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

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

Edited By Irshad Anwar | Updated on Sep 13, 2023 10:29 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Biology Chapter 2 - Sexual Reproduction in Flowering Plants: The myriads of flowers that we enjoy gazing at, the scents and perfumes that we swoon over, and the rich colours that attract us are all there as an aid to sexual reproduction in flowering plants. In the solutions for NCERT Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants, you will get questions and answers related to the male and female reproductive systems, as in the flower, the androecium, consisting of stamens, represents the male reproductive organs, and the gynoecium, consisting of pistils, represents the female reproductive organs. Class 12 Biology NCERT Solutions will help students score well in the exams. In case you have any doubts or queries about solving these questions, then download and get help from NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants.

NCERT Solutions For Class 12 Biology Sexual Reproduction in Flowering Plants

In Class 12 Biology Ch 2 NCERT Solutions, you will get solutions related to those topics that you have already learned in the NCERT Textbook. If you are looking for the answers to any other class from 6–12, then NCERT Solutions are there for you. In order to help Class 12 students understand the concepts and develop the habit of including diagrams in their answers to earn bonus points, Sexual Reproduction in Flowering Plants Class 12 Solutions are direct, and diagrams have been used whenever necessary.

Ch 2 Bio Class 12 NCERT Solutions give students a clear idea of how to respond to questions in a way that will earn them the maximum marks. The Questions and answers provided in the NCERT Solutions for Class 12 for all subjects are provided by subject matter experts, and the students can refer to these for their board exam preparation.Students can also refer Sexual reproduction in flowering plants Class 12 NCERT solutions for competitive exams such as NEET.

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NCERT Solutions For Class 12 Biology Sexual Reproduction in Flowering Plants

According to the latest CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 1.

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NCERT Solutions for Class 12 Biology Chapter 2 - Sexual Reproduction in Flowering Plants (Solved Exercise)

Class 12 Biology Chapter 2 Question Answer are given below in the sequence as per NCERT data:

Sexual Reproduction in Flowering Plants Class 12 Questions and Answers

Q1. Name the parts of an angiosperm flower in which the development of male and female gametophytes takes place.

Answer:

In an angiosperm plant, the development of a male gametophyte occurs in the pollen chamber while that of female gametophyte takes place in nucellus.

Q2. Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs duringthese events? Name the structures formed atthe end of these two events.

Answer:

Differences between microsporogenesis and megasporogenesis are as follows:

Microsporogenesis Megasporogenesis
The process of formation of a functional haploid microspore is called microsporogenesis.The process of formation of a functional haploid microspore is called megasporogenesis.
Microsporogenesis takes place in pollen sac of the antherMegasporogenesis takes place inside the ovule
Four functional microspores are formed as a result of microsporogenesisOnly one functional megaspore is formed as a result of megasporogenesis, the other three gets degenerated
The four micropores produced as a result of microsporogenesis, are arranged in a tetradThe four micropores produced as a result of megasporogenesis, are arranged linearly

During microsporogenesis and megasporogenesis, meiosis occurs and results into formation of four microspores and megaspores respectively, Later, mitosis occurs to give rise to microgametophyte and megagametophyte.

At the end of microsporogenesis, four functional haploid microspores are formed whereas at the end of megasporogenesis, four megaspores are formed out of which three degenerates and only one functional megaspore remains.

Sexual Reproduction in Flowering Plants Class 12 Solutions:

Q3. Arrange the following terms in the correct developmental sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.

Answer:

The correct developmental sequence is as follows:

Sporogenous tissue \rightarrow Pollen mother cell \rightarrow Microspore tetrad \rightarrow Pollen grain \rightarrow male gametes

Q4. With a neat, labelled diagram, describe the parts of a typical angiosperm ovule.

Answer:

The diagram of a typical angiosperm ovule is as follows:

Angiosperm ovule

An ovule consists of funicle, micropyle, integuments, nucellus, embryo sac, hilum, chalaza etc.

  • Funicle - It is the stalk of ovule which connects ovule to the placenta. Funicle is short and multicellular.
  • Hilum - It is the point where funicle connects the main body of the ovule.
  • Integuments - These are the layers that surround the ovule. There can be outer integument and inner integument. The main function of integuments is to protect the inner tissues.
  • Micropyle - The opening which is present at the top of integuments is called micropyle. The main function of micropyle is to allow the entry of the pollen tube into the ovule.
  • Nucellus - Integuments surround parenchymatous tissue which constitutes the main body of the ovule and is called nucellus. The primary function of nucellus is to provide nutrition to the developing embryo.
  • Chalaza - The swollen part of ovule present opposite to micropyle is called chalaza.
  • Embryo sac - It is present in the nucellus and it contains female gamete called egg.

