Category: Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.4

Question 1.
Without actually performing the long divison, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:


Solution:

Question 2.
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:

Multiplying and dividing by 2⁵, we have

Question 3.
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form \(\frac{p}{q}\), what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000….
(iii) \(43 . \overline{123456789}\)
Solution:
(i) 43.123456789
∴ The given decimal expansion terminates.
∴ It is rational of the form \(\frac{p}{q}\)

Hence, p = 43123456789 and q = 2⁹ × 5⁹
Prime factors of q are 2⁹ and 5⁹.

(ii) 0.120120012000120000…
∵ The given decimal expansion is neither terminating nor repeating.
∴ It is irrational number, hence cannot be written in p/q form.

(iii) \(43 . \overline{123456789}\)
∵ The given decimal expansion is non-terminating repeating.
∴ It is rational number.

Multiplying both sides by 1000000000, we have
1OOOOOOOOOx = 43123456789.123456789…
… (2)
Subtracting (1) from (2), we have
(1000000000x) – x
= (43123456789.123456789 ) – 43.123456789…
⇒ 999999999x = 43123456746

Here, p = 4791495194 and q = 111111111, which is not of the form 2^m × 5ⁿ i.e., the prime factors of q are not of the form 2^m × 5ⁿ.

MP Board Class 10th Science Solutions Chapter 8 How do Organisms Reproduce?

MP Board Class 10th Science Chapter 8 Intext Questions

Class 10th Science Chapter 8 Intext Questions Page No. 128

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
The chromosomes in the nucleus of a cell contain information for inheritance of features from parents to next generation in the form of DNA (Deoxyribo Nucleic Acid) molecules. The DNA in the cell nucleus is the information source for making proteins. Hence DNA copying is important in reproduction.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:

If a population of reproducing organisms were suited to particular niche and if the niche were drastically altered, the population could be wiped out. However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive.

Thus, if there were a population of bacteria living in temperature waters and if the water temperature were to be increased by global warming, most of these bacteria would die, but the few variants resistant to heat would survive and grow further. Variation is thus useful for the survival of species over time. Variation is not useful for all organisms.

Class 10th Science Chapter 8 Intext Questions Page No. 133

Question 1.
How does binary fission differ from multiple fission?
Answer:
Binary fission: It is a simple kind of division which formate new individual. In binary fission, a single cell divides into two equal halves but it is possible only with very simple single cell kind. Amoeba and Bacteria divide by binary fission.

Multiple fission: Another type of simple division is multiple fission, in this, a single cell divides into many daughter, cells, e.g., Plasmodium divide by multiple fission.

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
The spores are covered by thick walls that protect them until they come into contact with another moist surface and can begin to grow. Thus organism be benefited if it reproduces through spores.

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Multicellular organisms are not simply a random mass of cells but a carefully organized entity of tissues and organs are placed at definite positions in the body to form organ systems. These systems are well coordinated to perform specific functions. Hence complex organisms cannot reproduce through fragmentation.

Question 4.
Why is vegetative propagation practised for growing some types of plants?
Answer:
Advantages of vegetative propagation:

• Used in methods such as layering or grafting, to grow many plants like sugarcane, roses or grapes for agricultural purposes.
• Plants raised can bear more flowers and fruits in comparison to plants produced from seeds.
• Plants such as banana, orange, rose and jasmine which have lost the capacity to produce seeds can be propagated.
• All plants produced by vegetative propagation are genetically similar enough to the parent plant.

Question 5.
Why is DN Acopying an essential part of the process of reproduction?
Answer:
The consistency of DNA copying during reproduction is important for the maintenance of body design features that allow the organism to use that particular niche. Because of this DNA copying is an essential part of the process of reproduction.

Class 10th Science Chapter 8 Intext Questions Page No. 140

Question 1.
How is the process of pollination different from fertilization?
Answer:
Pollination is movement of pollens from one plant to another plant’s or its own plant’s stigma. It may require certain agents called pollinators such as air, water birds or some insects to perform. Fertilization, is a complex process, it involves the fusion of the male and female gametes. It occurs inside the ovule and leads to the formation of zygote.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
Along the path of the vas deferens, gland like the prostrate and the seminal vesicles add their secretions so that the sperms are now in a fluid which makes their transport easier and this fluid also provides nutrition.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
The changes seen in girls at the time of puberty are:

1. Development of secondary sexual characteristics.
2. Growth in breast size and darkening of skin of the nipples.
3. Growth of hair in the genital area and other areas of skin like underarms, face, hands and legs.
4. Growth in the size of uterus and ovary hence, start of menstrual cycle periodically.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
The embryo gets nutrition form the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue on the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances which can be removed by transferring them into the mother’s blood through the placenta.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
Copper-T will helps in protecting her from sexually transmitted diseases by helping to prevent infections of diseases.

MP Board Class 10th Science Chapter 8 NCERT Textbook Exercises

Question 1.
Asexual reproduction takes place through budding in:
(a) amoeba
(b) yeast
(c) plasmodium
(d) leishmania
Answer:
(b) yeast

Question 2.
Which of the following is not system in human beings? a part of the female reproductive
(a) ovary
(b) uterus
(c) vas deferens
(d) fallopian tube
Answer:
(c) vas deferens

Question 3.
The anther contains:
(a) sepal
(b) ovules
(c) carpel
(d) pollen grains
Answer:
(d) pollen grains

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:
In case of asexual reproduction, new generations are produced by one organism. But in sexual reproduction, new generations are produced by two organisms (male and female). In case of sexual reproduction germ cells are produced in testes and these secrete a hormone testosterone. In human beings also develop special tissues for this purpose.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
They are the glands where sperm and testosterone are generated and present in male body. The testes are contained in the scrotum and are composed of dense connective tissue. Functions of testes are as follows:

• It produces sperms, which contain haploid set of chromosomes of
• It produces testosterone, which initiate secondary sexual characteristics

Question 6.
Why does menstruation occur?
Answer:
Since the ovary releases one egg every month, the uterus also prepares itself every month to receive a fertilised egg. Thus its lining becomes thick and spongy. This would be required for nourishing the embryo if fertilisation had taken place. Now, however, this lining is not needed any longer. So the lining slowly breaks and comes out through the vagina as blood and mucous. This cycle takes place roughly every month and is known a menstruation. It usually lasts for about two to eight days.

Question 7.
Draw a labelled diagram of the longitudinal section of a flower.
Answer:

Longitudinal section flower.

Question 8.
What are the different methods of contraception?
Answer:

Many ways have been devised to avoid pregnancy. These contraceptive methods fall in a number of categories. One category is the creation of a mechanical barrier so that sperm does not reach the egg. Condoms on the penis or similar coverings worn in the vagina can serve this purpose.

Another category of contraceptives acts by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. These drugs commonly need to be taken orally as pills. However, Since they change hormonal balances, they can cause side effects too. Other contraceptive devices such as the loop or the copper-T are placed in the uterus to prevent pregnancy. Again, they can cause side effects due to irritation of the uterus. Surgery can also be used for removed of unwanted pregnancies.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:
In unicellular organisms, reproduction occurs by the division of the entire cell. The modes of reproduction in unicellular organisms can be fission, budding etc. whereas in multi cellular organisms, specialised reproductive organs are present. Therefore, they can be reproduced by complex reproductive methods such as vegetative propagation, spore formation etc. In more complex multicellular organisms such as human beings and plants, the mode of reproduction is sexual reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Reproduction is the process of producing new individuals of the same species by existing organisms of a species. So, it helps in providing stability to population of species by giving birth to new individuals as the rate of birth must be at par with the rate of death to provide stability to population of a species.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:
Contraceptive methods are mainly adopted because of the following reasons:

• It prevent unwanted pregnancies.
• It control rise in population and birth rate.
• It prevent sexually transmitted diseases.

MP Board Class 10th Science Chapter 8 Additional Important Questions

MP Board Class 10th Science Chapter 8 Multiple Choice Questions

Question 1.
All individuals produced by an organism are:
(a) Genetically similar
(b) Non-identical
(c) Fission
(d) Moneociuos
Answer:
(a) Genetically similar

Question 2.
Sexual reproduction is completed by _______ division:
(a) Mitotic
(b) Meiotic and mitotic both
(c) Meiosis
(d) Mitotic at some stages .
Answer:
(c) Meiosis

Question 3.
In yeast cell, division results in:
(a) Offspring
(b) Bud
(c) Clone
(d) Branch
Answer:
(b) Bud

Question 4.
Which of the following organisms undergo multiple fission?
(a) Paramecium
(b) Plasmodium
(c) Amoeba
(d) All of the above
Answer:
(b) Plasmodium

Question 5.
Hydra reproduces asexually through:
(a) Budding
(b) Binary fission
(c) Multiple fission
(d) Vegetative propagation
Answer:
(a) Budding

Question 6.
In which plant, the site of origin of new plants is node?
(a) Potato tuber
(b) Onion bulb
(c) Rhizome ginger
(d) All of the above
Answer:
(d) All of the above

Question 7.
In which of the following, asexual reproduction takes place through binary fission?
(a) Amoeba
(b) Yeast
(c) Plasmodium
(d) Leishmania
Answer:
(b) Yeast

Question 8.
Which of the following is in human beings?
(a) Ovary
(b) Uterus
(c) (a) and (b)
(d) Fallopian tube
(e) (a), (b) and (d)
Answer:
(e) (a), (b) and (d)

Question 9.
The anther, a part of male flower have:
(a) Sepals
(b) Ovules
(c) Carpel
(d) Pollen grains
Answer:
(d) Pollen grains

Question 10.
The information for making proteins is provided by:
(a) Rough endoplasmic reticulum
(b) DNA
(c) Hormones
(d) Enzymes
Answer:
(b) DNA

Question 11.
Nature of gametes are usually:
(a) Haploid
(b) Diploid
(c) Both (a) and (b)
(d) None of the above
Answer:
(a) Haploid

Question 12.
With the help of which tissues embryo gets nutrition from the mother’s blood?
(a) Zygote
(b) Uterus only
(c) Placenta
(d) None of these
Answer:
(c) Placenta

Question 13.
Which of the following is not a part of the male reproductive system in human beings?
(a) Testes
(b) Uterus
(c) Vas deferens
(d) Urethra
Answer:
(b) Uterus

Question 14.
Binary fission in some organisms occurs in definite orientation in relation to the cell structures. One such organisms is:
(a) Leishmania
(b) Plasmodium
(c) Amoeba
(d) Bacteria
Answer:
(c) Amoeba

Question 15.
Plants that have lost their capacity to produce seeds, reproduce by:
(a) Spores
(b) Vegetative propagation
(c) Fission
(d) Regeneration
Answer:
(a) Spores

