kseebsol

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.

KSEEB SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources.

Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources

KSEEB SSLC Class 10 Science Chapter 16 Intext Questions

Text Book Part I Page No. 128

Question 1.
What changes can you make in your habits to become more environment friendly?
Answer:

1.  We must refuse to buy products that harm us and the environment.
2.  We should minimise the use of electricity and water.
3.  We should encourage recycling of things.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
With the human population increasing at a tremendous rate due to improvement in health-care, thedem and for all resources is increasing at an exponential rate. The management of natural resources requires a long term perspective so that these will last for the generations to come and will not merely be exploited to the hilt for short term gains.

Question 3.
How would these advantages differ from the advantages of using a long term perspective in managing our resources?
Answer:
If resources are used in accordance with short term aims, present generation will be able to utilize the resources properly for overall development. But if we plan to use resources with long term aims, not only the present generation is benefited but also the future generations will also be able to utilize resources for fulfilling its necessities. Thus it would be better to use our natural resources with a long term perspective so that it could be used by the present generation as well as conserved for future use.

Question 4.
Why do you think that there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Nature shows no partiality. Natural resources belong to all and these resources should be used judiciously. Equitable distribution of resources will benefit both poor as well as rich people.
Human greed, corruption, and the lobby of the rich and powerful are the forces working against an equitable distribution of our resources.

Text Book Part I Page No. 132

Question 1.
Why should we conserve forests and wildlife?
Answer:

1. Destruction of forests not only affect on forest products but it affects the water resources and it is soil pollution.
2. Destruction of forest leads to shortage of fodder for animals, shortage of medicinal plants, shortage of fruits and nuts.
3. There is a shortage of valuable timbers such as sal, sandal wood etc.
4. Wild animals are also useful to us in many ways, hence we should conserve these by building sanctuaries and prohibited hunting.

Question 2.
Suggest some approaches towards the conservation of forests.
Answer:

1. Cutting valuable trees should be prevented.
2. We should use forest products, such that there should not be damage to the environment.
3. All people should participate in the conservation of forest and wild animals.

Text Book Part I Page No. 135

Question 1.
Find out about the traditional systems of water harvesting/ management in your region.
Answer:
We must dug small pits and lakes, put in place simple water shed systems, built small earthen dams, constructed dykes, sand and limestone reservoirs, set up root top water collecting units. These are the traditional systems of water harvesting/management in our region.

Question 2.
Compare the above system with the probable systems in hilly/ mountainous areas or plains or plateau regions.
Answer:
In the above mentioned places check dams are built because here water harvesting is difficult.

Question 3.
Find out the source of water in your region/locality. Is water from this source available to all people living in that area?
Answer:
Tube wells and river water (Tungabhadra) are the water sources available to all people in our area. There are different sources in different places. In some places there is too much shortage of water because of failure of rain recently.

KSEEB SSLC Class 10 Science Chapter 16 Textbook Exercises

Question 1.
What changes would you suggest in your home in order to be environment-friendly?
Answer:

1. We must save water and electricity.
2. We should not waste food.
3. We should encourage reuse and recycling.
4. We must minimise the use of plastics.

Question 2.
Can you suggest some changes in your school which would make it environment friendly?
Answer:

1. Enough plants and trees can be planted in the school.
2. Water should not be wasted but should be used judiciously.
3. Students should be taught to keep their classrooms and immediate surroundings neat and tidy.
4. Compost pits may be made in safe comers of the school, where biodegradable wastes may be dumped to prepare compost.
5. Minimising the usage of loudspeakers.
6. Organising seminars, quiz, essay competition, drawing competitions for spreading environmental awareness, celebrating Vanamahotsava… etc.

Question 3.
We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?
Answer:
The local people need large quantities of firewood, small timber and thatch. Bamboo is used to make slats for huts, and baskets for collecting and storing food materials. Implements for agriculture, fishing and hunting are largely made of wood, also forests are sites for fishing and hunting.

In addition to the people gathering fruits, nuts and medicines from the forests, there cattle also graze in forest areas or food on the fodder which is collected from forests.

