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

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:
We should switch off the electrical appliances when not in use. Water and food should not in wasted. Close the tap when not in use, Dump the objects made of plastic and glass in designated recycling boxes. Plastic, paper or glass must be recycled or reused and not dumped with other wastes. This is because objects made of plastic do not get decomposed easily. Besides soil fertility, they badly affect our environment. We should dispose the wastes safely and not disperse in public places. These are a few things that can be done to become more environment-friendly.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
There should be a judicious use of natural resources as they are limited in nature. We should not exploit resources for our short term gains as this would only lead to depletion of natural resources for the present generation as well as generations to come. Hence, we say that there are hardly any advantages of exploiting natural resources 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:
In the case of a long-time perspective in managing our resources, these resources will last for the generations to come. This management ensures uniform distribution among the people. It conserves the ‘ natural resources for many years and not just for a few years, as in the case of a short-term perspective in conserving natural resources.

Question 4.
Why do you think there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Natural resources of the Earth must be distributed among the people uniformly so that each and every one gets his share of the resource. Human greed, corruption and the lobby of the rich and powerful are thq forces working against an equitable distribution of 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:
One of the traditional systems of water harvesting used in our region is tanks. There may be other systems like ponds, water reservoirs etc.

Question 2.
Compare the above system with the probable systems in hilly/ mountainous areas or plains or plateau regions.
Answer:
In plains, the water harvesting structures are crescent-shaped earthen embankments. These are low, straight and concrete. In hilly regions, the system of canal irrigation called Kulyths is used for water harvesting. This involves a collection of rain water in a stream, which is then diverted into manmade channels down the hill sides.

Question 3.
Find out the source of w ater in your region/locality. Is water from this source available to all people living in that area?
Answer:
The source of water in our region is ground water. Water from the source is available to all the people living in that area.

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
(c) coal and petroleum?
Answer:
(a) Forest and wildlife:

  1. We should protest against the cutting of trees (deforestation).
  2. We should protest against the poaching of wild animals.
  3. We should stop the annexation of forest land for our use.

(b) Water resources:

  1. Turn the taps off while brushing or bathing and repair leaking taps.
  2. We should practice rainwater harvesting.
  3. We should avoid the discharge of sewage and other wastes into rivers and other water resources.

(c) Coal and Petroleum:

  1. We should take a bus or practice car pooling to avoid excessi ve use of petroleum.
  2. We should stop using coal.as a fuel (angithis).
  3. We should use alternative source of energy such as hydro¬energy and solar energy instead of depending largely on coal and petroleum.

Question 5.
What can on as an individual do to reduce your consumption of the various natural resources?
Answer:
Natural resources such as water, forests, coal and petroleum etc. are important for the survival of human beings. The ways in which we can reduce the

consumption of various natural resources are as follows:

  1. We should stop the cutting of trees (deforestation).
  2. We should use recycled paper to reduce the cutting down of trees
  3. We should not waste water.
  4. We should practice rainwater harvesting.
  5. We should practice car pooling to avoid the excessive use of petroleum.
  6. We should use alternative sources of energy such as hydro-energy and solar energy.

Question 6.
List five things you have done over the last one week to:
(a) conserve our natural resources.
(b) increase the pressure on our natural resources.
Answer:
(a) To conserve our natural resources:

  1. Travelled by a CNG bus for long distances and walk for short distances.
  2. Used recycled paper.
  3. Threw biodegradable and non-biodegradable waste into separate bins.
  4. Planted trees.
  5. Harvested rainwater.

(b) To increase the pressure on our natural resources:

  1. Used non-renewable resources of energy.
  2. Wasted water.
  3. Wasted electricity.
  4. Used plastics and polythene bags for carrying goods.
  5. Used escalators.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your life-style in a move towards a sustainable use of our resources?
Answer:
One should incorporate the following changes in life-style in a move towards a sustainable use of our resources:

  1. Stop cutting trees and practice plantation of trees.
  2. Stop using plastic and polythene bags for carrying goods.
  3. Use recycled paper.
  4. Throw biodegradable and non-biodegradable waste into separate bins.
  5. Waste minimum amount of water while using and repair leaking taps.
  6. Practice rainwater harvesting.
  7. Avoid using vehicles for short distances. Instead, one can walk or cycle to cover short distances. To cover long distances, one should take a bus instead of using personal vehicles.
  8. Switch off electrical appliances when not in use.
  9. Use fluorescent tubes in place of bulbs to save electricity.
  10. Take stairs and avoid using lifts.
  11. During winters, wear an extra sweater to avoid using heaters.

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 SSLC Class 10 Science Solutions Chapter 15 Our Environment

KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment.

Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment

KSEEB SSLC Class 10 Science Chapter 15 Intext Questions

Text Book Part I Page No. 140

Question 1.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
Several hierarchical level in an ecosystem are called trophic levels. They are based on following facts:

  • Energy and its transportation in form of food from one end to other.
  • Biomass balance in ecosystem.
  • Food chain of pond is a simple food chain. Example:

Weeds → Small pond animals → Fish (Rohu) → Man → Big fishes

Trophic level of a Pond:

  1. Producer: Weeds and plants of pond like ‘duck weed’, ‘lotus’ or ‘typha’ are at lowest level or bottom of a pond which produce food for pond animals and energy transfer.
  2. Primary consumers: Algae, bacteria, Phytoplankton eat producer and conserve the energy in their body.
  3. Secondary consumers: Fishes, crabs etc. are those which feeds on primary consumers.
  4. Top consumers: Bigger fishes and humans are top consumers.

Question 2.
What is the role of decomposers in the ecosystem?
Answer:
Decomposers has very important place in any ecosystem as every ecosystem contains dead and decaying materials which is of no use and most of the time harmful for the system too. Decomposers act over these materials and change or break them to the components easily invisible to the environment. Hence, decomposers act over dead and decayed materials and convert them to the components which can be easily mixed up with other environmental factors.

