NCERT Exemplar Class 11 Biology Solutions Chapter 15 Plant Growth and Development

NCERT Exemplar Class 11 Biology Solutions Chapter 15 Plant Growth and Development

Edited By Priyanka kumari | Updated on Aug 16, 2022 06:04 PM IST

NCERT Exemplar Class 11 Biology solutions chapter 15 is an important chapter important from an examination point of view. Professionals possessing immense knowledge in this field have prepared NCERT Exemplar Class 11 Biology chapter 15 solutions to provide accurate and precise information. While answering the questions, the experts have used simple language to prepare Class 11 Biology NCERT Exemplar solutions chapter 15.
Also, check NCERT Class 11 Biology Solutions

NCERT Exemplar Class 11 Biology Chapter 15 Solutions:

Multiple Choice Questions:

Question:1

Ethylene is used for

(a) Retarding ripening of tomatoes
(b) Hastening of ripening of fruits
(c) Slowing down the ripening of apples
(d) Both b and c

Answer:
The answer is option (b) Hastening the ripening of fruits Ethylene is a naturally composed hormone within fruits and monitors the ripening of the fruit by inducing and regulating many chemical and biochemical activities.

Question:2

Coconut water contains
(a) ABA
(b) Auxin
(c) Cytokinin
(d) Gibberellin

Answer:
The answer is the option (c) Cytokinin
Coconut water hosts an abundance of cytokinin, which helps regulate cell growth and cell division.

Question:3

The effect of apical dominance can be overcome by which of the following hormone:
(a) IAA
(b) Ethylene
(c) Gibberellin
(d) Cytokinin

Answer:

The answer is the option (d) Cytokinin
During apical dominance, cytokinin antagonises the auxin and triggers outgrowth.The presence of cytokinin can induce the movement of nutrients towards the site where it is present. In lateral buds, the cytokinin present encourages growth despite the lateral buds having apical buds.

Therefore, cytokinin is antagonistic to the auxin, and thus apical dominance can be overcome through the use of
cytokinin.

Question:4

Match the Following

Column IColumn II
A. IAAi) Herring sperm DNA
B. ABAii) Bolting
C. Ethyleneiii)Stomatal closure
D.GAiv) Weed-free lawns
E. Cytokinesv) Ripening of fruits


Options:
(a) A- (iv), B - (iii), C -(v), D - (ii), E- (i)
(b) A- (v), B - (iii), C -(iv), D - (ii), E- (i)
(c) A- (iv), B - (i), C -(v), D - (iii), E- (ii)
(d) A- (v), B - (iii), C -(ii), D - (i), E- (iv)

Answer:

The answer is the option (a) A- (iv), B - (iii), C -(v), D - (ii), E- (i)

Question:5

Apples are generally wrapped in waxed paper to

(a) Prevent sunlight from changing its colour
(b) Prevent aerobic respiration by checking the entry of O2
(c) Prevent ethylene formation due to injury
(d) Make the apples look attractive

Answer:
The answer is the option (b) Prevent aerobic respiration by checking the entry of O2 .
To avoid the ripening of the apple, in order for it to stay fresh for a much longer time,aerobic respiration is prevented because it makes the apple rot much faster.

Question:6

Growth can be measured in various ways. Which of these can be used as parameters to measure growth?

(a) Increase in cell number
(b) Increase in cell size
(c) Increase in length and weight
(d) All the above

Answer:

The answer is the option (d) All the above
The growth can be in lateral or apex regions and can cause an enlargement in height or weight; all of these changes are collectively termed as growth.

Question:7

The term synergistic action of hormones refers to

(a) When two hormones act together but bring about the opposite effects
(b) When two hormones act together and contribute to the same function
(c) When one hormone affects more than one function
(d) When many hormones bong about anyone function

Answer:

The answer is the option (b) When two hormones act together and contribute to the same function.

When two or more hormones function mutually to deliver a certain effect, their outcomes are called synergistic. These results can be additive or complementary.

