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KSEEB SSLC Class 10 Science Solutions Chapter 3 Metals and Non-metals are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 3 Metals and Non-metals.

Karnataka SSLC Class 10 Science Solutions Chapter 3 Metals and Non-metals

KSEEB SSLC Class 10 Science Chapter 3 Intext Questions

Text Book Part I Page No. 40

Question 1.
Give an example of a metal which

1. is a liquid at room temperature
2. can be easily cut with a knife.
3. is the best conductors of heat.
4. is a poor conductor of heat.

Answer:

1. Metal that exists in a liquid state at room temperature → Mercury.
2. Metal that can be easily cut with a knife → Sodium.
3. Metal that is the best conductor of heat → Silver.
4. Metals that are poor conductors of heat → Lead.

Question 2.
Explain the meanings of malleable and ductile.
Answer:

1. Malleable: Materials that can be beaten into thin sheets are called malleable.
2. Ductile: Materials that can be drawn into thin wires are called ductile.

Metals can be hammered into thin sheets. This property of a metal is called malleability and the metals showing this property are called malleable. Gold, Silver, Copper, aluminium etc are malleable metals. Metals can be drawn into wires. The ability of metals to be drawn into thin wires is called ductility. Gold is the most ductile metal. It is interesting to know that a wire of about 2 km length can be drawn from one gram of gold.

Text Book Part I Page No. 46

Question 1.
Why is sodium kept immersed in kerosene oil?
Answer:
Sodium is a highly reactive element. If kept in open, it can react with oxygen and cause an explosion which results in a fire. Hence, to prevent accidental damage sodium is immersed in kerosene oil.

Question 2.
Write equations for the reactions of

1. iron with steam
2. calcium and potassium with water

Answer:

3Fe(s) + 4H₂O(g) → Fe₃O₄(aq) + 4H₂(g)
Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g) + Heat
2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g) + Heat

Question 3.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Metal	Iron(II) Sulphate	Copper(II) sulphate	Zinc sulphate	Silver nitrate
A	No reaction	Displacement
B	Displacement		No reaction
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

Use the Table above to answer the following questions about metals A, B, C and D.

1. Which is the most reactive metal?
2. What would you observe if B is added to a solution of Copper(II) sulphate?
3. Arrange the metals A, B, C and D in the order of decreasing reactivity.

Answer:

1. Most reactive metal is B.
2. B will displace copper from copper sulphate.
3. Order of decreasing reactivity – B > A > C > D.

Question 4.
Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄.
Answer:
When dilute hydrochloric acid is added to a reactive metal, Hydrogen gas is liberated.
Fe(s) + H₂SO₄(aq) ➝ FeSO₄(aq)+H₂(g)
Iron reacts with dilute hydrochloric acid and forms Iron sulphate and Hydrogen gas.

Question 5.
What would you observe when zinc is added to a solution of iron(II) sulphate? Write the chemical reaction that takes place.
Answer:
When Zinc is added to a solution of iron (II) sulphate, Iron is displaced by FeSO₄.
Zn(s) + FeSO₄(aq) ➝ ZnSO₄(aq) + Fe(S)

Text Book Part I Page No. 49

Question 1.
(i) Write the electron-dot structures for sodium, oxygen and magnesium.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds?
Answer:

(iii) Ions in Na₂O are Na⁺ and O^2- Ions in MgO are Mg²⁺ and O^2-
Ions in MgO are Mg²⁺ and O^2-

Question 2.
Why do ionic compounds have high melting points?
Answer:
Ionic compounds have strong electrostatic forces of attraction between the ions. Therefore, it requires a lot of energy to overcome these forces. That is why ionic compounds have high melting points.

Text Book Part I Page No. 53

Question 1.
Define the following terms.

1. Mineral
2. Ore
3. Gangue

Answer:

1. Mineral: The naturally occurring compounds and elements are known as a mineral.
2. Ore: Minerals from which metals can be extracted profitably are known as ores.
3. Gangue: The impurities present in the ore such as sand, rocks etc. are known as gangue.

Question 2.
Name two metals which are found in nature in the free state.
Answer:
Gold and Platinum.

Question 3.
What chemical process is used for obtaining a metal from its oxide?
Answer:
Reduction process is the chemical process used for obtaining a metal from its oxide.

Text Book Part I Page No. 55

Question 1.
Metallic oxides of zinc, magnesium and copper were heated with the following metals.

Metal	Zinc	Magnesium	Copper
Zinc oxide Magnesium oxide Copper oxide

In which cases will you find displacement reactions taking place?
Answer:

Metal	Zinc	Magnesium	Copper
Zinc oxide	No dis­placement reaction	Displaces	No dis­placement reaction
Magnesium oxide	–	Displaces	–
Copper oxide	Displaces	Displaces	–

Question 2.
Which metals do not corrode easily?
Answer:
Metals which have low reactivity such as silver and gold does not corrode easily.

Question 3.
What are alloys?
Answer:
An alloy is a homogenous mixture of two or a metal and non-metal.

KSEEB SSLC Class 10 Science Chapter 3 Textbook Exercises

Question 1.
Which of the following pairs will give displacement reactions?
(a) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal.
Answer:
(d) AgNO₃ solution and copper metal.

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of the above.
Answer:
(c) Applying a coating of zinc.

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be
(a) calcium
(b) carbon
(c) silicon
(d) iron
Answer:
(a) calcium.

Question 4.
Food cans are coated with tin and not with zinc because
(a) zinc is costlier than tin.
(b) zinc has a higher melting point than tin.
(c) zinc is more reactive than tin.
(d) zinc is less reactive than tin.
Answer:
(c) zinc is more reactive than tin.

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.
(a) How could you use them to distinguish between samples of metals and non-metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Answer:

1. With the hammer, on beating the sample if it changes into thin sheets (that is, it is malleable), then it is a metal otherwise a non-metal. Similarly, we can use the battery, bulb, wires, and a switch to set up a circuit with the sample. If the sample conducts electricity, then it is a metal otherwise a non-metal.
2. The above tests are useful in distinguishing between metals and non-metals as these are based on the physical properties. No chemical reactions are involved in these tests.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
The oxides which behave as both acidic and basic oxides are called amphoteric oxides. Examples: Aluminium oxide (A₂O₃), zinc oxide (ZnO).

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:

1. Iron and aluminium will displace hydrogen from dilute acids because of being more reactive than
2. hydrogen. Mercury and copper cannot displace hydrogen from dilute acids because of being less reactive than hydrogen.

Question 8.
In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?
Answer:
In the electrolytic refining of a metal M, impure metal is connected to Anode and pure metal thin sheet is connected to cathode. Electrolyte is Acidified Copper sulphate.

1. Iron and aluminium will displace hydrogen from dilute acids because of being more reactive than
2. hydrogen. Mercury and copper cannot displace hydrogen from dilute acids because of being less reactive than hydrogen.

In the electrolytic refining of a metal M, impure metal is connected to Anode and pure metal thin sheet is connected to cathode. Electrolyte is Acidified Copper sulphate.

1. Anode → Impure metal, M.
2. Cathode → Thin strip of pure metal, M.
3. Electrolyte → Aqueous solution of a salt of the metal, M.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in figure below.

(a) What will be the action of gas on
(i) dry litmus paper?
(ii) moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.
Answer:
a) (i) No action of gas on diy litmus paper.
(ii) In case of moist litmus paper, it turns red. Because sulphur is a non metal. Oxides of Non metal are acidic.
b) S(s) + O₂(g) ➝ SO₂(g)

Question 10.
State two ways to prevent the rusting of iron.
Answer:

Two ways to prevent the rusting of iron are:

1. Oiling, greasing, or painting:
   By applying oil, grease, or paint, the surface becomes waterproof. The moisture and oxygen present in the air cannot come into direct contact with iron. Hence, rusting is prevented
2. Galvanization:
   An iron article is coated with a layer of zinc metal, which prevents the iron to come in contact with oxygen and moisture. Hence, rusting is prevented.

Question 11.
What type of oxides are formed when non-metals combine with oxygen?
Answer:
When non metals combine with oxygen, neutral oxides or Acidic oxides are formed.
Eg: NO₂, SO₄ are acidic oxides NO, CO are neutral oxides.

