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KSEEB SSLC Class 10 Science Solutions Chapter 14 Sources of Energy are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 14 Sources of Energy.

Karnataka SSLC Class 10 Science Solutions Chapter 14 Sources of Energy

KSEEB SSLC Class 10 Science Chapter 14 Intext Questions

Text Book Part I Page No. 110

Question 1.
What is a good source of energy?
Answer:
We could then say that a good source of energy would be one.

1. Which would do a large amount of work per unit volume or mass.
2. be easily accessible.
3. be easy to store and transport, and
4. perhaps most importantly, be economical.

Question 2.
What is a good fuel?
Answer:
A good fuel is one which

• produces more heat per unit mass. It has high calorific value.
• produces less harmful gases on combustion.
• is cheap and easily available.
• is every to handle safe to transport and convenient to store.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
We should select which is easily available and it should be cheaper. Bio-gas is an excellent fuel as it contains. It burns without some. Its heating capacity is high. This gas is convenient for consumption and transportation.

Text Book Part I Page No. 115

Question 1.
What are the disadvantages of fossil fuels?
Answer:

• The fossil fuels are non renewable source of energy. If we were to consuming these sources as such an  alarming rate, we would soon run out of energy.
• Air pollution is. caused bybuTnmg fossil fuels.
• The oxides of carbon, nitrogen and ‘ sulphur drat are released on burning fossil fuels am acidic oxides. This lead to acid rain which affects our water and soil resources.
• Carbon dioxide produced by burning these fuels produces green house effect

Question 2.
Why are we looking at alternate sources of energy?
Answer:
Fossil fuels are a non-renewable source of energy. So we need to conserve them. If we were to continue consuming these sources at such alarming rates, we would soon run out of energy. In order to avoid this, alternate sources of energy were explored.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
(1) Wind energy: The kinetic energy of the wind can be used to do work. This energy was harnessed by wind mills in the past to do mechanical work. For example in a water lifting pump, the rotatory motion of windmill is utilized to lift water from a well. Today wind energy is also used to generate electricity. A windmill essentially consists of a structure similar to a large electric fan that is erected at some height on a rigid support.

A number of windmills are erected over a large area, which is known as wind energy farm. The energy output of each windmill in a farm is coupled together to get electricity on a commercial scale wind energy farms can be established only at those places where wind blows for the greater part of a year. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine, since, the tower and blades are exposed to the vagaries of nature like rain, sun, storm and cyclone, they need a high level of maintenance.

(2) Water energy: In order to produce hydel electricity, high rise dams are constructed on the river to obstruct the flow of water and thereby collect water in larger reservoirs. The water level rises and in this process the kinetic energy of flowing water gets transformed into potential energy. The water from the high level in the dam is carried through pipes, to the turbine, at the bottom of the dam. Sine the water in the reservoir would be refilled each time. It rains (hydropower is a renewable source of energy) we would not have to worry about hydro electricity sources getting used up the way fossil fuels would get finished one day.

Text Book Part I Page No. 120

Question 1.
What kind of mirror – concave, convex or plain – would be best suited for use in a solar cooker? Why?
Answer:
A concave mirror would be best suited for use in a solar cooker. When a concave mirror reflector is. attached to a solar; cooker, it converges a large amount of sun’s be radiations at its focus due to which a high temperature is produced at die focus area.

Question 2.
What are the limitations of the energy that can be obtained from the oceans?
Answer:
Tidal energy wave energy and ocean thermal energy are the sources of the energy that can be obtained from the oceans. Their limitations are as follows:

1. Tidal energy: The locations where such dams can be built are limited.
2. Wave energy: Wave energy would be a viable proposition only where waves are very strong.
3. Ocean thermal energy: Efficient commercial exploitation is difficult for this energy.

Question 3.
What is geothermal energy?
Answer:
Due to geological changes, molten formed in the deeper hot regions of earth’s crust are pushed upward and trapped in certain regions called ‘hot spots’. This is called Geothermal energy.

Question 4.
What are the advantages of nuclear energy?
Answer:

• It produces huge amount of energy form very small amount of a nuclear’ fuel.
• It does not produce gases like carbon 2. dioxide which contribute to green house effect or sulphur dioxide which causes acid rain.
• Once the nuclear fuel is loaded into the reaction, the nuclear power plant can go on producing electricity for 3. two to three years at a stretch. Hence, there is no need for putting in the nuclear fuel again and again.

Text Book Part I Page No. 120

Question 1.
Can any source of energy be pollution-free? Why or why not?
Answer:
No source of energy be pollution free. Because fossil fuels cause air pollution. Nuclear energy causes more hazards for environment. In some cases, the actual operation of a device like the solar cell may be pollution-free, but the assembly of the device would have caused some environmental damage.

Question 2.
Hydrogen has been used as a rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not?
Answer:
Hydrogen gas burns and produce steam. Because of this environment is not polluted. Hence it is used in rockets as fuel. But this gas is explosive. It is not a cleaner fuel than CNG. Because CNG is cleaner than the other sources of energy.

Text Book Part I Page No. 121

Question 1.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
Solar energy and wind energy are the two renewable sources of energy.
i) Solar energy: The sun has been radiating an enormous amount of energy at the present rate for nearly 5 billion years and will continue radiating at that rate for about 5 billion year more. A large number of solar cells are, combined in an arrangement called solar cell panel that can deliver enough electricity for practical use.

The principal advantages associated with solar cells are that they have no moving parts, require little maintenance and work quite ossificatory without the use of any focussing device. Another advantage is that they can-be set up in remote and in accessible hamlets or very sparsely inhabited areas in which laying of a power transmission line may be expensive and not commercially viable.

ii) Wind energy: Unequal heating of the landmass and water bodies by solar radiation generates air movement and causes winds to blow. This kinetic energy of the wind can be used to work. Wind energy is an environment-friendly and efficient source of renewable energy. It requires no recurring expenses for the production of electricity.

Question 2.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:

• fossil fuels
• Nuclear fuels

Fossil fuels are present in a limited. amount in the earth. Once exhausted, they will not be available to us again. It takes millions of years for fossil fuel to be formed. The nuclear materials which can be conveniently extracted from earth 7. are limited and hence they will get exhausted one day.

KSEEB SSLC Class 10 Science Chapter 14 Textbook Exercises

Question 1.
A solar water heater cannot be used to get hot water on
(a) a sunny day
(b) a cloudy day
(c) a hot day
(d) a windy day
Answer:
(b) a cloudy day.

Question 2.
Which of the following is not an example of a bio-mass energy source?
(a) wood
(b) gobar-gas
(c) nuclear energy
(d) coal
Answer:
(c) nuclear energy.

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?
(a) geothermal energy
(b) wind energy
(c) nuclear energy
(d) bio-mass.
Answer:
(b) wind energy.

Question 4.
Compare and contrast fossil fuels and the Sun as direct sources of energy.
Answer:
Air pollution is caused by burning of coal or petroleum products. The oxides of carbon, nitrogen and sulphur that are released on burning fossil fuels are acidic oxides. These lead to acid rain which affects our water and soil resources. In addition to the problem of air pollution, recall the green house effect of gases like carbon dioxide.

Sun is the main source Of energy, we get all sources of energy by solar energy. Hence this a renewable source of energy which we get in plenty and free of cost.

Question 5.
Compare and contrast bio-mass and hydro electricity as sources of energy.
Answer:
Bio-mass: Cow-dung, various plant materials like the residue after . harvesting the crops, vegetable waste and sewage are decomposed in the absence of oxygen to give biogas. Since the starting material is mainly cow-dung, it is popularly known as ‘gober gas’. Biogas is produced in biogas plant.

Bio gas is an excellent fuel as it contains up to 75% Methane. It burns without smoke, leaves no residue like ash in wood, charcoal and coal burning. The slurry left behind is removed periodically and used as excellent manure. This is one of the renewable source of energy.

Hydro Electricity: In order to produce hydel electricity, high rise dams are constructed on the river to obstruct the flow of water and thereby collect water level rises and in this process the kinetic energy of flowing water, gets transformed into potential energy. The water from the high level in the dam is carried through pipes to the turbine, at the bottom of the dam. Since the water in the reservoir would be refilled each time it rains. We would not have to worry about hydro electricity sources getting used up the way fossil fuels would get finished one day.

Question 6.
What are the limitations of extracting energy from—
(a) the wind?
(b) waves?
(c) tides?
Answer:
(a) The wind:

1. Wind energy farms can be established only at those places where wind blows for the greater part of a year.
2. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine.
3. Establishment of wind energy farms require large are of land.

These are the limitations of extracting energy from the wind.

(b) Limitations of extracting waves energy: The waves are generated by strong winds blowing across the sea. Wave energy would be a viable proposition only where waves are very strong.

(c) Limitations of extracting tidal energy: The locations where such dams can be built are limited.

Question 7.
On what basis would you classify energy sources as
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:
The options given in (a) and (b) are the same.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
The qualities of an ideal source of energy are as follows:

1. Which would do a large amount of work per unit volume or mass.
2. Be easily accessible.
3. Be easy to store and transport, and
4. Perhaps most importantly, be economical.