Because of these reasons the people who live in or around forests have authority to decide the management of forest produce.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) forests and wildlife
(b) water resources and
(c) coal and petroleum?
Answer:
a) Forests and wild animals.

1. cutting valuable trees should be avoided by destroying forest affects the quality of soil and water resources.
2. Hunting should be prohibited.
3. There should be wild sanctuaries which gives protection for wild animals.

b) Water Resources:
Answer:

1. Water resources should be free from pollution.
2. Excess usage of water should be avoided.

c) Coal and Petroleum:
Answer:
We should minimise the use of coal and petroleum, because these are fossil fuels. By burning these there are ill effects such as air pollution and acid rainfall etc.

Question 5.
What can you as an individual do to reduce your consumption of the various natural resources?
Answer:

1. We must have come across the five R’s to save the environment: Refuse, Reduce, Reuse, Repurpose and Recycle.
2. We should encourage tree plantation programmes.
3. We must reduce the burning of fossil fuels.
4. Encouragement should be given for harvesting the water.

Question 6.
List five things you have done over the last one week to —
(a) conserve our natural resources.
Answer:
We should travel in bus instead of using own vehicles or we should practice walking, we must use LED bulbs or fluorescent tubes in our homes. We must use the lift or taking the stairs, wearing an extra sweater or using a heating device (heater or sign) on cold days.

(b) increase the pressure on our natural resources.
Answer:

1. We should grow Number of trees around our house.
2. Reducing own vehicles by using public transport system or by, walking.
3. There should not be more factories.
4. We must prevent soil erosion.
5. We must reduce the usage of vehicles to avoid air pollution.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your lifestyle in a move towards sustainable use of our resources?
Answer:
We need to change our lifestyles so that we can use natural resources on a sustainable basis. The changes which can be brought about are as follows:

• Stop cutting trees and start planting trees.
• Use LED bulbs and fluorescent tubes.
• Take the stairs and avoid using lifts.
• During summers use bamboo made fans avoid air coolers and electricians.
• Use more of public transport.
• Let our conscience be always alert not to pollute the environment from any of our activities.

KSEEB SSLC Class 10 Science Chapter 16 Additional Questions and Answers

1. Fill in the blanks

Question 1.
…… and …… are followed as means of protection of nature and natural resources.
Answer:
traditions, customs and rituals.

Question 2.
Prticipation of the …… can indeed lead to the efficient management of forests.
Answer:
local people

Question 3.
Irrigation methods like have been used in various parts of India since ancient times.
Answer:
dams, tanks and canals.

Question 4.
…… and …… were formed from the degradation of biomass millions of years ago.
Answer:
Coal, petroleum

Question 5.
Fossil fuels contain carbon along with …… also.
Answer:
hydrogen, nitrogen and sulphur.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
i) x + 1
ii) \(x-\frac{1}{2}\)
iii) x
iv) x + π
v) 5 + 2x
Answer:
i) p(x) = x³ + 3x² + 3x + 1 g(x) = x – 1
Let x – 1 = 0, then
x = 1.
As per Remainder theorem, r(x) = p(x) = p(a)
p(x) = x³ + 3x² + 3x + 1
p(1) = (1)³ + 3(1)² + 3(1) + 1
= 1 + 3(1) + 3(1) + 1
= 1 + 3 + 3 + 1
P(1) = 8
∴ r(x) = p(x) = 8
∴ Remainder is 8.

ii) p(x) = x³ + 3x² + 3x + 1 g(x) = \(x-\frac{1}{2}\)
If \(x-\frac{1}{2}=0\) then \(x=\frac{1}{2}\)
p(x) = x³ + 3x² + 3x + 1

∴ r(x) = p(x) = p(a) = \(\frac{15}{8}\)
∴ Remainder is \(\frac{15}{8}\)

iii) p(x) = x³ + 3x² + 3x + 1
g(x) = x
If x = 0, then
p(x) = x³ + 3x² + 3x + 1
p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 3(0) + 3(0) + 1
= 0 + 0 + 0 + 1
p(0) = 0
Remainder r(x) = 1.