Question 12.
Give reasons:
(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.
Answer:
a) Platinum, gold and silver are used to make jewellery because they are shining metals and ress reactive and they do not corrode easily.

b) Sodium, Potassium and Lithium are highly reactive metals. These metals react so vigorously that they catch fire if kept in the open.
Hence to protect them and to prevent accidental fire, they are kept immersed in kerosene oil.

c) Aluminium is highly reactive metal. It does not rust. Because it combines with oxygen and forms Aluminium oxide. The protective oxide layer prevents the metal from further oxidation and Aluminium is light, good conductor of heat. Hence this is used to make utensils for cooking.

d) The metals in the middle of the activity series such as iron, zinc, lead, copper are moderately reactive. These are usually present as sulphides or carbonates in nature. It is easier to obtain a metal from its oxide, as compared to its sulphides and carbonates. Therefore prior to reduction, the Metal sulphides and carbonates must be converted into metal oxides.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
Copper-reacts with moist carbon dioxide in the air to form copper carbonate and as a result, copper vessel loses its shiny brown surface forming a green layer of copper carbonate. The citric acid present in the lemon or tamarind neutralises the basic copper carbonate and dissolves the layer. That is why tarnished copper vessels are cleaned with lemon or tamarind juice to give the surface of the copper vessel its characteristic lustre.

Question 14.
Differentiate between metal and non-metal on the basis of their chemical properties.
Answer:
Metals:

1. Metals are electropositive.
2. Metals combine with oxygen and forms metallic oxides.
3. Metallic oxides are bases.
4. Metallic oxides are insoluble in water. But some oxides soluble in water and forms alkalies.
5. Metals react with water and forms metallic oxides and Hydrogen gas.
6. Again these metallic oxides dissolve in water and forms metallic hydroxides.
7. Metals produce chlorides. These are electrovalent or Ionic compounds.

Non-metals:

1. Non-metals react with metals and gain electrons and forms cations.
2. Most of the Non-metals dissolve in water and forms. Acidic or Neutral oxides.
3. Non-metals do not displace Hydrogen from dilute acids.
4. Non-metals react with Hydrogen and forms Hydrides.

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
Aqua regia (Latin for royal water) is a freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3 : 1. It can dissolve gold, even though neither of these acids can do so alone. Aqua regia is a highly corrosive, fuming liquid. It is one of the few reagents that it is able to dissolve gold and platinum.

Question 16.
Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron).
Answer:
Copper does not react with cold water, hot water or steam. However, iron reacts with steam. If the hot water tanks are made of steel (an alloy of iron), then iron would react vigorously with the steam formed from hot water,
3Fe(s) + 4H₂O → Fe₃O₄(s) + H₂O(g)
Cu(s) + H₂O → No reaction
That is why copper is used to making hot water tanks and not steel.

KSEEB SSLC Class 10 Science Chapter 3 Additional Questions and Answers

I. Fill in the blanks:

Question 1.
Shining property of metals is called ……
Answer:
Metallic lustre.

Question 2.
The best conductors of heat are ……
Answer:
Silver and copper.

Question 3.
…… Metals have very low melting point.
Answer:
gallium and caesium.

Question 4.
Metals combine with …… metal oxides.
Answer:
oxygen.

Question 5.
…… is a strong oxidising agent.
Answer:
Nitric acid (HNO₃)

II. Answer the following questions:

Question 1.
What is the very good method of improving the properties of a metal?
Answer:
Alloying.

Question 2.
Name the metals present in Brass.
Answer:
Copper and Zinc.

Question 3.
Draw a neat diagram of Electrolytic refining of copper and label the parts.
Answer:

Question 4.
Draw a neat diagram showing Action of steam on a metal and label the parts.
Answer:

Question 5.
Name some alkali metals.
Answer:
Lithium, Sodium and Potassium.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 3 Metals and Non-metals will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 3 Metals and Non-metals, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 8 How do Organisms Reproduce? are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 8 How do Organisms Reproduce?.

Karnataka SSLC Class 10 Science Solutions Chapter 8 How do Organisms Reproduce?

KSEEB SSLC Class 10 Science Chapter 8 Intext Questions

Text Book Part II Page No. 38

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
DNA is a chemical or complex compound which works as genetic material found in every cell of. all organisms. Genetic informations are carried by DNA from parental generation to daughter generations. In reproduction, it is very important to create a DNA copy. The DNA in the cell nucleus is the information source for making proteins. Cells undergo different chemical reactions to build copies of their DNA. This creates two copies of the DNA in a single reproducing cell. It is therefore, possible for the organism to produce organism of similar type.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:
Variations are a kind of change in genotype of an individual. These changes are caused by sudden or slow change in surroundings and other associated factors with individuals or species. It is found that for a species, the environmental conditions change so drastically that their survival becomes difficult. Thus, their internal genotype changes to give offspring resistant to the surrounding and these variants help in the survival of the species. However, variations are not necessary for the individual organisms as preservation of specific species character. Variation has nothing to do with normal life processes.

Text Book Part II Page No. 43

Question 1.
How does binary fission differ from multiple fission?

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.

Binary fission	Multiple fission
In this fission, one cell split into two equal halves during cell division. Eg: Bacteria.	Here one organism divide into many daughter cells simultaneously. Eg: yeast.

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:

One among the best and common types of vegetative reproduction is reproduction through spores. Advantages of spore formation:

1. Huge numbers of spores are produced at a time.
2. Distribution of spores are very easy by air to far-off places to avoid competition at one place.
3. Spores can store genetic informations for very long time as they are covered by thick walls to prevent dehydration and other unfavourable conditions for new ones development.

Question 3.
Can you think of reasons why more complex organisms cannot . give rise to new individuals through regeneration?
Answer:
The ability to produce new individual organisms from their body parts or give rise to new individual from cut or broken up parts is called regeneration. For example, simple animals like Hydra and Planaria. Higher complex organisms cannot give rise to new individuals through regeneration because regeneration is carried out by very specialised cells which is particular to some very particular species. These cells proliferate and forms mass of cells, which undergo changes to become various cell types and tissues. Every complex organism have multiple level of organization, hence it is very difficult to regenerate complete body capable of growth and development. All the organ systems of their body work together as a unit. They can regenerate their lost body parts such as skin, muscles, blood etc. However, they cannot give rise to new individuals through regeneration.

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

1. Because plants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds. Such methods also make possible the propagation of plants such as banana, orange, rose and jasmine that have lost the capacity to produce seeds.
2. Another advantage of vegetative propagation is that all plants produced are genetically similar enough to the parent plant to have all its characteristics.

Question 5.
Why is DNA copying an essential part of the process of reproduction?
Answer:
Genetic informations are carried by DNA from parental generation to daughter generations. In reproduction, it is very important to create a DNA copy. It determines the body design of an individual. The reproducing cells produce a copy of their DNA through complicated chemical reactions and result in almost equal two copies of DNA. The copying of DNA always takes place along with the creation of additional cellular structure. This process is then followed by division of a cell to form two cells. While copying it preserves the genetic character or genotype of parents and hence, reproduce almost similar looking and working individuals.

Text Book Part II Page No. 50

Question 1.
How is the process of pollination different from fertilisation?
Answer:
Secretions of the seminal vesicles and the prostate gland make the

Pollination	Fertilisation
1. Transfer of, pollen grains to the stigma of a pistil is termed pollination.	1. Fusion of male germ cells or gametes (sperms) with female gametes (ova) is called fertilisation.
2. Pollinating agents like insects, wind or water assist the process.	2. Pollen tube carries male gametes to the female gamete in plants. In animals sperms swim through the body fluids in reproductive tract of female to reach egg.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
Role of the seminal vesicles and the prostate gland is very important in fertilization. Seminal vesicles and prostate glands r.rovide lubrication and a fluid medium for easy transport of sperms. These secretions are also rich in nutrients like form of fructose, calcium and some enzymes.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
The changes (secondary sexual characters) seen in girls at the time of puberty are as follows:

• A Breast size begins to increase, with darkening of the skin of the nipples at the tips of the breasts.
• A Menstrual cycle starts.
• A Thick hair starts growing in the genital area between the thighs and armpits.
• A Pimples develop on face and widening of hips occurs.
• The skin frequently becomes oily.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
After fertilization, the lining of uterus thickens and is richly supplied with blood to nourish the growing embryo. The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. It is embedded in the uterine wall. Placenta contains villi on the embryo’s side of the tissue and blood spaces on mother’s side surrounding the villi. This provides a large surface from mother to the embryo and waste products from embryo to mother.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
Copper-T, is inserted in female body to react with semen entering with egg in uterus. Copper ions prevent pregnancy by inhibiting the movement of sperm, because the copper-lon-containing fluids are toxic to sperm. And, if a spermatozoa fertilizes an egg, the copper ion prevents implantation of the fertilized egg, and thus check pregnancy. But, it do not stop entrance of fluids inside the female body. Hence, it will not protect women from sexually transmitted diseases.

KSEEB SSLC Class 10 Science Chapter 8 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 a part of the female reproductive system in human beings?
(a) Ovary
(b) Uterus
(c) Vas deferens
(d) Fallopian tube
Answer:
(c) Vas deferens.