Sexual Reproduction in Flowering Plants Class 12 NCERT Solutions:

Q5. What is meant by monosporic development of female gametophyte?

Answer:

The female gametophyte of flowers develops from a single functional megaspore after continuous mitotic divisions. This type of development of female gametophyte from a single uninucleate megaspore is called monosporic development of female gametophyte.

Q6. With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.

Answer:

The diagram of 7- celled 8- nucleate embryo sac of plants is as follows:

1651061454351

The female gametophyte of plants is formed from a single functional megaspore with the help of four unequal mitotic divisions. These mitotic divisions give rise to 8 nuclei. Later, these nuclei assemble into 7 cells. To both chalazal and micropylar end three cells each move. The three cells at chalaza are called antipodal cells while those at micropylar end constitute egg apparatus. The egg apparatus includes two larger cells called synergid while the egg is the smallest cell that acts as a female gamete.

Two nuclei migrate to the centre and together they form a single cell called the central cell. This central cell consists of two nuclei called polar nuclei. Thus, the mature embryo sac possesses 7 cells ( 3 antipodal cells, 2 synergid cells, 1 egg cell and 1 central cell) and 8 nuclei ( 3 of antipodal cell, 2 of synergid cells, 1 of egg cell and 2 of central cell).

Class 12 Biology Chapter 2 Question Answer:

Q7. What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.

Answer:

Flowers can be of two types i.e. chasmogamous flowers or cleistogamous flowers. Chasmogamous flowers are open flowers which have their anthers and stigma exposed. These flowers facilitate cross-pollination. On the other hand, cleistogamous flowers are closed flowers whose anthers and stigmas are not exposed. Since these flowers do not open at all, only self-pollination occurs in these and cross-pollination is never seen. Thus. cross-pollination cannot occur in cleistogamous flowers.

Q8. Mention two strategies evolved to prevent self-pollination in flowers.

Answer:

Self-pollination refers to the transfer of pollen grains from the anther of a flower to the stigma of the same flower. Continuous self-pollination can reduce the variations in the progeny and cause a reduction in the vigour and vitality. Thus, plants have developed certain mechanisms to avoid self-pollination and ensure cross-pollination. The two of these mechanisms to prevent self-pollination are as follows:

  1. Self sterility or self-incompatibility - In this mechanism, the pollen grains of a flower do not germinate on the stigma of the same flower. This occurs due to the presence of some self-sterile genes. This mechanism is genetic.
  2. Dichogamy- In this method, the timing of maturation of anthers and stigmas of a bisexual flower is different so as to prevent self-pollination.

Protandry- Anthers mature earlier than the stigma of the same flower. So, even if the pollens fall on these stigmas, they are unable to germinate. E.g. sunflower, Salvia.

Protogyny- Stigmas mature earlier than anthers and get pollinated by mature pollen grains of other flowers. E.g. Mirabilis jalapa.

NCERT Solutions for Class 12 Biology Chapter 2 - Sexual Reproduction in Flowering Plants:

Q9. What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?

Answer:

Self-incompatibility refers to the genetic inability of fully functional pollen grains and ovules to produce viable seeds. This is mainly due to the presence of S or self genes. If these genes are present in pollen as well as ovule, they will not be able to produce viable seeds. Self-pollination does not lead to the formation of seeds in self-incompatible species due to the presence of certain chemical substances that block the germination of pollen grains on stigma and also obstruct their passage to the ovary.

Q10. What is bagging technique? How is it useful in a plant breeding programme?

Answer:

Bagging technique - Bagging refers to covering of emasculated flowers with the help of butter paper or plastic bags to avoid contamination of stigma of these flowers to undesired pollens.

This technique is used during artificial hybridisation experiments. This technique is useful in breeding programmes because it prevents contamination of stigma with unwanted pollens. So, through this technique, we can allow the breeding of species according to our need and obtain superior species.

Class 12 Biology Chapter 2 NCERT Solutions:

Q11. What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.

Answer:

A pollen grain produces two male gametes. Out of these, one male gamete fuses with the egg cell and produces a zygote. The second male gamete fuses with the two polar nuclei of the central cell to produce primary endosperm nucleus. Since this fusion includes three haploid nuclei, it is called triple fusion.

The process of triple fusion occurs in the central cell of embryo sac. The pollen tube delivers one nucleus to the egg cell while the other one migrates to centre and fuses polar nuclei.

In triple fusion, two polar nuclei and one male gamete are involved.

Q12. Why do you think the zygote is dormant for sometime in a fertilised ovule?

Answer:

The zygote starts developing into an embryo only when a certain amount of endosperm has developed. This is because endosperm provides nourishment to the developing embryo. Thus, a zygote remains dormant in a fertilised ovule till the endosperm has developed.