Question 16.
A stamen consists of two parts namely:
(a) Anther and style
(b) Anther and filament
(c) Stigma and style
(d) Filament and style
Answer:
(b) Anther and filament

Question 17.
A bisexual flower contains:
(a) Stamens only
(b) Carpels only
(c) Either stamens or carpels
(d) Both stamens and carpels
Answer:
(d) Both stamens and carpels

Question 18.
Germinated seeds do not contains:
(a) Sepals
(b) Cotyledon
(c) Plumule
(d) Radicle
Answer:
(a) Sepals

Question 19.
A feature of reproduction that is common to amoeba, spirogyra and yeast is that:
(a) they reproduce asexually
(b) they are all unicellular
(c) they reproduce only sexually
(d) they are all multicellular
Answer:
(a) they reproduce asexually

Question 20.
Which of the part of flower ripens to form a fruit?
(a) Ovule
(b) Ovary
(c) Carpel
(d) Egg cell
Answer:
(b) Ovary

Question 21.
The testes perforin the following function/functions:
(a) Produce testosterone
(b) Produce sperms
(c) Produce male gametes and hormone
(d) Produce sperms and urine
Answer:
(b) Produce sperms

Question 22.
Where does fertilisation take place in human beings?
(a) Uterus
(b) Vagina
(c) Cervix
(d) Fallopian Tube
Answer:
(d) Fallopian Tube

Question 23.
Condom is a method of control that falls under the following category:
(a) Surgical method
(b) Hormonal method
(c) Mechanical method
(d) Chemical method
Answer:
(c) Mechanical method

Question 24.
The common passage for sperms and urine in the male reproductive system is:
(a) Ureter
(b) Seminal vesicle
(c) Urethra
(d) Vas deferens
Answer:
(c) Urethra

Question 25.
In sperm, which part dissociates after fertilization?
(a) Acrosome
(b) Tail
(c) Head
(d) Middle piece
Answer:
(b) Tail

MP Board Class 10th Science Chapter 8 Very Short Answer Type Questions

Question 1.
Which life process is not essential to maintain the life of an individual organism but important for the survival of species?
Answer:
Reproduction.

Question 2.
How a species can get a danger of being extinct?
Answer:
If individuals of any species stops reproducing, then that species can get a danger of being extinct.

Question 3.
How an individual is able to make a copy of itself?
Answer:
DNA copying is a process at cellular level which enables an individual to make copy of it self.

Question 4.
Write the name of process by which Hydra reproduces.
Answer:
Budding only.

Question 5.
Generally, how many individuals are involved in asexual reproduction?
Answer:
One.

Question 6.
Write the name of some common method of asexual reproduction.
Answer:
Vegetative propagation, budding, fragmentation and spore formation.

Question 7.
Which type of flower is called unisexual flowers?
Answer:
A flower which have either male or female reproductive parts is called unisexual flowers.

Question 8.
What is pollination?
Answer:
The transfer of pollen grains from the anther to the stigma of the same or of another flower of the same kind is known as pollination.

Question 9.
What do you understand by term fertilisation?
Answer:
The fusion of male and female gametes is termed as fertilisation.

Question 10.
How seed is dispersed?
Answer:
Seed dispersal takes place by means of wind, water and animals.

MP Board Class 10th Science Chapter 8 Short Answer Type Questions

Question 1.
How does plasmodium undergo fission?
Ans.
Plasmodium divides into many daughter cells through multiple fission.

Question 2.
How spirogyra reproduces by fragmentation?
Answer:
An individual spirogyra breaks up into many smaller pieces, each fragment grows into new individual.

Question 3.
Which cells are responsible for budding in hydra?
Answer:
Regenerative cells.

Question 4.
Name the structure into which following develops: the plumule and radicle?
Answer:
Plumule develops to shoot while radicle form root of a plant.

Question 5.
On which plant can you find buds on its leaves?
Answer:
Bryophyllum.

Question 6.
Write the scientific name of the bread mould.
Answer:
Rhizopus.

Question 7.
Where are the testes located in human beings?
Answer:
In abdominal cavity, in scrotum.

Question 8.
For what specific reason have the testes specific location?
Answer:
As testes, requires lesser temperature, to produce sperm than of abdominal cavity.

Question 9.
Correlate the rate of general body growth and maturation of reproductive tissue during puberty.
Answer:
When reproductive tissues (organs) begin to mature, body growth rate slows down.

Question 10.
Where does the zygote get implanted in human beings?
Answer:
In the wall of uterus.

Question 11.
Which two important substances are delivered to developing embryo through placenta?
Answer:
Glucose and oxygen.

Question 12.
How change in hormonal balance prevents pregnancy?
Answer:
It prevents the release of eggs.

Question 13.
Name the tissue in mother’s body that provides nutrition to developing embryo?
Answer:
Placenta provides nutrition to developing embryo.

Question 14.
Write one side effect of loop placed in uterus.
Answer:
It may cause permanent irritation and excessive and prolong bleeding in uterus.

Question 15.
Which structures need to be blocked in males and females respectively to prevent pregnancy?
Answer:
Vas deferens in male (vasectomy), fallopian tube in female (tubectomy).

Question 16.
Why is children sex ratio alarmingly declining in our country.
Answer:
Abortions based on sex selections.

Question 17.
Name the chemical methods of preventing pregnancy.
Answer:
Morning over oral pills.

Question 18.
Name some of the devices used as mechanical method for preventing pregnancy.
Answer:
Loop, copper T, condoms.

Question 19.
Name the only mammal(s) which lays eggs.
Answer:
Echidna and duck-billed platypus.

Question 20.
What is parthenogenesis?
Answer:
Parthenogenesis is a type of asexual reproduction. In this case, embryo development takes places without fertilisation. A few species of insects, bees, wasps, birds and lizards (e.gKomodo dragon lizard) reproduce this way.

Question 21.
Give an example of an organism which reproduces by:
(a) Fragmentation
(b) Spore formation
(c) Stems
Answer:
(i) Spirogyra.
(ii) Bacteria, fungi (rhizopus), moss, algae.
(iii) Plants like potato (tuber), onion (bulb) reproduce by vegetative propagation of stems.

Question 22.
Discuss various artificial vegetative propagation techniques.
Answer:
Various artificial vegetative propagation techniques are:

1. Cutting
2. Layering
3. Grafting
4. Tissue culture

Question 23.
What is grafting? What are different types of grafting techniques?
Answer:
In grafting, one part of a plant is inserted into another plant in a way that both of them will unite and grow together as a single plant. Different methods of grafting are:

• Approach grafting
• Cleft grafting
• Bud grafting
• Tongue grafting

Question 24.
Name some:

1. Plants which are reproduced by vegetative propagation.
2. Plants which have unisexual flowers.
3. Plants which have bisexual flowers.
4. Plants with self-pollination.
5. Plants that do cross-pollination.

Answer:

1. Rose, sweet potatoes, bryophyllum.
2. Coconut, papaya, watermelon.
3. Lily, rose, sunflower.
4. Beans, peas, tomatoes.
5. Grasses, catkins, maple trees.

Question 25.
What is germination?
Answer:
The seed contains the future plant or embryo which develops into a seedling under appropriate conditions. This process is known as germination.

Question 26.
What is cross-pollination?
Answer:
Cross-pollination is the process of transfer of pollen from the anther of a flower to stigma of a flower of another plant of the same species or closely related species.

Question 27.
Explain hormonal pills of contraception.
Answer:
Oral contraceptives: In this method, tablets or drugs are taken orally by females to check pregnancy These contain small doses of hormones in forms of pills that prevent the release of eggs and thus, fortilisation cannot occur.

MP Board Class 10th Science Chapter 8 Long Answer Type Questions

Question 1.
Why simply copying of DNA in a dividing cells not enough to maintain continuity of life?
Answer:
Copying of DNA preserve and pass specific characters of a generation to next generation offsprings. In reproduction, it is very important to create DNA copy. It determines the body design of an individual. But variation in genotype is also important,, because sometimes existing genotype don’t find its potential to survive in changing surroundings. So, genotype must have some alterations which are caused by variations only. Hence, simply copying of DNA in a dividing cells is not enough to maintain continuity of life.

Question 2.
Describe in brief the fragmentation mode of asexual reproduction.
Answer:
Fragmentation: Many lower organisms, use fragmentation mode of asexual reproduction for its growth e.g., algae. When water and nutrients are available in sufficient amount algae grow and multiply rapidly by fragmentation. An algae breaks up to multiple fragments. These fragments or pieces grow into new individuals.

Question 3.
Explain budding in yeast.
Answer:
The yeast is a single-celled organism. The small bulb-like projection come out from the yeast cell in favourable time and is called a bud. The bud gradually grows and gets dettached from the parent cell and forms a new yeast cell. The new yeast cell grows, matures and produces more yeast cells.

Question 4.
Describe the process of implantation.
Answer:
A week after the sperm fertilizes the egg, the fertilized egg (zygote) undergo development and become a multicelled blastocyst. The blastocyst fix itself into the lining of the uterus, called the endometrium. The hormone estrogen causes the endometrium to become thick and rich with blood. Progesterone and other hormone released by the ovaries, keeps the endometrium thick with blood so that the blastocyst can absorb nutrients from uterus. This process is called implantation.

Question 5.
Explain the following.

1. Hermaphrodites
2. Unisexual
3. Syngamy

Answer:

1. Hermaphrodites are bisexual organisms which possess both male and female reproductive organs. Examples: earthworm, leech, starfish.
2. Animals which have different male and female individuals as birds, mammals etc.
3. The process of fusion of male gamete with female gamete is called syngamy.

Question 6.
What is contraception? Discuss natural and barrier method of contraception.
Answer:
Contraception or birth control methods include: condoms, the diaphragm, the contraceptive pill, implants, IUDs (intrauterine devices), sterilization and the morning after pill and many more some of best methods are given below:

• Natural method: It involves avoiding the chances of meeting of sperms and ovum. In this method, the sexual intercourse is avoided by the couple from day 10th to 17th of the menstrual cycle of female as in this period, ovulation is expected and therefore, the chances of fertilisation are very high.
• Barrier method: In this method, the fertilisation of ovum and sperm is checked out with the help of artificially developed barriers. Barriers are developed for both males and females. Most common barrier available in market are condoms.

Question 7.
Describe implants and surgical methods of contraception
Answer:
Contraceptive devices are also developed as the loop or copper-T to prevent pregnancy. Surgical methods are also used to block the gamete transfer. It includes the blocking of vas deferens to prevent the transfer of Sperms known as vasectomy. Similarly, tubectomy in the fallopian tubes of the female can be blocked so that the egg will not reach the uterus.

