1st PUC Economics Question Bank Chapter 4 Poverty

Students can Download Economics Chapter 4 Poverty Questions and Answers, Notes Pdf, 1st PUC Economics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Economics Question Bank Chapter 4 Poverty

1st PUC Economics Poverty TextBook Questions and Answers

Question 1.
Why caloric – based norm is not adequate to identify the poor?
Answer:
The government uses Monthly Per Capita Expenditure (MPCE) as proxy for income of households to identify the poor. Poverty line is estimated by the monetary value (per capita expenditure) of the minimum calorie intake that was estimated at 2400 calories for a rural person and 2100 for a person in the urban area. But this calorie based norm is not adequate to identify the poor duo to following reasons:

1. This mechanism groups all the poor together and does not differentiate between the very poor and the other poor which makes it ’ difficult to identify who amongst them needs help the most.

2. Economists question, the basis of taking expenditure on food and a few select items as proxy for income.

3. This norm does not take into account the other factors associated with poverty such as accessibility to basic education, health care, drinking water, and sanitation.

4. This norm does not take social factors such as illiteracy, lack of access to resources, discrimination or lack of civil and political freedoms into consideration.

Question 2.
What is meant by the ‘Food for work programme’?
Answer:
‘Food for work’ programme was started in the 1970’s to raise the standard of living of poor. These poverty alleviation programme aimed to raise income and employment for the poor through the creation of incremental assets and by means of work generation.

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Question 3.
Why are employment generation programmes important in poverty alleviation in india?
Answer:
In India, twin problems exist i.e., poverty and unemployment. Poverty alleviation has been one of the guiding principles of the planning process in India. Poverty can effectively be eradicated only when the poor start contributing to growth by their active involvement in the growth process.

This can only be achieved by launching various employment schemes. Following points discussed the importance of Employment Generation Programmes to eradicate poverty.

1. Nexus between Unemployment and Poverty:
There exists a deep nexus between unemployment and poverty. If employment opportunities are generated, then more people will be employed leading to rise in their income which in turn will reduce poverty.

2. Availability of Basic Facilities:
With the rise in employment opportunities, income increase and poor people are able to get access to education, health facilities, proper sanitation, etc.

3. Creation of Assets:
The Employment Generation Programmes aim at creation of assets like water harvesting, irrigation facilities, construction of roads, construction of dams, etc. All these assets help in the social and economic development of the rural areas and hence eradication of poverty.

4. Creation of skills
An essential element of employment generation programmes is the formation of human capital by imparting skills to the unskilled labourers through training. This alleviation programmes like Prime Minister’s Rozgar Yojana, Swarna Jayanti Shahari Rogzar Yojana, National Food for Work Programme, Annapurna came into existence.

Question 4.
How can creation of income earning assets address the problem of poverty?
Answer:
By creating income earning assets, we can generate employment opportunities through which poor can raise their income which ultimately helpful in improving standard of living. Thereby, it address the problem of poverty.

Question 5.
The three dimensional attack on poverty adopted by the government has not succeded in poverty alleviation in India. Comment.
Answer:
Poverty alleviation has always been accepted as one of the major objectives of planned development process in India but even after vast spending on poverty alleviation programmes, the government has not succeeded in poverty alleviation in India.

Despite various strategies to alleviate poverty, problems like hunger, malnorishment illiteracy, and lack of basic amenities are prevalent in India. None of the poverty alleviation strategies resulted in any radical change in the ownership of assets, process of production and improvement of basic amenities to the needy.

Due to unequal distribution of assets, the benefits from poverty alleviation programmes have not actually reached the poor. The amount of resources allocated for the poverty alleviation programmes is not sufficient when we take the magnitude of poverty into consideration.

The implementation of the poverty alleviation programmes is the responsibility of government and bank officials who are ill motivated, inadequately trained, corruption prone and vulnerable to pressure from local elites.

The resources are thus used inefficiently. Government policies have also failed to address the various issues related to poverty due to non – participation of local level institutions in programme implementation. It is evident that high grouwth alone is not sufficient to reduce poverty without the active participation of the people.

Further, it is necessary to identify poverty stricken areas and provide infrastructure such as schools, roads, power, telecom, IT services, training institution s, etc. Institutional weaknesses abound and implementation failures are the biggest reasons that these programmes not succeeded.

Question 6.
What programme has the Government adopted to help the elderly people and poor and destitute women?
Answer:
National Social Assistance Programme’ is one of the programme started by Government to help the elderly people and poor and destitute women this programme targets elderly people, widows and the poor and destitute women who are alone and have no one to take care of them. Under this programme, these targeted people are given pension to sustain their livelihood.

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Question 7
Is there any relationship between unemployment and poverty? Explain.
Answer:
There exists a deep nexus between unemployment and poverty. Unemployment or underemployment and the casual and intermittent nature of work in both rural and urban areas drives unemployed people who do not have resources to make their ends meet into indebtedness and poverty.

If employment opportunities are generated, then more people will be employed leading to rise in their income which in turn will reduce poverty.

Due to unemployment, income of the people is reduced to a large extent and they are unable to get access to education, health facilities, proper sanitation, etc. This causes poor quality of living and hence poor human capital and skills which in turn lead to poverty making a vicious circle of poverty.

Question 8.
Suppose you are from a poor family and you wish to get help from the Government to set up a petty shop under which scheme will you apply for assistance and why?
Answer:
For setting up a petty shop, I would apply for financial assistance under the programme prime Minister’s Rozgar Yojana’ (PMRY). Under this programme, an unemployed educated person from low-income family in rural and urban areas can set up any kind of enterprise that can generate employment.

Question 9.
Illustrate the difference between rural and urban poverty. It is correct to say that poverty has shifted from rural to urban areas? Use the trends in poverty ratio to support your answer?
Answer:

Rural Poverty Urban Poverty
1. Open unemployment under & disguised unemployment are found simultaneously in rural areas Open unemployment is generally found
2. It is difficult to differentiate between open employment & under employment in rural areas One can differentiate between open and disguised unemployment
3. As the population increases dependent on agriculture increase. This leads to further rise in seasonal, open, under & distinguished unemployment in rural areas. In urban areas the main reason for increase in open unemployment is increase in education, health & other facilities
4. No facility of education, employment, social welfare etc. that lead to poverty among the masses Good facility of education, social welfare etc., that make the standard of living of urban people high
5. Rural people posses few assets Urban people posses many assets
6. Malnutrition among rural people is high Generally malnutrition is not found.

Question 10.
Suppose you are a resident of a village, suggest a few measures to tackle the problem of poverty.
Answer:
Following measures have been taken by the Government to remove poverty under five year plan:
1. Integrated rural development programme:
With a view to remove poverty in rural areas and making provision for full employment under this programme, attempts are being made to provide more employment by developing agriculture, animal husbandry, fisheries, small scale, and cottage industries etc.,

2. Jawahar Rozsar Yojana:
It was launched in 1989. Its aim was to provide employment to at least one member of a IPUC Economics INDIAN ECONOMIC DEVELOPMENT rural poor family for 50 to 100 days in a year.

3. Jawahar Gram Samriddhi Yojana (JGSY):
It was launched on 1st April 1999. It has two main objectives:

  1. Creation of durable productive community assets at the village level.
  2. Generation of supplementary employment for the unemployed poor in rural areas.

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4. Swarnaiavanthi Gram Swaroiear Yoiana (SGST):
It was started on 1st April 1999. Its aims are:

  • Focussed approach to poverty alleviation.
  • Capitalising advantages of group lending
  • Overcoming the problems.

5. Employment Assurance Scheme:
The programme is presently being implemented in all rural blocks.

  • Creation of additional wage employment opportunities during the period of acute shortage for the rural poor living below the poverty line.
  • Creation of durable community, social and economic assets for sustained economic development.

6. Pradhan Mantri Gramodava Yojana (PMGY):
It was introduced in 2000-2001 focussing on village development in five critical areas, health, primary education, drinking water, housing and rural roads with the objectives of improving the quality of life or people in the rural areas.

7. Sampoorna Grameen Roisar Yojana (SGRY):
The scheme aims at providing wage employment in rural areas as also food, security, creation durable, community, social and economic assets.

8. Grameena Roisar Guarantee Scheme:
It was launched in February 2005 to provide work for 100 days a year in rural areas.

1st PUC Economics Poverty Additional Questions and Answers

1st PUC Economics Poverty Very Short Answer Type Questions

Question 1.
Give the meaning of Poverty?
Answer:
Poverty is a situation in which, section of the people in a country are unable to get the minimum basic needs (like food, clothing, housing, education and health facilities)

Question 2.
What makes the poor physically weak?
Answer:

  • Malnutrition
  • 111 health & disability
  • makes the poor physically weak.

Question 3.
Mention two kinds of Poverty?
Answer:
The two kinds of poverty are

  1. Absolute poverty
  2. Relative poverty

Question 4.
On what basis the poor are identified?
Answer:
The poor people are identified on the basis of their occupation and ownership of assets. In both rural and urban areas, the people who are unable to get even minimum income from their present occupation to lead their life are considered as poor.

Question 5.
Who are called churning poor?
Answer:
The churning poor are those who regularly move in and out of poverty, ex: small farmers, seasonal workers

Question 6.
Define poverty line as per Planning Commission?
Answer:
Poverty line as per planning commission is measured on the basis of minimum nutrition calories required for rural and urban population. It is estimated that a person is said to be poor if he is unable to consume 2400 calories per day in rural areas and 2100 calories per day in case of urban areas.

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Question 7.
What do you mean by ‘Head Count Ratio’
Answer:
The number of poor estimated as the proportion of people below the poverty line is known as ‘Head Count Ratio’.

Question 8.
Name the National Programme which provides social security to aged and widows?
Answer:
National SocialAssistance Programme.

Question 9.
Write the meaning of Absolute Poverty?
Answer:
Absolute poverty is one of the kinds of poverty where consumption and expenditure of a person is not sufficient to maintain a minimum acceptable level of living, according to national standard, which is expressed in terms of food grains.

Question 10.
What is Relative Poverty?
Answer:
Relative Poverty refers to poverty which is measured in relation to levels of income of individuals. A section of people whose level of income is low when compared to other sections of people.

Question 11.
Who discussed the concept of poverty line first in India?
Answer:
‘Dadabhai Naroji’ was the first to discuss the concept of poverty line India.

Question 12.
Name the programme which guarantees minimum of 100 days of employment in rural areas?
Answer:
‘Mahatma Gandhi National Rural Employment Guarantee’ (MGNREGP).

Question 13.
Suppose you are from a poor family of a rural area and you wish to get help from the Government to set up a petty shop. Under which scheme will you apply for assistance.
Answer:
‘SwamajayanthiGramSwarojgarYojana’.

Question 14.
Mention the states which have 70% of India’s poor people?
Answer:

  1. Orissa
  2. Bihar
  3. West Bengal
  4. Uttar Pradesh
  5. Madhya Pradesh.

Question 15.
What programmes has the Government adopted to help the elderly people and poor and destitute women?
Answer:
National social assistance programme help the elderly people and poor and destitute women. This programme in Karnataka named as ‘Sandhya Suraksha Yojana’.

Question 16.
Name the programme introduced by the Government to provide insurance to the head of the family of rural landless households?
Answer:
AamAadmi Bima Yojana.

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Question 17.
Name the programme introduced by the Government of Karnataka to provide health insurance?
Answer:
‘Yashaswini Yojana’.

Question 18.
Name the programme introduced by the Government of Karnataka to provide financial support to female child of poor family.
Answer:
‘Bhagya LakshmiYojana’.