Text Book Part I Page No. 142

Question 1.
Why are some substances biodegradable and some non-biodegradable?
Answer:
Substances that are broken down by biological processes are said to be biodegradable. Eg: paper and peel of a fruit. But plastic leather etc. are not broken down. These are called Non-biodegradable.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.
Answer:

  1. Leaves of the plants decay and reduces soil fertility,
  2. Bio-degradable substances have carbon. When this is burnt, CO2 and CO are produced and causes air pollution.

Question 3.
Give any two ways in which non-biodegradable substances would affect the environment.
Answer:

  1. As these are not decomposing, they cause air pollution and water pollution.
  2. Plastic enters stomach of many animals and causes death of animals.

Text Book Part I Page No. 144

Question 1.
What is ozone and how does it affect any ecosystem?
Answer:
Ozone is an atmospheric gas and its chemical formula is 03. It is formed by the union of three atoms of oxygen.

Formation of Ozone:
1. In the presence of UV rays, oxygen molecules break down to its component or forms to nascent oxygen atom.
MP Board Class 10th Science Solutions Chapter 15 Our Environment 1
2. This oxygen atom combines with another molecule of oxygen and forms ozone molecule.
MP Board Class 10th Science Solutions Chapter 15 Our Environment 2

Effect of Ozone:

Ozone is not good for health or to be consumed as it is poisonous in nature but it plays an important role in environment protection. The living organisms and factors which prevails life at earth, like water and atmospheric gases get distracted by UV rays coming from sun. Ozone forms a layer over our atmosphere and absorb harmful UV rays.

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
Answer:
(a) Waste disposal and management is a big issue nowadays as new technologies produce many useful products but at the end :t turn to a garbage because all the things are non-biodegradable in nature. Waste management require 3R principle i.e., Reduce, reuse and recycle.

(b) Waste disposal can be managed by:

  1. Categorizing the waste and disposing accordingly.
  2. Collecting things which can be recycled after use and send them to recycling units.
  3. Making manure by unrequired, spoiled and left over food products.

KSEEB SSLC Class 10 Science Chapter 15 Textbook Exercises

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather.
(b) Grass, wood and plastic.
(c) Fruit-peels, cake and lime-juice.
(d) Cake, wood and grass.
Answer:
(b) Grass, wood and plastic.

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango.
(b) Grass, goat and human.
(c) Goat, cow and elephant.
(d) Grass, fish and goat.
Answer:
(b) Grass, goat and human.

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping.
(b) Switching off unnecessary lights and fans.
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above.
Answer:
(d) All of the above.

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
Ecosystem will get disturbed due to killing of organisms in one trophic level:
1. If producers are killed: No energy will be converted to food for primary consumers, so they will die, since primary consumers are not available, secondary and top consumers as well will collapse.

2. If primary consumers are killed: Secondary and top consumers will die immediately in absence of food (in form of primary consumer) and if all fauna will die, population will stop and humans will also be effected in absence of fauna: Similarly, CO2 balance will be reduced and cause depletion of flora,

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:

  • No, same effect will be caused and that is disturbance in ecosystem.
  • No, there is no way to remove a trophic level organism without harming ecosystem.

Question 6.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:
Some non-biodegradable substances (like pesticides) somehow enters the food chain or web and since, they do not get decomposed at any level and get accumulated in the bodies of top level consumers, this phenomenon is called biological magnification.

Yes, at lower level it is lesser and at higher it is more.

Question 7.
What are the pro Problems caused by the non-biodegradable wastes that we generate?
Answer:
Technological development leads to a better and comfortable life a way and garbage enhancement in other way. We always think of a product which last for long time or a durable product which do not get affected by atmospheric changes like waterproof heat resistance product etc. But after use due to its nature, it do not get degraded and remain garbage for very long time.

Problem generated by garbage:

  1. They react with other materials and atmospheric gases and form many harmful by products and chemicals.
  2. Harm the animals which mistakenly eat them.
  3. Acquire huge land and engage manpower to manage them all the time.
  4. Pollution rises.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Answer
Yes, biodegradable waste will be decomposed with lesser time and its component will be mixed with environment. Hence, it will leave no impact on environment.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
Ozone is an atmospheric gas and its chemical formula is O3.

Damage to the ozone layer will lead:

  1. More UV rays to enter the atmosphere.
  2. Damage to flora and fauna.
  3. Diseases to human beings related to exposure to sun.

Damage control. Protection to ozone layer can be initiated by:

  1. Main reason for ozone layer depletion is pollution. Hence, ceasing pollution is a must to stop the depletion.
  2. Coolant like CFC must be banned which are considered to be most harmful to ozone layer.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter Chapter 14 Sources of Energy

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

Karnataka SSLC Class 10 Science Solutions Chapter 14 Sources of Energy

KSEEB SSLC Class 10 Science Chapter 14 Intext Questions

Text Book Part I Page No. 110

Question 1.
What is a good source of energy?
Answer:
A good source of energy fulfills the following criteria:

  1. It does a huge amount of work per unit mass.
  2. It is easily accessible and economical.
  3. It is easy to store.
  4. It produces less amount of smoke.

Question 2.
What is a good fuel?
Answer:
A good fuel produces a huge amount of heat on burning, does not produce a lot of smoke, and is easily available.

Question 3.
If you could use any source of energy for heating your food which one would you use and why?
Answer:
I would prefer to use any renewable energy as solar energy because it will never be insufficient and will not harm atmosphere by generating pollution.

Text Book Part I Page No. 115

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

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

Question 2.
Why are we looking at alternate sources of energy?
Answer:
Alternative sources of energy of present time i.e., fossil fuel is very much required because:

  1. It is non renewable, so if not now, then alternatively a day will come when we have to look for its alternative.
  2. These sources of energy are limited and can not replenish on their own.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
Transformation of wind and water energy use:

  1. Windmills were used to harness wind energy to do mechanical work such as lifting/drawing water from well.
  2. Waterfalls were used as a source of potential energy which was converted to electricity with the help of turbines.

Text Book Part I Page No. 120

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

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

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

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

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

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

Text Book Part I Page No. 120

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

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

Text Book Part I Page No. 121

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

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

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

Question 2.
Give the names of two energy sources that you would consider to be exhaustible.Give reasons for your choices.
Answer:
Energy sources like coal and petroleum are exhaustible. These take millions of years to form and once all used up will not be available for present generation.