Class 12 Sexual Reproduction in Flowering Plants NCERT Solutions:

Q13(a). Differentiate between:

(a) hypocotyl and epicotyl

Answer:

The differences between hypocotyl and epicotyl are as follows:

Hypocotyl Epicotyl
This is the portion of the embryo present below the cotyledon.This is the portion of embryo present above the cotyledon.
Hypocotyl ends into radicleEpicotyl ends into plumule

Q13(b) . Differentiate between:

(b) coleoptile and coleorrhiza

Answer:

The differences between coleoptile and coleorrhiza are as follows:

ColeoptileColeorrhiza
Coleoptile refers to the protective covering that surrounds plumuleColeorrhiza is the sheath that surrounds the radicle and root cap
This sheath can come out of the soil, become green and perform photosynthesis.Coleorrhiza remains in the soil only.

Q13(c) . Differentiate between:

(c) integument and testa

Answer:

The differences between integument and testa are as follows:

Integument Testa
It is a pre fertilisation structure that covers the ovuleIt is a post-fertilisation structure that covers the seed
The integument is a thin layer consisting of living cellsTesta is a thick coat consisting of dead cells
The integument arises from chalazaTesta is derived from the integuments

Q13(d). Differentiate between:

(d) perisperm and pericarp

Answer:

The differences between perisperm and pericarp are as follows:

PerispermPericarp
Perisperm refers to the residual persistent nucellusThe pericarp is the wall of fruit derived from the ovary wall
Perisperm is non-functional to the seedThe function of the pericarp is the protection of inside structures
It is dryIt is fleshy.

Sexual Reproduction in Flowering Plants Class 12 Questions and Answers:

Q14. Why is apple called a false fruit? Which part(s) of the flower forms the fruit?

Answer:

The fruits are formed from the ovary of a flower after fertilisation. Such fruits that develop from the ovary are called true fruits. On the other hand, fruits formed from any part of the flower other than ovary are called false fruits. Apple is called a false fruit because it develops from the thalamus and not from the ovary.

The parts of flower that take part in fruit formation are ovary, thalamus etc.

Q15. What is meant by emasculation? When and why does a plant breeder employ this technique?

Answer:

The process of removal of anthers from the flower with the help of forceps without affecting the female reproductive parts is called emasculation. This technique is used by plant breeders to allow suggestive breeding between plants in order to produce plants with desired seeds. Hence, it can be used to prevent self-pollination or prevent cross-pollination according to the need of the breeder.

Class 12 Biology Ch 2 Question Answer:

Q16. If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?

Answer:

The fruits formed as a result of parthenocarpy are generally seedless. Hence, for parthenocarpy, we can select fruits which have a lot of seeds in them. Such fruits include watermelon, orange, muskmelon, lemon etc. So we can select watermelon to induce parthenocarpy because by this method we can have seedless watermelons.

Q17. Explain the role of tapetum in the formation ofpollen-grain wall.

Answer:

Tapetum is the nutritive innermost layer of microsporangium that provides nutrition to the developing microspores. The role of tapetum in the formation of pollen grain wall is as follows:

  • Tapetum secretes ubisch granules required for the formation of sporopollenin in the exine (outer wall) of pollen grain
  • Tapetum also secretes pollenkit substances required for the formation of pollen grain wall

Sexual Reproduction in Flowering Plants Class 12 Solutions

Q18 . What is apomixis and what is its importance?

Answer:

The process of development of seeds from diploid cells of nucellus tissue without the process of fertilisation is called apomixis. This process is a form of asexual reproduction that mimics sexual reproduction. Apomixis is very commonly seen in grasses.

Importance of apomixis

  • Apomixis is used by horticulturists to produce hybrid seeds.
  • Apomixis leads to the formation of infection-free embryos.
  • Apomixis can be used to produce multiple embryos at a time.
  • Apomictic embryos are genetically similar to their parent plants. For this reason, they can be used in order to conserve specific characters in crop plants.

Class 12 Biology Ch 2 NCERT Solutions Sexual Reproduction in Flowering Plants will help you in both preparations as it will help you in your board exams along with the NEET preparation. A cell of the archesporium, the megaspore mother cell divides meiotically and one of the megaspores forms the embryo sac (the female gametophyte). The mature embryo sac is 7-celled and 8-nucleate. And so many interesting things are there in this chapter which you have to learn and class 12 biology chapter 2 Sexual Reproduction in Flowering Plants will make you clear.