Question 8.
Discuss fertilization in flowering plants.
Answer:
There are two main procedures of completing fertilization in flowering plants, which are:
(i) Pollination
(ii) Fertilisation

(i) Pollination: Pollination is a very important part of the life cycle of a flowering plant which results in seeds that grow into new plants. It is part of the sexual reproduction process of flowering plants. Flowers are the structures of flowering plants that contain all the specialized parts needed for sexual reproduction. Plants have gametes, which contain half the normal number of chromosomes for that plant species. Male gametes are found inside tiny pollen grains on the anthers of flowers. Female gametes are found in the ovules of a flower. Pollination is the process that brings these male and female gametes together. The wind or animals, especially insects and birds, pick up pollen from the male anthers and carry it to the female stigma. Flowers have to encourage animals to pollinate them.

(ii) Fertilisation: After pollination, when pollen has landed on the stigma of a suitable flower of the same species, various process occurs in the making of seeds. A pollen grain on the stigma grows a tiny tube, all the way down the style to the ovary. This pollen tube carries a male gamete to meet a female gamete in an ovule. In a process called fertilization, the two gametes join. The fertilised ovule form a seed, which contains a food store and an embryo that grow into a new plant. The ovary develops into a fruit to protect the seed.

Question 9.
Inside womb, how does a child receive food, oxygen and water? Discuss.
Answer:
As a mother eats something the nutrient like glucose, proteins, fats, vitamins, etc. are absorbed into the mother’s blood by the small intestine. The nutrients flow to the placenta, and then transferred to the baby’s bloodstream through the umbilical cord. The baby’s waste products (like CO2) are disposed of in the mother’s blood stream as well. In the placenta, the mothers blood flows into a network of blood Vessels and capillaries. Molecules in the mother’s blood like glucose, proteins, fats, oxygen etc. flow out of the mother’s blood supply and are absorbed into another network of blood vessels and capillaries containing the baby’s blood supply. The baby’s blood then flows through the umbilical cord back to the baby. It is the complete process of baby’s nutrition inside womb.

Question 10.
Discuss the advantages and disadvantage of autogamy or self¬pollination.
Answer:
Advantages of autogamy:

It is a sure method of seed formation. Scent and Nectar are not needed by the flower to attract insects. Parent characteristics are preserved in off spring’s. Small quantity of pollen is required for pollination. Flowers need not be large or attractive. Disadvantages of autogamy plants lose their vigor in their future generations due to repeated self-pollination. Since, there is no variation, no genetic improvement occurs in offsprings. Weak characteristics of the plant are inherited by the next generations.

MP Board Class 10th Science Chapter 8 Textbook Activities

Class 10 Science Activity 8.1 Page No. 129

• Dissolve about 10 gin of sugar in 100 mL of water.
• lake 20 mL of this solution in a test tube and add a pinch of yeasl granules to it.
• Put a cotton plug on the mouth of the test tube and keep it in a warm place.
• Alter 1 or 2 hours, put a small drop of yeast culture from the test tube on a slide and cover it with a coverslip.
• Observe the slide under a microscope.

Observations:

• Formation of yeast cells can be seen. Some of them, shows chain budding.

Class 10 Science Activity 8.2 Page No. 129

• Wet a slice of bread, and keep it in a cool, moist and dark place.
• Observe the surface of the slice with a magnifying glass.
• Record your observation for a week.

Observations:

• A layer of while cottony mass is seen over the surface of slice. These inercase in size and number and after a week, the layer turns black show ing formation of spores or sporangia.

Class 10 Science Activity 8.3 Page No. 129

• Observ e a permanent slide of Amoeba under a microscope.
• Similarly observe another permanent slide of Amoeba show-ing binary fission.
• Now, compare the observation of both the slides.

Observations:

• The permanent slide of amoeba shows normal cytoplasm and nucleus. Nucleus can be seen dividing and construction in cytoplasm can also be seen. The binary fission with two daughter cells is observed in the other slide.

Class 10 Science Activity 8.4 Page No. 129

• Collect water from a lake or pond that appears dark green and contains filamentous structures.
• Put one or two filaments on a slide.
• Put a drop of glycerin on these filaments and cover it with a coverslip
• Observe the slide under a microscope.
• Can you identify different tissues in the Spimgyra filaments.

Observations:

• Spirogyra filament consists of many cells which are attached linearly to form a filament.

Class 10 Science Activity 8.5 Page No. 132

• Take a potato and observe its surface. Can notches be seen?
• Cut the potato into small pieces such that some pieces contain a notch or bud and some do not.
• Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
• Observe changes taking place in these potato pieces over the next few days. Make sure that the cotton is, kept moistened.
• Which arc the potato pieces that give rise to fresh green shoots and roots.

Observations:

• The,potato undergoes various changes in few days. The buds in notches show growth of young shoots and roots. The pieces which do not have eye buds do not show any growth.

Class 10 Science Activity 8.6 Page No. 132

• Select a money-plant.
• Cut some pieces such that they contain at least one leaf.
• Cut out some other portions between two leaves.
• Dip one end of all the pieces in water and observe over the next few days.
• Which ones grow’ and give rise to fresh leaves?
• What can you conclude from your observations ?

Observations:

• The leaves at the nodes show formation of fresh leaves. The formation of branch from axillary buds axil of leaf is also observ ed.
• The leaves that undergo photosynthesis show tendency to grow into a new plant through vegetative propagation.

Class 10 Science Activity 8.7 Page No. 135

• Soak a few seeds of Bengal gram (chana) and keep them overnight.
• Drain the excess water and cover the seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry.
• Cut open the seeds carefully and observe the different parts.
• Compare your observations with the Fig. 8.2 and sec if you can identify all the parts.

Observations:

• The parts identified includes- cotyledon which stores food, plumule which is a future shoot radicle that is a future root.

Germination.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3Pair of Linear Equations in Two Variables Ex 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:
At present: Let Aftab’s age = x years
His daughter’s age = y years
Seven years ago : Aftab’s age = (x – 7) years
His daughter’s age = (y – 7)years
According to the condition,
[Aftab’s age] = 7[His daughter’s age]
⇒ [x – 7] = 7[y – 7] = x – 7 = 7y – 49
⇒ x – 7y – 7 + 49 = 0 ⇒ x – 7y + 42 = 0 …. (1)
After three years : Aftab’s age = (x + 3) years
His daughter’s age = (y + 3) years
According to the condition,
[Aftab’s age] = 3[His daughter’s age]
⇒ [x + 3] = 3[y + 3]
⇒ x + 3 = 3y + 9 ⇒ x — 3y + 3 — 9 = 0
⇒ x — 3y – 6 = 0 …. (2)
Graphical representation of equation (1) and (2): From equation (1), we have :

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of a bat = ₹ x
and the cost of a ball = ₹ y
Cost of 3 bats = ₹ 3x
and cost of 6 balls = ₹ 6y
Again, cost of 1 bat = ₹ x
and cost of 3 balls = ₹ 3y
Algebraic representation:
Cost of 3 bats + Cost of 6 balls = ₹ 3900
⇒ 3x + 6y = 3900 ⇒ x + 2y = 1300 …. (1)
Also, cost of 1 bat + cost of 3 balls = ₹ 1300
⇒ x + 3y = 1300 …. (2)
Thus, (1) and (2) are the algebraic representations of the given situation.
Geometrical representation:
We have for equation (1),

We can also see from the obtained graph that the straight lines representing the two equations intersect at (1300, 0).

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
solution:
Let the cost of 1 kg of apples = ₹ x
And the cost of 1 kg of grapes = ₹ y
Algebraic representation:
2x + y = 160 … (1)
and 4x + 2y = 300
⇒ 2x + y = 150 … (2)
Geometrical representation:
We have, for equation (1),

The straight lines l₁ and l₂ are the geometrical representations of the equations (1) and (2) respectively. The lines are parallel.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 64 cm³

Let the edge of each cube = x
∴ x³ = 64 cm³
⇒ x = 4 cm
Now, Length of the resulting cuboid (l) = 2x cm = 8 cm
Breadth of the resulting cuboid (b) = x cm = 4 cm
Height of the resulting cuboid (h) = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm²
= 2 [32 + 16 + 32] cm² = 2 [80] cm²
= 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
For hemispherical part,
radius (r)= \(\frac{14}{2}\) = 7cm
∴ Curved surface area = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7cm²
= 308cm² 7
Total height of vessel = 13 cm

∴ Height of cylinder = (13 – 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 6cm² = 264cm² 7
∴ Inner surface area of vessel = (308 + 264)cm² = 572 cm²

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Let h be the height of cone and r be the radius of cone and hemisphere.

∴ h = [height of toy – radius of hemi sphere]
= (15.5 – 3.5) cm = 12 cm
Also l² = h² + r² = 12² + (3.5)² = 156.25 cm²
∴ l = 12.5 cm
Curved surface area of the conical part = πrl
Curved surface area of the hemispherical part = 2πr²
∴ Total surface area of the toy = πrl + 2πr²
= πr(l + 2 r)
= \(\frac{22}{7} \times \frac{35}{10}\) (12.5 + 2 × 3.5) cm²
= 11 × (12.5 + 7) cm² = 11 × 19.5 cm²
= 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Let side of the block, l = 7 cm

∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid = [Total surface area of the cubical block] + [C.S.A. of the hemisphere] – [Base area of the hemisphere]
= 6 × l² + 2πr² – πr²
[where l = 7 cm and r = \(\frac{7}{2}\) cm]

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let l be the side of the cube.

