Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 15 Plant Growth and Development Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

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

Part – I.

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant.
a) 3 inches
b) 6 inches
c) 12 inches
d) 30 inches
Answer:
options are wrong, (correct Ans = 1.5 inches)
Solution:
Step I:
Diameter of the Pulley=6 inches
Radius of the pulley \(=\frac{6}{2}\)= 3 inches
Actual growth in length= Distance travailed by pointer x Radius of the pulley Length of the pointer
=\(\frac{5 \times 3}{10}\) =1.5=1.5.
Answer:
1.5 inches

Question 3.
In unisexual plants, sex can changed by the application of
a) Ethanol
b) Cytokinins
c) ABA
d) Auxin
Answer:
c) ABA

Question 4.
Select the correctly matched one
A) Humanurine i) Auxin-B
B) Corn gram oil ii) GA₃
C) Fungs iii) Abscisic acid II
D) Herring fish sperm iv) Kinetin
E) Unripcrnaizegrains v) AuxinA
F) Young cotton boils vi) Zeatin

Answer:
b) A – v, B – i, C – ii, D – iv, E – vi, F – iii

Question 5.
Seed dormancy allows the plants to:
(a) overcome unfavorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:
Growth in plants can be measured in terms, of

• Increase in length or girth (roots and stems)
• Increase in fresh or dry weight
• Increase in area or volume (fruits and leaves)
• Increase in a number of cells produced.

Question 7.
What is plasticity?
Answer:
Plasticity refers to the environmental heterophylly seen in Butter cup plant (Ranunculus). In this aquatic plant, the leaves in the air is normal, where as the leaves submerged underwater are highly thin and hairy highly adapted to do carbon assimilation Developmental heterophlly seen in the juvenile plant leaves of cotton and corianter. Where the young leaves have a different shape from the mature leaves is not considered as plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

1. Cytokinin promotes cell division in the presence of auxin (IAA).
2. Induces cell enlargement associated with IAA and gibberellins
3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as the Richmond Lang effect.
6. Cytokinin:
   • increases rate protein synthesis
   • induces the formation of inter-fascicular cambium
   • overcomes apical dominance
   • induces the formation of new leaves, chloroplast and lateral shoots.
7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:
Mechanism of photoperiodic induction of flowering.

• The physiological change on flowering due to the relative length of light and darkness (photoperiod) is called Photoperiodism.
• The photoperiod required to induce flowering is called critical day length. Eg. 12 hours in Maryland Mammoth’s Tobacco Xanthium 15.05 hours.

Photoperiodic induction:

• An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants may require one or more inductive cycles for flowering.
• The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction. Example: Xanthium (SDP) -1 inductive cycle and Plantago (LDP) -25 inductive cycles.

Site of photoconductive perception:

• Leaves are the parts that receive photoperiodic stimulus (PPS), again it is only leaves that synthesize floral hormones and translocate them to the apical tip to promote flowering.
• This can be demonstrated by experiments conducted in the Cocklebur plant. Which is an SD plant. The nature of flower-producing stimulus has been elusive so far. It is believed by physiologists that a hormone is responsible for it, Chailakyan (1936) named it as Florigen It is not possible to isolate it.

Procedure	Observation	Inference
1. Take potted plant A and defoliate the plant subject it to SD – a condition	There is no induction of flowering	No leaf to receive stimulus or induction of flowering
2.  Take potted plant B – and defoliate all, except one leaf subject it to SD – condition.	There is the induction of flowering	One leaf is enough to receive stimulus or induction of flowering.
3.  Take potted plant C – and defoliate it and subject it to LD condition	There is no induction of flowering	no leaf to receive stimulus or induction of flowering
4. Take potted plant D and subject all leaves to LD but one leaf to SD	There is the induction of flowering	One leaf is enough to receive an induction in the SD condition

Question 10.
Give a brief account of programmed cell death (PCD).
Answer:
Senescence is controlled by plants’ own genetic program and the death of the plant or plants part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which enclose large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while the other aborts through PCD.

Part-II.

11th Bio Botany Guide Plant Growth and Development Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
The open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
An example of a De-Differentiating cell is ………………
a) Tracheary element
b) shoot apex
c) Cork cambium
d) root apex
Answer:
c) Cork cambium

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4
Choose the natural Auxin of the following
a) Anti Auxin
b) NAA
c) 2.4.D
d) IndoleAcetic Acid (IAA)
Answer:
d) Indole Acetic Acid (IAA)

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
The hormone present in Coconut milk is
a) Gibberellins
b) Ethylene
c) Cytokinin
d) Auxin
Answer:
c) Cytokinin

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Which of the following Phytóhormone does not occur naturally in plants?
a) 2. 4. D
b) GibberellicAcid
c) 6. Furfuryl amino purine
d) IAA
Answer:
a) 2.4.D

Question 9.
Absence of light may lead to the yellowish color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Apical dominance is caused when Auxin
a) Concentration is more than Cytokinins
b) Concentration is less than Cytokinins
c) and Cytokinin concentration are equal
d) and Cytokinin concentration are fluctuating
Answer:
a) Concentration is more than Cytokinins

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Which prevents premature fall of fruit?
a) NAA
b) Ethylene
c) GA₃
d) Zeatin
Answer:
a) NAA

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
The term Auxin was coined by
a) Went
b) Darwin
c) Smith
d) Garner
Answer:
a) Went

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
The term Gibberellin was coined by
a) Went
b) Kurosawa
c) Skoog
d) Yabuta
Answer:
d) Yabuta

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Naphthalene Acetic Acid
Answer:
(d) Naphthalene Acetic Acid

Question 18.
The mineral required for the synthesis of IAA is
a) Copper
b) Magnesium
c) Zinc
d) Boron
Answer:
c) Zinc

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The most widely occurring Cytokinin in plants is
a) ABA
b) NAA
c) TNT
d) IPA
Answer:
d) IPA

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross etal

Question 22.
The term Florigen was coined by
a) Maheswari
b) Chailakyan
c) R Gane
d) Richmond Lang
Answer:
b) Chailakyan

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Which of the following is a bioassay for Cytokinins?
a) Chlorophyll preservation test
b) Dwarf maize Assay
c) Seed germination Assay test
d) Neem cotyledon Assay
Answer:
d) Neem cotyledon Assay

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Avena curvature test as a BioAssay for
a) Auxins
b) GA₃
c) Cytokinin
d) Ethylene
Answer:
a) Auxins

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The Gibberellins have been commercially exploited for
a) increasing the size of grapefruits
b) inducing rooting in stem cuttings
c) breaking the dormancy in seeds
d) production of disease-resistant varieties
Answer:
c) breaking the dormancy in seeds

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from the pentose phosphate pathway.
(iv) ABA has anti-auxin and anti-gibberellin properties.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Biennials can be induced to flower in the first season itself by treatment with
a) Auxin
b) Kinetin
c) GA
d) ABA
Answer:
c) GA

Question 31.
Pea and barley are classified under:
(a) short-day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
Auxin a was isolated from human urine by
a) F.W. went
b) Charles Darwin
c) Kogl and Haugen Smith
d) Denny
Answer:
c) Kogl and Haugen smith

Question 33.
Usually, Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photo neutral condition
(d) short-day condition
Answer:
(d) short-day condition

Question 34.
The most widely occurring Cytokinin in plants is
a) Indole Acetic Acid (LAA)
b) Indole Butyric Acid (IBA)
c) Pentenyl Adenine (IPA)
d) Naphthalene  Acetic Acid (NAA)
Answer:
c) Pentenyl Adenine (IPA)

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
Scientists, those who are connected with Ethylene
(I) Denny
(II) R. Gane
(III) Kurosawa
(IV) Cocken
Options:
a) (I) (II) & (III)
b) (II) (III) & (IV)
c) (I) (II) & (IV)
d) (I) (III) & (IV)
Answer:
c) (I)(II)& (IV)

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increases the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
Day-neutral plants are
a) Sugarcane & Coleus
b) Bryophyllum& Night Jasmine
c) Wheat, rice & Oats.
d) Potato, Tomato & Cotton
Answer:
d) Potato, Tomato & Cotton

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
Xanthium (Cocklebur) requires …………….. hours of light to induce flowering,
a) 12
b) 9
c) 15.05
d) 13.05
Answer:
c) 15.05

Question 41.
The hormone that cannot be isolated
a) IAA
b) ABA
c) NAA
d) Florigen
Answer:
d) Florigen

Question 42.
The term Photoperiodism was coined by
a) Went
b) Butler
c) Gamer
d) Skoog
Answer:
c) Garner

Question 43.
ABA acts as antagonistic to
a) Ethylene
b) Cytokinin
c) Gibberellic acid
d) IAA
Answer:
c) Gibberellic acid

Question 44.
If a short-day plant, flowering is induced by
a) Long nights
b) Photo periods less than 12 hrs
c) Photoperiods shorter than critical value and uninterrupted long night
d) Short photoperiods and interrupted long nights
Answer:
c) Photoperiods shorter than critical value and uninterrupted long night.

Question 45.
Phytochrome is
a) Reddish phytohormone
b) Bluish biliprotein pigment
c) Photoreceptor of apical bud
d) Unstable pigment molecule
Answer:
b) Bluish biliprotein pigment

Question 46.
The growth & ripening is induced by Ethylene in
a) Tropical fruits
b) Temperate fruits
c) Climacteric fruits
d) Nonclimacteric fruits
Answer:
c) Climacteric fruits

Question 47.
The bioassay of ABA was done with
a) Rice
b) Wheat
c) Maize
d) Barley
Answer:
a) Rice

Question 48.
Four types of senescence were recognized by
a) Leopold
b) Gamer
c) Addicott
d) Cocken et al
Answer:
a) Leopold

Question 49.
The final stage of senescence is
a) PCD
b) Scarification
c) Yellowing
d) Abscission
Answer:
d) Abscission

Question 50.
Match & Find out the Correct Answer

Column I	Column II
1.  Yabuta&Sumiki	a) Identified Ethylene
2.  Lethan & Miller	b) Isolated Auxin from Human urine
3.  Cockenetal	c) Isolated and identified Zeatin
4.  Kogi & Haugen Smith	d) Isolated Gibberellin in Crystal form

Answer:
b) D C A B

Question 51.
Match the following and Find the Correct Answer

I. Auxin	a) Bolting
II. ABA	b) Induces Respiration
III. Gibberellin	c) Cell division
IV. Ethylene	d) Weedicide
V. Cytokinin	e) Closure of stomata

Answer:
b) D E A B C

II. Assertion (A) & Reason (R)

Question 52.
a. Both A & R are true and ‘R’ is the correct explanation of A
b. Both A & R are true but ‘R’ is not the correct explanation of A
c. A is true but R is False
d. Both A and ‘R’ are False
Assertion (A): The shoot Apical meristems are the only source of Auxin synthesis
Reason (R): Dormancy of lateral buds over Apical buds is due to Auxin
Answer:
C. A is true but R is False

Question 53.
Assertion (A): Hormones are also called Growth regulator
Reason (R): Hormones promote or inhibit plant growth
Answer:
A. Both Assertion (A) and Reason (R) are true and Reason is the correct explanation of Assertion.

Question 54.
Assertion (A): In many land Mammoth flowering occurred at different times at different latitude
Reason (R): Many land Mammoth is a tobacco variety
Answer:
b. Both Assertion (A) and, Reason (R) are true and Reason is not the correct explanation of Assertion.

III. 2 Mark Questions

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers, and fruits are limited in growth or of determinate or closed-form growth.

Question 2.
Compare between Absolute and Relative growth rates
Answer:

Absolute growth	Relative growth
An increase in the total growth of two organs measured and compared per unit time is called Absolute growth rate	The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

• Lag phase
• Log phase
• Decelerating phase
• Maturation phase

Question 4.
Mention the phase of growth in plants
Answer:
I. Formative phase
II. Elongation phase
III. Maturation phase

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
An increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
What is the Grand period of growth
Answer:
The total period from initial to the final stage of growth is called Grand period of growth.
When plotted against time the growth curve is ‘S’ shaped, (sigma curve) it is also known as Grand Period curie consists of 4 phases

1. Lag,
2. Log,
3. Decelerating,
4. Maturation.

Question 7.
What is meant by the dedifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 8.
Define Phytohormone.
Answer:
The chemical substances synthesized by plants and thus naturally occuring are known as Phytohormones. Eg. Auxin, Gibberellins.
Recently 2 groups – Brassinosteroids, Polyamines were also known to behave like hormones.

Question 9.
Mention any two synthetic auxins.
Answer:

• 2, 4 – Dichloro Phenoxy Acetic Acid (2, 4 – D)
• 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2, 4, 5 – T)

Question 10.
State 3 characteristic features of phytohormones.
Answer:

1. They are produced in root tips and stem tips and leaves (do not have specialized cells or organs for secretion)
2. The transfer of hormones takes place through the conducting system (xylem and phloem)
3. They are required in trace quantities
4. They either promote, inhibit or modify growth.

Question 11.
Name the natural auxins present in plants.
Answer:

• Indole Acetic Acid (IAA)
• Indole Propionic Acid (IPA)
• Indole Butyric Acid (IBA)
• Phenyl Acetic Acid (PAA)

Question 12.
Give the historial significance of Agent Orange
Answer:

• Mixture of two phenoxy herbicides – 2.4. D and 2.4.5 T together known as Agent orange.
• This Agent orange, was used by USA in Vietnam war as chemical warfare weapon to defoliate forests in Vietnam.

Question 13.
Does the trimming of plants in gardens have any scientific explanation?
Answer:

• Yes, trimming of plants removes apical buds and hence apical dominance is prevented the lateral buds sprout and give a beautiful bushy appearance and aesthetic value.
• Also in tea estates, this trimming develops more lateral branches and more tea leaves thus it has commercial significance.

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is bolting?
Answer:

• When treated with Gibberellins the rose the plants (genetic dwarf) exhibit excessive internodal growth.
• This sudden elongation of a stem followed by flowering is called bolting.

Question 16.
What is Richmond Lang effect?
Answer:
Application of Cytokinin delays the process of aging by nuitrient mobilization.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
Define photoperiodism & Critical day length.
Answer:

• The physiological change on flowering due to relative length of light and darkness (Photoperiod) is called
  Photoperiodism.
• The photoperiod required to induce flowering is called critical day length Eg.
  • Mary land mammoth (tobacco variety) requires 12 hours of light.
  • Cocklebur required 15.05 hours of light.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

• The knowledge of photoperiodism plays an important role in hybridisation experiments.
• Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
What is the importance of photoperiodism?
Answer:

• The knowledge of photoperiodism an important role in hybridization experiments.
• It is an excellent example of physiological preconditioning that is using an external factor to induce physiological changes in the plant.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination, cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl.
Eg: Castor and Bean.

Question 22.
Define Vernalization.
Answer:

• It is a process by which many annuals and biennials are induced to flower when subjected to low-temperature exposure.
• T.d. Lysenko first used the term.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Distinguish between Epigeal and Hypogeal germination.
Answer:

Epigeal	Hypogeal
Cotyledons pussed out of the soil	Cotyledons remain below the soil due to rapid elongation of epicotyls.
Happens due to the elongation of the hypocotyl Eg. Castor & Bean	Eg. Maize

Question 25.
Define seed dormancy and what are its types.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.
There are two types
(I) Innate dormancy (II) Imposed dormancy.

Question 26.
What is Scarification?
Answer:

• By mechanical and chemical treatments like cutting or chipping of hard tough sed coat and use of organic solvents to remove waxy or fatty compounds are called scarification.
• It is a method of breaking dormancy of the seeds.

Question 27.
Distinguish between Re differentiation and Devernalization.
Answer:

Redifferentiation	Devernalization
Differentiated cells after multiplication again lose the ability to divide and mature to perform specific functions, is called Re differentiation. Eg. Sec.Xylem & Sec.	Phloem The reversal of the effect of vernalization is called Devemalization.

Question 28.
Define Senescence.
Answer:

• Ageing or getting old is called senescence.
• It refers to all collective, progressive and deteriorative processes which ultimately lead to complete loss of organization and function (Eg. leaves turn yellow and fall off from plant).

Question 29.
What is impaction in seed dormancy.
Answer:

• In some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells.
• These seeds are shaken vigorously to remove the plug
• The process of removing the plug or block is called impactation.

Question 30.
What is called stratification in seed dormancy?
Answer:

• The break dormancy, some plant seeds have to be exposed to well aerated, moist conditions under low temperature (0°c to 10°c) for weeks to months.
• This kind of seed dormancy breaking treatment is known as stratification.
• The stratified soil layers should be given a low-temperature treatment for a certain period so as to induce germination.
• Eg. the Seeds of Rosaceae plants Apple, Plum, Peach, etc.

Question 31.
What are the 4 types of Senescence?
Answer:
Leopold (1961) explained 4 types they are

1. Overall senescence
2. Top senescence
3. Deciduous senescence
4. Progressive senescence.

Question 32.
What is the Abscission layer or Abscission Zone?
Answer:
Abscission is marked internally at the place of petiole by a distance zone of few layers of thin-walled cells arranged transversely. This zone is called Abscission Zone, which leads to Abscission of the leaf.

Question 33.
The photoperiodic response will not be possible in a defoliated plant. Give scientific reasons.
Answer:
Yes, a defoliated plant will not respond to photoperiodic change because the hormonal substance responsible for flowering is present in the leaves of the plant.

Question 34.
What is gas chromatography?
Answer:

• It is a bioassay technique by which Ethylene can be measured.
• It helps in the detection of the exact amount of ethylene from different plant tissues like lemon and orange.

Question 35.
Give the occurrence and precursors of Gibberellins and Cytokinins.
Answer:

Character	Gibberellins	Cytokinin
Occurrence	Produced by plant parts like an embryo, roots, and young leaves near the tip. Immature seeds are rich in Gibberellins.	Formed in root apex shoot apex like Auxin. Also formed in buds & young fruits.
Precursor	Formed by 5C precursor, Iso prenoidunit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates primary precursor – Acetate.	Derived from purine-Adenine.

IV. 3 Mark Questions

Question 1.
Explain Arithmetic growth rate and Geometric growth rate by diagrams.
Answer:

Question 2.
Explain stages in growth by drawing the sigmoid curve.
Answer:

• The total period from the initial to the final stage of growth is called the Grand period of growth.
• The graph that is drawn by taking time and rate of growth is ‘S’ shaped. It is known as a sigmoid curve.

It has 4 stages:

1. Lag phase
2. Log phase
3. Decelerating phase
4. Maturation phase

Question 3.
Mention the internal factors, that affect the growth of plants.
Answer:

• Genes are intracellular factors for growth.
• Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
• C/N ratio.

Question 4.
Mention the Agricultural role of Auxin.
Answer:

• Eradicate weeds: Eg. 2.4 D and 2.4.5.7
• Formation of seedless fruits: (Parthenocarpic fruits) Eg. Synthetic Auxin.
• Break dormancy.
• Induction of flowering: In pineapple NAA induce flowering
• Increase the number of female flowers: Eg. Cucurbita.

Question 5.
List out the agricultural applications of auxins.
Answer:

• It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
• Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
• It is used to break the dormancy in seeds.
• Induce flowering in Pineapple by NAA & 2,4 – D.
• Increase the number of female flowers and fruits in cucurbits.

Question 6.
What are the Precursors of Gibberellins?
Answer:

• Gibberellins are chemically related to terpenoids (natural rubber, Carotenoids, and steroids) formed by 5-C precursors and an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates.
• The primary precursors are Acetate.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in the cucumber crop produces female flowers and increases the yield.

Question 8.
Explain the mechanism of Vernalization by Hypothesis of hormonal involvement.
Answer:
I. Vernalization: According to Purvis

 

 

 

 

II. Devernalization

 

 

Question 9.
What are the practical applications of Vernalization?
Answer:

• It shortens the vegetative period and induces the plant to flower earlier.
• It increases the cold resistance of the plants.
• It increases the resistance of plants to fungal disease.
• Plant breeding can be accelerated.

Question 10.
What is meant by the viability of seeds?
Answer:

• Viable means the living condition of the seed
• The shelf life of the seed after which it cannot germinate is known as the viable period
• It varies from plant to plant

Name of the plant	Viability
1. Oxalis seeds	Few days
2.  Lotus seeds	More than 1000 years
3.  Judean Dale palm (Methuselah)	More than 2000 years

Question 11.
Differentiate between climacteric and Non-climacteric fruits.
Answer:

Climacteric fruits	Non-Climacteric fruits
1.There is a sharp rise in respiration rate near the end of the development of fruit. 2. The ripening on demand can be induced in these fruits by exposing them to normal air conditioning. 3. Epthan secrete Ethylene continuously about 1 ppm of ethylene Eg. Lemon, Apples, Banana, Mango	These fruits cannot be ripened by exposure to ethylene so-known as non-climacteric fruits. Eg. Grapes, Watermelon orange.

Question 12.
Differentiate between scarification & Stratification in breaking seed dormacy
Answer:

Scarification	Stratification
Mechanical and chemical treatment either by cutting, chipping or use of organic solvents to remove waxy or fatty compounds is called scarification.	Rosaceous plants (Apple, Plum Peach, and Cherry) will not germinate until they have been exposed to well derated, moist conditions under low temperature (1°c to 10°c) for weeks to months and this treatment is known as stratification.

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

• Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
• Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
• Lack of specific light requirement leads to seed dormancy.
• A range of temperatures either higher or lower cause dormancy.
• The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the factors that affect senescence?
Answer:

Name of the factor	Effect of senescence
ABA & Ethylene	Accelerates
Auxin & Cytokinin Nitrogen deficiency	reduces increases
Nitrogen supply	retards
High temperature in vernalized seeds	Accelerates
Low temperature	Retards
Water stress	Accumulation of ABA leading to senescence

Question 15.
What are the morphological and Anatomical changes due to Abscission?
Answer:

• Abscission Zone: formed at the base of petiole
• Greenish grey in colour by rows of 2 to 15 cells thick primary wall and middle lamella
• The dissolution of by pectinase & Cellulase
• Formation tyloses – that block conduction of vessels
• Degradation of chlorophyll – Colour of leaves changes and leaves fall off.
• After Abscission – Suberization of outer layer of cells by the development of periderm.

Question 16.
Write down the significance of Abscission.
Answer:
1. Abscission separates dead parts of the plant like old leaves and ripe fruits.
2. Helps in dispersal of fruits and continuing the life cycle.
3. Abscission of leaves (in deciduous plants) helps in water conservation during summer.
4. Helps in vegetative propagation (Shedding of gemmae or plantlets) Eg. Bryophyta.

V. 5 Mark Questions

Question 1.
Describe an experiment to measure the growth of a plant or By lever Auxanometer measure the rate of growth in stem tip.
Answer:
Experiment:
1. Arc auxanometer:
The increase in the length of the stem tip can easily by measured by an arc auxanometer. If consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another and to a weight passes over the pulley tightly. As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Refer Figure) The reading is taken. The acutal increase in the lengthwm stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:

Actual growth in length\(=\frac{\text { Distance travelled by the pointer radius of the }}{\text { Length of the pointer }}\)
For example, actual growth in length \(=\frac{10 \times 4 \text { inches }}{20 \text { inches }}\) = 2 inches

Question 2.
Explain the physiological effect of Auxin? Add a note on its agricultural applications.
Answer:
Cell elongation:
Promotes cell elongation in stem & Coleoptile

Root growth:
At extremely low concentration – promote root growth, at high concentrations it inhibits elongation of roots, but induce more lateral roots

Apical dominance:
Suppression of growth of lateral buds – by apical bud is known as Apical dominance
Prevents Abscission

Secondary growth:
Promotion of cell division in the cambium, responsible for secondary growth this property is exploited in tissue culture. (Callus foundation)

Respiration Stimulates respiration Induces Vascular differentiation.

Question 3.
Give the Agricultural application of Auxin.
Answer:
Weedicide
2.4.D & 2.4.5. T – Weedicides to remove weeds

Induce parthenocarpy
Synthetic auxins used to induce parthenocarpy (formation of seedless fruits).

Break dormancy
Used to break seed dormancy.

Induce flowering in Pineapple Eg. NAA & 2.4.D

Induce female flowers (numbers)
Eg. Cucumber.

Question 4.
Explain physiological effects of Gibberellins
Answer:

• Induction of cell division & cell elongation – Extraordinary stem elongation.
• Reversal of dwarfism & Bolting – Rosette (genetic dwarfism) plants when treated with Gibberellins exhibit excessive enter nodal growth – This sudden elongation of a stem followed by flowering is called Bolting.
• Breaks dormancy – in Potato tubers.
• Biennials flower in the 1 st year – Instead of cold exposure, if biennials treated with Gibberellins flower in the 1st year itself.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in sterft and roof and inhibits linear growth.
• It breaks the dormancy of buds, seeds and storage organs.
• It stimulates the formation of an abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
• Inhibition of stem elongation (shortening the internode).
• In low concentration, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

7. Agricultural role:

• Ethylene normally reduces flowering in plants except in Pineapple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in cucumber crops produces female flowers and increases the yield.

Question 6.
Explain the physiological Effects of Cytokinins.
Answer:

• With IAA – Promotes cell division With IAA & GA – Induces cell enlargement
• Breaks dormancy of light-sensitive seeds (tobacco) induces seed germination.
• Promotes growth of lateral
• buds even in the presence of apical bud.
• Delays the process of aging by nutrient mobilization known as Richmond Lang effect.
• Induces rate of protein synthesis.
• Induces the formation of interfascicular cambium Overcomes apical dominance
• Induces the formation of new leaves chloroplast and lateral shoots.
• Induces Accumulation of solutes.

Question 7.
Write down the physiological effects of Ethylene
Answer:

• Stimulates respiration and thereby ripening of fruits
• Stimulates radial growth in stem and root and inhibits linear growth.
• breaks dormancy of
  1. buds
  2. Seeds
  3. Storage Organs
• Stimulates abscission 2 one formation in
  1. leaves
  2. flowers
  3. fruits (so leaves shed prematurely)
• Prevents stem elongation by preventing internodal growth
• Root growth in low concentration
• Stimulates growth of lateral roots and root hairs and increase the absorptive surface
• Ripening of fruits – Increases ripening in climacteric fruits (Mango, banana) etc.
• It causes epinasty

Question 8.
Describe the methods of breaking the dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called Scarification.

2. impaction:
in some seeds, water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Impaction.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach, and Cherry) will not germinate until they have been exposed to well aerated, moist conditions under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoelastic seeds can be broken by exposing them to red light.

Question 9.
Define photoperiodism – Classify plants based on photoperiodism
Answer:
a. The physiological change on flowering due to the relative length of light and darkness is called photoperiodism.

• Gamer and Allard (1920) coined the term
• They studied photoperiodism in Biloxi variety of soybean (Glycine max) and Many land mammoth varieties of tobacco.

b. Depending on photoperiodic responses plants are classified into several types.
1. L.D. Plants (Long Day) The photoperiod required to induce flowering is called critical day length depending on critical day length if it is long (more than 12 hours) and with short nights. Eg. Pea Barley and Oats
Short LD Plants: These are Long day plants but need short day length during the early period of growth for flowering Eg. Wheat, Rye

2. SD Plants: Plants requiring short critical day length for flowering or a long night.
Eg. Tobacco, Cocklebur, Soya, Rice, and Chrysanthemum.
Long SD Plants: Actually SD plants but need long days during the early period of growth for flowering Eg. Some SPS of Bryophyllum & Night Jasmine.