Question:8

Plasticity in plant growth means that
(a) Plant roots are extensible
(b) Plant development is dependent on the environment
(c) Stems can extend
(d) None of the above

Answer:
The answer is the option (b) Plant development is dependent on the environment.
Plants take distinct paths in response to environmental changes and different stages of life to develop various types of compositions. This entire ability is called plasticity.

Question:9

To increase sugar production in sugarcanes, they are sprayed with
(a) IAA
(b) Cytokinin
(c) Gibberellin
(d) Ethylene

Answer:

(c) Gibberellin
Explanation: Gibberellin increases the length of stem and thus increases the amount of sugar in sugarcane.

Question:10

ABA acts antagonistically to

(a) Ethylene
(b) Cytokinin
(c) Gibberellic acid
(d) IAA

Answer:

The answer is the option (c) Gibberellic acid
ABA's function is to inhibit the growth and to be antagonistic to it; the correct option should encourage and promote growth and gibberellin does, so it is the correct option.

Question:11

Monocarpic plants are those which

(a) Bear flowers with One ovary
(b) Flower once and die
(c) Bear only one flower
(d) All of the above

Answer:

The answer is the option (b) Flower once and die.
Monocarpic plants are known to flower, set seeds and then die consequentially.

Question:12

The photoperiod in plants is perceived at

(a) Meristem
(b) Flower
(c) Floral buds
(d) Leaves

Answer:

The answer is the option (d)Leaves
Leaves are the primary producers and receive light energy and thus are key players in the photoperiod.

Question:3

A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.

Answer:
Absolute Growth rate can be calculated as follows:
W_1 = W_0 e^{rt}
Where W1 = final Size (weight, height, number etc.)
W0 = initial Size at the beginning of the period
r = growth rate
t= time of growth
e = base of natural logarithms
Here, r is the relative growth rate
Given W1 = 19 cm
W0 = 5 cm
t =7 days
or,
Using W1 = W0 ert
We \: \: get\: \: 19 = 5 \times (2.7)^{7� r}
Then
Then 3.8 = (2.7) ^{7� r}
Then taking log on both the sides,
\log 3.8 = 7r \times \log 2.7
Then, 0.5798 = 7r \times 0.4314

r=\frac{0.5798}{7\times 0.4314}
Or r = 0.1907
Relative growth rate can be calculated as follows:
r=\frac{final \: size}{initial \: size}
r=\frac{19}{5}
r = 3.8 cm
Absolute growth rate= 0.1907
Relative growth rate 3.8 cm

Question:4

Gibberellins were first discovered in Japan when rice plants were suffering from bakane (the foolish seedling disease) caused by a fungus Gibberena fujikuroi.

(a) Give two functions of this phytohormone.

(b) Which property of gibberellin caused foolish seedling disease in rice?

Answer:

(a) Two functions of the Gibberellin phytohormone is that encourages the sprouting of the seed by disruption of seed dormancy. In rosette plants, gibberellin induces the development of internodes; it encourages the process of bolting.

(b) The surplus quantity of gibberellin produced by the fungus causes foolish seedling disease in rice. The surplus of gibberellin thus induces the plant's development and hasten the growth in internodes. This thus results in the plant growing too tall and being
unable to bear its own weight.

Question:5

Gibberellins promote the formation of __________ flowers on genetically _____ plants in Cannabis, whereas ethylene promotes the formation of _________ flowers on genetically _______ plants.

Answer:
male, female, female, male

Gibberellins promote the formation of male flowers on genetically female plants in Cannabis, whereas ethylene promotes the formation of female flowers on genetically male plants.


Question:6

Classify the following plants into Long-Day plants (LDP), Short Day plants (SDP) and Day Neutral Plants (DNP): Xanthium. Henbane (Hyoscyamus niger), Spinach, Rice, Bryophyllum, Sunflower, Tomato, Maize, Strawberry

Answer:

Long-Day plants (LDP)Short Day plants (SDP)Day Neutral Plants (DNP)
BryophyllumXanthiumSunflower
Henbane (Hyoscyamus niger)RiceTomato
SpinachStrawberryMaize


Question:7

A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth regulator can be applied to achieve this?