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

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:

Advantages of sexual reproduction:

• Sexual reproduction is the process of combining two different genetic materials, resulting into offspring that share similar traits with their parents but are genetically diverse.
• In sexual reproduction, large number of species with variations are produced. Thus, it ensures survival of species in a population.
• The newly formed individual has characteristics of two genetically different parents.
• Variations are more viable in sexual reproduction.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
The male gametes sperms are produced by the gonads-the testes. Testes also produce a hormone called testosterone, which is responsible for secondary sexual characters developing at the time of puberty in males.

Question 6.
Why does menstruation occur?
Answer:

In female body, blood and mucous flows out every month through the vagina which is termed menstruation. If the egg present inside uterus does not get fertilised, then the lining of the uterus breaks down slowly and comes out in the form of blood and mucous from the vagina. This is complete process of menstruation.

This process occurs every month as one egg is released from the ovary every month and at the same time, the uterus (womb) prepares itself to receive the fertilised egg. Thus, the inner lining of the uterus gets thickened and is supplied with blood to nourish the embryo.

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

Question 8.
What are the different methods 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 methods: 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 methods: 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 barriers available in market are condoms.

• 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, fertilisation can not occur. .

• Implants and surgical methods: 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, in tubectomy in this fallopian tubes of the ’female can be blocked, so that the egg will not reach the uterus.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:

Unicellular organisms	Multicellular organisms
In Unicellular organisms reproduction takes place by fission, and budding etc.	In multicellular organisms reproduction takes place by regeneration budding and vegetative propagation, spore formation and sexual reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Populations of organisms fill well defined places or niches in the ecosystem. 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. Reproduction is therefore linked to the stability of population of species.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:
The reasons for adopting contraceptive methods are,

1. Couples are not interested to get issues early.
2. Pregnancy will make major demands on the body and the mind of the woman and if she is not ready for it, her health will be adversely affected.
3. There may be pressure to avoid having children from government agencies.

KSEEB SSLC Class 10 Science Chapter 8 Additional Questions and Answers

Question 1.
Draw a neat diagram showing
Answer:

Question 2.
Give an example for Fragmentation.
Answer:
Spirogyra.

Question 3.
Give examples where Regeneration takes place.
Answer:
Hydra and Planaria.

Question 4.
Name the plants where vegetative propagation take place.
Answer:
Sugarcane, roses or grapes etc.

Question 5.
Draw a neat diagram and label the parts showing germination of a seed.
Answer:

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 8 How do Organisms Reproduce? will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 8 How do Organisms Reproduce?, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3

(Take π \(=\frac{22}{7}\) unless stated otherwise)

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Let the height of cylinder be ‘h’ cm.
Radius of cylinder, r = 6 cm.
i) Volume of metallic sphere, V = \(\frac{4}{3} \pi r^{3}\)

ii) Volume of Cylinder, V = \(\pi r^{2} h\)
\(\frac{22}{7}\) × 36 × h = 310.46

∴ Height of cylinder is 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Let the radius of the resulting sphere be ‘r’ cm.
(i) Volume of 3 spheres having radii 8 cm and 10 cm.

(i) Volume of the resulting Sphere, V

∴ Volume of the resulting sphere = Volume of three spheres

∴ Radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with a diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Radius of well, ‘r’ = 3.5 m.
Height of well, h = 20 m.

(i) Volume of soil dug out of well, V = \(\pi r^{2} h\) (∵ cylinder sphere)
= \(\frac{22}{7} \times(3.5)^{2} \times 20\)
= \(=\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \times 20\)
= 770m³
(ii) Length of platform, l = 22 m.
Breadth, b = 14 m.
Height, h =?
Volume of Platform, V= l × b × h (∵ Cuboid)
= 22 × 14 × h
∴ Volume of Platform = Volume of soil dug out of well.

∴ Height of platform is 2.5 m

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Let the height of the embankment be ‘h’ m.

Dimeter of well, d = 3 m.
Radius of well, r =1.5 m.
Height of well, h = 14 m.
Volume of an embankment = Volume of soil dug from well.

∴ Height of embankment = 1.125 m.

Question 5.
A container shaped like a right cylinder diameter 12height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones cones which can be filled with ice cream.

Solution:
Diameter of right circular cylinder is 12 cm ➝ r = 6 cm.
Height: 15 cm. ➝ h = 15 cm.
Diameter of Cone, 6 cm. ➝ r = 3 cm.
Height 12 cm. ➝ h = 12 cm.
(i) Volume of the ice cream which has cone-shaped and hemisphere.

(ii) Volume of cylindrical vessel, V = \(\pi r^{2} h\)

Let number of cones be ‘n’,
n × 54π = π × 36 × 15
54n = 36 × 15
\(n=\frac{36 \times 15}{54}=\frac{540}{54}\)
∴ n = 10
Ice cream can be filled in 10 cones.

Question 6.
How many silver coins, 1.75 cm. in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
Diameter of the silver coins is 1.75 cm.
Thickness = 2 mm.
Measurement of cuboid: 5.5 × 10 × 3.5 cm.
∴ Let the required silver coins be ‘n’,
Volume of ‘n’ coins = Volume of cuboid
\(\pi r^{2} h\) = l × b × h

∴ n = 400
∴ 400 silver coins are required.

Question 7.
A cylindrical bucket, 32 cm high and with a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Height of cylindrical bucket, h = 32 cm.
Radius, r = 18 cm.
Height of conical heap of sand = 24 cm.
Radius, r =?
Slant height, l =?
Volume of the conical heap of sand = Volume of Cylinder.

Radius of Cone, r = 36 cm.
Slant height of cone, l = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)

∴ Radius of cone, r = 36 cm.
Slant height of cone, h = \(12 \sqrt{13}\) cm

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Breadth of a canal, b = 6 m.
Height, h = 1.5 m
Speed of water =10 km/hr.
Quantity of water = 8 cm.
Quantity of water flowing in 30 minutes?
Volume of water flowing in 60 minutes,
= (10 × 1000) × 6 × 1.5 cm³.
( ∵ Speed = 10 km/hr.; 10 × 1000 m/hr.)
Volume of water flowing in 30 minutes,
= \(10000 \times 9 \times \frac{30}{60} \mathrm{m}^{3}\)
∴ Let the area of land be ‘x’ sq.m.

x= 562500 sq.m.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Internal diameter of a pipe = 20 cm.
Radius of pipe =10 cm. = \(\frac{1}{10} \mathrm{m}\)
Diameter of cylindrical tank = 10 m; r=5 m.
Depth of cylindrical tank = 2 m.
Speed of water is 3 km/hr.
Time required to fill the water tank =?
Speed of water = 3 km/hr.
= \(\frac{3000}{60}\) m/minute
= 50 m/min.
Let the time required to fill the tank be n’ minutes.
∴ Flowing water in ‘n’ minutes = Volume of the water tank

∴ n = 100 minutes
∴ Time required to fill the water tank = 100 minutes

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. (see the fig. given)

Solution:
Length of rope is 20 m.
Height of the pole = ?, AB = ?
Let the height of the pole AB be ‘A’.
Length of rope AC = l = 20 m.
Angle of elevation, ∠BCA = 30°.
∴ Let ⊥∆AABC, ∠B = 90°.

∴ AB = 10 m.
∴ Height of the pole 10 m.

Question 2.
A tree breaks due to storm and the broken part bends, so that the tope of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:
Let the Height of the broken part be ‘x’ m.
BC is the distance between the foot of the tree to the point where the top touches the ground.
In ⊥∆ABC, ∠B = 90°.

∴ Height of broken part of the tree, AB = x

= 4.62 m
Let AC = AD = y m.

∴ y = 2x = 2 × 4.62
y = 9.24 m.
Full height of the tree = x + y
∴ Height of the tree = x + y
= 4.62 + 9.24
= 13.86 m.

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case ?

Solution:
i) AB = 1.5 m., AC = l₁ m., ∠ACB = 30°
In ⊥∆ABC, ∠B = 90°

∴ l₁ = 2 × 1.5 = 3 m
∴ Length of slide of younger children = 3 m
ii) PQ = 3 m, OP = 12 m, ∠POQ = 60°
In ⊥∆PQR, ∠Q = 90°

∴ Length of slide of elder children = \(2 \sqrt{3}\) m

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:
Let the height of a tower. PQ = h m.
Distance of QO from base of tower = 30 m.
Angle of elevation, ∠POQ = 30°
In ⊥∆APQO, ∠Q = 90°,

∴ Height of tower, PQ = h = \(10 \sqrt{3}\) m.

Question 5.
A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:
P is the position of Kite. It is flying at a height of 60 m. above the ground.
The inclination of the string with the ground, ∠POQ = 60°
Let length of string be OP = L

∴ Length of slide of elder children \(40 \sqrt{3}\) m.