3. Intermediate day plants:
These require a photoperiod between a long day and a short day for flowering Eg. Sugarcane and coleus.

4. Day Neutral plants:
There are a number of plants which can flower in all possible photoperiods, known as photo neutral or hiterterminate plants. Eg. Potato, Rhododendron, Tomato & Cotton.

Question 10.
Describe the role of phytochrome in inducing Flowering
Answer:
Definition:
It is a bluish biliprotein responsible for the perception of light in the photophysiological process, existing in two different forms is mainly involved in flower induction, (i.e) Pr and PFr.

• Butler et al(1959) named the pigment.
• It exists in two interconvertible forms

Pr	PFr
1. red light absorbing form 2. Absorbs red lgiht of wavelength 660 nm 3. Biologically inactive form & stable found in the diffused state in cytoplasm 4. Promotes flowering in SD plants and inhibits flowering LD plants.	1. Far-red light absorbing form 2. Absorbs far-red light of wavelength 730 nm 3. Biologically active and it is unstable Associated with a hydrophobic area of the membrane system 4. Promotes flowering in LD plants and inhibit flowering in SD plants.

Mechanism:

Other functions:
Play a role in seed germination and changes in membrane conformation.

Question 11.
Write an Essay on Vernalization
Answer:
Definition:

• Many biennials and perennials are induced to flower by low-temperature exposure (O°c to 5°c) This process is called Vernalization.
• T.D. Lysenko – Coined the term.

Mechanism of Vernalization:
2 theories explain the mechanism of vernalization.

1. Hypothesis of Phasic development (T.D. Lysenko),
The development of the annual plant has 2 phases.

1. Thermostate-Vegetatine stage requiring low temperature and suitable moisture.
2. Photo stage -high temperature needs to synthesize florigen.

2. Hypothesis of hormonal involvement (Purvis 1961)

Vernalization has several steps

The technique of Vernalization:

• Seeds soaked in water
• Allowed to germinate at 10°C to 12°C
• Transferred to low temperature for few days to 30 days (3°C to 5°C).
• Germinated seeds after the low temp, treatment are allowed to dry & then sown.
• Quickened flowering than untreated (control seedling)

Devernalization:
The reversal of the effect of vernalization is called Devemalization.

• Practical Applications:
• Vernalization shortens the vegetative period and induces the plant to flower earlier
• It increases cold resistance
• It increases fungal resistance
• It accelerates Plant Breeding.

Question 12.
Define Senescence and give its types
Answer:
Definition:
Getting old or Ageing is call d senescence in plants.
It refers to all collective, progressive, and deteriorative processes which ultimately lead to complete loss of organization and function.
Types – 4 types (Leopold -1961)

1. Overall senescence: When the entire plant gets affected and dies – Eg. Annuals – Wheat & Soybeans, Perennials – Agave & Bamboo
2. Top senescence: Occur in aerial parts only Eg. Parrennials – Banana and Gladiolus
3. Deciduous senescence: Occur only in leaves Eg. Decidual plants – Elm and Maple
4. Progressive Senescence: Occur in Annuals occur in old leaves first followed by new leaves than stem and finally root system.

Question 13.
Explain the physiology of senescence and Factors affecting senescence Physiology of Senescence :
Answer:

• Change in the structure of cells
• Vacuoles act like lysosome-secrete hydrolytic enzymes.
• Reducation in photosynthetic rate (due to loss of chlorophyll & accumulation of anthocyanin)
• The decrease in Starch content,  Protein content
• Decrease in …………. r RNA level due to increased activity of enzyme RNA ase
• Degeneration of DNA – by increased activity of enzyme DNA ase

Factors affecting senescence :

Name of the factor	Effect of senescence
ABA & Ethylene	Accelerates
Auxin & Cytokinin Nitrogen deficiency	reduces increases
Nitrogen supply	retards
High temperature in vernalized seeds	Accelerates
Low temperature	Retards
Water stress	Accumulation of ABA leading to senescence

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 14 Respiration Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration

11th Bio Botany Guide Respiration Text Book Back Questions and Answers

Part-I.

Question 1.
The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
(b) 13
(c) 14
(d) 15
Answer:
(a) 12

Question 2.
During oxidation of two molecules of cytosolic NADH + H⁺, number of ATP molecules produced in plants are
a) 3
b) 4
c) 6
d) 8
Answer:
b) 4

Question 3.
The compound which links glycolysis and Krebs cycle is:
(a) succinic acid
(b) pyruvic acid
(c) acetyl CoA
(d) citric acid
Answer:
(c) acetyl CoA

Question 4.
Assertion (A): Oxidative phosphorylation takes place during the electron transport chain in mitochondria.
Reason (R): Succinyl Co A is phosphorylated into succinic acid by substrate phosphorylation.
a) A and R is correct. R is correct explanation of A
b) A and R is correct but R is not the correct explanation of A,
c) A is correct but R is wrong
d) A and R is wrong
Answer:
c) A is correct but R is wrong

Question 5.
Which of the following reaction is not involved in Krebs cycle.
(a) Shifting of phosphate from 3C to 2C
(b) Splitting of Fructose 1,6 bisphosphate of into two molecules 3C compounds.
(c) Dephosphorylation from the substrates
(d) All of these
Answer:
(d) All of these

Question 6.
What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
Answer:
(i) Enzymes involved in phosphorylation are
a) Hexokinase and phospnofructio kinase.
(ii) Enzymes involved in dephosphorylation are
a) Phosphoglycerate Kinase
b) Pyruvate Kinase

Question 7.
The respiratory quotient is zero in succulent plants. Why?
Answer:
In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without the corresponding release of CO₂ but O₂ is consumed hence the RQ value will be zero.

Question 8.
Explain the reactions taking place in mitochondrial inner membrane.
Answer:
Electron and hydrogen (proton) transport takes place across four multiprotein complexes (I-IV). They are.
1. Complex-I (NADH dehydrogenase).
It contains a flavoprotein (FMN) and associated with non-heme iron Sulphur protein (Fe-S). This complex is responsible for passing electrons and protons from mitochondrial NADFI (Internal) to Ubiquinone (UQ)
NADH+H⁺UQ ⇌ NAD^–+UQH₂

2. In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H⁺.
Ubiquinone (UQ) or Coenzyme Quinone (CoQ) is a small, lipid-soluble electron, proton carrier located within the inner membrane of mitochondria).

3. Complex-II (succinic dehydrogenase) It contains FAD flavoprotein is associated with non-heme iron Sulphur (Fe-S) protein. This complex receives electrons and protons from succinate in Kerbs cycle and is converted into fumarate and passes to ubiquinone.
Succinate + UQ Fumaraic LQH₂

4. Complex-III (Cytochrome bcj complex) This complex oxidises reduced ubiquinone (ubiquinol) and transfers the electrons through Cytochrome bc1 Complex (Iron Sulphur centci bcl complex) to cytochrome c.

5. Complex IV (Cytochrome c oxidase) Complex IV is the terminal oxidase and brings about the reduction of 1/2 O₂ to H₂O. TWO protons are needed to form a molecule of H₂O (terminal oxidation).

Question 9.
What is the name of alternate way to glucose breakdown? Explain the process in involved in it?
Answer:

• Pentose phosphate pathway is the alternate pathway for breakdown of glucose.
• Pentose phosphate pathway was described by Warburg, Dickens and Lipmami (1938).
• It is also known as Hexose monophosphate shunt (HMP shunt) or Direct oxidative phase and non – oxidative phase.
• The oxidative phase convert six molecules of six carbon Glucose 6 phosphate to 6 molecules of five-carbon sugar Ribulose – 5 Phosphate with loss of 6CO₂ and generation of 12 NADPH + H⁺

Non oxidative pathway convert Ribulose – 5 – phosphate molecules to various intermediates such as
Ribose – 5 – phosphate (5C)
Xylulose – 5 – phosphate (5C)
Glyceraldehyde – 3 – phosphate (3C)
Sedoheptulose – 7 – phosphate (7C) and
Erythrose – 4 – phosphate (4C)

Finally five molecules of glucose 6 – phosphate is regenerated
6 x Glucose – 6 – phosphate + 12NADP⁺ + 6H₂O
↓
5 x glucose – 6 – phosphate + 6CO₂ + Pi + 12 NADPH + 2H⁺
The net result of complete oxidation of one glucose – 6 – phosphate yield 6CO₂ and 12 NADPH + H⁺. The oxidative pentose phosphate pathway is controlled by glucose – 6 – phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP^+.

Question 10.
How will you calculate net products of one sucrose molecule upon complete oxidation during aerobic respiration as per recent view?
Answer:
When the cost of transport of ATPs from the matrix into the cytosol is considered, the number will be 2.5 ATPs for each NADH + H⁺ and 1.5 ATPs for each FADH₂ oxidized during the electron transport system. Therefore, in plant cells net yield of 30 ATP molecules for complete aerobic oxidation of one molecule of glucose. But in those animal cells (showing malate shuttle mechanism) net yield will be 32 ATP molecules. Since the sucrose molecule gives, two molecules of glucose and net ATP in plant cell will be 30 × 2 = 60.
In an animal cell, it will be 32 × 2 = 64.

Part-II.

11th Bio Botany Guide Respiration Additional Important Questions and Answers

I. Choose The Correct Answer

Question 1.
The term respiration was coined by:
(a) Lamark
(b) Kerb
(c) Pepys
(d) Blackman
Answer:
(c) Pepys

Question 2.
Black man divided respiration into floating respiration and protoplasmic respiration based on respiratory ………..
a) Quotient
b) respiratory reaction
c) respiratory pathway
d) substrate
Answer:
d) substrate

Question 3.
The discovery of ATP was made by:
(a) Lipman
(b) Hans Adolt
(c) Warburg
(d) Karl Lohman
Answer:
(d) Karl Lohman

Question 4.
The type of respiration which is rare and liberates toxic ammonia
a) Protoplasmic respiration
b) floating respiration
c) Aerobic respiration
d) Anaerobic respiration
Answer:
a) Protoplasmic respiration

Question 5.
On hydrolysis, one molecule of ATP releases energy of:
(a) 8.2 Kcal
(b) 32.3 kJ
(c) 7.3 Kcal
(d) 7.8 Kcal
Answer:
(c) 7.3 Kcal

Question 6.
To convert Kcal to KJ multiply by 4.18(100 Kcal=418 KJ) calculate the amount KJ energy for 7.3 Kcal
a) 30.6 KJ
b) 32.06 KJ
c) 29.03 KJ
d) 5.01 KJ
Answer:
a) 30.6 KJ

Question 7.
Identify the link reaction:
(a) conversion of glucose into pyruvic acid
(b) conversion of glucose into ethanol
(c) conversion of acetyl CoA into CO₂ and water
(d) conversion of pyruvic acid into acetyl coenzyme – A
Answer:
(d) conversion of pyruvic acid into acetyl coenzyme – A

Question 8.
Food materials like carbohydrate, fat and proteins are completely oxidised into CO₂, H₂O and energy in ………………. respiration
a) anaerobic
b) aerobic
c) Bacterial respiration
d) Facultative
Answer:
b) aerobic

Question 9.
Kreb’s cycle is a:
(a) catabolic pathway
(b) anabolic pathway
(c) amphibolic pathway
(d) hydrolytic pathway
Answer:
(c) amphibolic pathway

Question 10.
Which process is occur in yeast and some bacteria and 2 ATP molecules are produced during this process.
a) Anaerobic respiration
b) aerobic respiration
c) mixed fermentation
d) CAC cycle
Answer:
a) Anaerobic respiration

Question 11.
The oxidation of one molecule of NADH + H⁺ gives rise to:
(a) 2 ATP
(b) 3 ATP
(c) 4 ATP
(d) 2.5 ATP
Answer:
(b) 3 ATP

Question 12.
Net product in Glycolysis are
a) 4ATP and 2NADH + H⁺
b) 2 ATP and 2NADH + H⁺
c) 6ATP
d) 24 ATP
Answer:
b) 2ATP and 2NADH + H⁺

Question 13.
Cyanide acts as electron transport chain inhibitor by preventing:
(a) synthesis of ATP from ADP
(b) flow of electrons from NADH + H⁺
(c) flow of electrons from cytochrome a₃ to O₂
(d) oxidative phosphorylation
Answer:
(c) flow of electrons from cytochrome a₃ to O₂

Question 14.
Which is the raw material for the formation of chlorophylls, cytochrome, phytochrome and pyrrole substance
a) acetyl COA
b) Pyruvic acid
c) Malic acid
d) Succinyl COA
Answer:
d) Succinyl COA

Question 15.
End products of fermentation in yeast is:
(a) pyruvic acid and CO₂
(b) lactic acid and CO₂
(c) ethyl alcohol and CO₂
(d) mixed acid and CO₂
Answer:
(c) ethyl alcohol and CO₂

Question 16.
Which cycle is considered as amphibolic pathway.
a) Calvin cycle
b) Glycolysic
c) ETS chain
d) Kreb cycle
Answer:
d) Kreb cycle

Question 17.
The external factors that affect respiration are:
(a) temperature, insufficient O₂ and amount of protoplasm
(b) temperature, insufficient O₂ and high concentration of CO₂
(c) temperature, high concentration of CO₂ and respiratory substrate
(d) temperature, high concentration of CO₂ and amount of protoplasm
Answer:
(b) temperature, insufficient O_2, and high concentration of CO₂

Question 18.
The complex system responsible for passing electrons and protons from mitochondria to ubiquinone is ………………..
a) Complex I
b) Complex II
c) Complex III
d) Complex IV
Answer:
a) Complex I

Question 19.
The oxidative pentose phosphate pathway is controlled by the enzyme:
(a) glucose, 1, 6 diphosphate dehydrogenase
(b) glucose 6 phosphate dehydrogenase
(c) fructose – 6 – phosphate dehydrogenase
(d) none of the above
Answer:
(b) glucose 6 phosphate dehydrogenase

Question 20.
Which are the high energy phosphate groups in ATP
a) adenine
b) Pentose sugar
c) Last two phosphate group
d) First two phosphate group
Answer:
c) Last two phosphate group

Question 21.
In-plant tissue erythrose is used for the synthesis of:
(a) Erythromycin
(b) Xanthophyll
(c) Erythrocin
(d) Anthocyanin
Answer:
(d) Anthocyanin

Question 22.
How many ATP molecules are produced when a molecule of glucose undergo fermentation?
a) TwoATPs
b) Six ATPs
c) Eight ATPs
d) one ATP
Answer:
a) Two ATPs

Question 23.
Identify the electron transport inhibitor:
(a) phosphoenol
(b) dinitrophenol
(c) xylene
(d) indol acetic acid
Answer:
(b) dinitrophenol

Question 24.
Enzymatic reaction for partial oxidation of glucose in the absence of oxygen is present in
a) Some Bacteria
b) Yeast fungus
c) A and B
d) Bryophytes
Answer:
c) A and B

Question 25.
Cyanide resistant respiration is known to generate heat in thermogenic tissues as high as:
(a) 35° C
(b) 38° C
(c) 40° C
(d) 51° C
Answer:
(d) 51° C

Question 26.
Match the Column I with the enzyme responsible for its production in column II
Answer:

Column I	Column II
A. Citric acid	1. Hexose Kinase
B. Glucose 6-Phosphate	2. Lactate dehydrogenase
C. Lactic acid	3. Pyruvate dehydrogenase
D. Acetvl CO.A	4. Citric acid Synthetase

Answer:
a) A-4,B -1,C-2,D-3.

Question 27.
Which one is wrongly matched

Column I	Column II
A. NADH +H+	Three ATP
B. Glycolysis	Twenty four ATP
C. FAD	Two ATP
D. Cytoplasmic NADH+H+	Two ATP

Answer:
B. Glycolysis – Twenty four ATP

II. 2 Marks Questions

Question 1.
Define respiration?
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins, and fats take place and as a result of this, energy is produced where O₂ is taken in and CO₂ is liberated.

Question 2.
Name some High energy compounds present in a cell
Answer:

• ATP → Adenosine Tri Phosphate
• GTP → Guanosine Tri Phosphate
• UTP → Uridine Tri Phosphate

Question 3.
What do you understand by compensation of point?
Answer:
The point at which CO₂ released in respiration is exactly compensated by CO₂ fixed in photosynthesis that means no net gaseous exchange takes place, it is called the compensation point.

Question 4.
What is Anaerobic respiration? What are its steps?
Answer:

• In the absence of molecular oxygen-glucose is incompletely degraded into either ethyl alcohol (or) Lactic acid.
• It includes two steps (i) Glycolysis (ii) Fermentation

Question 5.
What is anaerobic respiration?
Answer:
In the absence of molecular oxygen-glucose is incompletely degraded into either ethyl alcohol or lactic acid. It includes two steps:

1. Glycolysis
2. Fermentation

Question 6.
What is Link reaction?
Answer:
In aerobic respiration, Conversion of Pyruvic acid into acetyl coenzyme – A in the mitochondrial matrix with two molecules of NADH + H⁺ and 2 CO₂. This is called Link reaction (or) transition reaction.

Question 7.
Who is Sir Hans Adolf Krebs?
Answer:
Sir Hans Adolf Krebs was born in Germany on 25th August 1900. He was awarded Nobel Prize for his discovery of Citric acid cycle in Physiology in 1953.

Question 8.
Why Kreb cycle is called as citric acid cycle (or) Tri Carboxylic acid Cycle?
Answer:

• TCA cycle starts with condensation of acetyl COA with oxaloacetate in the presence of water to yield Citri acid (or) Citrate.
• So it is also known as citric acid cycle (or) Tricarboxylic acid cycle.

Question 9.
Mention the role of NADH dehydrogenase enzyme in the electron transport system.
Answer:
NADH dehydrogenase contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).

Question 10.
Which cycle is amphibolic pathway? Why? The Krebs cycle is called an amphibolic pathway.
Answer:

• Kreb cycle is primarily a catabolic pathway Later it is an anabolic pathway too.
• Hence it is called amphibolic pathway.

Question 11.
Mention any two electron transport chain inhibitors.
Answer:
Two electron transport chain inhibitors:

1. 2, 4 DNP (Dinitrophenol) – It prevents synthesis of ATP from ADP, as it directs electrons from Co Q to O₂.
2. Cyanide – It prevents flow of electrons from Cytochrome a₃ to O₂.

Question 12.
How many ATP molecules are produced in aerobic respiration present in plants?
Answer:
In aerobic respiration net gain of 36 ATP molecules produced in complete oxidation of glucose.

Question 13.
What are the significances of Respiratory Quotient?
Answer:
The significances of Respiratory Quotient:

1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
2. It also helps to know which type of respiratory substrate is involved.

Question 14.
Who was awarded the Nobel Prize for coupling of oxidation and phosphorylation in mitochondria?
Answer:
Peter Mitchell, a British Biochemist received Nobel Prize for chemistry in 1978.

Question 15.
Mention any two industrial uses of alcoholic fermentation.
Answer:
Two industrial uses of alcoholic fermentation:

1. In bakeries, it is used for preparing bread, cakes, biscuits.
2. In beverage industries for preparing wine and alcoholic drinks.

Question 16.
Define mixed acid fermentation.
Answer:

• Formation of Lactic acid, ethanol, formic acid and gases like CO₂ and H₂ from pyruvic acid.
• eg. Enterobacteriaceae.

Question 17.
Mention any two internal factors, that affect the rate of respiration in plants.
Answer:
Two internal factors, that affect the rate of respiration in plants:

1. The amount of protoplasm and its state of activity influence the rate of respiration.
2. The concentration of respiratory substrate is proportional to the rate of respiration.

Question 18.
Why microorganisms respire an anaerobically?
Answer:

• Bacteria are prokaryotes and they are devoid of membrane-bound organelle mitochondria.
• So they are respire anaerobically.

Question 19.
Write down any two significance of the pentose phosphate pathway.
Answer:
Two significance of pentose phosphate pathway:

1. HMP shunt is associated with the generation of two important products.
2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen-free radicals.

Question 20.
Complete the following Picture.

Answer:
A. Compensation point
B. Rate of Respiration

Question 21.
Write the missing A and B.

Answer:
A. Ribose
B. Adenine

III. 3 Mark Questions

Question 1.
In the biosphere how do plants and animals are complementary systems, which are integrated to sustain life?
Answer:
In plants, oxygen enters through the stomata and it is transported to cells, where oxygen is utilized for energy production. Plants require carbon dioxide to survive, to produce carbohydrates, and to release oxygen through photosynthesis, these oxygen molecules are inhaled by humans through the nose, which reaches the lungs where oxygen is transported through the blood and it reaches cells. Cellular respiration takes place inside or the cell for obtaining energy.

Question 2.
What is Respiration?
Answer:

• Breaking of C-C bonds of complex organic compounds through oxidation within the cells.
• The energy released during respiration is stored in the form of ATP and heat is liberated.
• It occurs in all the living cells of organisms.

Question 3.
What are the factors associated with the compensation point in respiration?
Answer:
The two common factors associated with compensation points are CO₂ and light. Based on this there are two types of compensation points. They are the CO₂ compensation point and light compensation point. C₃ plants have compensation points ranging from 40 – 60 ppm (parts per million) CO₂ while those of C₄ plants range from 1 – 5 ppm CO₂.

Question 4.
Differentiate floating respiration and protoplasmic respiration.
Answer:

Floating respiration	Protoplasmic respiration
Carbohydrate (or) fat (or) organic acid serves as a respiratory substrate	Whereas protein is a respiratory substrate.
It is a common mode of respiration and does not produce any toxic product.	It is rare and liberates toxic ammonia.

Question 5.
What is a redox reaction?
Answer:
NAD⁺ + 2e^– + 2H⁺ → NADH + H⁺
FAD + 2e^– + 2H⁺ → FADH₂
When NAD⁺ (Nicotinamide Adenine Dinucleotide – oxidized form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H⁺ and FADH₂ respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD⁺ and FAD gain (reduction) or lose (oxidation) electrons are called redox reactions (Oxidation-reduction reactions). These reactions are important in cellular respiration.

Question 6.
Define ETS (or) Electron transport chain (or) What is the importance of ETS and oxidative Phosphorylation in respiration.
Answer:

• Electron transport chain and oxidative phosphorylation remove hydrogen atoms from the products of glycolysis, link reaction, and Kreb cycle.
• It releases water molecule with energy in the form of ATP molecules in the mitochondrial inner membrane.

Question 7.
Mention the significance of Kreb’s cycle.
Answer:
The significance of Kreb’s cycle:

1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
2. It provides carbon skeleton or raw material for various anabolic processes.
3. Many intermediates of the TCA cycle are further metabolised to produce amino acids, proteins, and nucleic acids.
4. Succinyl CoA is the raw material for the formation of chlorophylls, cytochrome, phytochrome, and other pyrrole substances.
5. α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
6. It acts as a metabolic sink which plays a central role in intermediary metabolism.

Question 8.
Write the differences between ubiquinone and Cytochrome C.
Answer:

Ubiquinone	Cytochrome C
It is a small, lipid-soluble electron, proton carrier located within the inner membrane of mitochondria.	It is a small protein attached to the outer surface of inner membrane of mitochondria
It is associated with ETS – complex I	It is associated with ETS – complex IIII

Question 9.
Write down the characteristic of Anaerobic respiration.
Answer:
The characteristic of Anaerobic respiration:

1. Anaerobic respiration is less efficient than aerobic respiration.
2. A limited number of ATP molecules is generated per glucose molecule.
3. It is characterized by the production of CO₂ and is used for Carbon fixation in photosynthesis.

Question 10.
RQ will be less than one in Red colour Parts Present in Plants? Why?
Answer:

• Red colour parts present in plants is due to the presence of anthocyanin
• Synthesis of anthocyanin require more O₂ than CO₂ evolved.
• So RQ will be less than one.

Question 11.
Write down any three external factors, that affect respiration in plants.
Answer:
Three external factors, that affect respiration in plants:

1. The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
2. When a sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called the Extinction point.
3. The high concentration of CO₂ reduces the rate of respiration.

Question 12.
Define Lactic acid fermentation.
Answer:
Formation of Lactic acid from pyruvic acid is Lactic acid fermentation.
Eg. Bacillus bacteria, fungi, muscles of vertebrates.

Question 13.
What is the Pentose phosphate pathway?
Answer:

• It is an alternate pathway for break down of glucose.
• It takes place in the cytoplasm of mature plant cells.
• In this pathway glucose 6 phosphate molecule is converted to Ribulose 5 phosphate with CO₂ and NADPH^– + H⁺.

Question 14.
How alcoholic beverages like beer and wine is made?
Answer:

• The conversion of pyruvate to ethanol takes place in malted barley and grapes through fermentation.
• Yeast Carryout this process under anaerobic conditions and this conversion increases ethanol concentration.
• If the concentration increases It’s toxic effect kills yeast cells and the left out is called beer and wine respectively.

IV. 5 Mark Questions

Question 1.
Ambulate the differences between aerobic and anaerobic respiration.
Answer:

Aerobic respiration	Anaerobic respiration
1. It occurs in all living cells of higher organisms.	It occurs yeast and some bacteria.
2. It requires oxygen for breaking the respiratory substrate	Oxygen is not required for breaking the respiratory substrate.
3. The end products are CO2 and H2O	The end products are alcohol and CO2 (or) lactic acid
4. Oxidation of one molecule of glucose produces 36 ATP molecules	Only 2 ATP molecules are produced.
5. It consists of four stages – glycolysis, link reaction, TCA cycle and electron transport chain.	It consists of two stages – glycolysis and fermentation.
6. It occurs in cytoplasm and mitochondria	It occurs only in cytoplasm

Question 2.
Draw the flow chart diagram For Glycolysis (or) EMP pathway.
Answer:

Question 3.
Explain the Pay – off phase of EMP Pathway of Glycolysis (or) Explain the oxidative phase of Glycolysis (or) Triose phase of Glycolysis.
Answer:

• Two molecules of glyceraldehyde 3 – phosphate oxidatively phosphorylated into two molecules of 1-3
  bisphospho glycerate.
• During this reaction 2 NAD⁺ is reduced to 2NADH+ H⁺ by glyceraldehyde 3-phosphate dehydrogenase.
• Further reactions are carried out by different enzymes at the end two molecules of pyruvate are produced.
• In this phase 4 ATPS are produced (at step 7 and step 10)
• Through Direct transfer of phosphate from substrate molecule to ADP and is converted into ATP is called substrate Phosphotylation. (or) Direct Phosphorylation (or) transphosphorylation.
• During the reaction (at step 9)2 phospo glycerate dehydrated into phosphoenol pyurvate, a water molecule is removed by the enzyme enolase.
• As a result enol group is formed within the molecule. This process is called Enolation.