Answer:

Ethylene
Ethylene encourages the development and increases the number of female flowers in monoecious species as it is known to be a growth regulator that has a feminising effect.

Question:8

Where are the following hormones synthesised in plants
(a) IAA
(b) Gibberellins
(c) Cytokinins

Answer:
(a) IAA is synthesised mainly in the shoot apices, primarily in the growing apices of plant. This includes the leaf primordial and developing seeds.

(b) Gibberellin hormone is synthesised in plants primarily in apical shoot buds,root tips as well as developing seeds.

(c) Cytokinin is synthesised in plants primarily in roots, endosperm region in the seeds as well as developing shoot buds and growing embryos.

Question:9

In botanical gardens and tea gardens, gardeners trim the plants regularly so that they remain bushy. Does this practice have any scientific explanation?

Answer:
Regular trimming makes the plants bushy as this growth is primarily based on the apical dominance of the plants. The apical dominance thus does not allow and in fact, prevents the growth of nearby lateral buds. This happens through the release of certain
auxins. If the shoot apex or the apical bud is removed, then the lateral buds can grow and sprout. This can lead to a bushy growth.

Question:10

Light plays an important role in the life of all organisms. Name any three physiological processes in plants which are affected by light.

Answer:

The physiological processes in plants which are affected by light are: -

  • Photosynthesis,
    This is the process through which the plant manufactures its own food in the presence of sunlight by utilising inorganic substances.
  • Transpiration
    This is the process by which plants experience water loss in the form of vapour or dew formed on the leaves, primarily through the stomata. This process required sunlight as the stomata are active and functioning only in the presence of sunlight.
  • Photoperiodism
    This process typically decides the flowering of the plant, and the number of light hours affects this growth.


Question:11

In the figure of Sigmoid growth curve given below, label segments 1, 2 and 3.

Answer:

  1. Lag phase.
  2. Exponential phase and
  3. Stationary phase


Question:12

Growth is one of the characteristics of all living organisms? Do unicellular organisms also grow? If so, what are the parameters?

Answer:
Growth is a process which marks the distinction between living and non-living beings. Unicellular organisms do not experience technical growth in length or volume, unlike their multicellular counterparts. In unicellular organisms, growth is synonymous and characterised by reproduction. The process of cell synthesis divides the cell into two daughter cells.


Question:13

The rice seedlings infected with fungus Gibberella fujikuroi is called foolish seedlings? What was the reason behind it?

Answer:

The rice seedlings infected with fungus Gibberella fujikuroi are called foolish seedlings because the rice plants which get infected with the fungus grow excessively in length in comparison to the plant species that are not infected. The height of the plant becomes disproportionate and is too much for the plant to bear, and it cannot support the weight of the plant, which leads to the plant's death ultimately. Hence, the plant causes its own death due to the infection by the fungus and is called foolish seedlings.

Long Answer Type Questions:

Question:1

Some varieties of wheat are known as spring Wheat While Others are called winter wheat. The former variety is sown and planted in spring, and it is harvested by the end of the same season. However, winter varieties; if planted in spring, fail to flower, or produce mature grains within a span of a flowering season. Explain, why?

Answer:

Some plants require special extrinsic conditions in order to commence the flowering stage, and one of the requirements can be a low temperature. Thus, the plants remain vegetative during the summer season and wait till they receive a low temperature to begin bearing flowers. To waiting for lower temperatures helps prevent precocious for early active reproductive development. This helps the plant to gain proper maturity and develop within the vegetative state before advancing to the reproductive stage. This is called the vernalisation. Some plants have separate spring and winter varieties. For example, the spring wheat is planted in the spring season and come to flower and fruit right at the end of the season, but if the winter varieties are sown in the same season, then it will not bear flowers and fruits as compared to the spring variety.

