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SCERT AP Board 9th Class Social Solutions 7th Lesson Industries in India Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 7th Lesson Industries in India

9th Class Social Studies 7th Lesson Industries in India Textbook Questions and Answers

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Question 1.
Why government took up the responsibility to set up basic goods industries?
Answer:

• Basic industries produce essential goods that can form a base to support a large variety of factories.
• Huge amount of capital is required to set up basic goods industries.
• Moreover it takes a long time to set up basic industries.
  E.g.: For setting up a power plant for production of electricity, it would require five to ten years.
• Private industrial groups were not willing to invest in such industries.
• Hence the Government took up the responsibility to set up basic goods industries.

Question 2.
Why are industries located in specific areas?
Answer:

• All factories need raw materials from which goods can be produced.
• Transportation is needed to bring raw materials to factories and transfer finished goods from them.
• So industries are located in specific areas where raw materials like minerals are available.
• Industries are also concentrated where infrastructural facilities like transportation electricity, banking services, etc. are available.

Question 3.
What are the basic goods industries? How they are different from consumer goods industries?
Answer:

Basic Goods Industries	Consumer Goods Industries
1. Factories producing essential goods necessary for industries are called basic goods industries. E.g.: Machines, electricity, etc.	1. Factories producing goods which for direct consumption are called consumer goods. E.g.: Soaps, Furniture, TVs, etc.
2. These goods are not for direct consumption.	2. These goods are directly consumed by consumers.
3. Basic industries produce essential goods that can farm a base to support a large variety of factories.	3. Consumer goods industries did not support any other industries.

Question 4.
Give a list of towns/areas in which some conventional mineral resources are found and ask students to identify then possible industries which can be set up.

Minerals/ Resources	Towns/areas in which these resources are available	List the kind of industries that can be set up in this area
1. Iron ore
2. Coal
3. Jute
4. Oil
5. Natural Gas
6. Forests
7. Manganese
8. Bauxite

Answer:

Minerals/ Resources	Towns/areas in which these resources are available	List the kind of industries that can be set up in this area
1. Iron ore	1. Singhbhum in Jarkhand 2. Mayurbhanj, Koenjhar in Odisha 3. Raipur and Bastar in Chattisgarh. 4. Bellary and Chikmaglur in Karnataka. 5. Salem in Tamilnadu 6. Khammam in Telangana and Rayalaseema in Andhra Pradesh. 7. Ratnagiri and Chanda in Maharashtra.	1. Iron and steel industry 2. Heavy machines 3. Tools 4. Weapons 5. Constructions and transport vehicles, etc. 6. Railway coaches, etc.
2. Coal	Important coal fields are 1. Jharia, Chandrapura, Bokaro and Giridhi in Jarkhand. 2. Ranigunj and Asansal in West Bengal. 3. Singrauli and Korba in Madhya Pradesh. 4. Singareni in Telangana	1. Iron and steel industry 2. Thermal power industry 3. Used as fuel in so many other industries. 4. Largely used in railways.
3. Jute	West Bengal, Assom, Bihar and Odisha.	Jute industry is concentrated in West Bengal. 1. Jute textiles 2. Bag making 3. Carpet making 4. Door mats, etc.
4. Oil	Gujarat, Assom, Bombay High	Different types of industries.
5. Natural Gas	1. Cambay and Ankleshwar fields in Gujarat 2. Bombay High in Maharashtra.	Power industry Fertilizer industry Used as fuel
6. Forests	Forests are found in Madhya Pradesh, Telangana, Andhra Pradesh, Odisha, Maharashtra, Arunachal Pradesh and Uttar Pradesh.	Forests yield both softwood and hard wood. Forest based industries are- 1. Paper (softwood trees) 2. Furniture 3. Timber 4. Soaps 5. Match boxes 6. Handicraft 7. Turpentine and 8. Railway sleepers
7. Manganese	1. Koenjhar, kalahandi, Mayurbhanj Sundarghar and Thalcher in Odisha. 2. Chitradurg, Shimoga, Chikmanglore, Bellary and Dharwad in Karnataka. 3. Bhalghat, Seori, Jabalpur in Madhya Pradesh. 4. Singhbum in Jarkhand. 5. Panchamahal in Gujarat. 6. Srikakulam and Visakhapatnam districts in Andhra Pradesh.	1. Iron and steel industry 2. Chemical industry 3. Bleaching powder 4. Electrical 5. Glass 6. Leather 7. Photography and 8. Metal industry
8. Bauxite	1. Palmau, Ranchi, Monghyr and Shahabad in Jarkhand. 2. Balaghat, Jabalpur, Bilaspur and Rajgarh in Madhya Pradesh. 3. Amreli and Kutch in Gujarat. 4. Belgaum in Karnataka. 5. Madurai, Nilgiris and Coimbatore in Tamilnadu. 6. Visakhapatnam in Andhra Pradesh	Used in Aluminium making which in turn used in aircraft, automobiles, ships, utensils, railways, photographic material and electrical goods industry.

Question 5.
Why government in 1990s allowed private industries in many areas which were earlier restricted only to government?
Answer:

• Indian industry as a result of government’s control was not modernizing fast and was producing goods at high cost and not making technological improvement.
• Government used to allocate a specifc amount every year to operate these industries.
• It was expected that these industries become independent, generate revenue for the government.
• But these government run factories continually require government assistance.
• Their functioning was much below that was expected.
• So, in the 1990s a new industrial policy was announced. Many activities which were earlier restricted only for the government are now allowed for the private industries.

Question 6.
What is the impact of industrial development on employment generation?
Answer:

• An important goal of industrial policies in India was to generate employment opportunities in industrial activities.
• Raising the proportion of people employed in factories is also generally seen as an important indicator of economic development of a country.
• Today nearly 2 lakh large factories and 3 crore small manufacturing units are operating in India.
• These units employ nearly about one-fifth of India’s 460 million workers.
• Number of employees generated in industrial section has increased from 11% in 1972-73 to 22 % in 2009-10.
• But large industries began to replace workers with technology. More and more automation has taken place.
• This has led to almost zero additional employment in large factrories.

Question 7.
What is the impact of industrial development on revenue?
Answer:

• When goods made in factories and exported to other countries, we get revenue.
• A few decades before, three-fourths of income from goods exported from manufacturing industries alone.
• Even today, nearly two-third of goods exports is contributed by industries and particularly small industries.
• Factory goods exported range from gems, jewellery, chemicals, cars, machinery, cashewnuts, etc.
• The money or foreign exchange earned in this process enable us to import from other countries a large number of goods that we want from other countries.

Question 8.
“Industrial activities increase environmental problems” Discuss.
Answer:

• The production process in industries involves the use of electricity and application of different chemicals.
• In the course of production, these industries release a lot of other materials. They are causing pollution in the industrial locations.
• The industrialization led to the indiscriminate exploitation of minerals, forests, soils, air, etc.
• It leads to deforestation and damages the natural habitat of animals.
• Underground water is being contaminated due to the release of wastes by the industries.
• Some animals are endangered and some creatures have become extinct.
• The depletion of ozone layer, pollution of water, air, soil has increased ecological problems.

Question 9.
Write a few slogans on the prevention of environment pollution.
Answer:

• We were born to help the world, not to destroy it.
• Only when the last tree has died, the last river has been poisoned and the last fish has been caught, will we realize that we cannot eat money?
• Save the environment and you will save the life and future.
• Hungry and the Earth will serve, Thirst and the seas will water I Do you still want to cut that all?
• Man made global warming, the biggest scan in the history of mankind to fulfil his greediness? Understood this or nature will teach you.
• Stop pollution – Stop global warming.

Question 10.
Read the para 3 of page 83 and comment on it.

The electronics industrys covers a wide range of products from transistor sets to television, telephones, cellular telecom, pagers, telephone exchange, radars, computers and many other equipments required by the telecommunication industry. Bengaluru has emerged as the electronic capital of India. Other important centres for electronic goods are Mumbai, Delhi, Hyderabad, Pune, Chennai, Kolkata, Lucknow and Coimbatore. 18 software technology parks provide single window service and high data communication facility to software experts. A major impact of this industry has been on employment generation. Upto 31 March 2005, the IT industry employed over one million persons. This number is expected to increase eight-fold in the next 3 to 4 years. It is encouraging to know that 30 per cent of the people employed in this sector are women. This industry has been a major foreign exchange earner in the last two or three years ‘ because of its fast growing Business Processes Outsourcing (BPO) sector. The continuing growth in the hardware and software is the key to the success of IT industry in India.

Answer:
IT and electronics are the fast growing segments of Indian industry both in terms of production and exports. This sector is attracting considerable interest not only as a vast market but also as potential production base international companies.

In recent times, software development and IT enabled services have emerged as a niche opportunity for India in the global context.

The Government is taking all necessary steps to make India, a Global IT super power and a front runner in the age of information revolution. It earns a major share of foreign exchange.

Question 11.
Observe the map given on page 95 and locate the iron and steel plants in the India outline map.

Answer:

Question 12.
Select one agro-based and one mineral based industry in your area.
i) What are the raw materials they use?
ii) What are the other inputs in the process of manufacturing that involve transportation cost?
iii) Are these factories following environmental norms?
Answer:
1. Agro Based :
Nandini Foods.

2. Mineral Based :
Ramagundam Thermal Power Station

Agro Based	Mineral Based
1. Raw material: Sugar, powders, chillies, tamarind, oil, etc.	Coal, etc.
2. Inputs that involve transportation lost: raw materials, staff and finished goods to market.	Raw materials, staff
3. Environmental Norms : Yes, upto some extent	Yes, but the wastage is sent out into a canal.

9th Class Social Studies 7th Lesson Industries in India InText Questions and Answers

Question 1.
Can you make a list of products produced by factories for other factories? (Text Book Page No. 76)
Answer:

1. Moulds
2. Speakers
3. Spare parts
4. Tyres, Tubes, etc.

Question 2.
Have you seen machines used in a factory? Make a collage of different kinds of machines that are used. (Text Book Page No. 76)
Answer:
I have seen a welding factory.
Machines in the factory are :

1. arc welding
2. resistance welding
3. laser welding
4. electron bean welding
5. stud welding
6. orbital welding
7. wave soldering
8. hot dip brazing
9. torch brazing
10. indution brazing
11. ultrasonic
12. friction welding

Question 3.
Discuss what is meant by the word “basic”. What are the basic necessities for industries? (Text Book Page No. 76)
Answer:

• “Basic” means the things that are basically necessary.
• The basic necessities for setting up any industries are machines, electricity, minerals and ores and infrastructural facilities like transport, telephones, etc.

Question 4.
At the time of independence what were the objectives that were desired to be achieved through industrialization? (Text Book Page No. 76)
Answer:

• After 1947, India began many initiatives to promote industrial activities in the country.
• Major objectives were – to become self-sufficient in meeting our needs and to make the country an industrially developed nation.

Question 5.
Where should the sugar and jaggery mills be ideally located? (Text Book Page No. 80)
Answer:
The sugar and jaggery mills should be ideally located near the crop grown areas.

Question 6.
Where would it be economically viable to set up the cement manufacturing units? (Text Book Page No. 82)
Answer:
Cement manufacturing units can be established near the areas where the raw materials are available in bulk.
Raw materials : Limestone, silica, alumina, gypsum, coal and electric power and transportation facilities.

Question 7.
Can you point out some examples of increase in production of goods that are used in the production of many products by different factories? (Text Book Page No. 90)
Answer:

1. Steel
2. Engines
3. Cement
4. Bricks
5. Wood
6. Glass
7. Vegetables
8. Cotton, etc.

Question 8.
The industry has strategically located plants in Gujarat that have suitable access to the market in the Gulf countries. Find out where the plants are located in other states of India. Find their names. (Text Book Page No. 82)
Answer:

1. Nirman Cements Ltd – Bihar.
2. Grasim Cements Ltd – Madhya Pradesh
3. Sagar Cements – Telangana
4. The India Cements Ltd – Tamilnadu

Question 9.
Fill in the following table. For some industries, you may need to discuss with the teacher. (Text Book Page No. 84)

Industry	States in which they are currently concentrated	Why they are concentrated in those states?
Chemical Industry
Fertiliser Industry
Cement Industry
Automobiles Industry

Answer:

Industry	States in which they are currently concentrated	Why they are concentrated in those states?
Chemical Industry	Gujarat	Availability of raw material, skilled and unskilled labourers, electricity, water, financial assistance, transport facilities, etc.
Fertiliser Industry	Maharashtra
Cement Industry	Rajasthan, Telangana, A.P., M.P, Gujarat
Automobiles Industry	Tamilnadu

Question 10.
Observe the following pie charts. (Text Book Page No. 87, 88)

a) What are the differences in employment in the three kinds of economic activities that you notice from these pie charts?
Answer:

• The employment generation in agricultural sector has been decreased from 74% in 1972-73 to 53% in 2009-10.
• The employment generation in industrial sector has been increased from 11% in 1972-73 to 22% in 2009-10.
• The employment generation in service sector has shown an increase of 10%.

b) What is the percentage of change in employment by industry?
Answer:
There is a 11% increase in employment generated by industry.

c) Did we expect to see a greater change in employment by industry that did not happen?
Answer:
1) Yes, we expected a greater change in employment by industry. But it did not happen.
2) Large industries began to replace workers with technology. More and more automation has taken place.
3) Hence it did not happen.

Question 11.
Look at the following Graph and answer the following question. (Text Book Page No. 89)

What has been the increase in production of cloth over the past 30 years? What would be the impact of this? Discuss in your class. (Text Book Page No. 90)
Answer:
1. The production of cotton cloth was 8000 million square metres approximately in 1980-81. The production went up to 15000 million square metres by 1990-91, to 19000 million square metres by 2000-01. The production has been increased to 31000 million square metres approximately by 2010-11.

2. The production of other cloth material was approximately 2500 million square metres in 1980-81 and that went up to 7000 million square metres by 1990-91, to 20000 million square metres by 2000-01 and to 30000 million square metres approximately
by 2010-11.

Question 12.
Refer the chart that shows the production of cement and steel construct a table to show the increase from 80-81 to present times. Discuss some positive and negative effect of this increase in production. (Text Book Page No. 90)

Answer:

The positive and negative effects of the increase in production :
1. The increase in the production of cement has positive impact on the development of our country. The infrastructural facilities like buildings, dams, roads etc., are increased due to the increase of production of cement.

2. Negative impact of cement production: The heating of limestone and clay may release mercury into the air which affects the health of the people and pollutes the ground water too.

Steel Production:
1. The increase in the production of steei has impact on the production of other goods, iike heavy tools and machines etc.

2. The steel factories release poisonous gases into the atmosphere and release waste into the streams causing air and water pollution.

Question 13.
Iron is the basic requirement for a large number of goods produced by various factories. Explain this with examples that you see around. (Text Book Page No. 76)
Answer:

• Iron is used in making steel, machinery, tools and weapons.
• Iron is also used in shipbuilding industry.
• Iron is also used for many other purposes in constructions and transport.

Question 14.
Make a chart to show how petroleum is the basic requirement for a large number of products. (Text Book Page No. 76)
Answer:
Petroleum or crude oil is a naturally occurring liquid found in the earth. It is refined and used to make so many products. The byproducts are Petrol, Diesel, Kerosene, etc.
It is used in:

1. Petrochemical industry
2. Plastic making
3. Lubricating oils
4. Fertilizers
5. Asphalt – which is used in road construction
6. Pesticides
7. Detergents
8. Photographic film
9. Artificial fibres, etc.

Question 15.
Why is the per capita consumption of steel so low in India? (Text Book Page No. 81)
Answer:

• India is the developing country, the process of development is a little bit slow than other developed countries.
• Economy plays a vital role in the development of India economy is good but not best.
• The steel industry has limited factories in our country.
• The production is also limited. So the per capita consumption of steel is low in India.

Question 16.
Why did Mahatma Gandhi lay emphasis on spinning yarn and weaving khadi? (Text Book Page No. 79)
Answer:

• To spend time usefully with some other work,
• To fight against foreign cloths imported,
• To encourage village industries,
• To insist that everyone learn to do his own work like making the yarn for his cloth and
• To lead a simple life and minimising the needs for living.

9th Class Social Studies 7th Lesson Industries in India Activity

Collect the wrappers of a tea packets and tooth paste. Read the wrappers carefully and try to relate to the question below. (Text Book Page No. 77)
Answer:
Student’s Activity

…………..(1)………….. can be considered as a product of agro based industry. …………..(2)………….. can be considered a product of mineral based industry.
Answer:
1) Tea,
2) Toothpaste

Raw material for the tooth paste…………..(1)………….. and …………..(2)…………..
industry. That industry is called key or basic industry. Whereas the tooth paste is a consumer goods and the industry producing such goods is called consumer goods industry.
Answer:
1) Flouride,
2) Calcium

The ownership of industries could be lying with individuals or groups of individuals such as …………..(1)………….. (for the tea packets) and …………..(2)………….. (tooth paste). Such an industry is called a private sector industry whereas if the ownership belongs to the government, it will be called as public sector industry. Two examples of public sector industries are …………..(3)………….. and …………..(4)…………..
Answer:
1) Brookebond,
2) Dabur (Red),
3) Bharat Dynamics Ltd,
4) BHEL Ltd.

Some industries are also owned by large number of people who supply raw materials (milk/sugarcane) or supply their labour (coir) pool their resources to run them. Such industries are called cooperative industries.

Telangana & Andhra Pradesh SCERT AP State Board Syllabus 9th Class Physical Science Physics Study Material Guide Pdf free download, 9th Class PS Guide, TS AP 9th Class Physical Science Physics Textbook Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Board 9th Class Textbook Solutions.

Students can also go through AP Board 9th Class Physical Science Notes to understand and remember the concepts easily. Students can also read AP 9th Class Physical Science Important Questions (Physics & Chemistry) for board exams.

AP State Syllabus 9th Class Physical Science Guide Pdf | AP 9th Class Physics Study Material Pdf

AP State 9th Class Physical Science Textbook Pdf English Medium | AP 9th Class Physics Textbook Pdf English Medium

9th Class Physical Science Guide Pdf English Medium

• Chapter 1 Motion
• Chapter 2 Laws of Motion
• Chapter 3 Is Matter Pure?
• Chapter 4 Atoms and Molecules
• Chapter 5 What is inside the Atom?
• Chapter 6 Chemical Reactions and Equations
• Chapter 7 Reflection of Light at Curved Surfaces
• Chapter 8 Gravitation
• Chapter 9 Floating Bodies
• Chapter 10 Work and Energy
• Chapter 11 Sound
• Chapter 12 Units and Graphs

AP 9th Class Physical Science Study Material Telugu Medium | 9th Class Physics Guide

AP 9th Class Physics Study Material Pdf | AP Board Solutions Class 9 Physics | Physical Science 9th Class Guide

• 1st Lesson చలనం
• 2nd Lesson గమన నియమాలు
• 3rd Lesson మన చుట్టూ ఉన్న పదార్థం శుద్ధమేనా ?
• 4th Lesson పరమాణువులు-అణువులు
• 5th Lesson పరమాణువులో ఏముంది ?
• 6th Lesson రసాయన చర్యలు – సమీకరణాలు
• 7th Lesson వక్రతలాల వద్ద కాంతి పరావర్తనం
• 8th Lesson గురుత్వాకర్షణ
• 9th Lesson తేలియాడే వస్తువులు
• 10th Lesson పని మరియు శక్తి
• 11th Lesson ధ్వని
• 12th Lesson ప్రమాణాలు మరియు గ్రాఫులు

SCERT AP Board 9th Class Social Solutions 6th Lesson Agriculture in India Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 6th Lesson Agriculture in India

9th Class Social Studies 6th Lesson Agriculture in India Textbook Questions and Answers

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Question 1.
Name one important beverage crop and specify the geographical conditions required for its growth.
Answer:

• Tea is an important beverage crop.
• The tea plant grows well in tropical and sub-tropical climates.
• Fertile, well-drained soil, rich in humus and organic matter is essential.
• Tea bushes require warm and moist frost-free climate all through the year.
• Frequent showers are necessary.

Question 2.
The land under cultivation has got reduced day by day. Can you imagine its consequences?
Answer:

• The land under cultivation has been reducing due to the competition for land between non-agricultural uses such as housing, etc.
• As a result, the productivity of India has started showing a declining trend.
• That would lead to the scarcity of food grains in future and in turn poverty and drought situation would prevail in our country.

Question 3.
On an outline map of India show millet producing areas.
Answer:

Question 4.
What is a Minimum Support Price (MSP)? Why is a MSP needed?
Answer:
1) A Minimum Support Price is a price at which the farmers can sell their grain, if they want, to the Government.
2) The Government sets the MSP so as to cover the cost of cultivation and allow a little bit of profit to the farmer.
3) Thus the farmer, produced far markets within the country.
4) The farmers are not forced to sell their grains at cheaper prices to the traders.

Question 5.
Explain all the ways the Indian government supported the Green Revolution.
Answer:

• The Green Revolution was initially introduced in Punjab, Haryana, western Uttar Pradesh, and in some parts of Andhra Pradesh and Tamilnadu.
• The Government introduced new kinds of seeds, known as High Yielding Varieties to the Indian soils.
• It was also accompanied by use of chemical fertilizers, machinery such as tractors, and others besides irrigation facilities.
• A variety of cooperative banks were set up to provide credit to farmers so that they buy raw materials such as seeds, fertilizers, and pesticides, machinery required for modern farming.

Question 6.
Do you think it is important for India to be self-sufficient in food grain production? Discuss.
Answer:

• A large portion of our population especially children and poor communities are unable to get adequate nutrition.
• Whenever there were little rains, drought situations prevailed. This led to decrease in food production and forced government to import food grains.
• To avoid these kind of situations we should be self-sufficient in food grain production.
• A large stock of food grains has built-up with the government through Food
  Corporation of India that could be used in case of shortage and can avoid drought and famine situations in our country.

Question 7.
How is dry land agriculture different from agriculture in other areas?
Answer:

• 45% of the cultivable land which cannot easily be irrigated and depend solely on rainfall is known as dryland in our country.
• Unlike the cultivation of irrigated lands, dryland farming poses different challenges.
  a) Conserving rainfall that the area receives is the first step. This is done through watershed development programme which includes afforestation, bunding, building check-dams, and tanks.
  b) Fertility of the soil needs to be raised by adding organic manure.
  c) Farmers may also need new varieties of seeds suitable for different regions, knowledge about the best ways of growing a mix of crops on the same land etc.
  Hence farming in dryland is different from other areas.

Question 8.
Can you recall the incidents such as pesticides being found in soft drinks? How is this related to the use of pesticides? Discuss.
Answer:

• The use of pesticides is polluting underground water.
• The soft drink factories use this underground water for making soft drinks.
• Hence undissolved pesticides are seen in soft drink.

Question 9.
Why is chemical fertilizer used in new farming methods? How could use of fertilizers make soil less fertile ? What are the alternative ways of enriching soil?
Answer:

• The chemical fertilizers are used to increase production of crops.
• But the chemical fertilizers are basically made for petro-chemicals. And eventually remain in the soil.
• As a result, many micro organisms like earth worms are destroyed.
• These micro organisms increase the fertlity of the soil. But these are detroyed and hence this, in turn, affects the long run fertility of the soil.
• The alternative ways of enriching soil, like vermicompost, can be used to increase the fertility of the soil.

Question 10.
How has the Green Revolution in some areas resulted in short-term gains but longterm losses to farmers?
Answer:

• In Green Revoluton, the farmers are encouraged to pump ground water to water-intensive crops in low rainfall areas.
• This unsustainable pumping has reduced water storage in ground.
• Consequently, many wells and tubewells have run dry.
• More over the use of chemical fertilizers has also affected the long-term fertility of the soil. Hence we can say that Green Revolution has short-term gains and long term losses.

Question 11.
What could be the effects of foreign trade on farmers’ income?
Answer:

• Foreign trade could cause farmers’ income to fluctuate a lot.
• In certain years and for certain crops the farmers might gain from exports.
• In other years, farmers could lose because of cheap imports and fall in prices of farm products.
• Small farmers without much savings will not be able to bear this loss.
• They will be caught in debt trap and become poorer.
• Hence the government must be careful in allowing trade in farm products.

Question 12.
In earlier classes we have studied about land distribution. How does the following image reflect this idea? Write a paragraph about this in the context of Indian agriculture.

Answer:
In India more than 70% of the farmers are having only small and marginal holdings,
i. e., Less than 2 hectares and hence most of these farms are not viable. The small farmers cannot produce enough to meet even their basic requirements of their families. They can invest more in agriculture to increase production. They are unable to get financial assistances.
Caption : “A man of sixty acres :
A group of six feet”.

Question 13.
Read the para under the title ‘Fertilizer Problems’ on page 70 and comment on it.

Fertilizer Problems : Manure and compost contain humus and living organisms that slowly release minerals as they decompose. Chemical fertilizers provide minerals (usually nitrogen, phosphorus and potassium) which dissolve in water and are immediately available to plants, but may not be retained in the soil for long. They may be leached from the soil and pollute groundwater, rivers, and lakes. Chemical fertilizers (as well as pesticides) can also kill bacteria and other organisms in the soil. This means that some time after their use, the soil will be less fertile than ever before. Without micro-organisms, the soil will be dependent on frequent addition of more and more chemical fertilizers. The variety of nutrients which are normally produced by micro-organisms may also be reduced. Thus, in many areas, the Green Revolution has actually resulted in a loss of soil fertility and ever-increasing costs to farmers.

Answer:

• The main environmental problem associated with fertilizer use is contamination of water with nitrates and phosphates.
• Elevated nitrogen levels in drinking water.
• Environmental pollution is a significant problem. But while most of the focus is placed on polluting industries, toxins like mercury and small particle traffic pollution, a major source of environmental devastation is caused by modern food production. Far from being life sustaining, our modern chemical dependent farming methods.

– Strip soil of nutrients
– Destroy critical soil microbes.
– Contribute to desertification and global climate change, and
– Saturate form lands with toxic pesticides, herbicides and fertilizers that then migrate into ground water, rivers, lakes and oceans.

Question 14.
Observe the map given in the page 74 and locate the States where paddy is grown in the India outline map.

Answer:

1. Punjab
2. Haryana
3. Uttar Pradesh
4. Bihar
5. West Bengal
6. Chattisgarh
7. Odisha
8. Telangana
9. Andhra Pradesh
10. Karnataka
11. Tamilnadu

Question 15.
Which crops are grown in your area? Which of these are grown from HYV seeds and which ones are grown from traditional seeds? Compare the HYV seeds and the traditional seeds with regard to each of the following points :
a) duration of crop
b) number of times irrigated
c) production
d) fertilisers
e) diseases
f) pesticides
Answer:
1. Paddy, sugarcane, maize, pulses are grown in our area.
2. Paddy is grown from HYV seeds and the remaining are grown from traditional seeds.
3.

	HYV Seeds	Traditional Seeds
Duration of crop	Less	More
No. of times irrigated	Least	More
Production (comparatively)	More	Less
Fertilisers	Nominal quantity	Large quantity
Diseases	Least chancess	More chances
Pesticides	Right quantity	More quantity

Question 16.
Conduct a Debate : Make the students two teams. One team should support chemical fertilizers and another should support organic farming.
Answer:
Teacher conducts this Debate.
Example : In agriculture it is better to be followed organic fertilizers.
Reasons :
Pesticides can enter the human body.
There is a chance of groundwater damage.
The use of fertilizers can reduce the natural nutrients on the soil surface, etc.

9th Class Social Studies 6th Lesson Agriculture in India InText Questions and Answers

Question 1.
Name some of the states of India where such farming is practised? (Text Book Page No. 59)
Answer:
Punjab, Haryana, Odisha, Karnataka, Telangana, Andhra Pradesh, Madhya Pradesh, Uttar
Pradesh and some parts of Rajasthan.

Question 2.
Give some more examples of crops which may be commercial in one region and may provide subsistence in another region. (Text Book Page No. 60)
Answer:
1) In Haryana and Punjab, rice is a commercial crop whereas in Odisha it is a subsistence crop.
2) In Punjab, maize is a commercial crop whereas in Jharkhand it is a subsistence crop.

Question 3.
Distinguish which of these pulses are grown in the kharif season and which are grown in the rabi season? (Text Book Page No. 62)
Answer:
Pulses grown in the Kharif season :
Red gram, Black gram and Green gram.

Pulses grown in Rabi season :
Peas, Bean, Chickling vetch and Lentil.

Question 4.
Complete the bar diagram given on P.65 in Textbook and find out the percentage of cultivators and agricultural labourers in 1971 and 2001 respectively. (Text Book Page No. 65)
Answer:

1. Student’s activity.
2. Percentage of cultivators in 1971 = 62
3. Percentage of agricultural labourers in 2001 = 46

Question 5.
In which decades the food grains yields grow fast? What could be probable reasons for this? (Text Book Page No. 69)
Answer:
The food grains yields grew fast between 1980 – 1990, 2000 – 2010.
The green revolution helped farmers to produce higher level of food grains.

Question 6.
Can organic farming produce enough food for all? (Text Book Page No. 72)
Answer:
If some measures are taken, I think, organic farming can produce food for all.

Question 7.
How is organic farming especially suited for small and marginal farmers? Discuss. (Text Book Page No. 72)
Answer:
Large sections of farmers in India and their land holdings are small and marginal.
Their access to external inputs is limited and their ability to improve production is low. Globalisation and the opening up of the trade barriers among the nations have resulted in the decline of agricultural prices in the local markets to the detriment of the interests of these small and marginal farmers. Thus organic farming especially is suited for small and marginal farmers.

Question 8.
Read the following table. (Text Book Page No. 66)
Number of farmers and land they possess in India (2010 – 2011)

Complete the data in the table and the explanation in the following passage below.
Majority of farmers operate only small plots of lands. A typical Indian marginal farmer has only aboutacres to cultivate. There are 924 lakh farmers so that ..(2)..% of all farmers are marginal. If we add up the number of small and marginal farmers they form …(3)..% of all farmers. However, even though in percentage terms medium and large farmers is small, the number in absolute terms is large. ..(4)…lakh farmers can be together considered to be in this group. They have a powerful voice in rural areas. This group of large and medium farmers together operate …(5)…% of the land. Each large farmer for example on an average operates ..(6)..acres of land. Compare this with each marginal farmer who operates on an average ..(7)..acres of land. This inequality in distribution of land explains the inequalities in opportunities that they experience, the pov¬erty or growth opportunity that they face.
Answer:

1. 2.5
2. 67%
3. 85%
4. 69
5. 32%
6. 42.9 acres
7. 0.95 acres.

Question 9.
Discuss the differences between self employment and looking for work using examples from your region. (Text Book Page No. 65)
Answer:
1. In self employment, people create their own employment and earn their livelihood. They do not depend either on government or an any other persons for their livelihood.
e.g.: Working in their own farms, opening a shop or public telephone booth or establishing a small organization like manufacturing of candles, matchboxes, etc.

2. On the other hand, some people cannot create their own employment. They generally depend either on government or on any other organization or a person for earning their livelihood.
e.g.: Working in the fields or others, or in a shop or working as government servant, etc.

Question 10.
Do you think that some families who were earlier cultivators are becoming agricultural labourers now? Discuss. (Text Book Page No. 65)
Answer:
Reasons:

1. Small land holdings
2. Dependency on seasons for rainfall
3. High rate of interests on loans
4. Very low minimum support price
5. Lack of technology
6. Big families and
7. Some rituals.
   All these changed the cultivators as agricultural labourers.

Question 11.
In which areas were the new methods of agriculture first tried? Why was the whole country not covered? (Text Book Page No. 68)
Answer:
1. The new methods were initially implemented in Punjab, Haryana, western Uttar Pradesh and in some districts of Telangana, Andhra Pradesh and Tamilnadu.

2. The HYV seeds required a lot of water and these areas were already irrigated.
Hence the new methods were introduced in these areas.

Question 12.
Why are different methods necessary for dry land areas? (Text Book Page No. 68)
Answer:

• 45% of the land which cannot be irrigated are known as dry lands.
• It would be very difficult and expensive to irrigate these dry lands.
• These areas must depend solely on rainfall.
  Hence different methods are necessary for dry land areas.

Question 13.
How increase in buffer stock would help to avoid situations of drought and famines? (Text Book Page No. 69)
Answer:

• A large stock of food grains should be maintained with the government.
• That buffer stock could be used in case of shortage of food grains and can also be used to supply foodgrains to inaccessible areas.
• We can also avoid drought and famine like situations in our country.

Question 14.
How farmers were able to raise higher amount of food grains on the same plot of land over the years? (Text Book Page No. 69)
Answer:
The use of High Yielding Varieties of seeds, chemical fertilizers, machinery, etc. made the farmers to raise higher amount of food grains on the same plot of land over the years.

Question 15.
Why did not the Indian government allow farmers to export foodgrains during the Green Revolution years? (Text Book Page No. 71)
Answer:
There would be a shortage of food grains if the government allows farmers to export food grains. So the government did not allow farmers to export food grains.

Question 16.
Why should government ban exports /import ? How does this policy help Indian farmers? (Text Book Page No. 71)
Answer:
Foreign trade could cause farmers income to fluctuate a lot. In some years farmers could lose because of cheap imports and fall in prices of farm products. So the exports ad imports should be banned.

Small farmers will not be able to bear this loss. They will get caught in debt trap and become poorer. So this ban helps the Indian farmers in preventing them from debts and heavy losses.

Question 17.
Use an atlas of India to find the locations of the below mentioned dams and mark them on a map of India. Also label the names of the major rivers on which these dams were built. (Text Book Page No. 67)
Answer:

1. Bhakra-Nangal Project (Punjab)
2. Nagarjuna Sagar Project (Telangana & Andhra Pradesh)
3. Hirakund Project (Odisha)
4. Damodar Valley (West Bengal)
5. Gandhi Sagar (Madhya Pradesh)

Question 18.
In your opinion, what would be the minimum amount of land required to do viable farming which would give a farmer a decent earning. How many farmers in the table Page No. 66 are doing viable farming? (Text Book Page No. 66)
Answer:
The minimum amount of land required to do viable farming may be 2 acres in my opinion. There are 67% of farmers do not have this minimum required land for viable farming.

Question 19.
Why only a small section of farmers have a powerful voice? (Text Book Page No. 66)
Percentage of large and medium farmers is 5% but they hold 32% of total cultivable land. Their average land holding is also high. Hence they have a powerful voice.