Energy Budge of pay off phase:

• In the payoff phase totally 4 ATP and 2NADH + H⁺ molecules are produced.
• Since 2 ATP molecules are already consumed in the preparatory phase the net products in glycolysis are 2ATP and 2NADH + H⁺

Question 4.
Explain the preparatory phase of Glycolysis (or) EMP pathway (or) Describe the energonic phase phase of Glycolysis (or) EMP pathway. Describe the hexose phase of Glycolysis (or) EMP pathway.
Answer:
Glycolysis is a linear series of reactions in which 6- carbon glucose split into two molecules of 3 carbon pyruvic acid.
Preparatory phase:

• Glucose enters glycolysis which is the end product of photosynthesis.
• Glucose is phosphorylated into glucose 6 phosphate by the enzyme hexokinase and subsequent reactions are carried out by different enzymes.
• At the end of this phase fructose 1,6 – bisphote is cleaved into glyceraldehyde 3- phosphate and dihydroxyacetone phosphate by the enzyme aldolase.
• These two are Isomers.
• Dihydroxyacetone phosphate is isomerised into glyceraldehyde 3- phosphate by the enzyme triose phosphate isomerase.
• Now two molecules of glyceraldehyde 3 phosphate enter into pay off phase.

During the preparatory phase, two ATP molecules are àonsumed.

Question 5.
Explain pyruvate oxidation (or) Link reaction of Glycolysis.
Answer:

• Two molecules of pyruvate formed by glycolysis in the cytosol enter into mitochondnalrnatrxi.
• In aerobic respiration this pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. .
• It produces two molecules of NADH + H⁺ and 2CO₂

It is also called transition reaction (or) Link reaction.

The pyruvate dehydrogenase complex consists of three distinct enzymes.

1. Pyruvate dehydrogenase
2. Dihydroiipoyil transacetylase
3. Dihydrolipoyil dehydrogenase and 5 coenzymes TPP (thymine pyrophosphate)
   • NAD⁺
   • FAD
   • COA and lipoate.

Question 6.
Draw the flow chart diagram for the Kreb cycle (or) Citric acid cycle.
Answer:

Question 7.
Explain Kreb cycle (or) Citric acid cycle (or) TCA cycle.
Answer:

• Two molecules of acetyl CoA formed from link reaction now enter into Kreb Cycle.
• It is named after its discoverer German Biochemist Sir Hans Adolf Kreb (1937).
• It is takes place in the mitochondrial matrix and inner membrane of mitochondria.
• The enzymes needed for TCA cycle are found in the mitochondrial matrix except for succinate dehydrogenase which is found in the mitochondrial inner membrane.
• First step starts with condensation of acetyl CoA with oxaloacetate in the presence of water to yield citric acid (or) citrate.
• It is followed by the action of different enzymes in cyclic manner.
• During the conversion of succinyl CoA to succinate by the enzyme succinyl CoA synthetase a molecule of ATP Synthesis from Substrate without entering the electron transport chain is called substrate-level phosphorylation.
• Kreb Cycle is repeated twice for every glucose molecule.
• Where two molecules of pyruvic acid produces six molecules of CO₂, eight molecules of NADH+H⁺ two molecules of FADH₂ and two molecules of ATP.

Question 8.
Significance of Kreb Cycle.
Answer:

• TCA cycle is to provide energy in the form of ATP for metabolism in plants.
• It provides carbon skeleton or raw material for various anabolic process.
• many intermediates of TCA cycle are further metabolised to produce amino acids, proteins and nucleic acids.
• Succinyl CoA is raw material for formation of chlorophyll, cytochrome, phytochrome and other pyrroles
  substances.
• α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
• it acts as metabolic sink which plays a central role in intermediary metabolism.

Question 9.
Write four Electron transport chain in hibitors.
Answer:

• 2,4 DNP (Dinitrophenol) – It prevents the synthesis of ATP from ADP, as it directs electrons from CoQ to O₂
• Cyanide – It prevents the flow of electrons from Cytochrome a3 to O₂
• Rotenone – It prevents flow of electrons from NADH + H⁺ / FADH₂ to Co Q
• Oligomycin – It inhibits oxidative phosphorylation

Question 10.
Tabulate Net Products of ATP gained during aerobic respiration per glucose molecule.
Answer:

Question 11.
Experiment to demonstrate the production of CO₂ in aerobic respiration.
Answer:

• Take small quantity of any seed (groundnut or bean seeds) and allow them to germinate by imbibing them.
• While they are germinating place them in a conical flask.
• A small glass tube containing 4 ml of freshly prepared Potassium hydroxide (KOH) solution is hing into the conical flask with the help of a thread and tightly close the one holed cork.
• Take a bent glass tube, the shorted end of which is inserted into the conical flask through the hole in the cork.
• The longer end is dipped in a beaker containing water.
• Observe the position of initial water level in bent glass tube.
• This experimental setup is kept for two hours.
• After two hours, the level of water rises in the glass tube. It is because the CO₂ evolved during aerobic
  respiration by germinating seeds will be absorbed by KOH solution and the level of water will rise in the glass tube.
• CO₂ + 2KOH → K₂CO3 + H₂O

Question 12.
Compare Alcoholic fermentation.
Answer:

Alcoholic fermentation	Lactic acid fermentation
1. it produces alcohol and releases CO2 from pyruvic acid	It produces lactic acid and does not release CO2 from pyruvic acid
2. It takes place in two steps.	It takes place in single steps.
3. It involves two enzymes, pyruvate decarboxylase with Mg++ and alcohol dehydrogenase	It uses one enzyme, lactate dehydrogenase with Zn++
4. It forms acetaldehyde as an intermediate compound	Does not form an intermediate compound.
5. It commonly occurs in yeast.	Occurs in bacteria, some fungi, and vertebrate muscles.

Question 13.
Write the Industrial uses of alcoholic fermentation.
Answer:

• In bakeries, it is used for preparing bread, cakes, biscuits.
• In beverage industries for preparing wine and alcoholic drinks.
• In producing vinegar and in tanning, curing of leather.
• Ethanol is used to make gasohol (a fuel that is used for cars in Brazil).

Question 14.
Tabulate the comparison between glycolysis and fermentation.
Answer:

Glycolysis	Fermentation
1. Glucose is converted into pyruvic acid	Stars from pyruvic acid and is converted into alcohol or lactic acid.
2. It takes place in the presence or absence of oxygen.	it takes place in the absence of oxygen.
3. Net gain is 2ATP.	No net gain of ATP molecules.
4. 2NADH + H+ molecules are produced.	2NADH+ H+molecules are utilised
5. It commonly occurs in yeast.	Occurs in bacteria, some fungi and vertebrate muscles.

Question 15.
Explain the demonstration of alcoholic fermentation.
Answer:

• Take a Kuhne’s fermentation tube which consists of an upright glass tube with a side bulb
• Pour 10% sugar solution mixed with baker’s yeast into the fermentation tube the side tube is filled plug the mouth with lid.
  After some time, the glucose solution will be fermented. The solution will give out an alcoholic smell.
• The level of the solution in the glass column will fall due to the accumulation of CO₂ gas.
• It is due to the presence of zymase enzyme  yeast which converts the glucose solution into alcohol and CO₂
• Now introduce a pellet of KOH into the tube, the KOH will absorb CO₂ and the level of solution will rise in the upright tube.
• This experiment proves during fermentation CO₂ gas is evolved.

Question 16.
Write the Factors (Internal and External) which affect the process of respiration.
Answer:
External Factors:

• The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate, respiration decreases.
• When sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration  is completely stopped. This is called Extinction point.
• The high concentration of CO₂ reduces the rate of respiration.
• A plant or tissue transferred from water to salt solution wi li increase the rate of respiration. It is called silt respiration.
• Light is an indirect factor affecting the rate of respiration.
• Wounding of plant organs stimulates the rate of respiration in that region.

Internal Factors:

• The concentration of respiratory substrate is proportional to the rate of respiration
• The amount of protoplasm and its state of activity influence the rate of respiration.

Question 17.
Write about the alternate pathway for glucose break down (or) Write about pentose phosphate pathway. (or) Phosphogluconate pathway (or) War burg – Dickens Lipmann pathway (or)Hexose Monophosphate pathway (or) HMP Shunt pathway.
Answer:

• The pentose phosphate pathway was described by Warburg, Dickens, and Lipmann (1938). Hence, it is also called Warburg – Dickens Lipmann pathway.
• It takes place in the cytoplasm of mature plant cells. It is an alternate way for break4own of glucose.
• It consists of two phases, oxidative phase, and non-oxidative phase.
• The oxidative events concert six molecules of six carbon Glucose 6 phosphate to 6 molecules of five-carbon sugar Ribulose -5 phosphate with loss of 12 NADPH + H⁺ (not NADH).
• The remaining reactions known as non oxidative pathway, covert Rihulose 5phosphate molecules to various intermediates such as Ribose – 5 – phosphate (5C), Xylulose – 5 – phosphate (5C), Glyceraldehyde – 7 – Phosphate (7C), and Eiythrose -4- phosphate (4C).
• Finally, five molecules of glucose -6- phosphate is regenerated. The overall reaction is:
  6 x Glucose – 6 – Phosphate + 12NADP⁺ + 6H₂O
  ↓
  5 x Glucose-6- Phosphate + 6CO₂ + Pi + 12NADPH + 12H⁺
  
• The net result of complete oxidation of one glucose-6-phosphate yield 6CO₂ and12NADPH+H⁺

Question 18.
Draw the cycle for pentsoe phosphate pathway (or) Draw the flow chart for HMP Shunt.
Answer:

Question 19.
Write about the Significance of Pentose Phosphate pathway.
Answer:

• HMP shunt is associated with the generation of two important products. NADPH and pentsoe sugars, which play a vital role in anaholic reactions.
• Coenzyme NADPH generated is used by reductive bisynthesìs and counter damaging the effects of oxygen-free radicals.
• Ribose – 5 – phosphate and its derivatives are used in the synthesis of DNA, RNA, ATP, NAD, FAD and
  Coenzynie A. .
• Erythrose is used for the synthesis of anthocyanin Jignin and other aromatic compounds.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Computer Science Guide Pdf Chapter 14 Classes and Objects Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 14 Classes and Objects

11th Computer Science Guide Classes and Objects Text Book Questions and Answers

Book Evaluation

Part I

Choose The Correct Answer
Question 1.
The variables declared inside the class are known as data members and the functions are known as
a) data functions
b) inline functions
c) member functions
d) attributes
Answer:
c) member functions

Question 2.
Which of the following statements about member functions are True or False?
i) A member function can call another member function directly with using the dot operator.
ii) Member function can access the private data of the class.
a) i-True, ii-True
b) i-False, ii-True
c) i-True, ii-False
d) i-False, ii-False
Answer:
b) i-False, ii-True

Question 3.
A member function can call another member function directly, without using the dot operator called as
a) sub function
b) sub member
c) nesting of member function
d) sibling of member function
Answer:
c) nesting of member function

Question 4.
The member function defined within the class behave like
a) inline functions
b) Non inline function
c) Outline function
d) Data function
Answer:
a) inline functions

Question 5.
Which of the following access specifier protects data from inadvertent modifications?
a) Private
b) Protected
c) Public
d) Global
Answer:
a) Private

Question 6.
class x
{
inty;
public:
x(int z)
{
y=z;
}
} x1[4];
intmain( )
{
x x2(10);
return 0;
}
How many objects are created for the above program?
a) 10
b) 14
c) 5
d) 2
Answer:
c) 5

Question 7.
State whether the following statements about the constructor are True or False.
i) constructors should be declared in the private section.
ii) constructors are invoked automatically when the objects are created.
a) True, True
b) True, False
c) False, True
d) False, False
Answer:
c) False, True

Question 8.
Which of the following constructor is executed for the following prototype ?
add display (add &); // add is a class name
a) Default constructor
b) Parameterized constructor
c) Copy constructor
d) Non Parameterized constructor
Answer:
c) Copy constructor

Question 9.
What happens when a class with parameterized constructors and having no default constructor is used in a program and we create an object that needs a zero- argument constructor?
a) Compile-time error
b) Domain error
c) Runtime error
d) Runtime exception
Answer:
a) Compile-time error

Question 10.
Which of the following create a temporary instance?
a) Implicit call to the constructor
b) Explicit call to the constructor
c) Implicit call to the destructor
d) Explicit call to the destructor
Answer:
b) Explicit call to the constructor

Part – II

Very Short Answers

Question 1.
What are called members?
Answer:
The class comprises members. Members are classified as Data Members and Member functions. Data members are the data variables that represent the features or properties of a class. Member functions are the functions that perform specific tasks in a class.

Question 2.
Differentiate structure and class though both are user-defined data types.
Answer:
The only difference between structure and class is the members of the structure are by default public whereas it is private in class.

Question 3.
What is the difference between the class and object in terms of oop?
Answer:
Object:

• Object is an instance of a class.
• Object is a real-world entity such as pen, laptop, mobile, chair, etc.
• Object allocates memory when it is created.

Class:

• Class is a blueprint or template from which objects are created.
• Class is a group of similar objects.
• Class doesn’t allocate memory when it is created.

Question 4.
Why it is considered a good practice to define a constructor though a compiler can automatically generate a constructor?
Answer:
A user-defined constructor is the best method of initialise array of objects and normal objects.

Question 5.
Write down the importance of the destructor.
Answer:
The purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object.

Part – III

Short Answers

Question 1.
Rewrite the following program after removing the syntax errors if any and underline the errors:
#include<iostream>
#include<stdio.h>
classmystud
{ intstudid =1001;
char name[20];
public
mystud( )
{ }
void register ( ) {cin>>stdid;gets(name);
}
void display ( )
{ cout<<studid<<“: “<<name<<endl;}
}
int main( )
{ mystud MS;
register.MS( );
MS.display( );
}
Answer:
MODIFIED PROGRAM:
#include<iostream>
#include<stdio.h>
class mystud
{
int studid;
char name[20];
public:
mystud( )
{
studid=1001;
}
void register ( )
{
cin>>stdid;
gets(name);
}
void display ( )
{
cout<<studid<<“: “<<name<<endl;
}
};
int main( )
{
mystud MS;
MS.reqister( );
MS.display( );
}

Question 2.
Write with an example of how will you dynamically initialize objects?
Answer:
Dynamic initialization of Objects:
When the initial values are provided during runtime then it is called dynamic initialization.
Program to illustrate dynamic initialization
#include
using namespace std;
class X
{
int n;
float avg;
public:
X(int p,float q)
{
n=p;
avg=q;
}
void disp( )
{
cout<<“\n Roll numbe:-” <<n;
cout<<“\nAverage :-“<<avg;
}
};
int main( )
{
int a ; float b;
cout<<“\nEnter the Roll Number”;
cin>>a; .
cout<<“\nEnter the Average”;
cin>>b;
X x(a,b); // dynamic initialization
x.disp( );
return 0;
}
Output
Enter the Roll Number 1201
Enter the Average 98.6
Roll number:- 1201
Average :- 98.6

Question 3.
What are advantages of declaring constructors and destructor under public access ability?
Answer:

When constructor and destructor are declared under public:

1. we can initialize the object while declaring it.
2. we can explicitly call the constructor.
3. we can overload constructors and therefore use multiple constructors to initialize objects automatically.
4. we can destroy the objects at the end of class scope automatically (free unused memory).

However, some C++ compiler and Dev C++ do not allow to declare constructor and destructor under private section. So it is better to declare constructor and destructor under public section only.

Question 4.
Given the following C++ code, answer the questions (i) & (ii).
Answer:
class TestMeOut
{
public:
~TestMeOut( ) //Function 1
{
cout<<“Leaving the examination hall”<<endl;
}
TestMeOut( ) //Function 2
{
cout<<“Appearing for examination'<<endl;
}
void MyWork( ) //Function 3
{
cout<<“Attempting Questions//<<endl;
}
};
i) In Object-Oriented Programming, what is Function 1 referred to as and when does it get invoked/called?
Function 1 is called a destructor. It will be automatically invoked when the object goes out of scope (ie. at the end of a program).

ii) In Object-Oriented Programming, what is Function 2 referred to as, and when does it get invoked/called?
Function2 is called a constructor. It will be automatically invoked when an object comes into scope.(ie. at the time of object creation).

Question 5.
Write the output of the following C++ program code:
Answer:
#include<iostream>
using namespace std;
class Calci
{
char Grade; .
int Bonus;
public:
Calci( )
{
Grade=’E’;
Bonus=0;
}//ascii value of A=65
void Down(int G)
{
Grade-=G;
}
void Up(int G)
{
Grade+=G;
Bonus++;
}
void Show( )
{
cout<<Grade<<“#”<<Bonus<<endl;
}
};
int main( )
{
Calci c;
c.Down(3);
c.Show( );
c.Up(7);
c.Show( );
c.Down(2);
c.Show( );
return 0;
}
Output

Part – IV

Explain In Detail

Question 1.
Explain nested class with example.
Answer:
When one class becomes a member of another class then it is called Nested class and the relationship is called containership. When a class is declared within another class, the inner class is called a Nested class (i.e. the inner class) and the outer class is known as the Enclosing class. The nested class can be defined in private as well as in the public section of the Enclosing class.

Classes can be nested in two ways:

1. By defining a class within another class
2. By declaring an object of a class as a member to another class
3. By defining a class within another class

C++ program to illustrate the nested class
#include<iostream>
using namespace std;
class enclose
{
private:
int x;
class nest
{
private :
int y;
public:
int z;
void prn( )
{
y=3;z=2;
cout<<“\n The product of”
< <y< <‘*'< <z<<“= “< <y*z< <“\n”;
}
}; //inner class definition over
nest m1;
public:
nest n2;
void square( )
{
n2.prn( ); //inner class member function is called by its object
x=2;
n2.z=4;
cout<<“\n The product of” <<n2.z<<‘*'<<n2.z<<“=”n2.z*n2.z<<“/n”;
cout<<“\n The product of” <<x<<‘*'<<x<<“= “<<x*x;
}
}; //outer class definition over
int main( )
{
enclose e;
e.square( ); //outer class member function is called
}
Output
The product of 3*2=6
The product of 4*4=16
The product of 2*2=4

In the above program the inner class nest is defined inside the outer class enclose, nest is accessed by enclose by creating an object of nest

Question 2.
Mention the differences between constructor and destructor.
Answer:

CONSTRUCTOR	DESTRUCTOR
The name of the constructor must be same as that of the class.	The destructor has the same name as that of the class prefixed by the tilde character
A constructor can have parameter list.	The destructor cannot have arguments.
The constructor function can be overloaded.	Destructors cannot be overloaded i.e., there can be only one destructor in a class.
Constructor cannot be inherited but a derived class can call the base class constructor.	Destructor cannot be inherited.
The constructor is executed automatically when the object is created.	The destructor is executed automatically when the control reaches the end of class.
Allocated memory space for the object.	Destroy the object.

Question 3.
Define a class RESORT with the following description in C++ :
Answer:
Private members:
Rno // Data member to storeroom number
Name //Data member to store user name
Charges //Data member to store per day charge
Days //Data member to store the number of days
Compute ( ) // A function to calculate total amount as Days * Charges and if the
//total amount exceeds 11000 then total amount is 1.02 * Days *Charges

Public member:
getinfo( ) // Function to Read the information like name , room no, charges and days
dispinfo ( ) // Function to display all entered details and total amount calculated
//using COMPUTE function
PROGRAM
using namespace std;
#include<iostream>
class RESORT
{
private:
int Rno,Days,Charges;
char Rname[20];
int compute( )
{
if (Days * Charges >11000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
getinfo( )
{
cout<<“\nEnter customer name : “;
cin>>Rname;
cout<<‘nEnter charges per day : “;
cin>>Charges;
cout< <‘nEnter Number of days : “;
cin>>Days;
cout<<‘n Enter Room Number : “;
cin>>Rno;
}
dispinfo( )
{
cout<<‘nRoom Number :
“<<Rno;
cout<<‘nCustomer name :
“<<Rname;
cout<<‘nCharges per day :
“<<Charges;
cout<<‘nNumber of days :
“<<Days;
cout<<‘nTotal Amount :
“<<compute( );
}
int main( )
{
RESORT Obj;
Obj,getinfo( );
Obj.dispinfo( );
Output

Question 4.
WrIte the output of the following:
Answer:
#include<iostream>
#indude<stdio.h>
using namespace std;
class sub
{
int day, subno;
public :
sub(int,int); // prototype
void printsub( )
{
cout<<” subject number: “<<subno;
cout<<” Days : ” <<day;
}
};
sub::sub(int d=150,int sn=12)
{
cout<<endl<<“Constructing the object
“<<endl;
day=d;
sub no=sn;
}
class stud ‘
{
int rno;
float marks; public:
stud( )
{
cout<< “Constructing the object of
students “<<endl;
rno=0;
marks=0.0;
}
void getval( )
{
cout<<“Enter the roll number and the marks secured”; cin>>rno>>marks;
}
void printdet( )
{
cout<<“Roll no : “<<rno<<“Marks : “<<marks<<endl; .
}
};
class admission
{
sub obj;
stud objone;
float fees; ,
public :
admission ( )
{
cout<< “Constructing the object of admission “<<endl;
fees=0.0;
}
void printdet( )
{
objone.printdet( );
obj.printsub();
cout<<“fees : “<<fees<<endl;
}
};
int main( )
{
system (“cls”);
admission adm;
cout<<endl<< ?’Back in main ()”;
return 0;
}
Output

Question 5.
Write the output of the following.
Answer:
#indude<iostream>
#include<stdio,h>
using namespace std;
class P
{
public:
P( )
{
cout<< “\nConstructor of class P }
~P( )
{
cout< < “\nDestructor of class P
}
};
class Q
{
public:
Q( )
{
cout< <“\nConstructor of class Q “;}
~ Q( )
{
cout<< “\nDestructor of class Q
}
};
class R
{
P obj1, obj2;
Q obj3;
public:
R( )
{
cout<< “\nConstructor of class R “;}
~R( )
{
cout<< “\nDestructor of class R
}
};
int main ( )
{
R oR; :
Q oq;
Pop;
return 0;
Output

11th Computer Science Guide Classes and Objects Additional Questions and Answers

Choose The Correct Answer (1 Mark)

Question 1.
The most important feature of C++ is ………………..
(a) object
(b) class
(c) public
(d) All the above
Answer:
(b) class

Question 2.
How many features are commonly present in OOP languages?
a) 3
b) 2
c) 4
d) 5
Answer:
c) 4

Question 3.
Calling a member function of an object is also known as ……………….. to object.
(a) call function
(b) call by value
(c) call by reference
(d) sending message
Answer:
(d) sending message

Question 4.
……………… is a way to bind the data and its associated functions together,
a) Class
b) Array
c) Structure
d) All the above
Answer:
a) Class

Question 5.
When one class become a member of another class, the relationship is called ………………..
(a) containership
(b) partnership
(c) friendship
(d) all the above
Answer:
(a) containership

Question 6.
The body of the class is defined inside the ………………. brackets.
a) Angle < >
b) Square [ ]
c) Curly { }
d) None of these
Answer:
c) Curly { }

Question 7.
……………….. can be defined either in private or in the public section of a class.
(a) Object
(b) Data type
(c) Memory
(d) constructor
Answer:
(d) constructor

Question 8.
The members of the structure are by default ………………..
a) Private
b) Public
c) Protected
d) None of these
Answer:
b) Public

Question 9.
There are ……………….. ways to create an object using the parameterized constructor.
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(c) 1

Question 10.
The class body contains ………………….
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 11.
The class body has………………… access specifiers.
a) Three
b) Four
c) Two
d) Five
Answer:
a) Three

Question 12.
The class body has…………….. access specifiers.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 13.
…………………. is a visibility label.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 14.
……………….. allows preventing the functions of a program to access directly the internal representation of a class type.
a) Data Hiding
b) Data Capturing
c) Data Processing
d) None of these
Answer:
a) Data Hiding

Question 15.
The access restriction to the class members is specified by ……………. section within the class
body.
a) Private
b) Public
c) Protected
d) All the above
Answer:
d) All the above

Question 16.
A …………………. member is accessible from where outside the class but within a program.
a) Private
b) Public
c) Protected
d) All the above
Answer:
b) Public

Question 17.
We can set and get the value of public data members using ………………… function.
a) Member
b) Nonmember
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 18.
A …………….. member cannot be accessed from outside the class.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 19.
Only the class member functions can access ……………… members.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 20.
……………….. members can be accessed in child classes.
a) Private
b) Public
c) Protected
d) All the above
Answer:
c) Protected

Question 21.
If all members of the class are defined as …………….. then the class become frozen.
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 22.
If all members of the class are defined as ………………….. then the object of the class can not access anything from the class,
a) Private
b) Public
c) Protected
d) All the above
Answer:
a) Private

Question 23.
………………. are the data variables that represent the features or properties of a class.
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
a) Data members

Question 24.
………………… are the functions that perform specific tasks in a class.
a) Data members
b) Member functions
c) Both A and B
d) None of these
Answer:
b) Member functions

Question 25.
Member functions are called as …………………..
a) Methods
b) Attributes
c) Properties
d) None of these
Answer:
a) Methods

Question 26.
Data members are also called as ……………….
a) Methods
b) Attributes
c) Properties
d) None of these
Answer:
b) Attributes

Question 27.
Classes contain a special member function called as ……………………
a) Constructors
b) Destructors
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 28.
The member functions of a class can be defined in ……………….. ways.
a) Two
b) Three
c) Four
d) Five
Answer:
a) Two

Question 29.
The member functions of a class can be defined in ………………. way.
a) Inside the class definition
b) Outside the class definition
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 30.
When a member function is defined Inside a class, it behaves like ………………. functions.
a) Inline
b) General
c) Local
d) None of these
Answer:
a) Inline

Question 31.
If a function is inline, the compiler places a copy of the code of that function at each point where the function is called at ………………….
a) Run Time
b) Compile time
c) Both A and B
d) None of these
Answer:
b) Compile time

Question 32.
When Member function defined outside the class, and then it is be called as ………………….
member function.
a) Outline
b) Non-inline
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 33.
When Member function defined outside the class using …………….. operator.
a) Scope resolution
b) Membership
c) Reference
d) Conditional
Answer:
a) Scope resolution

Question 34.
The class variables are called ………………….
a) Object
b) Attributes
c) Procedures
d) None of these
Answer:
a) Object

Question 35.
Objects are also called as …………………. of class.
a) Instant
b) Instance
c) Attributes
d) None of these
Answer:
b) Instance

Question 36.
Objects can be created in ………………… methods.
a) Three
b) Four
c) Two
d) Five
Answer:
c) Two

Question 37.
Objects can be created as ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 38.
…………….. objects can be used by any function in the program.
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
a) Global object

Question 39.
If an object is declared outside all the function bodies or by placing their names immediately after the closing brace of the class declaration then it is called as ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
a) Global object

Question 40.
If an object is declared within a function then it is called ……………….
a) Global object
b) Local object
c) Either A or B
d) None of these
Answer:
b) Local object

Question 41.
……………….. object can not be accessed from outside the function.
a) Global
b) Local
c) Either A or B
d) None of these
Answer:
b) Local

Question 42.
No separate space is allocated for …………………. when the objects are created.
a) Member functions
b) Data members
c) Both A and B
d) None of these
Answer:
a) Member functions

Question 43.
Memory space required for the ……………….
a) Member functions
b) Data members
c) Both A and B
d) None of these
Answer:
b) Data members

Question 44.
The members of a class are referenced (accessed) by using the object of the class followed by the ………………. operator.
a) Scope resolution
b) Conditional
c) Dot (membership)
d) None of these
Answer:
c) Dot (membership)

