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NCERT Exemplar Class 12 Biology Solutions Chapter 2 Sexual Reproduction in Flowering Plants

NCERT Exemplar Class 12 Biology Solutions Chapter 2 Sexual Reproduction in Flowering Plants

Edited By Priyanka kumari | Updated on Aug 25, 2022 05:58 PM IST | #CBSE Class 12th

Studying the chapter Sexual Reproduction in Flowering Plants from NCERT exemplar Class 12 Biology solutions chapter 2 will help students get a better understanding of the process of sexual reproduction in flowering plants that produce both male and female gametes. Moreover, students will be able to understand the concepts of microsporogenesis and megasporogenesis, the differences between them, and their different stages of development. Class 12 NCERT exemplar Class 12 Biology solutions chapter 2 are highly beneficial for the students of Class 12 to know the right approach to deal with the questions. NCERT Exemplar Class 12 Biology solutions chapter 2 pdf download will also be made available soon for the convenience of students.
Also read - NCERT Solutions for Class 12 Biology

Multiple Choice Questions:

Question:1

Among the terms listed below, those that of are not technically correct names for a floral whorl are:
i. Androecium
ii. Carpel
iii. Corolla
iv. Sepal
(a) i and iv, (b) iii and iv (c) ii and iv (d) i and ii.

Answer:

The answer is the option (c), (ii) and (iv)
Carpel and sepal are individual parts and make gynoecium and calyx respectively.

Question:2

The embryo sac is to ovule as _______ is to an anther.
a. Stamen
b. Filament
c. Pollen grain
d. Androecium

Answer:

The answer is the option (c) Pollen grains
The embryo sac is present in the ovule, while pollen grains are present in anther.

Question:3

In a typically complete, bisexual and hypogynous flower, the arrangement of floral whorls on the thalamus from the outermost to the innermost is:
a. Calyx, corolla, androecium and gynoecium
b. Calyx, corolla, gynoecium and androecium
c. Gynoecium, androecium, corolla and calyx
d. Androecium, gynoecium, corolla and calyx

Answer:

Ans. The answer is the option (a) Calyx, corolla, androecium and gynoecium

Question:4

A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is:
a. The plant is dioecious and bears only pistillate flowers
b. The plant is dioecious and bears both pistillate and staminate flowers
c. The plant is monoecious
d. The plant is dioecious and bears only staminate flowers.

Answer:

The answer is the option (d) Plant is dioecious and bears only staminate flowers.
Option (a) is incorrect since a pistillate flower can be pollinated from pollen grains from another flower. A dioecious flower can show self or cross-pollination and hence can be transformed into fruit. So, option (d) is the correct answer.

Question:5

The outermost and innermost wall layers of microsporangium in an anther are respectively:
a. Endothecium and tapetum
b. Epidermis and endodermis
c. Epidermis and middle layer
d. Epidermis and tapetum

Answer:

Ans. The answer is the option (d) Epidermis and tapetum

Question:6

During microsporogenesis, meiosis occurs in:
a. Endothecium
b. Microspore mother cells
c. Microspore tetrads
d. Pollen grains.

Answer:

The answer is the option (b) Microspore mother cells
Microspore mother cell undergoes meiosis to produce haploid pollen grains.

Question:7

From among the sets of terms given below, identify those that are associated with the gynoecium.
a. Stigma, ovule, embryo sac, placenta
b. Thalamus, pistil, style, ovule
c. Ovule, ovary, embryo sac, tapetum
d. Ovule, stamen, ovary, embryo sac

Answer:

The answer is the option (a) Stigma, ovule, embryo sac, placenta
Thalamus, tapetum and stamen are not associated with gynoecium.

Question:8

Starting from the innermost part, the correct sequence of parts in an ovule is,
a. egg, nucellus, embryo sac, integument
b. egg, embryo sac, nucellus, integument
c. embryo sac, nucellus, integument, egg
d. egg, integument, embryo sac, nucellus.

Answer:

The answer is the option (b) egg, embryo sac, nucellus, integument

Question:9

From the statements given below, choose the option that is true for a typical female gametophyte of a flowering plant:
i. It is 8-nucleate and 7-celled at maturity
ii. It is free-nuclear during the development of
iii. It is situated inside the integument but outside the nucellus
iv. It has an egg apparatus situated at the chalazal end
(a) i and iv, (b) ii and iii (c) i and ii (d) ii and iv

Answer:

Ans. The answer is the option (c) (i) & (ii)

Question:10

Autogamy can occur in a chasmogamous flower if:
a. Pollen matures before the maturity of ovule
b. Ovules mature before the maturity of pollen
c. Both pollen and ovules mature simultaneously
d. Both anther and stigma are of equal lengths.