Question 19.
Expand‘VAMBAY’?
Answer:
‘Valmiki Ambedkar Awas Yojana’.

Question 20.
Name the states which are poorest in India?
Answer:

  1. Bihar and
  2. Orissa

Question 21.
Expand IAY?
Answer:
Indira Awas Yojana.

Question 22.
Expand PMRY?
Answer:
‘PradhanaManthriRojgar Yojana’.

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Question 23.
Who are chronic poor?
Answer:
These are the person who always remains as poor for a long period of time, ex: casual workers, agricultural labourer’s, ragpickers, beggars etc.,

Question 24.
Who are called Transient Poor?
Answer:
The transient poor are those people who are rich most of the time but may sometimes have a pitch of bad luck.

Question 25.
Expand SGRY?
Answer:
‘SampoomaGrameena Rojgar Yojana’.

Question 26.
Expand SGSY?
Answer:
‘Swamajayanthi Gram Swarozgar Yojana’ (SGSY).

Question 27.
Expand SJSRY?
Answer:
‘Swarna jayanthi Shahari Rozgar Yojana’ (SJSRY).

1st PUC Economics Poverty Short Answer Type Questions

Question 1.
Write the important features of poorest households?
Answer:
The important features of poorest households:

  • The poor people possess few assets and live in huts but the poorest of them do not even. have such huts also.
  • They will not have capacity to eat a square meal a day.
  • Starvation, hunger, lack of basic literacy and skills.

Question 2.
Name the categories of Poverty?
Answer:
The categories of poverty are :

  • Chronic poor
  • Transient poor
  • Never poor or non-poor.

Question 3.
Name the factors which are to be considered to develop poverty line other than income and assets?
Answer:

  • Accessibility to basic education
  • Health care
  • Drinking water and sanitation have to be considered to develop poverty line.

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Question 4.
Do you think the growth oriented approach is successful in reducing poverty? why?
Answer:
The growth oriented approach has not succeeded in reducing poverty. This is mainly because of rapid growth rate of population and lack of proper implementation of land reforms.

Question 5.
How are poor people are identified?
Answer:
The poor people are identified on the basis of their occupation and ownership of assets. In both rural and urban areas, the people who are unable to get even minimum income from their present occupation to lead their life are considered as poor.

Question 6.
How do you calculate ‘Head Count Ratio’?
Answer:
The HeadCount Ratio is calculated with following formula.
HCL = Number of people living below the poverty line/Total population of the country.

Question 7.
What are the approaches to reduce poverty in India?
Answer:

  • Growth oriented approach
  • Income and employment generation approach
  • Providing minimum basic needs to the people approach.

Question 8.
State any two self employment programmes?
Answer:
The major self employment programmes are :

  • Swamajayanthi Gram Swarozgar Yojana (SGSY)
  • Swamajayanthi ShahariRozgar Yojana (SJBRY)
  • Pradhana Manthri Rozgar Yojana (PMRY).

Question 9.
Mention any two.national programmes to generate wage employment?
Answer:
The major national programmes to generate wage employment are as follows :

  • National Food for Work Programme (NFWP)
  • Sampooma Grameena Rozgar Yojana (SGRY)
  • Mahathma Gandhi National Rural Employment Guarantee Programme (MGNREGP).

Question 10.
Mention the programmes aimed at improving food and nutritional status of poor?
Answer:
There are programmes introduced by the Government to improve the food and nutritional status of poor. They are as follows:

  • Public distribution system (PDS)
  • Integrated child development scheme (ICDS)
  • Mid-day meals scheme.

Question 11.
Mention any four causes for poverty?
Answer:
The main causes for poverty are :

  • Unemployment
  • Population explosion
  • Inflation
  • Exploitation under British rule

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Question 12.
State the programme introduced to provide basic infrastructure in rural areas?
Answer:
Programmes like:

  • Pradhan Mantri Gram Sadak Yojana
  • Basava Vasathi Yojana (Karnataka)
  • Public Distribution System
  • Integrated Child Development Scheme.

Question 13.
Illustrate the difference between rural and urban poverty. Is it correct to say that poverty has shifted from rural to urban areas?
Answer:
The difference between rural and urban poverty is the nature of poverty. In rural areas, poor people are those who are landless agricultural labourers, small and marginal farmers. While in urban areas, poor people are those who are unemployed, under employed or employed in low productivity occupation with low wages.
Poverty Ratio:
1st PUC Economics Question Bank Chapter 4 Poverty img2
Estimates source:
planning commission estimates (Uniform Reference Period).

Question 14.
Who are poor according to Tendulkar Committee?
Answer:
According to Tendulkar Committee Report (2009-10), the poverty line is defined on the basis of per capita consumption expenditure of Rs.672.80 in rural areas and Rs.859.60 in urban areas.

1st PUC Economics Poverty Long Answer Type Questions

Question 1.
Briefly explain the causes of Poverty?
Answer:
The causes of Poverty:
1. Exploitation under British rule :
Britishers ruthlessessly exploited our country during their colonial rule. They destroyed our traditional cottage and small-scale industries to encourage their modem industries.

2. Economic Inequalities:
In India inequalities in the distribution of wealth and income are also responsible for poverty. A large section of the people are forced to remain under poverty due to concentration of wealth and income in the hands of few people.

3. Low resource base:
The most important factor which is responsible for poverty in India is the low resource base of poor both in rural and urban area. At present a large section of rural poor are having very little land or no land at all. The income from this smallholdings is not sufficient to meet their basic needs.

4. Unemployment:
Unemployment and underemployment are much more wide spread problem in India. The degree of unemployment among the poor is very high.

5. Rapid growth in population :
The single most important reason of poverty in India is rapid growth population. This is responsible for low level of per capita income and consumption.

6. Inflationary pressures:
Inflationary rise in the prices of food grains and other essential commodities further intensifies the situation of poverty.

7. Vicious circle of poverty:
At present India is facing the challenge of vicious circle of poverty. Here poverty is both the cause and the effect.

8. Social factors:
In India, social factors are also causing poverty. Illiteracy, ignorance, backwardness, inadequancy of social structure, narrow outlook affecting the quality of life of the people and their employability.

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Question 2.
The three dimensional attack on poverty adopted by the Government has not succeeded in poverty alleviation in India. Comment.
Answer:
The three dimensional approach of economic growth, employment generation and alleviating poverty could not achieve the desired result. Although there has been a reduction in the percentage of absolute poor in some of the states still the poor people lack basic amenities, literacy and nourishment.
This is because of:

  • Unequal distribution of land and other assets among rich and poor farmers.
  • Improper implementation of poverty alleviation programmes by ill-motivated and inadequately trained bureaucrats further worsened the situation.
  • Corruption along with inclination towards interest of elites led to an efficient and misallocation of scarce resources.

Question 3.
Suppose you are a resident of a village, suggest a few measures to tackle the problem of poverty?
Answer:
Being a resident of a village, I would suggest the following measures to tackle the problem of poverty:

  • Identification of poor
  • Generating employment opportunities for the identified poor
  • Free access to education and health care facilities.
  • Establishment of small scale industries.
  • Re-distribution of income-earning assets.
  • Encouraging poor for their active participation.
  • Organising training camps and night classes for importing vocation training to unskilled labourers.
  • Advancing financial and technical assistance to establish small enterprises.
  • Upgradation of Agricultural practices to raise productivity.
  • Enforcement of measures to check population growth.
  • Development of infrastructure.
  • Motivating the poor to acquire skills information and knowledge.

Question 4.
What are the various policies and programmes towards poverty alleviation of the Government?
Answer:
The second five year plan (1956-61) has pointed out that the benefits of economic development must accure more and more to the relatively less privileged classes of society.
The Government’s approach to poverty reduction is of three dimensions:
1. Growth oriented approach:
It is based on the expectation that the effects of economic growth that is rapid increase in the gross domestic product and per capita income of a nation would spread to all sections of society. Rapid industrial development and transformation of agriculture through Green Revolution in select regions would benefit underdeveloped regions and more backward sections of the community.

2. Food for Work:
In 1970, Food for work programme was started. The policymakers started thinking that incomes and employment for the poor could be increased through the creation of incremental assets and by means of work generation through specific poverty alleviation programmes. Under self-employment programmes, financial assistance is given to families or individuals.

3. Provision of minimum basic amenities :
Through public expenditure on social consumption needs such as education, health, water supply and sanitation, people’s standard of living could be improved. Programmes under this approach are expected to supplement the consumption of the poor, create employment opportunities and improvement in health and education.

For this purpose ‘Pradhana Manthi Gram Saraj Yojana, Pradhana Mantri Gramodaya, Yojana, Valmiki Ambedkar Awas Yojana and National Social Assistance Programme’ were started.

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Question 5.
Write some measures to remove poverty in India?
Answer:
Following steps should be taken to remove poverty in India.
1. Creation of employment opportunities:
There should be greater encouragement to small scale industries in rural areas, poultry farming, dairy farming and piggeries should be developed.

2. Population Control:
Population growth is a hindrance in our coming the problem of population. It should be checked. Various family planning methods should be adopted by the people. The people should be educated so that they understand the charm of a small sized family.

3. Economic development:
There should be improvement in agricultural and industries. Their production should be increased.

4. Provision of minimum needs :
The Government should provide water, housing, sanitation, other facilities. The public distribution system should be regulated properly. So that the poor people may get essential commodities at cheaper rates.

5. Removal of economic inequalities:
The Government should encourage small industries and agriculture. It should give incentive to industries in rural areas. The taxation policy should be progressive. The money collected by the Government should be spent on the welfare of poor people.

Question 6.
How is the poverty line determined?
Answer:
In simple words the poverty line is defined as the per capita expenditure at which the average calorie in take was 2400 calories for a person in rural areas and 2100 calories for urban areas. Most of the studies use data collected by national sample survey organisation on various aspects including consumption expenditure and region.

1. In 1999-2000 the poverty line was defined for rural areas as consumption worth Rs.328 per month and for urban areas it was Rs.454.

2. In 1973-74 more than 321 million people were below the poverty line. In 1999-2000 this has come down to about 260 million.

3. In 1973,74, about 55 percent of the total population were below the poverty line. In 1999-2000 it has fallen to 26 percent. In rural areas, about 75 percent of the total poor are below the poverty line.

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Question 7.
Briefly explain the three dimensional attack on poverty adopted by the Government?
Answer:
The Government’s approach to poverty reduction is of three dimensions:
1. The first one is the growth-oriented approach. It is based on the expectation that the effects of economic growth – rapid increase in gross domestic product and per capita income would spread to all sections of society and will trickle down to poorer sections.

This was the major focus of planning in the 1950’s and early 1960s. It was felt that rapid industrial development and transformation of agriculture through green revolution in select (regions would benefit the underdeveloped regions).

2. The second approach has been initiated from the third five-year plan (1961 -1966) that suggest that incomes and employment for the poor could be raised through the creation of incremental assets and by means of work generation. This could be achieved through specific poverty alleviation programme. The example based on this approach are :

a. Food for work programme:
Minimum food was ensured to workers and their families instead of money under this programme.

b. Rural Employment Generation Programme:
It aims at creating self employment opportunities in rural areas and small towns.

c. Prime Minister’s Rozgar Yojana:
The educated unemployed from low income families in rural and urban areas can get financial help to setup any kind of enterprise that generate employment under this programme.

d. Swama Jayanthi Shasri Rozgar Yojana:
It mainly aims at creating employment opportunity both of self-employment and wage employment in urban areas.

e. National Food for work programme and Sampooma Grameen Rojgar Yojana:
These programmes aim at generating wage employment for the poor unskilled people living in rural areas.

f. In August 2005, the parliament has passed a new Act to provide guaranteed wage employment to every household whose adult volunteer to do unskilled manual work for a minimum of 100 days in a year.