Question 45.
Calling a member function of an object is also known as ……………….
a) Sending a message to object
b) Communication with the object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 46.
An array which contains the class type of element is called …………………
a) Array of objects
b) Structure Objects
c) Block of objects
d) None of these
Answer:
a) Array of objects

Question 47.
The ………………… of the outline member function given in a class specification, instructs the compiler about its visibility mode.
a) Name
b) Prototype
C) Data type
d) None of these
Answer:
b) Prototype

Question 48.
A member function can call another member function of the same class directly without using the dot operator is called ………………………
a) Nesting of the member function
b) Invariant Members
c) Variant Members
d) None of these
Answer:
a) Nesting of the member function

Question 49.
A member function can call another member function of the same class for that you do not
need a(n) …………………..
a) Member function
b) Data Member
c) Object
d) None of these
Answer:
c) Object

Question 50.
A member function can access ………………. functions.
a) Public
b) Private
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 51.
…………………. operator will reveal the hidden file scope(global) variable.
a) Membership
b) Conditional
c) Scope resolution
d) All the above
Answer:
c) Scope resolution

Question 52.
When an object is passed by ……………… the function creates its own copy of the object and works on it.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
a) Value

Question 53.
When an object is passed by …………….. changes made to the object inside the function do not affect the original object.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
a) Value

Question 54.
When an object is passed by ……………….. its memory address is passed to the function so the called function works directly on the original object used in the function call.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
b) Reference

Question 55.
When an object is passed by ………………………. any changes made to the object inside the function definition are reflected in the original object.
a) Value
b) Reference
c) Either A or B
d) None of these
Answer:
b) Reference

Question 56.
Member Functions can ………………….
a) Receive object as an argument
b) Return an object
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 57.
When one class become a member of another class then it is called ……………. class.
a) Nested
b) Inline
c) External
d) Global
Answer:
a) Nested

Question 58.
When one class becomes a member of another class then the relationship is called ………………….
a) Containership
b) Nesting
c) Parent-Child
d) None of these
Answer:
a) Containership

Question 59.
Classes can be nested in ……………….. ways.
a) Three
b) Two
c) Four
d) Five
Answer:
b) Two
Question 60.
Classes can be nested in ………………..way.
a) By defining a class within another class
b) By declaring an object of a class as a member to another class
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 61.
When a class is declared within another class, the inner class is called a Nested class (ie the inner class) and the outer class is known as …………………… class.
a) Enclosing
b) Abstract
c) Transit
d) None of these
Answer:
a) Enclosing

Question 62.
The nested class can be defined in ………………….. section of the Enclosing class.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
c) Either Private or Public

Question 63.
Whenever an object of a class is declared as a member of another class it is known as a _____ class.
a) Abstract
b) Container
c) Literal
d) None of these
Answer:
b) Container

Question 64.
Instantiating object is done using ……………….
a) Constructor
b) Destructor
c) Data abstraction
d) Data hiding
Answer:
a) Constructor

Question 65.
A(n) ………………..in C++ can be initialized during the time of their declaration.
a) Array
b) Structure
c) Array or Structure
d) None of these
Answer:
c) Array or Structure

Question 66.
Member function of a class can access all the members irrespective of their associated …………………..
a) Access specifier
b) Data type
c) Return type
d) Argument
Answer:
a) Access specifier

Question 67.
When an instance of a class comes into scope, a special function called the ……………. gets executed.
a) Constructor
b) Destructor
c) Data abstraction
d) Data hiding
Answer:
a) Constructor

Question 68.
The constructor function name has the same name as the …………….. name.
a) Object
b) Class
c) Data member
d) None of these
Answer:
b) Class

Question 69.
The constructors return ………………
a) int
b) char
c) float
d) nothing
Answer:
d) nothing

Question 70.
…………………. are not associated with any data type
a) Constructor
b) Data member
c) Data abstraction
d) Member functions
Answer:
a) Constructor

Question 71.
……………… can be defined either inside class definition or outside the Class definition.
a) Constructor
b) Destructor
c) Data abstraction
d) Member functions
Answer:
a) Constructor

Question 72.
A constructor can be defined in ……………… section of a class.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
c) Either Private or Public

Question 73.
If a constructor is defined in ………………… section of a class, then only its object Can be created in any function.
a) Private
b) Public
c) Either Private or Public
d) None of these
Answer:
b) Public

Question 74.
The main function of the constructor is ……………………….
a) To allocate memory space to the object
b) To initialize the data member of the class object
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 75.
A constructor that accepts no parameter is called …………………… constructor.
a) Null
b) Default
c) Empty
d) None of these
Answer:
b) Default

Question 76.
Identify the correct statement from following with respect to constructor.
a) If a class does not contain an explicit constructor (user defined constructor) the compiler automatically generate a default constructor implicitly as an inline public member.
b) In the absence of user defined constructor the compiler automatically provides the default constructor. It simply allocates memory for the object.
c) Parameterized constructor is achieved by passing parameters to the function.
d) All the above
Answer:
d) All the above

Question 77.
A constructor which can take arguments is called ……………….. constructor.
a) Parmeterized
b) Default
c) Empty
d) None of these
Answer:
a) Parmeterized

Question 78.
……………….. type of constructor helps to create objects with different initial values.
a) Parmeterized
b) Default
c) Empty
d) None of these
Answer:
a) Parmeterized

Question 79.
Declaring a constructor with arguments hides the ……………………
a) Data members
b) Compiler generated constructor
c) Member functions
d) None of these
Answer:
b) Compiler generated constructor

Question 80.
………………… Constructor is used to creating an array of objects.
a) Default
b) Parameterized
b) Overloaded
d) None of these
Answer:
a) Default

Question 81.
There are ………………. ways to create an object using the parameterized constructor.
a) Three
b) Two
c) Four
d) Five
Answer:
b) Two

Question 82.
……………… is a way to create an object using the parameterized constructor,
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
c) Either A or B

Question 83.
In …………….. method, the parameterized constructor is invoked automatically
whenever an object is created.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
a) Implicit call

Question 84.
In the………………….. method, the name of the constructor is explicitly given to invoking the parameterized constructor.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 85.
………………… method is the most suitable method as it creates a temporary object
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 86.
The chance of data loss will not arise in ………………….. method.
a) Implicit call
b) Explicit call
c) Either A or B
d) None of these
Answer:
b) Explicit call

Question 87.
A ……………….. object lives in memory as long as it is being used in an expression.
a) Temporary
b) Nested
c) Inline
d) None of these
Answer:
a) Temporary

Question 88.
A constructor having a reference to an already existing object of its own class is called ………………….. constructor.
a) Reference
b) Value
c) Copy
d) Move
Answer:
c) Copy

Question 89.
A copy constructor is called ……………..
a) When an object is passed as a parameter to any of the member functions
b) When a member function returns an object
c) When an object is passed by reference to an instance of its own class
d) All the above
Answer:
d) All the above

Question 90.
The constructors are executed in the …………………. of the object declared.
a) Order
b) Reverse order
c) Either A or B
d) None of these
Answer:
a) Order

Question 91.
When the initial values are provided during runtime then it is called ………………….. initialization.
a) Static
b) Dynamic
c) Run time
d) None of these
Answer:
b) Dynamic

Question 92.
…………………. constructor can have parameter list.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 93.
No return type can be specified for ………………….
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 94.
The ……………….. function can be overloaded.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 95.
The compiler generates a ………………… in the absence of a user-defined.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 96.
Compiler generated constructor is ………………….. member function.
a) private
b) protected
c) public
d) None of these
Answer:
c) public

Question 97.
The ………………. is executed automatically,
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Question 98.
The ……………. is executed automatically when the object is created.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
a) Constructor

Question 99.
When a class object goes out of scope, a special function called the ………………. gets executed.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 100.
The destructor has the same name as the class tag but prefixed with a …………………
a) ~ (tilde)
b) #
c) @
d) None of these
Answer:
a) ~ (tilde)

Question 101.
A …………………… is a special member function that is called when the lifetime of an object ends.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 102.
The ………………… cannot have arguments,
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
b) Destructor

Question 103.
There can be ……………… destructor in a class.
a) Two
b) Three
c) Only one
d) Four
Answer:
c) Only one

Question 104.
_____ cannot be inherited.
a) Constructor
b) Destructor
c) Both A and B
d) None of these
Answer:
c) Both A and B

Very Short Answer (2 Marks)

Question 1.
Define methods of a class and write its types.
Answer:
The class comprises members. Member functions are called methods. The member functions of a class can be defined in two ways.

1. Inside the class definition
2. Outside the class definition

Question 2.
Why classes are needed?
Answer:
Classes are needed to represent real-world entities that not only have data type properties but also have associated operations. It is used to create user-defined data types.

Question 3.
What is called as nesting of member functions?
Answer:
Only the public members of a class can be accessed by the object of that class, using the dot operator. However, a member function can call another member function of the same class directly without using the dot operator. This is called as nesting of member functions.

Question 4.
What are the visibility labels of a class body?
Answer:
The class body has three visibility labels viz., private, public, and protected. The Visibility labels are also called as access specifiers.

Question 5.
What is a parameterized constructor?
Answer:
A constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.

Question 6.
What happened if all the members of a class are defined as private?
Answer:
If all members of the class are defined as private, then the class becomes frozen.
The object of the class can not access anything from the class.

Question 7.
Write about objects.
Answer:
A class specification just defines the properties of a class. To make use of a class specified, the variables of that class type have to be declared. The class variables are called objects. Objects are also called an instance of the class.

For example:
student s;
In the above statement ‘s’ is an instance of the class student.

Question 8.
How many ways objects can be created for a class? Give its types.
Answer:
Objects can be created in two methods:

1. Global object
2. Local object

Question 9.
What do you mean by an array of objects?
Answer:
An array which contains the class type of element is called an array of objects. It is declared and defined in the same way as any other type of array.

Example:
class stock
{
int itemno;
float price; public:
}s[5];
Here s[5] is an array of objects.

Question 10.
What is a nested member function?
Answer:
A member function can cal! another member function of the same class directly without using the dot operator. This is called as nesting of member functions.

Question 11.
How many ways objects can be passed to function argument?
Answer:
Objects can also be passed in both ways

• Pass By Value
• Pass By Reference

Question 12.
What is a container class?
Answer:
Whenever an object of a class is declared as a member of another class it is known as a container class. In the container-ship, the object of one class is declared in another class.

Question 13.
What is the need for a constructor in a class?
Answer:
Instantiating object is done using constructor. An array or a structure in C++ can be initialized during the time of their declaration using constructor. The constructor function initializes the class object.

Question 14.
What are the functions of a constructor?
Answer:
The main functions of the constructor are:

• To allocate memory space to the object and
• To initialize the data member of the class object.

Question 15.
What is a default constructor?
Answer:
Default constructor:
A constructor that accepts no parameter is called the default constructor.
For example in the class Data program Data::Data( ) is the default constructor.
Using this constructor objects are created similar to the way the variables of other data types are created.

Example:
int num; //ordinary variable declaration
Data d1; // object declaration
If a class does not contain an explicit constructor the compiler automatically generates a default constructor implicitly as an inline public member.

Question 16.
What is the significance of default constructor?
Answer:
Default constructors are very useful to crate objects without having specific initial value. It is also used to create array of objects.

Question 17.
How many ways a constructor can be invoked?
Answer:
There are two ways to create an object using parameterized constructor:

1. Implicit call
2. Explicit call

Question 18.
What is copy constructor?
Answer:
A constructor having a reference to an already existing object of its own class is called copy constructor.
In other words Copy Constructor is a type of constructor which is used to create a copy of an already existing object of a class type.

Question 19.
What is the order of constructor invocation?
Answer:
The constructors are executed in the order of the object declared. (If it is in same statement left to right)

For example:
Test t1;
Test t2; // the order of constructor execution is first for t1 and then for t2.
Consider the following example
Sample s1,s2,s3 ; //The order of construction is s1 then s2 and finally s3

Question 20.
What do you mean by dynamic initialization of Object?
Answer:
When the initial values are provided during runtime then it is called dynamic initialization.

Question 21.
Write a note on the destructor.
Answer:

• When a class object goes out of scope, a special function called the destructor gets executed.
• The destructor has the same name as the class tag but prefixed with a ~(tilde).
• The destructor function also returns nothing and it does not associate with any data type

Question 22.
What is the need for a destructor in a class?
Answer:
The purpose of the destructor is to free the resources that the object may have acquired during its lifetime. A destructor function removes the memory of an object which was allocated by the constructor at the time of creating an object

Question 23.
Define destructor.
Answer:
A destructor is a special member function that is called when the lifetime of an object ends and destroys the object constructed by the constructor. Normally it is declared under the public visibility of a class.

Short Answers (3 Marks)

Question 1.
Explain the local object with an example.
Answer:
If an object is declared within a function then it is called a local object.
It cannot be accessed from outside the function.
# include
# include
using namespace std
class add  //Global class
{
int a,b; public:
int sum; void
getdata()
{
a = 5; b = 10; sum
= a + b;
}
} a1;
add a2;
int main()
{
add a3;
a1.getdata();  //global object
a2.getdata();  //global object
a3.getdata();
cout << a1 .sum;  //Local object for a global class
cout << a2.sum;
cout << a3.sum;
return 0;   //public data member accessed from outside the class
}
Output:
151515

Question 2.
Write about private, protected, and public members of a class.
Answer:
The Public Members:
A public member is accessible from anywhere outside the class but within a program.

The Private Members:
A private member cannot be accessed from outside the class. Only the class member functions can access private members. By default, all the members of a class would be private.

The Protected Members:
A protected member is very similar to a private member but they can be accessed in child classes which are called derived classes (inherited classes).

Question 3.
What is a constructor?
Answer:
The definition of a class only creates a new user-defined data type. The instances of the class type should be instantiated (created and initialized). Instantiating objects is done using the constructor. An array or a structure in C++ can be initialized during the time of their declaration.

The initialization of a class type object at the time of declaration similar to a structure or an array is not possible because the class members have their associated access specifiers (private or protected or public). Therefore Classes include special member functions called constructors. The constructor function initializes the class object.

Question 4.
Explain memory allocation of objects.
Answer:
Memory allocation of objects:
All the objects belonging to that class use the same member function, no separate space is allocated for member functions when the objects are created.
Memory space required for the member variables are only allocated separately for each object because the member variables will hold different data values for different objects.
Memory for Objects for p1 and p2 is illustrated:

Question 5.
Explain the default constructor with an example.
Answer:
A constructor that accepts no parameter is called the default constructor. For example in the class data program Data::Data() is the default constructor. Using this constructor objects are created similar to the way the variables of other data types are created.

Example:
int num; //ordinary variable declaration
Data d1; // object declaration

Question 6.
How will you refer members of the class? Give its syntax and an example.
Answer:
The members of a class are referenced (accessed) by using the object of the class followed by the dot (membership) operator and the name of the member.

The general syntax for calling the member function is:
Object_name.function_name (actual parameter); For example consider the following illustration:

Question 7.
Explain the different methods of passing an object to the function argument.
Answer:
Pass By Value:
When an object is passed by value the function creates its own copy of the object and works on it. Therefore any changes made to the object inside the function do not affect the original object.

Pass By Reference:
When an object is passed by reference, its memory address is passed to the function so the called function works directly on the original object used in the function call. So any changes made to the object inside the function definition are reflected in the original object.

Question 8.
Write about constructor.
Answer:
When an instance of a class comes into scope, a special function called the constructor gets executed. The constructor function name has the same name as the class name. The constructors return nothing. They are not associated with any data type. It can be defined either inside class definition or outside the class definition.

Question 9.
What is a parameterized constructor?
Answer:
Parameterized Constructors:
A constructor which can take arguments is called a parameterized constructor. This type of constructor helps to create objects with different initial values. This is achieved by passing parameters to the function.

Example:
class simple
{
private:
int a,b;
public:
simple(int m, int n)
{
a= m ;
b= n;
cout< < “\n Parameterized Constructor of class-simple
}
};

Question 10.
What do you mean by the implicit and explicit call of a constructor?
Answer:
Implicit call:
In this method, the parameterized constructor is invoked automatically whenever an object is created.
For example, simple s1(10,20); in this for creating the object si parameterized constructor is automatically invoked

Explicit call:
In this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.

For example:
simple s1=simple(10,20); //explicit call

Question 11.
When copy constructor Is executed? Give examples.
Answer:
A copy constructor is called

• When an object is passed as a parameter to any of the member functions
  Example: void simple: :putdata(simple x);
• When a member function returns an object
  Example: simple get data( ) { }
• When an object is passed by reference to an instance of its own class
  For example: simple1, s2(s1); // s2(s1) calls copy constructor

Explain in Detail (5 Marks)

Question 1.
Explain how to define class members?
Answer:
Definition of class members:
Class comprises of members. Members are classified as Data Members and Member functions.

• Data members are the data variables that represent the features or properties of a class.
• Member functions are the functions that perform specific tasks in a class.
• Member functions are called methods, and data members are also called attributes.

Example:

Defining methods of a class:
Without defining the methods (functions), class definition will become incomplete. The member functions of a class can be defined in two ways.

• Inside the class definition
• Outside the class definition

Inside the class definition:
When a member function is defined inside a class, it behaves like inline functions. These are called Inline member functions.

Outside the class definition:
When Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.

The syntax for defining the outline member function is:
return_type class_name :: function name (parameter list)
{
function definition

Class using Inline and Outline member function:
# include<iostream>
using namespace std;
class Box
{
// no access specifier mentioned
double width;
public:
double length;
//inline member function definition
void printWidth( )
{
cout<<“\n The width of the box is…”<<width;
}
//prototype of the function
void setWidth(double w);
};
// outline member function definition
void Box :: setWidth(double w)
{
width=w;
}
int main( )
{
// object for class Box
Box b;
// Use member function to set the width.
b.setWidth(10.0);
//Use member function to print the width.
b.printWidth( );
return 0;
Output
The width of the box is… 10

Question 2.
What are the ways to create an object using the parameterized constructor with an example?
Answer:
There are two ways to create an object using the parameterized constructor:
1. Implicit call: In this method, the parameterized constructor is invoked automatically whenever an object is created. For example, simple s1( 10,20); in this, for creating the object s1 parameterized constructor is automatically invoked.

2. Explicit call: In this method, the name of the constructor is explicitly given to invoking the parameterized constructor so that the object can be created and initialized.

#include
using namespace std;
class simple
{
private:
int a, b;
public:
simple(int m,int n)
{
a = m;
b = n;
cout << “\n Constructor of class – simple invoked for implicit and explicit call” << endl;
}
void putdata()
{
cout << “\n The two integers are…” << a << ‘\t’ << b << endl;
cout << “\n The sum of the variables” << a << “+” << b << “=” << a + b;
}
};
int main()
{
simple s1(10,20); //implicit call
simple s2 = simple(30,45); //explicit call
cout << “\n\t\tObject 1\n”;
s1.putdata();
s2.putdata();
return 0;
}
Output:
Constructor of class – simple invoked for the implicit and explicit call
Constructor of class-simple invoked for the implicit and explicit call

Object 1
The two integers are… 10 20
The sum of the variables 10 + 20 = 30

Object 2
The two integers are… 30 45.
The sum of the variables 30 + 45 = 75

Question 3.
What are the characteristics of a destructor?
Answer:
Characteristics of destructors:

• The destructor has the same name as that of the class prefixed by the tilde character
• The destructor cannot have arguments.
• It has no return type.
• Destructors cannot be overloaded i.e., there can be only one destructor in a class.
• In the absence of a user-defined destructor, it is generated by the compiler.
• The destructor is executed automatically when the control reaches the end of the class scope to destroy the object.
• They cannot be inherited.

Evaluate Yourself

Question 1.
Define a class in general and in C++’s context.
Answer:
Classes represent real-world entities that not only have data type properties but also have associated operations.
In C++ class is a way to bind the data and its associated functions together. It is a user-defined data type.

Question 2.
What is the purpose of a class specifier?
Answer:
Data hiding is one of the important features of Object Oriented Programming which allows preventing the functions of a program to access directly the internal representation of a class type.
The access restriction to the class members is specified by class specifies like public, private, and protected sections within the class body.

Question 3.
Compare a structure and a class in C++ context.
Answer:
The only difference between structure and class is the members of structure are by default public where as it is private in class.

Question 4.
Compare private and public access specifier.
Answer:
Public members:
A public member is accessible from anywhere outside the class but within a program. We can set and get the value of public data members even without using any member function.

Private members:
A private member cannot be accessed from outside the class. Only the class member functions can access private members. By default all the members of a class would be private.

Question 5.
What is a non-inline member function? Write its syntax.
Answer:
When Member function defined outside the class just like normal function definition (Function definitions you are familiar with) then it is being called as an outline member function or non-inline member function. Scope resolution operator (::) is used for this purpose.
The syntax for defining the outline member function is:
Syntax:
return_type class_name :: function_name (parameter list)
{
function definition
}

Activity – 1
State the reason for the invalidity of the following code fragment.

(i)	(ii)
class count { int first int second; public: int first };	class item { int prd; }; int prdno;

Answer:

• Data member first is duplicated and it is defined with two scopes( both private and public). It is invalid.
• Object name prefix with the only class name. No data type allowed in between class name and object name.

Activity – 2
class area
{
int s;
public:
void calc( );
};
Write an outline function definition for calc( ); which finds the area of a square
Answer:
int area :: calc( ) .
{
return(s * s);
}

Activity – 3
Identify the error in the following code fragment
class A
{
float x;
void init( )
{
Aa1;
X1.5=1; .
}
};
void main( )
{
A1.init( );
}
Answer:
Error:
Local object can not be accessed from outside the function. Al is the local object, so it can not be accessed in main( );

Activity – 4
What is the size of the objects s1, s2?
class sum
{
int n1,n2;
public:
void add( )
{
int n3=10;n1=n2=10;
}
} s1,s2;
Answer:
The size of the object SI and S2 is 8 bytes each in Dev C++, In Turbo C++ 4 bytes each.
Program to test the memory requirement:
class sum
{
int n1,n2;
public:
void add( )
{
int n3-10;
n1=n2=10;
}
} s1,s2;
using namespace std;
#include<iostream>
int main( )
{
cout<<sizeof(s1)<< “”<<sizeof(s2);
}
Output

Activity – 5
i) Write member function called display with no return.
class objects.
ii) Try the output of the above coding with the necessary modifications.
PROGRAM
#indude<iostream>
using namespace std;
class compute
{
int n1, n2;
public :
void init (int a, int b)
{
n1 = a;
n2 = b;
}
int n;
int add ( )
return (n1+n2);;
{
int prd ( )
{
return (n1*n2);
}
};
compute c1, c2;
void display(compute &objl,compute &obj2)
{
c1.init(12,15);
c2.init(8,4);
objlm = obj1.add( );
obj2.n = obj2.add( );
cout<<“\n Sum of object-1 “<<obj1.n;
cout<<“\n Sum of object-2 “<<obj2.n;
cout<<“\n Sum of the two objects are”<<obj1.
n+obj2.n;
c1.init(5,4);
c2.init(2,5);
obj1.n = obj1.prd( );
obj2.n = obj2.prd( );
cout<<“\n Product of object-1 “<<objl.n;
cout<<“\n Product of object-2 “<<obj2.n;
cout<<“\n Product of the two objects are “<<objl.n*obj2.n;
}
int main( )
{
display(c1,c2);
return 0;
}
output

Activity – 6
#include<iostream>
using namespace std;
class Sample
{
int i,j;
public :
int k;
Sample( )
{
i=j=k=0;//constructor defined inside the class
}
};
int main( )
{
Sample s1;
return 0;
}
Output
In the above program justify your reason for no output.
Answer:
Constructor alone is defined without output statement. When the above program is executed, the constructor executed. But no output on the screen because of missing cout

Hands-On Practice

Question 1.
Define a class employee with the following specification.
Answer:
private members of class Employee
empno- integer
ename – 20 characters
basic-float
netpay, hra, da, – float
calculate ( ) – A function to find the basic+hra+da with float return type

public member functions of class employee
havedata( ) – A function to accept values for empno, ename, basic, hra,
da and call calculate( ) to compute netpay
dispdata( ) – A function to display all the data members on the screen
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class Employee
{
private :
int empno;
char ename[20];
float basic,hra,da,netpay;
float calculate( )
{
return (basic+hra+da);
}
public:
void have data( )
{
cout<<setw(35)<<“Enter Employee number :”;
cin>>empno;
cout<<setw(35)<<“Enter Employee name :”;
cin>>ename;
cout<<setw(35)<<“Enter Basic pay :”;
cin>>basic; ,
cout<<setw(35)<<“Enter House Rent Allowance (HRA):”;
cin>>hra; .
cout<<setw(35)<<“Enter Dearness Allowance (DA):”;
cin>>da;
netpay = calculate( );
}
void dispdata( )
{
cout<<“\nEMPLOYEE DETAILS\n\n”;
cout<<setw(35)<<“Employee number :”<<empno<<endl;
cout<<setw(35)<<“Employee name :”<<ename<<endl;
cout<<setw(35)<<“Basic pay :”<<basic<<endl;
cout<<setw(35)<<“House Rent Allowance (HRA) :”<<hra<<endl; cout<<setw(35)<<“Dearness Allowance (DA) :”<<da<<endl;
cout<<setw(35)< <“Netpay :”<<netpay<<endl;
}
};
int main( )
{
Employee e;
e.havedata( );
e.dispdata( );
}
Output

Question 2.
Define a class MATH with the following specifications.
Answer:
private members:
num1, num2, result – float
init( ) function to initialize num1, num2 and result to zero .

protected members:
add( ) function to add num1 and num2 and store the sum in result
diff( ) function to subtract num1 from num2 and store the difference in the result

public members:
getdata( ) function to accept values for num1 and num2
menu( ) function to display menu
1. Add…
2. Subtract…
invoke add() when the choice is 1 and invoke prod when the choice is 2 and also display the result.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class MATH
{
private:
float num1,num2,result;
init( )
{
num1=0;
num2=0;
result=0;
}
protected: void add( )
{
result = num1+num2;
}
void diff( )
{
result = num1 – num2;
}
public:
getdata( )
{
cout<<“\nEntertwo numbers “;
cin>>num1>>num2;
} ‘ menu( )
{
int choice;
cout<<“\n1.Add …”;
cout<<“\n2.Subtract …….”;
cout<<“\nEnter your choice :”; cin>>choice;
switch(choice)
{
case 1: getdata( );
add( );
cout<<“\nAdded value is”<<result;
break;
case 2: getdata( );
diff( );
cout<<“\nSubtracted value is “<<result;
break;
default: cout<<“\End”;
}
}
};
int main( )
{
MATH m;
m. menu( );
}
Output

Question 3.
Create a class called Item with the following specifications.
Answer:
private members:
code, quantity- Integer data type
price – Float data type
getdata( )-function to accept values for all , data members with no return

public members:
taxt – float
dispdata( ) member function to display code,quantity,price and tax .The tax is calculated as if the quantity is more than 100 tax is 2500 otherwise 1000.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class Item
{
private:
int code,quantity;
float price;
void getdata( )
{
cout<<“\nEnter product code “; cin>>code;
cout<<“\nEnter quantity “; cin>>quantity;
cout<<“\nEnter price “; cin>> price;
}
public:
float tax;
void display( )
{
getdata( );
if(quantity>100)
tax = 2500;
else
tax = 1000;
cout<<endl<<setw(25)<< “Product code : “<<code<<endl<<endl;
cout<<setw(25)<<“Quantity : ” <<quantity<<endl<<endl;
cout<<setw(25)<<“Unit price :” <<price<<endl<<endl;
cout<<setw(25)<<“Total Amount: ” < cout<<setw(25)<<“Net Bill amount : ”
<<quantity* price+tax<<endl<<endl;
}
};
int main( )
{
Item i; i.display( );
}
Output