Andhra Pradesh SCERT AP State Board Syllabus 9th Class Physical Science Physics & Chemistry Chapter Wise Important Questions and Answers in English Medium and Telugu Medium are part of AP Board 9th Class Textbook Solutions.

Students can also read AP Board 9th Class Physical Science Solutions (Physics & Chemistry) for exam preparation.

AP State Syllabus 9th Class Physical Science Important Questions and Answers English & Telugu Medium

AP 9th Class Physical Science Important Questions in English Medium

• Chapter 1 Motion Important Questions
• Chapter 2 Laws of Motion Important Questions
• Chapter 3 Is Matter Pure? Important Questions
• Chapter 4 Atoms and Molecules Important Questions
• Chapter 5 What is inside the Atom? Important Questions
• Chapter 6 Chemical Reactions and Equations Important Questions
• Chapter 7 Reflection of Light at Curved Surfaces Important Questions
• Chapter 8 Gravitation Important Questions
• Chapter 9 Floating Bodies Important Questions
• Chapter 10 Work and Energy Important Questions
• Chapter 11 Sound Important Questions
• Chapter 12 Units and Graphs Important Questions

AP 9th Class Physical Science Important Questions in Telugu Medium

• 1st Lesson చలనం Important Questions
• 2nd Lesson గమన నియమాలు Important Questions
• 3rd Lesson మన చుట్టూ ఉన్న పదార్థం శుద్ధమేనా ? Important Questions
• 4th Lesson పరమాణువులు-అణువులు Important Questions
• 5th Lesson పరమాణువులో ఏముంది ? Important Questions
• 6th Lesson రసాయన చర్యలు – సమీకరణాలు Important Questions
• 7th Lesson వక్రతలాల వద్ద కాంతి పరావర్తనం Important Questions
• 8th Lesson గురుత్వాకర్షణ Important Questions
• 9th Lesson తేలియాడే వస్తువులు Important Questions
• 10th Lesson పని మరియు శక్తి Important Questions
• 11th Lesson ధ్వని Important Questions
• 12th Lesson ప్రమాణాలు మరియు గ్రాఫులు Important Questions

SCERT AP Board 9th Class Social Solutions 5th Lesson Biosphere Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 5th Lesson Biosphere

9th Class Social Studies 5th Lesson Biosphere Textbook Questions and Answers

Improve Your Learning

Question 1.
Life itself constitutes a separate sphere called ‘Biosphere’. Explain.
Answer:

• The Earth is a unique planet, in that it has life.
•  It is inhabited by countless forms of life from microscopic bacteria to great banyan trees and animals like lions and blue whales and of course human beings.
• The earth has a combination of land, air and water and a moderate temperature has made life possible on it.
• Various forms of life are not only related to the three spheres around them, but also to one another.
• They influence one another and they are part of a complete “food chain.”
• Hence geographers are of the opinion that “life itself constitutes a separate sphere called ‘Biosphere’.

Question 2.
Why is ecological crisis created in modern times? What are its effects?
Answer:

• These days the increase in population creates great pressure on earth as well as its resources.
• To satisfy the needs of ever growing population systematic mining, cutting of forests, building of forests and fields and roads took place all over the earth.
• Burning of fossil fuels and release of enormous amounts of waste materials by industries contaminate air, water, and soil.
• For providing food for the ever growing population, there is a need for increase in agricultural production.
• Modern agriculture uses a large amount of chemical fertilizers and pesticides.
• The most frightening aspect of pollution due to modern agriculture is the increase in the levels of toxins in our food.
• This increasingly proves to be very harmful not only to humans but also to all types of life on earth.
• Since all living and non-living things on the earth are in one way or the other connected to each other, change affecting one, in due time affects all the others too. All these factors lead to ecological crisis.

Effects of ecological Grisis :

1. If a particular species which feeds upon a particular kind of plant is exterminated, it would result in unchecked growth of that plant and it may intrude the area where other plants grow and disturb their growth.
2. Many industries use chemicals and metals which flow into the rivers and streams. The metals like mercury are consumed by microorganisms in water and in turn beome the food of fishes. When human beings consume these fishes they too absorb some amount of mercury which is detrimental to their health.
3. Farmers use Diclofenac to treat cattle; when the cattle die their flesh retains this chemical: When their flesh is consumed by vultures, diclofenac leads to kidney failure in them and they die.

Question 3.
Natural vegetation depends upon the climate of the place. Write different kinds of forests and climatic conditions of their existence?
Answer:
a) Tropical Deciduous Forests :

1. Tropical deciduous forests are the monsoon forests.
2. These regions experience seasonal changes.
3. Trees shed their leaves in the dry season to conserve water.

b) Tropical Evergreen Forests :

1. These thick forests occur in the regions near the equator and close to the tropics.
2. These regions are hot and receive heavy rainfall throughout the year.
3. As there is no dry season, these trees do not shed their leaves altogether.
4. This is why they are called evergreen forests.

c) Temperate Evergreen Forests :

1. These forests are located in the mid latitudinal costal region.
2. They are commonly found along the eastern margin of the continents.

d) Temperate Deciduous Forests :

1. The temperate deciduous forests are found in higher latitudes.
2. They shed their leaves in the dry season.

e) Mediterranean Vegetation :

1. The west and the south west margins of the continents are covered by these forests.
2. Mediterranean trees adapt themselves to dry summers with the help of their thick barks and wax coated leaves. These help them to reduce transpiration.
3. These regions are marked for hot dry summers and dry summers and mild rainy winters.

f) Coniferous Forests :

1. Coniferous forests are found in the higher latitudes of Northern hemisphere.
2. These forests are found in abundance in the Himalayas.

g) Tropical Grasslands :

1. These forests grow on either side of the equator up to tropics.
2. This vegetation grows in the areas of moderate to low amount of rainfall.

h) Temperate Grasslands :

1. These are found in mid-latitudinal zones and in the interior part of the continents.
2. Usually the grass is very short and nutritious.

i) Thorny Bushes :

1. These are found in the dry desert like regions.
2. This vegetation cover is scarce here because of scanty rain and scorching heat.

j) Tundra :

1. This vegetation is found in the polar region.
2. The growth of natural vegetation is very limited here.
3. Only mosses, lichens and very small shrubs are found here.

Question 4.
How can we protect natural resources?
Answer:
Preservation of the natural environment is essential for maintaining community sustainability.
1. Water:
Communities must work to assure an adequate water supply to meet future needs.

2. Energy :
Energy should be consumed reasonably.

3. Air and climate :
Communities can preserve air quality by limiting or eliminating the discharge of harmful chemicals into the air and by minimizing the sources of air pollution.

4. Biodiversity :
Healthy wildlife should be supported through integrative approaches for, managing, protecting, and enhancing wildlife populations and habitats appropriate to their area and the atmosphere.

5. Land, forests, ecosystems should be provided protective measures. Thus we can protect natural resources.

Question 5.
Read the lesson and fill up the table.

Answer:

Question 6.
Locate the following countries in the world map.
a) New Zealand
b) Brazil
c) Australia
d) North America
e) China
f) India
Answer:

Question 7.
Read the paragraph under the title ‘Depletion of Resources’ on page 57 and comment on it.

Industrialisation, rapid population growth and urbanisation all have meant unprecedented exploitation of natural resources like minerals, forests, soil, water, air etc. besides sources of energy (coal, petroleum etc.) stored in the earth for billions of years. This has resulted in rapid deforestation and decline of reserves of minerals, oil and ground water. Many scientists have argued that the present way of life is not ‘sustainable’for if we use so much natural resources, nothing will be left for our children and grand children.

Answer:
Depletion of resources is the consumption of a resource faster than it can be replenished.

Use of either of these forms of resources beyond their rate of replacement is considered to be resource depletion.

Resource depletion is most commonly used in reference to farming fishings mining, water and fossil fuels.

Question 8.
Visit any nearby industrial establishment and observe the different kinds of smoke, liquid and Solid Wastes come*6uf of the cdrrtp’ourtd. FincTout from the nearby’residents about their impact on plants and animals. Based on the information collected, prepare a report and present in the class.
Answer:
I have visited a cement factory and observed the following :
Cement manufacture causes environmental impacts at all stages of the process. These include air-borne pollution in the form of dust, gases, noise and vibration when operating machinery and during blasting in quarries, and damage to country side from quarrying. Environmental protection also includes the re-integration of quarries.

• CO₂ emissions : Cement manufacturing releases CO₂ in the atmosphere.
• Heavy metal emissions in the air.
• Heavy metals present in the clinker.
• Use of alternative fuels and by-products materials.
• Noise levels are easy to establish.
• Smoke released from the factory causes breathing problems in the local people.
• Waste liquids are sent to the nearest river and pollute the river water.

Question 9.
Collect the details of some endangered animals and birds in your area. Fill the following table. Discuss in the classroom. Take the help of your elders and teachers.

Answer:
Student Activity.

9th Class Social Studies 5th Lesson Biosphere InText Questions and Answers

Question 1.
Can you say how plants are dependent upon air and water and how they affect the two in return? (Text Book Page No. 51)
Answer:

• Plants need water. Plants use water to carry moisture and nutrients from the roots to the leaves and food from the leaves back down to the roots.
• Plants absorb carbon dioxide from air to form carbohydrates during the process of photosynthesis.
• Nitrogen gas molecules that are trapped and carried in the water droplets provide nourishment to plants.
• Plants need oxygen in small amounts in respiration.
• Plants reduce the carbon dioxide levels in atmosphere and reduce the greenhouse effect and in turn the global warming.

How do plants affect water and air in return?
Answer:

• Plants return water to the atmosphere through the process of transpiration.
• Transpiration is the evaporation of water from aerial parts of plants, especially from . leaves but also from stems and flowers.

Question 2.
In what ways are insects like mosquitoes and butterflies dependent upon rocks or soil and upon water ? How do they affect them in return? (Text Book Page No. 51)
Answer:
Many insects depend on soil for part of their life. They make their home in the soils. Some insects live some portion of their lives in water.

Insects get involved lot of biological processes. Among the most significant beneficial impacts of insects on soil are

1. Their role in regulation of pest populations,
2. The pollination of crops and
3. The soil engineering. Insects mostly pollute the water.

Question 3.
Identify the desert regions in the world map.
Answer:

1. The Sahara Desert
2. The Kalahari Desert
3. The Thar Desert
4. The Arabian Desert
5. The Australian Desert
6. The Atacama Desert
7. The Sonaran Desert

Question 4.
Look around in your surroundings and find out the articles made of hard wood and soft wood. (Text Book Page No. 55)
Answer:
A) Articles made of hard wood :

1. Hard woods are often used to make items including furniture, flooring and utensils. Spoons, chopsticks, bowls, plates and cups are the utensils made from hard wood.
2. Musical instruments such as violins, guitars, pianos and hard drums are also made of hard wood.
3. Sofas, chairs, beds, benches, tables and wardrobes are made of hard wood.

B) Articles made of soft wood :
Manufacturing paper, newsprint, decorative articles, carvings, dolls, match boxes and packing boxes, etc. are made from soft wood.

Question 5.
Find out and learn few names of trees of your locality. (Text Book Page No. 55)
Answer:
In our local we can see Neem trees, Mango trees, Banyan trees and the trees are known by their local names: Chinta, Tumma, Seema Tumma, Palm (Tati), Bandaru, Yegisa are the trees which give us wood.

The trees such as Bamboo give us pulp.

Mangoes, Guavas, Blackberry are fruit bearing trees. In addition to them we have coconut trees and trees which give us drumsticks.

Question 6.
Can you discuss how human beings would have impacted the land, water, plants and animals around them when they began agriculture and animal herding? (Text Book Page No. 56)
Answer:

• With the development of agriculture humans began to radically transform the environment in which they lived.
• They cleared the lands around their settlements and controlled the plants that grew and the animals that grazed on them.
• They devised ways of storing rain water and rechanneling river water to irrigate plants.
• They built reservoirs, dug canals, dikes and sluices that permitted water storage.
• They domesticated major food crops, plants such as flax and cotton and began to cultivate them.
• Men also took lead in taming, breeding and raising the large animals associated with both farming and pastoral communities.
• Thus human beings impacted the land water, plants and animals around them when they began agriculture and animal rearing.

Question 7.
What sources of energy would they have used and how would they haveobtained them? (Text Book Page No. 56)
Answer:

1. The growth of sedantary farming greatly accelerated the pace of technological and social change.
2. Still the man used the physical energies of human and animals to use the agricultural and manufacturing tools.
3. He used wheel to make pots, plough to plough his fields and boats to transport his goods and kiln of blacksmith to make his tools.
4. He burnt firewood to cook his food.
5. Thus when they began agriculture man used merely the physical energy of human beings and animals.

Question 8.
Can you say what kinds of changes will building of cities have on the land and water around them? (Text Book Page No. 56)
Answer:

• With the beginning of agriculture and use of better tools give rise to larger, more elaborate and commodious housing and the construction of community ritual centres.
• Sun-dried bricks, interwoven branches usually plastered with mud and stone structures were associated with early agricultural communities.
• Houses in early agricultural settlements included special storage areas. They were centered on clay or stone hearths that were ventilated by a hole in the roof.
• To defend their cities from the rival nomadic settlements, they fortified their cities.
• The people had devised ways of storing rain water and rechanneling river water to irrigate plants.
• The reservoirs and canals, dikes and sluices that permitted water storage were constructed.

Question 9.
In what way do you think this would have affected the land scape and water cycle? (Text Book Page No. 56)
Answer:

• The increase in population create great pressure on earth as well as its resources.
• The entire earth was reshaped to suit the needs of humans.
• The surface of the earth received a great push with the industrial revolution and process of colonization.
• Industrial production needs raw materials on an unprecedunted scale.
• Industrial countries began to search for diverse kinds of raw materials and sources of energy.
• They dug deep wells and tried to find out what lay underneath.
• Systematic mining, cutting of forests, building of factories and fields and roads took place all over the world.
• The cumulative impact of industrial pollution causes greenhouse effect and global warming.
• The greenhouse effect could cause decrease in precipitation and soil moisture content and decrease of snow cover and finally decrease in rainfall.
• The increased surface temperature would cause melting of glaciers and lead to flooding of coastal areas.

 

Andhra Pradesh SCERT AP State Board Syllabus 8th Class Social Studies Study Material Guide Pdf free download, 8th Class Social Textbook Solutions in English Medium and Telugu Medium are part of AP Board 8th Class Textbook Solutions.

Students can also go through AP Board 8th Class Social Notes to understand and remember the concepts easily. Students can also read AP 8th Class Social Important Questions for exam preparation.

AP State Syllabus 8th Class Social Studies Guide Study Material Pdf Free Download

AP 8th Class Social Study Material Pdf Download English Medium

8th Class Social Study Material Pdf Theme I Diversity on the Earth

• Chapter 1 Reading and Analysis of Maps
• Chapter 2 Energy from the Sun
• Chapter 3 Earth Movements and Seasons
• Chapter 4 The Polar Regions
• Chapter 5 Forests: Using and Protecting Them
• Chapter 6 Minerals and Mining

8th Class Social Studies Study Material Theme II Production, Exchange and Livelihoods

• Chapter 7 Money and Banking
• Chapter 8 Impact of Technology on Livelihoods
• Chapter 9 Public Health and the Government

AP 8th Class Social Study Material Pdf English Medium Theme III Political Systems and Governance

• Chapter 10 Landlords and Tenants under the British and the Nizam
• Chapter 11A National Movement: The Early Phase 1885-1919
• Chapter 11B National Movement: The Last Phase 1919-1947
• Chapter 12 Freedom Movement in Hyderabad State
• Chapter 13 The Indian Constitution
• Chapter 14 Parliament and Central Government
• Chapter 15 Law and Justice: A Case Study

AP Board Solutions Class 8 Social Theme IV Social Organisation and Inequities

• Chapter 16 Abolition of Zamindari System
• Chapter 17 Understanding Poverty
• Chapter 18 Rights Approach to Development

AP Board 8th Class Social Solutions Theme V Religion and Society

• Chapter 19 Social and Religious Reform Movements
• Chapter 20 Understanding Secularism

Social Study Material for 8th Class Theme VI Culture and Communication

• Chapter 21 Performing Arts and Artistes in Modern Times
• Chapter 22 Film and Print Media
• Chapter 23 Sports: Nationalism and Commerce
• Chapter 24 Disaster Management

AP 8th Class Social Textbook Solutions Telugu Medium Pdf

8th Class Social Guide Pdf Download Theme 1 భూమి – వైవిధ్యం

• Chapter 1 పటాల అధ్యయనం – విశ్లేషణ
• Chapter 2 సూర్యుడు – శక్తి వనరు
• Chapter 3 భూ చలనాలు – రుతువులు
• Chapter 4 ధృవ ప్రాంతాలు
• Chapter 5 అడవులు – వినియోగం, సంరక్షణ
• Chapter 6 ఖనిజాలు, గనుల తవ్వకం

Social Guide 8th Class Theme 2 ఉత్పత్తి, వినిమయం – జీవనాధారం

• Chapter 7 ద్రవ్యం, బ్యాంకింగ్
• Chapter 8 జీవనోపాధులు – సాంకేతిక విజ్ఞాన ప్రభావం
• Chapter 9 ప్రజారోగ్యం – ప్రభుత్వం

8th Class Social Studies Lessons Theme 3 రాజకీయ వ్యవస్థలు – పరిపాలన

• Chapter 10 బ్రిటిష్, నిజాంల పాలనలో భూస్వాములు, కౌలుదార్లు
• Chapter 11A జాతీయోద్యమం : తొలి దశ 1885 – 1919
• Chapter 11B జాతీయోద్యమం : మలి దశ 1919 – 1947
• Chapter 12 భారత ఎన్నికల వ్యవస్థ
• Chapter 13 భారత రాజ్యాంగం
• Chapter 14 పార్లమెంటు – కేంద్ర ప్రభుత్వం
• Chapter 15 చట్టం, న్యాయం – ఒక సన్నివేశ అధ్యయనం

AP 8th Class Social Textbook Pdf Theme 4 సామాజిక వ్యవస్థీకరణ – అసమానతలు

• Chapter 16 జమీందారీ వ్యవస్థ రద్దు
• Chapter 17 పేదరికం – అవగాహన
• Chapter 18 హక్కులు – అభివృద్ధి

8th Class Social Textbook State Syllabus Theme 5 మతం – సమాజం

• Chapter 19 సాంఘిక, మత సంస్కరణోద్యమాలు
• Chapter 20 లౌకికత్వం – అవగాహన

8th Class Social Textbook Pdf Andhra Pradesh Theme 6 సంస్కృతి – సమాచారం

• Chapter 21 ఆధునిక కాలంలో కళలు – కళాకారులు
• Chapter 22 సినిమా – ముద్రణా మాధ్యమాలు
• Chapter 23 క్రీడలు : జాతీయత, వాణిజ్యం
• Chapter 24 విపత్తులు – నిర్వహణ

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 11 Electric Current.

AP State Syllabus SSC 10th Class Physics Important Questions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current 1 Mark Important Questions and Answers

Question 1.

Find the quantity of current in the above circuit. (AP March 2017)
Answer:
R = 3 + 5 + 2 = 10 Ω
I = \(\frac{1.5}{10}\) = 0.15 A.

Question 2.

Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors. (TS March 2015)
Answer:

Question 3.
What happens if we use a fuse made up of same wire which is used to make the electric circuit? (TS March 2017)
Answer:
It doesn’t work as a fuse. If high voltage occurs fuse do not melt and circuit will not be opened / breaked. So home appliances will be damaged.

Question 4.
Write any two differences between ohmic and non- ohmic conductors. (TS June 2018)
Answer:

Ohmic conductors	Non-ohmic conductors
Ohmic conductors follow the Ohms law.	Non-ohmic conductors do not follow the Ohms law.
Ohmic conductors are electric conductors.	Non-ohmic conductors are semicon­ductors.
V-I graph of ohmic conductors is a straight line.	V-I graph of non-ohmic conductors is a curve.

Question 5.
Draw the electric circuit with the help of a Battery, Voltmeter, Ammeter, Resistance and connecting wires. (TS March 2018)
Answer:

Question 6.
What happens, if the household electric appliances are connected in series? (TS March 2019)
Answer:
If all household appliances are connected in series, then if one appliances stop working due to failure then all the appliances stops working due incomplete circuit.

Question 7.
Define lightning.
Answer:
Lightning is an electric discharge between two clouds or between cloud and earth.

Question 8.
Define drift speed or drift velocity.
Answer:
The electrons in the conductor move with a constant average speed called drift speed or drift velocity.

Question 9.
Define conductors.
Answer:
The materials which can conduct electricity are called conductors.
Eg : Copper, Silver, Aluminium.

Question 10.
Define insulators.
Answer:
The materials which can’t conduct electricity are called insulators or non – conductors.
Eg: Wood, Rubber.

Question 11.
Define semi-conductors.
Answer:
The materials whose resistivity is 10⁵ to 10¹⁰ times more than that of metals and 10¹⁵ to 10¹⁶ times less than that of insulators.
Eg: Silicon, Germanium,

Question 12.
How does a battery work?
Answer:
In a circuit, the battery stores chemical energy and this energy converts into electric energy. Thus a battery works.

Question 13.
Define lattice.
Answer:
Conductors like metals contain a large number of free electrons while the positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.

Question 14.
Define potential difference.
Answer:
Work done by the electric force on unit positive charge to move it through a distance is called potential difference.

Question 15.
Define electromotive force.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 16.
Define resistance of a conductor.
Answer:
The obstruction to the motion of the electrons in a conductor is called resistance of a conductor.

Question 17.
Define resistivity (ρ).
Answer:
Resistivity is a constant.

Question 18.
Write Ohm’s formula.
Answer:
V = IR, where V is the potential difference (voltage), I is the electric current and R is the resistance.

Question 19.
Define resistor.
Answer:
The material which offers resistance to the motion of electrons is called a resistor.

Question 20.
What is a multimeter?
Answer:
It is an electronic measuring instrument that combines several measurement functions in one unit.

Question 21.
Define electric power.
Answer:
The product of voltage and electric current is called electric power.

Question 22.
Define electric energy.
Answer:
The product of electric power and time is called electric energy.

Question 23.
What is lattice?
Answer:
The arrangement of positive ions in a conductor is known as lattice.

Question 24.
Name two special characteristics of fuse wire.
Answer:
High resistivity and low melting point.

Question 25.
Name two special characteristics of heating coil.
Answer:
High resistivity and high melting point.

Question 26.
How does resistivity vary with material of conductor?
Answer:
The resistivity is less for a good conductor and is large for a bad conductor.

Question 27.
If length of a particular conductor increased by two times and its area of crosssection decreased by four times, then what happens to its resistivity?
Answer:
The resistivity of the conductor does not change because resistivity does not depend on dimensions of conductor.

Question 28.
What does the slope of V -1 graph for an Ohmic conductor represent?
Answer:
For an Ohmic conductor the slope of V -1 graph represents the resistance.

Question 29.
Express the units of ohm in terms of volt and ampere.
Answer:
1) The SI unit of resistance is ‘Ohm’.
2) Ohm = \(\frac{\text { Volt }}{\text { Ampere }}\)

Question 30.
What is the resultant resistance of this combination?

Answer:
R, R, R Ω resistances are in parallel.
⇒ Resultant resistance = \(\frac{R}{3}\)

Question 31.
What are the quantities are conserved in Kirchhoff’s and Is’ laws?
Answer:
According to Ist law, charge, and 2^nd law, energy are conserved.

Question 32.
If the length and radius of a conductor are both halved. What happend resistance of wire?
Answer:
Length and radius are halved.

Resistance is doubled.

Question 33.
If work done is W and the charge that flows through is Q, then what is the equation 1 of potential difference?
Answer:

Question 34.
How many electrons constitutes a current of one ampere?
Answer:
6.25 × 10¹⁸ electrons in one second.

Question 35.
What are the maximum and minimum resistances are prepared by 30Ω, 30Ω, 30Ω?
Answer:
Maximum resistance = 30 + 30 + 30 = 90Ω
Minimum resistance = \(\frac{30}{3}\) = 10 Ω

Question 36.
Electric current I = nqA,v_d. Write the representation of letters.
Answer:
n = Electron density
A = Area of cross – section
v_d = drift velocity
q = charge of electron .

Question 37.
What is the resistance of bulb marked 60W and 120V?
Answer:

Question 38.
From figure, if V_A = 10V, then V_B = ?

Answer:
V_A – IR – E = V_B
10 – 1 × 5 – 3 = V_B
10 – 5 – 3 = V_B
2 = V_B
∴ V_B = 2 volts

Question 39.
Write the examples of non-ohmic conductors.
Answer:
Non-ohmic conductors are electrolytes, semi conductors, vacuum tubes.

Question 40.
What is meant by electric shock?
Answer:
When the current flows through the body the functioning of organs inside the body gets disturbed. This disturbance inside the body is felt as electric shock.

Question 41.
There is no electric shock on bird, when stand on the Electric wires. Why?
Answer:
When the bird stands on a high voltage wire, there is no potential differences between the legs of the bird, so no current passes through the bird.

Question 42.
Which factors are influence the resistivity of wire?
Answer:
a) Temperature,
b) Nature of material.

Question 43.
The length of wire is doubled and area of cross-section also doubled. What is the change in resistivity.
Answer:
Resistivity is independent of length and Area of cross-section.
Resistivity is not change.

Question 44.
A battery of 6V is applied across a resistance of 15Ω. Find the current flowing through the circuit.
Answer:

Question 45.
The formula V = I R is applicable for what substances?
Answer:
V = IR is applicable for
a) Ohmic conductors,
b) Non-ohmic conductor.

Question 46.
Resistance of a conductor of length 75 cm is 3.250. Calculate the length of a similar conductor, whose resistance is 16.25Ω.
Answer:

Question 47.
If four wires of each resistance R are joined to form a square, then what is the resistance between its opposite vertices?
Answer:

Resistance of ABC = 2R
Resistance of ADC = 2R
2R & 2R are parallel ⇒ Resultant = ‘R’

Question 48.
How much power consumption in a day of 100W television utilised 10 hours?
Answer:
Power consumption = \(\frac{100 \mathrm{~W} \times 10 \mathrm{~h}}{1000}\)
1 KWH or 1 unit.

Question 49.
How should we connect a voltmeter to measure voltage?
Answer:
The voltmeter must be connected in parallel to the electric device to measure the potential difference across the ends of the electric device.

Question 50.
How are ammeter and voltmeter connected in a circuit?
Answer:
Ammeter is always connected in series and voltmeter is always connected in parallel in a circuit.

Question 51.
Is the voltmeter connected in series or parallel in circuit? Why?
Answer:
Voltmeter should be connected parallel in the circuit to measure the potential difference between two points of conductor.

Question 52.
State whether the home appliances like Fridge, TV, Computer are connected in series or parallel Why?
Answer:
They are connected in parallel because if any one device is damaged, the rest will work as usual because the circuit does not break.

Question 53.
Why are copper wires used as connecting wires?
Answer:
Copper is a good conductor of electricity so copper wires are used as connecting wires.

Question 54.
Why is the fuse wire fitted in a porcelain casing?
Answer:
Porcelain is an insulator of electricity. So fuse wire is fitted in a porcelain casing.

Question 55.
Draw the diagram of potential – current of a copper conductor.
Answer:

Temperature T = 27°C
It is a straight line passing through origin.

Question 56.
Draw the shape of V – I graph for a silicon.
Answer:

Silicon is a semi conductor.
It is not obey the Ohm’s law.

Question 57.
Is there any application of Ohm’s Law in daily life?
Answer:

• Electrical device like electric bulb, iron box and regulators are some applications of Ohm’s Law.
• Fuse in household circuits is also another application of Ohm’s Law.

Question 58.
Two wires of the same material and same length have radii r₁ and r₂ respectively. Compare (i) their resistance, (ii) their resistivities.
Answer:

Resistivity for a same material is same. So their ratio =1:1.

Question 59.
A wire of resistance is doubled on itself, then what is its new resistance?
Answer:
Suppose length is l, area of cross-section is A and resistance is R.

Question 60.
Calculate effective resistance between points A and B.

Answer:
R₁ = 1 Ω, R₂ = 2 Ω are in series. So R’ = 1 + 2 = 3 Ω.
Now R₃ and R’ are in parallel.

Question 61.
A fuse is rated 8A. Can it be used with an electrical appliance of rating 5 KW and 200 V?
Answer:
Given P = 5 KW = 5 × 1000 = 5000 W; V = 200 V.

So a fuse of rate 8 A is not suitable because it uses current of 25 A.

Question 62.
A current of 2A is passed through a coil of resistance 75 Ω for 2 minutes. How much heat energy is produced?
Answer:
Given i = 2A, R = 75 Ω and t = 2 min. = 2 × 60 sec. = 120 sec.
Heat energy produced H = i² Rt = 2² × 75 × 120 = 36000 J = 36 KJ.

Question 63.
What is ,the resistance under normal working conditions of a 240 V electric lamp rated at 60 W?
Answer:
P = 60 W; V = 240V.

Question 64.
State the use of Ammeter. How should the Ammeter be connected in electric circuit?
Answer:
Ammeter should be connected in series in a circuit.

Question 65.
What is current value of x?

Answer:
2 + 3 = 5 of ‘A’
5 – 1 = 4 of ‘B’
2x = 4
x = 2 A at ‘C’

Question 66.
What is the value of V_A, when V_B = 8V?

Answer:
V_A – 6 × 1 – 3 = V_B
V_A – 6 – 3 = 8
V_A = 8 + 9
V_A = 17 volts

Question 67.
In the figure, how much current passing through 6Ω resistor?

Answer:
Answer:
R₁ : R₂ = 6 : 4
R₁ : R₂ = 3 : 2
i₁ : i₂ = R₂ : R₁ = 2 : 3
∴ i₁= 2A
i₂ = 3A

10th Class Physics 11th Lesson Electric Current 2 Marks Important Questions and Answers

Question 1.
Observe the graph of potential difference (V) drawn between two ends of a conductor and current (I) passing through it. Answer the following questions :

a) Which law is used to explain the graph? State it.
b) What is the resistance of the conductor? (AP June 2017)
Answer:
a) Ohm’s law is used to explain the given graph.
Ohm’s Law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Question 2.
Draw the experimental set-up to verify that V/I is constant for a conductor. (TS March 2016)
Answer:

Question 3.
A house has 3 tube-lights of 20 watts each. On the average, all the tube-lights are kept on for five hours. Find the energy consumed in 30 days. (AP SCERT : 2019-20)
Answer:
Number of tube lights = 3
Wattage = 20 watts each
Consumed hours = 5 hrs

Consumed energy for 30 days = 0.3 × 30 = 9 KWH (or) 9 units.

Question 4.
How do the resistors 6 Ω, 10 Ω to be connected in a circuit to get minimum resistance? Find the resultant resistant of the circuit. (TS June 2019)
Answer:

1. Resistors 6Ω, 10Ω should be connected in parallel connection in a circuit to get minimum resistance.
2. Resultant resistance
   

Question 5.
Write two examples for ohmiq and non-ohmic materials each. (TS June 2019)
Answer:
Ohmic materials : Copper, Aluminium
Non – ohmic materials : Germanium, Silicon

Question 6.
Give reasons for using lead in making fuses.
Answer:

• Lead is used in making fuses because it has low melting point EK resistivity.
• If the current in the lead wire exceeds certain value, the wire will heat up and melt, so the circuit in the households is opened and all the electric devices are saved.

Question 7.
How can we decide the direction of electric current in a conductor?
Answer:
We know I = nqAv_d. In this n and A are positive. Hence the direction of current is determined by the signs of the charge ‘q’ and drift speed v_d.

1. For electrons, q is negative and v_d is positive. Then the product of q and v_d is negative. So the direction of electric current is opposite to the flow of negative charge.
2. For positive charge, the product of q and v_d is positive. Hence, the direction of electric current can be taken as the direction of flow of positive charges.

Question 8.
What are the devices used in a circuit?
Answer:
1) Ammeter :
It is used to measure current.

2) Volt meter:
It is used to measure potential difference across the ends of conductor.

3) Rheostat:
It varies current in a circuit.

4) Switch :
It is useful to make a circuit or break a circuit.

5) Cell:
It is source of electric energy in the circuit.

6) Multimeter :
It is useful to measure current, voltage and resistance in the circuit.

Question 9.
A student says “Potential difference and Emf are same.” Justify your answer.
Answer:
Both are different because potential difference is the work done by the electric force on unit positive charge to move it through a distance between two points whereas emf is the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 10.
Define Ohmic and non-ohmic conductors and give two examples each of them.
Answer:
Ohmic conductors :
The conductors which obey Ohm’s law are called ohmic conductors, e.g, : Copper, Iron.

Non-ohmic conductors :
The conductors which do not obey Ohm’s law are called non-ohmic conductors, e.g. : Semiconductors, Electrolytes.

Question 11.
Explain the Junction law of Kirchhoff.
Answer:

1. At any junction point the sum of the currents into the junction must be equal to the sum of currents leaving the junction.
2. There is no accumulation of electric charges at any junction in a circuit.
   I₁ + I₄ + I₆ = I₅ + I₂ + I₃

Question 12.
Write differences between overloading and short circuiting.
Answer:
Current chooses a path which has least resistance. So sometimes electrical appliances get damaged by passage of excess of current due to short circuit.

When so many electrical appliances are connected to the same electrical main point, maximum current can be drawn from the mains which causes overheating and may cause a fire which is called overloading.

Question 13.
The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.

Answer:
i) For same change in I, change in V is less for the straight line A than for the straight line B (i.e., straight line A is less steeper than B).
ii) So the straight line A represents small resistance, while straight line B represents more resistance.
iii) In parallel combination, the resistance decreases, while in series combination, the resistance increases, so A represents a parallel combination.

Question 14.
Two resistors are joined with a battery such that
a) same current flows in each resistor.
b) potential difference is small across each resistor.
c) equivalent resistance is less than either of the two resistors.
d) equivalent resistance is more than either of the resistors.
State how are the resistors connected in each of the above case.
Answer:
a) Since same current is flowing in each resistor, they are connected in series.

b) Potential difference is same across same resistor that shows they are connected in parallel.

c) Equivalent resistance is less than the either of the two resistances which shows they are connected in parallel because in parallel connection resistance decreases.

d) Equivalent resistance is more than the individual resistances which indicate that they are connected in series. Since in series connection resistance increases.