∴ Diameter of the hemisphere = l
⇒ Radius of the hemisphere (r) = \(\frac{l}{2}\)
Curved surface area of hemisphere = 2πr²
= 2 × π × \(\frac{l}{2} \times \frac{l}{2}=\frac{\pi l^{2}}{2}\)
Base area of the hemisphere = πr²
= \(\pi\left(\frac{l}{2}\right)^{2}=\frac{\pi l^{2}}{4}\)
Surface area of the cube = 6 × l² = 6l²
∴ Surface area of the remaining solid = [Total surface area of cube + C.S.A. of hemisphere – base area of hemisphere]

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Radius of the hemispherical part

Curved surface area of one hemispherical part = 2πr²
∴ Surface area of both hemispherical parts
= 2(2πr²) = 4πr² = [4 × \(\frac{22}{7} \times\left(\frac{25}{10}\right)^{2}\)] mm²
= \(\left(4 \times \frac{22}{7} \times \frac{25}{10} \times \frac{25}{10}\right)\) mm²
Entire length of capsule = 14 mm
∴ Length of cylindrical part = [Length of capsule – Radius of two hemispherical part]
= (14 – 2 × 2.5)mm = 9mm Area of cylindrical part = 2πrh
= (2 × \(\frac{22}{7}\) × 2.5 × 9 ]mm² = (2 × \(\frac{22}{7} \times \frac{25}{10}\) × 9) mm²
Total surface area
= [Surface area of cylindrical part + Surface area of both hemispherical parts]

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:

For cylindrical part:
Radius (r) = \(\frac{4}{2}\) m = 2m and height (h) = 2.1 m
∴ Curved surface area = 2πrh = (2 × \(\frac{22}{7}\) × 2 × \(\frac{21}{10}\))m²
For conical part:
Slant height (l) = 2.8 m
and base radius (r) = 2 m
∴ Curved surface area
= πrl = (\(\frac{22}{7}\) × 2 × \(\frac{28}{10}\)) m²
∴ Total surface area = [Curved surface area of the cylindrical part] + [Curved surface area of conical part]

Cost of the canvas used :
Cost of 1 m² of canvas = ₹ 500
∴ Cost of 44 m² of canvas = ₹ (500 × 44)
= ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:

For cylindrical part :
Height (h) = 2.4 cm and diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
∴ Total surface area of the cylindrical part
= 2πrh + 2πr² = 2πr [h + r]
= 2 × \(\frac{22}{7} \times \frac{7}{10}\) [2.4 + 0.7]
= \(\frac{44}{10}\) × 3.1 = \(\frac{44 \times 31}{100}\) = \(\frac{1364}{100}\) cm²
For conical part:
Base radius (r) = 0.7 cm and height (h) = 2.4 cm

Base area of the conical part
= \(\pi r^{2}=\frac{22}{7} \times\left(\frac{7}{10}\right)^{2}=\frac{22 \times 7}{100} \mathrm{cm}^{2}=\frac{154}{100} \mathrm{cm}^{2}\)
Total surface area of the remaining solid = [(Total surface area of cylindrical part) + (Curved surface area of conical part) – (Base area of the conical part)]
= \(\left[\frac{1364}{100}+\frac{550}{100}-\frac{154}{100}\right] \mathrm{cm}^{2}=\frac{1760}{100} \mathrm{cm}^{2}\)
Hence, total surface area to the nearest cm² is 18cm².

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:
Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac{22}{7} \times \frac{35}{10}\) × 10cm² = 220cm²
Curved surface area of a hemisphere = 2πr²
∴ Curved surface area of both hemispheres
= 2 × 2πr² = 4πr² = 4 × \(\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}\) cm²
= 154 cm²
Total surface area of the remaining solid = (220 + 154) cm² = 374 cm².

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x – y = 4
(ii) s – f = 3; \(\frac{s}{3}+\frac{t}{2}\) = 6
(iii) 3x – y = 3; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y\) = 0; \(\sqrt{3} x-\sqrt{8} y\) = 0
(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2, \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 … (1),
x – y = 4 …. (2)
From (1) , we get x = (14 – y) …. (3)
Substituting value of x in (2) , we get
(14 – y) – y = 4 ⇒ 14 – 2y = 4 ⇒ -2y = -10 ⇒ y = 5
Substituting y = 5 in (3), we have
x = 14 – 5 ⇒ x = 9
Hence, x = 9, y = 5

(ii) s – t = 3 … (1)

From (1), we have s = (3 + t) … (2)
Substituting this value of s in (2), we get

Substituting, t = 6 in (3) we get,
S = 3 + 6 = 9
Thus, S = 9, f = 6

(iii) 3x – y = 3 … (1),
9x – 3y = 9 … (2)
From (1) , y = (3x – 3)
Substituting this value of y in (2),
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9 which is true,
Eq. (1) and eq. (2) have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 …. (2)
From the equation (1),

Substituting the value of y in (2), we have


Substituting the value of x in (1), we have

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:

Question 3.
Form the pair of linear equations for the following problems and find their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 13800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the two numbers be X and y such that x > y
It is given that
Difference between two numbers = 26
∴ x – y = 26 … (1)
Also one number = 3 [the other number]
⇒ x = 3y … (2)
Substituting x = 3y in (1) , we get 3y – y = 26 ⇒ 2y = 26
Now, substituting y = 13 in (2) , we have
x = 3(13) ⇒ x = 39
Thus, two numbers are 39 and 13.

(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18° (Given)
∴ x = y + 18°…. (1)
Also, sum of two supplementary angles = 180°
∴ x + y = 180° … (2)
Substituting the value of x from (1) in (2) , we get,
(18° + y) + y = 180°

Substituting, y = 81° in (1) , we get
x = 18° + 81° = 99°
Thus, x = 99° and y = 81°

(iii) Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800
⇒ 7x + 6y = 3800 … (1)
Also, [cost of 3 bats] + [cost of 5 balls] = ₹ 1750
3x + 5y = 1750 …. (2)

Substituting this value of y in (1) , we have

(iv) Let fixed charges = ₹ x
and charges per km = ₹ y
∵ Charges for the journey of 10 km = ₹ 105 (Given)
∴ x + 10y = 105 … (1)
and charges for the journey of 15 km = ₹ 155
∴ x + 15y = 155 … (2)
From (1) , we have, x = 105 – 10y …. (3)
Putting the value of x in (2) , we get
(105 – 10y) + 15y = 155
⇒ 5y = 155 – 105 = 50 ⇒ y = 10
Substituting y = 10 in (3) , we get
x = 105 – 10(10) ⇒ x = 105 – 100 = 5
Thus, x = 5 and y = 10
⇒ Fixed charges = ₹ 5
and charges per km = ₹ 10
Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255
∴ The charges for 25 km journey = ₹ 255

(v) Let the numerator = x
and the denominator = y
∴ Fraction = \(\frac{x}{y}\)
According to the given condition,

⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y + 4 = 0 … (1)

Substituting this value of x in (1) , we have

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
(i) Here x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1
∴ The required distance

(ii) Here x₁ = -5, y₁ = 7 and x₂ = -1, y₂ = 3
∴ The required distance

(iii) Here x₁ = a, y₁ = b and x₂ = -a, y₂ = -b
∴ The required distance

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns /I and B discussed below as following: ‘A town 6 is located 36 km east and 15 km north of the town A’.
Solution:
Part-I
Let the points be A(0, 0) and B(36, 15)

Part-II
We have A(0, 0) and B(36,15) as the positions of two towns.

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let the points be A( 1, 5), B(2, 3) and C(-2, -11). A, B and C are collinear, if
AB + BC = AC or AC + CB = AB or BA + AC = BC

Here, AB + BC ≠ AC, AC + CB ≠ AB, BA + AC ≠ BC
∴ A, B and C are not collinear.

Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the points be A(5, -2), 6(6, 4) and C(7, -2).

We have AB = BC ≠ AC.
∴ ∆ABC is an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A B, Cand D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution:
Let the horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.
∴ The points are A(3, 4), B(6, 7), C(9, 4) and D(6, 1)


i.e., BD = AC ⇒ Both the diagonals are also equal.
∴ ABCD is a square.
Thus, Champa is correct.

Question 6.
Name the type of quadrilateral formed if any, by the following points, and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let the points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) of a quadrilateral ABCD.

⇒ AB = BC = CD = AD i.e., all the sides are equal.
Also, AC and BD (the diagonals) are equal.
∴ ABCD is a square.

(ii) Let the points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).


We see that \(\sqrt{13}+\sqrt{13}=2 \sqrt{13}\)
i. e., AC + BC = AB
⇒ A, B and C are collinear. Thus, ABCD is not a quadrilateral.

(iii) Let the points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).

Since, AB = CD, BC = DA i.e., opposite sides of the quadrilateral are equal.
And AC ≠ BD ⇒ Diagonals are unequal.
∴ ABCD is a parallelogram.

Question 7.
Find the point on x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
We know that any point on x-axis has its ordinate = 0
Let the required point be P(x, 0)
Let the given points be A(2, -5) and B(-2, 9)


∴ The required point is (-7, 0)

Question 8.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
The given points are P(2, -3) and Q(10, y).

Squaring both sides, y² + 6y + 73 = 100
⇒ y² + 6y – 27 = 0
⇒ y² – 3y + 9y – 27 = 0
⇒ (y – 3)(y + 9) = 0
Either y – 3 = 0 ⇒ y = 3
or y + 9 = 0 ⇒ y = -9
∴ The required value of y is 3 or -9.

Question 9.
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:

Squaring both sides, we have x² + 25 = 41
⇒ x² + 25 – 41 = 0
⇒ x² – 16 = 0 ² x = \(\pm \sqrt{16}\) = ±4
Thus, the point R is (4, 6) or (-4, 6)
Now,

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let the points be A(x, y), B(3, 6) and C(-3, 4).
∴ AB = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
And AC = \(\sqrt{[(-3)-x]^{2}+(4-y)^{2}}\)
Since, the point (x, y) is equidistant from (3, 6) and (-3, 4).
∴ AB = AC
⇒ \(\sqrt{(3-x)^{2}+(6-y)^{2}}=\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
Squaring both sides,
(3 – x)² + (6 – y)² = (-3 – x)² + (4 – y)²
⇒ (9 + x² – 6x) + (36 + y² – 12y)
⇒ (9 + x² + 6x) + (16 + y² – 8y)
⇒ 9 + x² – 6x + 36 + y² – 12y – 9 – x² – 6x – 16 – y² + 8y = 0
⇒ – 6x – 6x + 36 – 12y – 16 + 8y = 0
⇒ – 12x – 4y + 20 = 0
⇒ -3x – y + 5 = 0
⇒ 3x + y – 5 = 0 which is the required relation between x and y.

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of boys = x and Number of girls = y
∴ x + y = 10 …. (1)
Also, Number of girls = [Number of boys] + 4
∴ y = x + 4 ⇒ x – y = – 4 … (2)
Now, from the equation (1), we have :
l₁ : x + y = 10 ⇒ y = 10 – x

And from the equation (2), we have
l₂ : x – y = -4 ⇒ y = x + 4

The lines l₁ and l₂ intersects at the point (3, 7)
∴ The solution of the pair of linear equations is : x = 3, y = 7
⇒ Number of boys = 3
and number of girls = 7
Let the cost of a pencil = ₹ x
and cost of a pen = ₹ y
Since, cost of 5 pencils + Cost of 7 pens = ₹ 50
5 x + 7y = 50 … (1)
Also, cost of 7 pencils + cost of 5 pens = ₹ 46
7x + 5y = 46 …. (2)
Now, from equation (1), we have

And from equation (2), we have

Plotting the points (10, 0), (3, 5) and (-4, 10), we get a straight line l₁ and plotting the points (8, -2), (3, 5) and (0, 9.2) we get a straight line l₂.