3. The third approach to tackle poverty is to provide minimum basic amenities to the people. Programmes under this approach are expected to supplement the consumption of the poor, create employment opportunities and bring about improvement in health and education.

Three major programmes that aim at improving the food and nutritional status of the poor are public distribution system, integrated child development scheme and mid-day meal scheme.

Question 8.
Critically examine the growth oriented approach to poverty reduction?
Answer:
Growth oriented approach to poverty reduction:
It is the first Government’s approach to poverty reduction. This approach was based on the expectation that the effects of economic growth would spread to all sections of society and trickle down to the poor section also. This approach was the major focus of planning in the 1950’s and early 1960’s.

It was felt that rapid industrial development and transformation of agriculture through green revolution would benefit the under developed regions and the more backward sections of the community. But this approach was failure. The benefits of economic growth has not trickled down to the poor.

The green revolution exacerbated the disparities regionally and between large and small farmers. There was unwillingness and inability to redistributed lands.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling

Students can Download Maths Chapter 10 Data Handling Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 5 Maths Chapter 10 Data Handling

KSEEB Class 5 Maths Data Handling Ex 10.1

1. Observe the pictograph and answer the questions

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 2

  • Number of Science books read Six
  • Number of Kannada books read Four
  • Number of Adventure books read Twelve
  • Number of novels read Eight
  • Total number of books read Thirty

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2. The following pictograph shows the number of cars in 5 villages of a taluk.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 1 Planet128

Observe the pictograph and answer the following questions:

  • Which village has the maximum number of cars? Village 3
  • Which village has the minimum number of cars? Village 1
  • What is the total number of cars in five villages? 150 cars
  • How many more cars are there in village 3 than in village 5? Twenty-five
  • Which two villages have the same number of cars? Village 2 & 4

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Data Handling Additional Questions and Answers

1. A survey of sale of cars made by an agency for a particular month is an follows:

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 3 Planet128

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Question
Represent this data through a pictograph.
Answer:
Scale = 5 cars.

Answer the following questions:

Question
In which week were the maximum number of cars sold?
Answer:
First week

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Question
In which week were the minimum number of cars sold?
Answer:
Second week

Question
What is the difference in the number of cars sold in the third and fourth week?
Answer:
Five years

Question
What is the total number of cars sold in the month?
Answer:
110 cars

KSEEB Solutions Planet128

Question
Arrange the number of cars sold in the descending order along with the week.
Answer:
Example:
35, 30, 25, 20
First Week

Question 3.
Ramu had 5 pens. 3 erasers, 6 books, 2 pencils and one sharpener in his bag.
Answer:
Write the data in tubular form and represent the same through a pictograph.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 4 Planet128KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 5

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4. The number of bulbs manufactured in a week by a factory is given below. Draw a pictograph to represent the data. (Choose a suitable scale)

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 6 Planet128KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 7 Planet128

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Bar Graphs

I. Read the bar graph and answer the given questions:

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 8 Planet128

Question
What is the information given in the bar graph?
Answer:
Days against Earnings of flourist in rupees

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Question
Mention the scale taken.
Answer:
50 flourist = 1 cm

Question
Mention the earnings on each day.
Answer:
First day – 300
Second day – 500
Third day – 200
Fourth day – 400
Fifth day – 350

Question
What is the total earnings in five days?
Answer:
1750 flourist in Rupees

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Question
What is the difference in the amount earned on the second day and the fifth day?
Answer:
150 flourist in Rupees

Question
Arrange the amount earned on each day in the descending order.
Answer:
500,400,350,300,200, 2nd day 4th day 5th day ! day 3rd day

2. Study the bar graph and answer the questions given.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 9 Planet128

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Question
What is the information given in the bar graph?
Answer:
Saplings planted by a farmer against week

Question
Mention the scale taken.
Answer:
100 Plants = 1 week

Question
Mention the number of saplings planted in each week.
Answer:
First week – 700
Second week – 900
Third week – 500
Fourth week – 800
Fifth week – 600

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Question
What is the total number of saplings planted in the five weeks?
Answer:
3500 saplings

Question
In which week did the farmer plant the maximum number of saplings?
Answer:
Second week

Question
In which week did he plant the minimum number of saplings?
Answer:
Third week

KSEEB Solutions

3. A survey of 120 school students was done to find the activity they prefer to do in their free time.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 10

Draw a bar graph to illustrate the above data.
Scale: 1 cm= 5 students

Question
Which activity is preferred by most of the students other than playing?
Answer:
KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 11 Planet128

KSEEB Solutions

4. The number of belts sold by a shopkeeper on six consecutive days of a week is as follows:

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 12 Planet128

Draw a bar graph to represent the data.
Scale : 1 cm = 5 belts

Question
What is the total number of belts sold in 6 days?
Answer:
175

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 13 Planet128

KSEEB Solutions

5. The number of hours spent by a student on different activities in a day is given below.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 14

Represent the above data by a bar graph.
Scale: 1 cm = 1 hour
KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 15 Planet128

KSEEB Solutions

6. The following table shows the number of bicycles manufactured in a factory during the year 2005 – 2009. Illustrate the data using a bar graph.
(choose a suitable scale)

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 16

Question a.
In which year was the maximum number of bicycles manufactured?
Answer:
2007

KSEEB Solutions

Question b.
In which year was the minimum number of bicycle manufactured?
Answer:
2008
KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 17 Planet128

KSEEB Solutions

1st PUC Maths Model Question Paper 1 with Answers

Students can Download 1st PUC Maths Model Question Paper 1 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 1 with Answers

Time : 3 hrs 15 min
Max. Marks : 100

Instructions:
1. The question paper has five parts A, B, C, D and E and answer all parts,
2. Part-A carries 10 marks, Part-B carries 20 marks, Part-C carries 30 marks, Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer any TEN questions : (10 × 1 = 10)

Question 1.
Given that the numbers of subsets of a set A is 16. Find the number of elements of A.
Answer:
Let n(A) = m
n [P(A)] = 16 ⇒ 2m = 16 ⇒ m= 4
n(A)= 4

Question 2.
If tan x = \(\frac { 3 }{ 4 }\) and x lies in the third quadrant, find sin X.
Answer:
tan x = \(\frac{3}{4}=\frac{-3}{-4}\)
1st PUC Maths Model Question Paper 1 with answers - 1
ordinate = -3, abscissa = 4, distance = 5 sin x = \(\frac { -3 }{ 5 }\)

Question 3.
Find the modulus of \(\frac{1+i}{1-i}\)
Answer:
\(\frac{1+i}{1-i}=\frac{(1+i)^{2}}{1-i}=\frac{2 i}{2}\) = i Modulus of i= 1

1st PUC Maths Model Question Paper 1 with Answers

Question 4.
Find ‘n’ if nC7 = nC6
Answer:
nCn-7 = nC6 ⇒ n – 7 = 6 ⇒ n = 13.

Question 5.
Find 20th term of GP. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}—-\)
Answer:
1st PUC Maths Model Question Paper 1 with answers - 2

Question 6.
Find the distance between 3r + 4y +5= 0 and 6x + 8y +2=0.
Answer:
3x + 4y + 5 = 0) ………….. (1)
3x + 4y + 1 = 0 …………. (2)
Required distance \(\left|\frac{5-1}{\sqrt{9+16}}\right|=\frac{4}{5}\) units

Question 7.
1st PUC Maths Model Question Paper 1 with Answers - 3
Answer:
\(\lim _{x \rightarrow 0+}\) f(x) = \(\lim _{x \rightarrow 0} \frac{x}{x}=\frac{1}{1}=2\) ∴ f(x) in Discontinuous

Question 8.
Write the negation of For all a, b ∈ I, a – b ∈ I’.
Answer:
“There exists a, b ∈ I, such that a – b € I’
or ‘∃ a, b ∈ I, a – b ∉ I’.

Question 9.
A letter is chosen at random from the word “ASSASINATION”. Find the probability that letter is vowel.
Answer:
No of ways of selecting one vowel out of six vowels (3A’S, 21’s, 10’s) = 6C1 = 6.
P(1 vowel) = \(\frac{^{6} \mathrm{C}_{1}}{^{13} \mathrm{C}_{1}}=\frac{6}{13}\)

Question 10.
Let A = {2,3,4} and R be a relation on A defined by
R={(x,y)|x,y ∈ A,x divides y}, find ‘R’.
Answer:
R= {(2, 2), (2, 4), (3, 3), (4,4)}.

1st PUC Maths Model Question Paper 1 with Answers

Part – B

Answer any TEN questions : (10 × 2 = 20)

Question 11.
If A and B are two disjoimt sets and n(A) = 15 and n(B) = 10 find n(A ∪ B), n(A ∩ B)
Answer:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∩ B) = 0, n(A ∪ B) = 15 + 10 = 25.

Question 12.
If U = {x:r s≤10, r ∈ N} A = {x : x ∈ N, x is prime) B = {x : x ∈ N, x is even} write A ∩ B’ in roster form.
Answer:
U = {1, 2, 3 … 10), A = {2, 3, 5, 7}
B = {2, 4, 6, 8, 10} B1 = {3, 5, 7,9}
[A ∩ B1 = {3,5,7)]

Question 13.
If A × B = {(a,1) (a,2) (a,3) (b,1) (b,2) (b,3)}, find the sets A and B and hence fimnnd B × A
Answer:
A = {a, b}, B = {1, 2, 3}
B × A= {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}

Question 14.
The difference between two acute angles of a right angled triangle is to \(\frac{3 \pi}{10}\) radiAnswer: Express the angles in degrees.
Answer:
Let A and B be acute angles
Given A + B = \(\frac{\pi}{2}\) = 90° : A – B = \(\frac{3 \pi}{10}\) = 54° ⇒ A = \(\frac{2 \pi}{5}\) and B = \(\frac{\pi}{5} \) ⇒ A = 72 and B = 18°.

Question 15.
Find sin \(\frac{x}{2}\) if tan x = \(-\frac{4}{3}\)and x lies in second quadrant,
Answer:
tan x = \(\frac{-4}{3}=\frac{4}{-3}\) ⇒ ∴ cos x = \(\frac{-3}{5}\)
1st PUC Maths Model Question Paper 1 with Answers - 5
1st PUC Maths Model Question Paper 1 with Answers - 4

Question 16.
\(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-5 x+6}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 6

1st PUC Maths Model Question Paper 1 with Answers

Question 17.
CV = 60 σ = 21
Answer:

Question 18.
Write the inverse, converse of ‘If a parallelogram is a square, then it is rhombus’.
Answer:
Inverse: If a parallelogram is not a square then it is not a rhombus
Converse: If a parallelogram is a rhombus then it is a square.

Question 19.
On her vacations Veena visits four cities A, B, C and D in random order. What is the probability that she visits A before B?
Answer:
n (S) = 24
P(visiting A before B) = \(\frac{12}{24}\) = 1/2

Question 20.
In a triangle ABC with vertex A(2,3), B(4, -1) and C(1,2). Find the Length of the alitude from vertex A.
Answer:
Equation BC = x + y – 3 = 0
Length of the alitude from A (2, 3) = \(\left|\frac{2+3-3}{\sqrt{2}}\right|=\left|\frac{2}{\sqrt{2}}\right|=\sqrt{2}\) = √2.