Question 15.
Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.
Answer:
i) To carry larger current, the resistance of wire should be low, so its area of cross-section should be large.

ii) Because the second cable carrying 15 A ampere current which indicates it has low resistance that is more surface area which implies it is thicker wire.

Question 16.
When a potential difference 30V is applied across a resistor, it draws a current of 3A. If 20V is applied across the same resistor, what will be the current?
Answer:
Situation : 1

Situation : 2

Question 17.
Is Ohm’s Law universally applicable for all conducting elements? If not, give example of elements which do not obey the Ohm’s Law.
Answer:

• Ohm’s Law is not applicable for all conducting elements.
• For example, some semi-conductors like silicon, germanium do not obey the Ohm’s Law.
• Those which do not obey Ohm’s Law are called non-ohmic materials.
• LED’s are non-ohmic materials.

Question 18.
Alloys are used in electrical heating devices rather than the pure metal. Why?
Answer:
Alloys are homogeneous mixtures of two or more metals. Generally alloys have more resistivity than the metals from which they have been prepared. As the resistivity increases heating effect of conductor also increases. So alloys are preferred in heating devices.

Question 19.
A switch should not be touched with wet hands. Why?
Answer:

• A switch should not be touched with wet hands.
• If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand, and the person may get a total shock.

Question 20.
Which material is used for power transmission? Why?
Answer:

• The wires which are used for connections and for power transmission are made of material such as copper or aluminium.
• The reason is their resistivity is very small and they are made thick so that their resistance can be considered to be negligible.
• Further, the loss of electrical energy due to heating is also negligible in them.

Question 21.
Which material is preferred for heating element? Why?
Answer:

• The heating elements or resistance wires (or standard resistors) are made of material such as nichrome, manganin, constantan, etc. for which the resistivity is quite large and the effect of change in temperature on their resistance is negligible.
• So electrical energy is converted into heat energy when current passes through the wire.

Question 22.
Draw the symbols of the following.
i) Battery
ii) Resistance
iii) Ammeter
iv) Voltmeter
v) Key
vi) Rheostat
Answer:

Question 23.
Draw V-I graphs of Ohmic and non-ohmic conductors.
Answer:

Question 24.
Draw a circuit diagram with a cell, an electric bulb, an ammeter and plug key.
Answer:

Question 25.
Draw a circuit diagram to verify the Ohm’s Law.
Answer:

Question 26.
If 60 C of charge passes through a conductor for 1 minute, find the current through the conductor.
Answer:

Question 27.
The work done in moving 6 C of charge through a circuit is 12 J. Find the potential difference in the circuit.
Answer:

Question 28.
Resistance of two resistors are 6 Ω, 12 Ω respectively. Find the resultant resistance if the resistors are connected (1) in series (2) in parallel.
Answer:

Question 29.
10 equal resistors of resistance 20 are connected in a circuit. Find the resultant resistance if they are connected in series or in parallel ?
n = 10; R = 20 Ω
In series connection resultant resistance R’ = nR = 10 × 20 = 200 Ω

Question 30.
From the figure find the current through 6 Ω, 12 Ω resistors and find the resultant resistance in the circuit and also find current in the circuit.

Answer:
R₁ =6 Ω; R₂ = 12 Ω; V = 6 V

Question 31.
Find which has greater resistance. 1 KW heater or a 100 W tungsten bulb, both marked for 230 V.
Answer:

Question 32.
Two wires (one is copper and other is aluminium) have equal area of cross-section and have the same resistance. Find which one is longer.
Answer:
Suppose the resistance of copper and aluminium wires are R₁ and R₂. Suppose their area of cross-section is A.
The resistivity of copper (ρ₁) = 1.68 × 10^-8
The resistivity of aluminium (ρ₂) = 2.82 × 10^-8

Question 33.
Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?
Answer:
Suppose the resistance of equal resistors is ‘R’. Suppose they are connected in series. Then their equivalent resistance R’ = R + R + R = 3R
If they are connected in parallel their equivalent resistance R” = \(\frac{R}{3}\)
∴ Ratio of resultant resistances = R’: R” = 3R : \(\frac{R}{3}\) = 9 : 1

Question 34.
How is Ideal earthing helpful during short circuiting?
Answer:

• During short circuiting an excessive current flows through the live wires.
• It will pass to earth through the earth wire if there is local earthing otherwise it may cause a fire due to overheating of the live wires.

Question 35.
You have three resistor values 2Ω, 3Ω and 5Ω. How will you join them Sd that the total resistance Is less than 2Ω? Find resultant resistance.
Answer:
Given R₁ = 2Ω, R₂ = 3Ω, R₃ = 5Ω.
The resistors should be joined in parallel.

Question 36.
An electric kettle Is rated 3 KW, 250 V. Give reason whether this kettle can be used in a circuit which contains a 13 A fuse?
Answer:
V = 250 V, P = 3KW = 3000 W

The fuse is suitable because safe limit of current for kettle is 12 A.

Question 37.
Two resistors of 4 Ω and 6 Ω are connected parallel. The combination is connected across 6 V battery of negligible resistance. Find, i) the power supplied by the battery and ii) the power dissipated in each resistance.
Answer:
i) R₁ = 4 Ω, R₂ = 6 Ω, V = 6V.

ii) In parallel connection the resultant resistance

Question 38.
In the circuit shown below calculate the value of x if the equivalent resistance between A and B is 4 Ω.

Answer:
Given R₁ = 4 Ω, R₂ = 8Ω, R₃ = x Ω and R₄ = 5 Ω.
And resultant resistance R = 4 Ω.
R₁ and R₂ are in series.
Resultant resistance R’ = R₁ + R₂ = 4Ω + 8Ω = 12 Ω.
R₃ and R₄ are also in series.
Resultant resistance R” = R₃ + R₄ = x + 5
R’ and R” are in parallel.

Question 39.
A wire of 9 Ω resistance having 30 cm length is tripled on itself. What is its new resistance?
Answer:
Given R = 9 Ω, l = 30 cm and suppose area of cross-section A.

10th Class Physics 11th Lesson Electric Current 4 Marks Important Questions and Answers

Question 1.
A house has four tube-lights, three fans and a television. Each tube-light draws 40 W. The fan draws 80 W and the television draws 100 W. On an average, all the tube-lights are kept on for 5 hours, all fans for 12 hours and the television for 6 hours everyday. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWH. (AP March 2015)
Answer:
The power used by
1) Four tube-lights of 40 W for 5 hours in 30 days
= 4 × 40 × 5 × 30 = 2400 WH

2) Three fans of 80 W for 12 hours in 30 days
= 3 × 80 × 12 × 30 = 86400 WH

3) One television of 100 W for 6 hours in 30 days
= 1 × 100 × 6 × 30 = 18000 WH

4) Total electric energy used
= 24000 + 86400 + 18000 = 128400 WH

W.H. converted into K.W.H. = 128.4 K.W.H. \(\left[\because \frac{128400}{1000}\right]\)
Cost of 1 unit = Rs. 3.00
Amount to be paid for 128.4 units = 128.4 × 3 = Rs. 385.20

Question 2.
Observe the given circuit. (AP June 2017)
R₁ and R₂ are two resistors and R₁ = R₂ = 4Ω. Emf of the battery E is 10 V.

Answer the following questions.
a) How are the resistors R₁ and R₂ connected in the circuit ?
b) What is the potential difference across R₁?
c) What is the effective resistance of the circuit?
d) What is the total current drawn from the battery?
Answer:
a) Resistors R₁ and R₂ are connected in parallel in the given circuit.
b) The potential difference across R₁ is ‘E’ volts = 10 V.

Question 3.
State Kirchhoff’s Loop law and explain. (AP June 2018)
Answer:
Loop law :
The algebraic sum of the increases and decreases in potential difference across various components of the circuit in a closed circuit loop must be zero.

Explanation :
Let us imagine in a circuit loop the potential difference between the two points at the beginning of the loop has a certain value. As we move around the circuit loop and measures the potential difference across each component in the loop, the potential difference may decrease or increase depending upon the nature of the element like a resistor or battery. But when we have completely traversed the circuit loop and arrive back at starting point. The net change in the potential difference must be zero. Thus the algebraic sum of changes in potential difference is to be zero.
(OR)
Let us apply loop law to a circuit as below.

for the loop ACDBA
– V₂ + I₂R2₂ – I₁R₁ + V₁ = 0

for the loop EFDCE
-(I₁ + I₂)R₃ – I₂R₂ + V₂ = 0

for the loop EFBAE
-(I₁ + I₂)R₃ – I₁R₁ + V₁ = 0

Question 4.

Observe the above diagram and answer the following. (AP March 2018)
a) Are all the resistors connected in parallel or series?
b) What is the equivalent resistance of the combination of three resistors?
c) In this system, which physical quantity is constant?
d) If R₁ = 2Ω, R₂ = 3Ω, R₃ = 4Ω. find equivalent resistance.
Answer:
a) Connected in series.
b) R_eq = R₁ + R₂ + R₃
c) Current (I)
d) R_eq = R₁ + R₂ + R₃
=2 + 3 + 4
= 9Ω

Question 5.
How do you verify that resistance of a conductor of uniform cross-section area is proportional to the length of the conductor at constant temperature? (TS March 2015)
Answer:
Aim :
To verify the relation between re.M+iance and lenath of the conductor.

1) Required material :
Wires or spokes of different lengths with same cross-section area of the same metal.
2) Battery
3) Ammeter
4) Key
5) Connecting wires.

Procedure :
1) Construct a circuit with Battery, Ammeter, Switch (key) and connecting wires, keeping some space at the both ends.
2) Connect the selected wires or spokes at the ends to complete the circuit.

3) Connect the wires or spokes individually and record the current using ammeter.

Conclusion :
If the current flowing in the circuit decreases with an increase in the length of the wire or spokes (Resistance increases), we can say that the resistance of the conductor is proportional to the length of the conductor.

Question 6.
What are the factors affecting the resistance of an electric conductor? Explain any two factors. (TS June 2015)
Answer:
The factors affecting the resistance of an electrical conductor are

1. Nature of material
2. Temperature
3. Length of the conductor
4. Area of cross-section of conductor

Explanation :

1. As the temperature increases the resistance increases and vice versa.
2. As the material changes resistance changes.
3. Resistance is directly proportional to length of the conductor (if T and A are kept constant). R ∝ l
4. Resistance is inversely proportional to area of cross-section (if 1 and T are kept constant). R ∝ \(\frac{1}{A}\)

Question 7.
What is the relationship between length of a conductor and its resistance? Write the experimental procedure to verify that relationship. (TS Junc 2017)
Answer:

• The resistance of a conductor is directly proportional to its lenght for a constant potential difference.
• Take iron spokes of different lengths with same cross-sectional arreas.
• Make a circuit by connecting an iron spoke with battery, ammeter and switch in series.
• Put the switch on and allow the current to pass in the circuit. Measure the ammeter reading.
• Repeat this for other lengths of the iron spokes. Note the corresponding values of currents.
• The resistance of each spoke increases with increase in the length of the spoke.

Question 8.
Find the resultant resistance for the following given arrangement. Find the current when this arrangement is connected with 9 V battery. (TS March 2017)

Answer:
The diagram is not clear so award 4 marks in the public examinations.

Question 9.
12 V battery is conncected in a circuit and to this 4Ω, 12Ω resistors are connected in parallel, 3Ω resistor is connected in series to this arrangement. Draw the electric circuit from this information and find the current in the circuit. (TS June 2018)
Answer:

Question 10.
In a circuit,60V battery, three resistances R₁ = 10 Ω, R₂ = 20 Ω and R₃ = x Ω are connected in series. If 1 ampere current flows in the circuit, find the resistance in R₃ by using Kirchhoffs loop law. (TS March 2018)
Answer:

Given
I = 1 amp
R₁ = 10Ω
R₂ = 20Ω
R₃ = xΩ
V = 60V
In ADCBA loop
– IR₃ – IR₂ – IR₁ + V = 0
– lx – 1 × 20- 1 × 10 + 60 = 0
– x – 30 + 60 = 0
– x + 30 = 0
x = 30Ω

Question 11.
List out the material required for the experiment “The effect of increasing of cross-section of a conductor upon its resistance” and write the experimental procedure. (TS June 2019)
Answer:
Aim :
To show that the effect of increasing of cross-section of a conductor upon its resistance.

Required Material :
Battery eliminator, Key, Ammeter, Manganin wires of equal lengths but different cross sectional areas, Copper connecting wires.

Procedure :

• Collect manganin wires of equal lengths but different cross sectional areas.
• Make a circuit as shown in the figure.

• Connect one of the wires between points ‘P’ and ‘Q’.
• Note the value of the current using the ammeter connected to the circuit and note it.
• Repeat this with other wires.
• Note the corresponding values of currents in each case and note them.

Conclusion :
We can notice that the current flowing through the wire increases with increasing their cross sectional area.

Question 12.
Derive an expression to find drift velocity of electrons.
Answer:

• Consider a conductor with cross sectional area A. Assume that the two ends of the conductor are connected to a battery to make the current flow through it.
• Let ‘v_d‘ be the drift speed of the charges and ‘n’ be the number of charges present in the conductor in a unit volume.
• The distance covered by each charge in one second is ‘v_d‘

• Then the volume of the conductor for this distance = Av_d
• ∴ The number of charges contained in that volume = n.Av_d
• Let q be the charge of each carrier.
• Then the total charge crossing the cross sectional area at position D in one second is ‘n q A_d‘.
  This is equal to electric current.
  ∴ Electric current I = n q Av_d.
  ∴ v_d = \(\frac{I}{nqA}\)

Question 13.
Derive an expression to measure emf of a battery.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

1. Let this chemical force be F_c.

2. This chemical force does some work to move a negative charge ‘q’ from positive terminal to negative terminal against the electric force Fe. Let this work be ‘W’.

3. ∴ The work done by the chemical force to move 1 coloumb of charge from the

This S.I unit of emf is ‘volt’ and is measured using voltmeter.

Question 14.
What are the factors on which the resistance of conductor depends? Give the corresponding relation.
Answer:

• The value of resistance of a conductor depends on temperature for constant potential difference.
• Resistance of a conductor depends on the material of the conductor.
• Resistance of a conductor is directly proportional to its length, i.e., R ∝ l.
• Resistance of a conductor is inversely proportional to the area of cross-section of the material, i.e., R ∝ \(\frac{1}{A}\)

Question 15.
What do you mean by (i) short circuit (ii) overloading? What are the safety precautions taken to avoid these problems in domestic electric circuits?
Answer:
Short circuit:
Sometimes current chooses a path which has least resistance which is called short circuit.

Overloading :
The over heating due to drawing excess of current from a single main is called overloading.
Precautions to avoid damage due to short circuit and overloading.

1. Using fuse
2. Connecting electrical appliances in various mains.
3. Earthing

Question 16.
A circuit is set up as shown in the figure. Calculate the current and potential difference across R₁, R₂ and R₃, when
a) keys K₁ and K₂ are both closed,
b) key K₁ is closed and K₂ is open,
c) key K₁ is open and K₂ is closed.

Answer:
a) When both the keys K₁ and K₂ are closed :
The resistors R₁ and R₃ are parallel.
So resultant resistance R_p = \(\frac{R_{1} R_{3}}{R_{1}+R_{3}}=\frac{6 \times 4}{6+4}\) = 2.4 Ω.
The resistor R₁ and R_p are in series as shown in figure.

Total resistance of circuit R_s= R₁ + R_p = 6 + 2.4 = 8.4 Ω
Current I = \(\frac{V}{R_{S}}=\frac{4.2}{8.4}\) = 0.5 =A.
Potential difference across R₁ is V₁ = IR₁ = 0.5 × 6 = 3V.
Potential difference across the combination of R₂ and R₃ is
V’ = V – V₁ = 4.2 – 3 = 1.2 V.
Now since R₂ and R₃ are in parallel,
Potential difference across R₂ = Potential difference atross R₃ = V’ = 1.2 V

b) When the key K₁ is closed and key K₂ is open : The resistor R₃ will not be in circuit.
The resistors R₁ and R₂ are in series.
Total, resistance R_s = R₁ + R₂ = 6 + 6=12 Ω.
Current I = \(\frac{V}{R_{s}}=\frac{4.2}{12}\) = 0.35 A
The same current will flow through each resistor R₁ and R₂.
Potential difference across R₁ is V₁ = IR₁ = 0.35 × 6 = 2.1 V.
Potential difference across R₂ is V₂ = IR₂ = 0.35 × 6 = 2.1 V.
The current and potential difference across R₃ will be zero.

c) When key K₁ is open and key K₂ is closed :
No current flows through R₁ R₂ and R₃ since the circuit is incomplete. Hence potential difference across R₁, R₂ and R₃ is zero.

Question 17.
For the combination of resistance shown in figure, find the equivalent resistance between (a) C and D, (b) A and B.

Answer:
a) Between C and D :
The resistors R₂, R₃ and R₄ are in series. They can be replaced by an equivalent resistance R_s where
R_s = R₂ + R₃ + R₄ = 3 + 3 + 3 = 9 Ω
The resistance R₅ and R_s are in parallel between the points C and D.
The equivalent resistance between C and D then

Thus the equivalent resistance between C and D is 2.25 Ω.

b) Between A and B :
Now the resistors R₁, R_p and R₆ are in series between the points a and B.

The equivalent resistance between A and B is
R_AB = R₁ + R_p + R₆ = 3 +2.25 + 3 = 8.25 Ω.

Question 18.
How does resistance and resistivity vary with temperature?
Answer:

• For a metallic conductor, the resistance increases w ith the increase in temperature. The resistance of filament of bulb is more w hen it is glowing than when it is not glowing. The specific resistance or resistivity also increases with increase in temperature.
• For alloys (such as constantan and manganin), the resistance and the resistivity remains practically unchanged with the increase in temperature.
• For semi conductors (such as silicon, germanium, etc.), the resistance decreases with the increase in temperature.
  eg.: the resistance of carbon also decreases with the increase in its temperature.

Question 19.
Why is a copper wire unsuitable for making fuse wire?
Answer:

• A fuse is a short piece of wire made up of a material of high resistivity and of low melting point so that it may easily melt due to overheating when current in excess to the prescribed limit passes through it.
• The thickness of wire is different in different fuses depending on the amount of current which is permitted to flow through them.
• Generally an alloy of lead and tin is used as the material of the fuse wire because it has a high resistivity and low’ melting point.
• A copper wire is unsuitable for using as fuse wire because copper has low resistivity and high melting point.
• Therefore the use of an ordinary’ wire as a fuse must be avoided.

Question 20.
Why is current rating of a fuse required?
Answer:
1) The electric wiring for light and fan circuit uses a thin fuse wire of low current carrying capacity because the line wire has a current carrying capacity of 5A.

2) Thicker fuse wires of higher current carry ing capacity (15 A) are used for large current consuming appliances such as air conditioner, geyser, washing machine, etc. because the line wire for such dev ices have current carrying capacity of 15A.
The current rating of a fuse in a circuit can be obtained by the following relation.

Question 21.
Describe the activity with the help of diagram to establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.
Answer:
Aim :
To establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.

Material required :
5 dry cells of 1.5 V each, conducting wires, an ammeter, a volt meter, thin iron spoke of length 10 cm, LED and key.
Diagram :

Procedure :

• Connect a circuit as shown in the figure.
• Solder the conducting wires to the ends of the iron spoke.
• Close the key.
• Note the readings- of current from ammeter and potential difference from volt meter in the given table,

• Now connect two cells (in series) instead of one cell in the circuit.
• Note the respective readings of the ammeter and voltmeter and record the values in the table.
• Repeat the same for three cells, four cells and five cells respectively.
• Record the values of V and I corresponding to each case in the table.
•  Find \(\frac{V}{I}\) for each set of values.
  Conclusion : The ratio of \(\frac{V}{I}\) is constant.
• From this activity we can conclude that the potential difference between the ends of the iron spoke (conductor) is directly proportional to the current passing through it. This means V ∝ I.

Question 22.
In an experiment to verify Ohm’s Law the following values are given below. Draw a graph of ‘I’ versus ‘V’. Show that the graph conforms Ohm’s Law and find the resistance of the resistor.

Answer:
1) Graph between ‘V’ and ‘I’.

2) From the above graph,
Straight line §hows that the relation between potential difference (V) and current (I) as \(\frac{V}{I}\) is constant.

3) V ∝ I.

4) The potential difference between the ends of a conductor is directly prbportional to the electric current passing through it at constant temperature.

5) This is the Ohm’s Law and the graph conforms it.
Resistance of the resistor

Question 23.
Identify the defects 1ft the circuit. Redraw it.

Answer:
Defects in the circuit:

1. Ammeter was connected in parallel in the circuit.
2. Volt meter was connected in series in the circuit.
3. Positions of resistor and battery were reversed.

Correct diagram :

Question 24.
What is the advantage of MCB over fuse?
Answer:

• These days instead of fuses, Miniature Circuit Breakers (MCB) are used for each lighting circuit.
• They switch off the circuit in a very short time duration in case of short-circuiting or some fault in the line.
• After repairing the fault in the circuit, the MCB is again switched on.
• Thus, the use of MCB is better than a fuse. It avoids the inconvenience of connecting a new fuse wire and it is much safer due to its quick response.

Question 25.
A circuit is shown in the picture.

The current passing through A is I.
a) What is the potential difference between A and B?
b) What is the equivalent resistance between A and B?
c) What amount of current is flowed through C and D?
Answer:
a) According to KirchhofPs loop law the algebraic sum of increase and decrease in potential difference across various components of the circuit in a closed circuit loop must be zero.
So the potential difference across CD is zero because it is a closed loop.

b) Here 20 Ω, 5 Ω are parallel to each other and resultants are in series to each other. Resultant resistance of 20 Ω and 5 Ω.

Question 26.

Observe the picture. The potential values at A, B, C are 70 V, 0 V, 10 V
a) What is the potential at D?
b) Find the ratio of the flow of current in AD, DB, DC.
Answer:

a) By following Ohm’s law potential difference is (V) = IR
In the given circuit we are applying junction laws.
‘D’ works as junction so, I = I₁ + I₂
Let p.d at D is V₀.

b)

Question 27.
Observe the circuit R₁ = R₂ = R₃ = 200 Ω. If reading of voltmeter is 100 V, resistance of voltmeter is 1000 Ω.
Find the Emf of the battery.

Answer:
Given values are R₁ = R₂ = R₃ = 200 Ω.
and Voltmeter reading = V = 100 V.
Resistance of Voltmeter = R_v = 1000 Ω.
In the given circuit R₁ and R₂ are in series R = R₁ + R₂ = 200 + 200 = 400 Ω
Resultant resistance (400 Ω) and voltmeter (1000 Ω) are always in parallel.

Question 28.
A circuit is made with a copper wire as shown in the diagram. We know that conductor’s resistance is directly proportional to its length. Calculate the equivalent resistance between points 1 and 2.

Answer:

Let the resistance of the wire be ‘R’ and length of the wire be ‘l’.
The shape of the circuit be square length of the side (l) = R
In a square, diagonal is \(\sqrt{2}\) times its length = \(\sqrt{2}\)l
Resistance towards diagonal is \(\sqrt{2}\)R
The circuit diagrams for the given arrangement are along PTR and QTS is ineffective as no current flows through it.
PQ and PS are in series so effective resistance is R₁ + R₂ = R + R = 2R.
QR and SR are in series so effective resistance is R₁ + R₂ = R + R = 2R.
Redrawn of the circuit again as resultant resistance between the points 1 and 2 is

Question 29.
From the adjacent figure,

i) Find the potential at D.
ii) Find the current that passes through AD, DB and DC.
Answer:
Suppose from A and C current is flowing through the circuit and at B current is flowing away from the circuit. Suppose potential at D is V.
∴ I₁ + I₂ = I₃ (where I₁ I₂ are current into the junction. I₃ is the current away from the junction)

Question 30.
Find the electric current drawn from the battery of emf 8V from the given circuit.
Answer:

According Kirchhoffs loop law
6I₁ + 3I₁ – 8 = 0
9I₁ = 8
I₁ \(\frac{8}{9}\) = 0.89 A
∴ Cuttent in 8 V is 0.89 A

Question 31.
A household uses the following electric appliances.
i) Refrigerator of rating 400 W each for ten hours each day.
ii) Two electric fans of rating 80 W each for 12 hours each day.
iii) Six electric tubes of rating 18 W each for 6 hours each day. Calculate the electric bill of the household in a month if the cost per unit electric energy is Rs. 3.00.
Answer:
i) Electrical energy consumed by Refrigerator in a month (in KWH)

ii) Electrical energy consumed by two electrical fans in a month (in KWH)

iii) Electrical energy consumed by electric tubes in a month (in KWH)

Total electrical energy consumed by all the electrical appliances
= 120 + 57.6 + 19.44 = 197.04 units
Cost of 1 unit = ₹ 3
Cost of 197.04 units = 197.04 × 3 = ₹ 591.12

Question 32.
What is the reason for connecting the fuse in the live wire?
Answer:

• The fuse is always connected in the live wire of the circuit.
• If the fuse is put in the neutral wire, due to excessive flow of current the fuse burns, current stops flowing in the circuit.
• But the appliance remains connected to the high potential point of the supply through the live wire.
• Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Question 33.
In the diagram given below, two resistors R₁ and R₂ of 3Ω and 6Ω respectively are connected in parallel across a battery of potential difference 12 V. Calculate the electrical energy consumed in 1 minute in each resistance.

Answer:
Given R₁ = 3 Ω, R₂ = 6 Ω, V= 12 V, t = 1 min 60 sec.
The resistors R₁ and R₂ are connected in parallel, so the voltage V across each resistor is equal to 12 V.

Question 34.
Calculate the electrical energy consumed in a month, in a house using 2 bulbs of 100 W each and 2 fans of 60 W each, if the bulbs and fans are used for an average of 10 hours each day.
If the cost per unit is ₹ 3, calculate the amount of electrical bill to be paid per month.
Answer:
Given power of each bulb = 100 W
Power of each fan = 60 W
Time t =10 hours each day
Power of 2 bulbs = 2 × 100 = 200 W
Power of 2 fans = 2 × 60 = 120 W
Total power = 200 + 120 = 320 W = \(\frac{320}{1000}\) = 0.32 KW
Time duration of consumption 10 hours per day per month
= 10 hours × 30 = 300 hours
∴ Total energy consumed = Total power x Time duration
= 0.32 KW × 300 h = 96 KWh
∴ Total cost = 96 x 3 = ₹ 288.

Question 35.
What resistance must be connected to a 15 Ω resistance to provide an effective resistance of 6 Ω?
Answer:
Given R₁ = 15 Ω, R₂ = ?, Effective resistance R_p = 6 Ω.
Since the effective resistance has decreased, R2 must be connected in parallel with R₁

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 4 Acids, Bases and Salts Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 4th Lesson Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Which property do you think of while suggesting the remedy from a problem of acidity?
Answer:
Neutralization Property. Antacid tablets neutralise acidity.

Improve your learning

Question 1.
Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11,7 and 9 respectively, which solution is : (AS1)
a) neutral
b) strongly alkaline
c) strongly acidic
d) weakly acidic
e) weakly alkaline
Arrange the pH in increasing order of hydrogen ion concentration.
Answer:
Solution – pH Value
A → 4
B → 1
C → 11
D → 7
E → 9
a) Solution ‘D’ is neutral
b) Solution ‘C’ is strongly alkaline
c) Solution ‘B’ is strongly acidic
d) Solution ‘A’ is weakly acidic
e) Solution ‘E1 is weakly alkaline
∴ Increasing order of Hydrogen ion concentration : C < E < D < A < B.

Question 2.
What is a neutralization reaction? Give two examples. (AS1)
Answer:
Neutralization reaction : When acid reacts with base, forms its salt and water. This reaction is called a neutralization reaction.
Examples :

Equation: HCl + NaOH → NaCl + H₂O
ii) Acetic Acid + Sodium Hydroxide → Sodium Acetate + Water
Equation : CH₃COOH + NaOH → CH₃COONa + H₂O
Formula : Acid + Base > Salt + Water

Question 3.
What happens when an acid or base is mixed with water? (AS1)
Answer:
When an acid or base is mixed with water it changes into dilute acid or dilute base.
(OR)
Dilute acid or dilute base will be formed when an acid or base is mixed with water. Mixing an acid or base with water results in decrease in the concentration of ions (H30+/ OH-) per unit volume. Such a process is called dilution and the acid or base is said to be diluted.

Question 4.
Why does tooth decay start when the pH of mouth is lower than 5.5? (AS1)
(OR)
Does the pH change tooth decay? Explain.
Answer:

1. Tooth enamel is the hardest substance in the body.
2.  It doesn’t dissolve in water but corroded when the pH in the mouth is below 5.5.
3. It happens due to the bacteria which produce acids by degradation of sugar and food particles remaining in the mouth.

Question 5.
Why does not distilled water conduct electricit? (AS2)
Answer:

1. Distilled water does not contain impurities.
2. It is also extremely weak electrolyte.
3. So it does not dissociate into ions.
4. It does not have charge carriers.
5. Because of that it does not conduct electricity.

Question 6.
Dry hydrogen chloride gas does not turn blue litmus to red whereas hydrochloric acid does. Why? (AS1)
Answer:
1. Dry hydrogen chloride gas is not an acid. Because it does not produce H⁺_(aq) ions. Hence it can’t turn blue litmus into red.
2. Hydrochloric acid is an aqueous solution. So it can produce H⁺_(aq) ions. Hence it can turn blue litmus into red.

Question 7.
Why pure acetic acid does not conduct electricity? (AS1)
Answer:
The reasons for pure acetic acid does not conduct electricity are :
i) Acetic acid is a weak acid.
ii) It gives fewer H₃O⁺ ions.

Question 8.
A milkman adds a very small amount of baking soda to fresh milk. (AS2)
a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
Answer:
1. By adding a very small amount of baking soda to fresh milk, the milkman keeps the milk unspoiled for little more time than usual time.
2. As the pH value increases the milk turns to slightly alkaline.

b) Why does this milk take a long time to set as curd?
Answer:

1. Curd form from the milk by the action of Lactic acid produced by bacteria in the milk.
2. If milk man add Baking soda (NaHCO₃) to the milk it neutralise acid, which is produced by the bacteria.
3. Excess acid is required to change the milk as curd.
4.  It takes long time.

Question 9.
Plaster of Paris should be stored in a moisture-proof container. Explain why? (AS2)
Answer:
Storing of Plaster of Paris :

1. Plaster of Paris is a white powder.
2. It easily absorbs water in air and forms hard gypsum.
3. So, it should be stored in a moisture-proof container.

Question 10.
Fresh milk has a pH of 6. Explain why the pH changes as it turns into curd.
Answer:
1. Fresh milk has a pH of 6. Hence it is a weak acid.
2. To turn the milk as curd, we have to add yeast in the form of some curd. The fermentation takes place during this process and lactose changes in lactic acid and the pH decreases as milk sets as curd.

Question 11.
Compounds such as alcohols and glucose contain hydrogen but are not categorized as acids. Describe an activity to prove it. (AS3)
(OR) (Activity – 7)
Write an activity to show that the solutions of compounds like alcohol and glucose do not show acidic character even though they are having Hydrogen.
(OR)
Write an activity which proves acids are good conductors of electricity.
(OR)
The acidity of acids is attributed to the H+ ions produced by them in solution explain the above statement with an activity.
List out the material for the experiment to investigate whether all compounds containing Hydrogen are acids or not and write the experimental procedure.
Answer:
List of the material required :

1. Glucose
2. Alcohol
3. Dil. HCl
4. Dil-H₂SO₄
5. Beaker
6. Connecting wires
7. 230 voltage AC supply
8. Bulb
9. Graphite rods.

Procedure :

1. Prepare glucose, alcohol, hydrochloric acid and sulphuric acid solutions.
2. Connect two different coloured electrical wires to graphite rods separately as shown in figure.
3. Connect free ends of the wire to 230 volts AC plug.
4. Complete the circuit as shown in the figure by connecting a bulb to one of the wires.
5. Now pour some dilute HCl in the beaker and switch on the current.

Observation :
The bulb starts glowing.

Repetition:
Repeat activity with dilute sulphuric acid, glucose and alcohol solutions separately.

Observation :

1. We will notice that the bulb glows only in acid solutions.
2. But the bulb does not glow in glucose and alcohol solutions.

Result:

1. Glowing of bulb indicates that there is flow of electric current through the solution.
2. Acid solutions have ions and the movement of these ions in solution helps for flow of electric current through the solution.

Conclusion :

1. The positive ion (cation) present in HCl solution is H⁺.
2. This suggests that acids produce hydrogen ions H⁺ in solution, which are, responsible for their acidic properties.
3. In glucose and alcohol solution the bulb did not glow indicating the absence of H⁺ ions in these solutions.
4. The acidity of acids is attributed to the H⁺ ions produced by them in solutions.

Question 12.
What is meant by “water of crystallization” of a substance? Describe an activity to show the water of crystallisation. (Activity – 16) (AS3)
Answer:
Water of Crystallization : Water of crystallization is the fixed number of water molecules present in one formula unit of a salt in its crystaline form.
Ex : CuSO₄ • 5H₂O.
It means that five water molecules are present in one formula unit of copper sulphate.

Activity to show the water of crystallization :

1. Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
2. We observe water droplets on the walls of the test tube and salt turns white.
3. Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
4. We observe, the blue colour of copper sulphate crystals is restored.

Reason :
1. In the above activity copper sulphate crystals which seem to be dry contain the water of crystallization, when these crystals are heated, water present in crystals is evaporated and the salt turns white.

2. When the crystals are moistened with water, the blue colour reappears.
Removing water of crystallization

Question 13.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid is added to test tube A, while acetic acid is added to test tube B. Amount and concentration of both the acids are same. In which test tube will the fizzing occur more vigorously and why? (AS4)
Answer:
1. The volatility of acetic acid (CH₃COOH) is more than that of hydrochloric acid.
2. But HCl solution has more strength than acetic acid.
3. Hence magnesium ribbon in test tube A will react more vigorously than in B.
4. So fizzing occurs more vigorously in test tube ‘A’.