Instead, the winter is planted in the autumn, and during the winter they form the seedlings. The seedlings resume their growth in the spring season and bear flowers during the summer. The winter wheat requires specific low temperatures to begin sprouting and maturing.

Question:2

It is known that some varieties of wheat are sown in autumn but are harvested around next mid-summer.
(a) What could be the probable reason for this?
(b) What term is used for this promotion of flowering under low temperature?
(c) Which plant hormone can replace the cold treatment?

Answer:
(a) These plants usually germinate and grow as seedlings throughout winter. They only resume growth when it is summer and is finally harvested around the upcoming midsummer. This is because these plants require low temperatures to bear flowers and
fruits and are able to finish their vegetative phase of growth over the period of winter and summer.

(b) This process mentioned here is called vernalisation.
(c) Gibberellin hormone can replace cold treatment effectively.

Question:3

Name a hormone which
(a) is gaseous in nature
(b) is responsible for phototropism
(c) induces femaleness in flowers of cucumber
(d) is used for killing weeds (dicots)
(e) induces flowering in long-day plants

Answer:
(a) Ethylene hormone is gaseous in nature.
(b) Auxin is responsible for phototropism.
(c) Ethylene hormone induces female characters in flowers of cucumbers.
(d) Synthetic auxin (2,4-D) is used to kill weeds
(e) Gibberellin hormone induces flowering in long-day plants

Short Answer Type Questions:

Question:1

Nicotiana tabacum, a Short-Day plant, when exposed to more than critical of light fails to flower. Explain.

Answer:

Nicotiana tabacum is a short-day plant, and these plants require much less exposure to light than the required amount in normal plants. They demand a continuous critical dark period, which if interrupted, can affect the flowering of the plant. These plants require a photoperiod less than the general required one and played a key role in flowering. This is due to the fact that flowering is regulated by a hormone which is present in the leaves and requires the dark photoperiod to migrate to the required places in the shoot apices.

Question:2

What are the structural characteristics of

(a) Meristematic cells near the root tip

(b) The cells in the elongation zone of the root

Answer:
(a) Meristematic cells are located in the zone, which is the growing point of the root, and in an active state of cell division. The cells are closely arranged and thin- walled and have dense protoplasm. They are essential for the growth of roots.

(b) In this zone, the cells are newly formed and lose the ability of division, which is why they elongate and rapidly increase the length of the root. They have a large number of small vacuoles. These vacuoles fuse to form one large central vacuole. This is what
causes the cells to elongate and longitudinally grow in this region.

Question:3

Does the growth pattern in plants differ from that in animals? Do all the parts of plant grow indefinitely? If not, name the regions of the plant, which can grow indefinitely.

Answer:

Animals and plants have different growth patterns and habits. Animals experience growth of definite, determinate form, which means they experience limited growth or stop growth after attaining maturity. Plants, on the other hand, can experience growth of definite or indefinite and indeterminate form. They can stop growing after attaining maturity of developing leaves and flowers or parts like the stem, and their branches can keep continuing to grow. Plants experience continuous growth since they contain meristematic areas with potential for cell-division, whereas animals have cell division only to replace and repair and not for the purpose of growth.

Question:4

Explain in 2-3 lines each of the following terms with the help of examples taken from different plant tissues:
(a) Differentiation
(b) De-differentiation
(c) Redifferentiation

Answer:

(a) To facilitate growth, the meristematic cells through mitotic division, divide into daughter cells. The cells ranging from the root to the shoot apical meristem separate into different categories as they have matured and can perform specific functions. This phenomenon is called differentiation. During this phase, the cells experience critical structural modifications in protoplasm and cell wall.

(b) Under certain conditions, the differentiated living cells can divide mitotically after recovering the ability to divide. This phenomenon is called de-differentiation. Sometimes, the de-differentiated cell may become the meristem; for example, fully differentiated parenchyma cells form the meristems and interfascicular cambium and cork cambium.

(c) The cells formed by de-differentiated cells can lose their capacity to reproduce and thus mature to perform specific functions. This is known as redifferentiation. For example, the secondary cortex and cork are prime examples as well as the cambium which de-differentiates to form the cortex.