iv) p(x) = x³ + 3x² + 3x + 1
g(x) = x + π
If x + π = 0, then x = -π
p(x) = x³ + 3x² + 3x + 1
p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1
p(-πt) = -π²³ – 3π² – 3π + 1
r(x) = -π³ – 3v² – 3π+1

v) p(x) = x³ + 3x² + 3x + 1
g(x) = 5 + 2x
If 5 + 2x = 0, then 2x = -5
\(x=-\frac{5}{2}\)
p(x) = x³ + 3x² + 3x + 1

Free handy Remainder Theorem Calculator tool displays the remainder of a difficult polynomial expression in no time.

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Answer:
p(x) = x³ – ax² + 6x – a
If g(x) = x – a, then r(x) = ?
Let x – a = 0, then x = a
p(x) = x³ – ax² + 6x – a
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
∴ p(a) = 5a
∴ r(x) = p(a) = 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
p(x) = 3x³ + 7x
Let g(x) = 7 + 3x = 0. then
If 7 + 3x = 0, then 3x = -7
\(x=-\frac{7}{3}\)
If p(x) is divided by p(z), remainder r(x) – 0, then g(x) is a factor.
p(x) = 3x³ + 7x

Here, \(r(x)=-\frac{590}{9}\). This is not equal to Zero.
Hence 7 + 3x is not a factor of p(x).

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Solutions Chapter 3 Lines and Angles Exercise 3.3.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.3

Question 1.
In Fig. 3.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Answer:

Arms of the ∆PQR QP and RP are produced to S and T and ∠PQT = 110°, ∠SPR = 135°. ∠PRQ =?
Straight-line PR is on straight line SQ.
∠SPR and ∠RPQ are Adjacent angles.
∴ ∠SPR + ∠RPQ = 180°
135 + ∠RPQ = 180°
∴ ∠RPQ =180 – 135
∠RPQ = 45° (i)
Similarly QP straight line is on straight line TR.
∠TQP and ∠PQR are Adjacent angles.
∴ ∠RQP + ∠PQR = 180°
110 + ∠PQR = 180°
∠PQR = 180 – 110
∴ ∠PQR = 70°
Now, in ∆PQR,
∠QPR + ∠PQR + ∠PRQ = 180°
45 + 70 + ∠PRQ = 180°
115 + ∠PRQ = 180°
∠PRQ = 180 – 115
∴ ∠PRQ = 65°.

Reference Angle Calculator. The reference angle is defined as the smallest possible angle made by the terminal side of the given angle with the x-axis.

Question 2.
In Fig. 3.40, ∠X = 62°. ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ

Answer:
In this figure, ∠X = 62°
∠XYZ = 54°
In ∆XYZ, YO and ZO are angular bisectors of ∠XYZ and ∠XZY.
Then, ∠OZY =?
∠OYZ =?

In ∆XYZ
∠X + ∠Y + ∠Z = 180°
62 + 54 + ∠Z= 180
116 + ∠Z= 180
∠Z= 180- 116
∴ ∠Z = 64°
YO is the angular bisector of ∠Y
∴ ∠OYZ = \(\frac{54}{2}\) = 27°
ZO is the angular bisector of ∠Z
∴ ∠OZY = \(\frac{64}{2}\) = 32°
∴ ∠OZY = 32°
Now, in ∆OYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
27 + 32 + ∠YOZ = 180
59 + ∠YOZ = 180
∠YOZ = 180 – 59
∴∠YOZ = 121°
∴ ∠OZY = 32°
∠YOZ = 121°

Question 3.
In Fig. 3.41, if AB||DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Answer:
If AB || DE, ∠BAC = 35, ∠CDE = 53 then ∠DCE = ?
AB || DE, AE is the bisector.
∴∠BAC = ∠DEC = 35° (∵ Alternate angles)
∴∠DEC= 35°

Now in ∆CDE,
∠DCE + ∠CDE + ∠CED = 180°
∠DCE + 53 + 35 = 180
∠DCE + 88 = 180
∠DEC = 180 – 88
∴ ∠DCE = 92°.