These two straight lines intersects at (3, 5).
∴ Cost of a pencil = ₹ 3 and cost of a pen = ₹ 5.

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
Comparing the given equations with
a₁x + b₁y + C₁ = 0, a₂x + b₂y + c₂ = 0, we have

(i) For, 5x – 4y + 8 = 0, 7x + 6y – 9 = 0,
a₁ = 5, b₁ = -4, C₁ = 8; a₂ = 7, b₂ = 6, c₂ = -9

So, the lines are intersecting, i.e., they intersect at a point.

(ii) For, 9x + 3y + 12 = 0,18x + 6y + 24 = 0,
a₁ = 9, b₁ = 3, C₁ = 12; a₂ = 18, b₂ = 6, c₂ = 24

So, the lines are coincident.

(iii) For, 6x – 3y +10 = 0, 2x – y + 9 = 0

So, the lines are parallel.

Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y = 8; 4x – 6y = 9
(iii) \(\frac{3}{2} x+\frac{5}{3} y\) = 7; 9x-10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
Solution:
Comparing the given equations with

So, lines are intersecting i.e., they intersect at a point.
∴ It is consistent pair of equations.

(ii) For 2x – 3y = 8, 4x – 6y = 9, 2x – 3y – 8 = 0 and 4x – 6y – 9 = 0

So, lines are parallel i.e., the given pair of linear equations has no solution.
∴ It is inconsistent pair of equations.

(iii)

⇒ The given pair of linear equations has exactly one solution.
∴ It is a consistent pair of equations.

(iv)

So, lines are coincident.
⇒ The given pair of linear equations has infinitely many solutions.
Thus, it is consistent pair of equations

(v)

So, the lines are coincident.
⇒ The given pair of linear equations have infinitely many solutions.
Thus it is consistent pair of equations.

Question 4.
Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5; 2x + 2y = 10
(ii) x – y = 8; 3x – 3y = 16
(iii) 2x + y – 6 = 0; 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
(i) For, x + y = 5, 2x + 2y = 10 ⇒ x + y – 5 = 0 and 2x + 2y – 10 = 0


So, lines l₁ and l₂ are coinciding. i.e., They have infinitely many solutions.

(ii) For, x – y = 8,
3x – 3y = 16
⇒ x – y – 8 = 0
and 3x – 3y = 16

so, line are parallel
∴ The pair of linear equations are inconsistent

(iii) For 2x + y – 6 = 0, 4x – 2y – 4 = 0


The pair of linear equations are inconsistent

Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the width of the garden = x m
and the length of the garden = y m
According to question,


The lines l₁ and l₂ intersect at (16, 20).
∴ x = 16 and y = 20
So, Length = 20m, and Width = 16m

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Let the pair of linear equations be 2x + 3y – 8 = 0
⇒ a₁ = 2, b₁ = 3 and C₁ = -8 and a₂x + b₂y + c₂ = 0.

(i) For intersecting lines, we have:

∴ We can have a₂ = 3, b₂ = 2 and c₂ = – 7
∴ The required equation can be 3x + 2y – 7 = 0

(ii) For parallel lines, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ Any line parallel to 2x + 3y – 8 = 0, can be taken as 2x + 3y -12 = 0

(iii) For coincident lines, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ Any line parallel to 2x + 3y – 8 = 0 can be 2(2x + 3y – 8 = 0)
⇒ 4x + 6y – 16 = 0

Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
Since, we have

The lines l₁ and l₂ intersect at (2, 3). Thus, co-ordinates of the vertices of the shaded triangular region are (4, 0), (-1, 0) and (2, 3).

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\) = -1 and \(x-\frac{y}{3}\) = 3
Solution:
(i) Elimination method :
x + y = 5 … (1)
2x – 3y = 4 …. (2)
Multiplying (1) by 3, we get
3x + 3y = 15 …. (3)
Adding (2) and (3) , we get
2x – 3y + 3x + 3y = 19
⇒ 5x = 19 ⇒ x = \(\frac{19}{5}\)
Now, putting x = \(\frac{19}{5}\) in (1) , we get

Substitution Method :
x + y = 5 ⇒ y = 5 – x … (1)
2x – 3y = 4 … (2)
Put y = 5 – x in (2), we get
2x – 3(5 – x)= 4 ⇒ 2x – 15 + 3x = 4

(ii) Elimination method :
3x + 4y = 10 … (1)
2x – 2y = 2 … (2)
Multiplying equation (2) by 2, we have
⇒ 4x – 4y = 4 … (3)
Adding (1)and (3) , we get
3x + 4y + 4x – 4y = 10 + 4
⇒ 7x = 14 ⇒x = \(\frac{14}{7}\) = 2
Putting x = 2 in (1) , we get,
3(2)+ 4y = 10

(iii) Elimination method :
3x – 5y – 4 = 0 … (1)
9x = 2y + 7 ⇒ 9x – 2y – 7 = 0 …. (2)
Multiplying equation (1) by (3) , we get
⇒ 9x – 15y – 12 = 0(3)
Subtracting (2)from (3),
⇒ 9x – 15y – 12 – 9x + 2y + 7 = 0

(iv) Elimination method :

Question 2.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces 1
to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i)Let the numerator = x
and the denominator = y

(ii) Let, present age of Nuri = x years
and the present age of Sonu = y years
Age of Nuri = (x – 5) years
Age of Sonu = (y – 5) years
According to question:
Age of Nuri = 3[Age of sonu]
⇒ x – 5 = 3[y – 5] ⇒ x – 5 = 3y – 15
⇒ x – 3y + 10 = 0 … (1)
10 years later :
Age of Nuri = (x + 10) years,
Age of Sonu = (y + 10) years,
According to question :
Age of Nuri = 2[Age of Sonu]
⇒ x + 10 = 2(y + 10) ⇒ x + 10 = 2y + 20
⇒ x – 2y -10 = 0 …. (2)
Subtracting (1) from (2),
x – 2y – 10 – x + 3y – 10 = 0 ⇒ y – 20 = 0 ⇒ y = 20
Putting y = 20 in (1), we get
x – 3(20) + 10 = 0 ⇒ x – 50 = 0 ⇒ x = 50
Thus, x = 50 and y = 20
∴ Age of Nuri = 50 years
and age of Sonu = 20 years

(iii) Let the digit at unit’s place = x
and the digit at ten’s place = y
∴ The number = 10y + x
The number obtained by reversing the digits = 10x + y
According to question,
9[The number] = 2 [Number obtained by reversing the digits]
9[10y+ x] = 2[10x + y]
90y + 9x = 20x + 2y
x – 8y = 0 …. (1)
Also sum of the digits = 9
x + y = 9 …. (2)
Subtracting (1) from (2), we have
x + y – x + 8y = 9
⇒ 9y = 9 ⇒ y = 1
putting y = 1 in (2), we get
x + 1 = 9 ⇒ x = 8
Thus, x = 8 and y = 1
∴ The required number = 10y + x = (10 × 1) + 8 = 10 + 8 = 18

(iv) Let the number of 50 rupees notes = x
and the number of 100 rupees notes = y
According to the condition,
Total number of notes = 25
∴ x + y = 25 …. (1)
Also, the value of all the notes = ₹ 2000
∴ 50x + 100y = 2000 ⇒ x + 2y = 40 …. (2)
Subtracting equation (1) from (2), we get
x + 2y – x – y = 40 – 25
⇒ y = 15
Putting y = 15 in (1),
x + 15 = 25
⇒ x = 25 – 15 = 10
Thus, x = 10 and y = 15
∴ Number of 50 rupees notes = 10
and number of 100 rupees notes = 15

(v) Let the fixed charge (for the three days) = ₹ x
and the additional charge for each extra day = ₹ y
According to question,
Charge for 7 days = ₹ 27
⇒ x + 4y = 27
[∵ Extra days = 7 – 3 = 4]
Also, charge of 5 days = ₹ 21
⇒ x + 2y = 21
[∵ Extra days = 5 – 3 = 2]
Subtracting (2) from (1), we get
x + 4y – x – 2y = 27 – 21
\(\Rightarrow \quad 2 y=6 \Rightarrow y=\frac{6}{2}=3\)
Putting y = 3 in (2), we have
x + 2(3) = 21
⇒ x = 21 – 6 = 15
So, x = 15 and y = 3
∴ Fixed charge = ₹ 15
and additional charge per day = ₹ 3

MP Board Class 10th Science Solutions Chapter 13 Magnetic Effects of Electric Current

MP Board Class 10th Science Chapter 13 Intext Questions

Class 10th Science Chapter 13 Intext Question Page No. 224

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle get deflected when brought near a bar magnet because a compass needle is in fact, a small bar magnet. The ends of the compass needle point approximately towards north and south directions.

Class 10th Science Chapter 13 Intext Questions Page No. 228

Question 1.
Draw magnetic field lines around a bar magnet.
Answer:
Magnetic field lines of a bar magnet emerge from the north pole and terminate at the south pole. Inside the magnet, the field lines emerge from the south pole and terminate at the north pole.

Question 2.
List the properties of magnetic field lines.
Answer:
The properties are:

• They travel from north pole to south pole outside the magnet and south pole to north pole inside the magnet.
• They are closed and continuous curves.
• Two magnetic field lines never intersect each other. If the lines intersect, then at the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic fields which is not possible.
• The number of field lines per unit area is the measure of the strength of magnetic field, which is maximum at poles. The magnetic field is strong, where the field lines are close together and weak where the lines are far apart.

Question 3.
Why don’t two magnetic field lines intersect each other?
Answer:
Two magnetic fields lines of force never intersect each other. If the lines intersect, then at [the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic field, which is not possible.

Class 10th Science Chapter 13 Intext Questions Pages No. 229. 230

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
Inside the loop = Pierce inside the table.

Outside the loop = Appear to emerge out from the table.

For downward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop. Similarly, for upward direction of current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging from the table outside the loop and merging in the table inside the loop, as shown in the given figure.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
The magnetic field lines are parallel and equidistant.

Question 3.
Choose the correct option.
The magnetic field inside a long, straight, solenoid-carrying current
(a) is zero
(b) decreases as we move towards its end
(c) increases as we move towards its end
(d) is the same at all points
Answer:
(d) The magnetic field inside a long, straight, current-carrying solenoid is uniform. It is the same at all points inside the solenoid.