Question 4.
Write the definition of a class FRAME in C++ with the following description.
Answer:
Private members:
FramelD – Integer data type
Height, Width, Amount – Float data type
SetAmount( ) -Member function to calculate and assign amount as 10*Height*Width

Public members:
GetDetail( ) Afunction to allow user to entervalues of FramelD, Height, Width. This function should also call SetAmount() to calculate the amount.
ShowDetail( ) A function to display the values of all data members.
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
class FRAME
{
private:
int FrameId;
float Height, Width, Amount;
void SetAmount()
{
Amount = 10 * Height * Width;
}
public:
void Getdetails( )
{
cout<<“\nEnter Frame Id : “; cin>> FrameId;
cout<<“\nEnter Frame Height: cin>> Height;
cout<<“\nEnter Frame Width : “; cin>>Width;
SetAmount( );
}
void ShowDetaiis( )
{
cout<<endl<setw(25)<<“Frame Id :” <<FrameId<<endl<<endl; cout<<setw(25)<<“Frame Height:” <<Height<<endl<endl;
cout<<setw(25)<<“Frame Width :”<<Width<<endk<endl;
cout<<setw(25)<<“Total Amount:”
<<Amount< <endl< <endl;
}
int main( )
{
FRAME F;
F.Getdetails( );
F.ShowDetails( );
}
Output

Question 5.
Define a class RESORT in C++ with the following description:
Answer:
Private Members:
Rno //Data member to store Room No
RName //Data member to store customer name
Charges //Data member to store per day charges Days //Data member to store a number of days of stay
COMPUTE( ) //A function to calculate and return Amount as
//Days*Chagres and if the value of Days*Charges is more than 5000 then as 1.02*Days*Charges

Public Members:
Getinfo( ) //A function to enter the content Rno, Name, Charges //and Days Displayinfo( ) //A function to display Rno, RName, Charges, Days and
// Amount (Amount to displayed by calling function COMPUTE())
PROGRAM
using namespace std;
#include<iostream>
class RESORT
{
private:
int Rno,Days,Charges;
char Rname[20];
int compute( )
{
if (Days * Charges >5000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
getinfo( )
{
cout<<“\nEnter customer name :”;
cin>>Rname;
cout<<“\nEnter charges per day :”;
cin>>Charges;
cout<<“\nEnter Number of days :”;
cin>>Days;
cout<<“\nEnter Room Number :”;
cin>>Rno;
}
dispinfo( )
{
cout<<“\nRoom Number : “<<Rno;
cout<<“\nCustomer name :
“<<Rname;
cout«”\nCharges per day :
“<<Charges;
cout<<“\nNumber of days :
“<<Days;
cout<<“\nTotal Amount :
“<<compute( );
}
};
int main( )
{
RESORT Obj;
Obj.getinfo( );
Obj.dispinfo( );
}
Output

Question 7.
Answer:
struct pno
{
int pin;
float balance;
}
Create a BankAccount class with the following specifications

protected members
pno_obj //array of 10 elements
init(pin) // to accept the pin number and initialize it and initialize
// the balance amount is 0

public members
deposit(pin, amount):
Increment the account balance by accepting the amount and pin. Check the pin number for matching. If it matches increment the balance and display the balance else display an appropriate message withdraw(self, pin, amount):
Decrement the account balance by accepting the amount and pin. Check the pin number for matching and balance is greater than 1000 and amount is less than the balance. If it matches withdraw the amount and display the balance else display an appropriate message
PROGRAM
using namespace std;
#include<iostream>
#include<iomanip>
struct pno
{
int pin;
float balance;
};
class BankAccount
{
public:
pno pno_obj[10];
void deposit(int pn,float amt)
{
for(int i=0;i<10;i++)
if(pno_obj[i].pin == pn)
{
pno_obj[i].balance = pno_obj[i].balance + amt;
cout<<“\nTransaction successful!
cout<<“\nBalance amount in your account is”<< pno_obj[i].balance; break;
}
void withdraw(int self,int pn,float amt)
{
for(int i=0;i<10;i++)
{
if(pno_obj[i].pin== pn)
{
if (pno_obj[i].balance>1000 && amt < pno_obj[i].balance)
{
pno_obj[i].balance=pno_obj[i]. balance- amt;
cout<<“\nTransaction successful”;
cout<<“\nBalance amount in your account is “<< pno_obj[i].balance; break;
}
} }
}
};
int main( )
{
int pin_no, tamt;
BankAccount b;
// initialization of objects with pin and balance amount as 0
for(int i=0;i<10;i++)
{
b.pno_obj[i].pin=i+l;
}
int choice;
while(choice !=3)
{
cout <<“\n1. Deposit”;
cout <<“\n2.Withdrawal”;
cout<<“\n3.Exit”;
cout<<“\nEnter your choice “; cin >>choice;
switch(choice)
{
case 1:
cout<<“\nEnter PIN
cin>>pin_no;
cout<<“\nEnter Deposit amount”;
cin>>tamt;
b.deposit(pin_no,tamt); break;

case 2:
cout<<“\nEnter PIN cin>>pin_no;
cout<<“\nEnter Withdrawal amount”;
cin>>tamt;
cout<<“\nEnter 1 for Self 2 for Others :”;
int type;
cin>>type;
b.withdraw(type,pin_no,tamt); break;
default: cout<<“\nTransaction completed”;
}
}
}
Output

Question 8.
Define a class Hotel in C++ with the following description:
Answer:
Private Members:
Rno //Data member to store Room No
Name //Data member t store customer name
Charges //Data member to store per day charges
Days //Data member to store number of days of stay
Calculate() //A function to calculate and return Amount as
//Days*Chagres and if the value of Days*Charges is more than 12000 then as 1.2*Days*Charges

Public Members:
Hotel( ) //to initialize the class members
Getinfo( ) //A function to enter the content Rno, Name, Charges //and Days
Showinfo( ) //A function to display Rno, RName, Charges, Days and
//Amount (Amount to displayed by calling function CALCULATE( ))
PROGRAM
using namespace std;
#include<iostream>
#include<string.h>
class Hotel
private:
int Rno,Days,Charges;
char Name[20];
1 int Calculate( )
{
if (Days * Charges >12000)
return (Days * Charges * 1.02);
else
return(Days * Charges);
}
public:
Hotel( )
{
Rno=0;
Days=0;
Charges=0;
strcpy(Name,””);
}
void Getinfo( )
{
cout<<“nEnter customer name :”;
cin>>Name;
cout<<“nEnter charges per day : “;
cin>>Charges;
cout<<“\nEnter Number of days :”;
cin>>Days;
cout<<“\nEnter Room Number :”;
cin>>Rno;
}
void Showinfo( )
{
cout<<“\nRoom Number: “<<Rno;
cout<<“\nCustomer name : “<<Name;
cout<<“\nCharges per day : “<<Charges;
cout<<“\nNumber of days : “<<Days;
cout<<“\nTota! Amount: “<<Calculate( );
}
};
int main( )
{
Hotel obj;
obj.Getinfo( );
obj.Showinfo( )
}
Output

Question 9.
Define a class Exam in C++ with the following description:
Answer:
Private Members:
Rollno – Integer data type
Cname – 25 characters
Mark – Integer data type

public:
Exam(int,char[],int) //to initialize the object ~Exam() // display message “Result will be intimated shortly”
void Display( ) // to display all the details if the mark is above 60 other wise display “Result Withheld”
PROGRAM
using namespace std;
#include<iostream>
#include<string.h>
class Exam
{
private:
int Rollno,Mark;
char Cname[25];
public:
Exam(int r,char n[25],int m)
{
Rollno = r;
Mark = m;
strcpy(Cname,n);
}
~Exam( )
{
cout<<“\n\nResult will be intimated shortly”;
}
void Display( )
{
if (Mark>60)
{
cout<<“\n\nRoll Number : “<<Rollno;
cout<<“\nCandidate name : “<<Cname;
cout<<“\nMark :”<<Mark;
}
else
{
cout<<“\n\nRoll Number : “<<Rollno;
cout<<“\nCandidate name : “<<Cname;
cout<<“\nResult Withheld”;
}
}
};
int main( )
{
Exam obj 1(1011,”SURYA”,78),obj2( 1012,”JOHN”,44);
objl.Display( );
obj2. Display( );
}
Output

Question 10.
Define a class Student in C++ with the following specification:
Answer:
Private Members:
A data member Rno(Registration Number) type long
A data member Cname of type string A data member Agg_marks (Aggregate Marks) of type float
A data member Grade of type char
A member function setGrade () to find the grade as per the aggregate marks obtained by the student. Equivalent aggregate marks range and the respective grade as shown below.
Aggregate Marks -Grade
>=90 – A
Less than 90 and >=75 – B
Less than 75 and >=50 – C
Less than 50 – D
Public members:
A constructor to assign default values to data members:
A copy constructor to store the value in another object
Rno=0, Cname=”N.A” Agg_marks=0,0
A function Getdata ( ) to allow users to enter values for Rno.Cname, Aggjnarks and call functionsetGrade ( ) to find the grade.
A function dispResult( ) to allow user to view the content of all the data members.
A destructor to display the message “END”
PROGRAM .
using namespace std;
#include<iostream>
#include<string.h>
#include<iomanip>
class Student
{
private:
long Rno;
char Cname[25],Grade;
float Agg_marks;
void Setgrade()
{
if (Ag g_ma rks >=90)
Grade = ‘A’;
else if(Agg_marks>=75)
Grade = ‘B’;
else if(Agg_marks>=50)
Grade = ‘C’;.
else
Grade = ‘D’;
}
public:
Student( )
{
Rno = 0;
Agg_marks = 0;
strcpy(Cname,””);
Grade=”;
}
Student(Student &s) .
{
Rno = s.Rno;
Agg_marks = s.Agg_marks;
strcpy(Cname,s.Cname);
Grade=s.Grade;

~Student( )
{
cout<<“\nEND”; >
void Getdata( )
{
cout<<“\nEnter Register Number”;
cin>>Rno;
cout<<“\nEnter Candidate Name
cin>>Cname;
cout<<“\nEnter Aggrigate Mark”;
cin>>Agg_marks;
Setgrade( );
}
void dispResult( )
{
cout<<setw(30)<<“Candidate Register Number
“<<Rno<<endl<<endl;
cout< <setw(30)<< “Candidate Name : “<<Cname<<endk<endI;
cout<<setw(30)<<“Aggrigate Mark : “<<Agg_marks«endk<endl;
cout<<setw(30)<<“Grade :
“<<Grade<<endk<endl;
}
};
int main( )
{
Student s1;
s1.Getdata( );
Student s2(s1);
COut<<“\nFIRST CANDIDATE DETAIL \n\n”; s1.dispResult( );
cout<<“\nSECOND CANDIDATE DETAIL \n\n”;
s2.dispResult( );
}
Output

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 7 Thermodynamics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 7 Thermodynamics

11th Chemistry Guide Thermodynamics Text Book Back Questions and Answers

Textual Questions:

I. Choose the best answer:

Question 1.
The amount of heat exchanged with the surrounding at constant temperature and pressure is given by the quantity
(a) ∆E
(b) ∆H
(c) ∆S
(d) ∆G
Answer:
(b) ∆H

Question 2.
All the naturally occurring processes proceed spontaneously in a direction which leads to
(a) decrease in entropy
(b) increase in enthalpy
(c) increase in free energy
(d) decrease in free energy
Answer:
(d) decrease in free energy

Question 3.
In an adiabatic process, which of the following is true?
(a) q = w
(b) q = 0
(c) ∆E = q
(d) P∆V = 0
Answer:
(b) q = 0

Question 4.
In a reversible process, the change in entropy of the universe is
(a) > 0
(b) ≥ 0
(c) <0
(d) = 0
Answer:
(d) = 0

Question 5.
In an adiabatic expansionof an ideal gas
(a) w = – ∆U
(b) w = ∆U + ∆H
(c) ∆U = 0
(d) w = 0
Answer:
(a) w = – ∆U

Question 6.
The intensive property among the quantities below is
(a) mass
(b) volume
(c) enthalpy
(d) mass/volume
Answer:
(d) mass/volume

Question 7.
An ideal gas expands from the volume of 1 × 10^-3 m³ to 1 × 10^-2 m³ at 300 K against a constant pressure at 1 × 10⁵ Nm^-2. The work done is
(a) -900 J
(b) 900 kJ
(c) 270 kJ
(d) – 900 kJ
Answer:
(a) -900 J

Question 8.
Heat of combustion is always
(a) positive
(b) negative
(c) zero
(d) either positive or negative
Answer:
(b) negative

Question 9.
The heat of formation of CO and CO2 are -26.4 kCal and -94 kCal, respectively. Heat of combustion of carbon monoxide will be
(a) + 26.4 kcal
(b) – 67.6 kcal
(c) – 120.6 kcal
(d) + 52.8 kcal
Answer:
(b) – 67.6 kcal

Question 10.
C(diamond) → C(graphite), ∆H = -ve, this indicates that
(a) graphite is more stable than diamond
(b) graphite has more energy than diamond
(c) both are equally stable
(d) stability cannot be predicted
Answer:
(a) graphite is more stable than diamond

Question 11.
The enthalpies of formation of Al₂O₃ and Cr₂O₃ are – 1596 kJ and – 1134 kJ, respectively.
∆H for the reaction 2Al + Cr₂O₃ → 2Cr + Al₂O₃ is
(a) – 1365 kJ
(b) 2730 kJ
(c) – 2730 kJ
(d) -462 kJ
Answer:
(d) -462 kJ

Question 12.
Which of the following is not a thermodynamic function?
(a) internal energy
(b) enthalpy
(c) entropy
(d) frictional energy
Answer:
(d) frictional energy

Question 13.
If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then
(a) ∆H > ∆U
(b) ∆H – ∆U=0
(c) ∆H + ∆U=0
(d) ∆H < ∆U
Answer:
(d) ∆H < ∆U

Question 14.
Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is given out by the system is
(a) +1 kJ
(b) -5 kJ
(c) +3 kJ
(d) -3 kJ
Answer:
(c) +3 kJ

Question 15.
The work done by the liberated gas when 55.85 g of iron (molar mass 55.85 g mol^-1) reacts with hydrochloric acid in an open beaker at 25°C
(a) – 2.48 kJ
(b) – 2.22 kJ
(c) + 2.22 kJ
(d) + 2.48 kJ
Answer:
(a) – 2.48 kJ

Question 16.
The value of ∆H for cooling 2 moles of an ideal monatomic gas from 1250°C to 250°C at constant pressure will be [given C_p = \(\frac{5}{2}\)R]
(a) – 250 R
(b) – 500 R
(c) 500 R
(d) + 250 R
Answer:
(b) – 500 R

Question 17.
Given that C_(g) + O_2(g) → CO_2(g) ∆H° = – a kJ;
2 CO_(g) + O_2(g) → 2CO_2(g) ∆H° = -b kJ; Calculate the ∆H° for the reaction C_(g) + ½O_2(g) → CO_(g)
(a) \(\frac{b+2 a}{2}\)
(b) 2a – b

(c) \(\frac{2 a-b}{2}\)

(d) \(\frac{b-2 a}{2}\)
Answer:
(d) \(\frac{b-2 a}{2}\)

Question 18.
When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0° C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat of released from this combustion in kJ is (∆H_C(CH4)) = – 890 kJ mol and ∆H_C(C3H8) = – 2220 kJ mol^-1)
(a) -889 kJ
(b) -1390 kJ
(c) -3180 kJ
(d) -632.68 kJ
Answer:
(d) -632.68 kJ

Question 19.
The bond dissociation energy of methane and ethane are 360 kJ mol^-1 and 620 kJ mol^-1 respectively. Then, the bond dissociation energy of C-C bond is
(a) 170 kJ mol^-1
(b) 50 kJ mol^-1
(c) 80 kJ mol^-1
(d) 220 kJ mol^-1
Answer:
(c) 80 kJ mol^-1

Question 20.
The correct thermodynamic conditions for the spontaneous reaction at all temperature is
(a) ∆H < 0 and ∆S > 0
(b) ∆H < 0 and ∆S < 0
(c) ∆H > 0 and ∆S = 0
(d) ∆H > 0 and ∆S > 0
Answer:
(a) ∆H < 0 and ∆S > 0

Question 21.
The temperature of the system, decreases in an
(a) Isothermal expansion
(b) Isothermal Compression
(c) adiabatic expansion
(d) adiabatic compression
Answer:
(c) adiabatic expansion

Question 22.
In an isothermal reversible compression of an ideal gas the sign of q, ∆S and w are respectively
(a) +, -, –
(b) -, +, –
(c) +, -, +
(d) -, -, +
Answer:
(d) -, -, +

Question 23.
Molar heat of vapourisation of a liquid is 4.8 kJ mol^-1 If the entropy change is 16 J mol^-1 K^-1. the boiling point of the liquid is
(a) 323 K
(b) 27°C
(c) 164 K
(d) 0.3 K
Answer:
(b) 27°C

Question 24.
∆S is expected to be maximum for the reaction
(a) Ca(S) + 1/2 O₂(g) → CaO(S)
(b) C(S) + O₂(g) → CO₂(g)
(c) N₂(g) + O₂(g) → 2NO(g)
(d) CaCO₃(S) → CaO(S) + CO₂(g)
Answer:
(d) CaCO₃(S) → CaO(S) + CO₂(g)

Question 25.
The values of ∆H and ∆S for a reaction are respectively 30 kJ mol^-1 and 100 JK^-1 mol^-1. Then the temperature above which the reaction will become spontaneous is
(a) 300 K
(b) 30 K
(c) 100 K
(d) 20°C
Answer:
(a) 300 K

II. Write brief answers to the following questions:

Question 26.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics, known as the law of conservation of energy, states that the total energy of an isolated system remains constant though it may change from one foim to another.
The mathematical statement of the First Law as:
∆U = q + w
where q – the amount of heat supplied to the system; w – work done on the system.

Question 27.
Define Hess’s law of constant heat summation.
Answer:
The heat changes in chemical reactions are equal to the difference in internal energy (∆U) or heat content (∆H) of the products and reactants, depending upon whether the reaction is studied at constant volume or constant pressure. Since ∆U and ∆H are functions of the state of the system, the heat evolved or absorbed in a given reaction depends only on the initial state and final state of the system and not on the path or the steps by which the change takes place.

Question 28.
Explain intensive properties with two examples.
Answer:
The property that is independent of the mass or the size of the system is called an intensive property. Refractive index, Surface tension, density, temperature, Boiling point, Freezing point, molar volume.

Question 29.
Define the following terms:
(a) isothermal process
(b) adiabatic process
(c) isobaric process
(d) isochoric process
Answer:
a. An isothermal process is defined as one in which the temperature of the system remains constant, during the change from its initial to final state. The system exchanges heat with its surrounding and the temperature of the system remains constant. For an isothermal process dT = 0.

b. An adiabatic process is defined as one in which there is no exchange of heat (q) between the system and surrounding during the process. Those processes in which no heat can flow into or out of the system are called adiabatic processes.

c. An isobaric process is defined as one in which the pressure of the system remains constant
during its change from the initial to final state.
For an isobaric process dP = 0.

d. An isochoric process is defined as the one in which the volume of system remains constant during its change from initial to final state. Combustion of a fuel in a bomb calorimeter is an example of an isochoric process.
For an isochoric process, dV= 0.

Question 30.
What is the usual definition of entropy? What is the unit of entropy?
Answer:
Entropy is a measure of the molecular disorder (randomness) of a system.
The thermodynamic definition of entropy is concerned with the change in entropy that occurs as a result of a process.
It is defined as, dS = dq_rev /T

Question 31.
Predict the feasibility of a reaction when
(i) both ∆H and ∆S positive
(ii) both ∆H and ∆S negative
(iii) ∆H decreases but ∆S increases
Answer:
(i) ∆H_r = + ve, ∆S = + ve; non-spontaneous at low temperature spontaneous at high temperature
(ii) ∆H_r = – ve, ∆S = – ve; spontaneous at low temperature non-spontaneous at high temperature
(iii) ∆H decreases but ∆S increases = Spontaneous at all temperature.

Question 32.
Define is Gibb’s free energy.
Answer:
G is expressed as G = H – TS, free energy change of a process is given by the relation
∆G = ∆H – T∆S.

Question 33.
Define enthalpy of combustion.
Answer:
The heat of combustion of a substance is defined as “The change in enthalpy of a system when one mole of the substance is completely burnt in excess of air or oxygen”. It is denoted by ∆H_C.

Question 34.
Define molar heat capacity. Give its unit.
Answer:
The heat capacity for 1 mole of substance, is called molar heat capacity (cm). It is defined as “The amount of heat absorbed by one mole of the substance to raise its temperature by 1 kelvin”.
The SI unit of molar heat capacity is JK^-1 mol^-1

Question 35.
Define the calorific value of food. What is the unit of calorific value?
Answer:
The calorific value is defined as the amount of heat produced in calories (or joules) when one gram of the substance is completely burnt. The SI unit of calorific value is J kg^-1. However, it is usually expressed in cal g^-1.

Question 36.
Define enthalpy of neutralization.
Answer:
The heat of neutralisation is defined as “The change in enthalpy when one gram equivalent of an acid is completely neutralised by one gram equivalent of a base or vice versa in dilute solution”.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O₁
∆H = – 57.32 kJ

Question 37.
What is lattice energy?
Answer:
Lattice energy is defined as the amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance. It is also referred as lattice enthalpy.

Question 38.
What are state and path functions? Give two examples.
Answer:
State function:
A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path (or manner) by which the particular state is reached.
Example:
Pressure (P), Volume (V), Temperature(T), Internal energy (U), Enthalpy (H), free energy (G) etc.
Path functions:
A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.
Example:
Work (w), Heat (q).

Question 39.
Give Kelvin statement of second law of thermodynamics.
Answer:
It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.

Question 40.
The equilibrium constant of a reaction is 10, what will be the sign of ∆G? Will this reaction be spontaneous?
Answer:
∆G = -2.303 RT logK_eq
Substituting the known values of R, T and K_eq
R = 8.314(JK^-1 mol^-1)
T = 300 K
K_eq = 10
∆G = –2.303 8.314(JK^-1) 300(K) log 10
= -5744.14 J/mol
= -5.744 KJ/mol
∆G < 0, then it is spontaneous.

Question 41.
Enthalpy of neutralization is always a constant when a strong acid is neutralized by a strong base: account for the statement.
Answer:
It is because in dilution solution all strong acids and strong bases are completely ionized. The neutralization of strong acid and strong bases are completely ionized. The neutralization of a strong acid and strong base simply involves the combination of H⁺ ions (from acid) and OH^– ions (from base) to form unionized water molecules with the evolution of 57.1 kJ heat.
H⁺_(aq) + OH^–_(aq) → H₂O(l),
∆₂H° = -57.1 kJ
Since the same reaction takes place during neutralization of all strong acids and strong bases, the value of enthalpy of neutralization is constant.

Question 42.
State the third law of thermodynamics.
Answer:
The third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero. Otherwise it can be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps. Mathematically,
lim_T→0 S = 0 for a perfectly ordered crystalline state.

Question 43.
Write down the Born-Haber cycle for the formation of CaCl₂.
Answer:

Step 1:
Solid calcium is converted to gaseous state (Enthalpy of atomization)
Ca_(S) → Ca_(g) Ca(g)                         ∆H°_a = 178 KJ/mol
Step 2:
The calcium is converted to ionic form
(divalent cation): (Ionisation enthalpy)
Ca → Ca⁺ + e^–                          ∆H°_IE = 590 kJ
Ca → Ca²⁺ + e^–                        ∆H°_IE = 590 KJ
Step 3:
Atomisation of chlorine molecule to chlorine atom: (Atomisation enthalpy)
\(\frac{1}{2}\)Cl₂ → Cl               ∆H°_Cl-Cl = 121 KJ
For two chlorine atoms required, atomistion energy = 2 × 121 = 242 KJ/mol
Step 4:
Convrsion of chlorine atom into ion:(Electron affinity)
Cl + e^– → Cl^–;                           ∆H°_ea = – 364KJ
Step 5:
Finally the two ions join together by lattice energy as: (here two Cl^– ions are involved)
Ca²⁺ + 2Cl^– → CaCl₂

Hence Lattice energy = Heat of formation – heat of atomization – dissociation energy – (sum of ionization energies) – (sum of electron affinities)
Since, heat of formation (i.e standard enthalpy of formation ) of CaCl₂ = – 796KJ/mol
Lattice energy = -796 -178 -242 -(590 + 1145) – (2 × – 364) = -2223KJ/mol

Question 44.
Identify the state and path functions out of the following: (a) Enthalpy (b) Entropy (c) Heat (d) Temperature (e) Work (f) Free energy.
Answer:
State function:
Enthalpy, Temperature, Free energy, Entropy
Path function:
Work, Heat.

Question 45.
State the various statements of second law of thermodynamics.
Answer:
Entropy statement:
The second law of thermodynamics can be expressed in terms of entropy, i.e “the entropy of an isolated system increases during a spontaneous process”.

Kelvin-Planck statement:
It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink.

Clausius statement:
It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work.

Question 46.
What are spontaneous reactions? What are the conditions for the spontaneity of a process?
Answer:

1. A reaction that occurs under the given set of conditions without any external driving force is called a spontaneous reaction.
2. The spontaneity of any process depends on three different factors.
3. If the enthalpy change of a process is negative, then the process is exothermic and may be spontaneous. (∆H is negative)
4. If the entropy change of a process is positive, then the process may occur spontaneously. (∆S is positive)
5. The gibbs free energy which is the combination of the above two (∆H – T∆S) should be negative for a reaction to occur spontaneously, i.e. the necessary condition for a reaction to be spontaneous is ∆H – T∆S < 0
6. For spontaneous process,
   ∆S_total > 0, ∆G < 0, ∆S < 0

Question 47.
List the characteristics of internal energy.
Answer:
The internal energy of a system is an extensive property. It depends on the amount of the substances present in the system. If the amount is doubled, the internal energy is also doubled.

The internal energy of a system is a state function. It depends only upon the state variables (T, P, V, n) of the system. The change in internal energy does not depend on the path by which the final state is reached.

The change in internal energy of a system is expressed as ∆U = U_f – U_i.
In a cyclic process, there is no internal energy change. ∆U_(cyclic) = 0

If the internal energy of the system in the final state (U_f) is less than the internal energy of the system in its initial state (U_i), then ∆U would be negative.
∆U = U_f – U_i = -ve(U_f < U_i)

If the internal energy of the system in the final state (U_f) is greater than the internal energy of the system in its initial state (U_i), then ∆U would be positive.
∆U = U_f – U_i = + ve(U_f > U_i).