Question 14.
Draw a neat diagram showing acid solution in water conducts electricity. (AS5)
(OR)
Draw a neat diagram which shows acids contains H⁺ ions.
(OR)
Draw a neat diagram showing how does dilute HCl solution conduct electricity.
Answer:

Question 15.
How do you prepare your own indicator using beetroot ? Explain. (AS5)
Aim : To prepare own indicator.
Materials required :
1) Beetroots-2 or 3
2) Knife
3) Bowls
4) Water
5) Spoon
6) Mixy
7) Orange juice

Procedure:
1) Take the beetroots and peel them with the help of a knife. (Firstly wash them).
2) Chop them into pieces.
3) Put those pieces into a mixy jar and make a paste.
4) Add some water to the paste. Now filter this and collect only juice from this.

Observation and Result:
1) Now add 5 to 6 drops of this juice, (beetroot juice (indicator)) to orange juice (5 to 6 drops) and mix it.
2) We can see the colour changes. This indicates the presence of acidic nature in orange juice.

Question 16.
How does the flow of acid rain into a river make the survival of aquatic life in a river difficult? (AS7)
(OR)
What are the harmful effects of acid rain?
Answer:
1) Acid rains are combination of carbonic acid, sulphuric acid and nitric acid with rain water.
2) The pH of acid rain is less than 5.6.
3) Living organisms can survive only in a narrow range of pH change.
4) When acid rain with pH value less than 5.6, flows into rivers, it lowers the pH of river water.
5) Due to less pH, the river water becomes acidic and hence the aquatic life in such rivers becomes difficult.

Question 17.
What is baking powder? How does it make the cake soft and spongy? (AS7)
Answer:
1) Baking Powder:
Baking powder is a mixture of baking soda (NaHCO₃) and a mild edible acid such as tartaric acid. COOH (CHOH)₂ COOH

2) Chemical reaction :
When baking powder is heated or mixed in water, the following reaction takes place.
NaHCO₃ + H⁺ → CO₂ + H₂O + Sodium salt of acid.

3) Carbondioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

Question 18.
Give two important uses of washing soda and baking soda. (AS7)
(OR)
Write the chemical formulae for washing soda and Baking soda and give their uses.
(OR)
Write any four uses of washing soda.
Answer:
Uses of washing soda (Na₂CO₃.10H₂O) :
1) Washing soda (sodium carbonate) is used in glass, soap and paper industries.
2) It is used in the manufacture of sodium compounds such as borax.
3) Sodium carbonate can be used as a cleaning agent for domestic purposes.
4) It is used for removing permanent hardness of water.

Uses of baking soda (NaHCO₃ 10H₂O) :
1) Baking soda (Sodium hydrogen carbonate) is used for faster cooking.
2) Baking powder (a mixture of baking soda and a mild acid) is used in preparation of cakes.
3) Sodium hydrogen carbonate is also an ingredient in antacids.
4) It is also used in soda – acid, fire extinguishers.
5) It acts as a mild antiseptic.

Fill in the Blanks

1. i) ………………….. taste is a characteristic property of all acids in aqueous solution.
ii) Acids react with some metals to produce ………………….. gas.
iii) Because aqueous acid solutions conduct electricity, they are identified as …………………..
iv) Acids react with bases to produce a ………………….. and water.
v) Acids turn methyle orange into ………………….. colour.
Answer:
1. i) Sour
ii) hydrogen
iii) electrolytes or conductors
iv) salt
v) red

2. i) Bases tend to taste ………………….. and feel ………………….. .
ii) Like acids, aqueous basic solutions conduct ………………….., and are identified as ………………….. .
iii) Bases react with ………………….. to produce a salt and
iv) Bases turn phenophthalein into ………………….. colour.
Answer:
2. i) bitter, soapy (slippery) to touch
ii) electricity, electrolytes
iii) acids, water
iv) pink

Match the following :

a) Plaster of Paris	1) CaOCl2
b) Gypsum	2) NaHCO3
c) Bleaching powder	3) Na2CO3
d) Baking soda	4) CaSO4.½H2O
e) Washing soda	5) CaSO4.2H2O

Answer:
3. a – 4,
b – 5,
c – 1,
d – 2,
e – 3.

Multiple Choice Questions

1. The colour of methyl orange indicator in acidic medium is
A) yellow
B) green
C) orange
D) red
Answer:
D) red

2. The colour of phenolphthalein indicator in basic solution is
A) yellow
B) green
C) pink
D) orange
Answer:
C) pink

3. Colour of methyl orange in alkali conditions
A) orange
B) yellow
C) red
D) blue
Answer:
B) yellow

4. A solution turns red litmus blue, its pH is likely to be
A) 1
B) 4
C) 5
D) 10
(OR)
If a solution converts red litmus into blue colour, then its pH value is …………….. .
A) 1
B) 4
C) 5
D) 10
Answer:
D) 10

5. A solution reacts with crushed egg-shells to give a gas that turns lime-water milky, the solution contains …………….. .
A) NaCl
B) HCl
C) LiCl
D) KCl
Answer:
B) HCl

6. If a base dissolves in water, by what name is it better known?
A) neutralization
B) basic
C) acid
D) alkali
Answer:
D) alkali

7. Which of the following substances when mixed together will produce table salt?
A) Sodium thiosulphate and sulphur dioxide
B) Hydrochloric acid and sodium hydroxide
C) Chlorine and oxygen
D) Nitric acid and sodium hydrogen carbonate
Answer:
B) Hydrochloric acid and sodium hydroxide

8. What colour would hydrochloric acid (pH = 1) turn universal indicator?
A) Orange
B) Purple
C) Yellow
D) Red
Answer:
D) Red

9. Which one of the following types of medicines is used for treating indigestion?
A) antibiotic
B) analgesic
C) antacid
D) antiseptic
Answer:
C) antacid

10. What gas is produced when magnesium is made to react with hydrochloric acid?
A) hydrogen
B) oxygen
C) carbon dioxide
D) no gas is produced
Answer:
A) hydrogen

11. Which of the following is the most accurate way of showing neutralization?
A) Acid + base → acid-base solution
B) Acid + base → salt + water
C) Acid + base → sodium chloride + hydrogen
D) Acid + base → neutral solution
Answer:
B) Acid + base → salt + water

10th Class Chemistry 1st Lesson Acids, Bases and Salts InText Questions and Answers

10th Class Chemistry Textbook Page No. 25

Question 1.
Is the substance present in antacid tablet acidic or basic?
A. The substance present in antacid tablet is basic.

Question 2.
What type of reaction takes place in stomach when an antacid tablet is consumed?
mrearx
A. Neutralization reaction takes place in stomach when an antacid tablet is consumed.

10th Class Chemistry Textbook Page No. 26

Question 3.
You are provided with three test tubes containing distilled water, an acid and a base solution respectively. If you are given only blue litmus paper, how do you identify the contents of each test tube?
Answer:
I know that acid turns blue litmus to red. With the help of this test I can find the acid. Distilled water and base don’t do so. Thus I identify each.

Question 4.
Which gas is usually liberated when an acid reacts with a metal? How will you test for the presence of this gas?
Answer:
Usually acids generate hydrogen gas on reacting with metals.

Test: When a burning splinder is brought near to the collected gas (H₂), it puts off with a pop sound.
This test proves that the gas is H₂.

Question 5.
A compound of a calcium reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle ; turns lime water milky. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
Equation is : CaCO₃ + 2 HCl → CaCl₂ + CO₂ + H₂O

10th Class Chemistry Textbook Page No. 30

Question 6.
Why do HCl, HNO₂ etc. show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
HCl, HNO₃, etc. show acidic characters in aqueous solutions as they liberate H⁺ ions. But alcohol and glucose don’t liberate H⁺ ions. So, they do not show acidic character.

Question 7.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
1) If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns.
2) The glass container may also break due to excessive local heating.

10th Class Chemistry Textbook Page No. 33

Question 8.
What will happen if the pH value of chemicals in our body increases?
Answer:
When pH value of chemicals in our body increases then the body will effect by some problems. They are
1) Digestion problems raise in the stomach.
2) pH changes as the cause of tooth decay.

Question 9.
Why do living organism have narrow pH range?
Answer:
Because increasing acidity is thought to have a range of possibly harmful consequences such as depressing metabolic rate and immune response in some organisms and causing coral bleaching

10th Class Chemistry 4th Lesson Acids, Bases and Salts Activities

Activity – 1

Question 1.
Observe the change in colour in each case and tabulate the results in the table.
Answer:
Procedure:
1) Collect the following samples from the science laboratory ;
i) Hydrochloric acid (HCl)
ii)Sulphuric acid (H₂SO₄)
iii) Nitric acid (HNO₃)
iv) Acetic acid (CH₃COOH)
v) Sodium hydroxide (NaOH)
vi) Calcium hydroxide[Ca(OH)₂]
vii) Magnesium hydroxide [Mg(OH)₂]
viii) Ammonium hydroxide(NH₄OH)
ix) Potassium hydroxide (KOH)
2) Prepare dilute solutions of the respective substances.
3) Take four watch glasses.
4) Put one drop of the first solution in each one of them and test the solution as follows.
i) Dip the blue litmus paper in the first watch glass.
ii) Dip the red litmus paper in the second watch glass.
iii) Add a drop of methyl orange to the third watch glass.
iv) Add a drop of phenolphthalein to the fourth watch glass.

Observation :
Observe the respective colour changes and note down in the chart below.

1. What do you conclude from the observations noted in the above table? (AS1)
Answer:
Conclusion : Acids turn blue litmus to red and bases turn red litmus to blue. Acids turn phenolphthalein to colourless and bases turn pink. Acids turn methyl orange to red and bases turn methyl orange to yellow.

2. Identify the above sample as acidic or basic solution. (AS4)
Answer:
Acids : HCl, H₂SO₄, HNO₃, CH₃COOH.
Bases : NaOH, KOH, Mg(OH)₂, NH₄OH, Ca(OH)₂.

Activity -2

Question 2.
What are Olfactory indicators? Write an activity to prove them.
(OR)
What is the name given to a substance which identifies an acid or base by virtue of smell? Write an activity to prove the fact with an example.
Answer:
Olfactory Indicators : There are some substances whose odour changes in acidic or basic media. These are called olfactory indicators.
Activity :
Aim : To check the olfactory indicator.
Required materials :
1) Onions
2) Knife
3) Plastic bag
4) Clean clothes.

Procedure :
1) Take some onions and finely chop them.
2) Put the chopped onions in a plastic bag along with some clean cloth.
3) Tie up the bag tightly and keep it overnight in the fridge.
4) Then remove onions from fridge and add some base. We observe it loses its smell. Observation : Check the odour of the cloth strips.

Result: It is used as the basic indicator.

LAB ACTIVITY Reaction of Acids with metals

Question 3.
Write an experiment showing the reaction of acids with metals. (AS3)
(OR)
Ramu added acid to active metal then what is the gas which has been liberated. What are the apparatus required to prove the experiment. Write the experimental acitivity.
(OR)
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’ pieces and liberates H₂“.
Answer:
Aim : To show the reaction of acids with metals.

Required Materials :

1. Test tube
2. Delivery tube
3. Glass trough
4. Candle,
5. Soap water
6. Dil. HCl
7. Zinc granules
8. One holed rubber stopper
9. Retard stand

Experimental procedure :

1. Take some zinc granules in a test tube and arrange the test tube to the retart stand.
2. Fix a delivery tube to the rubber stopper and immerse the second end of the delivery tube into the soap water.
3. Add about 10 m/ of dilute hydrochloric acid to Zn granules and fix rubber stopper to the test tube.
4. Evolved gas forms bubbles in soap water.
5. Bring a lightened candle near to the gas bubbles. We can observe the burning of gas bubble with pop sound.

Result: We can conform that the evolved gas is hydrogen.
Chemical reaction:
Acid + Metal → Salt + Hydrogen
2Hcl_(aq) + Zn_(s) → Zncl_2(aq) + H_2(g)

Additional Experiment :

• Repeat the above experiment with H₂SO₄ and HNO₃.
• We observe the same observation of the HCL experiment.

Conclusion : From the above activities we can conclude that when acid reacts with metal, H₂ gas is evolved.

Activity – 3 Reaction of Bases with metals

Question 4.
Write an activity to show the reaction of bases with metals.
(OR)
Write an activity which proves certain bases produce hydrogen gas when they react with metals.
Answer:
Aim : To show the reaction of bases with metals.

Required Materials :

1. Test tube,
2. Delivery tube
3. Glass trough
4. Candle
5. Soap water
6. Sodium hydroxide (NaOH) Solution
7. Zinc granules
8. One holed rubber stopper

Procedure :

1. Set the apparatus as shown in figure.
2. Take about 10 ml of dilute Sodium hydroxide (NaOH) solution in a test tube.
3. Add a few granules of zinc metal to it.
4. We will observe formation of gas bubbles on the surface of granules.
5. The gas will pass through delivery tube evolved from soap solution as bubbles.
6. Bring burning candle near the gas filled bubble.
7. The gas in the bubble puts off the candle with pop sound.

Result: The evolved gas is hydrogen.

Chemical reaction :
Base + Metal → Salt + Hydrogen

Note : It is better to use cone. NaOH solution for this reaction.

Activity – 4 Reaction of carbonates and metal hydrogen carbonates with Acids

Question 5.
Write an activity to show that all metal carbonates and hydrogen carbonates react with acids to give a corresponding salt. (AS3)
Answer:
Aim : To show that all metal carbonates and hydrogen carbonates react with acids to give a corresponding salt.

Required Materials :

1. Two test tubes
2. Sodium Carbonate (Na₂CO₃)
3. Sodium Hydrogen Carbonate (NaHCO₃)
4. Two holed rubber stopper
5. Thistle funnel
6. Stand
7. Dilute hydrochloric acid
8. Delivery tube
9. Calcium Carbonate (in a test tube)

Procedure :

1. Take a test tube A with 0.5 gm of sodium carbonate.
2. Close the test tube A with two holed rubber cork.
3. Insert a thistle funnel through one hole and insert a delivery tube through the other hole.
4. Pour 2 ml of dilute HC/ to the test tube A.
5. Do the same as above with test tube B with sodium hydrogen carbonate.

Observation :
Carbon dioxide is released from test tube A and B. Passing CO₂ gas through Ca(OH)₂ solution

Chemical Reaction :
Na₂CO_3(s) + 2 HCl_(aq) → 2 NaCl_(aq) + CO_2(g) + H₂O_(l)
Metal Carbonate + Acid → Salt + Carbon dioxide + Water

NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + CO_2(g) + H₂O_(l)
Metal Hydrogen Carbonate + Acid → Salt + Carbon dioxide + Water

Result : All metal carbonates and hydrogen carbonates react with acids to give a corresponding salt.

Activity – 5 Neutralization reaction

Question 6.
Write an activity to find the change of colour in the reaction of an acid with a base (Neutralization) reaction. (AS3)
(OR)
Explain neutralization reaction with an activity.
Answer:
Aim : To test the change of colour in the reaction of an acid with a base.

Required Materials :

1. 2 ml of dilute NaOH (Sodium Hydroxide) solution.
2. Phenolphthalein indicator solution.
3. dilute HCl (Hydrochloric) solution.

Procedure :

1. Take about 2 ml of dilute NaOH solution in a test tube.
2. Add two drops of phenolphthalein indicator solution.

Observation (i) :

1. It turns to red or pink colour.
2. It shows that NaOH is a base.

Experiment (1) : Add dilute HCl solution to the above solution drop by drop.
Observation (ii) : Pink colour disappears due to the reaction of NaOH (base) with HCl (acid).

Experiment (2) : Now add one or two drops of NaOH to the above mixture.
Observation (iii) : Pink colour reappears on adding NaOH.
NaOH + HCl → NaCl + H₂O
base + acid → salt + water
Result: This reaction is called a neutralization reaction.

Activity – 6 Reaction of metallic oxides with acids

Question 7.
Write an activity to show that metal oxide reacts with acid is a neutralization. (AS3)
(OR)
How can you prove metallic acids are basic in nature?
Answer:
1) Take a small amount of copper oxide (CuO) in a beaker.
2) Add dilute HCl slowly while stirring.
3) Copper oxide dissolves in dilute HCl and the solution becomes blueish green colour due to the formation of copper (II) chloride.

Equation : Metal oxide + Acid → Salt + Water
Result: This reaction is same as the reaction of base with acid, (neutralization)

Question 8.
Write an activity to show that non-metallic oxide reacts with base is a neutralization.
Answer:
1) Take a small amount of calcium hydroxide (base).
2) Add CO₂ into it.
3) Salt and water are produced.
Equation : Non-metallic oxide + Base → Salt + Water
Result: It is a neutralization reaction.

Question 9.
Repeat the activity – 7 using alkalis such as sodium hydroxide, calcium hydroxide solutions, etc. instead of acid solutions.
i) Does the bulb glow?
Answer:
Yes, the bulb will glow.

ii) What do you conclude from the results of this activity?
Answer:
Basic solutions are also good conductors of electricity due to released OH^– ions.

iii) What happens to an acid or a base in aqueous solution?
Answer:
Acids produce H⁺ ions and bases produce OH^– ions in aqueous solutions.

iv) Do acids produce ions only in aqueous solution?
Answer:
Yes.

Activity – 8

Question 10.
Do acids produce ions only in aqueous solution? Prove it. (AS3)
(OR)
Acids produce ions only in aqueous solution. Justify your answer with an activity.
Answer:
Procedure :

1. Take about 1.0 g of solid NaCl in a clean and dry test tube.
2. Add some concentrated sulphuric acid to the test tube. .

Observation :

1. A gas comes out of the delivery tube.
2. If we test the gas with dry and wet blue litmus paper, there is no change in colour.

Chemical equation : 2 NaCl_(s) + H₂SO_4(l) > 2 HCl + Na₂SO_4(s)

Conclusion :

1. We can conclude that dry HCl gas (hydrogen chloride) is not an acid.
2. Because we have noticed that there is no change in colour of dry litmus paper.
3. But HCl aqueous solution is an acid because wet blue litmus paper turned into red.

Activity – 9 Reaction of water with acids or bases

Question 11.
Write an activity to show that dissolving of an acid in water is an exothermic process (or) endothermic process. (AS3)
(OR)
What do you observe when water is mixed with acid or base?
Answer:
Experiment :

1. Take 10 ml water in a beaker.
2. Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
3. Touch the base of the beaker.
4. The base is hot.
5. Do this experiment with other concentrated acids like HCl, HNO₃ Result: This is an exothermic process called dilution.

Activity -10 Strength of acid or base

Question 12.
Write an activity to know whether the acid is strong or weak. (AS3)
Answer:

1. Take dilute HCl in a beaker.
2. Close it with a cardboard and introduce two different colour electrical wires through the holes made on it.
3. Connect a bulb and make the connection as shown in the figure.
4. Do the same replacing dilute HCl with dilute CH₃COOH (acetic acid).

Observation :
The bulb glows brightly in HCl solution, while the bulb’s intensity is low in acetic acid solution.

Result:
More ions are present in HCl solution which is a strong acid than in CH₃COOH solution which is a weak acid.

Activity – 11

Question 13.
Test the pH value of solution given in table. Record your observations. What is the nature of each substance on the basis of your observations?
Answer:

Activity – 12

Question 14.
Write an activity to check the colour change in dilute HCl and antacid solution in addition of methyl orange. (AS3)
Answer:
Procedure :

1. Take dilute HCl in a beaker.
2. Add two to three drops of methyl orange indicator to it.
3. The solution colour turns to red.
4. Now take the same solution and mix antacid tablet powder.

Observation :
Check the colour change.

Result:
The colour of the solution turns to light yellow.

Chemical equation:
2 HCl + Mg(OH)₂ → MgCl₂ + 2H₂O

Activity – 13

Question 15.
How can we test the pH value of the soil? (AS3)
Answer:

1. Take about 2g of soil in a test tube.
2. Add 5 ml water to it.
3. Shake it well.
4. Filter the content.
5. Collect the filtrate in a test tube.
6. Add 2 drops of universal solution to it.
7. Observe the colour.
8. Compare the colour with strip colour on the bottle and find the pH value.
9. In this way we can test the pH of the soil.

Activity – 14

Question 16.
Write the formulae of the following salts and classify them as families based on radicals.
Potassium Sulphate, Sodium Sulphate, Calcium Sulphate, Magnesium Sulphate, Copper Sulphate, Sodium Chloride, Sodium Nitrate, Sodium Carbonate and Ammonium Chloride. (AS4)
Answer:

Name of the Salt	Formula
1. Potassium Sulphate	K2SO4
2. Sodium Sulphate	Na2SO4
3. Calcium Sulphate	CaSO4
4. Magnesium Sulphate	MgSO4
5. Copper Sulphate	CuSO4
6. Sodium Chloride	NaCl
7. Sodium Nitrate	NaNO3
8. Sodium Carbonate	Na2CO3
9. Ammonium Chloride	NH4Cl

Sodium family : Na₂SO₄, NaCl, NaNO₃, Na₂CO₃, etc.
Family of chloride salts : NaCl, NH₄Cl, etc.
Family of sulphate salts : K₂SO₄, Na₂SO₄, CaSO₄, MgSO₄, CuSO₄, etc.
Family of carbonate salts : Na₂CO₃ MgCO₃ CaCO₃, etc.

Question 17.
Identify the acids and bases from which they are obtained. (AS4)
Answer:

Name of the Salt	Parent Acid	Parent Base
1. Potassium Sulphate	Sulphuric Acid	Potassium Hydroxide
2. Sodium Sulphate	Sulphuric Acid	Sodium Hydroxide
3. Calcium Sulphate	Sulphuric Acid	Calcium Carbonate
4. Magnesium Sulphate	Sulphuric Acid	Magnesium Hydroxide
5. Copper Sulphate	Sulphuric Acid	Copper Hydroxide
6. Sodium Chloride	Hydrochloric Acid	Sodium Hydroxide
7. Sodium Nitrate	Nitric Acid	Sodium Hydroxide
8. Sodium Carbonate	Carbonic Acid	Sodium Hydroxide
9. Ammonium Chloride	Hydrochloric Acid	Ammonium Hydroxide

Activity – 15 pH of Salts

Question 18.
Collect the salt samples like sodium chloride, aluminium chloride, copper sulphate, sodium acetate, ammonium chloride, sodium hydrogen carbonate and sodium carbonate. Dissolve them in distilled water. Check the action of these solutions with litmus papers. Find the pH using pH paper (universal indicator. Classify them into acidic, basic or neutral salts. Identify the acid and base used to form the above salts. Record your observations in table. (AS4)
Answer:

Salt	pH	Acid	Base	Neutral
Sodium Chloride	7	–	–	✓
Aluminium Chloride	7	–	–	✓
Copper Sulphate	< 7	✓	–	–
Sodium Acetate	> 7	–	✓	–
Ammonium Chloride	< 7	✓	–	–
Ammonium Chloride	> 7	–	✓	–
Sodium Carbonate	> 7	–	✓	–

 

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 7 Human Eye and Colourful World.

AP State Syllabus SSC 10th Class Physics Important Questions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World 1 Mark Important Questions and Answers

Question 1.
Define the power of lens. (AP June 2016)
Answer:
The reciprocal of focal length is called “power of lens”.

Question 2.
What physical quantity can be found in an experiment done with prism? (AP June 2017)
Answer:

1. Angle of deviation.
2. Refractive index of a prism.

Question 3.
What is the relation between Power and Focal length of the lens? (AP June 2018)
Answer:
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 4.
Draw the diagram of a lens which will be recommended by an eye doctor to a long sighted patient. (TS June 2015)
Answer:

The lens is convex lens.

Question 5.
What is the cause of Presbyopia? (TS March 2015)
Answer:
Presbyopia is vision defect when the ability of accommodation of the eye decreases with ageing.

Question 6.
Draw a ray diagram to show the angle of deviation when a ray of light passes through a glass prism. (TS March 2015)
Answer:

d = angle of deviation

Question 7.
Suggest reasons for the phenomenon associated with the following. The sky appearing blue. (TS March 2015)
Answer:
The reason for blue sky is due to the scattering of light by the molecules of N₂ and O₂ whose size is comparable to the wavelength of blue light.

Question 8.
+50 cm focal length bi-convex lens is recommended to correct the defect of vision of a mart. Find the power of the lens. (TS June 2016)
Answer:
f = +50 cm
Power (P) = \(\frac{100}{\mathrm{f}}\) D (in cm); P = \(\frac{100}{50}\) = 2 D
Power of the bi-convex lens is 2D.

Question 9.
What happens if the eye lens of a person cannot accommodate its focal length more than 2.4 cm? (TS March 2017)
Answer:
The person can able to see certain distance only he cannot see distance objects. For correction, he should use concave lens.

Question 10.
Write the reason for Sun appears red during the Sun-rise and Sun-set. (TS June 2018)
Answer:
Due to the high velocity (wave length) of red right, it reaches our eye without under go scattering. So, sun appears red during sunrise and sunset.

Question 11.
A person is unable to see distant objects. Show the defect of vision of the person with the help of ray diagram. (TS March 2018)
Answer:
1) His vision defect is myopia.
2) Ray diagram

Question 12.
Which molecules of atmosphere act as scattering centres are responsible for the blue sky? (TS June 2019)
Answer:
Oxygen and Nitrogen molecules.

Question 13.
Write any one application of a prism. (AP SA-I; 2019-20)
Answer:

1. Prism is used in scopes like binoculars, telescopes and light houses.
2. Prism is used to create artificial rainbow.

Question 14.
Mention the function of retina in a human eye.
Answer:
It acts as a screen, (which the image is formed) for image formed.

Question 15.
State the role of ciliary muscles in accommodation.
Answer:
It can adjust the focal length of the eye lens.

Question 16.
Why is normal eye not able to see clearly the objects kept closer than 25 cm?
Answer:
The maximum accommodation of a normal human eye is reached when the object is at a distance of 25 cm from the eyes. The focal length of the eye lens cannot be decreased this minimum limit.

Question 17.
What is “Power of accommodation of the eye” (or) “Least distance of clear vision”?
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

Question 18.
What is “Iris” and “Pupil”?
Iris :
The muscular diaphragm between the aqueous humour and the lens is called ‘iris’. Iris is the coloured part that we see in an eye.

Pupil:
The small hole in iris is called ‘pupil’.

Question 19.
What are three common defects of vision?
Answer:
The common defects of vision are i) Myopia ii) Hypermetropia iii) Presbyopia.

Question 20.
What is “Far point”?
Answer:
The point of maximum distance at which the eye lens can form an image on retina is called ‘far point’.

Question 21.
What is power of lens?
Answer:
The reciprocal of focal length is called power of lens. Power of lens focal length = \(\frac{1}{\text { focal length }}\)

Question 22.
What is the function of pupil in human eye?
Answer:
It allows the light falling on iris.

Question 23.
What is the purpose of human eye?
Answer:
The purpose of eye is to see and perceive the objects around us.

Question 24.
What is the principle of the working of human eye?
Answer:
It acts as camera having a lens system forming an invented real image on the light sensitive screen, retina.

Question 25.
What is the nature of the image formed on the retina?
Answer:
Real, inverted and same sized.

Question 26.
What is meant by dispersion of light?
The splitting of white light into its component colours when it passes through the prism is called dispersion of light.

Question 27.
On which factors does the colour of the scattered white light depend?
Answer:

1. Angle of scattering
2. Distance travelled by light
3. Size of the molecules.

Question 28.
Why is normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The focal length of eye lens cannot decrease below 25 cm.

Question 29.
A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?
Answer:
Myopia (or) short sightedness.

Question 30.
A person suffering from an eye-defect uses lenses of power – 1D. Name the defect he is suffering from and the nature of lens is used.
Answer:
Defect:
Myopia, Nature of lens, Concave/divergence.

Question 31.
Name the two phenomena involved in the formation of rainbow.
Answer:
Dispersion of light and internal reflection.

Question 32.
Identify the following part of the human eye.
1. Where is image of an object formed?
2. Which controls size of pupil?
Answer:

1. Image of an object is formed on retina.
2. Iris controls the size of pupil.

Question 33.
What is the relation between power of lens and focal length (f)?
Answer:
Power of lens (concave/convex)
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 34.
Explain about “Bending of light”.
Answer:

1. When a light travels from rarer to denser medium, it bends towards the normal.
2. And from denser to rarer medium, it moves away from the normal.

Question 35.
What is sclerotic?
Answer:
It is the outermost covering of the eye. It protects the vital internal parts of the eye.

Question 36.
What is “Retina”?
Answer:
Retina is the internal part and the light sensitive surface of the eye. It is equivalent to the photographic film in a camera.

Question 37.
What is “Atmospheric refraction”?
Answer:
When the light rays pass through the atmosphere having layers of different densities and refractive indices, refraction of light takes place. This refraction of light by the earth’s atmosphere is called atmospheric refraction.

Question 38.
What is a “Telescope”?
Answer:
The instrument which is used to see the distant objects such as a star, planet (moon, sun) or distant tree is called telescope.

Question 39.
Have you seen a rainbow in the sky after rain? How is it formed?
Answer:

1. A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on raindrops after the rain.
2. It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets present in the atmosphere.

Question 40.
“To look at the twinkling of stars is a wonderful experience.” How does it happen?
Answer:
The continuous changing atmosphere (due to varying atmospheric temperature and density) refracts the light from the stars by varying amounts and in different directions from one moment to the next.

Question 41.
Some things appear blue on a misty day. Give two examples.
Answer:

1. The long distance hills covered with thick growth of trees appear blue.
2. The smoke coming from a cigarette or an incense stick (agarbatti) appears blue on a misty day.

Question 42.
Which coloured suits do rescue workers wear?
Answer:
Rescue workers wear orange coloured suits during any rescue operations.

Question 43.
Which colour is best for school buses?
Answer:
Orange or yellow colour is best for school busses.

Question 44.
What is persistence of vision?
Answer:
The time for which the sensation of vision (of an object) continues in the eye is called persistence of vision. It is about 1/16^th part of a second.

Question 45.
Why can’t some people identify some colours?
Answer:
Rods identify the colours in the retina. If some rods are absent, the distinction of colours is not possible. In such cases, persons can’t identify some colours.

Question 46.
Write the reasons for colour blindness.
Answer:

1. Absence of colour responding rod cells in the retina.
2. Due to genetic disorder.

Question 47.
Why does it take some time to see objects in a dim room when we enter the room from bright sunlight outside?
Answer:
In bright light, the size of the pupil is small to control the amount of light entering the eye. When we enter a dim room, it takes some time so that the pupil expands and allows more light to enter and helps to see things clearly.

Question 48.
What is tyndall effect?
Answer:
The phenomenon of scattering of white light by colloidal particles is known as ‘Tyndall effect”.

Question 49.
Give two examples illustrating “Tyndall effect”.
Answer:

• A fine beam of sunlight entering a smoke filled room through a hole. Smoke particles scatter the white light and hence the path of light beam becomes visible.
• Sunlight passing through the trees in forest.
• Tiny water droplets through the trees in forest.

Question 50.
An eye camp was organised by the doctors in a village. What were the benefits to organise such camps in rural areas?
Answer:

1. To make people aware of eye diseases
2. To take proper and balanced diet.

Question 51.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
In the eye, the image distance (distance between eye lens and retina) is fixed and it cannot be changed. So when we increase the distance of an object, there is no change in the image distance.

Question 52.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is at a distance of 25 cm from the eye.
The focal length of the eye lens cannot be decreased below this. Thus an object placed closer than 25 cm cannot be seen clearly by a normal eye.

Question 53.
Write the relation between intensity of scattered light (I) and wavelength (λ).
Answer:
Light of short wavelength is scattered more than the light of long wavelength.
i.e., Intensity of scattered light (I) ∝ \(\frac{1}{\text { wavelength }(\lambda)}\)

Question 54.
Why would the sky look dark if the earth had no atmosphere?
Answer:
If the earth has no atmosphere, no particles present either. Thus no scattering of light. Then, the sky appears dark.

Question 55.
Why do different coloured rays deviate differently in the prism?
Answer:
Because the angle of refraction of different colours is different while passing through the glass prism.

Question 56.
What prevents rainbow from being seen as complete circles?
Answer:
The earth comes in the way of the rainbow and prevents it to form a complete circle.

Question 57.
When a monochromatic light passes through a prism will it show dispersion?
Answer:
No, it will not show any dispersion but show deviation.

Question 58.
Will a star appear to twinkle if seen from free space (say moon)?
Answer:
No, because there is no atmosphere in free space for refraction to take place.

Question 59.
A short-sighted person may read a book without spectacles. Comment.
Answer:
The statement is true, because a short-sighted person has difficulty in observing far off objects.

Question 60.
What is angle of vision?
Answer:
The maximum angle, at which we can see the whole object is called angle of vision.

Question 61.
What is cornea?
Answer:
The front curved portion of eye, which is covered by a transparent protective membrane is called the cornea.

Question 62.
Which lens do you use to correct the eye defect, Myopia?
Answer:
Bi-concave lens are used to correct the eye defect, Myopia.

Question 63.
What happens to power of lens if (i) focal length is increased, (ii) focal length is decreased?
Answer:

1. If focal length is increased, then power of lens decreases.
2. If focal length is decreased, then power of lens increases.

Question 64.
What type of image is formed by magnifying glass?
Answer:
It forms virtual, erect and magnified image.

Question 65.
How is power of lens related to its focal length?
Answer:
Power of lens is inversely proportional to its focal length.

Question 66.
What is “Scattering of light”?
Answer:
The process of re-emission of light in all directions with different intensity is called scattering of light. The re-emitted light is called scattered light.

Question 67.
Ramu is unable to see letters on the blackboard sitting at the last bench in the class¬room. What is the defect from which Ramu is suffering?
Answer:
Ramu is suffering from Myopia.

Question 68.
Vinay is able to read letters in a book beyond certain distance from least distance of distinct vision. What is the eye defect of Vinay?
Answer:
Vinay is suffering from Hypermetropia.

Question 69.
How can an eye lens accommodate its focal length?
Answer:
To see an object comfortably and distinctly, one must hold at a distance about 25 cm from his/her eyes. This distance is called least distance of distinct vision.

Question 70.
Write the uses of “rods” and “cones”.
Answer:
Retina contains about 125 million receptors called rods and cones. Rods identify the colour and cones identify the intensity of light.