Question 4.
In Fig. 3.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Answer:
PQ and RS straight lines intersect at T.
If ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, then ∠SQT =?

In ∆PRT,
∠RPT + ∠PRT + ∠PTR = 180°
95 + 40 + ∠PTR = 180°
135 + ∠PTR = 180
∠PTR = 180 – 135
∴ ∠PTR = 45°
∠PTR = ∠STQ = 45° (∵ Vertically opposite angles)
In ∆TSQ,
∠STQ + ∠TSQ + ∠SQT =180
45 + 75 + ∠SQT = 180
120 + ∠SQT = 180
∴∠SQT = 180 – 120
∴ ∠SQT = 60°.

Question 5.
In Fig. 3.43, PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°. then find the value of x and y.

Answer:
PQ ⊥ PS, PQ ⊥ SR, ∠SQR = 28°, ∠QRT = 65°, Then x = ?, y = ?

Solution: ∠QRT + ∠QRS = 180° (∵ Linear pairs)
65 + ∠QRS = 180
∠QRS = 180 – 65
∴∠QRS =115°
In ∆SRQ,
∠QRS + ∠SQR + ∠RSQ = 180°
115 + 28 + ∠RSQ = 180
∴∠RSQ =180 – 143
∴∠RSQ = 37
Now, ∠RSQ = ∠PQS
37° = x
∴x = 37
In ∆SPQ,
∠SPQ + ∠PSQ + ∠PQS = 180°
90 + y + 37 = 180
∴y = 180- 127
∴y = 53°.

Question 6.
In Fig. 3.44, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\)∠QPR.

Answer:
Data: Arm QR of ∆PQR is produced upto S. Angular bisectors of ∠PQR and ∠PRS meet at T.
To Prove: ∠QTR = \(\frac{1}{2}\) ∠QPR

Proof: Let ∠PQR = 50°, and ∠PRS = 120°
∠PQT = ∠TQR = 25°
∠PRT = ∠TRS = 60°
QR arm of ∆ PQR is produced upto S.
∴Exterior angle ∠PRS = ∠PQR + ∠QPR
120 = 50 + ∠QPR
∴ ∠QPR = 120 – 50
∠QPR = 70°
∴ ∠PRQ = 60°
Now, in ∆TRQ,
∠TQR + ∠TRQ + ∠QTR = 180°
25 + 120 + ∠QTR = 180°
145 + ∠QTR = 180°
∠QTR = 180 – 145
∴ ∠QTR = 35°
Now, ∠QTR = 35° ∠QPR = 70°
∠QTR = \(\frac{70}{2}\)
∴ ∠QTR = \(\frac{10}{2}\) x ∠QPR.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
i) x = 0
ii) x = -1
iii) x = 2
Answer:
i) f(x) = 5x – 4x² + 3 x = 0 then,
f(0) = 5(0) – 4(0)² + 3
=0 – 0 + 3
f(0) = 3

ii) f(x) = 5x – 4x² + 3 x = -1 then,
f(-1) = 5(-1) – 4(-1)2 + 3
= -5 – 4(+1) + 3
= -5 – 4 + 3
= -9 + 3
f(-1) = -6

iii) f(x) = 5x – 4x² + 3 x = 2 then,
f(2) = 5(2) – 4(2)² + 3
= 5(2) – 4(4) + 3
= 10 – 16 + 3
= 13 – 16
f(2) = -3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials :
i) p(y) = y² – y + 1
ii) p(t) = 2 + t + 2t² – t³
iii) p(x) = x³
iv) p(x) = (x – 1) (x + 1)
Answer:
i) (a) p(y) = y² – y + 1
p(0) = (0)² – 0 + 1
= 0 – 0 + 1
∴ p(0) = =1