Question 6.
A 1.5 m. tall boy is standing at some distance from a 30 m. tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:
Height of the building, PQ = 30 m.
Height of the boy, OA = 1.5 m.
If the position of Boy at OA, angle of elevation, ∠POR = 30°, His second position, OO’, ∠POR = 60°.
RQ = OA = 1.5 m.
∴ PR = 30 – RQ = 30 – 1.5
PR = 28.5 m.

Distance he walked towards the building:
OO’ = OR – O’R

Distance he walked towards the building:
OO’ = \(19 \sqrt{3}\) m

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:
Let the height of building, PQ = 20 m.
PR= h m.
Bottom of transmission tower, P, vertex point ‘R’.
∠POQ = 45°, ∠ROQ = 60°.


∴ Height of transmission tower is = \(20(\sqrt{3}-1)\) m

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:
Height of pedestal, PQ = h m.
PR = 1.6 m.
∠POQ = 45°
∠ROQ = 60°


∴ Height of pedestal, h = \(0.8 \times(\sqrt{3}+1)\) m

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high. find the height of the building.

Solution:
Height of the tower, PQ = 50 m.
Let Height of the building, AB = h m.
Angle of elevation to top of tower, = ∠PBQ = 60°
Angle of elevation to building. = ∠AQB = 30°

h = 16.6 m

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:
Height of AB and PQ be h m.
Breadth of Road, BQ = 80 m.
‘O’ is the point on road.
∠POQ = 60°, ∠AOB = 30°
OQ = ‘x’ m. OB = ‘y’ m.
x + y = 80 m.

∴ Height of pole, AB = PQ = h = \(20 \sqrt{3}\) m.
Distance of the points c from the poles 20 m. and 60 m.

Question 11.
A TV tower stands vertically on a bank Of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the figure given). Find the height of the tower and the width of the canal.

Solution:
Let the height of TV tower, AB = h m.
Breadth of a canal, BC = x m.
∠ACB = 60° ∠ADB = 30°,
DC = 20 m..

From eqtn. (ii).
\(\mathrm{h}=\mathrm{x} \sqrt{3}=10 \sqrt{3}\)
∴ Height of tower = \(10 \sqrt{3}\) m
Breadth of canal = 10 m.

Question 12.
From the top of a 7m. high building, the angle of elevation of the top of a cable tower is 60° and th angle of depression is its foot is 45°. Determine the height of the tower.
Solution:
Let the height of tower, PQ = h m.
The height of building, AB = 7 m.

Angle of elevation, ∠PAR = 60°
Angle of depression, ∠RAQ = 45°
∠AQB = 45° (∵ Alternate angle)
Let, BQ = AR = x m.

∴ Height of tower, PQ = \(7(\sqrt{3}-1)\) m.

Question 13.
As observed from the top of a 75 m. high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:
Height of lighthouse, PQ = 75 m.
A and B are positions of ship.
Px is Horizontal line.
∠APx = 30° and ∠BPx = 45°
∠PAQ = 30° and ∠PBQ = 45°
Let AB = x m, BQ = y m.
Then AQ = AB + BQ = (x + y) m

∴ Distance between the two ships, AQ = x + y
= \(75 \sqrt{3}\) m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see the figure given below). Find the distance travelled by the balloon during the interval.

Solution:
Height of the girl, OO’ = 1.2 m
∠AOB = 60°
∠POQ = 30°
Let OB = x m, BQ = d m, O’Q’ = y m.
AB = PQ = Q’P – Q’Q
= 88.2 – O’O
= 88.2 – 1.2 = 87 m.
Let OQ = ‘y’.
Distance balloon travelled, d = BQ
= (y – x)

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:
Let the height of tower, PQ = h m.
Px = Horizontal line
Let the positions of 1^st car and 2^nd car be A and B.
∠APx = 30° ∴ ∠PAQ = 30°
∠BPx = 60° ∠PBQ = 60°
Let speed of car be x m/s.
∴ Distance, AB = 6x m.
Time required to travel from B to Q be ‘n’ seconds,
∴ Distance, BQ = nx m

\(n+6=n \sqrt{3} \times \sqrt{3}\)
3n = n+6
2n=6
∴ n = 3 seconds

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.

Solution:
Let the height of tower, PQ = h m.
AQ = 4 m, BQ = 9 m.
∠PAQ = θ ∠PBQ = 90 – θ

From eqn. (i) and eqn. (ii),

∴ h² = 36
∴ h = 6 m
∴ Height of tower, h = 6 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 7 Control and Coordination are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination.

Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination

KSEEB SSLC Class 10 Science Chapter 7 Intext Questions

Text Book Part I Page No. 85

Question 1.
What is the difference between a reflex action and walking?
Answer:
A reflex action is an automatic reaction for each stimulation in our body initiated by our sense responses e.g., we move our hand immediately after a contact with hot object. It is a direct controlled action. Walking is completely controlled by our brain. On the other hand, is a voluntary action. It requires complete coordination of muscles, bones, eyes etc.

Reflex action	Walking
a) Spinalcordcontrols reflex action.	a) Brain controls walking.
b) It is a spontaneous immediate response to a stimulus. It happens without the will of individual.	b) It is a voluntary action which occurs with the will of individual.

Question 2.
What happens at the synapse between two neurons?
Answer:
Synapse allows delivery of impulses (in chemical form) from neurons to other neurons and the target point, such as muscle cells. Tiny gap between the last portion of axon of one neuron and the dendron of the other neuron is known as a synapse. At a time, it acts as a one way path and transmits impulses in one direction only.

Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
A part of hindbrain-cerebellum is responsible for maintaining posture and equilibrium of the body.

Question 4.
How do we detect the smell of an agarbatti (incense stick)?
Answer:
Forebrain is responsible for thinking work. It has separate areas that are specialized for hearing, smelling, sight, taste, touch etc. The _ forebrain also has regions that collect information or impulses from various receptors. When the smell of an incense stick reaches us, out forebrain detects it. Then, the forebrain interprets it by putting it together with the information received from other receptors and also with the information already stored in the brain.

Question 5.
What is the role of the brain in reflex action?
Answer:
Reflex actions are sudden responses, which do not involve any thinking. A connection of detecting the signal from the nerves (input) and responding to it quickly (output) is called\a reflex arc. The reflex arcs can be considered as connections present between the input and output nerves which meets in a bundle in the spinal cord. The brain is only responsible of the signal and the response.

Text Book Part I Page No. 88

Question 1.
What are plant hormones?
Answer:
Plant hormones or phytohormones are chemicals which are naturally present inside the plants similar to those substances as hormones in animals. These hormones have specific functions and reach the target point to conduct the messages and initiate proper functions like growth, coordinations etc. These are synthesized in one part of the plant body (in minute quantities) and are translocated to other parts when required. The five important phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene.

Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
The movement of shoot towards light is natural and called phototropism. This type of movement is directional and is growth oriented. But, the movement of leaves of the sensitive plants, like Mimosa pudica or “touch me not”, occurs in response to touch or contact stimuli. This movement is not growth oriented.

Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Auxin is a growth-promoting plant hormone.

Question 4.
How do auxins promote the growth of a tendril around a support?
Answer:
Auxin helps the cell grow longer and is synthesized at the shoot tip. When a tendril comes in contact with a support, auxin stimulates faster growth of the cells on the opposite side, so that the tendril forms a coil around the support.

Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
Take two troughs A and B. Fill them with soil. In the trough B place, a small clay pot Now .plant a small seeding in both troughs. Uniformly water the soil in trough A. but pour water in the clay pot of trough B. After a few days take out the seedling from the troughs. The seedling in trough A which used get uniform water has normal straight roots while roots of seedling planted in trough B show bent growth towards the clay pot containing water.

Text Book Part I Page No. 91

Question 1.
How does chemical coordination take place in animals?
Answer:
Hormones cause chemical coordination in animals and help in various functions. Hormone is the chemical messenger that regulates the physiological processes in living organisms. It is secreted by glands. The regulation of physiological processes and control and coordination by hormones comes under the endocrine system. The nervous system along with the endocrine system in our body controls and coordinates the physiological processes.

Question 2.
Why is the use of Iodised salt advisable?
Answer:
If there is deficiency of Iodine we get disease called goitre. Hence use of iodized salt is advisable Iodine is necessary for the thyroid gland to make thyroxin hormone.

Question 3.
How does our body respond when adrenaline is secreted into the blood?
Answer:
Adrenaline is a hormone secreted by the adrenal glands and it is responsible to control any kind of danger or emergency or any kinds of stress. It is secreted directly into the blood and is transported to different parts of the body. It fasten the heartbeat and the breathing rate, and provides more oxygen to the muscles. It also increases the blood pressure. All these responses enable the body to deal with any stress or emergency.