Question:5

Auxins are growth hormones capable of promoting cell elongation. They have been in horticulture to promote growth, flowering, and rooting. Write a line to explain the meaning of the following terms related to auxins.
(a) Auxin precursors
(b) Anti-auxins
(c) Synthetic auxins

Answer:
(a) Auxin precursors are the raw substances utilised in the process of initial making and synthesis of auxins. For example- IAA's auxin precursor is tryptophan.
(b) Anti-auxins are compound substances which restrict the action of auxins. For example, TIBA and p-chlorophenoxy isobutyric acid (PCIB).
(c) Synthetic auxins are auxins which are artificially synthesised and do not occur naturally in plants. For example, 2:4-D, NAA.

Question:6

The role of ethylene and abscisic acid is both positive and negative. Justify this statement.

Answer:

Both ethylene and abscisic acid have positive and negative effects.
Positive Effects of Ethylene:

  • Ethylene acts as a hormone which induces apical dominance and encourages horizontal growth
  • It interrupts the dormancy of seeds and buds
  • Ethylene aids in the initiation of the root
  • Ethylene also helps in the ripening of certain fruits

Negative Effects of Ethylene:

  • Ethylene causes abscission of leaves because it hastens the senescence of leaves
  • Ethylene delays longitudinal development in the plant
  • Ethylene extends the dormancy of lateral buds

Positive Effects of Abscisic Acid:

  • Helps in reduce water loss in the form of transpiration
  • Induces flowering in short-day plants
  • In certain plants such as Rosen, abscisic acid encourages parthenocarpic development as well as the rooting of stem-cutting
  • Helps plants form resistance to environmental stress
  • Promotes root initiation in certain plants

Negative Effects of Abscisic Acid:

  • Encourages the abscission of flowers and fruits and induces leaf senescence.
  • Inhibits and slows down the cambial activity
  • Retards the effect and functioning of growth hormones.
  • Promotes dormancy in seeds and

Question:7

While experimentation, why do you think it is difficult to assign any effect seen to any single hormone?

Answer:

Plant hormones mostly work together, complementary to each other. Plant hormones can have both synergistic and antagonistic interactions which affect the plant tissues. A single hormone can have multiple effects, while multiple hormones can target and aim to achieve just one effect. Therefore it can be exceedingly difficult to identify one single hormone according to the particular effect in plant metabolism and system. Moreover, plant hormones function in the presence of a lot of extrinsic factors which are needed for the reaction. Thus, it is difficult to assign any effect seen to any single hormone.

Question:8

What is the mechanism underlying the phenomenon by which the terminal/apical bud suppresses the growth of lateral buds? Suggest measures to overcome this phenomenon.

Answer:
This phenomenon by which the terminal/apical bud suppresses the growth of the lateral buds is called apical dominance. On removal of the apical bud, the lateral buds are allowed to sprout, which produces bushy growth. The apical bud restricts the growth of the lateral buds through the release of certain auxins. Thus the auxin causing apical dominance can be countered with cytokinin. This attracts nutrients towards and helps the lateral buds to grow by acting as an antagonistic.

Question:9

In animals, there are special glands secreting hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?

Answer:
Plant hormones are produced by individual cells in the regions where they are essential by the plants. To take an example, auxin is synthesised in the cells where there are growing apices. The vascular system of plants is where plant hormones are translocated through. Other examples include the Abscisic acid produced in many parts of plants, which is transported through diffusion and transport channel.

Question:10

Many discoveries in science have been accidental. This is true for plant hormones also. Can you justify this statement by giving an example? Also, what term is used for such accidental findings?

Answer:

Various plant hormones and other scientific research have been led by accidental means. Gibberellin was discovered accidentally, and Charles Darwin and his son Francis Darwin accidentally came across the auxin. They noticed that coleoptiles of canary grass reacted to unilateral illumination through growth in the direction of light. After some time, F W Went separated auxin from coleoptiles of oat seedlings.