(b) p(y) = y² – y + 1
p(1) = (1)² – 1 + 1
= 1 – 1 + 1
∴ p(1) = 1

(c) p(y) = y² – y +
p(2) = (2)² – 2 + 1
= 4 – 2 + 1
= 5 – 2
∴ p(2) = 3

ii) (a) p(t) = 2 + t + 2t² – t³
p(0) = 2 + 0 + 2² – (0)³
= 2 + 0 + 0 – 0
∴ p(0) = 2

(b) p(t) = 2 + t + 2t² – t³
p(1) = 2 + 1 + 2(1)² – (1)³
= 2 + 1 + 2(1) – 1
= 2 + 1 + 2 – 1
∴ p(1) = 4

(c) p(t) – 2 + t + 2t² – t³
p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 2(4) – 8
= 2 + 2 + 8 – 8
∴ p(2) = 4

iii) (a) p(x) = x³
p(0) = (0)³
p(0) = 0

(b) p(x) = x³
p(1) = (1)³
∴ p(1) = 1

(c) p(x) = x³
p(2) = (2)³
∴ p(2) = 8

iv) (a) p(x) = (x – 1)(x + 1)
p(x) = x² – 1 [∵ (a + b)(a – b) = a² – b]
p(0) = (0)² – 1
= 0 – 1
∴ p(o) = -1

(b) p(x) = (x – 1)(x + 1)
p(x) = x² – a [∵ (a + b)(a – b) = a² – b²]
p(1) = (1)² – 1
= 1 – 1
∴ p(1) = 0

(c) p(x) = (x – 1)(x + 1)
p(x) = x² – 1 [∵ (a + b)(a – b) = a² – b²]
p(2) = (2)² – 1
= 4 – 1
∴ p(0) = 3

Question 3.
Verify whether the following are zeroes of the polynomial, induced against them.
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)
ii) p(x) = 5x – π; \(x=\frac{4}{5}\)
iii) p(x) = x² – 1; x = 1, -1
iv) p(x) = (x + 1) (x – 2); x = -1, 2
v) p(x) = x²; x = 0
vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
vii) p(x) = 3x² – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)
Answer:
i) p(x) = 3x + 1; \(x=-\frac{1}{3}\)

Here value of polynomial is zero.
\(x=-\frac{1}{3}\) is not zero of the polynomial

ii) p(x) = 5x – π; \( x=\frac{4}{5}\)

Here value of polynomial is not zero.
\(x=\frac{4}{5}\) is not zero of the polynomial

iii) p(x) = x² – 1; x = 1, -1
p(1) = (1)² – 1
= 1 – 1
p(1) = 0
Here value of p(x) is zero.
hence its zero is 1.
p(x) = x² – 1; x = -1
p(-1) = (-1)² – 1
= 1 – 1
p(-1) = 0
Here value of p(x) is zero.
∴ -1 is zero.

iv) p(x) = (x – 1)(x – 2); x = -1, 2
p(x) = x² – 2x + x – 2
p(x) = x² – x + 2 x = -1
p(-1) = (-1)² – (-1)² + 2
= 1 + 1 + 2
p(-1) = 4
Here value of polynomila is not zero.
∴ -1 is not zero.
p(x) = x² – x + 2 x = 2
p(2) = (2)² – (2) + 2
= 4 – 2 + 2
p(-1) = 4
Here value of polynomial is not zero.
∴ 2 is not zero.

(v) p(x) = x²; x = 0
p(0) = (0)²
p(0) = 0
Here value of p(x) is zero.
∴ 0 is its zero.

vi) p(x) = lx + m; \(x=-\frac{m}{l}\)
\(\mathrm{p}\left(-\frac{\mathrm{m}}{l}\right)=l\left(-\frac{\mathrm{m}}{l}\right)+\mathrm{m}\)
= -m + m
= 0
Here p(x) is zero.
∴ \(-\frac{\mathrm{m}}{l}\) is its zero.

(vii) p(x) = 3x² – 1; \(x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

viii) p(x) = 2x + 1; \(x=\frac{1}{2}\)

Here value of p(x) is not zero.
∴ \(\frac{1}{2}\) is not its zero.