Question 21.
Represent the complex number z = 1 + i in polar form.
Answer:
r = √2 and θ = \(\frac{\pi}{4}\) Polar form is √2 \(\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\)

Question 22.
Obtain all pairs of consecutive odd natural numbers such that in each pair both are more than 50 and their sum is less than 120.
Answer:
Taking the pair as x, x + 2,
we get x > 50, 2x + 2 < 120 ⇒ x < 59 and writing the required pairs are (51, 53), (53, 55), (55, 57), (57, 59).

1st PUC Maths Model Question Paper 1 with Answers

Question 23.
A line cuts off equal intercepts on the co – ordinate axes. Find the angle made by the line with the positive x-axis.
Answer:
Slope =-1 ⇒ tanθ = -1
angle made = 135°.

Question 24.
If the origin is the centroid of the triangle PQR with vertices P (2a, 4, 6) Q(-4, 3b, -10) and R(8, 14, 2c) then find the values of a, b, c.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 9

Part-C

Answer any EIGHT of the following questions : (10 × 3 = 30)

Question 25.
Out of a group of 200 students (who know at least one language), 100 students know English, 80 students know Kannada, 70 students know Hindi. If 40 students know all the three languages, find the number of students who know exactly two languages.
Answer:
writing n(E)= 100, n(K)= 80, n(H)= 70 .
n(E ∪ K ∪ H) = 200, n (E ∩ K N ∩ H) = 40
We know that n(E ∪ K ∪ H) = n(E) + n (K) + n(H) –n (E ∩ K)
– n (K ∩ H) -n (H ∩ E) + n (E ∩ K ∩ H)
∴ n (E ∩ K) + n (K ∩ H) + n (H ∩ E) = 90

Question 26.
Let R: Z → Z be a relation defined by R = {(a,b) : a, b, ∈ Z, a – b ∈ z}. Show that
(i) ∀ a ∈ Z, (a, a) ∈ R.
(ii) (a, b) ∈ R ⇒ (b,a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
Answer:
∀ a ∈ Z, (a, a) ∈ R since a – a = 0 ∈ Z
(a, b) ∈ R ⇒ a – b ∈ Z ⇒ ∴ b – a ∈ Z ⇒ (b, a) ∈ R
(a, b) ∈ R, (b, c) ∈ R ⇒ a – b ∈ Z, b – C ∈ Z
∴ a – b + b – c ∈ Z ⇒ (a, c) ∈ Z

1st PUC Maths Model Question Paper 1 with Answers

Question 27.
Prove that (cos x + cos y)2 + (sin x – sin y)= 4cos2 \(\left(\frac{x+y}{2}\right)\)
Answer:
LHS = cos2x + cos2 y + 2 cos x cos y + sin2x + sin2y – 2sin x sin y
= 1 + 1 + 2 (cos x cos y – sin x sin y) = 2 [1 + cos (x + y)] = 4cos2 \(\left(\frac{x+y}{2}\right)\)

Question 28.
Solve: √2x2 + x + √2 = 0.
Answer:
√2x2 + x + √2 = 0; Here a = √2, b = 1,c = √2
∴ D = 52 – 4ac = (1)2 – 4(√2)(√2) = 1 – 8 = -7
1st PUC Maths Model Question Paper 1 with Answers - 10

Question 29.
How many letters words with or with out meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ü) all letters are used at a time
(iii) all letters are used but first letter is a vowel?
Answer:
There are 6 letters in the word MONDAY. So, the total number of words is equal to the number of arrangements of these letters taken four at a time.
The number of such arrangements = 6P4 = \(\frac{6 !}{(6-4) !}=\frac{6 !}{2 !}=\frac{6.5 .4 .3 .2 .1}{2.1}\) = 360
Hence, total number of words = 360
(ii) Total number of arrangements in this case = 6P6 = 6! = 720
(ii) Total number of arrangements when all letters are used but the first letter is a vowel
= 2 × 5P5 = 2 × 5! = 2 × 5 × 4 × 3 × 2 × 1 = 240.

Question 30.
If x + iy = \(\frac{2+i}{2-i}\) then prove that x2 + y2
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 11

Question 31.
Find the term independent of x in the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 12
The term will be independent of x if the index of x is zero, i.e., 12 – 3r= 0. Thus, r=4
Hence 5th term is independent of x and is given by (-1)4 6C4 \(\frac{(3)^{6-8}}{(2)^{6-4}}=\frac{5}{12}\)

1st PUC Maths Model Question Paper 1 with Answers

Question 32.
8, A1, A2, A3, 24
Answer:
Tn = a + (n − 1)d
24 = 8 + (5 – 1)d
16 = 4d ⇒ d = 4
3 Am’s are 12, 16, 20.

Question 33.
(i) At least one man.
Answer:
1 man and I woman
or 2
men and 0 women
= 2C1 × 2C1 + 2C2 × 2C0 = 2 × 2 + 1 × 1 = 4 + 1 = 5.

(ii) At most one man
Answer:
1 men and 1 women
or
0 men and 2 women
= 2C1 × 2C1 + 2C0 × 2C2 = 2 × 2 + 1 × 1 = 4 + 1 = 5.

Question 34.
Find the derivative of the function ‘cos x’ w.r.t ‘x’ from first principle
Answer:
Let f ‘(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
1st PUC Maths Model Question Paper 1 with Answers - 13

Question 35.
A parabola with vertex at origin has its focus at the centre of x2 + y2 – 10x + 9 = 0 Find the direction its directrix and latus rectum.
Answer:
The centre of the circle = (5,0).
The equation y2 = 20x ⇒ y2 = 4ax ⇒ a = 5
directrix is x = -5, LR = 20.

Question 36.
If an A.P. if mth term is n and the nth term is m, where m ≠ n, find the pth term
Answer:
We have
am = a + (m – 1) d = n, …… (1)
and an = a + (n – 1) d = m …….. (2)
Solving (1) and (2), we get
(m, n) d = n – m, or d = -1,
and a = n + m – 1 ……. (3)
Therefore ap = a + (p – 1)d ……. (4)
= n + m – 1 + (p – 1) (-1)= n + m – p
Hence, the pth term is n + m – p.

1st PUC Maths Model Question Paper 1 with Answers

Question 37.
Verify by the method of contradiction that √2 is irrational
Answer:
If possible Let √2 is rational
∴ √2 = \(\frac{p}{q}\), p, q ∈ z, q ≠ 0
We assume that p and q donot have any common factor
p = √2q ⇒ p = 2q2
p2 is a multiple of 2 ⇒ ∴ p is a multiple of 2
∴ p = 2k, k ∈ z ⇒ p2 = 4k
2q2 = 4k2 ⇒ q2 = 2k2
q2 = 2k2
q2 is a multiple of 2 ⇒ ∴ q is a multiple of z
∴ p and q are both multiple of 2 and hence has a common factor z which is a contradiction
∴ our assumption is wron
∴ √2 is irrational

Question 38.
Two students Anil and Sunil a appear in the examination The prpbability that Anil will qualify in the examination is 0.05 and that Sunil will qualify is 0.10. The probability that both will qualify in the examination is 0.02. Find the probability that Anil and Sunil will not qualify in the examination.
Answer:
Let A, B denote the events that Anil, Sunil qualify in the exam
P (A) = 0.05, P(B) = 0.1, P (A ∩ B) = 0.02
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.05 + 0.1 – 0.02 = 0.13
P(A’ ∩ B’) = 1 – P (A ∪ B) = 1 – 0.13 = 0.87.

Part – D

Answer any Six questions : (6 × 5 = 30)

Question 39.
Define signum function. Draw the graph of the signum function. Also write its domain and range.
Answer:
Singum function :
The function f:R → R defined by
1st PUC Maths Model Question Paper 1 with Answers - 14
1st PUC Maths Model Question Paper 1 with Answers - 15
is called the signum function. The domain of the signum function is R and the range is the set {-1, 0, 1}.

1st PUC Maths Model Question Paper 1 with Answers

Question 40.
Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\) (q being in radians) and hence show that \(\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\)
Answer:
Consider a circle with centre O and radius ‘r’ Mark two points A and B on the circle so that
1st PUC Maths Model Question Paper 1 with Answers - 16
Join AB. Draw BM ⊥ OA
From the figure it is clear that.
Area of ∆OAB < Area of sector OAB < sector of ∆OAC …. (1)
Area of ∆OAB = \(\frac { 1 }{ 2 }\) OA.BM
[In ∆OBM. Sin θ \(\frac{\mathrm{BM}}{r}\) ⇒ BM = r Sin θ ]
1st PUC Maths Model Question Paper 1 with Answers - 17
∴ Area of ∆OAB = \(\frac { 1 }{ 2 }\) r r. sin θ = \(\frac { 1 }{ 2 }\)r2 sin θ
Area of sector OAB = \(\frac { 1 }{ 2 }\) r2θ
Area of ∆OAC = \(\frac { 1 }{ 2 }\) OA.AC
= \(\frac { 1 }{ 2 }\)r2 tan θ [In ∆OAC, tan θ = \(\frac{A C}{O A}=\frac{A C}{r}\) ⇒ AC = r tan θ]
∴ (1) becomes
1st PUC Maths Model Question Paper 1 with Answers - 18
1st PUC Maths Model Question Paper 1 with Answers - 19

Question 41.
12 + 22 + 32 + ………… + n2 = \(\frac{n(n+1)(2 n+1)}{6}\)
Answer:
Let P(n) : 12 + 22 + 32 + ………… + n2 = \(\frac{n(n+1)(2 n+1)}{6}\)

step 1 : Prove that P(1) is true
when n= 1; L.H.S = 12 = 1
R.H.S = \(\frac{1(1+1)(2+1)}{6}\)
∴ L.H.S = R.H.S ⇒ ∴ P(1) is true

step 2: Assume that P(m) is true
i.e., 12 + 22 + 32 + ………… + m2 = \(\frac{m(m+1)(2 m+1)}{6}\) …… (i)

step 3: Prove that P(m + 1) is true
i.e., 12 + 22 + 32 + ………… + m2 + (m+1)2 = \(\frac{(m+1)(m+2)(2 m+3)}{6}\)
L.H.S = [12 + 22 + 32 + ………… + m2] + (m+1)2
1st PUC Maths Model Question Paper 1 with Answers - 20
= R.H.S ⇒ ∴ P(m+1) is true
Conclusion: P(1) is true ⇒ P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction, the result is true for all natural numbers ‘n’.

1st PUC Maths Model Question Paper 1 with Answers

Question 42.
A group consists of 7 boys and 5 girls. Find the number of ways in which a team of 5 members can be selected so as to have atleast one boy and one girl.
Answer:
Number of ways of selecting
1 B and 4 G = 7C1 × 5C4 = 35
2 B and 3G = 7C2 × 5C3 = 210
3 B and 2 G = 7C3 × 5C2 = 350
4. B and 1 G = 7C4 ×5C1 = 175
Total number of selections = 770.

Question 43.
State and prove binomial theorem for positive integers.
Answer:
Statement: “If’n’ is a +ve integer then
(x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 + ………….. + nCn x0 an.

Proof : (By Mathematical induction):
Let p(n) : (x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2a2 + ………….. + nCn x0 an.