Question 71.
Frame some questions, that you are going to ask your friend who is suffering from eye sight.
Answer:

1. Are you not able to see near objects or far objects?
2. Are you not able to see both near and far objects?

Question 72.
Write some more examples to find dispersion of light as VIBGYOR.
Answer:

1. Formation of rainbow.
2. When white light passes through water drop.

Question 73.
Why do stars twinkle?
Answer:
Due to change in atmospheric conditions, density changes so position keeps on changing.

Question 74.
“Sky appears dark to passengers flying at very high altitudes.” Why?
Answer:
At very high altitude, there is no atmosphere. So there is no scattering of light at such heights. So sky appears dark to passengers.

Question 75.
Why are danger signals red? (OR) Why are danger signals shown in red colour?
Answer:
Among the colours of visible light, red has more wavelength and least scattered. Thus, red colour can easily go through fog or mist or smoke without getting scattered. It can be seen from long distance. So red colour is used in universal danger signal.

Question 76.
How can you identify regarding the type of defect a person is suffering from by physically touching his spectacles?
Answer:
By touching the spectacles we can find out whether the lens is concave or convex lens and hence the defect from which he suffers.

Question 77.
A person cannot see objects beyond 1-2 m distinctly. What should be the nature of the defect and what type of lens should be used to correct the defect?
Answer:
The person is suffering from Myopia. It can be corrected by using concave lens.

Question 78.
How do you appreciate the working of “Retina”?
Answer:

1. It is the innermost delicate membrane having a large number of receptors called ‘rods’ and ‘cones’.
2. The rods identify the colour and the cones identify the intensity of light.
3. The retina is a part on which the image of an object is formed.

Question 79.
How do you appreciate the working of “Optic nerve”?
Answer:

1. Optic nerve consists of a large number of fibres.
2. These optic nerve fibres are connected to the rods and cones.
3. Optic nerve fibers transmit the light signals to the brain.

Question 80.
We see advertisements for eye donation on television or in newspaper. Write the importance of such advertisements.
Answer:
Eye donation advertisements are important as :

1. The people aw;are about donation of organs after their death.
2. Sympathetic nature towards others.

Question 81.
An eye donation camp is being organised by social workers in your locality. How and why would you help in this cause?
Answer:

1. We can intimate other people to participate in the camp.
2. As a human being, we should also register our eyes for donation after death.

Question 82.
A short sighted person cannot see clearly beyond 2m. Calculate the power of lens required to correct his vision.
Answer:

Question 83.
How does the power of a lens change if its focal length is doubled?
Answer:
The power get halved.

Question 84.
A person cannot see distinctly objects kept beyond 2m. Which among these lens is useful to correct the defect
a) – 0.2D
b) – 0.5 D
c) + 0.2D
d) +0.5D?
Answer:
The person is suffering from Myopia.
The lens is concave and its focal length f = – 2m.
\(P=\frac{1}{5}=\frac{1}{-2}=-0.5 D\)
So to correct the defect concave lens of – 0.5 D power should be used (b).

Question 85.
Express the power of concave lens of focal length 20 cm with its sign.
Answer:

10th Class Physics 7th Lesson Human Eye and Colourful World 2 Marks Important Questions and Answers

Question 1.
How do you appreciate the working of iris in the eye? (AP June 2018)
Answer:
Iris helps in controlling the amount of light entering the eye through pupil.

Question 2.
What is the reason for the blue colour of the sky? (AP March 2018)
How do you appreciate the role of molecules in the atmosphere in this regard?
Answer:
Blue colour of sky :

1. Atmosphere contains more O₂ and N₂ molacules and they are caused to blue colour of the sky.
2. The size of molecules of O₂ and N₂ are comparable with the blue colour and scatter blue colour only.

Role of molecules in the atomosphere in scattering :

1. Light of certain frequency falls on that atom or molecule.
2. This molecule responds to the light whenever the size of the molecule is comparable to the wavelength of light and vibrates.
3. Due to these vibration, molecule reemits a certain fraction of absorbed energy in all directions. The emitted light is called scattered light.
4. The atoms or molecules are called scattering centres.
5. I appreciate the role of the molecules in scattering.

Question 3.
In which conditions does a rainbow form? Why? (TS June 2015)
Answer:
1) Rainbow forms and appears when,
i) tiny water droplets present in the atmosphere (after rain shower),
ii) sunlight falls on the droplets,
iii) observer watches the rainbow in a specific direction.
2) Rainbow forms due to dispersion of sunlight by tiny droplets, present in the atmosphere, which acts as small prisms.

Question 4.
Draw the ray diagram, showing the correction of defect of vision hyper metropia by using a convex lens. (TS June 2016)
Answer:

Question 5.
Least distance of distinct vision of a person is observed as 35 cm. What lens is useful for him to see his surroundings clearly? Why? (TS March 2016)
Answer:
The least distance of distinct vision is 35 cm. This is more than the least distance of distinct vision of an ordinary person. Hence the person is suffering from Hypermetropia. He has to use double convex lens to see his surroundings clearly.

Question 6.
What happens, if Ciliary muscles do not perform contraction and expansion? Guess and write. (TS June 2018)
Answer:

1. If Ciliary muscles do not perform contraction and expansion, foal length of eye lens do not change.
2. Human eye can see the objects at specific distance only, eye cannot see the object either nearer or far distance.

Question 7.
Write any two situations to observe dispersion of light in your daily life. (TS June 2018)
Answer:
We can observe the dispersion of light in the following situations in our daily life.

1. In the formation of rainbow.
2. When observing sun light through the triangular transparent material like prism, scale edge.
3. At the time of curing of walls of new houses with water.
4. Dispersion of light by inclained plane mirror which is in water.

Question 8.
Write the material that you use to find out the value of refractive index of a prism. What is the necessity of the graph in this experiment? (AP March 2019)
Answer:
The material used to find out the value of refractive index of a prism:
Prism, Piece of white chart, Pencil, Pins, Scale and Protractor.

Necessity of the graph :
To find the angle of minimum deviation graph is required.

Question 9.
Draw a ray diagram showing the correction of myopia eye defect. (TS March 2019)
Answer:
Diagram of Myopia correction :
Note : Draw the diagram using Bi Concave Lens and show the far point (M). Image should form on Retina.

Question 10.
What happen if dispersion and scattering of light do not occur? (TS March 2019)
Answer:
If dispersion does not occur in nature, then there is no rainbow formation and splitting of light into seven different colors. If there is no scattering then the oceans and sky appears to be black. The sun appears white all the time (Including Sunrise and Sunset).

Question 11.
When Mohan viewed white light through a transparent scale, he observed some colours. Predict and write the phenomenon involved in his observation. (AP SCERT: 2019-20)
Answer:

1. The phenomenon involved in his observation is dispersion of light.
2. Splitting of white light into different colours (VIBGYOR) is called dispersion.

Question 12.
A boy who is suffering from eye defect has been given a prescription as -2D. Based on the information given, answer the following questions.
a) Identify the eye defect he is suffering.
b) Write the nature and focal length of the lens. (AP SCERT: 2019-20)
Answer:
a) The boy is suffering from myopia.
b) Nature of the lens :
The lens is biconcave lens.
It is thin at the middle and thicker at the edges.
Focal length of the lens :
Given that power is – 2D

Question 13.
How does eye lens change its focal length? (AP SA-I:2019-20)
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 14.
Kishore wore spectacles. When you saw through his spects the size of his eyes seemed bigger than their original size.
a) Which lens did he use?
b) Explain that defect of vision.
Answer:
a) When we saw through Kishore’s spects the size of his eyes seemed bigger than their original size. This is possible with convex lens only because magnification of the lens is greater than T.

b) The defect he suffers is hypermetropia. This is also called farsightedness.

A person who suffers with this type of defect, he can’t see the objects clearly which are placed near distance because the image is formed beyond the retina. So by using convex lens the rays can be converged on retina.

Question 15.
“God has given the gift for us to see the sunrise and sunset.” Explain the feeling of it.
Answer:

• The sun is visible two minutes before the actual sunrise and remains visible two minutes after the actual sunset.
• The actual sunrise takes place when the sun is just above the horizon.
• The actual sunset takes place when the sun is just below the horizon.

Question 16.
“Smoke coming out of coal fired chimney appears blue on a misty day.” Why?
Answer:

• On a misty day, the air has large amount of tiny particles of water droplets, dust and smoke.
• These tiny particles present in the air scatter blue colour of the white light passing through it.
• When this scattered blue light reaches our eyes the smoke appears blue.

Question 17.
“Motorists use orange light on a foggy day rather than normal white light.” Why?
Answer:

• On a foggy day, the air has large amount of water droplets.
• If a motorist uses white light, the water droplets present in the air scatter large amount of the blue light.
• This on reaching our eyes decreases visibility and hence driving becomes extremely difficult.
• Whereas orange light has longer wavelength and hence it is least scattered.

Question 18.
A rainbow viewed from an airplane may form a complete circle. Where will the shadow of the airplane appear? Explain.
Answer:

• A rainbow viewed from an airplane form a complete circle because the earth does not come along the way of the airplane and rainbow.
• A rainbow is a three dimensional cone of dispersed light it appear as a complete circle.
• The shadow of the airplane appears within the circle of the rainbow.

Question 19.
How do we see colours?
Answer:

• The retina of human eye has a large number of receptors.
• These receptors are of two types i.e., rods and cones.
• The rod cells recognise the colour of light rays, while the cones identify the intensity of light.
• It is these cone cells, which make it possible for a men to see different colours and distinguish between them.

Question 20.
Why do we use lenses in spectacles to correct defects of vision?
Answer:
The process of adjusting focal length is called “accommodation”. This process has to be done by eye itself. Sometimes the eye may gradually lose its power of accommodation. In such condition, the person cannot see the object clearly and comfortably. In this situation, we have to use lenses in spectacles to correct defects of vision.

Question 21.
What is a) far point of the eye and b) near point of the eye?
Answer:
a) The farthest point up to which the eye can see objects clearly without strain (in the eye) is called the far point of the eye. For a normal eye, the far point is at infinity.

b) The minimum distance at which objects can be seen most clearly without strain (in the eye) is called the least distance of distinct vision or the near point of the eye.

Question22.
Write the difference between “Myopia” and “Hypermetropia”.
(OR)
Distinguish between Myopia and Hypermetopia.
Answer:
The eye defect in which people cannot see at long distances but can see nearby objects clearly is called Myopia. The eye defect in which people cannot see near distant objects but can see distant objects is called hypermetropia.

Question 23.
Define the following words.
a) Prism
b) Dispersion of light
c) Scattering of light
Answer:
a) Prism :
A prism is a transparent medium separated from the surrounding medium by at least two plane surfaces which are inclined at a certain angle.

b) Dispersion of light :
The splitting of white light into different colours is called dispersion of light.

c) Scattering of light:
The process of reemission of absorbed light in all directions with different intensities by atoms or molecules is called scattering of light.

Question 24.
Define the words associated with prism with the help of figure.

1) Angle of incidence :
The angle between incident ray and normal is called angle of incidence.

2) Angle of emergence :
The angle between normal and emergent ray is called angle of emergence.

3) Normal:
Perpendicular drawn to the surface of prism.

4) Angle of deviation:
The angle between extended incident ray and emergent ray is called angle of deviation.

Question 25.
State the cause of dispersion, when white light enters a glass prism. Explain with a diagram.
Answer:

• Light is made up of different colours. Each colour travels at its own speed inside a prism.
• Due to this different colours of light bends through different angles with respect to the incident ray, as it passes through a prism.
• The red light bends the least while the violet most.
• Thus, the rays of each colour emerge along different paths and become distinct.
• It is the bond of distinct colours that we see in a spectrum.

Question 26.
What happens to the lens and the ciliary muscles when you are looking at distant objects and near objects?
Answer:
a) The ciliary muscles become relaxed and the lens becomes thin, i.e. its radius of curvature increases. So focal length of eye lens increases for distant object.

b) The ciliary muscles contract and the lens becomes thick, i.e. its radius of curvature decreases. So focal length of eye lens decreases for near objects.

Question 27.
Why does it take some time to see objects in cinema hall when we just enter the hall from bright sunlight? Explain in brief.
Answer:

• The pupil regulates and controls the amount of light entering eye.
• In bright sunlight, the size of pupil is small and when we enter the cinema hall it takes some time for the pupil to expand in size due to dim light.

Question 28.
What are the factors which influence the total angle of deviation?
Answer:

• The angle of incidence at the first surface (i).
• The angle of prism (A).
• Refractive index of the material.

Question 29.
How does eye change its focal length take place in the eyeball?
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 30.
Stars twinkle while planets do not. Why?
Answer:
1) Continuously changing atmosphere refracts light from the stars by different amounts from one moment to the other, when atmosphere refracts more starlight towards us and the stars appear to be bright and when the atmosphere refracts less star-light then the stars appear to be dim.

2) However the planets are nearer to us than the stars, they appear to be comparatively bigger to us so they cannot be considered as a point source, hence no twinkling is seen.

Question 31.
How do earth and stars appear for a person who is on the moon?
Answer:

• For the person who is on the moon, the earth appears blue due to blue colour of sunlight scattered by the earth’s atmosphere reaching him.
• Stars and other heavenly bodies are seen as usual, but without twinkling.

Question 32.
Why does the sky appear dark and black to an astronaut instead of blue?
(OR)
Why does the sky appear dark to the passenger flying at high altitudes?
Answer:

• This is because there is no atmosphere containing air in the outer space to scatter light.
• Since there is no scattered light which can reach our eyes in outer space, the sky looks dark and black there.
• This is why the astronauts who go to outer space find the sky to be dark and black instead of blue.

Question 33.
A person is able to see objects clearly only when these are lying at distance between 60 cm and 250 cm from his eye. What kind of defect of vision is he suffering from?
Answer:
For a normal eye, the near point is at 25 cm and the far point is at infinity. The given person cannot see object clearly either close to the eye or far away from the eye. So he is suffering from presbyopia.

Question 34.
Write the material required in finding the refractive index of a prism.
Answer:
Materials required:
Prism, piece of white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Question 35.
Draw the graph between angle of incidence and angle of deviation.
Answer:

Question 35.
Draw the diagram of scattering of sunlight.
Answer:

Question 35.
How do you appreciate the “Eye Donor”?
Answer:
The human eye is one of the most important sense organs. Without eye we are unable to see the beautiful world. So we have to appreciate “eye donor” for his kindness to give sight to blind people.

Question 36.
How can we appreciate the working of “Iris”?
Answer:

• Iris consists of muscles which help in controlling the amount of light entering the eye through pupil.
• In case of low light the iris makes the pupil to expand and allow more light to enter the eye.
• In case of bright or excess light the iris makes the pupil to contract in order to decrease the amount of light entering the eye. The iris consists of muscles that expand and contract the pupil.

Question 37.
The power of lens is 2.0 D. Find its focal length and state what kind of lens it is.
Answer:
P = \(\frac{1}{f}\) (f in metres)
Given P = 2D
∴ f= \(\frac{1}{P}\) = \(\frac{1}{2}\) = 0. 5 m = 50 cm.
The focal length positive indicates it is a convex lens.

Question 38.
Two convex lenses of powers 1D and 2D are combined together to form a new lens. Then what is the resultant power and focal length of lens?
Answer:
P₁ = 1 D ; P₂ = 2D
Resultant power P = P₁ + P₂ = 1 + 2 = 3D
P = \(\frac{1}{f}\) (f in metres)
∴ f= \(\frac{1}{P}\) = \(\frac{1}{3}\) = 0.3333 m = 33.33 cm.

Question 40.
Figure shows the refraction of light through an equilateral prism. Incident at an angle of 30°. The ray suffers a deviation of 37°. What are the angles marked at A, e and f respectively?
Answer:

Since the prism is an equilateral prism,
A = 60°, also D = 37° and i = 30° (i.e., i₁), e = i₂ = ?
We know that i₁ + i₂ = A + D
30° + e = 60 + 37
e = 97 – 30 = 67°
Also A + f = 180°
f = 180°-60° = 120°

10th Class Physics 7th Lesson Human Eye and Colourful World 4 Marks Important Questions and Answers

Question 1.
Kavya can see distant objects clearly but cannot see objects at near distance. With what eye defect is she suffering? Draw the diagrams showing the defected eye and its correction. (AP June 2016)
Answer:
Kavya is suffering from hypermetropia.
The following diagram shows the defective eye and its correction.

Question 2.
Revathi is a front bench student. She is unable to draw the picture drawn on the blackboard. She got permission from the teacher and sat in the back row. What could be the defect that Revathi is suffering from? Draw the diagram, which shows the correction of the above defect? (AP Mareh 2017)
Answer:
Hypermetropia

Question 3.
An eye specialist suggested a + 2D lens to the person with defect in vision. Which kind of defect in vision does he have? Draw the diagrams to show the defect of vision and its correction with a suitable lens. (TS June 2017)
Answer:
Eye specialist suggested +2D lens, that is convex lens is used to correct Hypermetropia. So the person has Hypermetropia.
Deffect of Vision:

Correction :

Question 4.
How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia? Explain it mathematically. (TS March 2017)
Answer:
The person who has hypermetropia cannot see near objects. He can see the objects those are beyond near point (H). For correction of this eye defect the image of the object placed at “least distance of distinct vision (L)” should be at near point (H).
u = -25 cm; v = -d cm

here d > 25, so ‘f gets positive value.
Hence, convex lens should be used.

Question 5.
Mention the required material and chemicals for the experiment of “scattering of light.” Write the experiment procedure. (TS March 2018)
(OR)
Write the required apparatus and chemicals to show the scttering of light experimentally and write the experimental process.
(OR)
How can you demonstrate scatteing of light by an experiment?
Answer:
Aim :
To show the scattering of light.

Material required :
Beaker, sodium thiosulphate, sulphuric acid.

Procedure:

1. Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sun light is available.
3. Watch the formation of grains of sulphur and observe the changes in the beaker.
4. We will notice that sulphur participates as the reaction is in progress. At the beginning, the grains of sulphur are smaller in size and as the reaction progress, their size increases due to precipitation.
5. Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases.
6. The reason for this is scattering of light.
7. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light. Hence they appear blue in colour.
8. As the size of grains increases, their size becomes comparable to wavelength of other colours.
9. As a result, they acts as scattering centres of all colours.

Question 6.
A boy has been playing games in mobile phone and is suffering from eye defect. The doctor prescribed him to use spectacles of power – 5D. What eye defect is he suffering from? (AP SA-I:2018-19)
Draw a neat diagram which shows the correction of above eye defect.
Answer:
Doctor suggested the boy – 5D lens, that is concave lens. Concave lens is used to correct myopia. So the boy has myopia.
Defect of Vision :

Correction:

Question 7.

Answer the following questions from the above information. (TS June 2019)
i) What is the defect of vision in ‘D’ suffering from? Why does it happen?
ii) Whose defect of vision can be corrected by using Biconcave lens?
iii) Who is suffering with similar defect of vision as of ‘B’?
iv) Who among the above do not have any defect of vision?
Answer:
i) 1) The vision defect of person ‘D’ is presbyopia.
2) Presbyopia happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens. This effect can be seen in aged people.
ii) Person A’ defect of vision can be corrected by using Biconcave lens.
iii) Person ’C’ is suffering with similar defect of vision as of ’B’.
iv) Person ‘E’ has no defect of vision.

Question 8.
A prism causes dispersion of white light while a rectangular glass block does not. Explain.
Answer:

• In a prism the refraction of light occurs at two plane surfaces.
• The dispersion of white light occurs at the first surface of prism where its constituent colours are deviated through different angles.
• At the second surface, these split colours suffer only refraction and they get further separated.
• But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces.
• At the first surface, although the white light splits into its constituent colours on refractions, but they split colours on suffering refraction at the second surface emerge out in the form of a parallel beam, which gives an impression of white light.

Question 9.
A convex lens of power 4D is placed at a distance of 40 cm from a wall. At what distance from the lens should a candle be placed so that its image is formed on the wall?
Answer:

So candle should be placed 66.66 cm from the lens.

Question 10.
Explain briefly the reason for the blue of the sky.
Answer:

• Our atmosphere contains different types of molecules and atoms.
• The reason for blue sky is the molecules N₂ and O₂.
• The sizes of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.

Question 11.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:

• For a normal eye, image distance in the eye is fixed.
• This is equal to the distance of retina from the eye lens.
• When we increase the distance of the object from the eye, focal length of eye lens is changed on account of power of accommodation of the eye so as to keep the image distance constant.

Question 12.
A person is able to see objects clearly only when they are lying at distance between 60 cm and 250 cm from his eye. What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain.
Answer:
A bi-focal lens consists of concave and convex lens of suitable focal lengths will be required to correct the defect and to increase his range of vision from 25 cm to infinity. In bi-focal lens, the upper half of the lens is concave lens which corrects distant vision and the lower half is convex which corrects near vision.

Question 13.
Discuss why sun is visible before actual sunrise and after actual sunset?
Answer:

• The sun is visible to us about 2 minutes before the actual sunrise and 2 minutes after the actual sunset because of atmospheric refraction.
• By actual sunrise we mean the actual crossing of the horizon by the sun.

• Figure shows the actual and apparent positions of the sun with respect to the horizon.
• The time difference between actual sunset and the apparent sunset is about 2 minutes.
• The apparent flattering of the sun’s disc at sunrise and sunset is also due to the same phenomenon.

Question 14.
When does the colour of sky appear black for an observer?
Answer:

• In the absence of atmosphere, there will not be any scattering of light and so light will reach our eye, i.e. the sky will appear black instead of blue at night in the absence of light.
• On the moon, since there is no atmosphere, there is no scattering of sunlight reaching the moon surface. Hence to an observer on the surface of moon, no light reaches except the light directly from sun. Thus sky will have no colour and will appear black to an observer on the moon surface. This is applicable for any planet which does not have atmosphere.
• When an astronaut goes above the atmosphere of the earth in rocket he sees the sky black.

Question 15.
Why does the colour of clouds appear black?
Answer:

• The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light.
• Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light does not reach our eye, the clouds seem black.

Question 16.
Give daily life examples of scattering of light by earth’s atmosphere.
Answer:
Some daily life effects of scattering of sunlight by earth’s atmosphere are

1. Red colour of sun at sunrise and sunset.
2. White colour of sky at noon.
3. Blue colour of sky is due to molecules N₂ and O₂.
4. Black colour of sky is due to the absence of atmosphere.
5. Use of red light for the danger signal because it is least scattered by particles due to its greater wavelength.
6. White colour of clouds is due to rise in temperature.

Question 17.
What is the relation between scattering and wavelength of light? Explain.
Answer:

• Scattering is the process of absorption and then re-emission of light energy.
• The air molecules of size smaller than the wavelength of incident light absorb the energy of incident light and then re-emit it without change in its wavelength.
• The intensity of scattered light is found to be inversely proportional to fourth power of wavelength of light.
• The wavelength of violet is least and red light is most, therefore from the incident white light, violet light is scattered the most and red light is scattered the least.

Question 18.
How can we get this (in human eye) same image distance for various positions of objects?
Answer:

• The ciliary muscle attached with eye lens helps to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• F_(max) – 2.5 cm,u = oo, v = 2.5 cm. The parallel rays coming from a distant object are focussed on the retina with 2.5 cm image distance.
• When the eye is focussed on a nearer object the muscles are strained. So the focal length of the eye lens decreases its minimum value.
• F_(min)= 2.27 cm, u = 25 cm, v = 2.5 cm. The rays from an object (u = 25 cm) at L (point of least distance of distinct vision) are focussed on the retina with 2.5 cm image distance.

Question 19.
Does eye lens form a real image or virtual image? Explain it.
Answer:

• Eye lens forms a real image.
• The light that enters the eye forms an image on the retina.

• The image is obtained on a screen (retina) and it is an inverted image.
• So, we can say eye lens forms a real image.

Question 20.
How does the image formed on retina help us to perceive the object without change its shape, size and colour?
Answer:

• The eye-lens forms a real and inverted image of an object on the retina.
• The retina is a delicate membrane which contains about 125 million receptors called ‘rods’ and ‘cones’ which receive the light signals.
• Rods identify the colour.
• Cones identify the intensity of light.
• These signals are transmitted to the brain through about 1 million optic nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 21.
How do you prove that a prism does not produce colours itself?
Answer:

• A white light from a slit ‘S’ is made to pass through prism P which forms spectrum on a white screen AB.
• A narrow slit H is made on the screen AB, parallel to slit S to allow the light of particular colour to pass through it.
• The light of a particular colour is made to fall on the second prism Q placed with its base in opposite direction to that of the prism P.
• The light after passing through the second prism Q is received on another white screen M.
• It is observed that the colour of light obtained on the screen M is same as that of the light incident on the second prism Q through the slit H.

Question 22.
Draw the structure of human eye and explain its parts.
Answer:

Question 23.
What is the part indicated by an arrow mark? What is its working function?
Answer:

• The part indicated by the arrow mark is ciliary muscle.
• Ciliary muscles help to change the focal length of the eye lens.
• When it relaxes the focal length of the eye lens is maximum.
• When it strains, the focal length of the eye lens is minimum.
• In this way the ciliary muscles to which eye lens is attached help to give us clear vision.

Question 24.
Two observers standing apart from each other do not see the “same” rainbow. Explain.
Answer:

• All the rain drops that disburse the light to form rainbow lie within a cone of semi vertical angle 40° to 42°.
• If two observers are standing at a distance apart, they will observe rainbow at different parts on the surface of the cone.
• So the portion of the rainbow observed by an observer depends on the position of the observer.
• Two different observers will form two different cones with the observer standing at the vertex of the cone, therefore rainbow seen by them will be different.

Question 25.
A prism with an angle A = 60° produces an angle of minimum deviation of 30°. Find the refractive index of material of the prism.
Answer:
Given, A = 60° and D = 30°

Question 26.
A person cannot see the objects distinctly, when placed at a distance less than 50 cm. Calculate the power and nature of the lens he should be using to see clearly the object placed at a distance of 25 cm from his eyes.
Answer:

Question 27.
A person cannot see the objects distinctly, when placed beyond 2 m.
Calculate the power and nature of the lens he should be using to see the distant objects clearly.
Answer:
For myopia the focal length = – far point distance = – 2m
Power = \(\frac{1}{\mathrm{f}}=\frac{1}{-2}\) = 0.5 D.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 6 Refraction of Light at Curved Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Suppose you are inside the water in a swimming pool. Your friend is standing on the edge. Do you find your friend taller or shorter than his actual height? Why? (AP June 2018)
Answer:
Friend is seemed to be taller. Because of refraction of light.

Question 2.
What happens to the image, if a convex lens is made up of two different transparent materials as shown in figure? (TS March 2016)

Answer:
The convex lens is made up of two different materials. So the refractive i these two materials will be different. Hence two images will be formed.

Question 3.
Write the list of materials required for the experiment to find the focal length of a convex lens. (TS June 2017)
Answer:
Convex Lens, Scale, Piece of paper, Sunrays.
(OR)
Convex Lens, V-Stand, Candle, Match box, Screen, Scale.

Question 4.
Complete the following ray diagram. (TS March 2019)

Answer:

The parallel rays coming with some angle to principal axis meet on focal plane.

Question 5.
If the object is placed between the focal point and the optical centre of a convex lens, what will be the characteristics of the image formed? (AP SA-1:2019-20)
Answer:
Object is placed between F₂ and optic centre P :

Nature :
Virtual, erect and magnified.

Position :
Same side of the lens where object is placed.

Question 6.
For a concave lens, what type of image will be formed if the object is placed at the centre of curvature? (AP SA-1:2019-20)
Answer:
Same size of object, inverted and real image will be formed.

Question 7.
Write lens formula. (AP SA-I:2019-20)
Answer:
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Question 8.
What is a lens? (or) Define lens.
Answer:
A lens is formed when a transparent material is bounded by two spherical surfaces.

Question 9.
What is a double convex lens?
The lens having two spherical surfaces bulging outwards is called double convex lens.

Question 10.
What about the thickness of double convex lens?
Answer:
It is thick at the middle as compared to edges.

Question 11.
What is a double concave lens?
Answer:
The lens having two spherical surfaces curved inward is called a double concave lens.

Question 12.
Write about the thickness of concave lens.
Answer:
It is thin at the middle and thicker at the edges.

Question 13.
What is centre of curvature?
Answer:
The centre of sphere which contains the part of curved surface is called centre of curvature.

Question 14.
What is radius of curvature?
Answer:
The distance between the centre of curvature and curved surface is called radius of curvature.

Question 15.
What is the mid point of lens called?
Answer:
The mid point of lens is called pole (or) optical centre.

Question 16.
What is a focus?
Answer:
The point where rays converge or the point from which rays seem emanate is called focal point (or) focus.

Question 17.
What is the distance between pole and focal point called?
Answer:
Focal length.

Question 18.
What happens if the ray passes through principal axis?
Answer:
It will be undeviated.

Question 19.
Where do light rays travelling to principal axis converge?
Answer:
They converge at focus.

Question 20.
What happens to light rays passing through focus?
Answer:
The path of the rays is parallel to principal axis after refraction.

Question 21.
What is a focal plane?
Answer:
A plane which is perpendicular to principal axis at the focus is called focal plane.

Question 22.
What is lens formula?
Answer:
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Question 23.
On what factor does focal length of a lens depend?
Answer:
It depends on refractive index of the medium, object distance and image distance.

Question 24.
What is lens maker formula?
Answer:
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)

Question 25.
What happens to be image formed by a convex lens if its lower part is blackened?
Answer:
Every part of a lens forms a complete image. If the lower part of the lens is blackened the complete image will be formed but its intensity will be decreased.

Question 26.
From which point of lens are all the distances are measured?
Answer:
The optical centre of lens.

Question 27.
Is it possible for a lens to Act as a convergent lens in are medium and a divergent lens in another?
Answer:
Yes. A convergent lens is placed in a higher refractive index of medium the nature of the lens changes i.e., it acts as divergent lens.

Question 28.
What are paraxial rays?
Answer:
The rays which move very close to the principal axis which can be treated as parallel are called paraxial rays.

Question 29.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in air to speed of light in any medium.

Question 30.
Give mathematic expression for power lens and explain the terms in the formula.
Answer:
Power (P) = \(\frac{1}{f}\)
where f is focal length of lens.

Question 31.
If the size of image is same as object through a convex lens, then where is the object placed?
Answer:
The object is placed at centre of curvature.

Question 32.
How will you identify a concave lens by touching it?
Answer:
A concave lens is thinner at centre and thicker at edges.

Question 33.
How will you identify a convex lens by touching it?
Answer:
A convex lens is thicker at centre and thinner at edges.

Question 34.
Give the sign conventions for lenses with regard to the object and image distance.
Answer:
The distance measured in the direction of incident ray is taken as positive.
The distance measured against the direction of incident ray is taken as negative.

Question 35.
Give the sign conventions for lenses with regard to the height of objects and images.
Answer:
All the heights of objects and images above principal axis are positive and below the axis are negative.

Question 36.
When light of two colours A and B passes through a plane boundary, A is bent more than B. Which colour travels more slowly in the second medium?
Answer:
Colour A travels slowly.

Question 37.
What type of lens behaviour will an air bubble inside water show?
Answer:
It will act as a concave lens.

Question 38.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:
Yes. A lens is placed in a medium of a high refractive index than that of the lens then nature of lens changes (M_L > M_g).

Question 39.
The image formed by a lens is always erect and diminished. What is the nature of lens?
Answer:
Given that the lens is forming an image which is always erect and diminished. So it is virtual also. Such type of image is formed by concave lens.

Question 40.
If a student observed an image of same size with a convex lens of focal length 20 cm, then where should he keep the object in front of lens?
Answer:
Because the student got image of same size the object should be placed at a distance of twice the focal length, i.e. 40 cm.

Question 41.
For an object placed at a distance of 20 cm in front of convex lens, the image formed is at a distance of 20 cm behind the lens. Find the focal length of lens.
Answer:
The object distance and image distance are same. So the object is kept at twice the focal length. So the focal length of the convex lens is 10 cm.

Question 42.
A doctor suggested spectacles for a student which has negative focal length. Which type of lens is that?
Answer:
Focal length negative indicates that it is a concave lens.

Question 43.
What happens to a light ray which passes through optical centre?
Answer:
The light ray which passes through optical centre does not deviate.

Question 44.
When do you get image at infinity with a convex lens?
Answer:
When the object is at the focal point.

Question 45.
When do you get a virtual image with a convex lens?
Answer:
When the object is placed between focus and pole.

Question 46.
Is focal length of a lens zero? If not, why?
Answer:
No, focal length of lens never equals to zero because it is the distance between focal point and optical centre.

Question 47.
A thin lens has a focal length of 12 cm. Is it a convex leps or a concave lens?
Answer:
It is a convex lens, because f is positive.

Question 48.
Name the different apparatus where we are using the convex and concave lenses.
Answer:
The magnifying lense, telescope, microscope.

Question 49.
Draw the given diagram in your answer book and complete it for the path of ray of light beyond the lens.

Answer:

Question 50.
The diagram below shows two incident rays P and Q which emerge as parallel rays R and S respectively. The appropriate device used in the box is ……..

Answer:
The rays are diverging and they produced the parallel from the device after refraction. So the device is concave lens.

Question 51.
The following figure shows the incident and refracted rays pass through a lens kept in the box. Draw the lens and complete the path of rays.

Answer:
The incident rays 1 and 2 have converged after refraction. So the lens is convex.

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
A convex lens is made of five different materials as shown in the figure. How many images does it form? Why? (AP March 2017)

Answer:
The given convex lens is made up of five different materials.
So they have different refractive indices / different focal lengths.
Hence they form five different images.

Question 2.
The focal length of a converging lens is 20cm. An object is 60cm from the lens. Where will be image be formed and write characteristics of the image. (AP March 2018)
Answer:
Focal length = f = 20 cm (+ 20cm)
Object distance = u = 60 cm (- 60cm)
Image distance = v = ?

Here, (-) indicates inverted images.
m = \(\frac{1}{2}\) < 1 indicates diminished image.
Image forms between F₁ and 2F₁.
Characteristics of the image :

1. real
2. inverted
3. diminished.