Question 48.
Explain how heat absorbed at constant volume is measured using bomb calorimeter with a neat diagram.
Answer:
Calorimeter is used for measuring the amount of heat change in a chemical or physical change. In calorimetry, the temperature change in the process is measured which is directly proportional to the heat capacity. By using the expression C = \(\frac{q}{m \Delta T}\), we can calculate the amount of heat change in the process. Calorimetric measurements are made under two different conditions
(i) At constant volume (qV)
(ii) At constant pressure (qp)
(A) ∆U Measurements:
For chemical reactions, heat evolved at constant volume, is measured in a bomb calorimeter.

The inner vessel (the bomb) and its cover are made of strong steel. The cover is fitted tightly to the vessel by means of metal lid and screws.

A weighed amount of the substance is taken in a platinum cup connected with electrical wires for striking an arc instantly to kindle combustion. The bomb is then tightly closed and pressurized with excess oxygen. The bomb is immersed in water, in the inner volume of the calorimeter. A stirrer is placed in the space between the wall of the calorimeter and the bomb, so that water can be stirred, uniformly. The reaction is started by striking the substance through electrical heating.

A known amount of combustible substance is burnt in oxygen in the bomb. Heat evolved during the reaction is absorbed by the calorimeter as well as the water in which the bomb is immersed. The change in temperature is measured using a Beckman thermometer. Since the bomb is sealed its volume does not change and hence the heat measurements is equal to the heat of combustion at a constant volume (∆U)_c.

The amount of heat produced in the reaction (∆U)_c is, equal to the sum of the heat abosrbed by the calorimeter and water.
Heat absorbed by the calorimeter
q₁ = k.∆T
where k is a calorimeter constant equal to mc Cc ( me is mass of the calorimeter and Cc is heat capacity of calorimeter)
Heat absorbed by the water q₂ = m_w C_w ∆T
where mw is molar mass of water
C_w is molar heat capacity of water (4,184 kJ K^-1 mol^-1)
Therefore ∆U_c = q₁ + q₂
= k.∆T + m_w C_w ∆T
= (k + m_w C_w)∆T

Calorimeter constant can be determined by burning a known mass of standard sample (benzoic acid) for which the heat of combustion is known (-3227 kJ mol^-1). The enthalpy of combustion at constant pressure of the substance is calculated from the equation
∆H°_c (pressure) = ∆U°_c (Vol) + ∆n_gRT

Question 49.
Calculate the work involved in expansion and 1 compression process.
Answer:
Work involved in expansion and compression processes:
In most thermodynamic calculations we are dealing with the evaluation of work involved in the expansion or compression of gases. The essential condition for expansion or compression of a system; is that there should be difference between external pressure (P_ext) and internal pressure (P_int).

For understanding pressure-volume work, let us consider a cylinder which contains V moles of an ideal gas fitted with a frictionless piston of cross sectional area A. The total volume of the gas inside is V and pressure of the gas inside is P_int. If the external pressure P_ext is greater than P_int, the piston moves inward till the pressure inside becomes equal to P_ext. Let this change be achieved in a single step
and the final volume be V_f.In this case, the work is done on the system (+w). It can be calculated as follows
w = -F.∆x ……….(1)
where dx is the distance moved by the piston during the compression and F is the force acting on the gas.

F = P_extA ……….(2)
Substituting (2) in (1)
w = – P_ext. A. ∆x
A .∆x = change in volume = V_f – V_i
w = -P_ext.{V_f – V_i) ……….(3)
w = -P_ext. (-∆V) …………(4)
= P_ext.∆V

Since work is done on the system, it is a positive quantity.
If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work donp on the gas by therelation
W_rev = – \(\int_{V_{i}}^{v_{f}}\) p_ext dV.
In a compression process, Pext the external pressure is always greater than the pressure of the system.
i.e., P_ext = (P_int + dP)
In an expansion process, the external pressure is always less than the pressure of the system
i.e., P_ext = (P_int – dP)

When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from V_i to V_f. the P-V plot looks like in figure. Work done on the gas is represented by the shaded area.
In general case we can write,
P_ext = (P_int ± dP). Such processes are called reversible processes. For a compression process work can be related to internal pressure of the system under reversible conditions by writing equation
W_rev = – \(\int_{V_{i}}^{v_{f}}\) P_int dV
For a given system with an ideal gas
P_int V = nRT
p_int = \(\frac{n R T}{V}\)
W_rev = – \(\int_{V_{i}}^{v_{f}}\) \(\frac{n R T}{V}\) dV
W_rev = – nRTln(\(\frac{V_{f}}{V_{i}}\))
W_rev= -2.303nRTlog\(\frac{V_{f}}{V_{i}}\) ………….(5)

If V_f > V_i (expansion), the sign of work done by the process is negative.
If V_f < V_i (compression) the sign of work done on the process is positive.

Question 50.
Derive the relation between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
Answer:
When the system at constant pressure undergoes changes from an initial state with H₁, U₁ and V₁ to a final state with H₂, U₂ and V₂ the change in enthalpy
∆H, can be calculated as follows:
H =U + PV
In the initial state
H₁ = U₁ + PV₁ ………………(1)
In the final state
H₂ = U₂ + PV₂ ……………..(2)
change in enthalpy is (2) – (1)
(H₂ – H₁) = (U₂ – U₁) + P(V₂ – V₁)
∆H = ∆U + P∆V ………………..(3)
As per first law of thermodynamics,
∆U = q + w
Equation (3) becomes
∆H = q + w+P∆V
w = -P∆V
∆H = qp – P∆V + P∆V
∆H = qp ………….(4)
qp – is the heat absorbed at constant pressure and is considered as heat content.
Consider a closed system of gases which are chemically reacting to form gaseous products at constant temperatureand pressure with V_i and V_f,as the total volumes of the reactant and product gases respectively, and niand nfas the number of moles of gaseous reactants and products, then,
For reactants (initial state):
PV_i = n_iRT ………..(5)
For products (final state):
PV_f = n_fRT ……………..(6)
(6) – (5)
P{V_f – V_i) = (n_f – n_i)RT
P∆V = ∆n_(g) RT ………..(7)
Substituting in (7) in (3)
∆H = ∆U+ ∆n_(g)RT …………….(8)

Question 51.
Suggest and explain an indirect method to calculate lattice enthalpy of sodium chloride crystal.
Answer:
The formation of NaCl can be considered in five steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.
Let us calculate the lattice energy of sodium chloride using Bom-Haber cycle

∆H = heat of formation of sodium chloride = -411.3 kJmol^-1
∆H₁ = heat of sublimation of Na_(s) = 108.7 kJmol^-1
∆H₂ = ionisation energy of Na_(s) = 495. kJ mol^-1
∆H₃ = dissociation energy of Cl_2(g) = 244 kJ mol^-1
∆H₄ = Electron affinity of Cl_(g) = -349 kJ mol^-1
U = lattice energy of NaCl
∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + ∆H₄ + U
∴ U = (∆H_f ) – (∆H₁ + ∆H₂ + \(\frac{1}{2}\)∆H₃ + ∆H₄)
=» U = (-411.3) – (108.7 + 495.0 + 122 – 349)
U = (-411.3) – (376.7)
∴ U = -788kJ mol^-1

Question 52.
List the characteristics of Gibbs free energy.
Answer:
(i) G is defined as (H-TS) where H and S are the enthalpy and entropy of the system respectively. T = temperature. Since H and S are state functions, G is a state function.
(ii) G is an extensive property while ∆G = (G₂ – G₁) which is the free energy change between the initial (1) and final (2) states of the system becomes the intensive property when mass remains constant between initial and final states (or) when the system is a closed system.
(iii) G has a single value for the thermodynamic state of the system.
(iv) G and ∆G values correspond to the system only. There are three cases of ∆G in predicting the nature of the process. When, ∆G < 0 (negative), the process is spontaneous and feasible; ∆G = 0. The process is in equilibrium and ∆G > 0 (positive), the process is nonspontaneous and not feasible.
(v) ∆G = ∆H – T∆S. But according to I law of thermodynamics,
∆H = ∆G + P∆V and ∆U = q + w
∴ ∆G = q + w + P∆V – T∆S
But ∆S = \(\frac{T}{q}\) and T∆S = q = heat involved in the process.
∴ ∆G = q + w + P∆V – q = w + P∆V
(or) – ∆G = – w – P∆V = network.
The decrease in free energy – ∆G, accompanying a process taking place at constant temperature and pressure is equal to the maximum obtainable work from the system other than work of expansion.
This quantity is called as the “net work” of the system and it is equal to (- w – P∆V).
∴ Network = – w – P∆V
-∆G represents all others forms of work obtainable from the system such as electrical, chemical or surface work etc other than P-V work.

Question 53.
Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.
Solution:
Given
n = 2 moles
V_i = 500 ml = 0.5 lit
V_f = 2 lit
T = 25°C = 298 K
w = -2303 nRTIog (\(\frac{V_{f}}{V_{i}}\))
w = -2.303 × 2 × 8.314 × 298 × log (4)
w = -2.303 × 2 × 8.314 × 298 × 0.6021
w = -6871 J
w = -6.871 kJ.

Question 54.
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5kJ K^-1. Calculate the enthalpy of combustion of the gas in kJ mol^-1.
Solution: Given
T_i = 298 K
T_f = 298.45 K
k = 2.5 kJ K^-1
m = 3.5g
Mm = 28
heat evolved = k∆T
= k(T_f – T_i)
= 2.5KJ K^-1 (298.45 – 298)K = 1.125 kJ
∆H_c = \(\frac{1.125}{3.5}\) × 28 kJ mol^-1

∆H_c = 9 kJmol^-1

Question 55.
Calculate the entropy change in the system, and surroundings, and the total entropy change in the universe during a process in which 245 J of heat flow out of the system at 77°C to the surrounding at 33°C.
Solution:
Given
T^sys = 77° C = (77 + 273) = 350 K
T^surr = 33°C = (33 + 273) = 360 K
q = 245J

∆S_sys = \(\frac{q}{T_{s y}}=\frac{-245}{350}\) = -0.7 JK^-1

∆S_surr = \(\frac{q}{T_{s \mathrm{sr}}}=\frac{+245}{306}\) = +0.8 JK^-1

∆S_univ = ∆S_sys + ∆S_surr

∆S_univ = -0.7 JK^-1 + 0.8 JK^-1

∆S_univ = 0.1 JK^-1

Question 56.
1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710/and expands to 2 litres. Calculate the entropy change in expansion process.
Solution:
Given
n = 1 mole
P = 4.1 atm
V = 2 Lit
T = ?
q = 3710 J
∆S = \(\frac{q}{T}\)

∆S = \(\frac{q}{\frac{p v}{n R}}\)

∆S = \(\frac{n R q}{P V}\)

∆S = \(\frac{1 \times 0.082 \text { lit atm } K^{-1} \times 3710 J}{4.1 \text { atm } \times 2 \text { lit }}\)

∆S = 37.10 JK^-1

Question 57.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride.
Answer:
Given,
∆H_f(NaCl) = 30.4 kJ = 30400 J mol^-1
∆S_f(NaCl) = 28.4 JK^-1 mol^-1
∆S^-1 = \(\frac{\Delta H_{f}}{T_{f}}\)

T_f = \(\frac{\Delta H_{f}}{\Delta S_{f}}\)

T_f = \(\frac{30400 J m o l^{-1}}{28.4 J K^{-1} m o l^{-1}}\)

T_f = 1070.4 K

Question 58.
Calculate the standard heat of formation of propane, if its heat of combustion is -2220.2 kJ mol^-1. The heats of formation of CO₂(g) and H₂O(1) are -393.5 and -285.8 kJ mol^-1 respectively.
Solution:
Given:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
∆H°C = 2220.2kJ mol^-1 ……………..(1)
C + O₂ → CO₂
∆H°_f = -393.5 kf mol^-1 ……….(2)
H₂ + \(\frac{1}{2}\)O₂ → H₂O
∆H°_f = -285.8 kJ mol^-1 ……………(3)
3C + 4H₂ → C₃H₈
∆H°_c =?
(2) × (3)
⇒ 3C + 3O₂ → 3CO₂
∆H°_f = -1180.5 kJ ………..(4)
(3)× (4)
⇒ 4H₂ + 2O₂ → 4H₂O
∆H°_f = 1143.2 kJ ……..(5)
(4) + (5) – (1)
⇒ 3C + 3O₂ + 4H₂ + 2O₂ + 3CO₂ + 4H₂O → 3CO₂ + 4H₂O + C₃H₈ + 5O₂
∆H°_f = -1180.5 – 1143.2 – (-2220.2) kJ
3C + 4H₂ → C₃H₈
∆H°_f = -103.5 kJ
Standard heat of formation of propane is ∆H°_f(C3H8) = -103.5 kJ

Question 59.
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below.

Answer:
For ethanol
Given:

Question 60.
For the reaction Ag₂O(s) → 2Ag(s)+12O₂(g) ∆H = 30.56 kJ mol^-1 and ∆S = 6.66JK^-1 mol^-1 (at 1 atm). Calculate the temperature at which G is equal to zero. Also predict the direction of the reaction (I) at this temperature and (ii) below this temperature.
Solution:
Given,
∆H = 30.56 kJ mol^-1
∆S = 6.66 x 10^-3 kJK^-1 mol^-1
T = ? at which ∆G =0
∆G = ∆H – T∆S
0 = ∆H – T∆S
T = \(\frac{\Delta H}{\Delta S}\)

T = \(\frac{30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}}{66.6 \times 10^{-3} \mathrm{~kJ} K^{-1} \mathrm{~mol}^{-1}}\)
T = 4589 K

(i) At 4589K;
∆G = 0 the reaction is in equilibrium.
(ii) at temperature below 4598 K
∆H > T∆S
∆G = ∆H – T∆S > 0, the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.

Question 61.
What is the equilibrium constant K_eq for the following reaction at 400K.
2NOCl(g) ⇌ 2NO(g) + Cl₂(g), given that ∆H° = 77.2 kJ mol^-1 ∆S° = 122 JK^-1 mol^-1
Answer:

Question 62.
Cyanamide (NH₂CN) is completely burnt in excess oxygen in a bomb calorimeter, ∆U was found to be -742.4 kJ mol^-1, calculate the enthalpy change of the reaction at 298K.
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O (l) ∆H = ?
Solution:
Given
T = 298 K;
∆U = -742.4 kJ mol^-1
∆H =?
∆H = ∆U + ∆n_(g)RT
∆H = ∆U + (n_p – n_r)RT
∆H = 742.4 + 2 – \(\frac{3}{2}\) × 8.314 × 10^-3 × 298
= -742.4 + (0.5 × 8.314 ×10^-3 × 298)
=-742.4 + 1.24
= -741.16 kJ mol^-1

Question 63.
Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C – H, C – C, C = C and H – Hare 414, 347, 618 and 435 kJ mol^-1.
Solution:
Given
E_C-H= 414 kJ mol^-1
E_C-C= 347 kJ mol^-1
E_C-C= 6l8 kJ mol^-1
E_H-H = 435 kJ mol^-1

∆H_r = Σ(Bond energy)_r – Σ(Bond energy)_p
∆H_r = (E_C=C + 4E_C-H + E_H-H) – (E_C-C + 6E_C-H)
∆H_r = (618 + (4 × 414) + 315) – (347 + (6 ×414))
∆H_r = 2709 – 2831
∆H_r = -122 kJ mol^-1

Question 64.
Calculate the lattice energy of CaCl2 from the given data
Ca(s) + Cl₂(g) → CaCl₂(s) ∆H°_f = – 795 kJ mol^-1
Atomisation:
Ca(s) → Ca(g) ∆H°₁ = -795 kJ mol^-1

Ionisation:
Ca(g) → Ca²⁺(g) + 2e^-1
∆H°₂ = 2422 kJ mol^-1

Dissociation:
Cl₂(g) → 2Cl(g)
∆H°₃ = + 242.8 kJ mol^-1

Electron Affinity:
Cl(g) + e^– → Cl^-1
∆H°₄ = -355 kJ mol^-1
Answer:
∆H_f = ∆H₁ + ∆H₂ + ∆H₃+∆H₄ + u
-795 = 121 + 2422 + 242.8 + (2 × -355) + u
-795 = 2785.8 – 710 + u
-795 = 2075.8 + u
u = -795 – 2075.8
u = -2870.8 kJ mol^-1

Question 65.
Calculate the enthalpy change for the reaction Fe203 + 3C0 – 2Fe + 3C02 from the following data.
2Fe + \(\frac{3}{2}\)O₂ → Fe₂O₃; ∆H = -741 kJ
C + \(\frac{3}{2}\)O₂ → CO; ∆H=-137kJ
C + O₂ → CO2; ∆H = – 394.5 kJ
Answer:

Question 66.
When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne(A), 95.2% 2-pentyne(B), and 3.5% of 1,2 pentadiene (C) the equilibrium was maintained at 175°C, calculate ∆G° for the following equilibria.
B ⇌ A; ∆G°₁ =?
B ⇌ C; ∆G°₂ =?
Solution:
Given
T = 175°
C = 175 + 273 = 448 K
Concentration of 1-pentyne [A] = 1.3%
Concentration of 2-pentyne [B] = 95.2%
Concentration of 1, 2-pentadiene [C] = 3.5%
At equilibrium
B ⇌ A
95.2% 1.3% ⇒
B ⇌ C
95.2% 3.5% ⇒
K₂ = \(\frac{3.5}{95.2}\) = 0.0367
⇒ ∆G°₁ = -2.303 RTlogK
∆G°₁ = – 2.303 × 8.314 × 448 × log 0.0136
∆G°₁ = +16010 J
∆G°₁ = +16 kJ

⇒ ∆G°₂ = – 2.303 RT log K₂
∆ G°₂ = -2.303 × 8.314 × 448 × log 0.0367
∆ G°₂ = +12312J
∆G°₂ = +12.312 kJ

Question 67.
At 33K, N₂O₄ is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Solution:
Given
T= 33 K
N₂O₄ ⇌ 2NO₂
Initial concentration: 100%  0
Concentration dissociated 50% – Concentration remaining at equilibrium 50% 100%
K_eq = \(\frac{100}{50}\) = 2
∆G° = -2.303 RT log K_eq
∆G°= -2.303 × 8.314 × 33 × log 2
∆G° = -190.18 Jmol^-1

Question 68.
The standard enthalpies of formation of SO₂ and SO₃ are – 297 kJ mol^-1 and – 396 kJ mol^-1 respectively. Calculate the standard enthalpy of reaction for the reaction: SO +4-0 — SO
Solution:
Given
∆H°_f(SO₂) = – 297 kJ mol^-1
∆H°_f(SO₃) = – 396 kJ mol^-1
SO₂ + 12O₂ → SO₃;
∆H°_r =?
∆H°_r = (∆H°_f)_compound – Σ( ∆H_f)_elements
∆H°_r =∆H°_r(SO₃) – ∆H°_f
∆H°_r = 396 kJ mol^-1– (- 297 kJ mol^-1 + 0)
∆H°_r = – 396 kJ mol^-1 + 297
∆H°_r = -99 kJ mol^-1

Question 69.
For the reaction at 298 K: 2A + B → C
∆H = 400 Jmol^-1; ∆S = 0.2 JK∆ mol^-1 Determine the temperature at which the reaction would be spontaneous.
Solution:
Given,
T = 298 ∆T
∆H = 400 J mol^-1 = 400 J mol^-1
∆S = 0.2 JK^-1 mol^-1
∆G = ∆H – T∆S
if T = 2000 K
∆G = 400 – (0.2 × 2000) = 0
if T > 2000 K
∆G will be negative. The reaction would be spontaneous only beyond 2000 K

Question 70.
Find out the value of equilibrium constant for the following reaction at 298K,
2NH₃ + CO₂ ⇌ NH₂CONH₂ (aq) + H₂O (l) Standard Gibbs energy change, ∆G°_r at the given temperature is -13.6 kJ mol^-1.
Solution:
Given:
T = 298 K
∆G°_r = -13.6 kJ mol^-1
= – 13600 J mol^-1
∆G° = – 2.303 RT log K_eq
log K_eq = \(\frac{-\Delta G^{0}}{2.303 R T}\)

log K_eq = \(\frac{13.6 \mathrm{~kJ} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \times 10^{-3} \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}}\)

log K_eq = 2.38
K_eq = antilog (2.38)
K_eq = 239.88

Question 71.
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25°C and at 1 atm pressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in kJ, during this combustion. (∆H_C(CH4)) = -890 kJmol^-1 and ∆H_C(C2H4) = -1423 kJ mol^-1
Solution:
Given
∆H_C(CH4) = – 890 kJ mol^-1
∆H_C(C2H4) = -1423 kJ mol^-1
Let the mixture contain x lit of CH₄ and (3.67 – x) lit of ethylene.
CH₄ + 2O₂ → CO + 2H₂O
x lit               x lit
C₂H₄ + 3O₂ → 2 CO₂ + 2H₂O
(3.67 -x) lit 2 (3.67 -x) lit
Volume of Carbondioxide formed = x + 2(3.67 – x) = 6.11 lit
x + 7.34 – 2x = 6.11
7.34 – x = 6.11
x = 1.23 lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence

11th Chemistry Guide Thermodynamics Additional Questions and Answers

I. Choose the best answer:

Question 1.
_______is a part of universe which is separated from the rest of the universe by real or imaginary boundaries.
(a) Surroundings
(b) boundary
(c) system
(d) matter
Answer:
(c) system

Question 2.
Which one of the following is a closed system?
(a) Hot water in a closed beaker
(b) Hot water in a thermos flask
(c) Hot water in a open beaker
(d) Chemical reactions
Answer:
(a) Hot water in a closed beaker

Question 3.
Which of the following is/are extensive properties?
1. volume
2. Surface tension
3. mass
4. internal energy
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 1 and 3
(d) 1, 2 and 3
Answer:
(b) 1, 3 and 4

Question 4.
Which of the following is/are intensive properties?
1. refractive index
2. density
3. number of moles
4. molar volume
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 1, 2 and 3
(d) 2, 3 and 4
Answer:
(b) 1, 3 and 4

Question 5.
For an isothermal process _______.
(a) dT = 0
(b) dV = 0
(c) dq = 0
(d) dP = 0
Answer:
(a) dT = 0

Question 6.
Which of the following is/are state function?
1. Pressure
2. work
3. internal energy
4. Free energy
5. heat
(a) 1, 2 and 4
(b) 1, 3 and 4
(e) 1, 2 and 3
(d)2, 3 and 4
Answer:
(b) 1, 3 and 4

Question 7.
Which of the following is/are path function?
1. pressure
2. work
3. internal energy
4. Free energy
5. heat
(a) 1, 2 and 4
(b) 1, 3 and 4
(c) 2 and 5
(d) 2, 3 and 4
Answer:
(c) 2 and 5

Question 8.
Which is the correct about internal energy, U?
(a) It is an extensive property
(b) It is a state function
(c) For a cyclic process. ∆U = 0.
(d) Change in internal energy is ∆U = U_i – U_f
Answer:
(a) It is an extensive property

Question 9.
Work done by’ a given system with an ideal gas in a reversible process is w_rev =
(a) – nRln (V_f/V_i)
(b) – nRTln (V_i/V_f)
(c) – nRTln (V_f/V_i)
(d) – nTln (V_f/V_i)
Answer:
(c) – nRTln (V_f/V_i)

Question 10.
For a cyclic process involving isothermal expansion of an ideal gas
(a) ∆U = 0
(b) ∆U = q
(c) ∆U = q + w
(d) ∆U = q – w
Answer:
(a) ∆U = 0

Question 11.
Choose the correct statement about enthalpy
(a) Enthalpy is a path function
(b) Enthalpy change ∆H = ∆U + V∆P
(c) In an endothermic process, ∆H = 0.
(d) In an exothermic process. ∆H is negative
Answer:
(d) In an exothermic process. ∆H is negative

Question 12.
Which ofthc following reactions correctly indicates the process of atomization?
(a) CH₄ → CH₃(g) + H(g)
(b) CH₄(g) → CH(g) + 4H(g)
(c) CH(g) + O₂(g) → CO₂(g)
(d) N₂(g) + 3H₂(g) → 2NN₃(g)
Answer:
(b) CH₄(g) → CH(g) + 4H(g)

Question 13.
A reaction has both H and S negative. The rate of reaction
(a) increases with increase of temperature
(b) increases with decrease of temperature
(e) remains unaffected by change of temperature
(d) cannot be predicted for change in temperature
Answer:
(b) increases with decrease of temperature

Question 14.
Evaporation of water is
(a) an exothermic change
(b) an endothermic change
(e) a process where no heat changes occur
(d) a process accompanied by chemical reaction
Answer:
(b) an endothermic change

Question 15.
The entropy change for a non spontaneous reaction is 140 JK^-1 mol at 298 K. The reaction is
(a) reversible
(b) irreversible
(c) exothermic
(d) endothermic
Answer:
(d) endothermic

Question 16.
The enthalpy and entropy change for the reaction: Br₂(l) +Cl₂(g) → 2BrCl(g) are 30 kJ mol^-1 and 105 kJ mol^-1 respectively. The temperature at which the reaction will be in
equilibrium is
(a) 300 K
(b) 285.7 K
(c) 273 K
(d) 450 K
Answer:
(b) 285.7 K

Question 17.
Three moles ofan ideal gas expanded spontaneously into vaccum. The work done will be
(a) Infinite
(b) 3 JouIes
(c) 9 Joules
(d) zero
Answer:
(d) zero

Question 18.
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^-1 at 27°C, the entropy change for the process would be
(a) 10 J mol^-1K^-1
(b) 1.0 J mol^-1K^-1
(c) 0.1 J mol^-1K^-1
(d) 100 J mol^-1K^-1
Answer:
(d) 100 J mol^-1K^-1

Question 19.
Enthalpy change for the reaction,
4H(g) → 2H₂(g) is – 869.6 kJ
The dissociation energy of H-H bond is
(a) -434.8 kJ
(b) -869.6 kJ
(c) + 434.8 kJ
(d) +217.4 kJ
Answer:
(c) + 434.8 kJ

Question 20.
Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?
(a) q = 0, ∆T ≠ 0, w = 0
(b) q ≠ 0, ∆T = 0, w = 0
(c) q = 0, ∆T = 0, w = 0
(d) q = 0, ∆T < 0, w ≠ 0
Answer:
(b) q ≠ 0, ∆T = 0, w = 0

Question 21.
The total entropy change for a system and its surroundings increases, if the process is
(a) reversible
(b) irreversible
(c) exothermic
(d) endothermic
Answer:
(b) irreversible

Question 22.
∆S is positive for the change
(a) mixing of two gases
(b) boiling of liquid
(c) melting of solid
(d) all of these
Answer:
(d) all of these

Question 23.
If w₁, w₂, w₃ and w₄ are work done in isothermal, adiabatic, isobaric and isochoric reversible processes, the correct order (for expansion) will be
(a) w₁ > w₂ > w₃ > w₄
(b) w₃ > w₂ > w₁ > w₄
(c) w₃ > w₂ > w₄ > w₁
(d) w₃ > w₁ > w₂ > w₄
Answer:
(d) w₃ > w₁ > w₂ > w₄

Question 24.
The subject matter of thermodynamics comprises
(a) energy transformations in a system
(b) mass changes in molecular reactions
(c) total energy of system
(d) rates of chemical reactions
Answer:
(a) energy transformations in a system