Question 9.
What are the advantages and disadvantages of using solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Solar cookers have limited utility at places which remain cloudy or have larger winters eg, hilly areas. Advantages of using a solar cooker

• It cooks food without causing any kind of pollution.
• It is economical to use solar cooker because nothing is to be paid for using solar energy.
• It is easy to handle solar cooker and there is no chance of any kind of accident.
• It nutrients in the food do not get destroyed.

Disadvantages of using a solar cooker.

• Solar cooker cannot be used at night and during cloudy weather.
• It takes more time to cook food.
• The direction of solar cooker is to be changed

Question 10.
What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
They are:

• Burning of fossil fuels to meet the increasing demand for energy causes air pollution.
• Construction of dams and rivers to generate hydroelectricity destroys large ecosystems which get submerged underwater in the dams further, a large amount of methane [which is a green house gas] is produced when submerged vegetation rots under anaerobic conditions.

In order to reduce energy consumption

• Fossil fuel should be used with care and caution to derive maximum benefit out of them.
• Fuel saving devices such as pressure cookers etc should be used.
• Efficiency of energy sources should be maintained be getting them regularly serviced.
• And last of all, we should be economical in our energy consumption as energy saved is energy produced.

KSEEB SSLC Class 10 Science Chapter 14 Additional Questions and Answers

Question 1.
Draw a neat diagram showing a model to demonstrate the process of thermoelectric production and label the parts.
Answer

Question 2.
Draw a neat diagram of a solar cooker and label the parts.
Answer:

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 14 Sources of Energy will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 14 Sources of Energy, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements.

Karnataka SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements

KSEEB SSLC Class 10 Science Chapter 5 Intext Questions

Text Book Part II Page No. 25

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer:
Yes, Dobereiner’s triads also exist in the columns of Newlands’ Octaves. One such column is Li, K, Na.

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:
Limitation of Dobereiner’s classification:
All known elements could not be classified into groups of triads on the basis of their properties.

Question 3.
What were the limitations of Newlands’ Law of Octaves?
Answer:
Limitations of Newlands’ law of octaves:

1. It was not applicable throughout the arrangements. It was applicable up to calcium only. The properties of the elements listed after calcium showed no resemblance to the properties of the elements above them.
2. Elements discovered after Newlands’ octaves did not follow the law of octaves.
3. The position of cobalt and nickel in the group of the elements (F, Cl) of different properties could not be explained. Similarly, properties of iron also could not be explained.

Text Book Part II Page No. 29

Question 1.
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements:
K, C, AI, Si, Ba.
Answer:

1. K belongs to group 1. Therefore, the oxide will be K₂O.
2. C belongs to group 4. Therefore, the oxide will be CO₂.
3. Al belongs to group 3. Therefore, the oxide will be Al₂O₃.
4. Si belongs to group 4. Therefore, the oxide will be SiO₂.
5. Ba belongs to group 2. Therefore, the oxide will be BaO.

Question 2.
Besides gallium, which other elements have since been discovered that were left by Mendeldev in his Periodic Table? (any two)
Answer:
Scandium and Germanium.

Question 3.
What were the criteria used by Mendeleev in creating his Periodic Table?
Answer:
Mendeleev considered the atomic mass of the elements as the unique criteria of the elements. He proposed that the chemical properties of elements are the periodic function of their atomic masses. And thus, he arranged the elements in the increasing order of their atomic masses.

Question 4.
Why do you think the noble gases are placed in a separate group?
Answer:
Noble gases are very inert and they are different when compared to other elements. Hence, these are placed in separate group.

Text Book Part II Page No. 34

Question 1.
How could the Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:

Various anomalies of Mendeleev’s Periodic Table are removed in the Modern Periodic Table as follows:

1. Elements are arranged in the increasing order of their atomic number in Modern Periodic Table, thus one element is placed in one position.
2. In Modern Periodic Table, there was no problem with the place of isotopes, as isotopes have the same atomic mass with different atomic numbers.
3. Elements having the same valence electrons are kept in the same group.
4. Elements having the same number of shells were put under the same period.
5. Position of hydrogen became clarified as it is kept in the group with the elements of same valence electrons.

Question 2.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer:
Calcium and strontium are the two elements we would expect to show chemical reactions similar to magnesium. Because these resemble magnesium in chemical properties, All three elements have two valence electrons.

Question 3.
Name
(a) three elements that have a single electron in their outermost shells.
(b) two elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Lithium, sodium and potassium are the 3 elements which have single electron in their outermost orbit.
(b) Magnesium and calcium are the two elements which have 2 electrons in their outermost orbit.
(c) Noble gases such as Helium, Neon and Argon are the three elements whose outermost shells are completely filled.

Question 4.
Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements? Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
Yes. The atoms of all the three elements lithium, sodium and potassium have one electron in their outermost shells. Both helium (He) and neon (Ne) have filled outermost shells. Helium has a duplet in its K shell, while neon has an octet in its L shell.

Question 5.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
Among the first ten elements, Lithium (Li) and Beryllium (Be) are metals.

Question 6.
By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic?
Ga Ge As Se Be
Answer:
Berylium has maximum metallic characteristic.

KSEEB SSLC Class 10 Science Chapter 5 Textbook Exercises

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of Periodic Table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be
in the same group of the Periodic Table as
(a) Na
(b) Mg
(c) AI
(d) Si
Answer:
(b) Mg.

Question 3.
Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon
(b) Magnesium
(c) Silicon
(d) Boran
(e) Carbon

Question 4.
(a) What property do all elements in the same column of the Periodic Table as boron have in common?
(b) What property do all elements in the same column of the Periodic Table as fluorine have in common?
Answer:
(a) All elements in Boron group have valency 3.
(b) All elements in fluorine group have valency 1.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)
N(7) F(9) P(15) Ar(18)
Answer:
(a) Atomic Number is 17.
(b) N and P are chemically similar.

Question 6.
The position of three elements A, B and C in the Periodic Table are shown below –

(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
a) A is a Non metal.
b) C is less reactive comparing to A.
c) C is smaller in size comparing to A.
d) Cation is formed by A.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
Nitrogen (7): 2,5
Phosphorus (15): 2, 8, 5.
Since electronegativity decreases with moving from top to bottom in a group, thus nitrogen will be more electronegative.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?
Answer:
In the modern periodic table, elements having some electronic configuration are arranged in the same group.

Question 9.
In the Modern Periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?
Answer:
Atomic number 12 = 2, 8, 2
Atomic number 20 = 2, 8, 8, 2
Atomic Number 19 = 2,8, 8, 1
Atomic number 21 = 2, 8, 9, 2
Atomic number 38 = 2, 8, 18, 8, 2
Calcium will have similar physical and chemical properties as element with atomic numbers 12 and 38.

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s Periodic Table and the Modern Periodic Table.
Answer:

Mendeleev s Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic mass.
   2. This table has 8 groups and 6 periods. And each group is subdivided as an A and B.
   3. In this table, Hydrogen has no position.
   4. No position for isotopes, because in Mendeleev period these are not discovered.

Modern Periodic table:

1. 1. Elements are arranged in the increasing order of their atomic number.
   2. It has 18 groups and 7 periods.
   3. Inert gases are placed in separate groups.
   4. In this table, a zigzag line separates Metals from Non-metals.

(or)

Mendeleev’s Periodic table vs Modern Periodic table:

1. Elements are arranged in the increasing order of their atomic masses, while in Modern Periodic table elements are arranged in the increasing order of their atomic numbers.
2. There are a total of 7 groups (columns) and 6 periods (rows) while in Mendeleev’s’ Periodic Table, there are a total of 18 groups (columns) and 7 periods (rows).
3. Elements having similar properties were placed directly under one another, while in Mendeleev’s’ Periodic Table elements having the same number of valence electrons are present in the same group.
4. In Mendeleev’s Periodic Table the position of hydrogen could not be explained, while in Modern Periodic table hydrogen is placed above alkali metals.
5. No distinguishing positions for metals and non-metals in Mendeleev’s Periodic Table while in Modern Periodic Table metals are present at the left-hand side of the periodic table whereas nonmetals are present at the right-hand side.

KSEEB SSLC Class 10 Science Chapter 5 Additional Questions and Answers

Fill in the blanks:

Question 1.
At present …….. elements are known to us.
Answer:
118.

Question 2.
Law of triads was given by ……..
Answer:
Dobereiner.

Question 3.
……. was introduced by Newlands’
Answer:
Law of Octaves.

Question 4.
…….. was the most important contributor to the early development of a periodic table.
Answer:
Mendeleev.

Question 5.
When Mendeleev started his work ……. elements were known.
Answer:
63.

Question 6.
Hydrogen combines with metals and non-metals to form ………
Answer:
Covalent compounds.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction.