Question 3.
When a light rays enters a medium with refractive index n2 from a medium with refractive index n, at curved interface with radius of curvature R is given by
\(\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}\)
Now assume that the surface is plane and rewrite the formula with suitable changes.
Answer:
Assume that the interface is plane surface
Then R becomes infinity
R = ∞ (or) R = 1/0
Substitute the above value in the given equation

Question 4.
Two convex lenses of same focal length are fixed in a PVC pipe at a distance double to their focal length. What happens if a boy sees the moon with that arrangement? (TS March 2017)
Answer:

• The rays coming from moon are parallel. The first lens converges the rays at focus.
• The converging point is the focus of second lens. So the second lens convert the diverging rays into parallel.
• Hence, in the rays of moon, there will be no change when we see moon with this arrangement or without this arrangement.

(OR)

This arrangement does not make any difference in the rays coming from moon.
The moon appears same if we see directly or with this arrangement.

Question 5.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 6.
Complete the ray diagram given below (TS March 2018)

Answer:

Question 7.
The refractive index of convex lens material is 1.46. The refractive index of Benzene and water is 1.5 and 1.0 respectively. How does the lens behaves when it is kept in Benzene and water? Given and write. (TS March 2018)
Answer:

• When the convex lens with refractive index 1.46 is kept in Benzene with refractive index 1.5, then the lens acts as a diverging lens.
• If the same lens is kept in water whose refractive index is 1, then it acts as a converging lens.

Question 8.
Write the applications of lenses in day to day life. (AP SCERT: 2019-20)
Answer:
Uses of lenses in day to day life :

• Lenses are used for correcting eye defects.
• They are used as magnifying lenses.
• They are used in microscopes, telescopes, binoculars.
• They are used in cinema projectors and cameras.

Question 9.
Water lens is made of double convex lens of radius of curvature “R”. Write lens makers formula for water lens. (AP SA-I: 2019-20)
Answer:
1) Radius of curvatures of water lens are R₁ = R₂ = R and n = 1.5.
2) Sign conversion R₁ = + R₁, R₂ = -R₂.
3) Lens makers formula

Question 10.
Find the focal length of plane convex dens if its radius of curvature is R and its refractive index is n.
Answer:
Given lens is plano-convex lens; radius of curvature = R
Refractive index = n

Question 11.
In a classroom, four friends found out the focal length of a lens by conducting an experiment. The value came out to be 12.1cm, 12.2cm, 12.05 cm, 12.3 cm. The friends discussed the reasons for the differences or defects. Mention those reasons.
Answer:
Students got different focal lengths.

1. By observing the values they got all positive values. This indicates they are given by convex lens.
2. All the students got exact interger but different decimal value.

Reasons :
The difference in values is due to least count errors, parallax errors, random errors and systematic errors, etc.

Question 12.
How will you decide whether a given piece of glass is a convex lens, concave lens or a plane plate?
Answer:
Hold the given piece of glass over some printed matter.

1. If the letters appear magnified, then the given piece of glass is convex lens.
2. If the letters appear diminished, then the given piece of glass is concave lens.
3. If the letters appear to be same size, then it is a plane glass piece.

Question 13.
State the type of lens used as a magnifying glass. Draw a labelled ray diagram to show that the image of the object is magnified.
Answer:

A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified. When the object to be seen in between the focus and optical centre of the lens, a vertical, erect magnified image of the object is formed as shown and convex lens is and to act as magnifying glass AB’ is the magnified image of AB.

Question 14.
Give conventions used in lenses.
(OR)
Write the signs of convex and concave lens using in drawing ray diagrams.
Answer:

• All distances are measured from optical centre of the lens.
• Distances measured along the direction of the incident light are taken as positive.
• The distances against the incident light are taken as negative.
• The heights measured vertically above the axis, are taken as positive.
• The heights measured vertically down the axis, are taken as negative.

Question 15.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Answer:
Yes, the statement is correct.
For magnified virtual image :
The object should be placed between optic centre (C) and focus (F) (< 20)

For magnified real image:
Placed between focus (F) and centre of curvature (2F) (20 – 40)

Question 16.
Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building ? What is the approximate focal length of this lens?
Answer:
As the image is real, therefore the lens use is convex lens. The distance of the real image formed by a convex lens from the lens decreases as the object distance from the lens increases. Hence, the screen has to be moved towards the lens to obtain the sharp image of the building. Approximate focal length of the lens =15 cm as the rays of light coming the window pane are considered to come from infinity. These rays of light are focussed by the convex lens at its focus, (i.e. on the screen).

Question 17.
What do you see when your friend brings a sheet of paper on which arrow was drawn behind the empty cylindrical shaped transparent vessel? Why do you see a diminished image?
Answer:
We will see a diminished image of the arrow.

When the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass and enters air, then it again undergoes refraction curved surface of the vessel and comes out into the air. In this way, light travels in two media, comes out of the vessel and forms a diminished image.

Question 18.
Using the formula of refraction at curved surfaces, write the formula for plane surfaces.
Answer:
For curved surfaces the formula for refraction is \(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)}{\mathrm{R}}\)

For plane surface, the radius of curvature (R) approaches infinity. Hence 1/R becomes zero.
\(\frac{\mathrm{n}_{2}}{\mathrm{v}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=0 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{v}}=\frac{\mathrm{n}_{1}}{\mathrm{u}}\)

Question 19.
Explain how a convex lens behaves on converging lens and diverging lens.
Answer:
The convex lens behaves as a converging lens, if it is kept in a medium with refractive index less than the refractive index of the lens. It behaves like a diverging lens when it is kept in transparent medium with greater refractive index than that of lens.
e.g. : Air bubble in water behaves like a diverging lens.

Question 20.
When does Snell’s law fail?
Answer:

• Snell’s law fails when light is incident normally on the surface of refracting medium.
• Both media have same refractive index.

Question 21.
If on applying sign convention for lens the image distance obtained is negative, state the significance of negative sign.
Answer:

• Negative sign of image distance means the image is virtual and erect.
• It is formed on the same side of the object with respect to lens.

Question 22.
Magnification of lens is found to be +2. What type of lens is that?
Answer:
Magnification +ve indicates the image is erect and virtual.
Magnification 2 indicates it is magnified.
Magnified virtual image is formed by only convex lens.

Question 23.
For same angle of incidence in media A, B and C the angle of refractions are 30°, 25° and 20° respectively. In which medium will the velocity of light be minimum?
Answer:
In medium R the velocity of light is minimum because it has greater refractive index. Refractive index and velocity of light in a medium are inversely proportional. So in medium R the velocity is minimum.

Question 24.
The radius of curvature (twice the focal length) of a convex lens is 40 cm. A student wants to get various images of following types (a) enlarged virtual image, (b) enlarged real image, (c) image of same size, (d) diminished image.
In order to get these images where should the object should be kept in front of convex lens?
Answer:
a) The object should be kept in less than 20 cm (i.e. less than focal length).
b) In order to get enlarged real image, object should be kept between 20 cm to 40 cm in front of lens.
c) In order to get image of same size object should be kept at a distance of 40 cm from the lens.
d) In order to get diminished image the object should be kept beyond 40 cm from the lens.

Question 25.
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images.
Answer:

• Yes, the statement is correct.
• A convex lens of focal length 20 cm will produce a magnified virtual image if object is placed at a distance less than 20 cm from the lens.
• A convex lens of focal length 20 cm will produce magnified real image if the object is placed at a distance more than 20 cm and less than 40 cm.

Question 26.
A concave mirror and a convex lensare held in water. What changes, if any, do you expect in their focal length?
Answer:
The focal length of a concave mirror independent of the medium and A convex lens depends on medium when they are placed in water.
The focal length of the mirror – Does not change.
The focal length of the convex lens – Changes (means increases 4 times).

Question 27.
When does a convex lens behave like a diverging lens? Given example.
Answer:
A Convex lens behaves like a diverging lens when it is kept in a tranparent medium with greater refractive index than that of the lens.
Eg : An air bubble in water behaves like a diverging lens.

Question 28.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays are originated from the bottom of the pond.
• As these rays get refracted into air, they bend away from the normal.
• When these two refracted rays produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus the pond appears to be shallower.

Question 29.
What happens to the image formed by a convex lens if its lower part is blackened?
Answer:

1. Every part of lens forms complete image.
2. If lower part of the lens is blackened, the complete image will be formed.
3. But its intensity will decrease.

Question 30.
Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?
Answer:

• Yes, the type of lens changes, if it is placed in medium having higher refractive index that of lens.
• For example, convex lens acts as converging lens when it is placed in a medium of lower refractive index otherwise it behaves like a diverging lens.

Question 31.
Draw the different types of convex and concave lens.
Answer:

Question 32.
Complete the ray diagram and give reason.

Answer:

The light ray which passes through optical centre does not undergo refraction. So it goes straight.

Question 33.
How do you appreciate the refraction at plane surfaces and at curved surfaces?
Answer:
The refraction of curved surfaces are used in various aspects such as

1. In microscope to enlarge microscopic objects.
2. In telescopes to see celestial objects.
3. To correct eye defects like myopia, hypermetropia and presbyopia.

So refraction at curved surfaces is thoroughly appreciated.

Question 34.
An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.
Answer:

Image distance is negative that indicates it is a virtual and erect image.

Question 35.
A bird is flying at the height 3m above the river surface while a fish is 4 m below the surface. At what depth would the fish appear to the bird ? At what height the bird would appear to the fish? (given a/w = 4/3)
Answer:
Given that refractive index of air / water = \(\frac{4}{3}\)
The height the bird would appear to fish = 4 + \(\frac{4}{3}\) × 3 = 4 + 4 = 8m

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Complete the ray diagram when an object is placed between F₂ and 2F₂. (AP June 2017)

Answer:

Question 2.
An object is placed at the following distances from a convex lens of focal length 10 cm.
(a) 8 cm.
(b) 15 cm.
(c) 20 cm.
(d) 25 cm.
Which position of the object will produce……. (TS March 2015)
(i) a diminished, real and inverted image?
(ii) a magnified, real and inverted image?
(iii) a magnified, virtual and erect image?
(iv) an image of same size as the object?
Justify your answer in each case.
Answer:
i) d (or) 25 cm
Reason : Object placed between centre of curvature and focal point.

ii) b (or) 15 cm
Reason : Object placed beyond centre of curvature.

iii) a (or) 8cm
Reason : Object placed between focal point and optic centre.

iv) c (or) 20 cm
Reason :Object placed at centre of curvature.

Question 3.
The ray diagrams showing the image formed by a convex lens are given in the following table. From these diagrams complete the table. (TS June 2016)

Answer:

Question 4.
Explain the behaviour of light rays in any four situations of their incidence on a convex lens. (TS March 2016)
Answer:
1) A ray passing along the principal axis is undeviated.

2) Any ray passing through optic centre is also undeviated.

3) The rays passing parallel to principal axis converge at focus.

4) The rays passing through the focus will take a path parallel to principal axis.

Question 5.
Draw the ray diagrams to find the images when an object is placed in front of the lens (i) at a distance of 8 cm, and (ii) at a distance of 10 cm on the principal axis of a convex lens whose focal length is 4 cm. Write the characteristics of images in both the cases. (TS June 2017)
Answer:
(i) Ray diagram :

Characteristics of Image :
i) Size of the image equal to the size of the object,
ii) Inverted image,
iii) Real image,
iv) Image formed at C.

(ii) Ray diagram

Characteristics of Image :
i) Image size is less than that of object size,
ii) Inverted image,
iii) Real image,
iv) Image is formed between F & C.

Question 6.
A double concave lens with the refractive index (n) = 1.5 is kept in the air. Its two spherical surfaces have radii R₁ = 20 cm and R₂ = 60 cm. Find the focal length of the lens. Write the characteristics of the lens. (TS March 2017)
Answer:

Hence f = – 30 cm (Here minus indicates that the lense is divergent)
Characteristics of the biconcave lens :

1. It is a diverging lens.
2. It is thin at the middle and thicker at the edges.

Question 7.
Draw ray diagrams for a double concave lens of focal length 4 cm, when objects are placed at 3 cm and 5 cm on principal axis. Write characteristics of images. (TS June 2018)
Answer:
i)

ii)

Characteristics of images :

1. Image formed between P and F
2. Diminished image
3. Errected image
4. Virtual image

Question 8.
Write the characteristics of the images which are formed when objects are placed at 50cm and 75cm on the principle axis of a convex lens with focal length of 25 cm. (TS March 2018)
Answer:
i) Object is placed at 50cm.

Characteristics of the image :

1. Image forms at 2Fp
2. Image is real,
3. Image is inverted,
4. Image is same size

ii) Object is placed at 75cm
f = +25cm; u = -75cm; v = ?

Characteristics of the image :

1. Image forms between F₁ and 2F₁
2. Image is real
3. Image is inverted
4. Image is diminished.

Question 9.
Write the role of lenses in our daily life. (AP March 2019)
Answer:
The role of lenses in our daily life :

1. Used for correcting eye defects.
2. Used as magnifying lens.
3. Used in Microscopes.
4. Used in Telescopes.
5. Used in Binoculars.
6. Used in Cinema Projectors.
7. Used in Cameras.

Question 10.
Draw the ray diagrams for the following positions of objects in front of a convex lens mention the characteristics of the image. (AP SCERT: 2019-20)
a) Object is placed beyond 2F₂.
b) Object is placed between focal point and opint center.
Answer:
a)

b)

a) Characteristics of the Image :

1. real
2. inverted
3. diminished.

b) Characteristics of the Image :
If we place an object between focus and optic centre, we will get an image which is virtual, erect and magnified.

Question 11.
The focal length of a convex lens is 2 cm. Draw the ray diagram of an object placed on principal axis at the ‘C’ of lens and at a distance of 3 cm from its optic centre.
Answer:
1) Object is placed on principal axis of a convex lens at ‘C’.

2) Object is placedon principal axis at a distance of 3 cm from its optic center.

Question 12.
Using biconvex lens, a point image is made on its principal axis S. Let us assume that we know optical centre P and its focus F. We also know PF > PS. Draw the ray diagram to identify the point source and give reasons.
Answer:

Given lens is biconvex lens and given condition is PF > PS’ means image is formed between optic centre (P) and Focus (F).

According to Snell’s law this condition is possible when the object is also placed between P and F. Because reflected rays are divergent.

Question 13.
Write about the behaviour of light rays when they incident on a lens.
Answer:
1) Situation I:
Ray passing through the principal axis.
⇒ It is not deviated.

2) Situation II:
Ray passing through the pole.
⇒ It is also undeviated.

3) Situation III:
Rays travelling parallel to the principal axis.
⇒ They converge at focus or diverge from the focus.

4) Situation IV :
Ray passing through focus.
⇒ It will take a path parallel to principal axis after refraction.

5) Situation V :
Parallel rays fall on a lens making some angle with principal axis.
⇒ They converge at a point or diverge from a point lying on a focal plane.

Question 14.
Write characteristics of image formed due to convex lens at various distances.
Answer:

Question 15.
Write characteristics of image formed by a concave lens at various distances.
Answer:

Question 16.
You are given a convex lens of focal length 10 cm. Where would you place an object to get a real inverted and highly enlarged image of the object? Draw a ray diagram.
Answer:
If an object is placed at the focus of the lens it forms real, inverted and highly enlarged image. Thus, the distance of the object from the optical centre of the lens is equal to the focal length of the lens =10 cm.
The ray diagram is as shown

Question 17.
Derive the formula of image formation in refraction at curved surfaces.
Answer:
1) Object at infinity :
The rays coming from the object at infinity are parallel to principal axis and converge to the focal point after refraction. So, a point-sized image is formed at the focal point.

2) Object placed beyond the centre of curvature on the principal axis :
When an object is placed beyond the centre of curvature 2F₂, a real, inverted and diminished image is formed on the principal axis between F₁ and 2F₁

3) Object placed at the centre of curvature :
When an object is placed at the centre of curvature (2F₂) on the principal axis, a real, inverted image is formed at 2F₁ which is same size as that of the object.

4) Object placed between the centre of curvature and focal point:
When an object is placed between centre of curvature (2F₂) and focus (F2), we will get an image which is real, inverted and magnified. This image will form beyond 2F₁.

5) Object located at focal point:
When an object is placed at focus (F₂), the image will be at infinity.

6) Object placed between focal point and optic centre :
If we place an object between focus and optic centre, we will get an image which in virtual, erect and magnified.

Question 18.
Distinguish between convex lens and concave lens.
Answer:

Convex lens	Concave lens
1. Objects appear to be big in the lens.	1. Objects appear to be shrink in the lens.
2. It generally forms real image, (except object is placed between optical centre and focal point)	2. It always forms virtual image.
3. Light rays tend to converge after refraction from lens.	3. Light rays tend to diverge from lens after refraction.
4. The image due to lens may be enlarged or same size or diminished.	4. The image is always diminished.
5. The image due to lens may be inverted or erect.	5. The image is always erect.
6. It is used to correct hypermetropia.	6. It is used to correct myopia.

Question 19.
The ray diagram given below illustrates the experimental set up for the determination of the focal length of a converging lens using a plane mirror.

1) State the magnification of the image formed.
2) Write the characteristics of the ipiage formed.
3) What is the name given to the distance between the object and optical centre of the lens in the following diagram?
Answer:

1. The magnification of the image formed is unity (or 1).
2. The image is a) real and b) inverted is at the same position of the object.
3. The distance between the object and the optical centre of the lens is called object distance.

Question 20.
A concave lens made of a material of refractive index n₁ is kept in medium of refractive index n₂. A parallel beam of light incident on the lens. Complete the path of rays of light emerging from the concave lens if
i) n₁ > n₂
ii) n₁ = n₂
iii) n₁ < n₂.
Answer:
i) When n₁ > n₂, light goes from rarer to denser medium. Therefore, in passing through a concave lens it diverges.

ii) When n₁ = n₂, there is no change in medium. Therefore no bending or refraction occurs.

iii) When n₁ < n2, light goes from a denser to rarer medium. Therefore, in passing through a concave lens it converges.

Question 21.
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:

• Every part of a lens forms an image.
• For formation of image we require only two light rays to converge.
• Therefore, if the lower half of the lens is covered, it will still form a complete image.
• However the intensity of the image will be reduced.
• This can be verified experimentally by observing the image of distant object like tree on a screen, when lower half of the lens covered with a black paper.

Question 22.
Complete the following table if the object is placed at various positions in front of a convex lens.

Position of object	Position of image	Nature of image
1. At infinity
2.	Between F1 and 2F1
3.		Same size, real and inverted
4.	Seen in the lens

Answer:

Position of object	Position of image	Nature of image
1. At infinity	On Fj (focal point)	Highly diminished, real and inverted
2. Beyond 2F2	Between F1 and 2F1	Diminished, real and inverted
3. On 2F1	On 2F2	Same size, real and inverted
4. Between focus and optical centre	Seen in the lens	Enlarged, virtual and erect

Question 23.
A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observations.

Distance of flame from the lens (cm)	Distance of the screen from the lens (cm)
1. 60	20
2. 40	24
3. 30	30
4. 24	40
5. 15	70

a) From the above table, find the focal length of lens without using lens formula.
b) Which set of observations is incorrect and why?
c) In which case the size of the object and image will be same? Give reason for your answer.
Answer:
a) From the observations, it is clear that for u = 30, v = 30 cm. This means this value must be equal to twice the focal length of the convex lens.
∴ Focal length of convex lens = 30/2 = 15 cm

b) The observation (5) is not correct because if u = 15 cm i.e., the object is kept at focus so the image should be at infinity and not at 70 cm.

c) For twice the focal length we know size of object = size of image. So when object is kept at 30 cm the size of object and image are same.

Question 24.
Draw the ray diagrams when incident ray striking a convex surface or a concave surface moving from one medium to another medium.
Answer:

Question 25.
Draw ray diagrams of image formed by a convex lens at various distances.
Answer:
1) Object at infinity.

2) Object placed beyond the centre of curvature (2F₂).

3) Object placed at the centre of curvature.

4) Object placed between 2F₂ and F₂.

5) Object at the focal point.

6) Object placed between F and P

Question 26.
Write about the focal length of the lens with diagram.
Answer:

• A parallel beam of light incident on a lens converges to a point as shown in figure (a) or seems to emanate from a point on the principal axis as shown in figure (b).
• The point of convergence (or) the point from which rays seem to emanate is called focal point or focus (F).
• Every lens has two focal points.
• The distance between the focal point and optic centre is called the focal length of lens denoted by ‘f’
• To draw ray diagram for lenses, we need two more points in addition to focal points F₁ and F₂.
• These points are equidistant from centre of the lens and also equal to double the focal length. So we call them 2F₁ and 2F₂.
• For drawing ray diagrams related to lenses, we represent convex lens with a symbol £ and concave lens with J as shown in the figure c and d.

Question 27.
The diagram shows an object OA and its image IB formed by a lens. The image is same size as the object.
a) Complete the ray diagram and locate the focus of lens by labelling it as F.
b) State whether the lens is convex or concave.
Show it in the diagram.

Answer:

a)

1. Optical centre goes undeviated, therefore to find the optical centre, join A to B to meet the line 01 at the point P which gives the position of optical centre of the lens.
2. Draw a line CP through P perpendicular to the line 01 to represent lens.
3. Draw another ray AC from the point A parallel to the principal axis 01 to meet the lens line CP at a point C.
4. This ray AC will reach the image point B while passing through the focus, therefore join C to B to meet line 01 at a point F which is the focus of the lens. The completed ray diagram is shown above.

b) Since the image is real and inverted, the lens is convex.

c) Since the size of object and image of equal, the object must be at a distance of twice the focal length, i.e., at 2F₂.

Question 28.
The diagram shows an object AB placed on the principal axis of the lens L. The two foci of lens are F₁ and F₂. The image formed by the lens is erect, virtual, and diminished.

i) Draw the outline of the lens L used and name it.
ii) Draw a ray of light starting from B and passing through ‘O’. Show the same ray after refraction by the lens.
iii) Draw another ray from B which is incident and parallel to the principal axis. Show how it emerges after refraction from the lens.
iv) Locate the final image formed.
Answer:
i) Since the image formed by the lens is erect, virtual, and diminished, the lens is concave.
ii) The light ray BO incident at the optical centre ‘O’ of the lens, passes undeviated as OC after refraction by the lens.
iii) The light ray BP is incident and parallel to the principal axis. It emerges as PQ after refraction which appears to diverge from the second focus F₂ of the lens.
iv) The refracted rays OC and PQ do not actually meet, but they appear to diverge from a point B’ (i.e. they meet at a point B’ when they are produced backwards).
v) Thus, B’ is the complete image A’B’ is obtained by drawing perpendicular from B’ on the line F₂OF₁. The image is formed between the optical centre O and focus F₂ of the lens.

Question 29.
Figure below shows the refracted ray BC through a concave lens and its foci marked as F₁ and F₂. Complete the diagram by drawing the corresponding incident ray and also give reason.

Answer:

1. Figure shows the refracted ray parallel to the principal axis. Therefore, the incident ray must be travelling towards the focus F₂.
2. Thus, to find incident ray, F₂ is joined to the starting point B of the refracted ray and produced backward as BA.
3. Then AB is the required incident ray.
4. The completed diagram is shown below.

Question 30.
State the type of lens used as a simple magnifying glass. Draw a labelled diagram to show that the image of the object is magnified.
Answer:

• A single convex lens is used as a magnifying glass, i.e. for seeing small object magnified.
• When the object to be seen between the focus and optical centre of the lens, a virtual, erect and enlarged image is formed.
• So a convex lens acts as magnifying glass for object AB as shown in the figure.

Question 31.
Radii of biconvex lens are equal. Let us keep an object at one of the centres of curvature. Refractive index of lens is ‘n’. Assume lens is in the air. Let us take R as the radius of the curvature.
a) How much is the focal length of the lens?
b) What is the image distance ?
c) Discuss the nature of the image.
Answer:
Radii of curvatures (R) of biconvex lens are equal, so R₁ = R₂ = R


c) The nature of the image is inverted and v < u.

Question 32.
Refractive index of a lens is 1.5. When an object is placed at 30 cm, image is formed at 20 cm. Find its focal length. Which lens is it? If the radii of curvature are equal, then what is its value?
Answer:
Object distance = u = – 30 cm (Infornt of the lens)
Image distance = v = 20 cm

Question 33.
A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall?
Answer:

Question 34.
A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. If image distance is thrice the focal length, find object distance, image distance and nature of image.
Answer:
Given that
Focal length of lens = + 20 cm
Object distance = -u
Image distance (v) = + 3u

So the object is between F₂ and 2F₂. So the image beyond 2F₁ it is real, inverted, and magnified.

Question 35.
What are the various applications of lens?
Answer:

• The objective lens of a telescope, camera, slide projector, etc. is a convex lens which forms real and inverted image of object.
• Our eye lens is also a convex lens. The eye lens forms the inverted image of the object on the retina.
• The eye defects are corrected by lenses.
• A magnifying glass is nothing but a convex lens of short focal length fitted in a steel (or plastic) flame provided with a handle.
• In spectroscope, convex lenses are used for obtaining a pure spectrum.
• A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 4 Acids, Bases and Salts.

AP State Syllabus SSC 10th Class Chemistry Important Questions 4th Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts 1 Mark Important Questions and Answers

Question 1.
Take some water in a test tube and add concentrated H₂SO₄ to it. Shake the test tube well. If you touch the bottom of the test tube, you feel it as hot. Now, instead of H₂SO₄, if you add NaOH pellets to water in another test tube and touch the bottom, what do you observe? (TS June 2015)
Answer:
The bottom of test tube is also hot because reactions of acids, bases with water are exothermic reactions.

Question 2.
What happens if the copper sulphate crystals taken into dry test tube are heated? (TS June 2016)
Answer:

• When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white.
• Evaporated water appears as droplets on the walls of the test tube.
• Blue coloured copper sulphate (CuSO₄ 5H₂O) is turned into white colour because 5H20 molecules are evaporated from crystals.

Question 3.
Why does the soil of agricultural lands get tested for pH?
Answer:
Plants require a specific pH range for their healthy growth. So, finding pH of a soil suggested the farmers to treat the fields with acidic or basic substances to maintain the required pH range.

Question 4.
Write the molecular formulae of common salt and baking soda which are widely used at home. (TS June 2017)
Answer:
Common Salt: NaCl; Backing soda : NaHCO₃

Question 5.
Mention the precautions to take while conducting an experiment to prove acids produce ions only in aqueous solutions. (TS June 2018)
Answer:

• Testing of the evolved gas by using dry litmus paper first. Then with wet litmus paper.
• Use gaurd tube containing calcium chloride.

Question 6.
What are antacids?
Antacids are mild alkalies. These are used for getting relief from acidity and indigestion and sometimes even headache. When taken orally, it reacts with hydrochloric acid present in the stomach and reduces its strength by consuming some of it.
Ex: Milk of magnesia.

Question 7.
Tap water conducts electricity whereas distilled water does not. Why?
Answer:
Tap water contains some impurities in the form of salts. Due to presence of salts, it conducts electricity. Distilled water is free from all kinds of salts and hence does not conduct electricity.

Question 8.
What do you mean by dilution of an acid or base? Why is it done?
Answer:
Dilution of an acid or base means mixing an acid or base with water. This is done to decrease the concentration of ions per unit volume. In this way the acid or the base is said to be diluted.

Question 9.
What is a universal indicator?
Answer:
An indicator which passes through a series of colour changes over a wide range of H₃O⁺ ions concentration is called universal indicator.

Question 10
What is tooth decay?
Answer:
Tooth enamel is chemically calcium phosphate Ca₃(PO₄)₂. It starts corroding when pH falls below 5.5. Food particles left in the mouth degrade to produce acid which lower the pH of the mouth. This is called tooth decay.

Question 11.
Define Alkalis and give some examples.
Answer:
Alkalis : An alkali is a base that dissolves in water.

Examples :
i) Sodium Hydroxide (NaOH),
ii) Potassium Hydroxide (KOH),
iii) Magnesium Hydroxide (Mg(OH)₂)

Question 12.
Why should we not taste or touch alkalis?
Answer:
We should not taste or touch alkali. Because they are corrosive.

Question 13.
Salts conduct electricity. Why?
Answer:
Salts contains ions. So they conduct electricity.

Question 14.
Why are calcium sulphate hemihydrates called Plaster of Paris?
Answer:
Calcium sulphate hemihydrates are used as plaster for supporting fractured bones in the right position. So, it is called Plaster of Paris.

Question 15.
Why does an aqueous solution of acid conduct electricity?
Answer:
An aqueous solution of acid liberates H⁺ ions. This makes the aqueous solution of acid to conduct electricity.

Question 16.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
When a solution of an acid is diluted, concentration of H₃O⁺ ions decreases.

Question 17.
What is pH?
A. pH is a scale for measuring hydrogen ion concentration in a solution. It is the negative logarithm of H⁺ concentration.
pH = – log [H⁺].

Question 18.
How is pH of a solution related to the [H₃O⁺] of that solution?
Answer:
The presence of H₃O⁺ ions indicate us whether it is a strong acid or weak acid.

Question 19.
There are two solutions of pH values 6 and 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:

• The solution whose pH value 6 is acid and has more hydrogen ion concentration.
• The solution of pH value 8 is basic and has less hydrogen ion concentration.

Question 20.
Can you give example for use of olfactory indicators in daily life?
Answer:
Examples of olfactory indicators : Onion, vanilla extract.

Question 21.
How do acids neutralize bases?
(OR)
How do acids and bases react with each other?
Answer:
According to Arrhenius theory acids produce H⁺ ions and bases produce OH^– ions in aqueous media.
The combination of H⁺ and OH^– ions is called ‘neutralization’.
Thus acids neutralize bases.

Question 22.
How strong are acids and base solutions?
Answer:
The acids of pH value as much less as possible have more concentration [pH < 7], Basic nature increases as pH value increases.

Question 23.
What do you say about salts of both weak acid and weak base?
Answer:
The pH of aqueous solutions of salt obtained from both weak acid and weak base is nearly 7.

Question 24.
Which base is used for removing permanent hardness of water?
Answer:
Sodium carbonate is used for removing permanent hardness of water.

Question 25.
Name two antacids used to get rid of our indigestion problem.
Answer:
Magnesium hydroxide and a mild base (baking soda).

Question 26.
Under what soil conditions would a farmer would treat the soil of his fields with quicklime (calcium hydroxide) or calcium carbonate?
Answer:
When the field has acidic nature, the farmer uses quicklime or calcium carbonate to neutralize it.

Question 27.
Write the formulas of Gypsum and Plaster of Paris.
Answer:
The formulae of Gypsum is CaSO₄ . 2H₂O and Plaster of Paris is CaSO₄ . ½H₂O.

Question 28.
Write any two Acid Base indicators.
Answer:

1. Methyl orange
2. Phenolphthalein.

Question 29.
Which salt is used in the manufacture of borax?
Answer:
Washing soda (Na₂CO₃.10H₂O)

Question 30.
What is family of salts? Give examples.
Answer:
Salt having the same positive or negative radical is called family of salts.
Eg : Family of sodium salts : NaCl, Na₂SO₄
Family of chloride salts : NaCl, KCl.

Question 31.
‘A’ is a substance which is acidic and it is added in solution to preserve pickles. What is A and what is the name given to its dilute solution?
Answer:
‘A’ is acetic acid and its dilute solution is called vinegar.

Question 32.
What are the chemical names of the following?
1) Baking soda
2) Gypsum
Answer:

• The chemical name of baking soda is sodium hydrogen carbonate (NaHCO₃).
• The chemical name of gypsum is calcium sulphate dihydrate (CaSO₄ • 2 H₂O).

Question 33.
Write the water of crystallisation of following compound.
a) Hydrous copper sulphate
b) Washing soda
c) Gypsum
d) Plaster of Paris
Answer:

Compound	Formula	Water of crystallisation
1) Hydrous copper sulphate	CuSO4 . 5H2O	5
2) Washing soda	Na2CO3 . 10 H2O	10
3) Gypsum	CaSO4 . 2 H2O	2
4) Plaster of Paris	CaSO4 . ½H2O	½

Question 34.
Why don’t we use a strong base like NaOH as antacid?
Answer:
Strong bases like potassium hydroxide (KOH), sodium hydroxide (NaOH) are corrosive in nature. So they can harm the internal organs. Therefore we should not use them as antacid.

Question 35.
What do you mean by HsO⁺ ion?
Answer:
Hydrogen ions cannot exist as base ions. They associate with water molecules and exist as hydrated ions with each H⁺ attached by 4 to 6 water molecules. For this we represent H⁺ as hydronium ion, H³O⁺.
H⁺ + H₂O → H₃O⁺

Question 36.
Which indicator is useful at all pH? Why?
Answer:
Universal indicator is useful to test solutions of all pH because it gives different colours at different pH range.

Question 37.
Which substance is useful in removing permanent hardness of water?
Answer:
The permanent hardness of water is due to chloride and sulphate salts of magnesium and calcium, which can be removed by adding washing soda.

Question 38.
Given two examples for strongest bases.
Answer:
Sodium hydroxide – NaOH
Potassium hydroxide – KOH

Question 39.
What is the confirmation test for hydrous and anhydrous salt?
Answer:

• On heating hydrous salt in a test tube it will form water droplets on the sides of test tube.
• On heating anhydrous salt in a test tube it will not form water droplets on the sides of test tube.

Question 40.
P.O.P, cement, calcium chloride should be stored in moisture proof containers. Why?
Answer:

• P.O.P, cement and calcium chloride react with moisture (H₂O) in the atmosphere and set into hard solid masses.
• To avoid availability of moisture they should be stored in moisture proof containers.

Question 41.
Give some examples for hydrous and anhydrous salts.
Eg : For hydrous salts :

1. CuSO₄ . 5H₂O
2. Na₂CO₃ . 10H₂O
3. CaSO₄ . 2H₂O

Eg : for anhydrous salts :

1. NaCl
2. MgCl₂
3. Na₂CO₃

Question 42.
How are bitter and sour taste substances tested without testing?
Answer:

• Sour taste substances turn blue litmus to red.
• Bitter taste substances turn red litmus to blue. By these tests we can test them as acids and bases.

Question 43.
Do the metallic oxides react with acids?
Answer:
Yes, metallic oxides are basic in nature. They react with acids and form salt and water.

Question 44.
Does non-metallic oxide react with base?
Answer:
Yes, non-metallic oxide is acidic in nature. It reacts with base and forms salt and water.