KSEEB SSLC Class 10 Science Chapter 14 Textbook Exercises

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

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

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

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

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

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

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

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

Question 6.
What are the limitations of extracting energy from:
(a) The wind?
(b) Waves?
(c) Tides?
Answer:
Limitation are of as follows:

From: (a) Wind energy:

  1. Periodic, not continuous.
  2. Small output.
  3. Large area required

From: (b) Wave energy:

  1. Periodic not continuous
  2. Small availability
  3. Large area required

From: (c) Tide energy:

  1. Again periodic.
  2. Small availability
  3. Lower output.

Question 7.
On what basis would you classify energy sources as:

  1. Renewable and non-renewable?
  2. Exhaustible and inexhaustible?

Are the options given in (i) and (ii) the same?
Answer:
Energy sources would be classified as (i) renewable and (ii) non-renewable.
Renewable sources of energy are inexhaustible.
Non-renewable sources of energy are inexhaustible.
Therefore, options (i) and (ii) are the same.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
Qualities of an ideal source of energy:

  1. It must produce large amount of energy per unit volume or mass.
  2. It must not leave more residue and harmful gases on burning.
  3. It must be convenient and economical in use.
  4. It must be easy to handle and portable.

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

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

Disadvantages of using a solar cooker.

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

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

  1. Burning of fossil fuels produces acid rain which damages plants, soil and aquatic life.
  2. Felling of trees in the forests to obtain firewood is causing soil erosion and destroying wild life.
  3. Nuclear power plants increases harmful radio active rays which affect the environment.
  4. Construction of dams on rivers is destroying large ecosystems which get submerged under water in dams.

Steps to reduce energy consumption:

  1. Reduce wastage of electricity by switching off unnecessary electrical appliances.
  2. Using fuel saving devices such as pressure cookers etc. for cooking. Solar cookers should be used wherever possible.
  3. Reduce the use of petrol by walking or cycling for short distances.
  4. Send discarded items of paper, plastic, glass and metal for recycling to respective industries.
  5. Use of bio-gas as a fuel should be encouraged in rural areas.

KSEEB SSLC Class 10 Science Chapter 14 Additional Questions and Answers

Question 1.
Draw a neat diagram showing a model to demonstrate the process of thermoelectric production and label the parts.
Answer
KSEEB SSLC Class 10 Science Solutions Chapter Chapter 14 Sources of Energy Ad Q 1

Question 2.
Draw a neat diagram of a solar cooker and label the parts.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter Chapter 14 Sources of Energy Ad Q 2

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

KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1

KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1

Question 1.
Complete the following statements :

  1. Probability of an event E + Probability of the event ‘not E’ = _____
  2. The Probability of an event that cannot happen is _____. Such an event is called ____.
  3. The probability of an event that is certain to happen is ____. Such an event is called ____.
  4. The sum of the probabilities of all the elementary events of an experiment is ____.
  5. The probability of an event is greater than or equal to ____ and less than or equal to ____.

Answers:

  1. 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
  2. 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
  3. 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
  4. 1: The sum of the probabilities of all the elementary events of an experiment is 1.
  5. 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) It depends on various factors such as whether the car will start or not. So, the probability of car will start does not equal to the probability of car will not start.
∴ The outcomes are not equally likely.
(ii) It depends on the player’s ability. So, probability that the player shot the ball is not the same as the probability that the player misses the shot.
(iii) The outcomes are equally likely as the probability of answer either right or wrong is \(\frac{1}{2}\)
(iv) The outcomes are equally likely as the probability of ‘newly born baby to be either bay or girls’ is \(\frac{1}{2}\).

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer:
Because Tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game. By this coin may be head or tail (only one). It has an equally likely outcome.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
Since, the probability of an event cannot be negative.
∴ -1.5 cannot be the probability of an event.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’ ?
Solution:
IfP(E) = 0.05, then P(\(\overline{\mathrm{E}}\))=?
But, P(E) + P(\(\overline{\mathrm{E}}\)) = 1 .
∴ P(\(\overline{\mathrm{E}}\)) = 1 – 0.05
∴ P(\(\overline{\mathrm{E}}\)) = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since there are only lemon flavoured candies in the bag.
∴ Taking out orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Probability of taking out a lemon flavoured candy = 1.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let P(E) be an event 2 students have the same birthday i.e P(E) =?
P(\((\bar{E})\)) is an event 2 students not have same birth day P(\((\bar{E})\)) = 0.992
P(E) + P(\((\bar{E})\)) = 1
P(E) = 1 – P(\((\bar{E})\)) = 1.000 – 0.992 = 0.008
The probability of 2 students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Total number of balls = 3 + 5 = 8
∴ umber of possible outcomes = 8
(i) ∵ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P (red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac{3}{8}\)
(ii) Probability of the ball drawn which is not red = 1 – P(red) = \(1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}\)

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ?
(ii) white ?
(iii) not green ?
Solution:
A box contains 5 red marbles, 8 white marbles and 4 green marbles.
∴ Total number of marbles, n(S)
= 5 + 8 + 4
= 17
(i) Probability that the 1 red marble drawn is n(A) = 5
∴ Probability, P(A) = \(\frac{n(A)}{n(S)}=\frac{5}{17}\)
(ii) Possibility that 1 white marble drawn, n(B) = 8
∴ Probability, P(B) = \(\frac{n(B)}{n(S)}=\frac{8}{17}\)
(iii) Possibility that 1 not green marble ?
P(C) = 17 – 4 = 13 (∵ Except 4 green marbles)
∴ Probability, P(C) = \(\frac{n(C)}{n(S)}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin ?
(ii) will not be a Rs. 5 coin ?
Solution:
Number of 50 ps coins = 100
Number of Re. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
∴ Total number of coins, n(S) = 180
(i) Possibility of one 50 ps coin :
n(A) = 100
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14 Q 10
(ii) Possibility of one Rs. 5 coin:
n(B) = 180 – 10 = 170
(∵ 10 coins are Rs. 5)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14 Q 10.1