Class 10th Science Chapter 13 Intext Questions Pages No. 231,232

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity
(d) momentum.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if

1. current in rod AB is increased;
2. a stronger horse-shoe magnet is used; and
3. length of the rod AB is increased?

Answer:

1. displacement of A is increased.
2. If a stronger horse-shoe magnet is used magnetic field is increasing.
3. current flows is more.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward.
Since the positively charged particle alpha particle projected towards west, so the direction of current is towards west. Now the deflection is towards north, so the force is towards north. Now hold the forefinger, centre finger and thumb of our left – hand at right angles to one another. Let us adjust the hand in such a way that our centre finger points towards west and thumb points towards north. If we look at our forefinger, it will be pointing, upward. Thus, the magnetic field is in the upward direction. So, the correct answer is (d).

Class 10th Science Chapter 13 Intext Questions Page No. 233

Question 1.
State Fleming’s left-hand rule.
Answer:
According to this rule, stretch the thumb, forefinger, and middle finger of your left hand such that they are mutually perpendicular. If the first finger points in the direction of the Magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
A current-carrying conductor when placed in a magnetic field experiences a force. This is the principle of an electric motor.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring reverse the direction of current in the armature coil after every half rotation i.e., it acts as a commutator. The reversal of current reverses, the direction of the forces acting on the two arms of the armature after every half rotation. This allows the armature coil to rotate continuously in the same direction.

Class 10th Science Chapter 13 Intext Question Page No. 236

Question 1.
Explain different ways to induce current in a coil.
Answer:
The different ways to induce current in a coil are as follows:

(a) If a coil is moved rapidly between the two poles of a horse-shoe magnet, then an electric current is induced in the coil.
(b) If a magnet is moved relative to coil, then an electric current is induced in the coil.

Class 10th Science Chapter 13 Intext Questions Page No. 237

Question 1.
State the principle of an electric generator.
Answer:
An electric generator works on the principle of electromagnetic induction. It generates electricity by rotating a coil in a magnetic field.

Question 2.
Name some sources of direct current.
Answer:
Some sources of direct current are cell, Dc generator, etc.

Question 3.
Which sources produce alternating current?
Answer:
AC generators, power plants etc., produce alternating current.

Question 4.
Choose the correct option:
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution
Answer:
(c) When a rectangular coil of copper is rotated in a magnetic field, the direction of the induced current in the coil changes once in each half revolution. As a result, the direction of current in the coil remains the same.

Class 10th Science Chapter 13 Intext Questions Pages No. 238

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

1. Electric fuse
2. Earthing wire.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
P = VI
Here P = 2 KW = 2000 W
V = 220

The current drawn by this electric oven is 9 A whereas the fuse in the circuit is ( only 5 A capacity. When a high current of 9 A flows through the 5 A fuse, the fuse wire will get heated too much, melt and break, the circuit. Therefore, when a 2 kW power rating electric oven is operated in a circuit having a 5 A fuse will blow off cutting off the power supply in this circuit.

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:

1. Each appliance has a separate switch to ON/OFF the flow of current through it.
2. The use of an electric fuse prevents the electric circuit and the appliance from possible damage by stopping the flow of unduly high electric current.
3. We should not connect too many appliances to a single socket to prevent overloading.

MP Board Class 10th Science Chapter 13 NCERT Textbook Exercises

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire
(b) The field consists of straight lines parallel to the wire
(c) The field consists of radial lines originating from the wire
(d) The field consists of concentric circles centered on the wire.
Answer:
(d) The magnetic field lines, produced around a straight current-carrying conductor are concentric circles. Their centres lie on the wire.

Question 2.
The phenomenon of electromagnetic induction is
(a) the process of charging a body
(b) the process of generating magnetic field due to a current passing through a coil
(c) producing induced current in a coil due to relative motion between a magnet and the coil
(d) the process of rotating a coil of an electric motor.
Answer:
(c) When a straight coil and a magnet are moved relative to each other, a current is induced in the coil. This phenomenon is known as electromagnetic induction.

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electro-magnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(c) AC generator will generate a higher voltage.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.
Answer:
(a) False
(b) true
(c) true
(d) False.

Question 7.
List two methods of producing magnetic fields.
Answer:

1. Permanent magnet
2. Electromagnet.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:
A solenoid is a long coil of circular loops of insulated copper wire. Magnetic field lines are produced around the solenoid when a current is allowed to flow through it. The magnetic field produced by it is similar to the magnetic field of a bar magnet. The field lines produced in a current¬carrying solenoid is shown in the following figure.

Fig. 13.4: Field lines of the magnetic field through and around a current carrying solenoid.

In the above figure, when the north pole of a bar magnet is brought near the end connected to the negative terminal of the battery, the solenoid repels the bar magnet. Since like poles repel each other, the end connected to the negative terminal of the battery behaves as the north pole of the solenoid and the other end behaves as a south pole. Hence, one end of the solenoid behaves as a north pole and the other end behaves as a south pole.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
The force experienced by a current -carrying conductor is the maximum when the direction of current is perpendicular to the direction of the magnetic field.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
According to Fleming’s left-hand rule, the magnetic field acts in the vertically downward direction.
Note that the direction of current will be opposite to that of the electron beam.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:
An electric motor converts electrical energy into mechanical energy. It works on the principle of the magnetic effect of current. A current-carrying coil rotates in a magnetic field. The following figure shows a simple electric motor. When a current is allowed to flow through the coil MNST by closing the switch, the coil starts rotating anti-clockwise. This happens because a downward force

acts on length MN and at the same time, an upward force acts on length ST. As a result, the coil rotates anti-clockwise.

Current in the length MN flows from M to N and the magnetic field acts from left to right, normal to length MN. Therefore, according to Fleming’s left hand rule, a downward force acts on the length MN. Similarly, current is the length ST. flows from S to T and the magnetic fields acts from left to light, normal to the flow of current. Therefore, an upward force acts on the length ST. These two forces cause the coil to rotate anti-clockwise.

After half a rotation, the position of MN and ST interchange. The half¬ring D comes in contact with brush A and half-ring C comes in contact with brush B. Hence, the direction of current in the coil MNST gets reversed.

The current flow’s through the coil in the direction TSNM. The reversal of current through the coil MNST repeats after each half rotation. As a result, the coil rotate unidirectional ly. The split rings help to reverse the direction of current in the circuit. These are called the commutator.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motor is used as an important component in electric fans, refrigerators, mixers, washing machines, computers, MP3 players etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is

1. pushed into the coil
2. withdrawn from inside the coil
3. held stationary inside the coil?

Answer:

1. There is a momentary deflection in the needle of the galvanometer.
2. Now the galvanometer is deflected towards the left showing that the current is now set up in the direction opposite to the first.
3. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in coil B? Give reason.
Answer:
If the current in the coil A is changed there is a change in its magnetic field. By this electricity is induced in B. This is called Electromagnetic induction.

Question 15.
State the rule to determine the direction of a

1. magnetic field produced around a straight conductor carrying current,
2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
3. current induced in a coil due to its rotation in a magnetic field.

Answer:
(i) Right-hand thumb rule: If the current-carrying conductor is held in the right hand such that the thumb points in the direction of the current, then the direction of the curl of the fingers will be given the direction of the magnetic field.

(ii) Fleming’slefthandrule: Stretch the forefinger, the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger in the direction of the current, then the thumb points in the direction of a force in the conductor.

(iii) Fleming’s right-hand rule: Stretch the thumb/ forefinger and the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, thumb in the direction conductor, then the central finger points in the direction of current induced in the conductor.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?
Answer:
An electric generator converts mechanical energy into electrical energy. The principle of working of an electric generator is that when a loop is moved in a magnetic field, an electric current is induced in the coil, ft generates electricity by rotating a coil in a magnetic field. The following figure shows a simple AC generator.

• MNST → Rectangular coil
• C and D → Two slip rings
• A and B → Brushes
• X → Axle, G → Galvanometer

If axle X is rotated clockwise, then the length MN moves upwards while length ST moves downwards. Since the lengths MN and ST are moving in a magnetic field, a current will be induced in both of them due to electromagnetic induction. Length MN is moving upwards and the magnetic field acts from left to right. Hence, according to Fleming’s right hand rule, the direction of induced current will be from M to N. Similarly, the direction of induced current in the length ST will be from S to T.

The direction of current in the coil is MNST. Hence, the galvanometer shows a deflection in a particular direction. After half a rotation, length MN starts moving down whereas length ST starts moving upward. The direction of the induced current in the coil gets reversed as TSNM. As the direction of current gets reversed after each half rotation the produced current is called an alternating current (AC).

To get a unidirectional current, instead of two slip rings, two split rings are used, as shown in the following figure.

In this arrangement, brush ‘A’ always remains in contact with the length of the coil that is moving up whereas brush B always remains in contact with the length that is moving down. The split rings C and D act as a commutator.

The direction of current induced in the coil will be MNST for the first rotation and TSNM in the second half of the rotation. Hence, unidirectional current is produced from the generator called DC generator. The current is called AC current.

Question 17.
When does an electric short circuit occur?
Answer:
If the resistance of an electric circuit becomes very low. Then the current flowing through the circuit becomes very high. This is caused by connecting too many appliances to a single socket or connecting – high power rating appliances to the light circuits. This results in a short circuit when the insulation of live and neutral wires undergoes wear and tear and then touches each other, the current flowing in the circuit increase abruptly. Hence, a short circuit occurs.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
The metallic body of electric appliances is connected to the earth by means of earth wire so that any leakage of electric current is transferred to the ground. This prevents any electric shock to the user. That is why earthing of the electrical appliances is necessary.