Question 45.
Why does dry HCl gas not change the colour of the dry litmus paper?
(OR)
Prove that dry HCl gas is not an acid but HCl aqueous solution is an acid using an activity.
Answer:

• Dry hydrogen chloride gas is not an acid. Hence it can’t turn blue litmus into red.
• Hydrochloric acid is an aqueous solution. Hence it can turn blue litmus into red.

Question 46.
Do you know that the atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid? Do you think life can exist on this planet?
Answer:

1. No, it is not possible.
2. When pH value decreases, the survival of living organisms becomes difficult.
3. Hence there is not any possibility of life on Venus.

Question 47.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids don’t show acidic behaviour in the absence of water as H⁺ ions are absent in them.

Question 48.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When a base like NaOH (Sodium hydroxide) is dissolved in water, it liberates (OH^–) ions.
Equation :

OH^– ion concentration increases.

Question 49.
How does the nature of the solution change with change in concentration of H⁺_(aq) ions?
Answer:
The concentration of H⁺ ions is responsible for the acidic nature of a substance.
If [H⁺] > 1 × 10^-7 mol/lit the solution is acidic.
If [H⁺] < 1.0 × 10^-7 mol/lit the solution is basic.

Question 50.
Do basic solutions also have H⁺_(aq) ions? If yes, then why are these basic?
Answer:
Yes. Basic solutions have OH^–_(aq) ions and bases have less number of H₃O⁺ ions. H⁺_(aq) ions are less in base. In basic solution [OH^–] > [H⁺].

Question 51.
What do acids have in common?
Answer:
(Acids have sour taste and conduct electricity. They release H₂ (Hydrogen) gas on reacting with metals). All acids have H⁺_(aq) ions.

Question 52.
What do bases have in common?
Answer:
Bases are slippery to touch and bitter to taste. All bases have OH^–_(aq) ions.

Question 53.
Why does pure acetic acid not turn blue litmus to red?
Answer:
Pure acetic acid is a weak acid so it does not have sufficient H⁺_(aq) ion to change the colour of blue litmus to red.

Question 54.
What will happen if the pH value in your body increases?
Answer:
It affects our digestion system.

Question 55.
A student checked pH of a salt solution and found that its pH is more than 7. How is that type of salt formed?
Answer:

• When a strong base reacts with weak acid then the solution is basic in nature. So its pH is more than 7.
• For example when acetic acid reacts with sodium hydroxide the salt formed has basic nature.

Question 56.
Explain the procedure that you follow to reduce water from a given salt.
Answer:
Procedure to reduce water from a given salt :

1. Take a boiling test tube.
2. Drop given salt in the test tube.
3. Heat the test tube gently.
4. Water from salt evaporates.
5. In this way we can reduce the water from salt.

Question 57.
Write the observations, when the hydrated salt or unhydrated salt is heated.
Answer:

• When hydrated salt is heated water droplets form inside the walls of test tube and sometimes blue or green colour salt turns into white colour.
• When unhydrated salt is heated it does not form water droplets inside the test tube walls and colour also does not change.

Question 58.
On heating the hydrated salt it loses water molecules present in it. To show this what are the equipment required?
Answer:
1) On heating the hydrated salt it loses water molecules present in it.
2) To show this the given equipment are required

1. Boiling tube
2. Test tube holder
3. Burner

Question 59.
Try to collect the information to reasons for calling calcium sulphate hemihydrates as Plaster of Paris (POP).
Answer:

• Gypsum plaster (or) Plaster of Paris (POP) is produced by heating gypsum to about 300°F.
• A large gypsum deposit is found at Montmartre in Paris (France).
• This gave the name Plaster of Paris to calcium sulphate hemihydrates.
• The term plaster can refer to gypsum.

Question 60.
Is the substance present in antacid tablet acidic or basis?
Answer:
The substance present antacid is weakly basic.

Question 61.
Give pH of neutral, acid and base.
Answer:

Nature of substance	pH range
Neutral	7
Acid	0 – 7
Base	7 – 14

Question 62.
Which nature of Plaster of Paris makes its importance? Appreciate it.
Answer:
Plaster of Paris is a white powder. It is very soft and can be used to make toys, materials for decoration and to make surfaces smooth.

But on mixing with water, it changes to a hard solid mass (Gypsum). This is the important character of Plaster of Paris (POP).

Question 63.
What is acid rain? How does it affect our aquatic life?
Answer:
When the pH of rain water is less than 5.6 it is called acid rain. When acid rain flows into the rivers, it lowers the pH of the river water. Since our body works within a narrow pH range close to 7 the survival of aquatic life in river water mixed with rain water becomes difficult.

Question 64.
Why are pickles and sour substances not kept in brass and copper vessels?
Answer:
Pickles and sour substances contain acidic nature which may react with brass and copper vessels to produce toxic substances.
So, we don’t keep them in brass arid copper vessels.

Question 65.
Can you suggest some examples of use of pH in everyday life?
Answer:
Uses of pH in everyday life :

1. pH value helps us to identify acids, bases and neutrals.
2. If pH value is less in our mouth, it leads to tooth decay. We can find it as the reason for our tooth decay.
3. pH value helps us to know about acid rain.

Question 66.
Write any two uses of Bleaching powder.
Answer:

• It is used for disinfecting drinking water to make it free of germs.
• It is used as a reagent in the preparation of chloroform.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 2 Marks Important Questions and Answers

Question 1.
What value of pH in the mouth leads to tooth decay? Why? (TS June 2015)
Answer:

• Tooth decay starts when the pH of the mouth is lower than 5.5.
• Tooth enamel, made of calcium phosphate is the hardest substance in the body.
• It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5.
• Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth.

Preventions :

1. Clean the mouth after eating food.
2. Using tooth pastes, which are generally basic neutralize the excess acid and pre¬vent tooth decay.

Question 2.
Equal lengths of Magnesium ribbons are taken in two test-tubes X and Y. Hydro¬chloric acid is added to test-tube X and Acetic acid is added to test-tube Y. In which test-tube, the reaction will be more vigorous? Why? (TS March 2015)
Answer:
The speed of the reactions is higher in X test tube than Y test tube.

Reason :
Due to strong acidic nature, Hydrochloric acid reacts very fast with magnesium ribbon.

Question 3.
Name the four chemicals that are obtained from common salt and write their molecular formulae. (TS March 2015)
Answer:
Chemicals that can be obtained from common salt are

1. Sodium Hydroxide – NaOH
2. Baking soda / Cooking soda / Caustic soda / Sodium bicarbonate / Sodium Hydrogen carbonate. – NaHCOv
3. Washing soda / Sodium carbonate – Na₂CO₃ 10H₂O
4. Bleaching powder / Calcium Oxy Chloride – CaOCl₂

Question 4.
Observe the information given in the table and answer the questions given below the table. (TS March 2017)

Substance (in aqueous solution)	Colour change with Blue Litmus	Colour change with Red Litmus
A	Red	No change
B	No change	Blue
C	No change	No change

i) Which one of them may be the neutral salt among A, B, C?
ii) What may happen when some drops of phenolphthalein is added to the substance B?
Answer:
i) C
ii) Pink Colour

Question 5.
Why do we use antacids? Write it’s nature. (TS March 2018)
Answer:
Pain and irritation will be caused in stomach during the acidity problem/indigestion problem. Antacids used to neutralize the excess acid in the stomach and gives relief from acidity Antacids are basic in nature.

Question 6.
Which product will form when CaO is dissolved in water? How do you find the nature of product? (TS March 2018)
Answer:
CaO reacts with water and gives calcium hydroxide [Ca(OH₂)]. The nature of the calcium hydroxide will be tested with red litmus paper or pH paper.

Calcium Hydroxide turns red litmus into blue. Thus we can say that ca(OH)₂ is basic in nature.

Ca(OH)₂ shows pH value more than 7. Thus we can say that Ca(OH)₂ is basic in nature.

Question 7.
How do you test the nature of the solution formed by dissolving CaO in water? What is the nature of the solution? (TS June 2019)
Answer:

• The solution formed by dissolving CaO in water is tested with red litmus paper, it turns into blue colour, (or) It is tested with methyl orange, it turns into yellow in colour.
• The solution of CaO and water is basic in nature.

Question 8.
Write the experimental procedure to test carbon dioxide gas. (AP SCERT: 2019-20)
Answer:

1. Pass the CO₂ gas through lime water [Ca(OH)₂].
2. The lime water appears as milky white.
3. The reaction is Ca(OH)₂ + CO₂ → CaCO₃ ↓ + H₂O.
4. The milky white is caused by CaCO₃.

Question 9.
Write two reactions of acids with carbonates and metal hydrogen carbonates. (AP SA-I : 2019-20)
Answer:
Reactions :

Question 10.
What do acids have in common?
Answer:
Common characteristics of acids :

1. Similar chemical properties.
2. Acids generate hydrogen gas on reacting with metals.
3. Hydrogen is common to acids.
4. Acids are sour in taste and turn blue to red when react with bases form salt and water.

Question 11.
What do bases have in common?
Answer:
Common characteristics of bases :

1. Bitter in taste.
2. Soapy in nature.
3. Turn red to blue colour.
4. On heating decompose into metal oxides and water.
5. React with acids to form salt and water.
6. Produce OH^– ions in aqueous solution.

Question 12.
How is bleaching powder produced?
Production of bleaching powder :
Bleaching powder is produced by the action of chlorine on dry slaked lime (Ca(OH)₂).
Equation :

Question 13.
What can you conclude about the ideal soil pH for the growth of plants in your region?
Answer:
1) Soil is considered the ‘skin of the earth’. The soil pH plays a vital role in the growth of a plant and it influences plant nutrition.
2) Soil pH strongly affects the nutrients required for the plant growth.
3) The nutrients may be stored on soil colloids, and live or dead organic matter, but may not be accessible to plants due to extremes of pH.

Conclusion :
For optimum plant growth, the generalized content of soil components by volume should be roughly 50% solids (45% mineral & 5% organic matter), and 50% voids of which half is occupied by water and half by gas.

Question 14.
How can you prepare turmeric indicator? What is the use of it?
Answer:
i) Turmeric indicator is prepared from turmeric.
ii) It has red colour in basic solution.

Question 15.
Name two salts and write their formulae which possess water of crystallization.
Answer:

1. Hydrous copper sulphate. Its formula is CuSO₄ . 5H₂O
2. Gypsum. Its formula is CaSO₄ . 2H₂O

Question 16.
What is neutralization reaction? Give two examples.
Answer:
When an acid reacts with base it forms salt and water. This reaction is called neutralisation reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
H₂SO_4(aq) + 2Na0H_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Question 17.
All alkalis are bases but all bases are not alkalis. Do you agree with the statement? If yes, why?
Answer:
Yes, I agree with the statement. Because alkalis are those bases which are soluble in water. So all alkalis are bases but all bases are not alkalis.

Question 18.
Why are solutions of acids, bases and salts good conductors of electricity?
Answer:
For passage of electricity through a material or substance charged particles are required. In metals charged particles are electrons whereas in solutions ions are charged particles which carry electrical energy. Solutions of acids, bases and salts undergo ionisation and produce ions. So they are good conductors of electricity.

Question 19.
What is strength of acid ? What are the factors that influence strength of acid?
Answer:
The extent which an acid undergoes ionisation is called strength of acid.
Factors influence strength of acid :

1. Degree of ionisation.
2. Concentration of hydronium ions produced by acid.

Question 20.
Why are organic acids weak acids when compared with mineral acids?
Answer:

• Strength of acid depends on extent of ionisation.
• Organic acids do not undergo 100% ionisation. Their ionisation is less than 30%. There is equilibrium between ionised and unionised molecules whereas mineral acids undergo complete ionisation.
• So mineral acids behave like strong acids when compared with organic acids.

Question 21.
“Acids do not contain OH^– ions”. Do you agree with this statement? If not, why?
Answer:
No. I do not agree with the statement because all acids also contain OH^– ions but in acid solutions, H⁺ ions are more than OH^– whereas in bases OH^– ions are more than H⁺.

Question 22.

If B is calcium chloride, what are A, C, D and E?
Answer:
A is calcium carbonate or calcium hydrogen carbonate.
C is water and D is carbon dioxide.
E is calcium carbonate and F is water.

Question 23.
Some salts are given below. Classify them into hydrous and anhydrous salts. Sodium carbonate, Sodium chloride, Sodium hydrogen carbonate, Copper sulphate, Hypo, Magnesium Sulphate (epsum salt)
Answer:
Hydrous salts :

1. Hypo (Na₂S₂O₃ • 2H₂O)
2. Epsum (MgSO₄ • 7H₂O)
3. Copper sulphate (CuSO₄ • 5H₂O)

Anhydrous salts :

1. Sodium chloride (NaCl)
2. Sodium carbonate (Na₂CO₃)
3. Sodium hydrogen Carbonate (NaHCO₃)

Question 24.
If someone in the family is suffering from a problem of acidity, which of the following would you suggest as a remedy : lemon juice, vinegar or baking soda solution? Which property do you think of while suggesting the remedy?
Answer:

• I suggest baking soda solution. As acidity can be neutralized by baking soda solution, we can use it.
• Neutralizing property of baking soda solution.

Question 25.
Why are curd and sour substances not kept in copper vessels?
Answer:
Curd and sour substances contain acids which react with copper vessels and form poisonous substances. So curd and sour substances should not be kept in copper vessels.

Question 26.
Which gas is liberated when acids react with metals? Give one example.
Answer:
When acids react with active metals they release hydrogen gas.
Zn_(s) + 2HCl_(aq) → Zncl_2(aq) + H₂ ↑_(g)

Question 27.
Why should pickles not be stored in metallic containers?
Answer:
Pickles contain acids which react with metallic containers and form poisonous substance. So they are kept in plastic containers.

Question 28.
Solution x turned blue litmus red and Solution y turned red litmus blue.
a) What products could be formed when x and y are mixed?
b) Which gas is released when we put magnesium pieces in solution x?
c) Will any chemical reaction take place when zinc pieces are put in solution y?
d) Which of the above solutions contain more hydrogen ions?
Answer:
Given solution Y turned blue litmus into red so, Y is an acid.
Given solution ‘y’ turned red litmus into blue so, ‘y’ is a base.
a) The reaction of an acid (x) with a base to give a salt and water.
b) When we put magnesium pieces in solution releases hydrogen gas.
c) When zinc pieces are put in solution y, a chemical reaction will take place there.
d) Acids contain more H+ ions in the given solutions, Y has more H+ ions because it is an acid.

Question 29.
Acid should be added to water but not water to the acid. Why?
Answer:

• The dissolving of an acid or base in water is an highly exothermic process. Care must be taken while mixing concentrated HNO₃ or concentrated H₂SO₄ with water.
• The acid must always be added slowly to water with constant stirring.
• If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause bums.
• The glass container may also break due to excessive local heating.

Question 30.
Explain the procedure to confirm the given salt is a hydrous or anhydrous.
Answer:

1. Take given salt in a test tube
2. Observe the colour of salt
3. Heat the test tube gently
4. Observe the colour of salt and also moisture (droplets) inside of the test tube walls.
5. If its colour changes or forms water droplets, it is hydrous salt.
6. Otherwise, it is anhydrous salt.

Question 31.
Categorize the following as acids, bases, and salts :
Lemon juice, salt water, soap water, tamarind juice, surf water, lime water.
Answer:
Acids :

1. Lemon juice
2. Tamarind juice

Bases :

1. Soap water
2. Surf water
3. Lime water

Salts :

1. Salt water

Question 32.
Classify the following salts as family of salts having same cation or anion and prepare a table.
Potassium sulphate, Sodium sulphate, Calcium sulphate, Magnesium sulphate, Copper sulphate, Sodium chloride, Sodium nitrate, Sodium carbonate and Ammonium chloride.
Answer:

Question 33.
Observe the table and answer the following questions.

Solutions	pH value
Blood	7.3
Pure water	7.0
Gastric fluid	1.2
Sodium hydroxide	13

1) Which of the solutions among these is a strongest base?
2) Which body fluid has slightly basic nature?
3) What is the nature of pure water?
4) Which body fluid is strongest acid?
Answer:

1. Sodium hydroxide because its pH is 13.
2. Blood because its pH is 7.3.
3. Pure water is neutral in nature because its pH is 7.
4. Gastric juice because its pH is 1.2.

Question 34.
The diagram given below shows the removal of water crystallisation. Find error in the diagram.

Answer:
Error in the diagram

1. Test tube is placed towards observer. It causes bums on his hands.
2. So, test tube should be placed away from the observer.

Question 35.
What are the uses of Plaster of Paris?
Answer:

• The substance which doctors use as plaster for supporting fractured bones in the right position.
• Plaster of Paris is used for making toys, materials for decoration and for making surfaces smooth.

Question 36.
What are the applications of pH in daily life?
Answer:
1) In medical science :
The pH values of urine and blood are taken for diagnosis of various diseases.

2) In dairies :
Milk has pH of 6.6. A change in the pH of milk indicates that milk has turned sour.

3) In agriculture :
For better growth of crops the pH of the soil is regularly tested.
For examples :

1. Citrus fruits require slightly alkaline soil.
2. Rice requires acidic medium.
3. Sugarcane requires neutral soil.

4) In technology:
Organic and biochemical reactions are carried out under control pH.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 4 Marks Important Questions and Answers

Question 1.
Draw a neat diagram showing a base solution in water conducts electricity. Why the solution of sugar/glucose in water do not conduct electricity? (AP March 2017)
Answer:

The solution of sugar/glucose in water do not conduct electricity because there is no H⁺ ions in the solution.

Question 2.
Explain an activity to show the water of crystallisation in CuSO₄ • 5H₂O. (AP June 2018)
Answer:

• Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
• We observe water droplets on the walls of the test tube and salt turns white.
• Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
• We observe the blue colour of copper sulphate crystals is restored.

Question 3.
Read the information given in the table and answer the following questions. (TS March 2016)

a) List out the acids in the above table.
Answer:
The acids are HCl and lemon juice.

b) What is the nature of the solution which gives pink colour with Phenolphthalene solution?
Answer:
The nature of the solution which turns pink colour with phenolphthalene solution is basic.

c) List out the neutral solutions in the above table.
Answer:
The neutral solutions are distilled water and NaCl.

d) Name the strongest acid and the strongest base among the given solutions.
Answer:
The strongest acid is HCl and the strongest base is NaOH.

Question 4.
Observe the following table and answer the questions given below. (TS June 2o17)
The table contains the aqueous solutions of different substances with the same concentrations and their respective pH values.

i) Which one of the above acid solutions is the weakest acid? Give a reason.
Answer:
Weakest acid is ‘C’. Because its pH value is less than 7 and it is nearer to 7.

ii) Which one of the above solutions is the strongest base? Give a reason.
Answer:
Strongest base is ‘D’. Because it’s pH value is near to 14.

iii) Which of the above two produce maximum heat when they react ? What does that heat energy called?
Answer:
B, D produce maximum heat when they react. This heat energy is known as neutralization energy.

iv) Which one of the above solutions has the pH equal to that of the distilled water? What is the name given to solutions of that pH value?
Answer:
‘G’ has the pH equal to that of the distilled water. These type of solutions are known as neutral solution.

Question 5.
List out the materials required to test whether the solutions of given acids and bases contain ions or not. Explain the procedure of the experiment. (TS March 2017)
Answer:
Required Materials :
Beaker, Bulb, Graphite rods, connecting wires, 230 V AC current, water, different acids, bases.

Experimental Procedure :

1. Connect the two connecting wires to the graphite rods.
2. Keep the graphite rods into the beaker, take care that two graphite rods do not touch each other.
3. Arrange a bulb in the circuit.
4. Pour dilute acid into the beaker.
5. Connect the ends of the connectors to 230 V AC.
6. In this way, change the acid / base and do the experiment.

The bulb glows in the experiment when the beaker contains acid or base. Hence, when the bulb glows we can say that acid or base contain ions.

Question 6.
List out the material for the experiment “when Hydrochloric acid reacts with NaHCO₃ and evolves CO₂“. Write the experimental procedure. (TS March 2018)
Answer:
Required material : Stand, test tubes, delivery tube, thistle funnel, two hole rubber corks, Ca(OH)₂, NaHCO₃, HCl.

Experimental procedure :

1. Take NaHCO₃ in a test tube and fix two hole cork to the test tube.
2. Fix thistle funnel in one hole of cork and insert delivery tube in the second hole of the cork. Insert the second end of the delivery tube in the other test tube which is containing Ca(OH)₂/lime water.
3. Pour dil. HCl into the test tube using thistle funnel.
4. Due to chemical reaction, gas is evolved and pass into the Ca(OH)₂ through delivery tube. It turns in to milky. We can conclude that it is CO₂ gas.

Question 7.
Prepare a table based on the colour responses of acid, base and salt with indicators such as indicators. (AP SA-I:2018-19)
Answer:

Question 8.
Draw universal pH value indicator and identify different substances. (AP SA-I : 2019-20)
(OR)
Draw a neat diagram showing variation of pH with the chage in concentration of H⁺_(aq) ion and OH^–_(aq) ions.
Answer:

Variation of pH with the change in concentration of H⁺_(aq) ions and OH^–_(aq) ions.

Question 9.

Answer the following questions by using above information. (TS June 2019)
1) Which of the above is neutral solution?
2) Which of the above is used to neutralize the acidity in stomach?
3) Which is the strong acid among the above solutions?
4) What is the colour of Phenolphthalein indicator in NaOH solution?
Answer:

1. Distilled water
2. Milk of Magnesia
3. Gastric juice
4. Pink

Question 10.
If the pH values of solutions X, Y and Z are 13, 6 and 2 respectively, then
a) Which solution is a strong acid? Why?
b) Which solution contains ions along with molecules of solution?
c) Which solution is a strong base? Why?
d) Does the pH value of a solution increase or decrease when a base is added to it? Why?
Answer:
The strength of an acid (or) an alkali can be tested by using pH value of a solution. If the value of a pH of a solution is less, then that solution exhibits acidic nature.

If the value of a pH of a solution is more, then that solution exhibits basic nature.
pH value of a solution “X” is 13
pH value of a solution “Y” is 6
pH value of a solution “Z” is 2

a) Solution ‘Z’ is strong acid because its pH is 2.
b) Among given solutions, solution X is weakest acid. Weak solution contains ions along with molecules of solution. So X exhibits like this character.
c) Solution X is strong base. Because its pH is 13.
d) If base is added to solution ‘Z’, then its pH will increase.

Question 11.
Distinguish between acids and bases.
Answer:

Acids	Bases
1) They are sour to taste.	1) They are better to taste and soapy to touch.
2) When non-metallic oxides dissolved in water they form acids.	2) When metallic oxides dissolved in water they form bases.
3) They react with bases to form salt and water.	3) They react with acids to form salt and water.
4) They produce aqueous H+ ions.	4) They produce aqueous OH- ions.
5) They turn blue litmus into red.	5) They turn red litmus into blue.
6) They turn methyl orange indicator to red.	6) They turn methyl orange indicator to yellow.
7) The turn phenolphthalein indicator to colourless.	7) They turn phenolphthalien indicator to pink.

Question 12.
Explain chlor-alkali process.
Answer:
When electricity is passed through an aqueous solution of sodium chloride, it decomposes to form sodium hydroxide. The process is called chloralkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide.
2 NaCl (aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Chlorine gas is given off at the anode and hydrogen gas at the cathode and sodium hydroxide is formed near the cathode.

Question 13.
Define the following. Give one example for each.
a) Strong acid
b) Strong base
c) Weak acid
d) Weak base.
Answer:
a) Strong acid :
The acid which undergoes 100% ionisation is called strong acid.
e.g.: HCl, H₂SO₄

b) Strong base :
The base which undergoes 100% ionisation is called strong base.
e.g.: NaOH, KOH

c) Weak acid:
The acid which undergoes less than 100% ionisation is called weak acid.
e.g.: CH₃COOH, H₂CO₃

d) Weak base:
The base which undergoes less than 100% ionisation is called weak base.
e.g.: NH₄OH, Mg(OH)₂

Question 14.
Write any four chemical properties of acids.
Answer:
Chemical properties of acids :
1) Active metals react with acids and liberate hydrogen gas.
Zn + HCl → ZnCl₂ + H₂ ↑

2) Acids react with bases to form salt and water.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

3) Acids react with metallic oxides to form salt and water.
MgO_(s)+ 2HCl_(aq) → MgCl_2(aq) + H₂O_(l)

4) Acids react with carbonates and hydrogen carbonates and release carbon dioxide gas.
CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq)+ H₂O_(l) + CO_2(g) ↑
Ca(HCO₃)_2(l) + 2HCl_(aq) → CaCl_2(aq) + 2H20((| + 2CO_2(g) ↑

Question 15.
Write the formulae of the following salts.
a) Sodium sulphate
b) Ammonium chloride
Identify the acids and bases for which the above salts are obtained. Also write chemical equations for the reactions between such acids and bases. Which type of chemical reactions are they?
Answer:
a) Formula of sodium sulphate is Na2S04. When sulphuric acid reacts with sodium hydroxide it forms sodium sulphate.
H₂SO_4(aq)+ 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

b) Formula of Ammonium chloride is NH4C/. When Ammonium hydroxide reacts with hydrochloric acid it forms Ammonium chloride.
NH₄OH_(aq) + HCl_(aq) → NH₄Cl_(aq) + H₂O_(l)

Question 16.
Write balanced equations to satisfy each statement,
a) Acid + Active metal → Salt + Hydrogen
Answer:
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

b) Acid + Base → Salt + Water
Answer:
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

c) Acid + Carbonate / Hydrogen carbonate → Salt + Water + Carbon dioxide
Answer:
CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) ↑
NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g) ↑

d) Metal oxide + Acid → Salt + Water
Answer:
CaO_(s) + 2 HCl_(aq) → CaCl_2(aq) + H₍₂₎O_(l)

e) Non metal oxide + base → Salt + Water
Answer:
CO_2(g) + 2 NaOH_(aq) → Na₂CO_3(aq) + H₂O_(l)

Question 17.
Give important products obtained from chloralkali process.

Answer:
Result:
Five water molecules are present in one formula unit of copper sulphate. Water of crystallization proves that the crystals contain a fixed quantity of water in them.

Question 18.
Give the equations for the preparation of each of the following.
i) Copper sulphate from copper (II) oxide.
Answer:
CuO + H₂SO₄ → CuSO₄ + H₂O

ii) Potassium sulphate from potassium hydroxide solution.
Answer:
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

iii) Lead chloride from lead carbonate.
Answer:
PbCO₃ + 2 HCl → PbCl₂ + H₂O + CO₂

Question 19.
How are the following salts prepared?
1) Calcium sulphate from calcium carbonate
Answer:
When calcium carbonate is treated with sulphuric acid it forms calcium sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

2) Lead carbonate from lead nitrate
Answer:
When lead nitrate is treated with carbonic acid we will get lead carbonate.
Pb(NO₃)₂ + H2CO₃ → PbCO₃ + 2 HNO₃

3) Sodium nitrate from sodium hydroxide
Answer:
When sodium hydroxide is reacted with nitric acid it will form sodium nitrate.
NaOH + HNO₃ → NaNO₃ + H₂O

4) Magnesium carbonate from magnesium chloride
Answer:
When magnesium carbonate is reacted with hydrochloric acid it forms magnesium chloride.
MgCO₃ + 2 HCl → MgCl₂ + H₂O + CO₂

Question 20.
Which of the following reactions are considered as neutralization reactions? Why?
1) NaOH + HCl → NaCl + H₂O
2) CaO + 2 HCl → CaCl₂ + H₂O
3) CO₂ + 2 NaOH → Na₂CO₃ + H₂O
4) SiO₂ + CaO → CaSiO₃
Answer:
All of them are considered as neutralization reactions.

1. An acid (HCl) reacts with base (NaOH) and forms salt and water. So it is a neutralization reaction.
2. Here metallic oxide which is basic in nature reacts with acid and forms salt and water. So it is also a neutralization reaction.
3. In third case non-metallic oxide (acidic oxide) reacts with base (NaOH) and forms salt and water. So it is also a neutralization reaction.
4. In fourth case a metallic oxide (CaO) reacts with non-metallic oxide (SiO₂) and forms salt. So it is also a neutralization reaction in the absence 6f water.

Question 21.
Which metals produce hydrogen gas when they are reacted with bases like NaOH and KOH? Write the chemical equations for the reactions.
Answer:
Zinc, aluminium and lead react with bases like NaOH and KOH and produce hydrogen gas.

Question 22.
i) A solution has a pH of 7. How would you increase its pH and decrease its pH? Explain.
Answer;
We can increase the pH of a solution by adding base because we know that bases have pH > 7. We can decrease the pH of a solution by adding an acid because acidic solution have pH < 7.

ii) If a solution changes the colour of litmus from red to blue, then what can you say about its pH?
Answer:
Bases can change red litmus into blue. So the pH of the solution is greater than 7.

iii) What can you say about pH of a solution that liberates carbon dioxide from sodium carbonate?
Answer:
Acids react with carbonates and liberate hydrogen gas. So the pH of the solution is less than 7.

Question 23.
Write the pH values of some solutions.
Answer:

pH value	Solutions
0	Battery Acid
1	Con. H2S04
2	Lemon juice
3	Orange juice
4	Tomato juice
5	Black coffee, Bananas
6	Milk, urine
7	Pure water
8	Sea water, eggs
9	Baking soda
10	Milk of magnesia
11	Ammonia solution
12	Soapy water
13	Bleach oven cleaner
14	Liquid drain cleaner

Question 24.
Fill the following table of results of reactions between some substances (acids, bases, neutral substances) and indicators.

Answer:

Question 25.
The pH values of six solutions A, B, C, D, E, F are given as 5,2,1,3,7 and 9 respectively. Which solution is
a) Neutral
b) Strongly alkaline
c) Strongly acidic
d) Weakly acidic?
Arrange the pH in increasing order of Hydrogen ion concentration.
Answer:
a) Solution E is neutral.
b) Solution F is Alkaline.
c) Solution C is strongly acidic.
d) Solution A is weakly acidic.
e) Solution B is strongly acidic.
f) Solution D is strongly acidic.
g) Ascending order of increase of Hydrogen ion concentration is F, E, A, D, B, C.

Question 26.
Collect information about various organic acids different occurring naturally and prepare a table.
Answer:

1. Acetic acid	Vinegar (obtained from fruits after fermentation).
2. Citric acid	Citrus fruits like orange and lemons.
3. Butyric acid	Butter gone bad or rancid
4. Lactic acid	Curd
5. Malic acid	Apples
6. Oleic acid	Olive oil
7. Tartaric acid	Fruits such as grapes, apples and tamarind
8. Stearic acid	From fats
9. Succinic acid	From vegetables like lettuce and unripe fruits
10. Uric acid	From urine

Question 27.
Complete the table.

Answer:

Question 28.
Fill the table.

Answer:

Question 29.
Draw a diagram to show the reaction of acids with metals.
Answer:

Reaction of Zinc granules with dil. HCl and testing hydrogen gas by a burning candle

Question 30.
Draw a diagram to show that all metal carbonates and react hydrogen carbonates
Answer:

Question 31.
What are the uses of Bleaching powder?
Answer:

• It is used for bleaching cotton and linen in the textile industry for bleaching wood pulp in paper industry and for bleaching washed clothes in laundry.
• Used as an oxidizing agent in many chemical industries.
• Used for disinfecting drinking water to make it free of germs.
• Used as a reagent in the preparation of chloroform.

Question 32.
What are the uses of Baking soda?
Answer:
1) Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place.

Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

2) Sodium hydrogen carbonate is also an ingredient in antacids. Being alkaline, it neutralizes excess acid in the stomach and provides relief.

3) It is also used in soda-acid, fire extinguishers.

4) It acts as mild antiseptic.

Question 33.
What are the uses of Washing soda?
Answer:

• Sodium carbonate (washing soda) is used in glass, soaps and paper industries.
• It is used in the manufacture of sodium compounds such as borax.
• Sodium carbonate can be used” as a cleaning agent for domestic purposes.
• It is used for removing permanent hardness of water.

Question 34.
Write the chemical formulae of the following :
i) Bleaching powder
ii) Sodium Chloride
iii) Slaked lime
iv) Baking Soda
v) Washing Soda
vi) Gypsum
vii) Plaster of Paris
viii) Acetic acid
ix) Sodium Hydroxide
x) Limestone
Answer:
i) Bleaching powder = CaOCl₂
ii) Sodium Chloride = NaCl (Common Salt)
iii) Slaked lime (or) lime water = Ca(OH)₂
iv) Baking Soda = NaHCO₃
v) Washing Soda = Na₂CO₃, 10H₂O
vi) Gypsum = CaSO₄ . 2H₂O
vii) Plaster of Paris = CaSO₄ . ½H₂O
viii) Acetic acid = CH₃COOH
ix) Sodium Hydroxide = NaOH
x) Limestone = CaCO₃

Question 35.
What are the various applications of neutralization?
Answer:

• The acidity of soil is reduced by adding slaked lime.
• The sting of yellow wasps contains alkalis. If acetic acid is rubbed on affected area, they are neutralized.
• Ants and bees have formic acid in their stings which can be neutralised by applying soap and some other alkali.
• Antacids tablets contain magnesium hydroxide, persons suffering from acidity are administered these tablets.
• The affect of nettle plant leaves is neutralized by leaves of dock plant.

Question 36.
Mention two situations where you use hydrated and unhydrated salts in your daily life.
Answer:

• NaCl is unhydrated salt. It flows freely when filled in a container.
• NaHCl (Baking soda) is unhydrated salt. It flows freely when filled in a container.
• NaCO₃ . 10H₂O (washing soda) is hydrated salt. It leaves wetness inside the container.
• CaSO₄ . 2H₂O (Gypsum) is also hydrated salt. On careful heating of gypsum it loses water molecules partially to become (CaSO₄ . ½H₂O) P.O.P. It is used in hospitals as plaster for supporting fractured bones in right position.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 12 Electromagnetism.

AP State Syllabus SSC 10th Class Physics Important Questions 12th Electromagnetism

10th Class Physics 12th Lesson Electromagnetism 1 Mark Important Questions and Answers

Question 1.
Draw the diagram showing the magnetic field lines of bar mannet. (TS June 2016)
Answer:

Question 2.
Correct the diagram according to Lenz law and draw it again. (TS March 2017)
Answer:

Question 3.
What is the use of slip – ring in AC motor? (TS June 2018)
Answer:
Uses of slip rings :
Slip rings are used to change the direction of current in the coil continuously.