Karnataka SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

KSEEB SSLC Class 10 Science Chapter 10 Intext Questions

Text Book Part II Page No. 78

Question 1.
Define the principal focus of a concave mirror.
Answer:
The number of rays parallel to the principal axis are falling on a concave mirror which meat at a point is called principal focus of the concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 2f Here R = 20 cm
20 = 2f
∴ \(f=\frac { 20 }{ 2 } =10\)
∴ Focal length = 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Because these mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving.

Text Book Part II Page No. 81

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature, R = 32 cm
Radius of curvature = 2f
\(R=2f=\frac { R }{ 2 } =\frac { 32 }{ 2 } =16\)
∴ Convex mirror focal length is = 16cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
\(M=\frac { Height\quad of\quad image }{ Height\quad of\quad object }\)
\(=\frac{h_{1}}{h_{0}}=\frac{-u}{v}\)
Let the height of object be h then height of image h = – 3h
\(=\frac{3 h}{h}=\frac{-v}{u}=\frac{v}{u}=3\)
∴ Distance of object, u = – 10 cm
v = 3 × (10) = – 30 cm
Here – sign indicates, image is real and it is 30 cm in front of concave mirror.

Text Book Part II Page No. 86

Question 1.
A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Lightray bend towards normal. Because when a ray of light enters from rearer medium to denser medium, it changes its direction in the second medium.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vaccum is 3 10⁸ ms^-1
Answer:
Refractive index, n_m
\(=\frac { Velocity\quad of\quad light\quad in\quad vaccum\quad }{ Refractive\quad Index\quad of\quad glass } \)
\(=\frac{3 \times 10^{8}}{1.50}=2 \times 10^{8} \mathrm{m} / \mathrm{s}\)

Question 3.
Find out, from Tabel 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
Diamond is having highest optical density.
Air is having lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table 10.3
Answer:
Light travel faster in water because Refractive index of water is lesser than kerosene and turpentine.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means Ratio of velocity of light in air and velocity of air in diamond is 2.42.

Text Book Part II Page No. 94

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of lens whose focal length is 1 metre 1 D = 1 m^-1

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Image of Needle is real and inverted means this is real image it is 2f
Image is at a distance of 50 cm
Hence needle is kept 50 cm in front of convex lens.
Distance of object, u = – 50 cm.
Distance of image v = 50 cm
Focal length f = ?
As per lens formula.

f = 25 cm = 0.25 m
Power of the lens

Power of the lens P = + 4D.

KSEEB SSLC Class 10 Science Chapter 10 Textbook Exercises

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) At twice the focal length.

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
(a) both concave.

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
(c) A convex lens of focal length 5cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror?
What is the nature of the image? Is the image larger or smaller than the object?
Draw a ray diagram to show the image formation in this case.
Answer:
Distance of the object = o to 15 cm
Nature of image = virtual, erect and bigger than object

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.
(b) A convex mirror is used as a side/ rear-view mirror of a vehicle because,

• A convex mirror always forms an erect, virtual, and diminished image of an object placed anywhere in front of it.
• A convex mirror has a wider field of view than a plane mirror of the same size.

(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
This lens gives full image, though one-half of this lens is covered with black paper as shown in below figure.

As shown in figure light ray moves in half part and image is formed in another part of the lens.
If black paper is covered in lower part: Following figure explain this

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Height of object, h = 5 cm
Distance of object from converging lens u = 25 cm
Focal length of lens f = 10 cm
As per lens formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Converging lens, \(\frac{h_{1}}{h_{0}}=\frac{v}{u}\)

= – 3.3 cm
Images is inverted and it is formed it is formed behind the lens about 16.7 cm. Its height is 3.3 cm.
Diagram is as follows:

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Focal length (F₁) of concave lens
f = 15 cm
Image distance, v = – 10 cm
As per lens formula

u = -30 cm
Negative sign indicates, image is front of the lens about 30 cm.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror,
f = +15 cm
Object distance, u = -10 cm
As per lens formula

Magnification \(=\frac{v}{u}=\frac{-6}{-10}=0.6\)
Virtual image is formed at the distance of 6 cm and it is erect.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
The positive [+] sign of magnification [m] indicates that the image is virtual and erect. The magnification m = 1 indicates that the image is of the same size as the object. Thus, the magnification of +1 produced by a plane mirror means the image formed in a plane mirror is virtual, erect and of the same size as the object.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = 20 cm
Height of object h = 5 cm
Radius of curvature R = 30 cm
R = 2f, f = 15 cm
As per mirror formula

Positive sign indicates image is formed behind the mirror
Magnification \(=\quad \frac { Image\quad distance }{ Object\quad distance } \)
\(=\frac{-8.57}{-20}=0.428\)
Image is behind the mirror because magnification is positive
Magnification \(=\frac{\text { Image distance }}{\text { Object distance }}\)
\(=\frac{h^{1}}{h}\)
h¹ = m × h = 0.428 × 5 = 2.14 cm

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focussed image can be obtained? Find the size and nature of the image.
Answer:
Objective distance, u = 27 cm
Object height, h = 7 cm
Focal length, f = 30 cm
R = 2f, f = -18 cm
As per mirror formula

Screen should be placed in front of mirror at the distance of = 54 cm

Negative sign of magnification indicates image is real Magnification,

h₁ = 7 × (2) = -14 cm
Image is inverted because of negative sign.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Power of lens, P \(=\frac{1}{f}\)
P = -2D
f \(=\frac{-1}{2}=-0.5 \mathrm{cm}\)
Negative sign indicates this is concave lens.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Power of lens, P \(=\frac{1}{f}\)
P = 1.5 D
F \(=\frac{1}{1.5}=\frac{10}{15}=0.66 \mathrm{m}\)
This is converging lens means convex lens.

KSEEB SSLC Class 10 Science Chapter 10 Additional Questions and Answers

Question 1.
What are spherical mirrors?
Answer:
Mirrors, whose reflecting surfaces are spherical, are called spherical mirrors.

Question 2.
What is pole of the mirror?
Answer:
The centre of the reflecting surface of a spherical mirror is a point called the pole.

Question 3.
What is principal axis?
Answer:
The line passing through the pole and the centre of curvature of a spherical mirror is called principal axis.

Question 4.
Draw a ray diagram of concave mirror and convex mirror.
Answer:


(a) Concave mirror (b) Convex mirror

Question 5.
Draw a neat diagram showing Refraction of light through a rectangular glass slab
Answer :

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current.

Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

KSEEB SSLC Class 10 Science Chapter 13 Intext Questions

Text Book Part I Page No. 118

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

Text Book Part I Page No. 122

Question 2.
Draw magnetic Held lines around a bar magnet.
Answer:

Field lines around a bar magnet

Question 3.
List the properties of magnetic field lines.
Answer:
The properties are:

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

Question 4.
Why don’t two magnetic field lines intersect each other?
Answer:
Two magnetic fields lines of force never intersect each other. If the lines intersect, then at [the point of intersection there would be two directions [the needle would point towards two directions] for the same magnetic field, which is not possible.

Text Book Part I Page No. 123, 124

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

Since the current passes through the loop in a clockwise direction, therefore the front face of the loop will be the south pole and the back face, ie, the face touching the table will be north pole. According to right-hand rule, the direction of the magnetic field inside the loop will be pointing downward. Outside the loop, the direction of the magnetic field will be upward.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

The figure indicates that the magnetic field is the same at all points in the solenoid. That is field is uniform inside the solenoid.

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

Text Book Part I Page No. 125, 126

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

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

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

Answer:

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

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

Text Book Part I Page No 127

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

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

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

Text Book Part I Page No. 130

Question 1.
Explain different ways to induce current in a coil.
Answer:
We can induce current in a coil either by moving it in a magnetic field or by changing the magnetic around it. It is convenient in most situations to move the coil in a magnetic field.

Text Book Part I Page No. 131

Question 1.
State the principle of an electric generator.
Answer:
Based on the phenomenon of electromagnetic induction, electric generator are prepared. In an electric generator, Mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. This is the principle of an electric generator.

Question 2.
Name some sources of direct current.
Answer:
Dry cell, Battery and D.C. generator.

Question 3.
Which sources produce alternating current?
Answer:
A.C. generator and D.C. generator.

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

Text Book Part I Page No. 132

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

1. Electric fuse
2. Earthing wire.

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

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

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

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

KSEEB SSLC Class 10 Science Chapter 13 Textbook Exercises

Question 1.
Which of the following correctly describes the magnetic Held near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.

Question 2.
The phenomenon of electro-magnetic induction is
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

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

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

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

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

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

1. Permanent magnet
2. Electromagnet.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:

One end of the solenoid behaves as a magnetic north pole, while the other behaves as the south pole. The field lines inside the solenoid are in the form of parallel straight lines. This indicates that the magnetic field is the same at the points inside the solenoid.

As shown in figure a strong magnetic field produced inside a solenoid can be used to magnetise a piece of Magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
If the direction of magnetic field and flow of electric current are mutually perpendicular then force experienced by a current-carrying conductor in a magnetic field is largest.