Step1 : Prove that P(1) is true
when n = 1, L.H.S = (x + a)1 = x + a; RHS = 1c0 x1 a0 + 1c1 x0 a1 = 1.x.1 + 1.1.a = x+a
L.H.S. = R.H.S. ⇒ ∴ P(1) is true

Step 2: Assume that P(m) is true
i.e., (x + a)m = mc0 xm a0 + mc1 xm-1 a1 + mc2  xm-2 a2 + ………….+ mcmx0am …….(i)

Step 3 : Prove that P(m + 1)m+1 is true
i.e. (x + a)m+1 = m+1c0 xm+1 a0 + m+1c1 xm a1 + m+1c2 xm-1 a2 + …….. + m+1cm+1 x0 am+1
Multiply both sides of equation (i) by (x+a)
∴ (x + a)m (x + a) = (x + a) [mc0 xm a0 + mc1 xm-1 a1 + mc2 xm-2 a2 + ………….+ mcmx0am]
(x + a)m+1 = x [mc0 xm a0 + mc1 xm-1 a1 + mc2 xm-2 a2 + ………….+ mcmx0am] + a [mc0 xm a0 + mc1 xm-1 a1 + mc2 xm-2 a2 + ………….+ mcmx0am]
(x + a)m + 1 = mc0xm+1 a0 + mc1xma1 + mc2 xm-1 a2 + ……. + mcm x0 am+1
= mc0xm+1a0 + (mc1 + mc0)xm a1 + (mc2 + mc1) xm-1 a2 + ………… + (mcm + mcm-1) x1 am + mcm x0 am+1

(x + a)m+1 = m+1c0 xm+1 a0 + m+1c1 xm a1 + m+1c2 xm-1 a2 + …….. + m+1cm+1 x0 am+1
[∴ mcm = m+1c0
mc1 + mc0 = m+1c1
mc2 + mc1 = m+1c2
m+1cm]
⇒ ∴ P(m + 1) is true
Conclusion: P(1) is true, P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction the result is true for all natural numbers n.

Notes: 1. The number of terms in the expansion of (x + a)n is n + 1.
2. The Gen term of binomial expansion is Tr+1 = ncr xn-r ar.

1st PUC Maths Model Question Paper 1 with Answers

Question 44.
Derive an expression for the coordinates of a point that divides the line joining the points A(x1, y1, z1,) and B (x2, y2, z2.) internally in the ratio m : n. Hence, find the coordinates of the midpoint of AB where A = (1, 2, 3) and B = (5, 6, 7).
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 21
Let the two given points be P (x1, y1, z1,) and Q (x2, y2, z2). Let the point R (x, y, z) divide PQ in the given ratio m : n internally, Draw PL,
QM and RN perpendicular to the XY-plane. Obviously PL ∥ RN ∥ QM and feet of these perpendiculars lie in a XY-plane. The points L, Mand N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-Plane. Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig 12.5.

Also note that quadrilaterals LNRS and NMTR are parallelograms.
The triangles PSR and QTR are similar. Therefore,
1st PUC Maths Model Question Paper 1 with Answers - 22
1st PUC Maths Model Question Paper 1 with Answers - 23
Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get
1st PUC Maths Model Question Paper 1 with Answers - 24
Hence, the coordinates of the point R which divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2,) internally in the ratio m: n are
1st PUC Maths Model Question Paper 1 with Answers - 25

1st PUC Maths Model Question Paper 1 with Answers

Question 45.
Derive a formula for the angle between two lines with slopes m1 and m2. Hence find the slopes of the lines which make an angle \(\frac{\pi}{4}\) with the line x – 2y + 5 = 0.
Answer:
m1 = tan θ1, m2 = tan θ2
1st PUC Maths Model Question Paper 1 with Answers - 26

Question 46.
Prove that \(\frac{\cos 4 x+\cos 2 x+\cos 3 x}{\sin 4 x+\sin 2 x+\sin 3 x}=\cot 3 x\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 27

Question 47.
Solve graphically 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Answer:
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6 …..(1)
Points A(2, 0) and B(0,4) lie on 2x + y = 4
Plot the points and join them to get line AB
(0,0) does not satisfy 2x + y ≥ 4
⇒ Half plane given by 2x + y ≥ 4 is away from origin …..(2)
Points C(3, 0) and D(0, 3) lie on x + y= 3
Plot the points and join them to get line CD.
(0,0) satisfies x + y ≤ 3 ….. (3)
⇒ Half plane given by x + y ≤ 3 is towards origin
Points C(3, 0) and E(0, -2) lie on 2x – 3y = 6
1st PUC Maths Model Question Paper 1 with Answers - 28
Plot the points and join them to get line CE.
(0,0) satisfies 2x – 3y ≤ 6
⇒ Half plane given by 2x – 3y ≤ 6 is towards origin
From (2), (3), (4) common region shown shaded in figure represents solution of (1).

1st PUC Maths Model Question Paper 1 with Answers

Question 48.
Find the mean deviation about the mean for the following data.
1st PUC Maths Model Question Paper 1 with Answers - 29
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 30
1st PUC Maths Model Question Paper 1 with Answers - 31

Part – E

Answer any ONE questions : (1 × 10 = 10)

Question 49.
(a) To cos (A + B) = cos x. cos y – sin x sin y and hence find cos 75°.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 32
(i) becomes
∴ cos (A + B)= cos A.cos B – sin B
cos 75° = cos (45 +30)
= cos 45 cos 30 – sin 45 sin 30
1st PUC Maths Model Question Paper 1 with Answers - 33

1st PUC Maths Model Question Paper 1 with Answers

(b) Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + ……..
Answer:
nth term Tn = 12 + (12 + 22) + (12 + 22 + 32) + ……. + n2 = \(\frac{n(n+1)(2 n+1)}{6}=\frac{2 n^{3}+3 n^{2}+n}{6}\)
∴ Tn = \(\frac{2 n^{3}+3 n^{2}+n}{6}\)
Sum to n terms, Sn = ΣTn
= \(\frac{1}{6}\) [2Σn3 +3Σn2 + Σn]
1st PUC Maths Model Question Paper 1 with Answers - 34
∴ Sn = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Question 50.
(a) An elipse is the set of all ninte in a plane tha e of whose distance from two fived points in the need of all points in a plane the sun of whose distance points in the plane is a constant.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 35
Let F1 and F2 be the foci and O be the mid point of the line segment F1F2 Let O be the origin and the line from 0 through F2 be the positive x-axis and that through F1 as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F1 be (-c, 0) and F2 be (c, 0).

Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a
i.e., PF1 + PF2 = 2a. …(1)
Using the distance formula, we have
\(\sqrt{(x+c)^{2}+y^{2}}+\sqrt{(x-c)^{2}+y^{2}}=2 a\)
i.e., \(\sqrt{(x+c)^{2}+y^{2}}=2 a-\sqrt{(x-c)^{2}+y^{2}}\)
Squaring both sides, we get
(x + c)2 + y2 = 4a2 – 4a \(\sqrt{(x-c)^{2}+y^{2}}\) + (x – c)2 + y2
which on simplification gives
\(\sqrt{(x-c)^{2}+y^{2}}=a-\frac{c}{a} x\)
Squaring again and simplifying, we get
1st PUC Maths Model Question Paper 1 with Answers - 36

1st PUC Maths Model Question Paper 1 with Answers

(b) \(y=\frac{x^{5}-\cos x}{\sin x}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 37

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2

Students can Download Maths Chapter 2 Bijoktigalu Ex 2.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2 1
KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2 2

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2 3

2nd PUC Biology Model Question Paper 3 with Answers

Students can Download 2nd PUC Biology Model Question Paper 3 with Answers, Karnataka 2nd PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

  1. This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
  2. All the parts are Compulsory.
  3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

I. Answer the following questions in One Word or One Sentence each : ( 10 × 1 = 10 )

Question 1.
Name the site of fertilization in female human beings.
Answer:
Upper limb (Ampullary – isthmic junction) of fallopian tube.

Question 2.
Define adaptive radiation
Answer:
It is the evolution of different animals in different direction from a common ancestor adapting to different ecological niches. It is also called as divergent evolution.

Question 3.
Name the organism that causes typhoid.
Answer:
Salmonella typhi.

KSEEB Solutions

Question 4.
What is green revolution?
Answer:
The dramatic increase in food production due to improvement of high yielding and disease resistant crop varieties is called green revolution.

Question 5.
Name the microbe that causes big holes in the Swiss cheese.
Answer:
Propioni bacterium sharmanii

Question 6.
Why does the logistic growth curve becomes S shaped?
Answer:
In sigmoid growth unlimited resources are rare, population growth is, therefore becomes stable due to environmental resistance.

Question 7.
Breast feeding during the initial period of infant growth is recommended. Give reason?
Answer:
The yellowish milk (colostrum) secreted from mammary glands during the initial days contains antibodies (IgA) to boost the development of immune system of the child.

Question 8.
Give an example for chemical carcinogen that causes cancer.
Answer:
Alcohol or tobacco or coal tar or coal gas or asbestos cement etc

Question 9.
What is totipotency?
Answer:
The capacity to generate a whole plant from any cell or explant is called totipotency

KSEEB Solutions

Question 10.
How do birds cope with temporary fluctuations in their environmental conditions?
Answer:
They migrate to hospitable areas for food and breeding purposes during unfavourable conditions in their habitat.

Part – B

II. Answer any FIVE of the following questions in 3 – 5 sentences each, wherever applicable : ( 5 × 2 = 10 )

Question 11.
Name the plant that shows unusual flowering phenomenon in hilly tracks of Karnataka, Kerala and Tamilnad.
Answer:
Strobilanthus kuluthiana ( Neelakuranji)

Question 12.
What is Biochemical Oxygen Demand (BOD)? Mention its significance.
Answer:
BOD refers to the amount of oxygen that would be consumed if all the organic matter in one liter of water were oxidized by bacteria.

Significance

  • The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water.
  • Indirectly BOD is a measure of the organic matter present in the water.
  • The greater the BOD of waste water more is its polluting potential.

Question 13.
Differentiate test cross and out cross.
Answer:
Test cross
It is a cross between F1 hybrid and homozygous recessive parent.

Out cross
It is a cross between F1 hybrid and homozygous dominant parent.

KSEEB Solutions

Question 14.
Give two examples for natural selection in organisms.
Answer:

  1. Industrial melanism in Biston betularia
  2. Antibiotic resistance in bacteria

Question 15.
What are single cell proteins? Mention their importance?
Answer:
Dried biomass of a single species of microbes that can be used as protein source in the diet is known a single cell protein (SCP).

Single cell proteins (SCP) are alternative source of proteins used to feed the ever increasing population with protein rich diet to overcome the problems of protein malnutrition.

Question 16.
Why is India said to have greater ecosystem diversity than Norway?
Answer:
India with diverse ecosystems such as rain forests, coral reefs, wet lands, coastal areas, mangroves, estuaries and deserts has greater ecosystem diversity than Norway.

Question 17.
Draw a diagram of sparged tank bioreactor and label the parts.
Answer:
2nd PUC Biology Model Question Paper 3 with Answers 1

KSEEB Solutions

Question 18.
Explain the mechanism of action of restriction endonucleases.
Answer:
Restriction endonucleases recognize a specific base sequence of a DNA molecule.
They cut double stranded DNA at different sites. This generates protruding (51 or 31) ends. Such ends are called cohesive or sticky ends. These ends can join with single stranded sequences of other DNA having complementary base pairs.
E.g. EcoRI, Bam H1
2nd PUC Biology Model Question Paper 3 with Answers 2
Cuts symmetrically placed around the line of symmetry
2nd PUC Biology Model Question Paper 3 with Answers 3

Part – C

III. Answer any FIVE of the following Questions in 40-80 words each, wherever applicable. ( 5 × 3 = 15 )

Question 19.
Describe the different phases of life cycle.
Answer:
Juvenile phase
The period of growth from birth till individual attains sexual maturity is called Juvenile phase. It is known as vegetative phase in plants.