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Number of male fishes = 5
Number of female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 3

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see the Figure) and these are equally likely outcomes.
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12
What is the probability that it will point at
(i) 8?
(ii) an odd number
(iii) a number greater than 2 ?
(iv) a number less than 9 ?
Solution:
Total numbers in the spinning = 8
∴ n(A) = 1
(i) Possibility that points one number is 8 : n(A) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.1
(ii) Possibility that points one odd number : 1, 3, 5, 7
∴ n(B) = 4
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.2
(iii) A number greater than 2 :
3, 4, 5, 6, 7, 8
∴ n(C) = 6
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.3
(iv) A number less than 9 :
1, 2, 3, 4, 5, 6, 7, 8
∴ n(D) = 8
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 12.4

Question 13.
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Since, numbers on a die are 1, 2, 3, 4, 5 and 6.
∴ Total number of possible outcomes = 6
(i) Since 2, 3 and 5 are prime number.
∴ Favourable outcomes = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q13

(ii) Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 10

(iii) Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 11

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour,
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds.
Solution:
Total number of cards = 52
(i) P(One red king) = \(\frac{2}{52}=\frac{1}{26}\)
(ii) Number of a face card :
4 king, 4 queen, 4 jack = Total 12 cards
∴ P(1 face card) = \(\frac{12}{52}=\frac{3}{13}\)
(iii) One red colour face card = 6
∴ P(1 red face card) = \(\frac{6}{52}=\frac{3}{26}\)
(iv) P(Heart Jack) = \(\frac{1}{52}\)
(v) P(1 spade) = \(\frac{13}{52}=\frac{1}{4}\)
(vi) P(Diamond card) = \(\frac{1}{52}\)

Question 15.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
We have five cards.
∴ Total number of possible outcomes = 5
(i) ∵ Number of queen = 1
∴ Number of favourable outcomes = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q15

(ii) The queen is drawn and put aside.
∴ Only 5 – 1 = 4 cards are left.
∴ Total number of possible outcomes = 4
(a) ∵ There is only one ace.
∴ Number of favourable outcomes = 1
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 19

(b) Since, the only queen has been put aside already.
∴ Number of favourable outcomes = 0
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 20

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens =12
Number of good pens =132
Total Number of pens = 12 + 132 = 144
The probability that the pen taken out is good one,
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 16

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Since, there are 20 bulbs in the lot.
Total number of possible outcomes = 20
(i) ∵ Number of defective bulbs = 4
∴ Favourable outcomes = 4
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q15 Q17

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Number of remaining bulbs = 20 – 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 – 4 = 15
⇒ Number of favourable outcomes = 15
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 23

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
Number of discs which are numbered from 1 to 30,
n(S) = 90
(i) A two-digit number :
Out of 90, one digit number = 9
∴ 2-digit numbers = 90 – 9 = 81
∴ 2-digit numbers, n(E) = 81
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18
(ii) A perfect square number :
1, 4, 9, 16, 25, 36, 49, 64, 81
∴ n(E) = 9
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18.1
(iii) A number divisible by 5 :
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 .
∴ n(E) = 18
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 18.2

Question 19.
A child has a die whose six faces show the letters as given below :
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19
The die is thrown once. What is the probbility of getting (i) A ?, (ii) D ?
Solution:
Number of faces in a die is 6.
∴ n(S) = 6
(i) The probability of getting face ‘A’:
∵ n(E) = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19.1
(ii) The probbility of getting face ‘D’ is 1.
n(E) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 19.2

Question 20.
Suppose you drop a die at random on the rectangular region as shown in the following figure. What is the probability that it will land inside the circle with diameter 1 m ?
(* Not from the examination point of view)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20
Solution:
(i) Length of a rectnagular region is 3m.
Breadth of a rectangular region is 2m.
∴ Area of rectangle = length × breadth
= 3 × 2 = 6 sq.m.
(ii) A circle with diameter 1 m.
∴ Radius, r = \(\frac{1}{2}\) m
Area of Circle = \(\pi r^{2}\)
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20.1
∴ The probability that die will land inside the circle
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 20.2

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
Solution:
Total number of ball pens = 144
⇒ Total number of possible outcomes = 144
(i) Since there are 20 defective pens.
∴ Number of good pens = 144 – 20 = 124
⇒ Number of favourable outcomes = 124
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 33

(ii) Probability that Nuri will not buy it = 1 – [Probability that she will buy it]
= \(1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}\)

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 22.
Refer to Example 13.
(i) Complete the following table:
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 22
(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore each of them has a probability \(\frac{1}{11}\). Do you agree with this argument ? Justify your answer.
Solution:

∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:
MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1 35

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
{H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H)
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game denoted by E.
∴ Favourable events are: {HHT, HTH, THH, THT, TTH, HTT}
⇒ Number of favourable outcomes = 6
∴ P(E) = \(\frac{6}{8}=\frac{3}{4}\)

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either Urne?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.)
Solution:
The possibility that one die is thrown:
n(S) = 6 × 6 = 36
(i) 5 will come up either time:
(1, 5), (2. 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3) (5, 4), (5, 6)
∴ Possible events = 11.
∴ Probability that 5 will not come up either time : 36 – 11 = 25
n(E) = 25
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 24
(ii) Possibility that 5 will come up at least ones: 11
∴ n(E) = 11
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 24.1

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{3}\)
Solution:

(i) Possible outcomes, if two coins are tossed simultaneously: HH, TT, HT, TH.
∴ n(S) = 4
Out of these possibilities of getting 2 heads: 1 ➝ HH
∴ n(A) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 25
Out of these possibility of getting Tail : 1 ➝ TT
∴ n(B) = 1
n(B) = 1
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 25.1
Out of these possibilities of getting Head or Tail : 2 ➝ HT, TH
∴ n(C) = 2
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 25.2
∴ This Argument (i) is not correct.
(ii) Outcomes when one die is thrown:
1, 2, 3, 4, 5, 6 ∴ n(S) = 6
Out of these, one is odd number ➝ 1, 3, 5
∴ n(E) = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 25.3
Out of these, one is even number ➝ 2, 4, 6
∴ n(E) = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 Q 25.4
∴ This Argument (ii) is correct.