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

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:

Principle:
An electric motor is a rotating device that converts electrical energy to mechanical energy working.

Working:
Current in the coil ABCD enters from the source battery through conducting brush X and flow back to the battery through brush Y. Notice that the current in the Arm AB of the coil flows from A to B. In arm CD it flows from C to D that is opposite to the direction of current through arm AB on applying Fleming’s left hand rule for the direction of force on a current¬carrying conductor in a magnetic field. We find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards.

Thus the coil and the Axle O, mounted free to turn about an axis, rotate anti-clockwise at half rotation. Q makes contact with the brush X and P with brush Y. Therefore the current in the coil gets reversed and flows along the path DCBA. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down, is now pushed up and the arm CD previously pushed up is pushed down. There is a continuous rotation of the coil and to the axle.
Split rings in electric motors acts as a commutator.

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

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

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

Answer:

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

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

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

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

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

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

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

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:

Principle: In an electric generator, mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. It is working on the principle of electromagnetic induction.

Working: When the Axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Let us say the coil ABCD is rotated clockwise in the arrangement. By applying Fleming’s right-hand rule, the induced currents are set up in these arms along with the directions AB and CD. Thus an induced current flows in the direction ABCD. If there are a larger number of turns in the coil, the current generated in each turn adds up to give a large current through the coil. This means that the current in the external circuit flows from B₂ and B₁.

After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. The current in the external circuit now flows from B₁ to B₂. Thus after every half rotation the polarity of the current in the respective arms changes.

There are two brushes and in the electric generator, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down. Because of these Brushes unidirectional current is produced.

Question 17.
When does an electric short circuit occur?
Answer:
Overloading can occur when the live wire and the neutral wire come into direct current (This occurs when the insulation of wires is damaged or there is a fault in the appliance) In such a situation, the current in the circuit abruptly increases. This is called short-circuiting.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
This is used as a safety measure, especially for those appliances that have a metallic body, for example, electric press, toaster, table fan, refrigerator, etc. The metallic body is connected to the earth wire which provides a low-resistance conducting path for the current. Thus earth wire ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth and the user may not get a severe electric shock.

KSEEB SSLC Class 10 Science Chapter 13 Additional Questions and Answers

1. Fill in the blanks:

Question 1.
Magnetic field is a quantity that has both …….. and ……
Answer:
Magnitude, direction.

Question 2.
An electric current through a metallic conductor produces a ……. around it.
Answer:
Magnetic field.

Question 3.
In electric motors, the …… acts as a commutator
Answer:
split ring.

Question 4.
…… current reverses its direction periodically.
Answer:
Alternating.

Question 5.
In our country the potential difference between two wires is ……
Answer:
220 V.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds.

Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

KSEEB SSLC Class 10 Science Chapter 4 Intext Questions

Text Book Part II Page No. 5

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂?
Answer:
In carbon dioxide molecule, the two oxygen atoms are bonded on either side with carbon atom be double bonds. These there are 2 double bonds in CO₂. Carbon shares its electrons in the formation of a double bond with one
oxygen atom and another two electrons with another oxygen atom. In this process, both the oxygen atoms and the carbon atom acquire the stable electronic configuration of the noble gas neon. The formation of CO₂ molecule is shown below.

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?
(Hint – The eight atoms of sulphur are joined together in the form of a ring.)
Answer:

Text Book Part II Page No. 12

Question 1.
How many structural isomers can you draw for pentane?
Answer:
Pentane has 3 structural isomers. We can write as follows.
i) CH₃-CH₂-CH₂-CH₂-CH₃

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

1. Catenation: It is the ability to form bonds with other atoms of carbon.
2. Tetravalency: With the valency of four, carbon is capable of bonding with four other atoms.

Question 3.
What will be the formula and electron dot structure of cyclopentane?
Answer:
Molecular formula of cyclopentane is: C₅H₁₀
Electron dot structure:

Question 4.
Draw the structures for the following compounds.

1. Ethanoic acid
2. Bromopentane*
3. Butanone
4. Hexanal.

* Are structural isomers possible for bromopentane ?
Answer:

Question 5.
How would you name the following compounds?
(i) CH₃—CH₂—Br


Answer:

1. Bromoethane
2. Methanol
3. Hexane.

Text Book Part II Page No. 15

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
Addition reaction means adding oxygen. Adding ethanol to potassium permanganate, we get ethanoic acid. Hence this reaction is called oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
Air, also contains other gases like nitrogen, carbon dioxide and few more gases apart from oxygen. When ethyne is burnt in air, it gives a sooty flame. This is due to incomplete combustion caused by the limited supply of oxygen. However, if ethyne is burnt with oxygen, it gives a clean flame with temperature 3000°C because of complete combustion. This oxy-acetylene flame is used for welding. It is not possible to attain such a high temperature without mixing oxygen. This is the reason why a mixture of ethyne and air is not used,
2HC ≡ CH + 5O₂ → 4CO₂ + 2H₂O + Heat.

Text Book Part II Page No. 18

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
All the carboxylic acids decompose sodium hydrogen carbonate giving brisk effervescence of carbon dioxide gas whereas ethanol does not react with sodium hydrogen carbonate

Experiment:

1. Take two test tubes, label them as A and B
2. Take about 0.5 g of sodium hydrogen carbonate (NaHco₃) in each test tube
3. Add 2 ml of ethanol in test tube A and 2ml of ethanoic acid in test tube B.
4. We can observe the gas bubbles in test tube B. No such bubbles are seen in test tube A. Pass the gas produced in test tube B through lime water taken in another test tube
5. We will find that lime water turns milky It is a test for carbon dioxide.

Hence, this experiment proves that when ethanoic acid reacts with sodium hydrogen carbonate, then carbon dioxide gas is produced with an effervescence (a rapid evolution of gas bubbles). Ethanol does not react with NaHCO₃.

Question 2.
What are oxidising agents?
Answer:
Oxidising agents are the substances that gain electrons in redox reaction and whose oxidation number is reduced. Examples: KMnO₄ or K₂Cr₂O₇. They have the ability to oxidize or give their oxygen to other substances.

Text Book Part II Page No. 20

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
Detergent gives lather both with hard and soft water, while a soap gives lather with soft water only. Thus, it is not possible to check if the water is hard by using a detergent.

Question 2.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
A soap molecule has two parts namely hydrophobic and hydrophilic. With the help of these particles, it attaches to the grease or dirt particle and forms a cluster called micelle. These micelles remain suspended as a colloid. To remove these micelles, it is necessary to agitate clothes.

KSEEB SSLC Class 10 Science Chapter 4 Textbook Exercises

Question 1.
Ethane, with the molecular formula C₂H₆ has
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.
Answer:
(b) 7 covalent bonds.

Question 2.
Butanone is a four-carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone.
(d) alcohol.
Answer:
(c) ketone.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH3Cl.
Answer:
Carbon has a valency of four. It shares one electron to each 3 hydrogen atoms and one more electron with chlorine.

Question 5.
Draw the electron dot structures for
(a) ethanoic acid.
(b) H₂S.
(c) propanone.
(d) F₂.
Answer:
a) Ethanoic acid

b) H₂S

c) propanone

(d) F₂

Question 6.
What is an homologous series? Explain with an example.
Answer:

A homologous series is a series of carbon compounds that have different numbers of carbon atoms but contain the same functional group. Every next member of a homologous series has a clear difference of 14 units of mass.

For example, methane, ethane, propane, etc., are all part of the alkane homologous series. The general formula of this series is C_nH_2n+2.

An example is explained with formula as below:

1. Methane, CH₄
2. Ethane, CH₃CH₃
3. Propane, CH₃CH₂CH₃
4. Butane, CH₃CH₂CH₂CH₃

It can be noticed that there is a difference of -CH₂ unit between each successive compound.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:

Ethanol and Ethanoic acid can be differentiated on the basis of their following properties by:

1. Ethanol is a liquid at room temperature with a pleasant smell. Ethanoic acid has a melting point of 17°C. Since it is below the room temperature so, it freezes during winter. Moreover, ethanoic acid has a smell like vinegar.
2. Ethanol does not react with metal carbonates while, ethanoic acid reacts with metal carbonates to form a salt, water and carbon dioxide.
   For example:
   2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ +H₂O
3. Ethanol does not react with NaOH while ethanoic acid reacts with NaOH to form sodium ethanoate and water.
   For example,
   CH₃COOH + NaOH → CH₃COONa + H₂O
4. Ethanol is oxidized to give ethanoic acid in the presence of acidified KMnO₄ while no reaction takes place with ethanoic acid in the presence of acidified KMnO₄.

Difference in physical properties:

Ethanol	Ethanoic acid
This is in liquid form at room temperature. Its melting point is 156° K.	Its melting point is 290K and hence it often freezes during winter in cold climates.
Difference in chemical properties
Ethanol will not react with metallic carbo­nates.	Ethanoic acid reacts with carbo­nates and Hydrogen carbonate and forms salts, carbon dioxide and water.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
A soap molecule has two ends. One end is hydrophilic and another end is hydrophobic. When soap is dissolved in water and clothes are put in the soapy solution, soap molecules converge in a typical manner to make a structure is called micelle. The hydrophobic ends of different molecules surround a particle of grease and make the micelle, which is a spherical structure.