Reproductive phase
The period of growth after Juvenile or vegetative phase during which individual reaches sexual maturity and capable of producing gametes is called reproductive phase. In human beings reproductive phase begins on the onset of puberty usually after 13 or 14 years. In plants flowering marks the beginning of reproductive phase.

Senescent phase
The phase begins after reproductive phase and marks the end of growth cycle in organisms.

KSEEB Solutions

Question 20.
Give a brief account of Human Origin and evolution.
Answer:

  • Human evolution states that humans developed from primate or ape like ancestors.
  • Human origin occurred in Central Asia and Africa, China, Java and India (Shivalik hills).
  • The first mammals were shrew like terrestrial, insectivores or rat like creatures.
  • The insectivorous mammals namely tree shrews gave rise to the primitive primates called prosomians which include tarsiers, lemurs and lories and anthropoids (marmosets, baboons, monkeys, apes and man).
  • Modem man Homo sapiens belongs to class mammalia, order primata and suborder anthropoidea.
  • Dryopithecus africanus is regarded as common ancestor of man and apes.
  • It is lived about 20 – 25 million years ago (Miocene).
  • Dryopithecus gave rise to Rama pithecus; he appeared 14-15 million years ago.
  • The fossil of Rama pithecus was discovered from Pliocene rocks of Shivalik hills of India by Edward Lewis.
  • Rama pithecus gave rise to Australo pithecus after a gap of 9 – 10 million years.
  • Rama pithecus the first ape man commonly called southern ape .He appeared about 5 million years ago (early Pleistocene).
  • Homo habillus lived in early Pleistocene about 2 million years ago.
  • He was the first tool maker or handy man.
  • Homo erectus is the direct ancestor of modern man.
  • Homo erectus evolved from homo habillus about 1-7 million years ago in middle Pleistocene.
  • Homo erectus include Java ape man, pieking man and Heidelberg man.
  • The Neanderthal man existed in the late Pleistocene he was roughly equal to the modem man.
  • Cro-Magnon is regarded as direct and most recent ancestor of the living modem man.
  • He lived about 34.0000 years ago in Holocene epoch.
  • The cranial capacity was about 1650cc and considered more intelligent than the man of today.
  • The modern man Homo sapiens arose in Africa about 25.0000 years ago in Holocene epoch.
  • It is believed that the man of today appeared about 11.000 or 10.000 years ago in the region around Caspian and Mediterranean seas from their members migrated west wards, eastwards and south wards respectively changing into present day white or Caucasoid, Monogoloid and black or Negroid races.
  • Modern man underwent cultural evolution. Cave art developed about 18.000 years ago.
  • Agriculture came around 10.000 years back and human settlement started.

Question 21.
List the various public health measures as a safeguard against infectious diseases?
Answer:
Following are the public health measures, considered safe against infectious diseases.

  1. Maintenance of personal and public hygiene like proper disposal of waste and excreta through proper sanitation.
  2. Provision of safe water supply, periodic cleaning and disinfection of water reservoirs, pools, and tanks, etc.
  3. Standard practices of hygiene in public health catering.
  4. Avoiding stagnation of water in and around residential areas to prevent breeding of mosquitoes and vectors.
  5. Spraying of insecticides like DDT in ditches, drainage areas, swamps, etc., to prevent mosquito breeding.
  6. Measures against vector borne diseases like dengue and chikungunya, include fixing of wire meshes to doors and windows to prevent the entry of mosquitoes.
  7. Regular health camps, immunization (Vaccination) programmes to prevent the spread of infectious diseases.

KSEEB Solutions

Question 22.
What is food chain? Describe the types of food chain.
Answer:
The transfer of food energy from producers to decomposers through a series of organisms with repeated eating and being eaten is called food chain.
1. Grazing food chain
The food chain begins with producers (green plants) at the first trophic level. It involves grazing Of grass by herbivores and flow of energy occurs from smaller organisms (herbivores) to larger animals (consumers). It is based on prey – predator relationship and is therefore, called predator food chain.
Grass → Deer → Tiger

2. Parasitic food chain
The food chain begins with producers and flow of energy goes from larger organisms (host) to smaller organisms (parasites) as parasites draw nourishment from the host. This is based on host – parasite relationship.
2nd PUC Biology Model Question Paper 3 with Answers 4

3. Detritus or saprophytic food chain
The food chain begins with dead and decaying organic matter (detritus). Which is acted upon by decomposers or saprophytes and energy from decomposers goes to detrivores.
Leaf litter → Earth worm → bird → hawk

Question 23.
Define the terms
(a) Emasculation
Answer:
The removal of anthers from the flower bud of a bisexual flower before the anther dehisces is called emasculation.

(b) Bagging
Answer:
The process of covering the emasculated flowers with a bag of suitable size to prevent contamination with unwanted pollen is called bagging.

KSEEB Solutions

Question 24.
What is transformation? Explain any two methods of introduction of recombinant DNA in to host organism.
Answer:
Transformation is the process by which plasmid or DNA can be introduced into a cell.

1. Gene gun or particle gun or Biolistics method
It is a popular and widely used direct gene transfer method for delivering foreign genes into any cells and tissues or even intact seedlings.

  • The foreign DNA is coated or precipitated onto the surface of minute gold or tungsten particles (1-3 pm).
  • It is bombarded or shot onto the target tissue or cells using the gene gun or micro projectile gun or shot gun.with a device much like a particle gun. Hence the term biolistics
  • The bombarded cells or tissues are cultured on selection medium to regenerate plants from the transformed cells.

2. Micro injection
In this technique a solution of DNA is directly injected in to the host nucleus using a fine micro capillary pipette or micro syringe, under a phase contrast microscope to aid vision.

Question 25.
Define trophic level. Show schematically the different tropic levels in a food chain.
Answer:
The amount of energy available at each step in food chain is called Trophic level.
2nd PUC Biology Model Question Paper 3 with Answers 5

KSEEB Solutions

Question 26.
What is pedigree? Show diagrammatic representation of pedigree for sickle cell anemia.
Answer:
Pedigree is a chart of graphic representation of record of inheritance of a trait through several generations in a family
2nd PUC Biology Model Question Paper 3 with Answers 6

Part – D

Section – I

IV. Answer any FOUR of the following questions on 200 – 250 words each, wherever applicable . ( 4 × 5 = 20 )

Question 27.
What are assisted reproductive technologies? Describe the various reproductive technologies to assist an infertile Couple to have children.
Answer:
The couple can be assisted to have children through certain special techniques known as assisted reproductive technologies (ART).
1. In vitro fertilization and embryo transfer (IVF-ET)

  • In this technique egg cells are fertilized by sperm (usually 100,000 sperm / ml) outside the body in almost similar conditions as that in the body.
  • In this process Ova from the wife or donor female and sperm from husband or donor male are allowed to fuse in shallow containers called Petri dishes. (Made of glass or plastic resins) Under laboratory condition.
  • The fertilized egg (zygote) is then transferred to the patient’s uterus this called embryo transfer (E.T.), Where she provides suitable conditions for the development of embryo.
  • The baby born by this technique is called test tube baby.

2. ZIFT ( Zygote intra fallopian transfer)
Zygote or early embryo up to eight blastomeres is transferred into the fallopian tube.

3. Gamete intra fallopian transfer (GIFT)
Transfer of an ovum collected from a donor to fallopian tube of another female who cannot produce ova, but provide suitable conditions for fertilization and further development up to parturition.

4. Intra uterine transfer (IUT)
Embryo with more than eight blastomeres is transferred in to the uterus.

5. Intra cytoplasmic sperm injection (ICSI)
The sperm is directly injected into the cytoplasm of the ovum to form an embryo in the laboratory and then embryo transfer is carried out.

6. Artificial insemination (AI)

  • This method is used in cases where infertility is due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates.
  • In this method, the semen collected from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (IUI – Intra uterine insemination).

KSEEB Solutions

Question 28.
What is dihybrid cross? State and explain Mendel’s second law of independent assortment.
Answer:
When a cross is made between two parents differing in two pairs of contrasting characters like round yellow and wrinkled green is called dihybrid cross.

Based on the results obtained from dihybrid cross, Mendel postulated his second law “The law of independent assortment”. It states that, when two pairs of contrasting characters are brought into the hybrid they behave independently of each other without contamination. The presence of one character is not contaminated with the other character. These characters are assorted independently of one another during the formation of gametes.
Mendel conducted dihybrid cross for which he selected two characters at a time.

2nd PUC Biology Model Question Paper 3 with Answers 7
Self crossed (F2)
2nd PUC Biology Model Question Paper 3 with Answers 13

KSEEB Solutions

Question 29.
What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Answer:

2nd PUC Biology Model Question Paper 3 with Answers 8
It is a process by which DNA produces daughter DNA molecules, which are exact copies of the original DNA. In each new DNA molecule, one strand is old (original) while the other is newly formed. Hence, Watson and Crick described this method as semi-conservative replication.

Meselson and Stahl conducted experiments on E.coli to prove that DNA replication is semi conservative.

  1. They grew E.coli cells in a culture medium containing radioactive isotopes N15 (NH4cl).
    The 15N was incorporated into the newly synthesized DNA and other nitrogen containing compounds.
  2. Bacteria labeled with 15N are transferred into a medium containing normal 14NH4cl.
  3. The analysis was done after every generation at regular intervals.
  4. The DNA samples were separated by centrifugation on CSCl2 gradient.
  5. The hybrid DNA separated from fresh generations possesses one heavier strand contributed by parental with 15N isotope which is radioactive and lighter strand is newly synthesized with 14N isotope, is non-radioactive.
  6. The DNA of the second generation showed that in one DNA molecule one strand was radioactive (15N) and other was non radioactive (14N), Whereas both the strands of other DNA molecule were non-radioactive (14N).
  7. It is clear that the DNA of the first generation is intermediate it contains both 15N and 14N.
  8. In the second generation there were two types of DNA one heavier DNA with 15N and 14N isotopes while the lighter DNA with 14N and 14N, the ratios of two types of DNA was 50:50.
  9. In the third generation the ratio of intermediate (15N and 14N) was reduced from 50% to 25% while the lighter fraction (14N and 14N) increased from (50% – 75%).
  10. In the 4th generation the percentage of intermediate still reduced to 12.5% while that of lighter fraction increased from 75% to 87.5%.
  11. These experiments prove that DNA shows semi conservative replication.

KSEEB Solutions

Question 30.
What is water pollution? Write short notes on the following with respect to water pollution.
a) Eutrophication
b) Biomagnifications
Answer:
Any undesirable change in the physical, chemical and biological characteristics of water, which makes water unfit for human use and harmfully affect aquatic organisms is called water pollution.

a) Eutrophication
It the process of nutrient enrichment of water bodies and subsequent loss of species diversity like fishes.
Harmful effects

  1. Excess nutrients cause algal bloom, which may cover the whole surface of water body and release toxins.
  2. It causes oxygen deficiency in water that leads, to the death of aquatic animals like fishes.

b) Biomagnifications
It refers to increase in concentration of toxic substances at successive trophic levels in a food chain. Example: Biomagnifications of DDT in an aquatic food chain.

Harmful Effect
High concentration of DDT disturbs calcium metabolism in birds, which causes thinning of egg shell and their premature breaking, causing decline in bird’s population.