 

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.1, drop a comment below and we will get back to you at the earliest.

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

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

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

KSEEB SSLC Class 10 Science Chapter 5 Intext Questions

Text Book Part II Page No. 25

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

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

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

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

Text Book Part II Page No. 29

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

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

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

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

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

Text Book Part II Page No. 34

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

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

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

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

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

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

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

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

KSEEB SSLC Class 10 Science Chapter 5 Textbook Exercises

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

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

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

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

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

Question 6.
The position of three elements A, B and C in the Periodic Table are shown below –
KSEEB SSLC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements Ex Q 6
(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
a) A is a Non metal.
b) C is less reactive comparing to A.
c) C is smaller in size comparing to A.
d) Cation is formed by A.

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

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

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

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

Mendeleev s Periodic table:

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

Modern Periodic table:

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

(or)

Mendeleev’s Periodic table vs Modern Periodic table:

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

KSEEB SSLC Class 10 Science Chapter 5 Additional Questions and Answers

Fill in the blanks:

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

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

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

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

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

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

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

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

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

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

KSEEB SSLC Class 10 Science Chapter 10 Intext Questions

Text Book Part II Page No. 78

Question 1.
Define the principal focus of a concave mirror.
Answer:
Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Radius of curvature, R = 20 cm.
Radius of curvature of a spherical mirror = 2 × Focal length (f)
R = 2f f = \(\frac { R }{ 2 } \) = \(\frac { 20 }{ 2 } \) = 10 cm
Hence, the focal length of the given spherical mirror is 10 cm.

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object when object is placed between the pole and principal focus.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror when fitted at rear-view position of vehicles, it gives a wider field of view, with which driver can see most of the traffic behind him. Convex mirrors give a virtual, erect and diminished image of the objects in front of it. So, we prefer a convex mirror as a rear-view mirror in vehicles.

Text Book Part II Page No. 81

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Given: Radius of curvature, R = 32 cm
We know, Radius of curvature = 2 × Focal length (f)
R = 2f = \(\frac { R }{ 2 } \) = \(\frac { 32 }{ 2 } \) = 16 cm
Hence, the focal length of the given convex mirror is 16 cm.

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

Text Book Part II Page No. 86

Question 1.
A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
Yes, the light ray will bends towards the normal. As water is more denser to air and also have high refractive index, its speed slows down and it bends towards the normal. Hence, a ray of light entering from air into water will bend towards the normal.

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

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

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

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
The refractive index of diamond is given 2.42. This means that the speed of light in diamond will reduce by a factor equal to 2.42 as compared to its speed in air. In other words, the speed of light in diamond is 1 (2.42) times the speed of light invacuum.

Text Book Part II Page No. 94

Question 1.
Define 1 dioptre of power of a lens.
Answer:
The SI unit of power of lens is dioptre which is denoted by the letter D
∴ P = \(\frac { 1 }{ f } \) (in metres)
1 dioptre is defined as the nower of a lens of focal length 1 metre.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
Since, the image is real and of same size. The position of image should be at 2F. It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Hence, the needle is placed in front of the lens at a distance of 50 cm.
Object-distance, u = – 50 cm
Image-distance, v = 50 cm
Focal length = f
According to the lens formula, = \(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ f } \) = \(\frac { 1 }{ 50 } \) – \(\frac { 1 }{ (-50) } \) = \(\frac { 1 }{ 50 } \) + \(\frac { 1 }{ 50 } \) = \(\frac { 1 }{ 25 } \)
f = 25 cm = 0.25 m
MP Board Class 10th Science Solutions Chapter 10 Light Reflection and Refraction 5

KSEEB SSLC Class 10 Science Chapter 10 Textbook Exercises

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

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

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

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

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

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror?
What is the nature of the image? Is the image larger or smaller than the object?
Draw a ray diagram to show the image formation in this case.
Answer:
Distance of the object = o to 15 cm
Nature of image = virtual, erect and bigger than object
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 7

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

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

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

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
This lens gives full image, though one-half of this lens is covered with black paper as shown in below figure.
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 9
As shown in figure light ray moves in half part and image is formed in another part of the lens.
If black paper is covered in lower part: Following figure explain this
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 9

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Height of object, h = 5 cm
Distance of object from converging lens u = 25 cm
Focal length of lens f = 10 cm
As per lens formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10
Converging lens, \(\frac{h_{1}}{h_{0}}=\frac{v}{u}\)
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10.3
= – 3.3 cm
Images is inverted and it is formed it is formed behind the lens about 16.7 cm. Its height is 3.3 cm.
Diagram is as follows:
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 10.2

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Focal length (F1) of concave lens
f = 15 cm
Image distance, v = – 10 cm
As per lens formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 11
u = -30 cm
Negative sign indicates, image is front of the lens about 30 cm.
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 11.1

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Focal length of convex mirror, f = + 15 cm
Object-distance, u = – 10 cm
According to the mirror formula,
\(\frac { 1 }{ v } \) + \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \) ⇒ \(\frac { 1 }{ v } \) = \(\frac { 1 }{ f } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ 15 } \) + \(\frac { 1 }{ 10 } \) = \(\frac { 25 }{ 150 } \)
v = 6 cm
The positive value of v indicates that the image is formed behind the mirror,
Magnification, m = \(\frac { -v }{ u } \) = \(\frac { -6 }{ -10 } \) = + 0.6
Image is located at a distance 6 cm from the mirror on the other side of the mirror.