In this, the hydrophilic end is outside the sphere and hydrophobic end is towards the centre of the sphere. This is why micelle formation takes place when soap is added to water. Since ethanol is not as polar as soap, micelles will not be formed in other solvents such as ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon in all its allotropic forms, burns in oxygen to give carbondioxide along with the release of heat and light. Most carbon compounds also release a large amount of heat and light on burning. Hence carbon and its compounds are used as fuels for most applications.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hard water often contains salts of calcium and magnesium. Soap molecules react with the salts of calcium and magnesium and form a precipitate. This precipitate begins floating as an off-white layer over water. This layer is called scum. Soaps lose their cleansing property in hard water because of the formation of scum.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Soap is basic in nature, hence red litmus changes to blue. Blue litmus is seen blue only.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:

Hydrogenation is a reaction between hydrogen and other compounds in the presence of the desired catalyst. Hydrogenation is used for reducing saturated hydrocarbons. Hydrogenation is an addition reaction. For example: When ethane is heated with the catalyst, nickel, it is reduced to ethane.

Industrial application:

1. In the petrochemical industry, hydrogenation is used to convert alkenes into alkanes (paraffin) and cyclo-alkanes.
2. It is also used to prepare vegetable cooking fat from vegetable oils.

Question 13.
Which of the following hydro-carbons undergo addition reactions:
C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄.
Answer:
C₂H₆ and C₂H₂ are unsaturated Hydrocarbons. Hence these undergo addition reactions.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Butter contains saturated fats. Therefore, it cannot be hydrogenated. On the other hand, oil has unsaturated fats. That is why it can be hydrogenated to saturated fats (solids).

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:

Soap are molecules in which the two ends have differing properties. One is hydrophilic, that is, it interacts with water, while the other end is hydrophobic, that is, it interacts with hydrocarbons. In the clusters of molecules in which the hydrophobic tails are on the surface of the cluster. This formation is called micelle.

Since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus the dirt suspended in the micelles is also easily rinsed away.

KSEEB SSLC Class 10 Science Chapter 4 Additional Questions and Answers

Question 1.
Write the electron dot formula of O₂.
Answer:

Question 2.
What is substitution Reaction? Give an example.
Answer:
If one type of atom or a group of atoms takes the place of another, it is called substitution reaction.
Eg: CH₄ + Cl₂ CH₃Cl + HCl (in the presence of sunlight)

Question 3.
Name 2 commercially important compounds.
Answer:
Ethanol and ethanoic acid.

Question 4.
Give an example for Esterification reaction.
Answer:

Question 5.
Write one use of ester.
Answer:
Esters are used in making perfumes and as flourishing agents.

Question 6.
What are detergents?
Answer:
Detergents are generally sodium salts of sulphonic acids or ammonium salts with chlorides or bromides etc.

Question 7.
Where is Ethanol used?
Answer:
It is used in medicines such as tincture iodine, cough syrups, and many tonics.

Question 8.
What is vinegar? Mention one of its use.
Answer:
5 to 8% solution of acetic acid in water is called vinegar. It is used as a preservative in pickles. We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity.

Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity

KSEEB SSLC Class 10 Science Chapter 12 Intext Questions

Text Book Part I Page No. 94

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path of an electric current is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
The SI unit of electric current is ampere (A).
When I coulomb of electric charge flows through any cross. Section of a conductor in I second, the electric current flowing through it is said to be 1 ampere.
∵ 1 ampere = 1C/1s

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
The SI unit of electric charge is column, which is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons.

Text Book Part I Page No. 96

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Voltmeter.

Question 2.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
One Volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

1 V = 1 Jc^-1

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
W = VQ
= 6 × 1 = 6 joules
Hence 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.

Text Book Part I Page No. 103

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of the conductor depends

1. on its length
2. on its area of cross section and
3. on the nature of its material.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
Resistance, \(\mathrm{R} \alpha \frac{1}{\mathrm{A}}\). The resistance of a conductor is inversely proportional to its area of cross-section. A thick wire has a greater area of cross-section whereas a thin wire has a smaller area of cross-section. Thus, thick wire has less resistance and a thin wire has more resistance therefore current will flow more easily through a thick wire.

Question 3.
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
As per ohm’s law, V = IR
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\) —— (i)
Potential difference is half

∴ Current flowing is also half of ts former value.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
a) Resistivity of iron = 10.0 × 10^-8
Resistivity of Mercury = 94.0 × 10^-8
Resistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.
b) Resistivity of silver is less, hence it is a good conductor of heat.

Text Book Part I Page No. 107

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.
Answer:
R_s = R₁ + R₂ + R₃
= 2V + 2V + 2V
= 6W

Question 2.
Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:

Resistance, R = 5 + 8 + 12 = 25 Ω
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{25}=0.24 \mathrm{A}\)
As per ohm’s law =0.24A
V₁ = IR
Potential difference
V₁ = IR = 0.24 × 12 = 2.88 V.
reading in ammeter = 0.24A
reading in voltmeter = 2.88 v.

Text Book Part I Page No. 110

Question 1.
Judge the equivalent resistance when the following are connected in parallel –
(a) 1 Ω and 10⁶ Ω,
(b) 1 Ω and 10³ Ω, and 10⁶ Ω.
Answer:
(a) If two resistances are connected in parallel

equivalent resistance = 1Ω.
(b) If three resistances 1Ω, 10³Ω and 10⁶Ω are connected in parallel,

equivalent resistance = 0.999Ω

Question 2.
An electric lamp of 100Ω, a toaster of resistance 50Ω and a water filter of resistance 500Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Resistance of Electric 1 amp R₁ = 100Ω
Resistance of Tosser, R₂ = 50Ω
Resistance of water filter, R₃ = 50Ω
Potential difference, V = 220V
When these are connected in parallel,


7.04A of electricity is obtained by three appliances
Resistance of an electric iron connected to the same source that takes as much current as all three appliances

∴ Resistance of iron box = 31.25Ω
Electricity flowing through this = 7.04A.

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
(i) In parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.

(ii) In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In series circuit, appliances do not get the same voltage, as that of the power supply line.

(iii) In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much due to which the current from the power supply is low.

Question 4.
How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of (a) 4Ω, (b) 1Ω?
Answer:
In this diagram 2 resistors of resistance 3W and 3 W are connected in parallel.

If 2Ω and 2Ω are connected in series 2Ω + 2Ω = 4Ω
∴ Total resistance = 4Ω
If resistors are connected in series

∴ Total resistance = 1Ω

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 8Ω, 12Ω, 24Ω?
Answer:
(a) If four resistors are connected in series then resistance
= 4 + 8 + 12 + 24
= 48Ω
(b) If these are connected in parallel

∴ We can get lowest resistance.

Text Book Part I Page No. 112

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
In cord of an electric heater, as current flows these become hot arid glows but in case of electric heater this will not happen.

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
As per Joule’s law H = VIt.
Here V = 50 V.

Question 3.
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
As per Joule’s law
H = VIt
H = IR = 5A × 20Ω = 100V
I = 5A, t = 30 sec.
∴ H = 100 × 5 × 30 J
= 15000 J = 1.5 × 10⁴ J.

Text Book Part I Page No. 114

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
The rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power.
P = VI.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
P = VI
V = 220V, and I = 5A.
Power P = 220 × 5 = 1100 W
Power of the motor = P × t
P = 1100 W.
t = 2 Hr 2 × 60 × 60 W
= 7200 S
∴ Energy consumed, E = 1100 × 7200 J
= 7920000
= 7.92 × 10⁶J.

KSEEB SSLC Class 10 Science Chapter 12 Textbook Exercises

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \(\frac { R }{ R` } \) is –
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac { { V }^{ 2 } }{ R } \)
Answer:
(b) IR²

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6.
A copper wire has diameter 0.5 mm and resistivity 1.6 10^-8Ω m. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?
Answer:
Area of the copper wire A \(=\pi\left(\frac{\mathrm{D}}{2}\right)^{2}\)
diameter = 0.5 mm = 0.0005 m
Resistance, R = 10W.

Diameter of copper wire is doubled, then diameter = 2 × 0.5 = 1 mm = 0.001 m

∴ Length of wire = 122.7 m
Resistance of wire = 2.5 W

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
In the following graph, voltage is taken along the x-axis and the current is taken along the y-axis. Different values are as follows:

V (volts)	1.6	3.4	6.7	10.2	13.2
I (amperes)	0.5	1.0	2.0	3.0	4.0

Slope indicates Resistance

∴ Resistance of the resistor = 3.4Ω.

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Here I = 2.5 mA = 2.5 × 10^-3 A.
∴ \(R=\frac{12}{2.5 \times 10^{-3}}\)
= 4.8 × 10³Ω
= 4.8 K cal.