Answer:

The answer is the option (c) Both pollen and ovules mature simultaneously
Relative lengths of stigma and anther are not the only factors; time of maturity of pollens and ovules is also important in deciding the type of pollination. If pollens mature before ovules; they will become ineffective by the time ovule matures. Hence, the option (c) is the correct answer.

Question:11

Choose the correct statement from the following:
a. Cleistogamous flowers always exhibit autogamy
b. Chasmogamous flowers always exhibit geitonogamy
c. Cleistogamous flowers exhibit both autogamy and geitonogamy
d. Chasmogamous flowers never exhibit autogamy

Answer:

The answer is the option (a) Cleistogamous flowers always exhibit autogamy
Cleistogamous flowers do not open at all, and thus entry of pollens from another flower is not possible. Hence, cleistogamous flowers exhibit autogamy.

Question:12

A particular species of the plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by:
a. Insects
b. Water
c. Wind
d. Animals

Answer:

The answer is the option (c) Wind
Light and not-sticky pollens are ideal to be blown away by the wind. Feathery stamens are able to sway with the wind, which helps in the release of pollens into the air.

Question:13

From among the situations given below, choose the one that prevents both autogamy and geitonogamy.
a. A monoecious plant bearing unisexual flowers
b. A dioecious plant bearing only male or female flowers
c. Monoecious plant with bisexual flowers
d. Dioecious plant with bisexual flowers

Answer:

The answer is the option (a) Monoecious plant bearing unisexual flowers
Autogamy can happen in case of bisexual flowers. Geitonogamy can happen in case of dioecious plants bearing only male or female flowers. Hence, option (a) is correct.

Question:14

In a fertilised embryo sac, the haploid, diploid and triploid structures are:
a. Synergid, zygote and primary endosperm nucleus
b. Synergid, antipodal and polar nuclei
c. Antipodal, synergid and primary endosperm nucleus
d. Synergid, polar nuclei and zygote.

Answer:

The answer is the option (a) Synergid, zygote and primary endosperm nucleus

Question:15

In an embryo sac, the cells that degenerate after fertilisation are:
a. Synergids and primary endosperm cell
b. Synergids and antipodals
c. Antipodals and primary endosperm cell
d. Egg and antipodals.

Answer:

The answer is the option (b) Synergids and antipodals
PEN provides food for the growing embryo, while egg develops into the embryo. Hence, option (b) is the correct answer.

Question:16

While planning for an artificial hybridisation programme involving dioecious plants, which of the following steps would not be relevant:
a. Bagging of female flower
b. A dusting of pollen on stigma
c. Emasculation
d. Collection of pollen

Answer:

The answer is the option (c) Emasculation
In the case of a dioecious plant, male and female flowers are not usually on a single plant. Hence, emasculation may not be necessary in certain cases.


Question:17

In the embryos of a typical dicot and a grass, true homologous structures are:
a. Coleorhiza and coleoptile
b. Coleoptile and scutellum
c. Cotyledons and scutellum
d. Hypocotyl and radicle

Answer:

The answer is the option (c) Cotyledons and scutellum
In monocot, the cotyledon is called scutellum

Question:18

The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilisation is called:

a. Parthenocarpy
b. Apomixis
c. Vegetative propagation
d. Sexual reproduction

Answer:

The answer is the option (b) Apomixis
When seeds are produced without fertilisation, this phenomenon is called apomixes. In parthenocarpy, seeds are not produced, which means embryos are not produced. Hence, option (b) is the correct answer

Question:19

In flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be:
a. Haploid
b. Diploid
c. A few haploid and a few diploid
d. With varying ploidy.

Answer:

The answer is the option (b) Diploid
Since no meiosis takes place, so no change in ploidy will take place.

Question:20

The phenomenon wherein, the ovary develops into a fruit without fertilisation is called:
a. Parthenocarpy
b. Apomixis
c. Asexual reproduction
d. Sexual reproduction

Answer:

The answer is the option (a) Parthenocarpy

Very Short Answer Type Questions:

Question:1

Name the component cells of the ‘egg apparatus’ in an embryo sac.

Answer:

The egg apparatus consists of two synergids and one egg cell.

Question:2

Name the part of gynoecium that determines the compatible nature of pollen grain.

Answer:

The pistil has the ability to recognize compatible pollen grains.

Question:3

Name the common function that cotyledons and nucellus perform.

Answer:

Cotyledons and nucellus contain abundant food that is utilized by the developing cells.