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

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object distance, u = 20 cm
Height of object h = 5 cm
Radius of curvature R = 30 cm
R = 2f, f = 15 cm
As per mirror formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 14
Positive sign indicates image is formed behind the mirror
Magnification \(=\quad \frac { Image\quad distance }{ Object\quad distance } \)
\(=\frac{-8.57}{-20}=0.428\)
Image is behind the mirror because magnification is positive
Magnification \(=\frac{\text { Image distance }}{\text { Object distance }}\)
\(=\frac{h^{1}}{h}\)
h1 = m × h = 0.428 × 5 = 2.14 cm

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focussed image can be obtained? Find the size and nature of the image.
Answer:
Objective distance, u = 27 cm
Object height, h = 7 cm
Focal length, f = 30 cm
R = 2f, f = -18 cm
As per mirror formula
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15
Screen should be placed in front of mirror at the distance of = 54 cm
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15.1
Negative sign of magnification indicates image is real Magnification,
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ex Q 15.2
h1 = 7 × (2) = -14 cm
Image is inverted because of negative sign.

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

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

KSEEB SSLC Class 10 Science Chapter 10 Additional Questions and Answers

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

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

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

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

KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction Ad Q 5.1
(a) Concave mirror (b) Convex mirror

Question 5.
Draw a neat diagram showing Refraction of light through a rectangular glass slab
Answer :
KSEEB SSLC Class 10 Science Solutions Chapter 10 Light Reflection and Refraction ad Q 5

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

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

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

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

KSEEB SSLC Class 10 Science Chapter 13 Intext Questions

Text Book Part I Page No. 118

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle is a small bar magnet. When it is brought near a bar magnet, its magnetic field lines interact with that of the bar magnet. Hence, a compass needle shows a deflection.

Text Book Part I Page No. 122

Question 1.
Draw magnetic Held lines around a bar magnet.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 122 Q 1
Field lines around a bar magnet

Question 2.
List the properties of magnetic lines of force.
Answer:
The properties of magnetic lines of force are as follows:

(a) Magnetic field lines emerge from the north pole.
(b) They merge at the south pole.
(c) The direction of field lines inside the magnet is from the south pole to the north pole.
(d) Magnetic lines do not intersect with each other.

Question 3.
Why don’t two magnetic lines of force intersect each other?
Answer:
If two field lines of a magnet intersect, then at the point of intersection, the compass needle points in two different directions. This is not possible. Hence. two field lines do not intersect each other.

Text Book Part I Page No. 123, 124

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right- hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 123,124 Q 1
Since the current passes through the loop in a clockwise direction, therefore the front face of the loop will be the south pole and the back face, ie, the face touching the table will be north pole. According to right-hand rule, the direction of the magnetic field inside the loop will be pointing downward. Outside the loop, the direction of the magnetic field will be upward.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current 123,124 Q 2
The figure indicates that the magnetic field is the same at all points in the solenoid. That is field is uniform inside the solenoid.

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

Text Book Part I Page No. 125, 126

Question 1.

Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.
(a) Mass
(b) Speed
(c) Velocity
(d) Momentum.
Answer:
(c) and (d)
When a proton enters in a region of magnetic field, it experiences a magnetic force. As a result of the force, the path of the proton becomes circular. Hence, its velocity and momentum change.

Question 2.
In Activity 13.7 of NCERT Book, how do we think the displacement of rod AB will be affected if (i) current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) length of the rod AB is increased?
Answer:
A current-carrying conductor placed in a magnetic field experiences a force. The magnitude of force increases with the amount of current, strength of the magnetic field, and the length of the conductor.

Hence, the magnetic force exerted on rod AB and its deflection will increase if, ’

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

Question 3.
A positively – charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is,
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) The direction of the magnetic field can be determined by the Fleming’s left hand rule. According this rule, if we arrange the thumb, the center finger, and the forefinger of the left hand at right angles to each other, then the thumb points towards the direction of the magnetic force, the center finger gives the direction of current, and the forefinger points in the direction of magnetic field. Since the direction of positively charged alpha particle is towards west, the direction of current will be the same i.e., towards west. Again, the direction of magnetic force is towards north. Hence, according to Fleming’s left hand rule, the direction of magnetic field will be upwards.

Text Book Part I Page No 127

Question 1.
State Fleming’s left-hand rule.
Answer:
Fleming’s left hand rule states that if we arrange the thumb, the center finger and the forefinger of the left hand at right angles to each other, then the thumb points towards the direction of the magnetic force, the center finger gives the direction of current and the forefinger points in the direction of magnetic field.

Question 2.
What is the principle of an electric motor?
Answer:
The working principle of an electric motor is based on the magnetic effect of current. A current-carrying loop experiences a force and rotates when placed in a magnetic field. The direction of rotation of the loop is given by the Fleming’s left-hand rule.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring in the electric motor acts as a commutator. The commutator reverses the direction of current flowing through the coil after each half rotation of the coil. Due to this reversal of the current, the coil continues to rotate in the same direction.

Text Book Part I Page No. 130

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

Text Book Part I Page No. 131

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

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

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

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

Text Book Part I Page No. 132

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Two safety measures commonly used in electric circuits and appliances are as follows:
(i) Each circuit must be connected with an electric fuse. This prevents the flow of excessive current through the circuit. When the current ‘passing through the wire exceeds the maximum limit of the fuse element, the fuse melts to stop the flow of current through that circuit, hence protecting the appliances connected to the circuit

(ii) Earthing is a must to prevent electric shocks. Any leakage of current in an electric appliance is transferred to the ground and people using the appliance do not get the shock.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
Current drawn by the electric oven can be obtained by the expression,
P = VI
Where, I = \(\frac { P }{ V } \)
Power of the oven, P = 2k W = 2000 W
Voltage supplied, V = 220 v
I = \(\frac { 2000 }{ 220 } \) = 9.09 A
Hence, the current drawn by the electric oven is 9.09 A, which exceeds the safe limit of the circuit. Fuse element of the electric fuse will melt and break the circuit.

Question 3.
What precautions should be taken to avoid the overloading of domestic electric circuits?
Answer:
The precautions that should be taken to avoid the overloading of domestic circuits are as follows:

(a) Too many appliances should not be connected to a single socket.
(b) Too many appliances should not be used at the same time.
(c) Faulty appliances should not be connected in the circuit.
(d) Fuse should be connected in the circuit.