Question 4.
Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Answer:
i) p(x) = x + 5
Let p(x) =0, then,
p(x) = x + 5 = 0
x = 0 – 5
∴ x = -5
-5 is zero of p(x).

ii) p(x) = x – 5
If p(x) = 0, then
p(x) = x – 5 = 0
x = 0 + 5
∴ x = 5
5 is the zero of p(x).

iii) p(x) = 2x + 5
If p(x)= 0, then
p(x) = 2x + 5 = 0
2x = – 5
∴ \(x=\frac{-5}{2}\)
\(\frac{-5}{2}\) is the zero of p(x).

iv) p(x) = 3x – 2
If p(x)= 0, then
p(x) = 3x – 2 = 0
3x = 2
∴ \(x=\frac{2}{3}\)
\(\frac{2}{3}\) is the zero of p(x).

v) p(x) = 3x
If p(x) = 0, then
p(x) = 3x = 0
∴ \(x=\frac{0}{3}\)
\(\frac{0}{3}\) is the zero of p(x)

vi) p(x) = ax, a ≠ 0
If p(x)= 0, then
p(x) = ax = 0
∴ \(x=\frac{0}{a}\) ∴ x = ∞(infinity)
∞ is the zero of p(x).

vii) p(x) = cx + d, c ≠ 0, c, d are real numbers
If p(x)= 0, then
p(x) = cx + d = 0
cx = 0 – d
cx = -d
∴ \(x=-\frac{d}{c}\)
\(-\frac{\mathrm{d}}{\mathrm{c}}\) is the zero of p(x).

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
i) 4x² – 3x + 7
ii) y² + \(\sqrt{2}\)
iii) \(3 \sqrt{t}+t \sqrt{2}\)
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
v) x¹⁰ + y³ + t⁵⁰
Answer:
i) 4x² – 3x + 7
Here polynomial has one variable, i.e. x
ii) y² + \(\sqrt{2}\)
Here polynomial has one variable, i.e. y
iii) \(3 \sqrt{t}+t \sqrt{2}\)
This is polynmomial with one variable, because T is only one variable.
iv) \(\mathrm{y}+\frac{2}{\mathrm{y}}\)
Here polynomial has one variable, ie. y.
v) x¹⁰ + y³ + t⁵⁰
This polynomial is not having one variable because here 3 variables means ‘x’, y and ‘t’ are there.

Question 2.
Write the coefficients of x in each of the followng :
i) 2 + x² + x
ii) 2 – x² + x³
iii) \(\frac{\pi}{2}\)x² + x
v) \(\sqrt{2} \mathrm{x}\) – 1
Answer:
i) 2 + x² + x
Here, coefficient of x² is 1.
ii) 2 – x² + x³
Here coefficient of x² is -1
iii) \(\frac{\pi}{2}\)x² + x
Here coefficient of x² is \(\frac{\pi}{2}\).
iv) \(\sqrt{2} \mathrm{x}\) – 1
Here coefficint of x² is -1.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100
Answer:
i) A Bionomial of degree 35
E.g. f(x) = – x³⁵ + 10
ii) A binomial of degree 100
E.g. f(y) = – y¹⁰⁰.

Question 4.
Write the degree of each of the following polynomials :
i) 5x³ + 4x² + 7x
ii) 4 – y²
iii) 5t – \(\sqrt{7}\)
iv) 3
Answer:
i) 5x³ + 4x² + 7x Highest power (degree) 3
ii) 4 – y² Highest power degree) 2
iii) 5t – \(\sqrt{7}\) Highest power (degree) 1
iv) 3 Highest power (degree) 0

Question 5.
Classify the folloiwng as linear, quadratic and cubic polynomials :
i) x² + x
ii) x – x³
iii) y + y² + 4
iv) 1 + x
iii) 3t
iv) r²
vii) 7x³
Answer:

Linear Polynomial	Quadratic Polynomial	Cubic Polynomial
iv) 1 + x	i) x2 + x	iii) y + y2 + 4
		ii) x – x3
(v) 3t	(vi) r2	(vii) 7x3

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.1, drop a comment below and we will get back to you at the earliest.