Question:4

Complete the following flow chart
Pollen mother cell → Pollen tetrad → Pollen grain (a) = vegetative cell (b) = ?

Answer:

Generative cell

Question:5

Indicate the stages where meiosis and mitosis occur (1, 2 or 3) in the flow chart.
Megaspore mother cell →Megaspores → Embryo sac →Egg

Answer:

1. Meiosis 2. Mitosis 3. Mitosis

Question:6

In the diagram given below, show the path of a pollen tube from the pollen on the stigma into the embryo sac. Name the components of the egg apparatus.

Answer:

Longitudinal section of a flower showing growth of pollen tube.
Synergids and egg cells are the components of the egg apparatus.



Question:7

Name the parts of the pistil which develop into fruit and seeds.

Answer:

The ovary develops into fruits, and ovules develop into seeds.

Question:8

In the case of polyembryony, if an embryo develops from the synergid and another from the nucellus, which is haploid, and which is diploid?

Answer:

The embryo developing from synergid will be haploid, and the embryo developing from nucellus will be diploid.

Question:9

Can an unfertilized, apomictic embryo sac give rise to a diploid embryo? If yes, then how?

Answer:

We know that when seeds develop without fertilization, it is known as apomixes. In this, the embryo can develop from nucellus, which is diploid. Thus, an apomictic embryo sac can give rise to a diploid embryo.

Question:10

Which are the three cells found in a pollen grain when it is shed at the three celled stages?Answer:

When a pollen grain is shed at the three celled stages, it contains 2 male gametes and a vegetative cell. The generative cell divides into two male gametes.

Question:11

What is self-incompatibility?

Answer:

When pollen from the same plant is incompatible for fertilization, this condition is called self-incompatibility.


Question:12

Name the type of pollination in self-incompatible plants.

Answer:

In the case of self-incompatible plants, cross-pollination takes place.

Question:14

Which is the triploid tissue in a fertilized ovule? How is the triploid condition achieved?

Answer:

PEN or Primary Endosperm Nucleus shows triploid conditions. When one of the male gametes fuses with the polar nuclei, it results in the formation of triploid PEN.

Question:15

Are pollination and fertilization necessary in apomixes? Give reasons.

Answer:

Pollination and fertilization are not necessary for apomixis. Apomixis is a condition in which an embryo develops without fertilization.

Question:16

Identify the type of carpel with the help of diagrams given below:

Answer:

Figure “a” shows the multicarpellary syncarpous condition, and figure “b” shows the multicarpellary apocarpous condition.


Question:17

How is pollination carried out in water plants?

Answer:

Water mediated pollination takes place in a selected number of plants. For example, in Vallisneria, the female flower reaches the surface of the water. Pollen grains are sprinkled on the water surface, and they are passively transported to the female flower for pollination.
Another example is that of seagrasses, the female flower remains submerged in water, and pollen grains are released below the water surface. In the case of pollination by water, pollen grains have a mucilaginous covering which prevents the pollens from becoming wet.

Question:18

What is the function of the two male gametes produced by each pollen grain in angiosperms?

Answer:

The male gametes fuse with the female gametes and form the embryo. The embryo subsequently develops into a new plant. Another male gamete fuses with polar nuclei and eventually forms endosperm. The endosperm supplies food to the developing embryo.

Short Answer Type Questions:

Question:1

List three strategies that a bisexual chasmogamous flower can evolve to prevent self-pollination (autogamy).

Answer:

Following are 3 strategies that a bisexual chasmogamous flower can evolve to prevent self-pollination or autogamy:

  1. Pollen release and stigma receptivity are not synchronised in many flowers. Either the pollen is released much before the maturity of stigma or stigma matures much before the release of pollen.

  2. In some flowers, anthers and stigma are placed at different places so that pollen grains from the same flower cannot reach the stigma.

  3. Self-incompatibility between pollen and stigma is seen in some flowers. This is a genetically mediated process that prevents autogamy in these flowers.

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Question:2

Given below are the events that are observed in an artificial hybridisation programme. Arrange them in the correct sequential order in which they are followed in the hybridisation programme.
(a) Re-bagging (b) Selection of parents (c) Bagging (d) Dusting the pollen on stigma (e) Emasculation (f) Collection of pollen from male parent.

Answer:

The correct sequence of steps being followed in hybridisation is:
Selection of parents-> Emasculation -> Bagging -> Collection of pollen from male parent -> Re-bagging.

Question:3

Vivipary automatically limits the number of offspring’s in a litter. How?