Question 4.
Why are some patients of diabetes treated by giving injections of insulin?
Answer:
Insulin is the hormone secreted by pancreas and helps regulating blood sugar levels. If it is not secreted in proper amounts, the sugar level in the blood rises causing many harmful effects. Hence some patients of diabetes treated by giving injections of Insulin.

KSEEB SSLC Class 10 Science Chapter 7 Textbook Exercises

Question 1.
Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin.
Answer:
(d) Cytokinin.

Question 2.
The gap between two neurons is called a
(a) dendrite
(b) synapse
(c) axon
(d) impulse
Answer:
(b) synapse.

Question 3.
The brain is responsible for
(a) thinking.
(b) regulating the heart beat.
(c) balancing the body.
(d) all of the above.
Answer:
(d) all of the above.

Question 4.
What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
The receptors are usually located in our sense organs such as the inner ear, the nose, the tongue and so on. So gustatory receptors will detect taste while olfactory receptors will detect smell. If these receptors are not working properly, it is harmful to our body.
Eg: If we are suffering from cold, we do not feel the taste of food.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:

Function of a neuron:
At the end of the axon, the electrical impulse sets off the release of some chemicals. These chemicals cross the gap or synapse, and start a similar electrical impulse in a dendrite of the next neuron. A similar synapse finally allows delivery of such impulses from neurons to other cells, such as muscle cells or gland.

Question 6.
How does phototropism occur in plants?
Answer:
The growth movement in plants in response to light stimulus is known as phototropism. The shoots show positive phototropism and the roots show negative phototropism. This means that the shoots bend towards the source of light whereas the roots bend away from the light source. For example: The flower head of sunflower is positively phototropic and hence, it moves from east to west along with the sun.

Question 7.
Which signals will get disrupted in case of a spinal cord injury?
Answer:
The reflex arc connections between the input and output nerves meet in a bundle in the spinal cord. In fact, nerves from all over the body meet in a .bundle in the spinal cord on their way to the brain. In case of any injury to the spinal cord, the signals coming from the nerves as well as the signals coming to the receptors will be disrupted.

Question 8.
How does chemical coordination occur in plants?
Answer:
In animals, control and coordination occur with the help of nervous system. However, plants do not have a nervous system. Plants respond to stimuli by showing movements. The growth, development and responses, to the environment in plants is controlled and coordinated by a special class of chemical substances known as hormones. These hormones are produced in one part of the plant body and are translocated to other needy parts. For example, a hormone produced in roots is translocated to other parts when required. The five major types of phytohormones are auxins, gibberellins, cytokinins, abscisic acid and ethylene. These phytohormones are either growth promoters (such as auxins, gibberellins cytokinins and ethylene) or growth inhibitors such as abscisic acid.

Question 9.
What is the need for a system of control and coordination in an organism?
Answer:
Controlled movement must be connected to the recognition of various events in the enviroment, followed by only the correct movement in response. In other words, living organisms must use systems providing control and coordination. In keeping with the general principles of body organisation. In multi cellular organisms, specilised tissues are used to provide these control and coordination activities. In animals such control and coordination are provided by nervous and muscular tissues.

Question 10.
How are involuntary actions and reflex actions different from each other?
Answer:

Involuntary actions	Reflections
i) These are actions which are not controlled by our will.	There are sudden actions.
ii) These are controlled by medulla of our brain.	These are under the control of Mid brain and hind brain.

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:

Nervous mechanism.	Hormonal mechanism.
1. Information is conducted in the form of electrical impulse	Information is transferred in the form of chemicals called hormones
2. Neurons provide point to point contact for transmission of message	Hormones are carried throughout the body through blood.
3. Message travels rapidly.	Message travels slowly
4. Effect of message continues for a very short period.	Effect of message remains for a longer duration

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
Answer:

Movement in sensitive plants

1. The movement that takes place in a sensitive plant such as Mimosa pudica occurs in response to touch (stimulus).
2. For this movement, the information is transmitted from cell to cell by electric chemical signals as plants do not have any specialised tissue for conduction of impulses.
3. For this movement to occur, the plant cells change shape by changing the amount of water in them.

Movement in our legs

1. Movement in our legs is an example of voluntary actions.
2. The signal or messages for these no action are passed to the brain and hence are consciously controlled.
3. In animal muscle cells, some proteins are found which allow the movement to occur.

KSEEB SSLC Class 10 Science Chapter 7 Additional Questions and Answers

Question 1.
Draw a neat diagram showing Reflex arc and label the parts.
Answer:

Question 2.
Which organ protects the spinal cord and brain?
Answer:
Vertebral column or backbone protects the spinal cord and bony box protects the brain.

Question 3.
Give one example of chemotropism.
Answer:
Growth of pollen tubes towards omles.

Question 4.
Which hormone promote cell division?
Answer:
Cytokinins.

Question 5.
Which hormone helps in regulating blood sugar levels.
Answer:
Insulin.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 7 Control and Coordination will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 7 Control and Coordination, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World.

Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

KSEEB SSLC Class 10 Science Chapter 11 Intext Questions

Text Book Part II Page No. 100

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing focal length or converging power of its lens is called accommodation. The normal eye has a power of accommodation which enables the object as close as 25cm & as far as infinity to be focused on its retina.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
A person with a myopic eye cannot see objects beyond 1.2 m distinctly because, In a myopic eye, the image of a distant object is formed in front of the retina. This defect can be corrected by using a concave lens.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Far point of the human eye with normal vision is near than infinity and near point is 1.2 m.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
A student has difficulty reading the black board while sitting in the last row means he is suffering from myopia. This defect can be corrected using concave lens of suitable power.

KSEEB SSLC Class 10 Science Chapter 11 Textbook Exercises

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) accomodation
(c) near-sightedness
(d) far-sightedness
Answer:
(b) accomodation

Question 2.
The human eye forms the image of an object at its
(a) cornea
(b) Iris
(c) pupil
(d) retina
Answer:
(d) retina

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer:
(c) 25 cm

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina
(c) ciliary muscles
(d) iris
Answer:
(c) ciliary muscles

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his distant vision. For correcting his near vision he needs a lens of power + 1.5 dioptre. What is the focal length of the lens required for correcting (i) distinct vision, and (ii) near vision?
Answer:
i) Lens required for correcting
distant vision = – 5.5
Focal length of lens F \(=\frac{1}{P}\)
\(\mathrm{F}=\frac{1}{-5.5}=0.181 \mathrm{m}\)
Lens required for correcting this defect =-0.181 M
ii) Lens required for correcting near vision = + 1.5 D
Focal length of lens F \(=\frac{1}{P}\)
\(F=\frac{1}{1.5}=0.667 \mathrm{m}\)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
A person is suffering from near vision. So image is formed in front of retina.
Distance of Image, V = – 80 cm
focal length = f
As per lens formula.

f = 80 cm = – 0.8 m,

∴ Power of -125 D is required to correct this problem.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:

Object distance, u = 25 cm
Image distance, v = 1m = -100 m
As per lens formula,

This defect should be corrected by +3.0 power.

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is reached when the object is at a distance of 25 cm from the eye. The focal length of the eye lens cannot be decreased below this minimum limit. Thus an object placed closer than 25cm [or very close to eye] cannot be seen clearly by a normal eye.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
The distance eye lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance inside the eye.

Question 10.
Why do stars twinkle?
Answer:
The twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. This apparent position of the star is not stationary, but keeps on changing slightly, as the physical conditions of the earths atmosphere are not stationary since the stars are very distant, they approximate point – sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the’ apparent position of the star fluctuates and the amount. Of starlight entering the eye flickers some other time, fainter, which is the twinkling effect.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the earth and are thus seen as extended sources. If we consider a plant as a collection of a large number of point sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
Light from the Sun near the horizon passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. However, light from the Sun overhead would travel relatively shorter distance. At noon, the Sun appears white as only a little of the blue and violet colours are scattered. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelength more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark. The sky appears dark to astronaut flying at very high attitude, as scattering is not prominent at such heights.

KSEEB SSLC Class 10 Science Chapter 11 Additional Questions and Answers

Question 1.
Draw a neat diagram showing
(a) neat point of a Hypermetropic eye
(b) Hypermetropic eye
(c) correction for Hypermetropic eye.
Answer:

Question 2.
Which causes presbyopia? Name the lens to correct this defect.
Answer:
This causes due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens.
A common type of bi-focal lense is necessary to correct this defect.

Question 3.
Who discovered spectrum of sunlight?
Answer:
Isaac Newton.

Question 4.
What is the reason for Advance sunrise and delayed sunset?
Answer:
The sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction.

Question 5.
At noon, the sun appears white. Give reason.
Answer:
Because at noon, the sun appears white as only a little of the blue and violet colours are scattered.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 6 Life Processes are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes.

Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes

KSEEB SSLC Class 10 Science Chapter 6 Intext Questions

Text Book Part I Page No. 61

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?
Answer:
Diffusion is a process in which unicellular or small multicellular and generally, aquatic animal’s skin absorb atmospheric oxygen due to concentration difference between internal and external medium of the body. Multicellular organisms like humans possess complex body structure. They have specialised cells and tissues for performing different functions of the body. Special permeable skin is also required for diffusion process. Therefore, diffusion cannot meet huge oxygen requirements of multicellular organisms.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
An indication of life, can be considered as Movement of various kind, breathing or growth etc. However, a living organism can also have movements, which are not visible by naked eye as movements at cellular level. Therefore, the presence of life processes is a fundamental criterion that can be used to decide whether something is alive or not.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Food, water and oxygen are outside raw materials mostly used by an organism. Depending on the complexity of the organism its requirement varies from organism to organism.

An organism needs various raw materials from outside which are as follows:

• Food – To supply energy and provide materials for growth and development of body.
• Water – To provide medium in the cells for all metabolic processes necessary for living condition.
• Oxygen – To oxidise food to release energy.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
The main life processes are nutrition, respiration, transportation, excretion, reproduction, movement etc. which are considered essential.

Text Book Part I Page No. 67

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic Nutrition

1. Food is produced by conversion of atmospheric gases and water in special conditions internally within their body.
2. Presence of chlorophyll is necessary.
3. Food is generally prepared during day time.
4. Almost all plants and some bacteria have this type of nutrition.

Heterotrophic Nutrition

1. Food is taken directly from autotrophs. Then, this food is broken down with the help of enzymes.
2. No pigment is required in this type of nutrition.
3. Food can be prepared at all times.
4. All animals and fungi have this type of nutrition.

Question 2.

Where do plants get each of the raw materials required for photosynthesis?
Answer:

1. Carbon dioxide: Plants get carbon dioxide through stomata.
2. Water: Roots absorb water from soil and transports to leaves.
3. Solar energy: Chlorophyll absorbs solar energy into chemical energy.

Question 3.
What is the role of the acid in our stomach?
Answer:
The hydrochloric acid present in our stomach dissolves bits of food and creates an acidic medium. In this acidic medium, enzyme pepsinogen is converted for pepsin, which is a protein-digesting enzyme.

Question 4.
What is the function of digestive enzymes?
Answer:
The digestive enzymes converts proteins to amino acids, complex carbohydrates into glucose and fat into fatty acids and glycerol.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
The small intestine has millions of tiny finger-like projections called villi. These villi increase the surface area for more efficient food absorption. Within these villi, many blood vessels are present that absorb the digested food and carry it to the blood stream. From the blood stream, the absorbed food is delivered to each and every cell of the body.

Text Book Part I Page No. 71

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
Terristrial animals can breathe the oxygen in the atmosphere, but animals that live in water need to use the oxygen dissolved in water. This oxygen is absorbed by different organs in different animals. All these organs have a structure that increase the surface area which is in contact with the oxygen-rich atmosphere.

Question 2.
What are the different ways in which glucose is oxidised to provide energy in various organisms?
Answer:
In all living organisms, firstly. Glucose is partially oxidised to form two molecules of pyruvate. This process occurs in the cytoplasm of the cell. Further breakdown of pyruvate takes place in different manners in different organisms.

a) Anaerobic respiration: Here the pyruvate is converted into ethanol and carbondioxide in the absence of oxygen. This process takes place in yeast during fermentation. Sometimes, during vigorous muscular activities, when oxygen is inadequate for cellular respiration. Pyruvate is converted into Lactic acid.

b) Aerobic respiration: It takes place in the mitochondria, where puruvate is completely oxidised in the presence of oxygen. At the end CO₂, water and large amount of energy is released.

Question 3.
How is oxygen and carbon dioxide transported in human beings?
Answer:
Haemoglobin transports oxygen molecule to all the cells of body for cellular respiration. The haemoglobin pigment present in the blood gets attached to four O₂ molecules that are obtained from breathing. It thus, forms oxyhaemoglobin and the blood turns oxygenated. This oxygenated blood is distributed to all the body cells by the heart. After giving away O₂ to the body cells, blood takes away CO₂ which is the end product of cellular respiration. Now, the blood becomes de-oxygenated. Since, haemoglobin pigment has less affinity for CO₂, CO₂ is mainly transported in the dissolved form. This de-oxygenated blood gives CO₂ to lung’s alveoli and takes O₂ in return.

Question 4.
How are the lungs designed in human beings to maximise the area for exchange of gases?
Answer:
The exchange of gases takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli. Thus, alveoli are the site for exchange of gases. The lungs get filled up with air during the process of inhalation as ribs are lifted up and diaphragm is flattened. The air that is rushed inside the lungs fills the numerous alveoli present in the lungs. Each lung contains 300-350 millions alveoli. These numerous alveoli increase the surface area for gaseous exchange making the process of respiration more efficient.

Text Book Part I Page No. 76

Question 1.
What are the components of the transport system in human beings? What are the functions of these components?
Answer:
Arteries, veins and capillaries are the components of the transport system in human beings.

1. Arteries: Arteries are the vessels which carry blood away from the heart to various organs of the body.
2. Veins: Veins collect the blood from different organs and bring it back to the heart under high pressure.
3. Capillaries: These are smallest vessels which have one cell thick. Exchange of material between the blood and surrounding cells takes place across this thin wall.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
Separation of pure and impure blood is necessary to keep oxygenated and deoxygenated blood away from mixing. Warm-blooded animals such as birds and mammals maintain a constant body temperature by cooling themselves when they are in a hotter environment and by warming their bodies when, they are in a cooler environment. Hence, these animals require more oxygen (O₂) for more cellular respiration so that they can produce more energy to maintain their body temperature. Such separation helps in a highly efficient supply of oxygen to the body.

Question 3.
What are the components of the transport system in highly organised plants?
Answer:
The transport system in highly organised plants includes xylem and phloem (vascular tissues). Xylem transports water and mineral ions. Phloem conducts food from leaves to other parts of plants.

Question 4.
How are water and minerals transported in plants?
Answer:
The components of xylem tissue (tracheids and vessels) of roots, stems and leaves are interconnected to form a continuous system of water – conducting channels that reaches all parts of the plant. Transpiration creates a suction pressure, as a result of which water is forced into the xylem cells of the roots. Then, there is a steady movement of water from the root xylem to all the plant parts through the interconnected water-conducting channels. Components of xylem tissue helps in the water transport.

Question 5.
How is food transported in plants?
Answer:
Phloem transports food materials to the plant body. The transportation of food in phloem is achieved by utilizing energy stored as ATP. As a result of this, the osmotic pressure in the tissue increases causing water to move into it. This pressure moves the material in the phloem to the tissues which have less pressure. This is helpful in moving materials according to the needs of the plant. For example, the food material, such as sucrose, is transported into the phloem tissue using ATP energy.

Text Book Part I Page No. 78

Question 1.
Describe the structure and functioning of nephrons.
Answer:
Each kidney has large numbers of filtration units called nephrons packed close together.

Functions of nephrons:
Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water, are selectively re-absorbed as the urine flows along the tube. The amount of water re-absorbed depends on how much excess water there is in the body, and on how much of dissolved waste there is to be excreted.

The urine forming in each kidney eventually enters a long tube, the ureter, which connects the kidneys with the urinary bladder. Urine is stored in the urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra. The bladder is muscular, so it is under nervous control. As a result, we can usually control the urge to urinate.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Plants can get rid of excess of water by transpiration. Waste materials may be stored in the cell vacuoles or as gum and resin, especially in old xylem. It is also stored in the leaves that later fall off.

Question 3.
How is the amount of urine produced regulated?
Answer:
Different organisms use varied strategies for excretion. Many unicellular organisms remove the wastes by simple diffusion from the body surface into the surrounding water. Multi cellular organisms use specialised organs to perform the same function. In human beings Nitrogenous waste products are removed by the nephrons in kidneys.

KSEEB SSLC Class 10 Science Chapter 6 Textbook Exercises

Question 1.
The kidneys in human beings are a part of the system for
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) excretion.

Question 2.
The xylem in plants are responsible for
(a) transport of water.
(b) transport of food.
(c) transport of amino acids.
(d) transport of oxygen.
Answer:
(a) transport of water.

Question 3.
The autotrophic mode of nutrition requires
(a) carbon dioxide and water.
(b) chlorophyll.
(c) sunlight.
(d) all of the above.
Answer:
(d) all of the above.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in
(a) cytoplasm.
(b) mitochondria.
(c) chloroplast.
(d) nucleus.
Answer:
(b) mitochondria.