The term used for accidental discovery is 'serendipity'

Question:11

To get carpet-like grass lawns are mowed regularly. Is there any scientific for this?

Answer:

Ans. Mowing regularly means the trimming of the apical parts of the plant, which thus causes the sprouting and faster growth of lateral branches. Thus apical dominance is prevented and restricts longitudinal growth of the plant. Since the apices are removed, the auxins released due to the apical buds to inhibit the lateral buds to get mobilised and encourage the growth of lateral branches. This helps in maintaining carpet-like lawns.

Question:12

In a slide showing different types of cells can you identify which type of the cell may be meristematic and the one which is incapable of dividing and how?

Answer:

The presence of the following characteristics would indicate meristematic cells in the slides showing different types of cells:

(a) Thin cellulose wall
(b) Dense cytoplasm with a large nucleus
(c) Cells displaying cell division, like mitosis and other various stages.
(d) Smaller sized vacuoles or even negligible vacuoles.

If the above features are absent, then it can be determined that the cells are incapable of division.

Question:13

A rubber band stretches and reverts to its original position. Bubble gum stretches, but it would not return to its original position.

Is there any difference between the two processes? Discuss it with respect to plant growth.

[Hint: Elasticity (reversible) plasticity (irreversible)]

Answer:

Referring to an experiment with oat coleoptile, the auxin hormone induces the shoot tip tending in the direction of the light source. This process is irreversible and is known as plasticity. It is the plant's ability to adjust to the surrounding environmental
conditions. While, if weight is attached along to the coleoptile segment, the weight causes bending and thus stretches the plant. If the weight is removed, the shoot of the plant returns to the original position meaning that this change is reversible. This is known as elasticity.

Question:14

Label the diagram

(a) This is which part of a dicotyledonous plant?

(b) If we remove part I from the plant, what will happen?

Answer:

(a) This diagram displays the shoot apex of the dicotyledonous plant.


(b) Removal of the shoot apex will prevent apical dominance. This will, in turn, help in the growth of lateral branches.


Question:15

Both animals and plants grow. Why do we say that growth and differentiation in plants are open and not so in animals? Does this statement hold true for sponges also?

Answer:
The growth of a plant can be said to be indeterminate. This is because a part of a plant can grow infinitely, i.e. always growing and forming new organs to replace the older ones. This is not the case for animals. Animal growth is limited and stops after a certain age when they get mature. This is why one can say that the growth and differentiation in plants and animals are different.

The sponges can be said to be an exception in the animal kingdom. This is because they are the animals which show a cellular level of organisation. They depict the characteristics of open-ended growth but cannot be called one since they cannot grow beyond a certain size.

Question:16

Define parthenocarpy. Name the plant hormone used to induce parthenocarpy.

Answer:

Parthenocarpy is the process of formation and development of fruits that occurs without the use of any kind of fertilisation. The fruit produced by this process does not contain seeds. Fruits such as pear, banana, pineapple, and some varieties of papaya are parthenocarpic fruits. In order to induce parthenocarpy, auxins, cytokinin and gibberellins are utilised.

Question:17

While eating watermelons, all of us wish it were seedless. As a plant physiologist can you suggest any method by which this can be achieved?

Answer:

To have seedless watermelons, there needs to be a cross hybridisation that occurs between a tetraploid female parent and a diploid male. This results in a triploid plant which has three sets of chromosomes. This particular genome thus produces the watermelon to not contain seeds.

Question:18

A gardener finds more broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently?

Answer:
The weed can be effectively eliminated by using weedicides. One can use the 2:4- D and 2:4: 5-T which do not affect mature monocotyledons. This is a synthetic auxin and is widely used by farmers but is to be used in caution and restraint so as to not hamper the crops.

Question:19

On germination a seed first produces shoots with leaves, flowers appear later,
(a) Why do you think this happens?
(b) How is this advantageous to the plant?