Question 25.
A hydrogen bond is very useful in determining the structures and properties of compounds. Its energy varies between
(a) 12 – 20 kJ /mol
(b) 50 – 100 kJ/mol
(c) 10 – 100 kJ/mol
(d) 75 – 200 kJ/mol
Answer:
(c) 10 – 100 kJ/mol

Question 26.
In which of the enlisted cases, Hess’s law is not applicable?
(a) Determination of lattice energy
(b) Determination of resonance energy
(c) Determination of enthalpy of transformation of one allotropic form to another
(d) Dtermination of entropy
Answer:
(d) Dtermination of entropy

Question 27.
The amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance is called
(a) lattice energy
(b) ionization energy
(c) internal energy
(d) free energy
Answer:
(a) lattice energy

Question 28.
Statement -1:
The allotropes of carbon, namely, graphite and diamond differ each other.
Statement -2:
They possess different internal energies and have different structures.
In the above statement/s
(a) 1 alone is correct
(b) 2 alone is correct
(c) both 1 and 2 are correct
(d) both 1 and 2 are incorrect
Answer:
(c) both 1 and 2 are correct

Question 29.
Which of the following statement is incorrect? According to thermodynamics, work
(a) is a path function
(b) appears only at the boundary of the system
(c) appears during the change in the state of the system.
(d) surroundings is so large that macroscopic changes occurs in the surroundings
Answer:
(d) surroundings is so large that macroscopic changes occurs in the surroundings

Question 30.
The work done by a force of one Newton through a displacement of one meter is called
(a) joule
(b) calorie
(c) erg
(d) tesla
Answer:
(a) joule

Question 31.
The enthalpy of neutralization of strong acid vs strong base is approximately equal to ________(in kJ).
(a) 57.32
(b) – 57.32
(c) 5.98
(d) – 5.98
Answer:
(b) – 57.32

Question 32.
The maximum efficiency of an automobile engine working between the temperatures 816°C and 21°C is
(a) 73 %
(b) 45%
(c) 67%
(d) 78%
Answer:
(b) 45%

Question 33.
A reaction that occurs under the given set of conditions without any external driving force is called a reaction.
(a) Reversible
(b) spontaneous
(c) irrversible
(d) cyclic
Answer:
(b) spontaneous

Question 34.
Which one of the following spontaneous reaction is endothermic?
(a) combustion of methane
(b) dissolution of ammonium nitrate
(c) acid-base neutralization reaction
(d) none of the above
Answer:
(b) dissolution of ammonium nitrate

Question 35.
The available energy in the system to do work is called
(a) Gibbs free energy
(b) internal energy
(c) potential energy
(d) kinetic energy
Answer:
(a) Gibbs free energy

Question 36.
Which one of the following is incorrect about Gibbs free energy?
(a) Extensive property
(b) path function
(c) ∆G < 0 for a spontaneous process (d) ∆G > 0 for a non-spontaneous process
Answer:
(b) path function

Question 37.
Match the following:

(A) Adiabatic	(i) dp = 0
(B) Isothermal	(ii) dV = 0
(C) Isobaric	(iii) dq = 0
(D) Isochoric	(iv) dT = 0

(a) A – iii, B – iv, C – i, D – ii
(b) A – ii, B – iv, C – iii, D – i
(c) A – iv, B – iii, C – i, D – ii
(d) A – i, B – iv, C – ii, D – iii
Answer:
(a) A – iii, B – iv, C – i, D – ii

Question 38.
For a cyclic process involving isothermal expansion of an ideal gas, q =
(a) 0
(b) P∆V
(c) w
(d) – w
Answer:
(d) – w

Question 39.
In Calorimeter, the expression used to calculate the amount of heat change in the process is
(a) C = qm∆T
(b) C = m/q∆T
(c) C = q/m∆T
(d) C = q∆T/m
Answer:
(c) C = q/m∆T

Question 40.
“It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work”. This statement is given by
(a) Clausius
(b) Kelvin
(c) Gibbs
(d) Joule
Answer:
(a) Clausius

II. Very short question and answers (2 Marks):

Question 1.
Define isolated system.
Answer:
A system which can exchange neither matter nor energy with its surroundings is called an isolated system. Here boundary is sealed and insulated.

Question 2.
What is a homogeneous system? Give an example.
Answer:
A system is called homogeneous if the physical state of all its constituents is the same. Example: a mixture of gases.

Question 3.
Illustrate Closed system with an example.
Answer:
Hot water contained in a closed beaker is an example for a closed system. In this system energy (heat) is transferred to the surroundings but no matter (water vapour) can escape from this system. A gas contained in a cylinder fitted with a piston constitutes a closed system.

Question 4.
Define irreversible process.
Answer:
The process in which the system and surrounding cannot be restored to the initial state from the final state is called an irreversible process. All the processes occurring in nature are irreversible processes.

Question 5.
Define standard entropy change.
Answer:
The absolute entropy of a substance at 298 K and one bar pressure is called the standard entropy S°.

Question 6.
What is Zeroth law of thermodynamics?
Answer:
The law states that’ If two systems are separately in thermal equilibrium with a third one, then they tends to be in thermal equilibrium with themselves’.

Question 7.
Define enthalpy.
Answer:
The enthalpy (H), is a thermodynamic property of a system, is defined as the sum of the internal energy (U) of a system and the product of pressure and volume of the system.
H = U + PV

Question 8.
Define specific heat capacity of a system.
Answer:
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a specified temperature.

Question 9.
What is a reversible process?
Answer:
The process in which the system and surrounding can be restored to the initial state from the final state without producing any changes in the thermodynamic properties of the universe is called a reversible process.

Question 10.
What is a thermodynamic process? Give two examples.
Answer:
The method of operation which can bring about a change in the system is called thermodynamic process.
Examples: Heating, Cooling, expansion.

Question 11.
Distinguish between state function and path function.
Answer:
A state function is a thermodynamic property of a system, which has a specific value for a given state and does not depend on the path by which the particular state is reached. A path function is a thermodynamic property of the system whose value depends on the path by which the system changes from its initial to final states.

Question 12.
Write any two characteristics of internal energy.
Answer:

1. The internal energy of a system is an extensive property.
2. The internal energy of a system is a state function.

Question 13.
Write the importance of internal energy.
Answer:
The internal energy possessed by a substance differentiates its physical structure. For example, the allotropes of carbon, namely, graphite C (graphite) and diamond C (diamond), differ from each other because they possess different internal energies and have different structures.

Question 14.
Write notes on heat.
Answer:
The heat (q) is regarded as an energy in transit across the boundary separating a system from its surrounding. Heat changes lead to temperature differences between system and surrounding. Heat is a path function.

Question 15.
Write the thermodynamic significance of work.
Answer:
The work

1. is a path function.
2. appears only at the boundary of the system.
3. appears during the change in the state of the system.
4. in thermodynamics, surroundings is so large that macroscopic changes to surroundings do not happen.

Question 16.
What is meant by pressure-volume work?
Answer:
In elementary thermodynamics the only type of work generally considered is the work done in expansion (or compression) of a gas. This is known as pressure-volume work, PV work or expansion work.

Question 17.
What is specific heat capacity of a system?
Answer:
The heat absorbed by one kilogram of a substance to raise its temperature by one Kelvin at a specified temperature is called specific heat capacity of a system.

Question 18.
Distinguish between Cv and Cp.
Answer:
The molar heat capacity at constant volume (Cv) is defined as the rate of change of internal energy with respect to temperature at constant volume. The molar heat capacity at constant pressure (Cp) can be defined as the rate of change of enthalpy with respect to temperature at constant pressure.

III. Short Question and Answers (3 Marks):

Question 1.
What is internal energy?
Answer:
The internal energy of a system is equal to the energy possessed by all its constituents namely atoms, ions and molecules. The total energy of all molecules in a system is equal to the sum of their translational energy (U_t), vibrational energy (U_v), rotational energy (U_r), bond energy (U_b), electronic energy (U_e) and energy due to molecular interactions (U_i).
Thus:
U = U_t + U_v + U_r + U_b + U_e + U_i
The total energy of all the molecules of the system is called internal energy.

Question 2.
Give applications of bomb calorimeter.
Answer:

1. Bomb calorimeter is used to determine the amount of heat released in combustion reaction.
2. It is used to determine the calorific value of food.
3. Bomb calorimeter is used in many industries such as metabolic study, food processing, explosive testing etc.

Question 3.
What is Molar heat of fusion?
Answer:
The molar heat of fusion is defined as “the change in enthalpy when one mole of a solid substance is converted into the liquid state at its melting point”.
For example, the heat of fusion of ice
H₂O(s) → H₂O(l)
∆H(fusion) = +5.98 kJ

Question 4.
Write any three statements of first law of thermodynamics.
Answer:

1. Whenever an energy of a particular type disappears, an equivalent amount of another type must be produced.
2. The total energy of a system and surrounding remains constant (or conserved)
3. “Energy can neither be created nor destroyed, but may be converted from one form to another”.

Question 5.
Describe the need for the second law of thermodynamics.
Answer:
From the first law of thermodynamics, the energy of the universe is conserved. Let us consider the following processes:
A glass of hot water over time loses heat energy to the surrounding and becqmes cold.
When you mix hydrochloric acid with sodium hydroxide, it forms sodium chloride and water with evolution of heat.

In both these processes, the total energy is conserved and are consistent with the first law of thermodynamics. However, the reverse process i.e. cold water becoming hot water by absorbing heat from surrounding on its own does not occur spontaneously even though the energy change involved in this process is also consistent with the first law. However, if the heat energy is supplied to cold water, then it will become hot. i.e. the change that does not occur spontaneously and an be driven by supplying energy.

Question 6.
Write notes on entropy statement of second law of thermodynamics.
Answer:
The second law of thermodynamics can be expressed in terms of entropy, i.e “the entropy of an isolated system increases during a spontaneous process”.
For an irreversible process such as spontaneous expansion of a gas,
S_total > 0
S_total > S_system + S_surrounding
i.e., S_universe > S_system + S_surrounding
For a reversible process such as melting of ice,
S_system = S_surrounding
S_universe = 0

IV. Long Question and Answers(5 Marks):

Question 1.
Explain steps to write thermochemical equations.
Answer:
A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (∆H). The following conventions are adopted in thermochemical equations:

1. The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
2. The enthalpy change of the reaction ∆Hr has to be specified with appropriate sign and unit.
3. When the chemical reaction is reversed, the value of ∆H is reversed in sign with the same magnitude.
4. The physical states (gas, liquid, aqueous, solid in brackets) of all species are important and must be specified in a thermochemical reaction, since ∆H depends on the physical state of reactants and products.
5. If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also multiplied by the same number.
6. The negative sign of ∆H_r indicates that the reaction is exothermic and the positive sign of ∆H_r indicates an endothermic reaction.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 13 Photosynthesis Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 13 Photosynthesis

11th Bio Botany Guide Photosynthesis Text Book Back Questions and Answers

Part-I

Questions 1.
Assertion (A): Increase in Proton gradient inside lumen responsible for ATP synthesis.
Reason (R): Oxygen evolving complex of PS I located on the thylakoid membrane facing Stroma, releases H⁺ ions.
(a) Both Assertion and Reason are True.
(b) Assertion is True and Reason is False.
(c) Reason is True and Assertion is False.
(d) Both Assertion and Reason are False.
Answer:
(a) Both Assertion and Reason are True.

Question 2.
Which chlorophyll molecule does not have a phytol tail?
a) Chl – a
b) Chl – b
c) Chl – c
d) Chl – d
Answer:
(c) Chl – c

Question 3.
The correct sequence of flow of electrons in the light reaction is:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.
(b) PS I, plastoquinone, cytochrome, PS II ferredoxin.
(c) PS II, ferredoxin, plastoquinone, cytochrome, PS I.
(d) PS I, plastoquinone, cytochrome, PS II, ferredoxin.
Answer:
(a) PS II, plastoquinone, cytochrome, PS I, ferredoxin.

Question 4.
For every CO₂ molecule entering the C₃ cycle, the number of ATP & NADPH required
a) 2ATP + 2NADPH
b) 2ATP + 3NADPH
c) 3ATP + 2NADPH
d) 3ATP + 3NADPH
Answer:
(c) 3ATP + 2NADPH

Question 5.
Identify true statement regarding light reaction of photosynthesis?
(a) Splitting of water molecule is associate with PS I.
(b) PS I and PS II involved in the formation of NDPH + H⁺.
(c) The reaction center of PS I is Chlorophyll a with absorption peak at 680 nm.
(d) The reaction center of PS II is Chlorophyll a with absorption peak at 700 nm.
Answer:
(b) PS I and PS II involved in the formation of NDPH + H⁺.

Question 6.
Two groups (A&B) of bean plants of similar size and same leaf area were placed in identical conditions. Group A was exposed to light of wavelength 400 – 450 nm & Group B to light of wavelength of 500 – 550nm. Compare the photosynthetic rate of the 2 groups giving reasons.
Answer:
‘A’ group of plants exposed to light of 400 – 450nm. Chlorophyll a shows maximum absorption peak at 450nm (blue region). Hence rate of photosynthesis was high.
‘B’ group of plants exposed to light of 500 – 550nm. This wavelength refers to green region of the spectrum. Chlorophyll does not absorb light in the green region but reflects green. So plants appear green rate of photosynthesis was negligible in these plants.

Question 7.
A tree is believed to be releasing oxygen during nighttime. Do you believe the truthfulness of this statement? Justify your answer by giving reasons?
Answer:
Yes, a tree is believed to be releasing O₂ during nighttime because at night CAM plants fix CO₂ with the help of phospho Enol Pyruvic acid and produce oxala acetic acid, which is converted into malic acid-like C₄ cycle.

Question 8.
Grasses have an adaptive mechanism to compensate for photorespiratory losses – Name and describe the mechanism.
Answer:
The photorespiratory losses are checked by certain grasses by having physiological adaptation. The process of photosynthesis occurs in mesophyll cells and bundle sheath cells.

Mesophyll cells:

• Initially, CO₂ is taken up by Phosphoenolpyruvate (PEPA) (3C) and changed to oxaloacetate (4C) in the presence of PEP carboxylase.
• Oxaloacetate is reduced to Malate/Aspartate. The product formed reaches the bundle sheath.

Bundle Sheath:

• The oxidation of Malate and Aspartate occurs with the release of carbon dioxide and the formation of Pyruvate (3C)
• Due to increased CO₂ concentration RUBISCO functions as a carboxylase and not as Oxygenase.
• The photosynthetic losses are prevented.
• RUBP operates now under the Calvin cycle and pyruvate transported back to Mesophyll cells is changed into Phosphoenolpyruvate to keep the cycle going.

Question 9.
In Botany class, teacher explains, Synthesis of one glucose requires 30 ATPs in C₄ plants and only 18 ATPs in C3 plants. The same teacher explains C₄ plants are more advantageous than C₃ plants. Can you identify the reason for this contradiction?
Answer:
C₄ Plants are more advantageous than C₃ plants because of the following reasons:

C4 Plants	C3 Plants
CO2 fixation occurs in mesophyll cells only	CO2 fixation occurs in mesophyll and bundle sheath cells
RUBP is the only CO2 acceptor	PEPA Phosphoenol pyruvate in mesophyll is the acceptor in the first phase
Fixation of CO2 occurs if the atmospheric concentration of C02 is 50 ppm only	It can fix carbon dioxide even if the atmospheric concentration of CO2 is below 10 ppm
Optimum temperature is 20° to 25°C	Optimum temperature is 30° to 45°C and is thus effective in tropical regions.
RUBP carboxylase enzyme also functions as oxygenase if the 0, concentration is higher than carbon dioxide	PEP carboxylase enzyme functions even at low carbon – dioxide concentrations.
Higher rate of photorespiration and hence rate of photosynthesis is reduced.	Minimal rate of photorespiration is seen is C4 plants.

Question 10.
When there is plenty of light and higher concentration of O₂, what kind of pathway does the plant undergo? Analyse the reasons.
Answer:
The rate of photosynthesis decreases when there is an increase in oxygen concentration. This Inhibitory effect of oxygen was first discovered by Warburg (1920) using green algae, Chlorella.

Part-II.

11th Bio Botany Guide Photosynthesis Additional Important Questions and Answers

I. Choose the Correct Answers

Question 1.
Photosynthesis is the major:
(a) endothermic reaction
(b) exothermic reaction
(c) endergonic reaction
(d) exergonic reaction
Answer:
(c) endergonic reaction

Question 2.
The physiological unit of photosynthesis is
a) 150-250 Chlorophyll molecules
b) 200-300 chlorophyll molecules
c) 440-660 chlorophyll molecules
d) 450- 650 chlorophyll molecules
Answer:
b) 200-300 chlorophyll molecules

Question 3.
How many million tonnes of dry matter produced annually by photosynthesis?
(a) 1700 million tonnes
(b) 1900 million tonnes
(c) 1400 million tonnes
(d) 2000 million tonnes
Answer:
(a) 1700 million tonnes

II. Match Correctly & Choose The Right Answer

Question 4.
I) Black Mann – A) the importance of Chlorophyll
II) Warburg – B) Law of limiting factor
III) Dustrochet – C) C4 cycle
IV) Hatch & Slack – Chlorella

Answer:
c) B-D-A-C

Question 5.
Thylakoid disc diameter is:
(a) 0.35 to 0.75 microns
(b) 0.25 to 0.8 microns
(c) 0.45 to 0.8 microns
(d) 0.50 to 0.9 microns
Answer:
(b) 0.25 to 0.8 microns

Question 6.
The pigment responsible for the yellowing of leaves during autumn season is
a) Violaxanthin
b) Fucoxanthin
c) Phycobillin
d) Lycopene
Answer:
d) Lycopene

Question 7.
The no of quanta of light required for the release of one oxygen molecule
a) 18 quanta
b) 8 quanta
c) 81 quanta
d) 19 quanta
Answer:
b) 8 quanta

Question 8.
Each pyrrole ring comprises of:
(a) six carbons and one nitrogen atom
(b) three carbons and one nitrogen atom
(c) four carbons and one nitrogen atom
(d) four carbons and two nitrogen atom
Answer:
(c) four carbons and one nitrogen atom

Question 9.
RUBISCO – Constitute …………. of chloroplast protein
a) 17%
b) 20%
c) 18%
d) 16%
Answer:
d) 16%

Question 10.
Pheophytin resembles chlorophyll ‘a’ except that it lacks:
(a) Fe atom
(b) Mn atom
(c) Mg atom
(d) Cu atom
Answer:
(c) Mg atom

Question 11.
According to Emerson the fall in quantum yield about 680 nm is called
a) Phoisynthtic drop
b) Emerson drop
c) Airburg effect
d) Red drop
Answer:
d) Red drop

Question 12.
Which one of the photosynthetic pigments is called shield pigment:
(a) carotenes
(b) chlorophyll ‘b’
(c) pheophytin
(d) carotenoids
Answer:
(d) carotenoids

Question 13.
Which of the following equation correctly sums up photosynthesis

Answer:

Question 14.
Photosynthetic rate of red light (650 nm) is equal to:
(a) 42.5
(b) 10.0
(c) 43.5
(d) 40.8
Answer:
(c) 43.5

Question 15.
Which photosystem is found to be located on the outer surface of the thylakoid?
a) PS I
b) P.S II
c) P. 890
d) Both (a) and (b)
Answer:
a) P.S I

Question 16.
Phosphorylation taking place during respiration is called as:
(a) Photophorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) None of the above
Answer:
(b) Oxidative phosphorylation

Question 17.
The term Quantosome was coined by
a) Emerson
b) Liebig
c) Calvin & Melvin
d) Park & Biggins
Answer:
d) Park & Biggins

Question 18.
In bioenergetics of light reaction, to release one electron from pigment system it requires:
(a) two quanta of light
(b) four quanta of light
(c) one quantum of light
(d) eight quanta of light
Answer:
(a) two quanta of light

Question 19.
Photosynthesis produces
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
b) 7100 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year
c) 1600 million tonnes of dry matter/week by fixing 56 x 12¹⁰ kg of carbon every week
d) 6100 million tonnes of dry matter/month by fixing 100 x 10¹² kg of carbon every month
Answer:
a) 1700 million tonnes of dry matter/year by fixing 75 x 10¹² kg of carbon every year

Question 20.
In C₄ plants, how many ATPs and NADPH + H⁺ are utilised for the release of one oxygen molecule:
(a) 3 ATPs and 2 NADPH + H⁺
(b) 4 ATPs and 3 NADPH + H⁺
(c) 2 ATPs and 2 NADPH + H⁺
(d) 5 ATPs and 2 NADPH + H⁺
Answer:
(d) 5 ATPs and 2 NADPH + H⁺

Question 21.
Products of light reaction in photosynthesis are
a) ATP & NADPH₂
b) ADP & glucose
c) Ferredoxin and cytochrome b6
d) Cytochrome
Answer:
a) ATP & NADPH₂

Question 22.
In the sugarcane plant, the dicarboxylic acid pathway was first discovered by:
(a) Hatch and Slack
(b) Kortschak, Hart and Burr
(c) Calvin and Benson
(d) Mitchell and Root
Answer:
(b) Kortschak, Hart and Burr

Question 23.
Photosynthetic pigments in chloroplasts lie embedded in
a) Chloroplast envelope
b) Plastogloblue
c) matrix
d) thylakoids
Answer:
d) thylakoids

Question 24.
Indicate the correct answer:
(a) C₄ plants are adapted to only rainy conditions
(b) C₄ plants are partially adapted to drought condition
(c) C₄ plants are exclusively adapted to desert condition
(d) C₄ plants are adapted to aquatic condition
Answer:
(b) C₄ plants are partially adapted to drought condition

Question 25.
Carotenoids and Xanthophylls are also known as
a) Respiratory pigments
b) Accessory pigments
c) Photosynthetic pigments.
d) Photolytic pigments
Answer:
b) Accessory pigments

Question 26.
Which metal ion is a
a) Iron
b) cobalt
c) Magnesium
d) Zinc
Answer:
c) Magnesium

Question 27.
The important external factors affecting photosynthesis are:
(a) light, chlorophyll, temperature
(b) light, stomatal opening, oxygen
(c) light, protoplasmic factor, oxygen
(d) light, CO₂ and oxygen
Answer:
(d) light, CO₂ and oxygen

Question 28.
The process of photophosphorylation was discovered by
a) Priestly
b) Calvin
c) Amon
d) Warburg
Answer:
c) Arnon

Question 29.
Which of the following is a C4 plant
a) Potato
b) Sugarcane
c) Pea
d) Papaya
Answer:
b) Sugarcane

Question 30.
Splitting of water molecule (photolysis) produces:
(a) hydrogen and oxygen
(b) electrons, protons and oxygen
(c) electrons and oxygen
(d) hydrogen, carbon dioxide and oxygen
Answer:
(b) electrons, protons and oxygen

Question 31.
Dimorphism in chloroplasts is seen in
a) C4 plants
b) C₂ plants
c) CAM – plants
d) C3 plants
Answer:
a) C4 plants

Question 32.
Energy required for ATP synthesis in PSII comes from
a) Proton gradient
b) Electron gradient
c) Reduction of glucose
d) Oxidation of glucose
Answer:
a) Proton gradient

Question 33.
The by-product of Photosynthesis is
a) O₂
b) CO₂
c) Carbohydrate
d) H₂O
Answer:
a) O₂

Question 34.
Which of the following process is called reverse of Glycolysis?
a) CO₂ reduction
b) RUBP carboxylation
c) RUBP regeneration
d) ATP synthesis
Answer:
a) CO₂ reduction

Question 35.
The Dark reaction of Photosynthesis occurs in
a) Matrix
b) Grana
c) Stroma
d) Cytoplasm
Answer:
c) Stroma

Question 36.
A granal chloroplasts are characteristics of
a) Mesophyll of pea leaves
b) Bundle sheath of Mango leaves
c) Mesophyll of maize leaves
d) Bundle sheath of sugar cane leaves
Answer:
d) Bundle sheath of Sugar cane leaves

Question 37.
Which of the following plant is a better photosynthesis?
a) Mango
b) Sugarcane
c) Wheat
d) Rice
Answer:
b) Sugarcane

Question 38.
The enzyme that is not found in a C3 plant is
a) RUBP carboxylase
b) PEP carboxylase
c) NADP reductase
d) ATP synthase
Answer:
b) PEP carboxylase

Question 39.
Which of the following factors affect the rate of photosynthesis?
I. Light
II. Protoplasmic factor
III. Hormones Codes
IV. Haemoglobin
a) only III
b) I and II
c) only IV
d) I, II, and III
Answer:
d) I, II and III

Question 40.
Match the following columns

Column I	Column II
I. The 5C sugar that	A. RUBIS Co
II. 3 C sugar that gives Calvin cycle  its nickname	B. Glyceraldehyde 3 phosphate
III. Activated form of  3 PGA	C. 3 phosphoglyceric acid
IV. Huge enzyme complex that brings CO2 and 5C sugar together.	D. RUBP

Question 41.
One complete light reaction involves light energy.
a) 30 quanta
b) 48 quanta
c) 40 quanta
d) 25 quanta
Answer:
b) 48 quanta

Question 42.
During photosynthesis which of the following event does not take place?
a) oxidation of CO₂
b) Reduction of CO₂
c) oxidation of H₂O
d) Light absorption
Answer:
a) Oxidation of CO₂

Question 43.
The site of light trapping in the chloroplast is
a) Thylakoid membrane
b) Stroma
c) Plasma fluid
d) Stromal lamellae
Answer:
a) Thylakoid membrane

Question 44.
Kranz anatomy is traced in the Leaves of
a) Wheat
b) Potato
c) Mustard
d) Sugarcane
Answer:
d) Sugarcane

Question 45.
The intermediate got from Kreb’s cycle that is used for chlorophyll synthesis is
a) Citric acid
b) Isocitric acid
c) Succinic acid
d) Fumaric acid
Answer:
c) Succinic acid

Question 46.
The existence of light and dark reaction of photosynthesis was proved by
a) Blackman
b) Emerson
c) Warburg
d) Arnon
Answer:
a) Blackman

Question 47.
Choose the wrong match.
a) Hatch & Slack – Dicarboxylic acid pathway
b) Decker – PCO cycle
c) Ruben, Kamen – CAM cycle
d) Calvin Benson – PCR cycle
Answer:
c) Ruben, Kamen – CAM cycle

Question 48.
I) Primary CO, acceptor – PEPA
II) 4C compound produced 1 st – OAA
III) 1st carboxylation occurs in – Bundle sheath cells
IV) 2nd carboxylation occurs in – Mesophy 11 cells
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer:
c) III & IV

Question 49.
Say True or False with respect to C₂ cycle
I) RUBISCO has the most abundant protein on earth.
II) Photorespiration does not yield any free energy in the form of ATP
III) The end product is a 2 – c compound. So the cycle is known as C₂ cycle
IV) Under certain conditions 50% of the photosynthetic potential is lost because of photorespiration.

Answer:
b) True – True – False – True

Question 50.
Say True or False and choose the right option from the given choice.
I) PAR is between – 400 – 700 mil.
II) Heliophytes (Beans) require higher light intensity than sccophytes (oxalis)
III) Red light induces lowest rate of photosynthesis.
IV) Green light induces highest rate of photosynthesis.