KSEEB SSLC Class 10 Science Chapter 13 Textbook Exercises

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

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

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor
Answer:
(a) An electric generator produces electric current. It converts mechanical energy into electricity.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) An AC generator has two rings called slip rings. A DC generator has two half rings called commutator. This is the main difference between both the types of generators.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially
(b) does not change.
(c) increases heavily.
(d) vary continuously.
Answer:
(c) When two naked wires of an electric circuit touch each other, the amount of current that is flowing in the circuit increases abruptly. This causes short-circuit.

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

(b) True
A generator is an electric device that generates electricity by rotating a coil in a magnetic field. It works on the principle of electromagnetic induction.

(c) True
A long circular coil is a long solenoid. The magnetic field lines inside the solenoid are parallel lines.

(d) False
Live wire has red insulation cover, whereas earth wire has green insulation colour in the domestic circuits.

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

  1. Permanent magnets
  2. Electromagnets

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 8

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

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

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

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
The direction of magnetic field is given by Fleming’s left hand rule. Magnetic field inside the chamber will be perpendicular to the direction of current (opposite to the direction of electron) and direction of deflection\force i.e., either upward or downward. The direction of current is from the front wall to the back wall because negatively charged electrons are moving from back wall to the front wall The direction of magnetic force is rightward. Hence, using Fleming’s left hand rule, it can be concluded that the direction of magnetic field inside the chamber is downward.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 11

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

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

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

Question 12.
Name some devices in which electric motors are used.
Answer:
Some devices in which electric motors are used as follows:

(a) Water pumps
(b) Electric fans
(c) Electric mixers
(d) Washing machines

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is
(i) pushed into the coil
(ii) withdrawn from inside the coil
(iii) held stationary inside the coil?
Answer:
A current induces in a solenoid if a bar magnet is moved relative to it. This is the principle to electromagnetic induction.
(i) When the bar magnet is pushed into a coil of insulated copper wire, a current is induced momentarily in the coil. As a result, the needle of the galvanometer deflects momentarily in a particular direction.

(ii) When the bar magnet is withdrawn from inside the coil of. the insulated copper wire, a current is again induced momentarily in the coil in the opposite direction. As a result, the needle of the galvanometer deflects momentarily in the opposite direction.

(iii) When a bar magnet is held stationary inside the coil, no current will be induced in the coil. Hence, galvanometer will show no deflection.

Question 14.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
Two circular coils A and B are placed close to each other. When the current in coil A is changed, the magnetic field associated with it also changes. As a result, the magnetic field around coil B also changes. This change in magnetic field lines associated coil B also changes. This change in magnetic field lines around coil B induces an electric current in it. This is called electromagnetic induction.

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

  1. magnetic field produced around 0 straight conductor-carrying current
  2. force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
  3. current induced in a coil due to its rotation in a magnetic field.

Answer:

  1. Maxwell’s right hand thumb rule.
  2. Fleming’s left hand rale.
  3. Fleming’s right hand rule.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:
KSEEB SSLC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Ex Q 16

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

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

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

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

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

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

KSEEB SSLC Class 10 Science Chapter 13 Additional Questions and Answers

1. Fill in the blanks:

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

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

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

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

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

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

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2

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

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of ‘n’.
Solution:
Radius of hemisphere, r = 1 cm.
Volume of hemisphere, V \(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2 \pi}{3} \times(1)^{3}\)
Radius of Cone, r = 1 cm,
Height of Cone, h = 1 cm.
∴ Volume of a cone, V \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 1
∴ Volume of Total cube \(=\frac{2 \pi}{3}+\frac{\pi}{3} \mathrm{cm}^{3}\)
= πcm3

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2
solution:
Volume of cylinder, V = πr2h
= π × (1.5)2 × 8
= 18π cm3.
Volume of Cylinder V \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \pi \times(1.5)^{2} \times 2\)
\(=\frac{3}{2} \pi \mathrm{cm}^{3}\)
∴ The volume of air contained in the model that Rachel made.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 2.1
= 66 cm3.

KSEEB Solutions

Question 3.
A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see figure given aside).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.1
Solution:
Length of cylindrical Gulab jamun,
l = 5 – (1.4 + 1.4) = 2.2 cm.
Radius, r = 1.4 cm.
∴ Volume of Cylinder (1 Gulab Jamun), = Left hemisphere + Volume of Cylinder + Right hemisphere
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 3.2
∴ Total volume of such 45 Gulab jamun,
\(=45 \times \frac{22}{3} \times 0.28 \times 12.2\)
= 22 × 0.28 × 183
= 1127.28 cm3.
∴ Gulab Jamun contains sugar syrup upto about 30% of its volume.
\(1127.28 \times \frac{30}{100}\)
= 338 cm3.

KSEEB Solutions

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see
Figure)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4
Solution:
Radius of conical depression, r = 0.5 cm.
(Depth) height, h = 1.4 cm.
Volume of 1 depression which is conical, \(=\frac{1}{3} \pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 4.1
Volume of such 4 depressions
\(=4 \times \frac{11}{30} \mathrm{cm}^{3}\)
\(=\frac{44}{30} \mathrm{cm}^{3}\)
∴ Total volume of wooden pen-stand, \(=(15 \times 10 \times 3.5)-\frac{44}{30}\)
= 525 – 1.47
= 523.53 cm3

KSEEB Solutions

Question 5
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Volume of water in a Cone,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 5
Volume of lead shot \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \pi \times(0.5)^{3}\)
\(=\frac{\pi}{6} \mathrm{cm}^{3}\)
Let the number of lead balls kept in vessel \(\frac{1}{4}\) of the water flows out,
\(\Rightarrow n \times \frac{\pi}{6}=\frac{1}{4} \times \frac{200 \pi}{3}\)
\(\Rightarrow n \times \frac{\pi}{6}=\frac{100}{6} \pi\)
∴ n = 100
∴ Number of lead balls =100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. (Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14).
Solution:
Height of solid iron pole, h = 220 cm.
Base diameter of pole, d = 24 cm.
∴ Radius, r = 12 cm.
Height of cylinder, h = 60 cm.
Radius of cylinder, r= 8 cm.
Mass of the pole = ?
Volume of 1st Cylinder, V = πr2h
= π × (12)2 × 220
Volume of 2nd cylinder, V = πr2h
= 77 × (8)2 × 60
∴ Total Volume = π × (12)2 × 220 + π × (8)2 × 60
= {144 × 220 + 64 × 60}π
= 35520π
= 111532.8 cm3.
Mass of iron about 1 ccm is 8 gm.
∴ Mass of iron 111532.8 ……. ??
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 6
= 111.5328 × 8 kg.
= 892.26 kg.