Question 9.
A battery of 9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0. 4Ω, 0.5Ω and 12Ω, respectively. How much current would flow through the 12Ω resistor?
Answer:
A battery of 9V is connected in series with resistors 0.2Ω, 0.3Ω, 0.4Ω,
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Ω.
V = 9V
\(I=\frac{9}{13.4}=0.671 \mathrm{A}\)
∴ 12 V battery, 0.671 A current flows.

Question 10.
How many 176Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Let the resistors be x’
Resistance = 176Ω
As per ohm’s law,
V = IR


∴ Four resistors (in parallel) are required to carry 5A on a 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of (i) 9Ω, (ii) 4Ω.
Answer:
If resistors are connected in series 6Ω + 6Ω + 6Ω =18Ω This ia not correct
When they are connected in parallel
\(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=3\) This is also wrong,
i) When they are connected in parallel

Two 6Ω resistors are connected in parallel

If 3rd resistor of 6Ω and 3Ω are connected in series, it becomes 6Ω + 3Ω = 9Ω.
ii) When they are connected in series

Resistance = 6Ω + 6Ω
= 12Ω
If 3rd resistor 6Ω is connected to 12Ω in parallel

Total resistance = 4Ω.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Here V = 220V
I – 5A

∴ 110 lamps can be connected in parallel with each other across the two wires.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
(i) If coils are connected separately V = 220 V
Resistance R₁ = 240
As per ohm’s law, V = IR

∴ If coils are connected separately 9.16A electricity flows in the coil.
(ii) If coils are connected in series
Resistance R₂ = 24Ω + 24Ω = 48Ω
As per ohm’s law V = IR

∴ If coils are connected in series 4.58A electricity flows.

Question 14.
Compare the power used in the 20 resistors in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel with 12Ω and 2Ω resistors.
Answer:
(i) Potential Difference V = 6V
If 1Ω and 2Ω resistors are connected in series, then Resistance
R = 1 + 2 = 3W.
As per ohm’s law
\(\mathrm{I}=\frac{6}{3}=2 \mathrm{A}\)
P(I²)R = (2)² × 2 = 8W.
(ii) Potential difference V = 4V
If 12Ω and 2Ω resistors are connected in parallel, voltage is equal
Voltage of resistance 2 W is 4 volts

∴ Power of 2Ω is 8 W.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Both lamps are connected in parallel potential difference = 220 V
Power = V × I.

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
250W TV set in used in 1 Hr, then its energy
= 250 × 3600 = 9 × 10⁵
energy of Toaster = 1200 × 600
If it is used in 10 minutes, then its power = 1200 × 600 × 7.2 × 10⁵ J
∴ Energy of 250 W TV set is used in 1 Hr is greater than 1200 W toaster used in 10 minutes.

Question 17.
An electric heater of resistance 8Ω draws 15 A from the service mains 2 hours.
Calculate the rate at which heat is developed in the heater.
Answer:
P = I²R
R = 8Ω, I = 15A
P= (15)² × 8 = 1800 J/s.
∴ Rate at which heat is developed in the heater
= 1800 J/s.

Question 18.
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Answer:
Tungstan is a strong and metal having high melting point. This will not melt in high temperature. Because of this tungsten is used almost exclusively for filament of electric lamps.

(b) Why are the conductors of electric heating devices, such as bread- toasters and electric irons, made of an alloy rather than a pure metal?
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. Because of this reason the conductors of electric heating devices, such as bread-toasters and electric irons made of an alloy rather than a pure metal.

(c) Why is the series arrangement not used-for domestic circuits?
Answer:
In case of series connection, when one component fails, the circuit it broken and none of the components work. But in case of parallel connection, circuit divides the current throughout the electrical gadgets. Because of this reason series arrangement is not used for domestic purposes.

(d) How does the resistance of a wire vary with its area of cross-section?
Answer:
Resistance is inversely proportional to its cross section. As area is increasing resistance is less. Thus resistance of wire is changing with area of cross section.

(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
Copper and Aluminium are good conductors of electricity and they have less resistance. Hence they are used in electricity transmission.

KSEEB SSLC Class 10 Science Chapter 12 Additional Questions and Answers

Question 1.
Draw a Neat diagram of Electric circuit for studying ohm’s law.
Answer:

Question 2.
What is the S.I. Unit of resistivity?
Answer:
ΩM

Question 3.
How is fuse wire made of?
Answer:
It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc.

Question 4.
Write the formula of Electric power.
Answer:
\(P\quad =\quad \frac { { V }^{ 2 } }{ R } \)

Question 5.
Give examples where we find heating of electric current.
Answer:
Electric heater, electric iron etc.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution.

Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution

KSEEB SSLC Class 10 Science Chapter 9 Intext Questions

Text Book Part II Page No. 53

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B have higher percentage hence, it is likely to have arisen earlier. In asexual reproduction, there would be only very minor differences generated due to small inaccuracies in DNA copying, so trait B, which exists in 60% of the same population may get inherited earlier while trait A which exists in 10% of the population may be originated late due to variations.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Variation is a process which occurs sometimes at cellular level of reproduction. It cause small changes in the genotype natural selection selects the individuals having useful variations which ensure their survival in the prevailing conditions of environment, variant individuals that can withstand or cope with prevailing environment will survive letter and will increase in number through reproduction.

Text Book Part II Page No. 57

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
The trait which appears in all the members of F1 generation and also in 75% numbers of F2 generation obtained by self fertilisation of F1 generation is dominant character.
The trait which does not appear in F1 generation but after selffertilisation of F1 generation, reappears in 25% of F2 generation is known as recessive.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel thought that if two different characteristics, rather than just one are bred with each other. What do the progeny of a tall plant with round seeds and a short plant with wrinkled seeds look like? They are all tall and have round seeds. Tallness and round seeds are thus dominant traits, But what happens when these F₁ progeny are used to generate F₂ progeny by self pollination? A Mendelian experiment will find that some F₂ progeny are tall plants with round seeds and some were short plants with wrinkled seeds. However there would also be some F₂ progeny that showed new combinations. Some of them would be tall, but have wrinkled seeds, while others would be short, but have round seeds, you can see as to how new combinations of traits are formed in F₂ offspring when factors controlling for seed shape and seed colour recombine to form zygote leading to form F₂ offspring. Thus the tall/short trait and the round seed/wrinkled seed trair are independently inherited.

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No this information is not sufficient to determine which of the traits: blood group A or O is dominant. Blood group A can be genotypically AA or AO. Hence, the information is incomplete.

Question 4.
How is the sex of the child determined in human beings?
Answer:

Most human chromosomes have a maternal and a paternal copy, and we have 22 such pairs. But one pair, called the sex chromosomes, is odd in not always being a perfect pair. Women have a perfect pair of sex chromosomes, both called x. But men have a mismatched pair in which one is a normal-sized x, while the other is a short one called y. So women are xx, while men are xy. As fig. shows half the children will be boys and half will be girls. And children will inherit an x chromosome from their mother regardless of whether they are boys of girls. Thus the sex of the children will be determined by what they inherit from their father. A child who inherits an x chromosome, from her rather will be a girl, and one who inherits a y chromosome from him will be a boy.

Text Book Part II Page No. 60

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:

1.  Natural selection,
2.  Genetic drift.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Traits acquired physically, emotionally are not a change which affects the genotype of an individual so it does not get inherited for coming generations and also acquired trait involves change in non-reproductive tissues which cannot be passed on to germ cells or the Progeny. Therefore, these traits cannot be inherited.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small number of tigers indicates that tiger variants are having many challenges and not capable to adopt the existing environment and may extinct soon. The small number of members in a population of tigers may cause small number of variations, which are essential for the survival of the species. A deadly disease, or calamity may be fatal to alhthe tigers. Since, tigers are among the top consumers in our ecosystem hence, their presence in our surrounding is a must.

Text Book Part II Page No. 61

Question 1.
What factors could lead to the rise of a new species?
Answer:
Genetic drift and Natural selection are the two factors which lead to the rise of a new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self pollinating plant species? Why or why not?
Answer:
No, geographical isolation cannot prevent speciation in this case, since the plants are self-pollinating, which means that the pollens are transferred from the anther of one flower to the stigma of the same flower or of another flower of the same plant. Geographical isolation can prevent the transfer of pollens among different plants only and not pollination.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, because geographical isolation does not affect much in asexually reproducing organisms. Asexually reproducing organisms pass on the parent DNA to offsprings that leaves no chance of speciation. However, geographical isolation works as a major factor in cross pollinated species. As it would result in pollinated species and accumulation of variation in the two geographically separated population.

Text Book Part II Page No. 66

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
Birds and reptiles are great example of two close species. Feathers in some ancient reptiles, as fossils indicate, they were evolved to provide insulation in cold weather. However, they cannot fly with these feathers, later on birds adapted the feathers to flight. This means that birds are very closely related to reptiles, since dinosaurs were reptile.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat are similar in function i.e., flying. They look similar because of common use for flying, but their origins are different. Since, they perform similar function, they are analogous organs and not homologous.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Usually when organisms die, their bodies will decompose and be lost. But every once in a while, the body or at least some parts may be in an environment that does not let it decompose completely. If a dead insect gets caught in hot-mud, for example, it will not decompose quickly, and the mud will eventually harden and retain the impression of the body parts of the insect. All such preserved traces of living organisms are called fossils.
Fossils explain about the extinct species every existed.