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
Fats are digested in the small intestine. The small intestine gets bile juice and pancreatic juice respectively from the liver and the pancreas. The bile salts (from the liver) break down the large fat globules into smaller globules, so that the pancreatic enzymes can easily act on them. This is referred to as emulsification of fats. It takes place in the small intestine.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
The saliva contains an enzyme called salivary amylase that breaksdown starch which is complex molecule to give simple sugar.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by products?
Answer:
Autotrophs absorbs solar energy and take CO₂ and water, prepare their own food.
Glucose, CO₂ and water are the by products of photosynthesis.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:

The differences between Aerobic and Anaerobic respiration are:

1. Aerobic respiration occurs in the presence of O₂ while anaerobic respiration occurs in the absence of O₂.
2. Aerobic respiration involves the exchange of gases between the organism and the outside environment while in anaerobic respiration exchange of gases is absent.
3. Aerobic respiration occurs in cytoplasm and mitochondria while anaerobic respiration occurs only in cytoplasm.
4. Aerobic respiration always releases CO₂ and H₂O while in anaerobic respiration end products vary.
5. Aerobic respiration yields 36 ATPs while anaerobic respiration yields only 2 ATPs.

Question 9.
How are the alveoli designed to maximise the exchange of gases?
Answer:
The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels.

Question 10.
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Red pigment present in our blood is haemoglobin. It supplies oxygen to all cells of our body Bloodlessness is caused by the deficiency of haemoglobin.

Question 11.
Describe double circulation of blood in human beings. Why is it necessary?
Answer:

A circulation system in which blood pumps through the heart twice during each trip around the body is called double circulatory system. Firstly, blood is pumped into the lungs, where it receives oxygen and becomes oxygenated, and is then pumped back into the heart, before it is finally pumped into the rest of the body.

The human heart has four chambers – the right atrium, the right ventricle, the left atrium and the left ventricle.

Pathway of blood in the heart:

• The heart has superior and inferior vena cava, which carries de- oxygenated blood from the upper and lower regions of the body respectively and supplies this de-oxygenated blood to the right atrium of the heart.
• The right atrium then contracts and passes the de-oxygenated blood to the right ventricle, through an auriculo – ventricular aperture.
• Then, the right ventricle contracts and passes the de-oxygenated blood into the two pulmonary arteries, which pumps it to the lungs where the blood becomes oxygenated. From the lungs, the pulmonary veins transport the oxygenated blood to the left atrium of the heart.
• Then, the left atrium contracts and through the auricular-ventricular aperture, the oxygenated blood enters the left ventricle.
• The blood passes to aorta from the left ventricle. The aorta gives rise to many arteries that distribute the oxygenated blood to all the regions of the body.
• Therefore, the blood goes twice through the heart. This is known as double circulation.

Importance of double circulation:

The separation of oxygenated and de-oxygenated blood allows a more efficient supply of oxygen to every single cells. This efficient system of oxygen supply is very useful in warm-blooded animals such as human beings. As we know, warm-blooded animals have to maintain a constant body temperature. Thus, the circulatory system of humans becomes more efficient because of the double circulation.

Question 12.
What are the differences between the transport of materials in xylem and phloem?
Answer:

Xylem	phloem
i) This tissue trans­ports water and mineral salts.	This tissue trans­ports only food.
ii) Roots absorb water and this is carried to all parts of the plant (upward)	Food prepared in to upwards and downwards in both direction.

Question 13.
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:

1. Structure of alveoli:
   • These are balloon like structures.
   • These provide a surface where the exchange of gases can take place.
     Function:
     Alveoli absorbs oxygen and give up carbon dioxide.
2. Structure of Nephrons:
   Each kidney has large numbers of filtration units called nephrons packed close together.
   Function:
   Blood is reaching kidneys by renal artery. Here filtration takes place. By this glucose, Amino acid and salts are reabsorbed.

KSEEB SSLC Class 10 Science Chapter 6 Additional Questions and Answers

Question 1.
Draw a neat diagram showing schematic representation of transport and exchange of oxygen and carbon dioxide.
Answer:

Question 2.
Write an equation which represents photosynthesis.
Answer:

Question 3.
Mention the function of Lymph.
Answer:
Lymph carries digested and absorbed fat from intestine and drains excess fluid from extra cellular space back into the blood.

Question 4.
Which are Heterotrophic organisms?
Answer:
Animals and Fungi.

Question 5.
Write three steps of photosynthesis.
Answer:
The following events occur during photosynthesis.

1. Absorption of light energy by chlorophyll.
2. Conversion of light energy to chemical energy and splitting of water molecule into hydrogen and oxygen.
3. Reduction of carbon dioxide to carbohydrates.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 6 Life Processes will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 6 Life Processes, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 1.
In Fig. 3.28. find the values of x and y and then shows that AB||CD.

Answer:
i) ∠x = ?
ii) ∠y = ?
iii) Show tht AB||CD.

(i) PQ Bisects AB at R.
∠PRA and ∠ARS are on straight line PQ.
∴ ∠PRA + ∠ARS = 180° Adjacent angles
50 + ∠ARS = 180°
∠ARS = 180 – 50
∴ x = ∠ARS = 130° (i)

(ii) Similarly, PQ straight line intersects straight line CD at S.
∴ ∠CSQ = ∠RSD . Vertically opposite angles
∴ y = ∠RSD = 130° (ii)
From (i) and (ii),
∠ARS = ∠RYD
x = y = 130°

(iii) Therse are mutual alternate angles
∴ x = y = 130°
∴ AB || CD.

Question 2.
In Fig. 3.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Answer:
PQ straight line intersects AB||CD]|EF at points R, S, T.
If y : z = 3 : 7, then value of ‘x’ = ?
PQ straight line intersects CD||EF at S and T.
∴ ∠CSR = ∠DST = y (∵ vertically opposite angle)
But, ∠DST + ∠STF = 180°
(∵ Sum of interior angles in the same side)
∴ y + z = 180°
But, y : z = 3 : 7.
Sum of ratio 3 + 7 = 10
If ratio 10 means 180
If ratio 9 means ?
\(\frac{180 \times 3}{10}=54^{\circ}\)
∴ y = 54°
y + z = 180
54 + z = 180
∴ z = 180 – 54
∴ z = 126°
But, x + y = 180
(∵ Sum of interior angles on one side)
x + 54 = 180
∵ x = 180 – 54
∵ x = 126°.

Question 3.
In Fig. 3.30, If AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Answer:
AB||CD and EF⊥CD and ∠GED = 126°

∠AGE =?
∠GEF =?
∠FGE =?
∠GED = 126°.
∴ ∠GEF + ∠ZFED = 126°
∠GEF + 90° = 126° ∵ EF ⊥ CD.
∴ GEF = 126-90
∴  GEF = 36°.
∴ ∠GFE = 90° . EF ⊥ AB
∴ In ∆GEF
∠FGE + ∠GFE + ∠GEF = 180°
∠FGE + 90° + 36° = 180°
∠FGE + 126° = 180°
∴ ∠FGE = 180-126
∴ ∠FGE = 54°.
∠AGE + ∠FGE = 180° (Adjacent angles)
∠AGE + 54° = 180°
∴ ∠AGE =180-54
∠AGE = 126°
∴ ∠AGE = 126°
∠GEF = 36°
∠FGE = 54°.

Question 4.
In Fig. 3.31, If PQ||ST, ∠PQR =110°, and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point R).

Answer:

Data: PQ||ST and ∠PQR= 110° and ∠RST= 130°.
To Prove: ∠QRS =?
Construction: Draw ST||UV through ‘R’
Proof: PQ||ST (Data)
ST || UV (Construction)
∴ PQ||ST||UV
PQ||UV.
∴ ∠PQR + ∠URQ = 180° (Sum of interior angles)
110 + ∠URQ = 180°
∠URQ = 180- 110
∴ ∠URQ = 70° (i)
Similarly, ST||UV.
∴ ∠RST + ∠SRV = 180°
130 + ∠SRV = 180
∴ ∠SRV = 180 – 130
∴ ∠SRV = 50° (ii)
But, URV is a straight line.
∴ ∠URV = 180°.
∠URQ + ∠QRS + ∠RSV = 180°
70 + ∠QRS + 50 = 180
∠QRS + 120 = 180°
∴ ∠QRS = 180 – 120
∴ ∠QRS = 60°.

Question 5.
In Fig. 3.32. if AB||CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Answer:
If AB || CD and ∠APQ= 50° and ∠PRD = 127°, then ∠x = ? ∠y = ?
AB || CD PQ straight line intersects these at P and Q.
∴ ∠APQ = ∠PQR . alternate angles.
50 = ∠PQR = x
∴ x = 50°
Similarly, AB || CD PR straight line intersects these at P and R.
∴ ∠APR = ∠PRD
∠APQ + ∠QPR = ∠PRD
50 + ∠QPR = 127
∠QPR = 127 – 50
∴∠QPR = 77 °
∠QPR = y = 77 °
∴ x = 50 °,
y = 77 °.