Question 31.
Explain the structure of an anatropous ovule with a neat labeled diagram.
Answer:
2nd PUC Biology Model Question Paper 3 with Answers 9
Ovule is the female gametophyte develops inside the ovary from a cushion like structure called placenta. It is attached to placenta by a stalk known as funicle. The nutritive tissue enclosed inside the ovule is the nucellus. Nucellus is surrounded by outer and inner integuments. These integu-ments leave a small pore at the anterior end called micropyle. Anterior end of the ovule is micropyle chalazal end is the posterior part.

KSEEB Solutions

Question 32.
Distinguish DNA and RNA with respect to their structure or Chemistry and function.
Answer:

DNA RNA
1. Present mainly in the chromatin the nucleus. 1. Most of RNA is present in the cell of cytoplasm (90%) and a little (10%) in the nucleolus
2. Normally double stranded and rarely single stranded. 2. Normally single stranded and rarely double stranded.
3. It contains deoxyribose sugar. 3. It has ribose sugar.
4. It contains bases like Adenine, Guanine, cytosine and Thymine. 4. It contains bases like, Adenine, Guanine cytosine and uracil.
5. DNA acts as template for its Synthesis. 5. RNA does not act as a template for its synthesis.

Section – II

Answer any THREE of the following questions in 200 – 250 words each, wherever applicable. ( 3 × 5 = 15 )

Question 33.
What is biocontrol? Name the principle behind biological method of pest control. Mention examples of biocontrol agents and their function.
Answer:
It is the use of micro organisms to control or eliminate insect pests. The micro organisms employed in biological control are called bio control agents.

Principle
It is based on prey – predator relationship.

Examples
1. Ladybird and Dragon flies useful to get rid of aphids and mosquitoes.

2. Bacillus thuringiensis (Bt) is used to control butterfly caterpillar. Spores available in sachets are mixed with water and sprayed on plants, eaten by insect larva, toxin released in gut kills larvae.
Example: Bt toxin genes are introduced into cotton plants and made resistant to insect pests such as cotton boll worms, stem borer, aphids and beetles.

3. Nucleo polyhedrovirus ( NPV) is a virus suitable for narrow spectrum insecticide applications. It has no negative impacts on plants, mammals, birds, fish or target insects. It is suitable for overall integrated pest Management programme (IPM) in ecologically sensitive areas.

Question 34.
What is logistic growth? Explain sigmoid curve with a suitable diagram.
Answer:
The rate of population growth slows as the population size approaches carrying capacity, leveling to a constant level.
2nd PUC Biology Model Question Paper 3 with Answers 10
Logistic growth curve or S-shaped growth curve (sigmoid growth curve) is characteristic of all higher animals including man and microbes like yeast cells growing in a natural environment.

Logistic growth is a pattern of growth in which the population density of few individuals introduced into the new habitat increases slowly initially, in a positive acceleration phase (lag phase) then increases rapidly, approaching an exponential growth rate (log phase) due to abundance of food, favourable environmental conditions such a climate, few predators and low levels of disease. Then declines in a negative acceleration phase.

This decline reflects increasing environmental resistance at higher population densities. This type of population growth is termed density-dependent, since growth rate depends on the numbers present in the population. The point of stabilization, or zero growth rates, is termed the saturation value or carrying capacity (K) of the environment for that organism.

When population growth rate is plotted graphically against time S shaped or sigmoid curve is obtained. This type of population growth is also called Verhulst- pearl logistic growth.
The mathematical equation for sigmoid growth is
dN / dt = rN(N – K) / k
Where
N = Population size at a given time
t = Time,
r = intrinsic rate of natural increase
k = Saturation value or carrying capacity for that organism in that environment.

KSEEB Solutions

Question 35.
What is animal breeding? Explain the controlled methods of animal breeding and mention their significance.
Answer:
The production of new varieties of animals which are superior to their parents through selective breeding or mating is called animal breeding.
There are two control breeding methods

  1. Artificial insemination
  2. Multiple ovulation embryo transfer technology (MOET)

Artificial insemination
Semen of superior male is collected and-injected unto the reproductive tract of selected female.

Significance:

  1. It helps to overcome several problems of normal mating.
  2. Semen of a desirable bull can be inseminated into cows.
  3. Semen can be stored by cryopreservation to use at later stages and can be easily transported to distant places.
  4. Semen of single ejaculation can be used to inseminate large number of cows.

Multiple Ovulatuion Embryo transfer technology (MOET)
Technique for herd improvement by successful production of hybrids

  1. Hormones (FSH) are administered to the cow for inducing follicular maturation and super ovulation.
  2. Cow produces 6-8 eggs instead of one egg and is either mated with, superior bull or artificially inseminated.
  3. Fertilized egg at 8-32 cell stage are recovered non-surgically & transferred to surrogate mother.

Significance:

  1. It is practiced in cattle, sheep, rabbits, buffaloes, etc.
  2. Genetically superior animals can be produced.
  3. It solves the problems of infertility.
  4. A single female can donate many ovules.
  5. Preservation and transport of embryos is convenient.

Question 36.
What is recombinant DNA technology? Draw a schematic representation of various steps involved in recombinant DNA technology.
Answer:
The branch of biotechnology that deals with altering the characters of an organism by introducing genes selected from desired organism is called genetic engineering or gene cloning or gene splicing.

2nd PUC Biology Model Question Paper 3 with Answers 11

KSEEB Solutions

Question 37.
What is gametogenesis? Mention the types. Explain spermatogenesis with a schematic representation.
Answer:
The process of formation of haploid male and female gametes in gonads is called gametogenesis.
It is of two types

  1. Spermatogenesis
  2. Oogenesis

2nd PUC Biology Model Question Paper 3 with Answers 12
1. Multiplication phase (mitosis)
Each testis is composed of numerous seminiferous tubules. Epithelial cells lining these tubules are called germinal epithelium. At the time of gametogenesis some cells of the epithelium become active, detach from the layer and start dividing mitotically to produce large number of primordial germ cells (2n), in turn primordial germ cells undergo repeated mitotic divisions to form sperm mother cells or spermatogonial cells which are diploid in nature.

2. Growth phase
Some of the spermatogonial cells (2n) stop dividing and enter into the seminiferous tubule where they grow in size by accumulating cytoplasm and duplication of DNA; sertoli or nurse cells provide necessary nutrients for the growing sperm mother cells, now sperm mother cells are termed as primary spermatocytes.

3. Maturation phase (meiosis)
Each primary spermatocyte (2n) enters into first meiotic division and forms two cells with haploid number of chromosomes called secondary spermatocytes. The secondary spermatocytes undergo second meiotic division to give rise to four haploid spermatids. The spermatids are unspecialized cells and do not undergo further division; all the four spermatids are transformed into four haploid sperms. This process of transformation of spermatids into mature and motile sperms is called spermiogenesis or spermateliosis.

1st PUC Economics Question Bank Chapter 1 Statistics for Economics Introduction

Students can Download Economics Chapter 1 Introduction Questions and Answers, Notes Pdf, 1st PUC Economics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Economics Question Bank Chapter 1 Statistics for Economics Introduction

1st PUC Economics Introduction TextBook Questions and Answers

Question 1.
Mark the following statements as true of false.
Answer:

  1. Statistics can only deal with quantitative data. False
  2. Statistics solves economic problems. True
  3. Statistics is of no use to economists without data. True

Question 2.
Make a list of activities that constitute the ordinary business of life. Are these economic activities?
Economics activities:
Answer:

  1. Working in a field.
  2. Working in a factory or office
  3. Shopkeeper
  4. Treatment to a patient by a doctor
  5. Teachers teaching in school

Non-Economic activities:

  1. Teacher teaching his own son
  2. Nurse serving her old mother
  3. Cooking of food by the housewife in her home.

KSEEB Solutions

Question 3.
The Government and policy makers use statistical development to formulate suitable policies of economic development. Illustrate with two examples.
Answer:
The statistical data is important for Government and policy makers to formulate suitable policies of economic development. It not only helps in analysing and evaluate the outcomes of the past policies but also assist them to take corrective measures and to formulate new policies accordingly. It is clear from examples.

1. It can be ascertained by using statistical techniques whether the policy of family planning is effective in checking the problem of rapidly growing population.

2. In preparing annual Government budgets, previous data of Government expenditures and Government resources are taken into consideration for estimating the allocation of funds among various projects.

Question 4.
‘You have unlimited wants and limited resources to satisfy them’. Explain by giving two examples?
Answer:
Every individual have unlimited wants but the resources for satisfying the wants are limited. Scarcity is the root of all economic problems. If there has been no scarcity there would have been no economic problem. This can be understood by examples.

1. Achild’s pocket money is limited so he/she has to choose only those things that they want the most. You can’t purchase almost all the things you want.

2. A land available should be put in use either in agricultural or industrial. We can’t use same land for both activities.

Question 5.
How will you choose the wants to be satisfied?
Answer:
Any individual fulfills his/her wants according to his/her needs satisfaction and priority attached to different wants. Moreover, the choice of want also depends on the need of the time and availability of means (money) to purchase that want.

Question 6.
What are your reasons for studying Economics?
Answer:
The reasons for studying economics are:

  • To study the theory of consumption
  • To study the theory of production
  • To study the theory of distribution
  • The study of economics also helps us to understand and analyse the root cause of basic problems faced by an economy.

KSEEB Solutions

Question 7.
Statistical methods are no substitute for common sense?
Answer:
This is true that statistical methods are no substitute for common sense. Statistical data should not be believed blindly as it can be misinterpreted or misused. The statistical data may involve personal bias or may undergo manipulations, also, statistical data and methods fail to reveal the errors committed by an investigator while surveying and collecting data. This can be understood by a story.

It is said that a family of four persons (husband, wife and two children) once set out to cross a river. So he calculated the average height of his families members since the average height of his family members was greater than the average depth of the river. He thought they could cross safely.

Consequently some members of the family (children) drowned while crossing the river. Thus commonsence must be used while applying statistical methods.

1st PUC Economics Introduction Additional Questions and Answers

1st PUC Economics Introduction Very Short Answer Type Questions

Question 1.
What is Economics?
Answer:
The term Economics is derived from two Greek words
1st PUC Economics Question Bank Chapter 1 Introduction Add Very Short img 1 Q 1
so totally it means ‘The art of house hold management’.

Question 2.
What is statistics in Singular Sense?
Answer:
In Singular Sense ‘Statistics is a science which concerns itself with the collection, presentation, analysis, and interpretation of numerical data.

Question 3.
What is Data?
Answer:
The term ‘data’ refers to ‘numerical information about social or economic activities’.

KSEEB Solutions

Question 4.
Who is called as the father of Economics?
Answer:
‘Adam Smith’ is called as the ‘father of Economics’.

Question 5.
Name the book written by Adam Smith?
Answer:
‘Wealth of Nations’ is the book written by Adam Smith, which was published in 1776.

Question 6.
What do you mean by scarcity in Economics?
Answer:
Scarcity means lack of something or anything which is in shortage of fulfill human wants.

Question 7.
What do you mean by ‘Ends’ and ‘Means’?
Answer:
According to Lionel Robbins ‘Ends’ refer to human wants and ‘Means’ refer to resources.

Question 8.
What do you mean by Production?
Answer:
The term production refers to the process of converting raw materials into finished goods. It also include raising of crops, forestry, animal husbandry, poultry farming etc.,

KSEEB Solutions

Question 9.
What do you mean by Consumption?
Answer:
The act of using goods and services of satisfy our wants is called consumption.

