
On substituting equations (2) in equation (1), we get.
λ’ = λ(1 – \(\frac{v_{s}}{v}\))
Since frequency is inversely proportional to wavelength, we have

Since, \(\frac{v_{s}}{v}\) << 1, by using the binomial expansion and retaining only first order in \(\frac{v_{s}}{v}\), we get
f ‘ = f(1 + \(\frac{v_{s}}{v}\)) … (4)

(b) Source moves away from the observer Since the velocity of the source is opposite in direction when compared to case (a), hence by changing the sign of the velocity of the source in the above case i.e., by substituting (v_s → – v ) in equation (1), we get
f ‘= \(\frac{f}{\left(1+\frac{v_{s}}{v}\right)}\) … (5)
Using binomial expansion again, we get
f ‘= f(1 – \(\frac{v_{s}}{v}\)) … (6)

(ii) Observer in motion and source at rest:
(a) Observer moves towards Source:

We can assume that the observer O moves towards the source S with velocity v_o. The source S is at rest and the velocity of sound waves (with respect to the medium) produced by the source is v.
From the Figure, It is observed that both v_o and v are in opposite direction. Then, their relative velocity is v_r = v + v_o. The wavelength of the sound wave is λ = \(\frac { v }{ f }\), which means the frequency observed by the observer O is f ‘ = \(\frac{v_{r}}{\lambda}\). Then

(b) Observer recedes away from the Source: If the observer O is moving away (receding away) from the source S, then velocity v₀ and v move in the same direction. Hence, their relative velocity is v_r = v – v₀. Hence, the frequency observed by the observer O is

(iii) Both are in motion:
(a) Source and observer approach each other:

Let v_s and v_o be the respective velocities of source and observer approaching each other as shown in Figure. In order to calculate the apparent frequency observed by the observer, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, the dummy (observer) observes the apparent frequency due to approaching source as given in equation f ‘ =\(\frac{f}{\left(1-\frac{v_{s}}{v}\right)}\)
f_d = \(\frac{f}{\left(1-\frac{v_{s}}{v}\right)}\) … (1)
The true observer approaches the dummy from the other side at that instant of time. Since the source (true source) comes in a direction opposite to the true observer, the dummy (source) is treated as a stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), f’ = f(1 + \(\frac{v_{0}}{v}\)).

Since this is true for any arbitrary time, therefore, comparing equation (1) and equation (2), we get

(b) Source and observer recede from each other:

It is noticed that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute (v_s → – v_s) and (v₀ → – v₀) in equation (3), and therefore, the apparent frequency observed by the observer when the source and observer recede from each other is f’ = \(\left(\frac{v-v_{0}}{v+v_{s}}\right)\)f

(c) Source chases the observer:

Only the observer’s velocity is oppositely directed when compared to case (a). Therefore, substituting (v₀ → – v₀) in equation (3), we get f’ = \(\left(\frac{v-v_{0}}{v-v_{s}}\right)\)f

(d) Observer chases the source:

Only the source velocity is oppositely directed when compared to case (a). Therefore, substituting (v_s → – v_s) in equation (3), we get f’ = \(\left(\frac{v+v_{0}}{v+v_{s}}\right)\)f

IV. Numerical Problems:

Question 1.
The speed of a wave in a certain medium is 900 m/s. If 3000 waves pass over a certain point of the medium in 2 minutes, then compute its wavelength?
Answer:
Given:

Question 2.
Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.
Answer:
Number of molecules of helium = 2
Number of molecules of oxygen = 4
When helium and oxygen are mixed, hence the molecular weight of the mixture of gases is given by

In addition, helium is monoatomic,
\(\mathrm{C}_{v_{2}}\) = \(\frac { 2R }{ 2 }\)
Oxygen is diatomic \(\mathrm{C}_{v_{1}}\) = \(\frac { 5R }{ 2 }\)

Ratio of specific heat capacitors of a mixture of gases is

According to Laplace, the speed of sound in a gas is

∴ The speed of sound = 400.9 m/s

Question 3.
A ship in a sea sends SONAR waves straight down into the seawater from the bottom of the ship. The signal reflects from the deep bottom bedrock and returns to the ship after 3.5 s. After the ship moves to 100 km it sends another signal which returns back after 2 s. Calculate the depth of the sea in each case and also compute the difference in height between two cases.
Answer:
In the first case
time = 3.5s
Velocity sound in sea water = 1450 m/s
Distance 2d = v x t
= 1450 x 3.5 = 5,075 m
Depth of the sea d = \(\frac { 5075 }{ 2 }\) = 2537.5 m
In the second case,
time = 2s
Velocity of sound in water = 1450 m/s
Distance 2d = vt
= 1450 x 2
= 2900 m
Depth of the sea d = \(\frac { 2900 }{ 2 }\) = 1450
Difference ∆d = 2537.5 – 1450 = 1087.5
= 1087.5 m