Answer:

(a) On germination of the seed, shoots with leaves are produced at first as the plant is gradually entering the vegetative growth phase of the plant. The leaves begin to manufacture food, and the photoperiod follows to determine the flowering of the plant.
The leaves are produced to regulate the process of photosynthesis and produce food, which could be utilised in sexual reproduction. This marks the beginning of the important flowering process as well.

(b) The vegetative phase marks immense growth and aids the plant to gain maturity.Then the process of reproduction commences, and flowering also occurs. The vegetative phase is necessary to prepare the plant for the upcoming process and for it to develop the necessary structures for it. Thug, the vegetative phase of growth is necessary for the survival of the plant species. The flowering phase depends on this phase to provide for nourishment.

More About NCERT Exemplar Class 11 Biology Solutions Chapter 15:

Plants are a very important part of the lives of both humans and animals. Plants absorb carbon dioxide and release oxygen which is important for us and animals to survive. Therefore when we’re considering ourselves in an environment, we can’t think without plants being involved.

Growth is an irreversible change in every living organism and is important for the environment as well. Plants have certain chemical procedures that go on in them which facilitate their growth and development along with providing us with food and oxygen.

Students can utilise the NCERT Exemplar Class 11 Biology solutions chapter 15 PDF Download function to make learning even more convenient. Students get access to quality study material effectively construed by experts for the best learning experience.

Major Subtopics in NCERT Exemplar Class 11 Biology Solutions Chapter 15 Growth and Development

  • 15.1 Growth
  • 15.1.1 Plant growth generally is indeterminate
  • 15.1.2 Growth is measurable
  • 15.1.3 Phases of growth
  • 15.1.4 Growth Rates
  • 15.1.5 Conditions for growth
  • 15.2 Differentiation, Dedifferentiation and Redifferentiation
  • 15.3 Development
  • 15.4Plant Growth Regulators
  • 15.4.1 Characteristics
  • 15.4.2 The discovery of plant growth regulators
  • 15.4.3 Physiological Effects of plant growth regulators
  • 15.4.3.1 Auxins
  • 15.4.3.2 Gibberellins
  • 15.4.3.3 Cytokinins
  • 15.4.3.4 Ethylenes
  • 15.4.3.5 Abscisic Acid
  • 15.5 Photoperiodism
  • 15.6 Vernalisation

What will the students Learn using the NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Growth and Development?

In Class 11 Biology NCERT Exemplar solutions chapter 15 the students will learn about the different topics that one needs to know when it comes to the growth and development of plants. Its growth and development are affected by light, temperature, nutrition, oxygen status, gravity and such external factors along with the internal factors called the plant growth regulators [PGR]. Every topic is given in NCERT Exemplar Class 11 Biology solutions chapter 15 is fairly detailed and accurately explained.

Everything we eat is directly or indirectly obtained from plants so it is important to have good knowledge about them. To understand the developments in our bodies we can’t forget about plants as we take nutrients from it as well as mentioned before. Overall, NCERT Exemplar Solutions for Class 11 Biology chapter 15 along with being informative is also fun and interesting and for all those nature lovers out there who love being around plants and in future want to work on something related to plants, this is for you.

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Important Topics in NCERT Exemplar Class 11 Biology Chapter 15 Solutions

· The growth and development of plants due to the effect of light, temperature, nutrition, oxygen status, gravity and such external factors along with the internal factors referred to as the plant growth regulators [PGR] are detailed in the NCERT Exemplar Class 11 Biology solutions chapter 15. Photosynthesis, capillary effect and external temperatures are also important.

· NCERT Exemplar Class 11 Biology chapter 15 solutions talks about the factors that help and control the development of these plants. These factors can be intrinsic (internal) and extrinsic (external) to the plants.

· It also provides information about the phases of its growth. It also contains relevant and fairly detailed diagrams for a better and clearer understanding of the students.

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Students will learn about plants growth, differentiation, dedifferentiation, growth regulators, vernalisation and photoperiodism.

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Application Date:07 October,2024 - 22 November,2024

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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