MP Board Class 10th Science Chapter 13 Additional Important Questions

MP Board Class 10th Science Chapter 13 Multiple Choice Questions

Question 1.
Two magnet when come closer:
(a) Attract each other
(b) Repei each other
(c) Sometimes attract and sometimes repel
(d) No reaction
Answer:
(c) Sometimes attract and sometimes repel

Question 2.
Permanent magnet can be made by:
(a) Gold
(b) Carbon
(c) Alnico
(d) Wood
Answer:
(c) Alnico

Question 3.
When we draw magnetic field lines of any magnet it:
(a) Begin from N pole and end at S pole
(b) Begin from S pole and end at N pole
(c) Form circles around a magnet
(d) Form box around a magnet
Answer:
(a) Begin from N pole and end at S pole

Question 4.
Strongest magnetic pole around a magnet is
(a) Near North pole
(b) Near south pole
(c) In center of magnet
(d) At both poles
Answer:
(d) At both poles

Question 5.
Electro magnetism was discovered by
(a) Newton
(b) Oersted
(c) Ohm
(d) Joule
Answer:
(b) Oersted

Question 6.
Electro magnets are used in
(a) AC
(b) Fridge
(c) Radio
(d) All of these
Answer:
(d) All of these

Question 7.
Magnetic field are lines forming
(a) Straight lines
(b) Closed curves
(c) Dotted lines
(d) Dotted triangles
Answer:
(b) Closed curves

Question 8.
Magnetic field of a straight conductor form:
(a) Straight lines
(b) Concentric lines
(c) Points
(d) None of above
Answer:
(b) Concentric lines

Question 9.
According to right hand thumb rule, current is generated in a system:
(a) Parallel to magnetic field
(b) Perpendicular to magnetic field
(c) Just above the magnetic field
(d) All of these.
Answer:
(b) Perpendicular to magnetic field

Question 10.
If a circular loop conductor has 5 turn, magnetic field produced by it will be to than single loop conductor.
(a) 5 times bigger
(b) 5 times lesser
(c) Similar
(d) 10 times lesser
Answer:
(a) 5 times bigger

Question 11.
Magnetic field generated due to a passing current in a solenoid forms pattern:
(a) Similar to circular loops
(b) Similar to straight conductor
(c) Similar to Bar magnet
(d) Which is unique
Answer:
(c) Similar to Bar magnet

Question 12.
According to Fleming’s left hand rule
(a) Field is perpendicular to current
(b) Field is perpendicular to force generated
(c) Current is perpendicular to force generated .
(d) All of these
Answer:
(d) All of these

MP Board Class 10th Science Chapter 13 Very Short Answer Type Questions

Question 1.
What we call the end, where north end of freely hanged magnet stops?
Answer:
North pole.

Question 2.
Where force of a magnet can be detected?
Answer:
Near a magnet, in its magnetic field.

Question 3.
How we mark a magnetic force?
Answer:
With the help of magnetic field lines.

Question 4.
What do closer field lines of a magnet presents?
Answer:
Closer lines present stronger magnetic strength.

Question 5.
What is reason for different kinds of pattern of a magnetic pattern around a conductor generated magnetic field?
Answer:
Shape of conductor.

Question 6.
Give some examples of daily use where electro magnet is being used.
Answer:
Radio and Television.

Question 7.
How an electromagnet is formed?
Answer:
Electro magnet is formed by wrapping coil of insulated copper wire over core of soft iron.

Question 8.
What is experienced by a current carrying conductor when placed in a magnetic field?
Answer:
Force.

Question 9.
Who gave the theory of induced electromagnetic induction?
Answer:
Michael Faraday.

Question 10.
What is a galvanometer?
Answer:
An instrument which detects current in a circuit.

Question 11.
Give examples of appliances using electric motor.
Answer:
Electric pan, mixer, washing machines.

Question 12.
Which rule of electromagnetism suits best to express working of an electric motor?
Answer:
Fleming’s left hand rule.

Question 13.
What is the basic principle used in electric generator?
Answer:
It is based on electromagnetic induction.

Question 14.
What is the permissible value of electricity use in India?
Answer:
220 V, 50Hz.

Question 15.
How short-circuit can be prevented?
Answer:
By using fuse.

MP Board Class 10th Science Chapter 13 Short Answer Type Questions

Question 1.
Write two properties of magnet.
Answer:

1. A magnet always points north or south when suspended freely.
2. Like pole repel each other while opposite pole attract each other.

Question 2.
Define magnetism.
Answer:
A magnet influence its near by object and attracts towards itself, if magnetic in nature. This phenomenon is called magnetism.

Question 3.
Define magnetic field.
Answer:
Area under which a magnet can influence other magnetic objects is called its magnetic field.

Question 4.
Where we find closer line and what does it indicate?
Answer:
Closer lines show high strength of magnetic field while wide lines represents weak strength of magnet.

Question 5.
State Right hand thumb rule.
Answer:
Right hand thumb rule is represented by a down thumb fist in which fist represents a magnetic field while thumb represents current movement.

MP Board Class 10th Science Chapter 13 Long Answer Type Questions

Question 1.
An electric heater rated 800 W operates 6h/day. Find the cost of energy to operate it for 30 days at ₹ 3.00 per unit.
Solution:
Power of the heater, P = 800 W Time, t = 6 hour /day
No; of days, n = 30 Cost per unit = ₹ 3.00
Total cost of its usage = ? Energy, E = P × t
Consumed in 1 day = 800 × 6 = 4800 Wh
Energy consumed in 30 days = 4800 × 30 = 144000 Wh

∴ Cost of 144 units = 3 × 144 = ₹ 432.

Question 2.
(a) Draw magnetic field lines produced around a current carrying straight conductor passing through cardboard. How will the strength of the magnetic field change, when the point where magnetic field is to be determined, is moved away from the straight wire carrying constant current ? Justify your answer.
(b) Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
(a) (i) The magnetic field lines around a straight conductor carrying current are concentric circles whose center lies on the wire.
(ii) When a point where magnetic field is to be determined is moved away from the straight wire, the strength of the magnetic field decreases because as we move away from a current carrying straight conductor, the concentric circles around it representing magnetic field lines become larger and larger indicating the decreasing strength of magnetic field.

Fig. 13.18: A pattern of concentric circles indicating the field lines of a magnetic field a straight conducting wire.

(b) Yes, current is induced in the coil B. Because as the current in the coil A changes, the magnetic field lines around the coil B also change. Therefore, the change in magnetic field lines associated with the coil B is the cause of induced electric current in it.

Question 3.
(a) Draw magnetic field lines of a bar magnet. “Two magnetic field lines never intersect each other.” Why?
(b) An electric oven of 1.5 kW is operated in a domestic circuit (220 V) that has a current rating of 5 A. What result do you expect in this case? Explain.
Answer:
(a)

Two magnetic field lines do not intersect one another. The direction of magnetic field lines is always from north pole to south pole. If the two magnetic field line do intersect, it means at the point of intersection the compass needle is showing two different directions which is not possible.

(b) Power, P = 1.5 kW = 1.5 × 1,000 = 1,500 W
Voltage, V = 220 V, I = ?
P = V × I I = \(\frac { P }{ V } \) = \(\frac { 1,500 }{ 220 } \) = 6.8 A
Now, the current drawn by the oven is 6.8 A which is very high but the fuse in this circuit is only 5 A capacity. When a very high current of 6.8. A flows through 5 A fuse, the fuse wire will get heated too much, melt and break the circuit, cutting off the power supply.

Question 4.
A circuit has a line of 5 A. How many lamps of rating 40W; 220V can simultaneously run on this line safely?
Answer:

Question 5.
A bulb is rated at 200 V, 100 W. Calculate its resistance. Five such bulbs burn for 4 hours daily. Calculate the units of electrical energy consumed per day. What would be the cost of using these bulbs per day at the rate of ₹ 4.00 per unit?
Solution:

Electrical energy consumed, E = P × t
Energy consumed by 1 bulb = 0.1 × 4 = 0.4 kWh
∴ Energy consumed by 5 bulb = 5 × 0.4 = 2 kWh = 2 units
Cost of electrical energy:
Cost of 1 unit of electricity = ₹ 4
∴ Cost of 2 units of electricity = 4 × 2 = ₹ 8

Question 6.
(a) Describe an activity to show with the help of a compass that magnetic field is strongest near poles of a magnet.
(b) Mention the direction of magnetic field lines (i) inside a bar magnet and (ii) outside a bar magnet.
Answer:
(a) A bar magnet is placed on a sheet of paper and its boundary is marked with a pencil. A magnetic compass is brought near the N-pole of the bar magnet. It is observed that N-pole of magnet repels the N-pole of compass needle due to which the tip of the compass needle moves away from the N-pole. Thus, a magnetic field pattern is obtained around a bar magnet.

(b) Each magnetic field line is directed from the north pole of a magnet to its south pole. The field lines are closest together at the two poles of the bar magnet.

1. Inside a bar magnet, the lines of forces start from south pole and end on north pole.
2. Outside a bar magnet, magnetic lines of forces start from north pole and end on south pole.

Question 7.
(a) With the help of a labelled diagram, describe an activity to show that a current carrying conductor experiences a force when placed in a magnetic field. Mention the position when this force is maximum.
(b) Name and state the rule which gives the direction of force acting on the conductor.
Answer:
(a) Activity:

• A small aluminium rod (AB) about 5 cm is suspended with two connecting wires horizontally from a stand.
• A strong horse-shoe magnet is placed in such a way that the rod lies between the two poles with the magnetic field directed upwards, the north pole of the magnet vertically below and south pole vertically above the aluminium rod.
  
  Fig. 13.20: A current-carrying conductor experiences force in magnetic field.

• The aluminium rod is connected in series with a battery, a key and a rheostat.
• When a current is allowed to pass through aluminium rod. Form end B to end A, it is observed that the rod is displaced towards the left.
• When the direction of the current is reversed from A to B, it is observed that the direction of displacement- of the rod is towards the right.

This activity shows that when a current carrying conductor is placed in a magnetic field, a mechanical force is exerted on conductor which makes it move.

The maximum force is exerted on a current carrying conductor only when it is perpendicular to the-direction of magnetic field.

(b) The direction of force acting on the current carrying conductor can be found out by using Fleming’s left-land rule.

According to Fleming’s left-hand rule, hold the fore finger, the center finger and the thumb of your left hand at right angles to one another. If the first finger of your left hand points in the direction of magnetic field and sound in the direction of current, then the thumb will point in the direction of motion or the force acting on conductor.

Question 8.
Study the following current-time graphs from two different sources:

1. Use above graphs to list two differences between the current in the two cases.
2. Name the type of current in the two cases.
3. Identify one source each for these currents.
4. What is meant by the statement that “ the frequency of current in India is 50 Hz”?

Answer:

1. D.C. – Direct Current.
2. A.C.- Alternating Current.
3. Source of D.C. → a cell, battery, solar cell, D.C. generator.
   Source of A.C. → A.C. Generator.
4. The frequency of current in India is 50 Hz means the direction of current in India changes 50 times in 1 second.

Question 9.
Explain two disadvantages of series arrangement for household circuit.
Answer:
Disadvantages of series circuits for domestic wiring:

1. In series circuit, if one electrical appliance stops working due to some defect then all other appliances also stop working because the whole circuit is broken.
2. In series circuit, all the electrical appliances have only one switch , due to which they cannot be turned off or turned on separately.

Question 10.
(i) State Maxwell’s right-hand thumb rule.
(ii) PQ is a current carrying conductor in the plane of the paper as shown in the figure. Mention the direction of magnetic fields produced by it at points A and B.
Given r₁ < r₂ where will the strength of the magnetic field be larger?
Answer:
(i) Maxwell’s right hand thumb rule: The direction of the current is given by Maxwell’s right hand thdmb rule, “If the current carrying conductor is gripped with the right hand in such a way that the thumb gives the direction of the current, then the direction of the fingers gives the direction of the magnetic field produced around the conductor”.