Question 4.
A sound wave is transmitted into a tube as shown in the figure. The sound wave splits into two waves at point A which recombine at point B. Let R be the radius of the semi-circle which is varied until the first minimum. Calculate the radius of the semi-circle if the wavelength of the sound is 50.0 m.
Answer:

Path difference = πR – 2R
[Here AB = 2R; Semicircle path = πR]
Formula:
Path difference = (2n – 1)\(\frac { λ }{ 2 }\) (for minimum n – 1)

Question 5.
N tuning forks are arranged in order of increasing frequency and any two successive tuning forks give n beats per second when sounded together. If the last fork gives double the frequency of the first (called as octave), Show that the frequency of the first tuning fork is f = (N – 1)n.
Answer:
f,f + n,f+ 2n, … f + (m – 1)n
f + (N – 1)n = 2f
n = \(\frac { f }{ N-1 }\)
(or) f = n(N – 1)

Question 6.
Let the source propagate a sound wave whose intensity at a point (initially) be I. Suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point?
Answer:
Given:

Question 7.
Consider two organ pipes of the same length in which one organ pipe is closed and another organ pipe is open. If the fundamental frequency of closed pipe is 250 Hz. Calculate the fundamental frequency of the open pipe.
Answer:

Formula:

Question 8.
Police in a siren car moving with a velocity of 20 ms^-1 chases a thief who is moving in a car with a velocity v₀ms^-1. The police car sounds at frequency 300 Hz, and both of them move towards a stationary siren of frequency 400 Hz. Calculate the speed at which the thief is moving.
Answer:
Velocity of sound v = 330 m/s
Velocity of a police siren car v_s = 20 m/s
Frequency of a police siren car f = 300 Hz
Let the velocity of the thief be v m/s
The frequency of police siren heard by a thief is
f₁ = \(\left(\frac{330-v}{330-20}\right)\) x 300
= \(\left(\frac{330-v}{310}\right)\) x 300 Hz
Frequency of stationary siren = 400 Hz
Frequency of stationaiy siren heard by a thief
f₂ = \(\left(\frac{330+v}{330}\right)\) x 400
If there are no beats then

Question 9.
Consider the following function,
(a) y = x² + 2 α tx
(b) y = (x + vt)²
which among the above function can be characterized as a wave?
Answer:
Given:
Formula:
For the function to be a wave function \(\frac { (dy/dx) }{ (dy/dt) }\) should be a constant.
For function (a):

Hence, function
(a) does not describe a wave.
(b) satisfies wave function.

V. Conceptual Questions:

Question 1.
Why is it that transverse waves cannot be produced in a gas? Can the transverse waves be produced in solids and liquids?
Answer:
They travel in the form of crests and troughs and so involve a change in shape. They can be produced in a medium that has elasticity in shape. As gas has no elasticity of shape, hence transverse waves cannot be produced in a gas. Transverse waves can be produced in solids anion the surface of liquids.

Question 2.
Why is the roar of our national animal different from the sound of a mosquito?
Answer:
The pitch of the sound of our national animal lion is higher than that of mosquitoes. In addition, the frequency of sound waves generated by our national animal is more than the frequency of sound waves generated by mosquitoes.

Question 3.
A sound source and listener are both stationary and a strong wind is blowing. Is there a Doppler effect?
Answer:
When both source and listener are stationary, there is no relative motion between the source and the observer. Hence there is no Doppler effect.

Question 4.
In an empty room, why is it that a tone sounds louder than in the room having things like furniture, etc.
Answer:
When a room has furniture reverberation time can be suitably decreased since furniture has a large absorption coefficient of sound. So a tone sounds with lesser amplitude and intensity. Whereas in an empty room, reverberation time will be more than a room having things like furniture.

Question 5.
How do animals sense the impending danger of hurricanes?
Answer:
Hurricane produces a shock wave which has a speed greater than the speed of sound. It travels with a supersonic sound that can be easily sensed by animals using the Doppler effect. The multiple reflections of sound can be easily sensed by animals.

Question 6.
Is it possible to realize whether a vessel kept under the tap is about to fill with water?
Answer:
Yes. The frequency of a note generated by an air column is inversely proportional to its length. Consequently, as the length of the air column decreases, the frequency increases, i.e., the note becomes more shrill. In our case, when a vessel kept under the top is about to fill with water, as the water level rises, the length of the air column in the vessel goes on decreasing and the emitted sound becomes more and more shrill. Hence it is realised.