KSEEB Solutions

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm. and its height is 180 cm.
Solution:
(i) Radius of cylinder GHEF, r = 60 cm.
Height, h = 120 + 60
= 180 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7
Volume of cylinderical vessel \(=\pi r^{2} h\)
= π × (60)2 × 180
(ii) Volume of hemisphere + Volume of cone
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.1
(iii) Quality of water remained in cylinder.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 7.2

KSEEB Solutions

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and n = 3.14.
Solution:
Diameter of spherical glass vessel = 8.5 cm.
∴ r = 4.25 cm.
Length of cylindrical neck = 8 cm.
Diameter = 2 cm.
∴ Radius, r = 1 cm.
Volume of water she found = 345 cm3.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8
(i) Volume of Cylinder neck, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.1
(ii) Volume of cylindrical vessel \(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \pi \times(4.25)^{3}\)
∴ Total volume of water, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 Q 8.2
∴ Her answer is not correct.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4

Question 1.
The following distribution gives the dialy income of 50 workers of a factory.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.1
n = 50, ∴ \(\frac{n}{2}\) = 25
On a graph paper mark the following points :
(120, 12), (140, 26), (160, 34), (180, 40), (200, 50).
For the Ogive graph,
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 1.2

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.4 4
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:
Here, the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals.
We plot the points (ordered pairs) (38, 0), (40, 3), (42, 5), (44, 9), (46,14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them by a free hand to get a smooth curve.
The curve so obtained is the less than type ogive.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q2
∵ N = 35
∴ \(\frac{N}{2}=\frac{35}{2}\) = 17.5
The point 17.5 is on y-axis.
From this point (i.e., from 17.5) we draw a line parallel to the x-axis which cuts the curve at P. From this point P, draw a perpendicular to the x-axis, meeting the x-axis at Question The point Q represents the median of the data which is 47.5.
Verification:
To verify the result, let us make the following table in order to find median using the formula :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q2 1
Thus, the median = 46.5 kg is approximately

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3
Change the distribution to a more than type distribution, and draw its ogive.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.1
We can draw Ogive graph by plotting ordered pairs:
(50, 100), (55, 98), (60, 90), (65, 78). (70, 54), (75, 16).
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 Q 3.2

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.1
(i) n = 68, ∴ \(\frac{n}{2}\) = 34
The median is in class interval (125 – 145)
l = 125, n = 68, f = 20, cf = 22, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.2
= 125 + 12
∴ Median = 137 units.

(ii) To find out Mode:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.3
Here, Mode is in class interval (125 – 145)
Maximum frequency, l = 125, f<sub>1</sub> = 20, f<sub>0</sub> = 13, f<sub>2/sub> = 14, h = 20
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 1.4
= 125 + 10.76
= 135.76
∴ Mode = 135.76 units.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 5
Solution:
Here, we have N = 60
Now, cumulative frequency table is:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q2
Since, median = 28.5 (Given)
∴ Median class is 20 – 30 and l = 20, f = 20, cf = 5 + x, N = 60
∴ l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
⇒ 28.5 = 20 + \(\left[\frac{30-(5+x)}{20}\right]\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
⇒ 57 = 40 + 25 – x
⇒ x = 40 + 25 – 57 = 8
Also, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 45 – 8 = 7.
Thus x = 8, y = 7

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.1
n = 100 ∴ \(\frac{n}{2}\) = 50
Median in which C.I. is (35 – 40)
l = 35, n = 100, f = 33, cf = 45, h = 5
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 3.2
= 35 + 0.76
∴ Median = 35.76 years

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 10
Find the median length of the leaves.
[Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5-126.5,
126.5 – 135.5 ………… 171.5 – 180.5.]
Solution:
After changing the given table as continuous classes we prepare the cumulative frequency table as follows:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q4
The cumulative frequency just above 20 is 29 and it corresponds to the class 144.5 – 153.5.
So, 144.5 – 153.5 is the median class.
We have: \(\frac{N}{2}\) = 20, l = 144.5, f= 12, cf = 17 and h = 9
∴ Median = l + \(\left[\frac{\frac{N}{2}-c f}{f}\right] \) × h
= 144.5 + \(\left[\frac{20-17}{12}\right]\) × 9
= 144.5 + \(\frac{3}{12}\) × 9 = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Median length of leaves = 146.75 mm.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5
Find the median life time of a lamp.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.1
n = 400 ∴ \(\frac{n}{2}=\frac{400}{2}=200\)
Class interval having median is = (3000 – 3500)
l = 3000, n = 400, f = 86, cf = 130, h = 500
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 5.2
= 3000 + 406.98
∴ Median = 3406.98 hours

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.1
(i) n = 100, ∴ \(\frac{n}{2}\) = 50
Class interval having median is (7 – 10)
l = 7, n = 100, f = 40, cf= 36, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.2
= 7 + 1.05
∴ Median = 8.05 Letters

(ii) Clall interval which has mode is (7 – 10)
Maximum frequency, l = 7, f1 = 40, f0 = 30, f2 = 16, h = 3
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.3
= 7 + 0.88
∴ Mode = 7.88

(iii) Mean(\(\overline{X}\)) : Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.4
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q 6.5
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32
∴ (i) Median = 8.05 letters
(ii) Mode = 7.88
(iii) Mean = 8.32.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 18
Solution:
We have cumulative frequency table as follows:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 Q7
The cumulative frequency just greater than 15 is 19, which corresponds to the class 55 – 60.
So, median class is 55-60 and we have \(\frac{N}{2}\) = 15,
l = 55, f = 6, cf = 13 and h = 5
MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.3 20
Thus, the required median weight of the students = 56.67 kg.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.3, drop a comment below and we will get back to you at the earliest.