Text Book Part II Page No. 68

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Because all humans are a single species.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:

Evolution cannot always be equated with progress or better body designs. Evolution simply creates more complex body designs. However, this does not mean that the simple body designs are inefficient. In fact, bacteria having a simple body design are still the most cosmopolitan organisms found on earth. They can survive in hot springs, deep sea and even freezing environment.

Therefore, bacteria, spiders, fish and chimpanzees are all different branches of evolution.

KSEEB SSLC Class 10 Science Chapter 9 Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW.

Question 2.
An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) all of the above.

Question 3.
In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
This information is not complete. We cannot say anything about whether the light eye colour trait is dominant or recessive only one generation is there.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:

Classification shows how clearly organisms are closely related with respect do evolution. As we know that each organism has descended from its ancestral type with some modification.

Classification involves grouping of organism based on similarities in internal and external structure or evolutionary history.

Two species are more closely related, if they have more characteristics in common. Different organisms would have common features if they are inherited from a common ancestor. And, if two species are more closely related, then it means they have a more recent ancestor. With subsequent generations, the variations make organisms more different than their ancestors. This discussion clearly proves that we classify organisms according to their resemblance which is similar to creating an evolutionary tree.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Homologous organs are those organs of different organisms which have the same basic structural design and origin but perform different functions. Example: the forelimbs of humans and the wings of birds look different externally but their skeletal structure is similar.

Analogus organs are those organs of different organism which have the different basic structural design and origin but have similar functions. Example: the wings of birds and bat.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Dog has 11 genes A….T. It acquires one chromosome from his father or mother. As per genetics Dog may be black or brown. They have 25% of BB and 50% Bb and 25% bb of genes. It is as follows.

	B	b
B	BB	Bb
b	Bb	bb

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:

The fossils are the remains of organisms that once existed on earth.

Fossil provide us evidence about

• The organisms that lived long ago , their structure etc.
• Evolutionary development of particular species i.e., line of their development with time and other environmental factors.
• Connecting links between groups. For example, feathers present in reptiles means that birds are very closely related to reptiles.
• Which organisms evolved earlier and which later.
• Development of complex body designs from the simple body designs.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
J.B.S. Haldane, a British scientist (who became a citizen of India later), suggested in 1929 that life must have developed from the simple in organic molecules which were present on earth soon after it was formed. How did these organic molecules arise? An answer was suggested by the experiment conducted by Stanley L. Miller and Harold C Urey in 1953.

They assembled an atmosphere similar to that thought to exist on early earth (this had molecules like ammonia, methane) and Hydrogen sulphide, but no oxygen) over water. This was maintained at a temperature just below 100° c and sparks were passed through the mixture of gases to stimulate lighting. At the end of a week, 15% of the carbon (from methane) had been converted to simple compounds of carbon including amino acids which make up protein molecules. This is the evidence we have for the life from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexusal reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Sexual reproduction causes more viable variations due to the following reasons:

Two different types of gametes meet to form new individuals which have better possibilities of combinations of traits. Due to the presence of multiple traits, error in copying of DNA are highly significant. At the time of gamete formation, random aggregation of paternal and maternal chromosome occurs. Exchange of genetic material may occur between , homologous chromosomes during formation of gametes. Due to sexual reproduction over generation after generation and selection by nature created wide diversity. In case of asexual reproduction, only the very small changes due to inaccuracies in DNA copying pass on the progeny. Thus, offsprings of asexual reproduction are more or less genetically similar to their parents. So, it can be concluded that evolution in sexually reproducing organisms proceeds at a faster pace than in asexually reproducing organisms.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Each cell will have two copies of each chromosome, one each from the male and female parents. Every germ cell will take one chromosome from each pair and these may be of either maternal or paternal origin. When two germ cells combine, they will restore the normal number of chromosomes in the progeny, ensuring the stability of the DNA of the species. Hence there is equal genetic contribution of male and female parents ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, we agree with the statement that only variations that confer an advantage to an individual organism will survive in a population. Genetic variations are part of natural selection. Genetic variations are one form of variation. Selection is done with respect to the environment. Every variation do not have a survival in the environment in which they exist. The chances of surviving depend upon adaptation by surrounding and the nature of variations. Selection of variants by environmental factors forms the basis for revolutionary process.

KSEEB SSLC Class 10 Science Chapter 9 Additional Questions and Answers

Question 1.
How do we know how old the fossils are?
Answer:
There are two components to this estimation one is relative. If we dig into the earth and start finding fossils, it is reasonable to suppose that the fossils we find closer to the surface are more recent than the fossils we find in deeper layers. The second way of dating fossils is by detecting the ratios of different isotopes of the same element in the fossil material. It would be interesting to find out exactly how this method works!

Question 2.
What is Evolution?
Answer:
Evolution is simply the generation of diversity and the shaping of diversity by environmental selection.

Question 3.
What is scientific name of man?
Answer:
Home sapients.

Question 4.
What does evolution of human beings indicate?
Answer:
It indicates that all of us belong to a single species that evolved in Africa and spread across the world in stage.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 9 Heredity and Evolution, drop a comment below and we will get back to you at the earliest.

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 Science Solutions Chapter 1 Chemical Reactions and Equations are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations.

Karnataka SSLC Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations

KSEEB SSLC Class 10 Science Chapter 1 Intext Questions

Text Book Part I Page No. 6

Question 1.
Why should a magnesium ribbon be cleaned before burning in air?
Answer:
Magnesium metal is highly reactive. In stored conditions, it reacts with oxygen to form magnesium oxide over its outer layer. To remove this layer and to expose the underlying metal into air, the magnesium ribbon is cleaned by sandpaper.

Question 2.
Write the balanced equation for the following chemical reactions.
i) Hydrogen + Chlorine ➝ Hydrogen Chloride
Answer:
H₂(g) + Cl₂(g) ➝ 2HCl(g)

ii) Barium chloride + Aluminium sulphate ➝ Barium sulphate + Aluminium chloride
Answer:
3BaCl₂(s) + Al₂(SO₄)₃(s) ➝ 3BaSO₄ + 2Al₂Cl₃(s)

iii) Sodium + water ➝ Sodium hydroxide + Hydrogen
Answer:
2Na(s) + 2H₂O(l) ➝ 2NaOH(aq) + H₂(g)

Question 3.
Write a balanced chemical equation with state symbols for the following reactions.
i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
Answer:
Balanced chemical equations with state symbols for the required reactions are as follows:

1. BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
2. NaOH(aq) + HCl(aq) + NaCl(aq) + H₂O(l)

Text Book Part I Page No. 10

Question 1.
A solution of a substance ‘X’ is used for whitewashing.
i) Name the substance ‘X’ and write its formula.
Answer:
The substance X is calcium Hydroxide. Its formula is Ca(OH)₂

ii) Write the reaction of the substance ‘X’ named in (i) above with water.
Answer:
Cao + H₂O ➝ Ca(OH)₂

Question 2.
Why is the amount of gas collected in one of the test tubes in Activity
Answer:

Water (H₂O) contains two parts of hydrogen and one part of oxygen. The ratio of water components i.e., hydrogen and oxygen is 2:1. Therefore, the amount of hydrogen and oxygen produced after water electrolysis is in a ratio of 2:1. This is why during electrolysis the amount of gas collected in hydrogen’s test tubes is double the amount collected in the oxygen’s test tube.

1. Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in Figure.
2. Connect these electrodes to a 6-volt battery.
3. Fill the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water.
4. Take two test tubes filled with water and invert them over the two carbon electrodes.
5. Switch on the current and leave the apparatus undisturbed for some time.
6. You will observe the formation of bubbles at both electrodes. These bubbles displace water in the test tubes.
7. Is the volume of the gas collected the same in both the test tubes?
8. Once the test tubes are filled with the respective gases. Remove them carefully.
9. Test these gases one by one by bringing a burning candle close to the mouth of the test tubes.

Caution:

1. This step must be performed carefully by the teacher.
2. What happens in each case?
3. Which gas is present in each test tube?

Text Book Part I Page No. 13

Question 1.
Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
Answer:
Iron is more reactive than copper. When an iron nail is dipped in copper sulphate solution, iron forms its sulphate (iron sulphate) solution by displacing copper of copper sulphate. The colour of iron sulphate is green. So, colour change in solution appears.

Question 2.
Give an example of a double displacement reaction other than the one given in Activity 1.10.
Answer:
2KBr(aq)+BaI₂(aq) ➝ 2KI(aq) + BaBr₂ (aq)

Question 3.
Identify the substances that are oxidised and the substances that are reduced in the following reactions.
i) 4Na(s) + O₂(g) ➝ 2Na₂O(s) 4Na(s)
Answer:
Sodium is oxidized because it gets oxygen and forms sodium oxide.

ii) CuO(s) + H₂(g) ➝ Cu(s) + H₂O(l)
Answer:
Copper oxide reduced to copper H₂ changes to water.