Answer:
c) True – True – False – False

Question 51.
Choose the wrongly matched pair
a) Stephen hales – Father of plant physiology
b) Lavoisier – Purifying gas oxygen is produced in sun light
c) Vonmayer – Green plants convert solar energy into chemical energy
d) Emerson &Amoid – C4 cycle
Answer:
d) Emerson & Arnold – C4 cycle

Question 52.
Choose the rightly matched pair
a) Chlorophyll a – Accessory pigments and trap solar energy
b) Chlorophyll b – Differs from Chlorophyll a in having CH3 instead of CHO – at 3rd C atom
c) Chlorophyll c – Differs from Chlorophyll a by lacking a phytol tail
d) Chlorophyll d – It has CHO at 3rd at the 3rd carbon atom at 11 – pyrrole ring
Answer:
c) Chlorophyll c – Differs from chlorophyll a by lacking a phytol tail.

Question 53.
Choose the right matched pairs from the given options.
I) Green non sulphur bacteria – Clostridium & Lynbya
II) Green sulphur bacteria – Chlorobacterium & Chlorobium
III) Purple sulphur bacteria – Thiospirillum & Chromatium
IV) Purple non sulphur bacteria – Rhodopseudomonas & Rhodospirillum
a) I, II, & III
b) II, III & IV
C) I, II & IV
d) I, III & IV
Answer:
b) II, III & IV

II. Assertion (A) & Reason (R)

a) Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion
b) Both Assertion (A) and Reason (R) are true – Reason is not the correct explanation of Assertion.
c) Assertion (A) is false but Reason (R) is true.
d) Both Assertion (A) and Reason (R) are false.

Question 1.
Assertion (A): Chlorophyll appears green.
Reason (R): It absorbs light mainly in the region of green part of light spectrum.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 2.
Assertion (A): Red of spectrum contains high energy.
Reason (R): Green light of Visible spectrum contain low energy than red light.
Answer:
c) Assertion (A) is False but Reason (R) is true.

Question 3.
Assetion (A): Non cyclic photo phosphorylation occurs in the stroma of chloroplasts
Reason (R): There is a continuous flow of electrons in this process.
Answer:
d) Both Assertion (A) and Reason (R) are false.

Question 4.
Assetion (A): Carotenes and Xanthophylls are soluble in either
Reason (R): These are accessory pigments of photosynthesis
Answer:
b) Both Assertion (A) and Reason (R) are true, Reason is not the correct explanation of Assertion

Question 5.
Assetion (A): Carotenoids are accessory pigments
Reason (R): Absorbed light energy is transferred to reaction centre by carotenoids.
Answer:
a) Assertion (A) and Reason (R) True and Reason is the correct explanation of Assertion.

III. 2 Marks Questions

Question 1.
What is the function of the plant in the universe?
Answer:
Plants are the major machinery which produces organic compounds like carbohydrates,lipids, proteins, nucleic acids and other biomolecules.

Question 2.
What is PAR?
Answer:
It refers to Photosynthetically Active Radiation, which is between 400 – 700 nm photosynthetic rate is maximum in blue and red light – Green light induces lowest rate of photosynthesis.

Question 3.
What is the site of photosynthesis?
Answer:
Chloroplasts are the main site of photosynthesis and both the energy-yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in the chloroplast.

Question 4.
Name the photosynthetic pigments of Algae?
Answer:

• Chlorophyll b – Green Algae
• Chlorophyll c – Dianoflagellates, Diatoms & Brown Algae
• Chlorophyll d – Red Algae
• Chlorophyll e – Xantho phycean Algae.

Question 5.
Endosymbiotic hypothesis says that chloroplasts evolved from bacteria. Substantiate the statement.
Answer:
Presence of 70S ribosome and DNA gives them status of semi-autonomy and proves endosymbiotic hypothesis which says chloroplast evolved from bacteria.

Question 6.
Write down the significance of photorespiration.
Answer:

• Glycine and Serine synthesized during this process are precursors of many biomolecules like Chlorophyll, Proteins, Nucleotides.
• It consumes excess NADH + H⁺ generated.
• Glycolate protects cells from Photooxidation.

Question 7.
What are the conclusions of Hill’s Reaction?
Answer:

• During Photosynthesis oxygen is evolved from water.
• Electrons for the reduction of CO₂ are obtained from water.
• A reduced substance produced, later helps to reduce CO₂
• 2H₂O + 2A→ 2 AH₂ + O₂

Question 8.
What are Xanthophylls?
Answer:
Yellow (C₄₀H₅₆O₂) pigments are like carotenes but contain oxygen. Lutein is responsible for yellow colour change of leaves during autumn season. Examples: Lutein, Violaxanthin and Fueoxanthin.

Question 9.
Notes on Phycobillins.
Answer:

• They are proteinaceous pigments.
• They are soluble in water.
• Lack ‘Mg’ and phytol tail.
• There are 2 forms 1. Phycocyanin 2. Phycoerythrin.
• Phycocyanin occur in Cyanobacteria.
• Phyco erythrin occur in Rhodophycean Algae.

Question 10.
Define absorption spectrum.
Answer:
Pigments absorb different wavelengths of light. A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called its absorption spectrum.

Question 11.
Why do we call carotenoids shield pigments?
Answer:

• Carotenoids are yellow to orange pigments mostly tetraterpens and absorb light strongly in the blue to violet region of the visible spectrum.
• These pigments protect chlorophyll from photosynthetic oxidative damage.

Question 12.
What is known as substrate-level phosphorylation?
Answer:
Phosphorylation taking place during respiration is called oxidative phosphorylation and ATP produced by the breakdown of substrate is known as substrate-level phosphorylation.

Question 13.
What are Quantosomes?
Answer:

• They are physiological photosynthetic units, located on the inner membrane of thylakoid lamellae of size 180A X 160 A length & breadth.
• It was named by Park &Pickins( 1964).
• One quantosome contains about 230 chlorophyll molecules.
• It constitutes a photosynthetic unit responsible for the production of one O₂ molecule or reduction of one CO₂ molecule.

Question 14.
What is Bioluminescence?
Answer:
It is the special aspect of few living organism, in which there are some biochemical substances production is responsible for the emission of light by a living organism.

Question 15.
What is the significance of photorespiration?
Answer:
Significance of photorespiration:

1. Glycine and Serine synthesized during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
2. It consumes excess NADH + H⁺ generated.
3. Glycolate protects cells from Photooxidation.

Question 16.
Explain water oxidizing clock or S state mechanism.
Answer:
The splitting of water molecule, mechanism was studied by KoK et, al (1970).
It consists of a series of 5 states so, s₁, s₂, s₃, s₄.
Each sate acquires positive charge by a photon (hv) and after the state s₄ – if acquires 4 positive charges 4 electron and evolution of oxygen.

Two molecules of water go back to the so.
At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.
4H₂O → 4H⁺+ + 40H^–
40H^{–  →} 2H₂O+O₂+4e^–
2H₂O → 4H⁺+ O₂ + 4e^–.

Question 17.
Distinguish between photophosphorylation and oxidative phosphorylation.
Answer:

Photophosphorylation	Phosphorylation
1. It is the process of synthesis of ATP from ADP by the addition of phosphate takes place with the help of photosynthesis light generated electron, which is known as photophosphorylation. It is of two types Cyclic and Non-cyclic photophosphorylation.	1. The process of production of ATP via terminal oxidation of reduced coenzymes during respiration is known as oxidative phosphorylation.

Question 18.
What are the air pollutants, that affect the rate of photosynthesis?
Answer:
Pollutants like SO₂, NO₂, O₃ (Ozone) and Smog affect the rate of photosynthesis.

Question 19.
What are the conditions for the occurrence of Non-cyclic photophosphorylation?
Answer:

• Non-Cyclic Photo Phosphorylation occurs, when.
• There is the availability of NADP⁺ for reduction.
• Two molecules of water go back to the so.
• When there is the splitting of water molecules.
• When both PSI and PS II are activated.

Question 20.
Name any three photosynthetic bacteria.
Answer:
Three photosynthetic bacteria:

1. Chlorobacterium
2. Thiospirillum
3. Rodhospirillum

Question 21.
What are the 2 phases of light reaction?
Answer:
The light reaction has 2 phases

1. Photooxidation phase
2.  Photochemical phase.

I) Photooxidation phase (POP):

•  Absorption of light energy.
• Transfer of energy from accessory pigments to the reaction centre.
• Activation of chlorophyll ‘a’ -molecule.

II) Photochemical phase (PCO):

• Photolysis of water and evolution of oxygen.
• Electron transport and synthesis of assimilatory power.

Question 22.
Greenlight induces lowest rate of photosynthesis justifies.
Answer:

• Yes green light induces the lowest rate of photosynthesis because it is not coming under photosynthetically
• Active reduction – (400 – 700 nm) known as PAR.
• PAR – (Photosynthetic rate is maximum in blue and red light not in green Light.

Question 23.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:

• Complet light reaction releases 6 oxygen molecules.
• If one molecule of oxygen evolution requires 8 quanta means for 6 oxygen molecules (6 x 8 = 48) quanta of light reaquired for a complete light reaction.

Question 24.
Give the balance sheet of Calvin or C3 cycle.
Answer:
One molecule of CO₂ is fixed in one turn of the Calvin or C³ cycle.
So 6 turns of cycle will be required to fix 6 molecules of C0₂ – (i.e) to form one molecule of Glucose C6H₁₂O6

In	Out
6CO2	1 Glucose
18 ATP	18 ADP
12 NADPH	12 NADP

IV. 3 Mark Questions

Question 1.
Mention any three significance of photosynthesis.
Answer:
Three significance of photosynthesis:

1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
3. Photosynthesis balances the oxygen and carbon cycle in nature.

Question 2.
What are the properties of light.
Answer:

• Light is a transverse electromagnetic wave.
• It consists of ocillating electric and magnetic fields that are perpendicular to each other and perpenticular to the direction of propagation of the light.
• Light moves at a speed of 3 x 108 ms^-1
• Wave length is the distance between successive crests of the wave.
• Light as a particle is called photon. Each photon contains an amount of energy known as quantum.
• The energy of a photon depends on the frequency of the light.

Question 3.
Distinguish between Absorption spectrum & Action Spectrum.
Answer:

Absorption spectrum	Action spectrum
A curve obtained by plotting the amount of absorption of different wavelengths of light by a pigment is called the Absorption spectrum.	The curve showing the rate of photosynthesis at different wavelengths of light is called the action spectrum.

Question 4.
Distinguish between fluorescence and phosphorescence.
Answer:

Fluorescence	Phosphorescence
1. Immediate emission of absorbed radiations in the form of radiation energy (light) in the red region.	1. This is the delayed emission of absorbed radiations in the form of light in the red region.
2. The electrons move from S1 →SO	2. The electrons pathway is from S2 →S1 →T1→ SO

Question 5.
What is meant by the ground state?
Answer:
The action of photon plays a vital role in the excitation of pigment molecules to release an electron. When the molecules absorb a photon, it is in an excited state. When the light source turned off, the high-energy electrons return to their normal low-energy orbitals as the excited molecule goes back to its original stable condition known as the ground state.

Question 6.
Write down the significance of Photosynthesis.
Answer:

• Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly All other organism depend on them for energy.
• It liberates oxygen in the atmosphere and balances.
• Fuels such as coal, petroleum, and other fossil fuels are preserved forms got only from photosynthetic plants.
• It also provides fodder, fibre, firewood, timber useful medicinal products and these sources come by the act of photosynthesis.

Question 7.
What is DCMU?
Answer:

• It is a chemical herbicide having an inhibiting effect on photosynthesis.
• It is Dichloro phenyl D₁ Methyl Urea.
• It can inhibit electron flow during light reactions of photosynthesis.
• It is a herbicide that blocks the plastoquinone binding site of P.S II and inhibits electron flow from plastoquinone to cytochrome.

Question 8.
State black man’s law of limiting factor.
Answer:
It is a modified law proposed by Liebig’s law of minimum.

Definition:
According to Blackman at any given point of time, the lowest factor among essentials will limit the rate of Photosynthesis.

Example:
When in a condition, if light intensity is low also C0₂ concentration low, in this situation among the two factors which ever is the lowest is considered as the limiting factor here among the essentials CO₂ concentration is the limiting factor.

Question 9.
What is meant by dicarboxylic acid pathway?
Answer:
C₄ pathway is completed in two phases, first phase takes place in the stroma of mesophyll cells, where the CO₂ acceptor mblecule is 3 – Carbon compound, phosphoenolpyruvate (PEP) to form 4 – carbon Oxalo acetic acid (OAA). The first product is a 4 – carbon and so it is named as C₄ cycle. Oxalo acetic acid is a dicarboxylic acid and hence this cycle is also known as a dicarboxylic acid pathway.

Question 10.
State some interesting facts about C4 cycle.
Answer:

• C4 cycle is an alternative path way for CO2 fixation.
• It occur in nearly 1000 plant species 300 dicots but mostly 700 monocots (tropical and sub tropical grasses)
• It represent about 5% earths biomass and 1% of its known plants
• 30% terrestrial carbon fixation on earth is due to C4 So, if C4 plants on earth is increased, then by carbon sequestration by thus strategy severe climate change would be avoided in the near future.

Question 11.
What is Kranz Anatomy or what is meant by Dimorphism of Chloropiasts in C4 plants.
Answer:

C3 plants	C4 plants
C3 plants kranz Anatomy not seen	C4 plants show kranz Anatomy
C3 plants only one type of chloroplasts seen both in bundle sheath and mesophyll cells.	C4 plants Bundle sheath surrounding the vascular bundles have larger chloroplast and have thylakoids are free, not arranged in granum.
Thylakoids are arranged in granum as coins.	Mesophyll cells have smaller chloroplasts thylakoid arranged in granum

Question 12.
what is the significance of the CAM cycle?
Answer:
The significance of the CAM cycle:

1. It is advantageous for succulent plants to obtain CO₂ from malic acid when stomata are closed.
2. During daytime, stomata are closed and CO₂ is not taken but continues their photosynthesis.
3. Stomata are closed during the daytime and help the plants to avoid transpiration and water loss.

Question 13.
Compare the C3 and C4 on the basis of ATP production.
Answer:

C3 plants	C4 plants
The evolution of one oxygen molecule (4 electrons required) requires 8 quanta of light
C3 plants utilise 2 ATPs and 2 NAD PH+H+ to evolve one oxygen molecule	C4 plants utilise 5 ATPs and 2NADPH + H+ to evolve one oxygen molecule
To evolve 6 molecules of oxygen or 1 molecule of Glucose 8 ATPs and 12 NADPH + H+ are utilised.	To evolve 6 molecules of oxygen 30 ATPs and 12 NADPH + H+ are utilised.

Question 14.
What will be the quanta requirement for the complete light reaction which releases 6 oxygen molecules?
Answer:
The complete light reaction releases 6 oxygen molecules if one molecule of oxygen evolution, requires 81 quanta means for 6 oxygen molecules 6 x 8 = 48 quanta of light required for the complete light reaction.

Question 15.
Draw the Graphical representation of any 3 factors affecting photosynthesis.
Answer:

1. Light Intensity
2. CO₂ Concentration
3. Temperature

V. 5 Mark Questions

Question 1.
Distinguish between Photosystem – I and photosystem – II
Answer:
Photosystem – I:

1. The reaction centre is P700.
2. PS I is involved in both cyclic and non – cyclic.
3. Not involved in the photolysis of water and evolution of oxygen.
4. It receives electrons from PS II during non – cyclic photophosphorylation.
5. Located in unstacked region granum facing chloroplast stroma.
6. Chlorophyll and Carotenoid ratio is 20 to 30 : 1.

Photosystem – II:

1. Reaction centre is P680.
2. PS II participates in Non – cyclic pathway.
3. Photolysis of water and evolution of oxygen take place.
4. It receives electrons by photolysis of water.
5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
6. Chlorophyll and Carotenoid ratio is 3 to 7 : 1.

Question 2.
Structure of Chloroplast.
Answer:

Question 3.
Notes on photosystem and Reaction centre.
Answer:

• Thylakoid membrane contains photosystem I (PSI) and photosystem II (PSII)
• PS I is unstacked region of granum tàcing stroma ofchÍoroplast.
• PS II is found in stacked region of thylakoid membrane facing lumen of thylakoid.
• Each photosystem consists of central core complex (CC) and light harvesting complex (LHC) or Antenna molecules.
• The core complex consists of respective reaction centre associated with proteins, electron donors and acceptors.
• PSI – CCI consists of reaction centre P 700 and LHC – I
• PS II- CC II consists of reaction centre P680 and LHC – II
• Light harvesting complex consists of several chiorophylls, carotenoids and xanthophyll molecules.
• The main function of LHC is to harvesting light energy and transfer it to their respective reaction centre.

Question 4.
Difference between photosystem I and Photosystem II.
Answer:

Photosystem I	Photosystem II
1. The reaction centre is P700	1. Reaction centre is P 680.
2. PSI is involved in Photolysis of water and evolution.	2. PS II participates in Non – Cyclic pathway
3. Not involved in photolysis of water and evolution of oxygen.	3. Photolysis of water and evolution of oxygen take place.
4. It receives electrons from PSII during non – cyclic photophosphorylation.	4. It receives electrons by photolysis of water.
5. Located in unstacked region granum racing chloroplast stroma.	5. Located in stacked region of thylakoid membrane facing lumen of thylakoid.
6. Chlorophyll and carotenoid ratio is 20 to 30:1	6. Chlorophyll and carotenoid ratio is 3 to 7:1

Question 5.
Differences between Cyclic Photophosphorylation and Non – cyclic photophosphorylation.
Answer:

Cyclic Photophosphorylation	Non – Cyclic Photophosphorylation
1. PSI only involved	1. PS I and PS II involved.
2. Reaction centre is P 700	2. Reaction centre is P 680.
3. Electrons released are cycled back	3. Electron released are not cycled back.
4. Photolysis of water does not take place	4. Photolysis of water takes place
5. Only ATP Synthesized	5. ATP and NADPH + H+ are synthesized.
6. Phosphorylation takes place at two places	6. Phosphorylation takes place at only one place
7. It does not require an external electron donor.	7. Requires external electron donor like H2O or H2S
8. It is not sensitive to dichloro dimethyl urea (DCMU)	8. It is sensitive to DCMU and inhibits electron flow

Question 6.
Compare and contrast the photosynthetic processes in C₃ and C₄ plants.
Answer:
Contrast the photosynthetic processes in C₃ and C₄ plants:
C₃ Plants:

• CO₂ fixation takes place in mesophyll cells only.
• CO₂ acceptor is RUBP only.
• First product is 3C – PGA.
• Kranz anatomy is not present.
• Granum is present in mesophyll cells.
• Normal Chloroplast.
• Optimum temperature 20° to 25° C.
• Fixation of CO₂ at 50 ppm.
• Less efficient due to higher photorespiration.
• RUBP carboxylase enzyme used for fixation.
• 18 ATPs used to synthesize one glucose.
• Efficient at low CO₂.
• eg: Paddy, Wheat, Potato and so on.

C₄ Plants:

• CO₂ fixation takes place mesophyll and bundle sheath.
• PEP in mesophyll and RUBP in bundle sheath cells.
• First product is 4C – OAA.
• Kranz anatomy is present.
• Granum present in mesophyll cells and absent in bundle sheath.
• Dimorphic chloroplast.
• Optimum temperature 30° to 45° C.
• Fixation of CO₂ even less than 10 ppm.
• More efficient due to less photorespiration.
• PEP carboxylase and RUBP carboxylase used.
• 30 ATPs to produce one glucose.
• Efficient at higher CO₂.
• eg: Sugar cane, Maize, Sorghum, Amaranthus and so on.

Question 7.
Explain Non cyclic photophosphorylation.
Answer:
When PS II (P680) gets activated, electrons from a high energy state pass through a series of electron carriers like pheophytin, plastoquinone cytochrome complex, plastocyanin, and finally accepted by PS I (P700).

During this flow ATP is generated:

• PS. I (P 700) is activated by light electrons moved to high energy state and accepted by electron acceptor (FRS) Ferredoxin Reducing Substance, during downhill passes through Ferredoxin. During this process NADPH is reduced by H+ formed during photolysis.
• Electrons released from PS II are not cycled back but used in the reduction of NADPH+ into NADPH+H⁺.
• During electron transport, it generates ATP and this type of Phophorylation is called.

Non Cyclic Photophosphorylation:
The electron flow looks like the letter ‘Z’ so known as Z scheme. It has 3 stages.

1. Electron transport from water to P 680: Electrons lost by the PS II are replaced by electrons from splitting of the water molecule, producing electron, protons, and oxygen.
2. Electron transport from P680 to P 700: The flow, through various electron carrier molecules, like pheophytin, plastoquinone (PQ) cytochrome b6 – F complex, plastocyanin (PC) finally reaches P 700 (P.S.I)
3. Electron transport from P 700 to NADP⁺: PSI (P700) is excited now and the electrons pass through ferredoxin, NADP is reduced to NADPH + H⁺

Question 8.
Explain Calvin cycle or C₃ cycle?
Answer:

• It follows a light reaction.
• Utilises ATP and NADPH + H⁺ produced during the light reaction, and reduce carbon dioxide carbohydrate.
• These reactions do not require light so named as Dark reactions.
• The first formed product is a 3 carbon compound (Phospho Glyceric Acid) and so-known as C₃ cycle.
• It was found by Melvin, Calvin, and Benson – so known as the Calvin cycle.
• Occur in the stroma of the chloroplast.
• It is temperature-dependent, and so it is also called a thermochemical reaction.

Phase I carboxylation (Carbon fixation):

• The 5 C compound Ribulose 1 – 5 Bisphosphate (RUBP) with the help of (RUBISCO) enzyme accepts one molecule of carbon dioxide → 6 carbon compound (unstable)
• The 6c compound is broken into → 2 molecules of 3 c compound.

Phase II – Glycolytic Reversai/Reduction :

Phase III – Regeneration:

• The regeneration of RUBP involves several intermediate compounds of 6c, 5c, 4c, and 7c compounds.
• Fixation of one CO₂ require 3 ATPs + 2NADPH⁺ + H⁺
• Fixation of six CO₂ require I8 ATPs+ 12NADPH⁺ + H⁺
• 6 c compound is the net gain to form hexose sugar.

Question 9.
Draw the flow Chart of C₄ pathway?
Answer:

Question 10.
Explain CAM Cycle?
Answer:

• It is one of the carbon path ways in succulent plants growing in semi arid or xerophytic condition.
• The stomata are closed during day (scoto active) and open during night.
• This reverse rhythm help to conserve water loss through transpiration and will stop the fixation of CO₂ during day.
• At night time CAM plants fix CO, with help of (PEP) phospho Enol Pyruvic acid and produce (OAA) Oxalo Acetic Acid.
• Subsequently OAA is converted into a Malic acid-like C4 cycle and get accumulated in the vacuole, increasing the acidity.
• During daytime stomata, are closed and Malic acid is decarboxylated into pyruvic acid resulting in the decrease of acidity.
• CO₂ thus formed enters into the Calvin cycle and produces carbohydrates.

Question 11.
Give the flow chart of Photo respiration or C₂ cycle.
Answer:

Question 12.
Differentiate Photorespiration and Dark respiration.
Answer:

Photorespiration	Dark respiration
1. It takes place in photosynthetic green cells	It takes place in all living cells
2. It takes place only in the presence of light	It involves only Mitochondria
3. It involves Chloroplast, Peroxisome, and Mitochondria	It involves only Mitochondria
4. It does not involve Glycolysis, Kreb’s cycle, and ETS	It involves Glycolysis, Kreb’s cycle and ETS
5. Substrate is Glycolic acid	Substrate is Carbohydrates protein or fats
6. It is not essential for survival	Essential for survival
7. No phosphorylation and yield of ATP Phosphorylation	produces ATP energy
8. NADH2 is oxidised to NAD+	NAD+ is reduced to NADH2
9. Hydrogen peroxide is produced	Hydrogen peroxide is not produced
10. End products are CO2 and PGA	End products are CO2 and water

Question 13.
Give any 5 External factors affecting photosynthesis.
Answer:
1. Carbon dioxide:
330ppm or 0.3% of CO₂ is available in the atomsphere If there is an increase in CO₂ concentration the rate of Photosynthesis increases -If it increases beyond 500 PPm rate of photosynthesis will be inhibited.

2. Oxygen:
When there is increase in oxygen concentration there is unhibition of photosynthesis Warburg – studied this in chlorellain 1920. This effect is known as Warburg effect.

3. Temperature:

• Optimum temperature for photosynthesis vary from plant to plant
• Normally it is 25°C to 3 5°C
• In Opuntia, it is 55°C
• In Lichens it is 20°C
• In Algae growing in hot spring it is 75°C
• At high and low temperature the stomata will close also the enzymes get inactivated.

4. Water:

• Pholysis of water provide electrons and protons for the reduction of NADP – directly.
• Affect stomatal movement and hydration of protoplasm – indirectly.
• During water stress, supply of NADPH + H⁺ affected

5. Minerals:

Deficiency	Effect
Mg, Fe and N	Synthesis of chlorophyll
P	phosphorylation reactions
Mn, Cl-	photolysis of water
CU	Formation of plastocyanin

Question 14.
Explain the test tube funnel experiment.
Answer:
AIM: To proove that oxygen is evolved during Photosynthesis.
Procedure:
Take some hydrilla plant and place them at the bottom of a beaker containing water – Add, little NaHCO3 in to the water. Cover plant with an inverted funnel Invert a test tube over the funnel keep this set up in sun light.

Observation: Air bubbles are released from Hydrilla plant and collected in the test tube by downward displace ment of water. Take the test tube carefully by closing with a finger and then introduce a burning match stick, it bum brightly.
Inference: Hydrilla plant perform photosynthesis and oxygen is liberated during photosynthesis.

Question 15.
Explain the experiment to determine rate of photosynthesis by Witmott’s bubbler.
Answer:
Procedure:

• Wilmott’s bubbler consists of a wide mouth bottle fitted with a single holed cork, a glass tube with lower and having wider opening to insert hydrilla plant.
• The upper end is fitted to a narrow bottle with water.
• Fill the bottle with water and insert hydrilla living into wider part of the tube.
• Hydrilla plant should be cut inside the water to avoid entry of
  air bubbles.
• Fix the tube with jar which acts as water reservoir.
• Keep the apparatus in sunlight count the bubbles when they are in same size.

Question 16.
Differentiate photosynthesis in plants and Bacterial photosynthesis.
Answer:

Photosynthesis	Bacterial photosynthesis
1. Cyclic and Non – Cyclic phosphorylation takes place	Only cyclic phosphorylation takes place
2. Photosystem I and II involved	Photosystem I only involved
3.  Electron donor is water	Electron donor is H9S
4.  Oxygen is evolved	Oxygen is not evolved
5. Reaction centres are P700 and P680	Reaction centre is P890
6.  Reducing agent is NADPH + H+	Reducing agent is NADH + H+
7. PAR is 400 to 700 nm	PAR is above 700nm
8.  Chlorophyll, Carotenoid and Xanthophyll	Bacterio chlorophyll and Bacterio viridin
9. Photosynthetic apparatus – chloroplast	It is chromosomes and Chromatophores