Question 6.
In Fig. 3.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
Prove that AB||CD.

Answer:
Data: PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B. the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
To Prove: AB || CD.
Proof: PQ is a mirror.
AB is an incident ray, BC is a reflected ray. ∠ABC is the angle of incidence.
∠BCD is an angle of reflection.
The incident ray formed in the PQ mirror is equal to the reflected ray formed in the RS mirror.
∴ ∠ABC = ∠BCD.
These are pair of alternate interior angles, then AB || CD.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.4.

Question 1.
Determine which of the following polynomials has (x+1) a factor :
i) x³ + x² + x + 1
ii) x⁴ + x³ + x² + x + 1
iii) x⁴ + 3x³ + 3x² + x + 1
iv) x³ – x² – (2 + \(\sqrt{2}\) )x + \(\sqrt{2}\)
Answer:
i) x- 1 is a factor of p(x)
x + 1 = x – a
a = -1
For the value of p(a),
value of r(x) = 0.
∴ p(x) = x³ + x² + x + 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 -1 + 1
p(-1) = 0
Here, p(a) = r(x) = 0
∴ x + 1 is a factor.

ii) If x + 1 = x – a, then a = -1
p(x) = x⁴ + x³ + x² + x + 1
p(-1)= (-1)⁴ + (-1)³ +(-1)² + (-1)+ 1
= 1 – 1 + 1 – 1 + 1
= 3 – 2 p(-1)= 1
Here, r(x) = p(a)= 1 Reminder is not zero.
∴ x+1 is not a factor.

iii) If x + 1 = x – a then
a = -1
p(x) = x⁴ + 3x³ + 3x² + x + 1
p(-1) = (-1)⁴ + 3(-1 )³ +3(-1 )² + (-1) + 1
= 1 + 3(-1) + 3(1) + 1(-1) + 1
= 1- 3 + 3 – 1 + 1
= 5 – 4
P(-1)= 1
Here, r(x) = p(a)=l Remainder is not zero
∴ x+1 is not a factor.

iv) If x + 1 = x – a then,
a = -1
p(x) = x³ – x² – (2 + \(\sqrt{2}\))x+ \(\sqrt{2}\)
p(-1) = (-1)³ – (-1)² -(2 + \(\sqrt{2}\))(-1) + \(\sqrt{2}\)
= -1 -(+1) – (2 – \(\sqrt{2}\))+ \(\sqrt{2}\)
= -1 – 1 + 2 + \(\sqrt{2}\) + \(\sqrt{2}\)
= -2 + 2 + \(2 \sqrt{2}\)
= = + \(2 \sqrt{2}\)
p(-1) = \(2 \sqrt{2}\)
Here, r(x) = p(a) = \(2 \sqrt{2}\) Value of remainder r(x) is not zero.
∴ x + 1 is not a factor.

Question 2.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
i) p(x) = 2x³ + x² – 2x – 1, g(x) = x+1
ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3
Answer:
i) p(x) = 2x³ + x² – 2x – 1
g(x) = x + 1
If x + 1 = 0, then x = -1
p(x) = 2x³ + x² – 2x – 1
p(-1)= 2(-1)³ + (-1)² -2(- 1) – 1
= 2(-1) + (1) + 2 – 1
= -2 + 1 + 2 – 1
P(-1)= 0
Here, r(x) = p(a) = 0,
∴ g(x) is the a factor f(x)

ii) p(x) = x³ + 3x² + 3x + 1
g(x) = x + 2
If x + 2 = 0, then
x = -2
∴ p(x) = x³ + 3x² + 3x + 1
p(-2) = (-2)³ + 3(-2)² + 3 (-2) + 1
= -8 + 3(4) + 3(-2) + 1
= -8 + 12 – 6 + 1
= 13 – 24
p(-2)= -11
Here we have r(x) = p(a) =-11.
Value of r(x) is not equal to zero.
∴ g(x) is not a factor of f(x).

iii) p(x) = x³ – 4x² + x + 6
g(x) = x – 3
Let, x – 3 = 0, then
x = 3
p(x) = x³ – 4x² + x + 6
p(3) = (3)² – 4(3)² + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6
= 36 – 36
p(3) = 0
Here, we have r(x) = p(a) = 0
∴ (x – 3) is the factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases :
i) p(x) = x² + x + k
ii) p(x) = 2x² + kx + \(\sqrt{2}\)
iii) p(x) = kx² – \(\sqrt{2} \mathrm{x}\) + 1
iv) p(x) = kx² – 3x + k
Answer:
i) p(x) = x² + x + k
g(x) = x – 1
k = ?
If x – 1 = 0, then
x = 1
p(x) = x² + x + k
p(1) = (1)² + 1 + k
p( 1) = 1 + 1 + k
p( 1) = 2 + k
If g(x) is a factor, then we have r(x) = 0
∴ p(1) = 0
2 + k= 0
∴ k = 0 – 2
k = -2

ii) p(x) = 2x² + kx + \(\sqrt{2}\)
g(x) = x – 1 k = ?
If x – 1 = 0, then x = 1
p(x) = 2x² + kx + \(\sqrt{2}\)
p(1) = 2(1)² + k(1) + \(\sqrt{2}\)
= 2(1) + k(l) + \(\sqrt{2}\)
p(1) = 2 + k + \(\sqrt{2}\)
If (x – 1) is the factor of p(x), then we have p(1) = 0.
∴ 2 + k + \(\sqrt{2}\) = 0
k = -2 – \(\sqrt{2}\)

iii) p(x) = kx² – \(\sqrt{2} x\) + 1
g(x) = x – 1 k = ?
If x – 1 = 0, then
x = 1
p(x) = kx² – \(\sqrt{2} x\) + 1
p(1) = k(1)²– \(\sqrt{2}\)(1) + 1
= k( 1) – \(\sqrt{2}\) + 1
p(1) = k- \(\sqrt{2}\) + 1
If (x – 1) is the factor of p(x) then we have p(1) = 0.
∴ p(1) = k – \(\sqrt{2}\) + 1 = 0
∴ k= \(\sqrt{2}\) – 1

iv) p(x) = kx² – 3x + k
g(x) = x – 1 k = ?
If x – 1 = 0, then x – 1
p(x) = kx² – 3x + k
p(1) = k(1)² – 3(1) + k
= k(1) – 3(1) + k
= k – 3 + k
p(1) = 2k – 3
If (x – 1) is the factor of p(x), then we have p(1) = 0.

Question 4.
Factorise :
i) 12x² – 7x + 1
ii) 2x² + 7x + 3
iii) 6x² + 5x – 6
iv) 3x² – x – 4
Answer:
i) 12x² – 7x + 1

= 12x² – 4x – 3x + 1
= 4x (3x – 1) – 1(3x – 1)
= (3x – 1) (4x – 1)

ii) 2x² + 7x + 3

= 2x² + 6x + x + 3
= 2x(x + 3) + 1 (x + 3)
= (x + 3) (2x + 1)

iii) 6x² + 5x – 6

= 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2)

iv) 3x² – x – 4

= 3x² – 4x + 3x – 4
=x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)

Question 5.
Factorise :
i) x² – 2x² – x + 2
ii) x² – 3x² – 9x – 5
iii) x³ + 13x² + 32x + 20
iv) 2y³ + y² – 2y – 1
Answer:
i) x³ – 2x² – x + 2
= x²(x – 2) – 1 (x – 2)
= (x – 2) (x² – 1)
= (x – 2) (x + 1) (x- 1)

ii) x³ – 3x² – 9x – 5
= x³ – 5x² + 2x² – 10x + x – 5
= x²(x – 5) + 2x(x – 5) + 1 (x – 5)
= (x – 5) (x² + 2x + 1)
= (x – 5) {x² + x + x + 1}
= (x – 5) (x(x + 1) + 1(x + 1)}
= (x- 5)(x + 1) (x + 1)

iii) x³ + 13x² + 32x + 20
= x³ + 10x² + 3x² + 30x + 2x + 26
= x²(x + 10) + 3x(x + 10) + 2(x + 10)
= (x + 10) (x² + 3x + 2)
= (x + 10) {x² + 2x + x + 2)
= (x + 10) {x(x + 2) + 1 (x + 2))
= (x + 10) (x + 2) (x + 1)

iv) 2y³ + y² – 2y – 1
= y²(2y + 1) – 1(2y + 1)
= (2y + 1) (y²– 1)
= (2y + 1) {(y)² – (1)²}
= (2y+ 1) (y + 1) (y- 1)

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.6.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.6

Question 1.
Find:

Answer:

Question 2.
Find:

Answer:

Question 3.
Simplify:

Answer:

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.6, drop a comment below and we will get back to you at the earliest.