KSEEB SSLC Class 10 Science Chapter 1 Textbook Exercises

Question 1.
Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) ➝ 2Pb(s) + CO₂(g)
a) Lead is getting reduced.
b) Carbon dioxide is getting oxidised.
c) Carbon is getting oxidised.
d) Lead oxide is getting reduced.
i) (a) and (b)
ii) (a) and (c)
iii) (a), (b) and (c)
iv) all
Answer:
(i) (a) and (b).
Lead is getting reduced. & Carbon dioxide is getting oxidised.

Question 2.
Fe₂O₃ + 2Al ➝ Al₂O₃ + 2Fe
The above reaction is an example of a
a) combination reaction
b) double displacement reaction.
c) decompoistion reaction.
d) displacement reaction.
Answer:
(d) The reaction is an example of a displacement reaction.

Question 3.
What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.
a) Hydrogen gas and iron chloride are produced.
b) Chlorine gas and iron hydroxide are produced.
c) No reaction takes place.
d) Iron salt and water are produced.
Answer:
(a) Hydrogen gas and iron chloride are produced.
Fe(s) + 2HCl(aq) ➝ FeCl₂(aq) + H₂l

Question 4.
What is a balanced chemical equation? Why should chemical eqations be balanced?
Answer:
The total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants. In other words, the number of atoms of each element remains the same, before and after a chemical reaction. This is called Balancing equation. We must balance the chemical equation, otherwise it becomes skeletal chemical equation.

Question 5.
Translate the following statements into chemical equations and then balance them.
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer;
(a) 3H₂(g) + N₂(g) → 2NH₃(g).
(b) 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g).
(c) 3BaCl₂(aq) + Al₂(SO₄)₃(aq) → 2AlCl₃(aq) + 3BaSO₄(s).
(d) 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).

Question 6.
Balance the following chemical equations.
a) HNO₃ + Ca(OH)₂ ➝ Ca(NO₃)₂ + H₂O
b) NaOH + H₂SO₄ ➝ Na₂SO₄ + H₂O
c) NaCl + AgNO₃ ➝ AgCl + NaNO₃
d) BaCl₂ + H₂SO₄ ➝ BaSO₄ + HCl
Answer:
a) 2HNO₃ + Ca(OH)₂ ➝ Ca(NO₃)₂ + 2H₂O
b) 2NaOH + H₂SO₄ ➝ Na₂SO₄ + 2H₂O
c) NaCl + AgNO₃ ➝ AgCl + NaNO₃
d) BaCl₂ + H₂SO₄ ➝ BaSO₄ + 2HCl

Question 7.
Write the balanced chemical equations for the following reactions.
a) Calcium hydroxide + Carbon dioxide ➝ Calcium carbonate + Water
b) Zinc + Silver nitrate ➝ Zinc nitrate + Silver
c) Aluminium + Copper chloride ➝ Aluminium chloride + Copper
d) Barium chloride + Potassium sulphate ➝ Barium sulphate + Potassium chloride
Answer:
a) Ca(OH)₂ + CO₂ ➝ CaCO₃ + H₂O
b) Zn + 2AgNO₃ ➝ Zn(NO₃)₂ + 2Ag
c) 2Al + 3CuCl₂ ➝ AlCl₃ + 3Cu
d) BaCl₂ + K₂SO₄ ➝ BaSO₄ + KCl

Question 8.
Write the balanced chemical equation for the following and identify the type of reaction in each case.
a) Potassium bromide(aq) + Barium iodide(aq) ➝ Potassium iodide (aq) + Barium bromide(s)
b) Zinc carbonate(s) ➝ Zinc oxide(s) + Carbon dioxide(g)
c) Hydrogen(g) + Chlorine(g) ➝ Hydrogen chloride(g)
d) Magnesium(s) + Hydrocholoric acid(aq) ➝ Magnesium chloride(aq) + Hydrogen(g)
Answer:
a) 2KBr(aq) + Bal2(aq) ➝ 2KI(aq) + BaBr₂(s) ➝ double displacement reaction.
b) ZnCO₃(s) ➝ ZnO(s) + CO₂(g) ➝ decompoistion reaction.
c) H₂(g) + Cl₂(q) ➝ 2HCl(g) ➝ combination reaction
d) Mg(s) + 2HCl(aq) —> MgCl₂(aq) + H₂(g) ➝ displacement reaction.

Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:

1.  Reactions in which heat is released along with the formation of products are called exothermic chemical reactions.
   Eg: CH₄(g)+2O₂(g) ➝ CO₂(g) + 2H₂O(g)
2.  Reactions in which energy is absorbed are known as endothermic reactions.
   6CO₂ + 6H₂O(l) ➝ C₆H₁₂O₆(aq) + 6O₂(g)

Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:
We get from the food we eat. During digestion complex molecules of food are broken into simpler molecule such as glucose. This glucose combines with oxygen in the cells of our body and provides energy. Therefore respiration is considered an exothermic reaction.

Question 11.
Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.
Answer:
Decomposition reactions involve breaking down of compounds to form two or more substances. These reactions require energy to proceed. Thus, they are the exact opposite of combination reactions in which two or more substances combine to give a new substance.

Examples:

1. ZnCO₃(s) → ZnO(s) + CO₂(g); Decomposition reaction.
2. H₂(g) + Cl₂(g) → 2HCl(g); Combination reaction.

In the first equation, since ZnCO₃ is broken down into ZnO and CO₂ it is a decomposition reaction. In the second equation, H₂ and Cl₂ combine to give a new substance HCl. Therefore, it is a combination reaction.

Question 12.
Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Ans:
a) Thermal decomposition

b) Decomposition by light

c) Decomposition by electricity

Question 13.
What is the difference between displacement and double displacement reactions Write equations for these reactions.
Answer:
In a displacement reaction, a more reactive element displaces a less reactive element from a compound.

A + BX → AX + B; where A is more reactive than B.
In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.
AB + CD → AD + CB
For example:
Displacement reaction:
CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)

Double displacement reaction:
Na₂SO₄(aq) + BaCl₂ (aq) → BaSO₄(s) + 2NaCl(aq)

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reactions involved.
Answer:

Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
Any reaction that produces a precipitate can be called a precipitation reaction.
Eg:

In this reaction Barium chloride obtained as precipitate.
The white precipatate of BaSO₄ is formed by the reaction of SO₄^2- and Ba²⁺. The other product is formed is Sodium Chloride which remains in the solution.

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each.
a) Oxidation (Gain of Oxygen)
Answer:

In equation (i) From H₂, H₂O is oxidised. In eqn (ii) From Cu, CuO is oxidised
b) Reduction is loss of Oxygen:

In eqn (i) CO₂ is reduced to CO. in eqn (ii) CuO is reduced to Cu.

Question 17.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:
‘X’ means copper (Cu) and black coloured compound copper oxide

Question 18.
Why do we apply paint on iron articles?
Answer:
Iron articles are painted to prevent them from rusting. When painted, the contact of iron articles with atmospheric moisture and the air is cut off. Hence, rusting is prevented.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:

Oil and fat containing food items are flushed with nitrogen to prevent the items from getting oxidised which may result in rancidity of such products. When fats and oils are oxidised, they become rancid and their smell and taste change. Nitrogen provides an inert atmosphere for them.

Question 20.
Explain the following terms with one example each.
a) Corrosion
b) Rancidity
Answer:

(a) Corrosion:
When a metal is attacked by substances around it such as moisture, acids etc. it gets corroded and the process is called corrosion. For example, rusting of iron products.

(b) Rancidity:
The Process in which fats and oils or food products made from fats or oils get oxidised resulting in a change of smell and the taste is called rancidity. For example, food items made from oil like chips becomes rancid if kept open for some time.

To prevent rancidity antioxidants (which prevent oxidation) are added to food containing fats and oils. Rancidity can also be prevented by flushing out oxygen with a gas like nitrogen. For example packets of food items like chips are flushed with nitrogen so that those can be used even after long duration.

KSEEB SSLC Class 10 Science Chapter 1 Additional Questions and Answers

Question 1.
How do you determine whether a chemical reaction has taken place?
Answer:
Following observations helps us to determine whether a chemical reaction has taken place

1.  Change in state
2.  Change in colour
3.  evolution of a gas
4.  Change in temperature.

Question 2.
Write the Balanced equation for the reaction of iron with steam.
Answer:
3Fe(s) + 4H₂O(g) ➝ Fe₃O₄(s) + 4H₂(g)

Question 3.
What is quicklime? Write one use of quicklime.
Answer:
Calcium oxide is called lime or quicklime. It is used in the manufacture of cement.

Question 4.
What are antioxidants?
Answer:
The substances which prevent oxidation are called antioxidants.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations, drop a comment below and we will get back to you at the earliest.