(ii) Since the direction of current in the straight conductor is from Q to P, then according to Maxwell’s right hand thumb rule, magnetic field at point A is inside the paper and at point B is outside the paper. Since r₁ < r₂ the strength of the magnetic field at A is more than at B because greater the
distance of a point from the current carrying wire, weaker will be the magnetic field produced at that point.

MP Board Class 10th Science Chapter 13 NCERT textbook activities

Class 10 Science Activity 13.1 Page No. 223

• Take a straight thick copper wire and place it between the points X and Y in an electric circuit, as shown in Fig. 13.8. The wire XY is kept perpendicular to the plane of paper.
  
  Compass needle is deflected on passing an electric current through a metallic conductor.
• Horizontally place a small compass near to this copper wire. See the position of its needle.
• Pass the current through the circuit by inserting the key into the plug.
• Observe the change in the position of the compass needle.

Observations:

• The needle is deflected showing that electric current through the copper wire has produced a magnetic effect.

Class 10 Science Activity 13.2 Page No. 224

• Fix a sheet of white paper on a drawing board using some adhesive material.
• Place a bar magnet in the centre of it.
• Sprinkle some iron filings uniformly around the bar magnet (Fig. 13.9). A salt – sprinkler may be used for this purpose.
• Now tap the board gently.
• What do you observe?
  
  Fig. 13.9: Iron filings near the bar magnet align themselves along the field lines

Observations:

• It is observed that iron filings arrange themselves in a pattern of concentric circles. This shows that iron filings experience a force due to magnetic effect of a magnetic. The lines along which the iron filings align themselves are magnetic field lines.

Class 10 Science Activity 13.3 Pages No. 224,225

• Take a small compass and a bar magnet
• Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.
• Mark the boundary of the magnet.
  
  Fig. 13.10: Drawing a magnetic field line with the help of a compass needle.
• Place the compass near the north pole of the magnet. How does it behave? The south pole Fig 13 10: Drawing a magnetic field line of the needle points with the help of a compass needle, towards the north pole of the magnet. The north pole of the compass is directed away from the north pole of the magnet.
• Mark the position of two ends of the needle.
• Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole.
• In this way, proceed step by step till you reach the south pole of the magnet as shown in Fig. 13.10.
  
  Fig. 13.11: Field lines around a bar magnet.
• Join the points marked on the paper by a smooth curve. This curve represents a field line.
• Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in Fig. 13.11. These lines represent the magnetic field around the magnet. These are known as magnetic field lines.
• Observe the deflection in the compass needle as you move it along a field line. The deflection increases as the needle is moved towards the poles.

Observations:

• The magnetic field is strong at the poles due to which deflection increases at the poles as the needle move towards it.

Class 10 Science Activity 13.4 Page No. 226

• Take a long straight copper wire, two or three cells ofl .5 V each, and a plug key. Connect all of them in scries as shown in Fig. 13.12 (a)
• Place the straight wire parallel to and over a compass needle.
• Plug the key in the circuit.
• Observe the direction of deflection of the north pole of the needle. If the current flows from north to south, as shown in Fig. 13.12 (a), the north pole of the compass needle would move towards the east.
• Replace the cell connections in the circuit as shown in Fig. 13.12 (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.
• Observe the change in the direction of deflection of the needle You will see that now the needle moves in opposite direction, that is, towards the west (Fig. 13.12 (b)). It means that the direction of magnetic field produced by the electric current is also reversed.
  Fig. 13.12: A simple electric circuit in which a straight copper wire is placed parallel to and over a compass needle. The deflection in the needle becomes opposite when the direction of the current is reversed.

Observations:

• As current flow changes its direction from south to north, the needle in the compass moves in a opposite direction that is towards the west. This shows the direction of magnetic filed produced by the electric current is also reversed.

Class 10 Science Activity 13.5 Pages No. 226, 227

• Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0 5 A), a plug key, connecting wires and a long straight thick copper wire.
• Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.
  Fig. 13.13: (a) A pattern of concentric circles indicating the field lines of a magnetic field around a straight conducting wire. The arrows in the circles show the direction of the field lines. (b) A close up of the pattern obtained.

• Connect the copper wire vertically between the points X and Y, as shown in Fig. 13.13 (a), in series with the battery, a plug and key.
• Sprinkle some iron filings uniformly on the cardboard. (You may use a salt sprinkler for this purpose).
• Keep the variable of the rheostat at a fixed position and note the . current through the ammeter.
• Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.
• Gently tap the cardboard a few times. Observe the pattern of the iron filings. You would find that the iron filings align themselves showing a pattern of concentric circles around the copper wire (Fig. 13.13).
• What do these concentric circles represent? They represent the magnetic field lines.
• How can the direction of the magnetic field be found? Place a compass at a point (say P) over a circle. Observe the direction of the needle. The direction of the north pole of the compass needle would give the direction of the field lines produced by the electric current through the straight wire at point P. Show the direction by an arrow.
• Does the direction of magnetic field lines get reversed if the direction of current through the straight copper wire is reversed? Check it.

Observations:

• The deflection is the needle changes. If the current is increased, the deflection also increases. It indicates the magnitude of the magnetic field produced at a given point increases as the current through the wire increases.

Class 10 Science Activity 13.6 Page No. 229

• Take a rectangular cardboard having two holes. Insert a circular coil having large number of turns through them, normal to the plane of the cardboard.
• Connect the ends of the coil in series with a battery, a key and a rheostat, as shown in Fig. 13.14.
• Sprinkle iron filings uniformly on the cardboard.
• Plug the key.
• Tap the cardboard gently a few times. Note the pattern of the iron filings that emerges on the cardboard.

Observations:

Fig. 13.14: Magnetic field produced by a current carrying circular coil.

Class 10 Science Activity 13.7 Page No. 230

• Take a small aluminium rod AB (of about 5 cm). Using two connecting wires suspend it horizontally from a stand, as shown in Fig. 13.15.
• Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and south pole vertically above the aluminium rod (Fig. 13.15).
  
  Fig. 13.15: A current carrying rod, AB, experiences a force perpendicular to its length and the magnetic field.
• Connect the aluminium rod in series with a battery, a key and a rheostat.
• Now pass a current through the aluminium rod from end B to end A.
• What do you observe? It is observed that the rod is displaced towards the left. Yr- will notice that the rod gets displaced.
• Reverse the direction of current flowing through the rod and observe the direction of its displacement. It is now towards the right.
  Why does the rod get displaced?

Observations:

• The rod is displaced due to force exerted on the current – carrying aluminium rod when placed in a magnetic field.
• The direction of force is recedes as the direction of anent through the induction is reversed.

Class 10 Science Activity 13.8 Pages No. 233,234

• Take a coil of wire AB having a large number of turns.
• Connect the ends of the coil to a galvanometer as shown in Fig. 13.16.
• Take a strong bar magnet and move its north pole towards the end B of the coil. Do you find any change in the galvanometer needle?
• There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.
• Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.
• Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. Wc sec that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly the needle moves toward left when the coil is moved aways.
  
  Fig. 13.16: Moving a magnet towards a coil sets up a current in the coil circuit, as indicated by deflection in the galvanometer needle.

• When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero. What do you conclude from this activity?

Observations:

• From this activity, it can be concluded that the motion of a magnet with respect to the coil produces an induced potential difference, which sets up an induced electric current in the circuit.

Class 10 Science Activity 13.9 Page No. 235

• Take two different coils of copper wire having large number of turns (say 50 and 100 turns respectively). Insert them over a non-conducting cylindrical roll, as shown in Fig. 13.17. (You may use a thick paper roll for this purpose.)
• Connect the coil-1, having larger number of turns, in series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.
• Plug in the key. Observe the galvanometer. Is there a deflection in its needle? You will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil-2.
• Disconnect coil-1 from the battery. You will observe that the needle momentarily moves, but to the opposite side. It means that now the current flows in the opposite direction in coil-2
  Fig. 13.17: Current is induced in coil-2 when current in coil-1 is changed.

Observations:

• We will observe that the needle of th e galvanometer instantly jumps to one side and just quickly returns to its initial position.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Solution:
(i) Here, dividend p(x) = x³ – 3x² + 5x – 3, and divisor g(x) = x² – 2
∴ We have

Thus, the quotient = (x – 3) and remainder = (7x – 9)

(ii) Here, dividend p(x) = x⁴ – 3x² + 4x + 5 and divisor g(x) = x² + 1 – x = x² – x + 1
∴ We have

Thus, the quotient = (x² + x – 3) and remainder = 8

(iii) Here, dividend, p(x) = x⁴ – 5x + 6 and divisor, g(x) = 2 – x² = – x² + 2
∴ We have

Thus, the quotient = -x² – 2 and remainder = -5x +10

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t² – 3; 2t⁴ + 3t³ -2t² – 9t – 12
(ii) x² + 3x + 1; 3x⁴ + 5x³ – 7x² + 2x +2
(iii) x³ – 3x + 1; x⁵ – 4x³ + x² + 3x + 1
Solution:
(i) Dividing 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we have

∵ Remainder = 0
∴ (t² – 3) is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) Dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we have

∵ Remainder = 0
∴ x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) Dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

∵ Remainder = 2, i.e., remainder = 0
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x +1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution:
We have p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5.
Given \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of p(x).

Thus, the other zeroes of the given polynomial are -1 and -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Solution:
Here, dividend, p(x) = x³ – 3x² + x + 2, divisor = g(x), quotient = (x – 2) and remainder = (-2x + 4)
Since, (Quotient × Divisor) + Remainder = Dividend
∴ [(x – 2) × g(x)] + [(-2x + 4)] = x³ – 3x² + x + 2
⇒ (x – 2) × g(x)
= x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
= x³ – 3x² + 3x – 2

Thus, the required divisor g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x) = 3x² – 6x + 27,
g(x) = 3 and q(x) = x² – 2x + 9.
Now, deg p(x) = deg q(x)
r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)

(ii) p(x) = 2x³ – 2x² + 2x + 3,
g(x) = 2x² – 1, and
r(x) = 3x + 2, deg q(x) = deg r(x)
⇒ p(x) = q(x) × g(x) + r(x)

(iii) p(x) = 2x³ – 4x² + x + 4,
g(x) = 2x² + 1,
q(x) = x – 2 and r(x) = 6,
deg r(x) = 0
⇒ p(x) = q(x) × g(x) + r(x)