Answer:

Viviparity is present in both animals and plants. In the case of animals, it means an animal gives birth to young ones. In the case of plants, viviparity means germination of the embryo on the plant itself, without the normal sequence of development of the seed. Viviparity involves too much drain of resources on the mother.
In the case of animals, a female has to constantly supply the nutrients and oxygen to the growing foetus, if the foetus developing in the womb. Enough resources are not available to support a large litter, and hence viviparity automatically limits the number of offspring’s in a litter. This is true in the case of plants also because a germinating embryo on the plant would require resources from the mother plant.

Question:4

Does self-incompatibility impose any restrictions on autogamy? Give reasons and suggest the method of pollination in such plants.

Answer:

Self- incompatibility is the condition in which pollen from the same plant cannot pollinate the flower. Thus, self-incompatibility imposes complete restriction on autogamy.
This evolution might have occurred in order to prevent too much inbreeding because continuous inbreeding prevent variations. In such plants, cross pollination is the normn and pollen from a plant pollinates the flower on another plant. This ensures accumulation of gene pools from 2 different plants.

Question:7

Are parthenocarpy and apomixis different phenomena? Discuss their benefits.

Hint: Yes, parthenocarpy, and different apomixis phenomena. Parthenocarpy leads to the development of seedless fruits. Apomixis leads to embryo development.

Answer:

Parthenocarpy is the condition in which fruits develop without seeds, while apomixis is a condition in which seeds develop without fertilization. Fertilization is absent in both the case, but seeds are present in apomixis only. Benefits of parthenocarpy: Seedless fruits are easier to consume, especially those fruits that naturally contain too many seeds, e.g., papaya, watermelon, and banana.
Benefits of Apomixis include the fact that it can be used to produce apomicts hybrid seeds every year. This will help in drastically cutting the cost for farmers.

Question:6

What is polyembryony, and how can it be commercially exploited?

Answer:

In some varieties of citrus and mango, the nucellar cells start dividing and protrude into the embryo sac. They eventually produce multiple embryos. This condition is called polyembryony. Embryos develop without fertilization in this apomictic condition.
Polyembryony can be commercially exploited by producing seeds of hybrid varieties at a lower cost. In the case of hybrid plants, a farmer needs to buy seeds every year because plants from hybrid seeds fail to produce hybrid seeds due to the law of inheritance.
Buying fresh seeds every season is very costly. If hybrid seeds are produced with a polyembryonic condition, then it would be possible for the farmers to utilize those seeds for the next year and the subsequent years. This is still at the research stage, but there are bright prospects for the future.

Question:8

Why does the zygote begin to divide only after the division of Primary endosperm cell (PEC)?

Answer:

The endosperm plays an important role in supplying food to the developing embryo. Once the division of PEC or Primary Endosperm Cell is complete, there is sufficient availability of food for the embryo.
In the absence of food, the zygote won’t be able to get the necessary raw materials for making new cells. Hence, the division of zygote beings only after the division of Primary Endosperm Cell (PEC).

Question:9

The generative cell of two-celled pollen divides into the pollen tube but not in three-celled pollen. Give reasons.

Answer:

In three-celled pollen, one of the cells is a vegetative cell that has no role to play in fertilization. The remaining two cells are the male gametes, and they are the actual participants in fertilization. Rather than the vegetative cells, the generative cells divide inside the pollen tube.
In 60% of the cases, the generative cell divides inside the pollen tube. In the remaining case, the generative cell divides much before pollination.

Question:2

What are the possible types of pollinations in chasmogamous flowers? Give reasons.

Answer:

Chasmogamous flowers are open flowers in which stigma and anthers are exposed, and flowers are similar to other species. Possible types of pollinations in chasmogamous flowers are as follows:

  1. Geitonogamy: The situation in which pollen grains from the same plant but different flower reaches the stigma is called geitonogamy. This is similar to autogamy because the zygote gets the gene pool from the same plant.

  2. Xenogamy: The situation in which the pollen grains from a different plant reaches the stigma is called xenogamy. This can be termed as the true cross-pollination because the zygote gets the gene pool from 2 different plants.

Most of the plants produce hermaphrodite flowers, and thus, self-pollination is a clear cut eventuality. However, continuous self-pollination can result in inbreeding depression. The variation will not be possible in case of self-pollination. Hence, plants have evolved in various ways and means to facilitate cross-pollination even in dioecious flowers. One of the strategies followed by plants is a loss of synchronization between pollen release and stigma maturity. Another strategy is self-incompatibility between pollens and the stigma of the same flower.
A third strategy is a positional difference between anthers and stigma so that pollens from the same flower are unable to reach the stigma.

Question:3

With a neat, labeled diagram, describe the parts of a mature angiosperm embryo sac. Mention the role of synergids.