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

Topics and Subtopics in NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants:

Section Name
Topic Name
2Sexual Reproduction in Flowering Plants
2.1Flower – A Fascinating Organ of Angiosperms
2.2Pre-fertilisation : Structures and Events
2.3Double Fertilisation
2.4Post-fertilisation: Structures and Events
2.5Apomixis and Polyembryony
2.6Summary

NCERT Solutions For Class 12 Biology Sexual Reproduction in Flowering Plants

In Ch 2 Bio Class 12 NCERT Solutions, you will understand that some questions are based on topics related to the pollen grains that develop inside the microsporangia of the flower. There are four wall layers, the epidermis, endothecium, middle layers, and tapetum surrounding the microsporangium. In the Chapter 2 Biology Class 12 NCERT Solutions, you will get all the solutions, but to understand them, you should be clear about every concept, like how the cells of the sporogenous tissue lying in the centre of the microsporangium, undergo meiosis (microsporogenesis) to form tetrads of microspores, etc. Pollen grains-related questions are also mentioned in the Sexual reproduction in flowering plants Class 12 NCERT PDF.

Let’s talk about the pistil. In Sexual Reproduction in Flowering Plants NCERT PDF, you will get questions and solutions about the pistil of a flower that has three parts: the stigma, style, and ovary. Ovules are present in the ovary. The ovules have a stalk called a funicle, a protective integument(s), and an opening called the micropyle. You will also get questions about the central tissue, the nucleus, in which the archesporium differentiates from Biology Class 12 Chapter 2. After going through the NCERT Solutions for Class 12 Biology Chapter 2 Sexual reproduction in Flowering Plants, you will be aware of the important topics of this chapter.

Highlights of NCERT Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

The main features of Sexual Reproduction in Flowering Plants Class 12 PDF are:

  • Solutions for Biology Chapter 2 Class 12 are provided by the subject experts in the best possible way.

  • Class 12 Biology Chapter 2 NCERT Solutions has a list of important topics

  • Class 12 Biology Ch 2 question answer are explained in simple language that is easier to read and understand.

  • Ch 2 Biology Class 12 Strengthen your fundamental knowledge.

  • Sexual reproduction in flowering plants Class 12 Solutions are according to the latest syllabus provided by the CBSE.

  • Chapter 2 Biology Class 12 NCERT Solutions are also helpful for competitive exams.

  • Labelled diagrams are used in NCERT Solutions for Class 12 Biology Chapter 2.

    NCERT Solutions for Class 12 Biology: Chapter wise

NCERT Exemplar Class 12 Solutions

NCERT Solutions for Class 12- Subject wise

We hope you will ace Your board examinations with the help of NCERT Solutions for Class 12 - Sexual Reproduction in Flowering Plants.

Frequently Asked Questions (FAQs)

1. What are the three fertilisation mentioned in ncert solutions for class 12 biology chapter 2?

The three fertilisation mentioned in class 12 biology chapter 2 :

  • Pre-fertilisation  
  • Double fertilisation  
  • Post-fertilisation 
2. What are the important topics which are mentioned in NCERT solution for class 12 biology chapter 2 sexual reproduction in flowering plants?

These are the important topics which are mentioned in NCERT solutions for class 12 biology chapter 2 sexual reproduction in flowering plants:   

  • Flower - A Fascinating Organ of Angiosperms  
  • Pre - Fertilisation: Structures and Events  
  • Stamen, Microsporangium and Pollen Grain  
  • The Pistil, Megasporangium (ovule) and Embryo sac  
  • Pollination  
  • Double Fertilization  
  • Post - Fertilisation: Structures and Events  
  • Endosperm  
  • Embryo  
  • Seed  
  • Apomixis And Polyembryony 
3. Why you should use NCERT solutions for class 12 biology chapter 2 sexual reproduction in flowering plants?
  • NCERT is the base, especially for CBSE students.  
  • In NCERT solutions for class 12 biology chapter 2 sexual reproduction in flowering plants, you will get all the answers to this chapter and it will help you to score good marks in the exam.  
  • This will also help you to score good marks in competitive exams like NEET.  
  • NCERT solutions also help you in your 12th board exam.  
  • class 12 biology ch 2 ncert solutions sexual reproduction in flowering plants will also boost up your knowledge.  
  • To Score Well in the examination, follow NCERT syllabus and exercise given in the NCERT Book. To practice more problems, students can refer to NCERT exemplar  
4. What are the difference between coleoptile and coleorrhiza given in class 12 biology chapter 2

The differences between coleoptile and coleorrhiza given in class 12 biology ch 2 ncert solutions are as follows:  

  Coleoptile  
  Coleorrhiza  
Coleoptile refers to the protective covering that surrounds plumule  
Coleorrhiza is the sheath that surrounds the radicle and root cap  
This sheath can come out of the soil, become green and perform photosynthesis.  
Coleorrhiza remains in the soil only.  

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Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

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  4. Focus on NEET 2025 Preparation:

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  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

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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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