Question 4.
Draw the direction of magnetic lines force, assuming that the current is flowing into the page. (AP SCERT: 2019-20 )
Answer:

Question 5.
What happens when a current carrying coil is placed in a uniform magnetic field? (TS June2019)
Answer:
The rectangular coil comes into rotation in clock – wise direction because of equal and opposite pair of forces acting on the two sides of the coil.

1. If the direction of the current in the coil unchanged it rotates half clock – wise and comes to half and rotates in anticlock – wise direction.
2. If the direction of the current in the coil changed after the first half rotation, the coil continuously rotates in a same direction.

Question 6.
Write the name of the device that converts mechanical energy into electrical energy. (AP June 2019)
Answer:
Generator.

Question 7.
Name some sources of direct current.
Answer:
Dry cell, lead-acid battery.

Question 8.
Which sources produce alternating current?
Answer:
A.C generator, thermal power station, hydroelectric stations.

Question 9.
What is the role of split ring in an electric motor?
Answer:
The split rings are used to change the direction of current flowing through the coil.

Question 10.
Write one method of inducing current in the coil.
ANswer:
By pushing or pulling a bar magnet into or away from the coil we can induce current. Name two safety measures commonly used in electric circuit, i) Fuse ii) Earthing

Question 11.
On what factors does the magnetic induction at the centre of the coil depend?
Answer:
It depends on current, number of turns and radius of the coil.

Question 12.
Which is more dangerous AC or DC?
Answer:
AC is more dangerous.

Question 13.
State two serious hazards of electricity.
Answer:

• If a person touches the live wire, he gets severe shock which may prove fatal.
• Short-circuiting can cause a spark which may lead to fire in a building.

Question 14.
Why is earthing of electrical appliances recommended?
Answer:
To protect the user from any accidental electrical shock caused due to leakage of current.

Question 15.
Why is a spark produced at the place of short circuit? Why is the spark in white colour?
The resistance of circuit decreases, and a sudden flow of large current heats up the live wire and vapourises the metal. This causes spark. The metal of wire becomes very hot and naturally emits white light.

Question 16.
What is electromagnetic induction?
Answer:
Mechanical energy can be converted into electrical energy by moving a magnet inside a coil.

Question 17.
What is Maxwell’s right hand screw rule?
Answer:
The direction of current is the direction in which the tip of the screw advances and direction of ration of the screw gives the direction of magnetic lines of force.

Question 18.
What type of energy transformation takes place in electric generator?
Answer:
Electrical energy from mechanical energy.

Question 19.
Where are the electromagnets used?
Answer:
In electric generators and televisions.

Question 20.
What is electromagnet?
Answer:
When current carrying conductor is wound over a magnetic material like soft iron it gets magnetized.

Question 21.
What are the different types of power stations?
Answer:
Electrical energy is produced in different power stations from mechanical energy of water, meat energy, and nuclear energy.

Question 22.
If the current in the coil is in anti-clockwise, then what would be the face of the coil?
Answer:
It behayes as north pole.

Question 23.
If the current in the ceil is in clockwise, then what would be the face of the coil?
Answer:
It behaves as south pole.

Question 24.
What is the frequency of the A.C. supplied in your house?
Answer:
It is approximately 50 Hz.

Question 25.
What type of current is generated in electric power station?
Answer:
Alternating current.

Question 26.
What is the shape of magnetic lines due to straight current carrying conductor?
Answer:
They are concentric circles.

Question 27.
What is a transformer?
Answer:
It is a device which increases or decreases the voltage.

Question 28.
State two ways by which speed or rotation of electric motor can be increased.
Answer:

1. By increasing strength of the current.
2. By increasing number of turns in the coil.

Question 29.
What happens if an iron piece is dropped between two poles of strong magnet?
Answer:
Eddy current is produced in it. These eddy currents oppose the motion of the piece of iron. So it falls as it is moving through a viscous liquid.

Question 30.
If a copper rod carries a direct current, then where will be the magnetic field in the conductor?
Answer:
It will be both inside and outside the rod.

Question 31.
In what form is the energy in a current carrying coil stored?
Answer:
It is stored in the form of magnetic field.

Question 32.
What is solenoid?
Answer:
A solenoid is a long wire wound in a close packed helix.

Question 33.
What is the pattern of field lines inside a solenoid around when current carrying solenoid?
Answer:
Parallel to each other.

Question 34.
List any two properties of magnetic field lines.
Answer:

• Inside the magnet they start from south pole and end at north pole whereas outside the magnet they start at north pole and end at south pole.
• Two magnetic lines of force never intersect each other.

Question 35.
Why does the picture appear distorted when the bar magnet is brought close to the screen of a television?
Answer:
Picture on a television screen is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet.

Question 36.
What is meant by electromagnetic induction?
Answer:
Whenever there is continuous change of magnetic flux linked with a closed coil, current generated in the coil is called electromagnetic induction.

Question 37.
State the Lenz’s principle.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 38.
What is induced emf?
Answer:
Change in magnetic flux produces emf in the circuit called induced emf.

Question 39.
What do you mean by magnetic effect of current?
Answer:
Current carrying conductor produces a magnetic field around it. This is called magnetic effect of current.

Question 40.
What is the direction of magnetic field at the centre of the coil carrying current in (i) clockwise, (ii) anti-clockwise direction?
Answer:
i) Along the axis of coil inwards.
ii) Along the axis of coil outwards.

Question 41.
Why does a current carrying freely suspended solenoid rest along a particular direction?
Answer:
A current carrying solenoid behaves like a bar magnet.

Question 42.
What effect will there be on a magnetic compass when it is brought near a current carrying solenoid?
Answer:
The needle of the compass will rest in the direction of magnetic field due to solenoid at that point.

Question 43.
How is magnetic field due to solenoid carrying current affected, if a soft iron bar is introduced inside the solenoid?
Answer:
The magnetic field increases when iron bar is introduced inside the solenoid.

Question 44.
What happens to magnetic field if we reverse the current direction?
Answer:
The magnetic field also gets reversed.

Question 45.
How do magnetic field lines inside a current carrying solenoid appear?
Answer:
They form along the axis and parallel to each other.

Question 46.
In which of the following cases does the electromagnetic induction occur?
i) A current is started in a wire held near a loop of wire.
ii) The current is stopped in a wire held near a loop of wire.
iii) A magnet is moved through a loop of wire.
iv) A loop of wire is held near a magnet.
Answer:
In first three cases there is a change in magnetic flux. So electromagnetic induction occurs in first three cases.

Question 47.
Why must an induced current flow in such a direction so as to oppose the change producing it?
Answer:
So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Question 48.
What is the maximum force acting on the current (i) carrying conductor of length (l) in the presence of magnetic field (B)?
Answer:
F = Bil

Question 49.
A charged particle q is moving with a speed v perpendicular to the magnetic field of induction B?
Find the equation of time period of the particle.
Answer:
\(\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{Bq}}\)

Question 50.
Name two safety measures commonly used in electric circuit.
Answer:
a) Fuse
b) Earthing

Question 51.
On what factors the magnetic induction at the centre of coil depends?
Answer:
a) Number of turns
b) Radius of the coil.

Question 52.
Write the formula for magnetic flux passing through an Area A with an angle θ.
Answer:
Flux ΦV= BA cos θ

Question 53.
Write the Lenz’s law.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 54.
What is the difference between AC and DC generator?
Answer:
In AC generator, ends of coil are connected to two slip rings.
In DC generator ends of coil are connected to two half split rings.

Question 55.
What are the uses of electromagnet?
Answer:
It is used in electric bells, electric motors, telephone diaphragms, etc.

Question 56.
What is the principle of Electric motor?
Answer:
When a rectangular coil is placed in magnetic field and current is passed through it, two equal and opposite forces act on the coil which rotates it continuously.

Question 57.
What factors are influence the speed of rotation of the motor?
Answer:

1. Strength of current
2. Number of turns
3. Area of the coil
4. Strength of magnetic field

Question 58.
Which two physical quantities are interrelated in Oerstead experiment?
Answer:
Electricity and magnetism are interrelated.

Question 59.
Which property of proton can change while it moves freely in a magnetic field?
Answer:
When a proton (the charge) moves in a magnetic field, then magnetic force is acting on proton. So its momentum changes.

Question 60.
Which type of conductors are producing magnetic field?
Answer:

1. Long straight current carrying conductor.
2. Circular loop.
3. Solenoid.

Question 61.
How much force acting on a neutron particle is moving with velocity V in a mag¬netic field with induction B?
Answer:
Zero, because neutron is charge less particle.

Question 62.
What are the instruments used in A.C Generator?
Answer:
Rectangular coil, brushes, slip rings, and magnetic poles.

Question 63.
What are the ways to produce the induced current in a coil?
Answer:
When a magnet is moved towards or away from coil or there is a relative motion between coil and magnet a current is induced in the coil.

Question 64.
At the time of short circuit, what happens to the current?
Answer:
At the time of short circuit, the current in the circuit increases heavily becomes the resistance of the conductor becomes almost zero.

Question 65.
A wire with green insulation is usually the line wire of an electric supply. Is it true?
Answer:
It is false, the wire with green insulation is the earth wire, not the line wire.

Question 66.
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:
Yes. current will be induced in coil ‘B’, because flux linked the coil ‘B’ changes with respect to time.

Question 67.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Electric fuse and Miniature Circuit Breaker (MCB).

Question 68.
An attemating current has frequency of 50Hz. How many times does it change its direction in one second?
Answer:
Frequency of AC = 50 Hz ⇒ 50 cycles in one sec. So it reverses its direction 100 times in one second.

Question 69.
Under what orientation, the induced current produced in moving conductor in a magnetic field can be maximum?
Answer:
The current induced in a conductor is maximum when direction of motion of conductor is at right angle to the magnetic field.

Question 70.
How could you make the coil rotate continuously in motor?
Answer:
The direction of current through the coil is reversed every half rotation, the coil will rotate continuously and the same direction.

Question 71.
What is the formula for induced cmf when change the magnetic flux?
Answer:
Induced EMF = \(-\frac{\Delta \phi}{\Delta t}\)

Question 72.
The magnetic flux is varying with time. Which cases E.M.F is induced?

Answer:
In OA and BC cases, E.M.F is induced.

Question 73.
In above problem, how much EMF is induced in BC curve?
Answer:

Question 74.
What is the equation of motional E.M.F?
Answer:
Motional E.M.F = Blv;
where B = magnetic induction,
l = length of rod,
v = velocity of rod.

Question 75.
When a magnet and a coil are moving same direction with same speed. Then induced E.M.F in coil is zero. Why?
Answer:
E.M.F = \(\frac{-\Delta \phi}{\Delta \mathrm{t}}\), but both moving same direction, so change in flux linked with coil is zero i.e., ∆Φ = 0

Question 76.
What is shape of conductor is drawn when current is passing through conductor?
Answer:

Question 77.
Draw the symbols of North and South pole when depends on current.
Answer:

Question 78.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

Question 79.
Draw the diagram of AC and DC current varying with time.
Answer:

Question 80.
Where are electric motors used?
Answer:
Electric fans, water-pumps, coolers.

Question 81.
Mention two uses of solenoid.
Answer:
It is used in electric bells, fans, and motors.

Question 82.
Mention applications of electromagnetic induction.
Answer:
It is used in devices which convert mechanical energy into electrical energy.

Question 83.
What are the advantages of an electromagnet over permanent magnet?
Answer:

1. An electromagnet can produce a strong magnetic field.
2. The strength of electromagnet can be changed.
3. The polarity of electromagnet can be changed.

10th Class Physics 12th Lesson Electromagnetism 2 Marks Important Questions and Answers

Question 1.
Anand appreciated the law behind the making of ‘generator’. Name the law and state it. (AP June 2017)
Answer:
1) The law behind the making of ‘generator’ is Faraday’s law.

2) Faraday’s Law :
“Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Question 2.
Explain Oersted experiment to show that Electricity and Magnetism were related phenomena. (AP June 2018)
Answer:

• Place a compass needle underneath a wire and then turn on electric current.
• Immediately the needle of compass shows the deflection. By this we can conclude that electricity and magnetism are related phenomena.

Question 3.
With the help of the given figure, the teacher explained that magnetic field lines are closed lines and not open lines. Write the questions which you will ask to rest whether the given statement is right or wrong. (TS June 2015)

Answer:

• Are there any magnetic field lines inside the magnet?
• If magnetic field lines are there inside the magnet, what is the direction of field lines inside the magnet?
• What is the direction of field lines outside the magnet?

Question 4.
State Right-hand rule with a labelled diagram. (TS March 2015)
Answer:

(OR)

• Right hand rule indicates the direction of magnetic force acting on a moving charge.
• It is used when velocity and field are perpendicular to each other. If the fore finger points towards the direction of velocity of charge or current (I), middle finger points to the direction of field (B), then thumb gives the direction of force (F).

Question 5.
A coil of insulated Copper wire is connected to a Galvanometer. (TS March 2015)
What happens, if a bar magnet is ………….
1) pushed into the coil?
2) withdrawn from inside the coil?
3) held stationary inside the coil?
Answer:

1. If a bar magnet is pushed into the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
2. If a bar magnet is withdrawn from inside the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
3. If a bar magnet is held stationary inside the coil, then the needle in Galvanometer does not get deflected. Because current is not generated in the coil.

Question 6.
Compare the magnetic field lines of force formed around a current carrying solenoid with the magnetic field lines of force of a bar magnet.
Answer:

Magnetic field lines of a bar magnet	Magnetic field lines of a solenoid
2) The direction of the field lines of the outside the magnet is from north to south pole.	2) The direction of the field lines of the outside the solenoid is from north to south pole.
3) The direction of the field lines of the inside of the magnet field looks like south to north pole.	3) The direction of the field lines of the inside of the solenoid is from south to north pole.
4) These lines are closed loops.	4) These lines are also closed loops.
5) We cannot find the field lines inside the bar magnet.	5) We can find the field lines inside the solenoid.
6) These field lines are same as field lines formed by a solenoid.	6) These field lines are also same as field lines formed by a bar magnet.
7) More field lines are found at poles.	7) More field lines are found at poles.

Question 7.
Which energy we get from an electric motor? Write two daily life applications of the electric motor. (TS June 2017)
Answer:

• We get mechanical energy from electric motor.
• In our daily life we use motor in
  i) Mixies
  ii) Grinders
  iii) Water Pumps
  iv) Fans / Coolers, etc.

Question 8.
List out the material required for Oersted experiment and mention the precautions to be taken in the experiment. (TS March 2019)
Answer:
Materials required for Oersted experiment are :

1. Thermocol sheet.
2. Wooden sticks.
3. Copper wire of 24 gauge.
4. Battery.
5. Key.
6. Magnetic compass.

Precautions to be taken are :

1. Copper Wire is made through the slits of the wooden sticks tightly.
2. Arrange/complete the circuit correctly.

Question 9.
What happens when magnetic flux passing through a coil changes continuously? Where does this process is used? (TS June 2019)
Answer:
1) When there is a continuous change of magnetic flux passing through a coil a current is generated in the coil.

2) This process is used in

1. Induction stoves
2. Security checking entrance/exit doors.
3. ATM cards scanners.

Question 10.
Distinguish between AC and DC.
Answer:

AC	DC
1) AC means alternate current.	1) DC means direct current.
2) The current direction changes	2) The current direction does not
always.	change. It is always a single direction.
3) The magnitude of current changes from minimum to maximum.	3) Always it has maximum value.

Question 11.
Distinguish between AC motor and DC motor.
Answer:

AC motor	DC motor
1) It works with alternate current.	1) It works with direct current.
2) It does not require change in current direction.	2) It requires change in current direction which is provided by split rings which act as commutator.

Question 12.
Distinguish between AC generator and DC generator.
Answer:

AC generator	DC generator
1) It generates alternate current.	1) It generates direct current.
2) It has two slip rings.	2) It has two half slip rings.

Question 13.
What is Faraday’s law of electromagnetic induction? Write its expression.
(OR)
State the Faraday’s law of electromagnetic induction. Write the equation of this law.
Answer:
The induced EMF generated in a closed loop is equal to the rate of change of magnetic flux passing through it.
Induced EMF = Change in flux / time
ε = ∆Φ / ∆t

Let Φ₀ be flux linked with single turn. If there are N turns of the coil, the flux linked with the coil is NΦ₀.
∴ ε = NAΦ₀ / ∆t

Question 14.
State the right-hand thumb rule. How does the rule help us?
Answer:
When you curl your right hand fingers in the direction of current thumb gives the direction of magnetic field.

It is useful to find the magnetic field direction as well as current direction.

Question 15.
A flat rectangular coil is rotated between the pole pieces of a horseshoe magnet. In which position of coil with respect to magnetic field, will the emf (i) be maximum, (ii) be zero, (iii) change its direction.
Answer:
i) The emf is maximum when the plane of coil is parallel to the magnetic field.
ii) The emf is zero when the plane of coil is normal to the magnetic field.
iii) The emf will change its direction when the plane of coil passes from the position normal to the magnetic field.

Question 16.
State two factors on which the magnitude of induced emf depend?
Answer:
The magnitude of induced emf depends on the following two factors.

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes i.e., the rate of change of magnetic flux.

More the change in magnetic flux, more is the emf induced. Further if more rapid the magnetic flux changes, more is the emf induced.

Question 17.
How do you increase the magnetic field of solenoid?
Answer:
The magnetic field of solenoid can be increased by the following two ways.

1. by increasing the number of turns of winding in the solenoid.
2. by increasing the current through the solenoid.

Question 18.
State the function of split ring in a DC motor.
Answer:

• The split ring acts as a commutator in a DC motor.
• With the split ring, the direction of current through the coil is reversed after every half rotation of coil.
• Thus the direction of couple rotating the coil remains unchanged and the coil continues to rotate in the same direction.

Question 19.
A DC motor is rotating in a clockwise direction. How can the direction of rotation be reversed?
Answer:
The direction of rotation of motor can be reversed by interchanging the connections at the terminals of the battery joined to the brushes of the motor.

Question 20.
State the effect of inserting a soft iron core within the coil of DC motor.
Answer:

• On inserting a soft iron core within the coil of a DC motor, the speed of rotation of coil increases.
• The reason is that the strength of magnetic field between the pole pieces of magnet increases due to which the deflecting couple on coil increases.

Question 21.
State condition when magnitude of force on a current-carrying conductor placed in a magnetic field is (a) zero, (b) maximum.
Answer:
a) When current in conductor is in direction of magnetic field,
b) When current in conductor is normal to the magnetic field.

Question 22.
A flat coil ABCD Is freely suspended between the pole pieces of a U – shaped permanent magnet with the plane of coil parallel to the magnetic field.
a) What happens when current is passed in the coil?
b) When will the coil come to rest?
c) Name the instrument which makes use of the principle stated above.
Answer:
a) The coil will experience a torque due to which it will rotate.
b) The coil will come to rest when its plane becomes normal to the magnetic field.
c) DC motor.

Question 23.
Why is it more difficult to move a magnet towards a coil which has a large number of turns?
Answer:

1. Emf induced in the coil becomes more when the number of turns in the coil are made large.
2. So it is difficult to move a magnet towards a coil which has a large number of turns.

Question 24.
A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.

a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil.
b) Why is commutator necessary for continuous rotation of the coil?
Answer:
a) Anti-clockwise direction.
b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

Question 25.
Draw the magnetic field lines in uniform magnetic field.
Answer:

Question 26.
Draw a labelled diagram to make an electromagnet from soft iron bar AB. Mark the polarity at its ends. What precaution would you observe?
Answer:

The labelled diagram is shown in figure. The polarity at the end A where the current is clockwise, is south (S), while at the end B where the current is anti-clockwise is north (N).
Precaution :
The source of current must be the DC source.

Question 27.
State the principle of an electric motor. Name some appliances in which the Electric motor is used.
Answer:
Current carrying coil rotates when it is kept in a uniform magnetic field. It is the working principle of electric motor.
Appliances containing electric motor are :

1. Fans,
2. Mixies,
3. Grinders,
4. Machines, etc.

10th Class Physics 12th Lesson Electromagnetism 4 Marks Important Questions and Answers

Question 1.
How can you verify with experiment “The magnetic field lines are closed loops”? (AP March 2017)
Answer:

• Place a retort stand on the plank as shown in the figure.
• Pass a copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged in a vertical position and not touch the stand.
• Connect the two ends of the wire to a battery via switch. Allow the current flows through wire.
• By sprinkling the iron fillings around the wire, we can observe the magnetic field lines are in circular shape.

Conclusion :
Hence it is proved that “Magnetic field lines are closed loops”.

Question 2.
Name the device that converts electrical energy into mechanical energy. Draw its diagram and label the parts. (AP March 2018)
Answer:
1) The device that converts electrical energy into mechanical energy is motor.
2) Diagram of motor

Question 3.
List out the materials required for the Oersted experiment of electromagnetism. Write the procedure of the experiment. What do you understand by this experiment? (TS March 2016)
Answer:
List of material required for Oersted experiment:
A thermocol sheet, two small sticks, insulated copper wires, 9 V battery, switch, magnetic compass.

Procedure :

1. Take a thermocol sheet and fix two thin wooden sticks of height 1 cm which have small slit at the top of their ends.
2. Arrange a copper wire so that it passes through these slits and makes a circuit.
3. The circuit consists of a 9 V batten’, key, and copper wires connected in series.
4. Now keep a magnetic compass below the wire.
5. Now switch on the circuit and observe the compass needle.
6. Change the directions of current and observe the compass needle.

Observation :

1. When current is passed through circuit, we observe deflection of compass needle in one direction.
2. When the direction of current is changed, the compass needle deflects in another direction.
3. This shows that a current carrying conductor possesses magnetic field around it.

Question 4.
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid. (TS June 2017)
Answer:

• Fix a white paper on a wooden plank.
• Make two holes to the plank at appropriate distance.
• Make some more holes parallel to these two holes.
• Insert a copper wire through these holes. It looks as a coil.
• Connect the two ends of the coil to a battery and a switch in series.
• When swich is on current flows through the wire.
• Sprinkle some iron filings around the coil and tap the plank.

Observation :
An orderly pattern of iron filing is seen on the paper. These are magnetic field lines. The magnetic field lines set up by solenoid resemble those that of a bar magnet.

Question 5.
Why the current-carrying straight wire which is kept in a uniform magnetic field, perpendicularly to the direction of the field bends aside? Explain this process with a diagram showing the direction of forces acting on the wire. (TS March 2017)
Answer:

Uniform magnetic field (due to horse shoe magnet)

Magnetic field due to current carrying straight wire

Net field formed due to the above two fields

Explanation :
The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around wire. Therefore the wire tries to move to the weaker field region.

Question 6.
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field. (TS June 2018)
Answer:
Required apparatus :
i) Horseshoe magnet,
ii) Conducting wire,
iii) Battery, switch

Experimental procedure :
1) Arrange the circuit :
Take a wooden plank and arrange two wooden sticks with slits and arrange a conductor through the sticks and make a circuits with switch, battery.

2) Put horse shoe magnet:
Arrange the horse shoe magnet on the conductor in such a way that the conductor should be in between the two poles of the magnet.

3) Deflections in conductor :
Allow the current to pass through the circuit. We can find that conductor deflects up wards or down wards.

4)

Question 7.

Answer the following questions by observing above diagram. (TS March 2018)
1. Which device function of working does the above figure gives?
Answer:
Motor

2. What is the angle made by AB and CD with magnetic field?
Answer:
90°

3. What are the directions of magnetic forces on sides AB and CD?
Answer:
By applying right hand rule to get the directions of magnetic force. At ‘AB’, the magnetic force acts inward perpendicular to field of the magnet and on ‘CD’, it acts outwards.

4. What is the net force acting on the rectangular coil?
Answer:
The net force on ‘AB’ is equal and opposite to the force on ‘CD’ due to external magnetic field because they carry equal currents in the opposite direction. Sum of these forces is zero. Similarly, the sum of the forces on sides ‘BC’ and ‘DA’ is also zero.

So, net force on the rectangular coil is zero.

Question 8.
Explain the working process of induction stove. (TS March 2019)
Answer:

• An induction stove works on the principle of electromagnetic induction.
• A metal coil is kept just beneath the cooling surface.
• It carries alternating current (AC) so that AC produce an alternating magnetic field.
• When you keep a metal pan with water on it, the varying magnetic field beneath it crosses the bottom surface of the pan, and EMF is induced in it.
• Induced EMF produces induced current in the metal pan.
• The pan has a finite resistance.
• The flow of induced current produces heat in it.
• That heat is conducted to the water. In this way, induction stove works and water will be heated.

Question 9.
Which device is used to convert mechanical energy into electrical energy? Draw a neat diagram and label the parts of this device. (TS March 2019)
Answer:
1) Dynamo is used to convert mechanical energy into electrical energy. They are also called Generators.

2) AC Generator (or) DC Generator

AC Generator

DC Generator

Question 10.
A coil is hung as shown in the figure. A bar magnet with north pole facing the coil is moved perpendicularly

a) How does the magnetic flux passing through the coil change?
b) State the direction of the flow of the current induced in the coil, keeping the direction of bar magnet in view.
c) Draw the diagrams showing the magnetic field formed due to bar magnet at the surface of the coil and the magnetic field formed due to induced current.
d) Explain the reason for induced current.
Answer:
a) A bar magnet with north pole facing the coil is moved perpendicularly, the magnetic flex increases when passing through the coil.
b) The direction of the flow of the. current induced in the coil, keeping the direction of bar magnet is anti-clockwise due to north pole. KT
c) Φ = 0

Plane of coil is parallel to ‘B’.
d) Electromagnetic induction is the reason for induced current.

Question 11.
Conductor of length T moves perpendicular to its length with the speed V. Length of the conductor is perpendicular to the magnetic field of the conductor. Let us assume that electrons could move freely in the conductor and the charge of an electron is ‘e’.

a) What is the magnetic force acting on electron in the conductor?
b) In which direction does the above force act?
c) What effect does this force have on motion of electrons?
Answer:
a) Magnetic field acting on the electron inside the conductor is = F_m = e (\(\bar{V} \times \bar{B}\)) = BeV
This field acts from P to Q.

b) Consider in the field P and Q are ends of a conductor. ‘Q’ will act as negative end and ‘P’ will act as positive end then flow passes from P to Q means downwards.

c) The force on electrons shows an effect creates a potential difference at the ends of the rods.
∴ BeV = eE ⇒ E = BY

Question 12.
A charged particle q is moving with a speed V perpendicular to the magnetic field induction B. Find the radius of the path and time period of the particle.
Answer:

1. Let us assume that the field is directed into the page.
2. Then the force experienced by the particle F = qvB.
3. We know that this force is always directed perpendicular to velocity.
4. Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
5. Let r be the radius of the circular path.

Question 13.
Explain different ways to induced current in a coil.
Answer:

• Moving a north pole of a magnet into a coil.
• Withdrawing north pole from a coil.
• Moving a south pole of magnet into a coil.
• Withdrawing a south pole of a magnet from a coil.
• Moving a coil towards a magnet.

Question 14.
What are the similarities between a current-carrying solenoid and a bar magnet?
Answer:
1) The magnetic field lines due to current carrying solenoid are identical to those of bar magnet. Thus a current-carrying solenoid behaves just like a bar magnet with fixed polarities at the ends. The end at which the direction of current is clockwise behaves like a south pole and the end at which current is anti-clockwise behaves like a north pole.

2) A current-carrying solenoid when suspended freely, will set itself in the north south direction exactly in the same manner as a bar magnet does.

3) A current-carrying solenoid also acquires the- attractive property of magnet. If iron filings are brought near the solenoid, it attracts them when current flows through the solenoid.

Question 15.
What are the dissimilarities between a current-carrying solenoid and a bar magnet?
Answer:

• The magnetic field strength due to solenoid can be altered by altering current in it, while the magnetic field strength of a bar magnet cannot be changed.
• The direction of magnetic field due to solenoid can be reversed by reversing the direction of current in it, but the direction of magnetic field of the bar magnet cannot be changed.

Question 16.
Compare electromagnet with a permanent magnet.
Answer:

Electromagnet	Permanent Magnet
1) It is made of soft iron.	1) It is made of steel.
2) It produces the magnetic field so long as current flows in its coils.	2) It produces permanent magnetic field.
3) The magnetic field strength can be changed.	3) The magnetic field strength cannot be changed.
4) The electromagnet can be made as strong as needed.	4) The permanent magnets are not so strong.
5) The polarity of an electromagnet can be reversed.	5) The polarity of permanent magnet cannot be reversed.
6) It can be easily de-magnetised by switching off the current.	6) It cannot be easily de-magnetised.

Question 17.
What are the characteristics of magnetic field lines due to current in a loop (or circular coil)?
Answer:

• The magnetic lines are nearly circular in the vicinity of coil.
• Within the space enclosed by the wire the magnetic field lines are in the same direction.
• Near the centre of loop, the magnetic field lines are nearly parallel and the magnetic field may be assumed to be uniform in a small space near the centre.
• At the centre, the magnetic lines are along the axis of the loop and at right angles to the plane of the loop.
• The magnetic field lines become denser if the strength of current in the loop is increased and there are more number of turns in the loop.

Question 18.
A straight conductor passes vertically through a cardboard having some iron filings sprinkled on it.
a) Show the setting of iron filings when current is passed in the downward direction and then the cardboard is gently tapped. Draw arrows to represent the direction of magnetic field lines.
b) What changes occur if
i) current is increased ?
ii) the single conductor is replaced by several parallel conductors each carrying same current flowing in the same direction?
c) Name the law used by you to find the direction of magnetic field lines.
Answer:

a) The figure shoy/s, the pattern in which iron filings will set themselves. The arrows show the direction of magnetic field lines.
b) i) The arrangement of iron filings remains unchanged, but they become denser and Cardboar get arranged up to a larger distance from the conductor when the strength of current is increased and it is effective up to a larger distance from the conductor.
ii) The magnetic field at a point due to each conductor will be in same direction, so they will be added up. Thus the magnetic field strength is increased and it is effective up to a large distance so the magnetic field lines come closer and iron filings get arranged up to a larger distance.
c) Right hand thumb rule.

Question 19.
In figure A and B represent two straight wires carrying equal currents in a direction normal to the plane of paper inwards.
a) Sketch separately the magnetic field lines produced by each current.
b) Give a reason why the magnetic field at K (mid point of the line joining A and B) will be zero.
c) What will be the effect on the magnetic field at the point K if the current in wire B is reversed?

Answer:
a) Figure shows the sketch of magnetic field A and B.

b) The point K is equidistant from the wires A and B, and the wires A and B carry equal currents. So the magnetic fields at K due to wires A and B are equal in magnitude, but opposite in direction. Due to the wire A, it is downwards in the plane of paper, while due to the wire B, it is upwards in the plane of paper. So the net magnetic field at the point K is zero as the two fields cancel each other.
c) On reversing the direction of current in the wire B, the direction of magnetic field due to current is reversed at the point K, i.e. it becomes downwards in the plane of paper.

Question 20.
The diagram given below shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G.

When the key K is closed state the polarity.
i) At the end B of the coil X.
ii) At the end C of the coil Y.
iii) At the end C of the coil Y if the coil Y is (a) moved towards the coil X, (b) moved away from the coil X.
Answer:
i) On closing the key K, the current at the end B of the coil X is anti-clockwise, therefore at this end there is a north pole.

ii) While closing the key, polarity at the end C of the coil Y will be north. But there will be no polarity at the end C of the coil Y when the current becomes steady in the coil X

iii) a) With the key K closed, while the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
b) With the key K closed, while the coil Y is moved away from the coil X, the polarity at the end C of the coil Y is south.

Question 21.
How do you increase the speed of rotation of coil in a DC motor?
Answer:
The speed of rotation of coil can be increased by

1. Increasing the strength of current.
2. Increasing the number of turns in coil.
3. Increasing the strength of magnetic field. To increase the strength of magnetic field a soft iron core can be inserted within the coil.

Question 22.
Explain Lenz’s law with an activity.
Answer:
Lenz law :
The induced current will appear in such a direction that it opposes the change in the flux in the coil.

Explanation :

1. We know that when a bar magnet is pushed towards a coil with its north pole facing the coil an induced current is set up in the coil.
2. Let the direction be clockwise then current carrying loop behaves like a magnet with its south pole facing the north pole of bar magnet.
3. In such a case, the bar magnet attracts the coil. Then it gains kinetic energy. This is contradictory to conservation of energy.
4. Hence our assumption is wrong. So correct induced current direction is anti-clockwise.
5. Let us see a case where the bar magnet is pulled away from the coil with the north pole facing the coil. In such case, the coil opposes the motion of bar magnet to balance the conversion of mechanical energy into electric energy.
6. It happens only when the north pole of the magnet faces the south pole of the coil.
7. So, the direction of induced current in the coil must be in anti-clockwise direction.
8. In simple terms, when flux increases through coil, the coil opposes the increase in the flux and when flux decreases through coil, it opposes the decrease in the flux. This is Lenz law.

Question 23.
What are the uses of electromagnet?
Answer:
Electromagnets are mainly used for the following purposes.

1. For lifting and transporting the large masses of scrap, girders, plates, etc. especially to the places where it is not convenient to take the help of human labour.
2. For loading furnaces with iron.
3. For separating the magnetic substances such as iron from other debris.
4. For removing pieces of iron from wounds.
5. In several electrical devices such as electric bell, telegraph, electric thumb, electric motor, relay, microphone, loudspeaker, etc.
6. In scientific research, to study the magnetic properties of a substance in a magnetic field.

Question 24.
Write the differences between AC generator and DC motor.
Answer:

AC Generator	DC Motor
1) A generator is a device which converts the mechanical energy into the electrical energy.	1) A DC motor is a device which converts electrical energy into the mechanical energy.
2) A generator works on the principle of electromagnetic induction.	2) A DC motor works on the principle of force acting on a current carrying conductor placed in a magnetic field.
3) In a generator, the mechanical energy is used in rotating the armature coil in a magnetic field so as to produce electricity.	3) In a DC motor electrical energy is provided by the DC source to flow current in the armature coil placed in a magnetic field due to which coil rotates.
4) A generator makes use of two separate coaxial slip rings.	4) A DC motor makes use of two parts of a slip ring (i.e., split rings) which acts as commutator.