Answer:


Structure of a mature Embryo Sac: A mature embryo sac is a 7-celled structure and has 8 nuclei. The end near the micropyle is called the micropylar end, while the opposite end is called the chalazal end. The following are the main parts of the embryo sac:

  • Egg Apparatus: The egg apparatus is composed of two synergids and an egg. There are special thickenings at the micropylar end of synergids. These thickenings are known as the filiform apparatus.

  • The function of Synergids: The synergids provide a channel to the pollen tube to enter through the filiform apparatus.

  • Polar Nuclei: The two nuclei enclosed in the central cell are called polar nuclei.

  • Antipodals: The three cells at the chalazal end are called antipodals.

Question:4

Draw the diagram of a microsporangium and label its wall layers. Write briefly on the role of the endothelium.

Answer:


Role of Endothecium: Endothecium, along with the epidermis and the middle layer, provides protection to the pollens during development. Once pollen grains are mature, the three layers (including endothecium) rupture and thus facilitate the dehiscence of pollens.

Embryo sacs of some apomictic species appear normal but contain diploid cells. Suggest a suitable explanation for the condition.

Answer:

The condition in which seeds are produced without fertilization is called apomixis. It is a kind of asexual reproduction, but it mimics sexual reproduction. There are several mechanisms for apomixis. One of them is seen in citrus and mango fruits. In this case, the nucellus beings to divide and intrude into the embryo sac. It eventually develops into a seed. Since nucellus is composed of diploid cells, the embryo sac in such a case has diploid cells.
It is also important to recall that fertilization cannot happen in the cells which were not formed after meiosis. Moreover, haploid cells are never involved in apomixes. It is the diploid cells which bring about apomixes. Hence, embryo sacs apomictic species appear normal but contain diploid cells.

Important Topics in NCERT Exemplar Class 12 Biology Solutions Chapter 2

● Process of formation and evolution of male and female gametes in plants

● Pollination, pre-fertilization events

● Fusion of male and female gametes, and the post-fertilization events

● Apomixis and Polyembryony

Major Subtopics in NCERT Exemplar Class 12 Biology Solutions Chapter 2

  • Flower- A Fascinating Organ of Angiosperms
  • Pre-fertilization: Structures and Events
  • Stamen, Microsporangium and Pollen Grain
  • The Pistil, Megasporangium (ovule) and Embryo sac
  • Pollination-
  • Double Fertilization
  • Post-Fertilization: Structures and Events
  • Endosperm
  • Embryo
  • Seed
  • Apomixis and Polyembryony
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NCERT Exemplar Class 12 Biology Solutions Chapter 2 Learning Outcome

  • In NCERT exemplar solutions for Class 12 Biology chapter 2 students will learn about the process of sexual reproduction in flowering plants.

  • Students will also learn the process of pollination, pre-fertilization events, the fusion of male and female gametes, and the post-fertilization events and apomixis and polyembryony in detail.

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NCERT Exemplar Class 12 Biology Chapter Wise:

Important Topics in NCERT Exemplar Class 12 Biology Solutions Chapter 2

  • Topics like, how sexual production works in a flowering plant alongside developmental sequences of microsporogenesis and megasporogenesis with diagrams are explained in NCERT exemplar Class 12 Biology chapter 2 solutions.

  • Information on pollination, pre-fertilization events, a fusion of male and female gametes, and post-fertilization events are crucial from the exams point of view. It is also listed in detail in NCERT exemplar Class 12 Biology solutions chapter 2.

  • Furthermore, students will be able to comprehend between Apomixis and Polyembryony.

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NCERT Exemplar Class 12 Solutions Subject Wise:

Check NCERT solutions for questions given in the book:

Also, Read NCERT Solution subject wise

Must Read NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. How NCERT exemplar solutions chapter 2 is beneficial for students?

Students would be able to answer all major questions that were asked in previous question papers.

2. How are the NCERT exemplar solutions for Class 12 Biology chapter 2 prepared?

Our Experts use the chapters and given information alongside their practical knowledge to solve the questions and prepare the Class 12 Biology NCERT exemplar solutions chapter 2 in a more detailed way.

3. What are the topics covered in NCERT Exemplar solution Chapter 2?
  • Flower- A Fascinating Organ of Angiosperms
  • Pre-fertilization: Structures and Events
  • Stamen, Microsporangium and Pollen Grain
  • The Pistil, Megasporangium (ovule) and Embryo sac
  • Pollination-
  • Double Fertilization
  • Post-Fertilization: Structures and Events
  • Endosperm
  • Embryo
  • Seed
  • Apomixis and Polyembryony

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  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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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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