Question 10.
Wh at do you mean by Distribution?
Answer:
In Economics, distribution refers to the rewards paid to the four factors of production (Land, Labour, Capital and Organisation). It includes rent, wages, interest, and profit.

Question 11.
What is Exchange?
Answer:
The term ‘Exchange’ refers to purchase and sale of goods and services. It involves transfer of products from the seller to the buyer for price.

Question 12.
Who is the father of Statistics?
Answer:
Sir Ronald AFisher is called as father of modem statistics.

KSEEB Solutions

Question 13.
Mention the two types of data?
Answer:
The two types of data are:

  1. Quantitative data
  2. Qualitative data

Question 14.
Why does the problem of choice arise?
Answer:
The problem of choice arises due to the following reasons:

  1. Human wants are unlimited
  2. To satisfy unlimited wants, the resources are limited
  3. The resources can be put to alternative uses, though they are limited in supply.

OR

The problem of choice arises due to the unlimited wants and limited resources.

Question 15.
Who is a Consumer?
Answer:
A consumer is one who consumes goods & services.

Question 16.
Who is a Producer?
Answer:
A producer is one who produces and sells goods and services for the generation of income.

Question 17.
Give two examples of Quantitative Data?
Answer:

  1. Price of goods
  2. Income of individuals

KSEEB Solutions

Question 18.
Give two examples of Qualitative Data?
Answer:

  1. Intelligence of individuals
  2. Tastes and preferences of individuals.

1st PUC Economics Introduction Short Answer Type Questions

Question 1.
Name the four factors of Production?
Answer:
The factors of production are:

  1. Land
  2. Labour
  3. Capital
  4. Organisation

Question 2.
Write any two forms of scarcity as you see in your daily life?
Answer:
The two forms of scarcity as we see in our daily life are :

  1. Scarcity of water
  2. Scarcity of petrol
  3. Scarcity of agricultural resources etc.,
  4. Scarcity of power.

Question 3.
What is Exchange? What does it Study?
Answer:
The term ‘Exchange’ refers to purchase and sale of goods and services for something in return. It studies – market, prices, finance, and trade, etc.,

KSEEB Solutions

Question 4.
What is meant by statistical methods? What are the important statistical methods?
Answer:
Statistical methods refers to general principles and techniques which are commonly used in the collection, analysis, and interpretation of data.
Following are the important statistical methods:

  1. Collection of data
  2. Organisation of data
  3. Presentation of data
  4. Analysis of data
  5. Interpretation of data.

Question 5.
Write the important functions of Statistics?
Answer:
The important functions of Statistics are:

  1. Presentation of data or facts
  2. Simplification of facts
  3. Comparison of related facts
  4. It helps in correlating the data
  5. Statistics helps in forecasting the future behaviour

Question 6.
Briefly explain the three distinct component of Economics?
Answer:
Three distinct components of economics are consumption, production, and distribution,

1. Consumption – Using of goods for satisfying human needs is called consumption.
2. Production – Adding or increasing the utility of a commodities is known as production
3. Distribution – It means the distribution of the national income or the total income arising from what has been produced in the country called (GDP).

1st PUC Economics Introduction Long Answer Type Questions

Question 1.
Distinguish between ‘Quantitative’ and ‘Qualitative’ data and give some examples of both?
Answer:
Quantitative data:
Quantitative data is referred to as data measured with help of numerical facts or terms For example

  1. Daily temperature
  2. Fleight
  3. Weights of individuals
  4. Price
  5. Income etc.,

are quantitative variables. Their values can be expressed numerically.

Qualitative data :
When data is measured with the help of attributes or character it is Qualitative data. Qualitative data is the data which can not be measured with help of numerical values.

Example:

  1. Attitudes of people to a political system.
  2. Intelligence of individuals
  3. Music and art
  4. Beauty of individual cannot measured numerically.

KSEEB Solutions

Question 2.
What are your reasons for studying statistics?
Answer:
Statistics is widely used in modem times. Now-a-days statistics is useful to bankers, workers, labourers, workers, trade associations, insurance companies, auditor, social workers, chamber of commerce, politics etc.,

Uses of Statistics:
1. Useful to Politicians:
The politicians and their supporters are immensely interested in knowing their prospects of
winning election. It helps them to know the percentage of vote may candidate will receive in the election can be worked out.

2. Useful to insurance companies:
Statistics is also very useful to insurance companies. Insurance companies premium rates are determined with help of mortality tables which are based on mortality experience statistics.

3. Useful to economics :
Statistics serves as a raw materials to the economists in making certain economic laws.

4. Useful of finance minister:
The finance minister is responsible for preparing the annual budget of the country. For preparation of better annual budget plan there will be high necessity of statistics.

5. Bankers:
Statistics is very much useful for bankers and banking industry. Every bank has a statistical department. This department helps the bank to make its own policy.

6. Useful to Business man:
It is necessary for an efficient businessman to have the full knowledge of the factors which influence his business decisions. He must know the probable demand in future and we must calculate the cost of production. All this can be done with the help of statistics and statistical methods.

Question 3.
Give the limitations of Statistics?
Answer:
The following are the limitations of statistics:
1. It does not study the qualitative aspect of a problem.

2. It does not study the individuals:
It is study of a mass data and deals with aggregates of facts.

3. Statistical laws are true only on an average:
They are not universally applicable like the laws of chemistry, physics, and mathematics. They are true on an average because the results are affected by a large number of causes.

4. Statistics can be manipulated:
The results obtained can be manipulated according to one’s own interest and such manipulated results can mislead the community.

5. Statistical results lack mathematical accuracy:
The results drawn from statistical analysis are normally approximates. As the statistical analysis is based on observation of mass data, number of inaccuracies may be present and it is difficult to rectify them.

KSEEB Solutions

Question 4.
Describe the function of Statistics?
Answer:
The functions of statistics are:

1. Presentation of facts in figures:
Statistics presents the various facts and figures, in such a manner so that various complicated facts become easy, brief and concrete and they may be understood easily.

2. Establishment of relationship:
With the help of statistical methods (correction) relationship between two or more facts may be established.

3. Comparison:
Two or more effects may be compared with the help of statistics. Various averages, rates, percentages, and co-efficients are used for this purpose.

4. Interpretation of problems:
With the help of statistical devices, interpretation of various problems may be done.

5. Formation of laws and principles of other sciences:
Statistics helps to other social and natural sciences in forming the laws and with its help the various rules and principles of those sciences may be tested.

6. Helpful in deciding the policy:
With the help of statistics various policy decisions may be taken logically and successfully.

Question 5.
Name the stages of statistical study and statistical tools for presentation of data.

OR

Describe the main stages of statistics and their related tools.
Answer:
Stages of statistics:
Following are main stages of statistics.

1. Collection data:
It refers to gather some statistical facts by different methods.

2. Organisation of data:
In tills stage the organised data are presented in the form of graphs, diagrams, tables etc.,

3. Presentation of data:
In this stage the organised data are presented in the form of graphs, diagrams, tables etc.,

4. Analysis of data:
In the fourth stage we analyse the data in terms of averages of percentages.

5. Interpretation of data:
This is the last stage of statistics. In this stage, we interpret the data to find some conclusion.

Stage statistical study statistical tools
Stage-1 Collection of data techniques census or sample
Stage-2 Organisation of data Array of data and tally bar
Stage-3 Presentation of data Tables, graphs & diagrams
Stage-4 Analysis of data Percentage, averages
Stage-5 Interpretation of data Magnitude of percentage, average, degree of relationship between variables.

2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3

Students can Download Maths Chapter 5 Continuity and Differentiability Ex 5.3 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3

2nd PUC Maths Continuity and Differentiability NCERT Text Book Questions and Answers Ex 5.3

Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\) in the following

Question 1.
2x + 3y = sin x
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.1

KSEEB Solutions

Question 2.
2x + 3y = sin y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.2

Question 3.
ax + by2 = cos y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.3

Question 4.
xy + y2 = tan x + y
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.4

Question 5.
x2 + xy + y2 = 100
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.5
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.6

KSEEB Solutions

Question 6.
x3 + x2y + xy2 + y3 = 81
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.7

Question 7.
sin2y + cos xy = π
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.8

Question 8.
sin2 x + cos2 y = 1
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.9

KSEEB Solutions

Question 9.
\(y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.10

Question 10.
\(y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.11

Question 11.
\(y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.12

KSEEB Solutions

Question 12.
\(y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1\)
Answer:

Question 13.
\(y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.14
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.15

KSEEB Solutions

Question 14.
\(y=\sin ^{-1}(2 x \sqrt{1-x^{2}}),-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.16

Question 15.
\(y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0<x<\frac{1}{\sqrt{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Ex 5.3.17

1st PUC Chemistry Model Question Paper 2 with Answers

Students can Download 1st PUC Chemistry Model Question Paper 2 with Answers, Karnataka 1st PUC Chemistry Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Model Question Paper 2 with Answers

Time: 3.15 Hours
Max Marks: 70

Instruction:

  1. The questions paper has five parts A, B, C, D and E. All parts are compulsory.
  2. Write balanced chemical equations and draw labeled diagram wherever allowed.
  3. Use log tables and simple calculations f necessary (use of scientific calculations is not allowed).

Part – A

I. Answer ALL of the following (each question carries one mark): ( 10 × 1 = 1 )

Question 1.
Express 0.001023 into scientific notation.
Answer:
1.023 x 10-3.

Question 2.
Define critical temperature.
Answer:
The temperature above which gas cannot be liquified.

Question 3.
Give the example which acts as Lewis base as well as Bronsted base.
Answer:
NH3 or Ammonia.

KSEEB Solutions

Question 4.
How does electronegativity related to atomic size?
Answer:
Atomic size ∝ \(\frac{1}{\text { Electronegativity }}\)
Electronegativity

Question 5.
What is the oxidation state of oxygen in peroxide?
Answer:
-1 or minus 1

Question 6.
Which alkali metal is act as strong reducing agent?
Answer:
Lithium.

Question 7.
Write the formula of inorganic benzene.
Answer:
B3N3H3

Question 8.
What is dry ice?
Answer:
Solid CO2.

KSEEB Solutions

Question 9.
Complete the reaction NH4CNO
1st PUC Chemistry Model Question Paper 2 with Answers - 1
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 2

Question 10.
Write the name of the chain isomer of n-Butane.
Answer:
CH3-CH2-CH2-CH3

Part – B

II. Answer any FIVE of the following questions carrying TWO marks: ( 5 × 2 = 10 )

Question 11.
Calculate the average atomic mass of chlorine using the following data.
1st PUC Chemistry Model Question Paper 2 with Answers - 3
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 4

Question 12.
Draw the graph of pressure versus volume of a gas at a different temperatures to illustrate the Boyle’s law.\
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 5

Question 13.
Give any two conditions for hybridization of atomic orbitals.
Answer:
(i) Should contain half filled atomic orbitals.
(ii) Energy of the combining atomic orbitals should be equivalent.

Question 14.
Explain the reactivity of second group elements towards hydrogen.
Answer:
Hydrogen combines with second group elements to form metallic hydrides,
i.e. M + H2 → MH2
OR
Ca + H2 → CaH2
when M = Mg, Ca, Ba, etc.

KSEEB Solutions

Question 15.
How many sigma and pi bonds are in carbon monoxide?
Answer:
2σ bonds and one 7t-bond.

Question 16.
Write the structural isomers of an alkene with molecular formula C2H8.
Answer:
H3C-H2C-HC=CH2 , CH3-CH=CH-CH3

Question 17.
What are the characteristics for any ring system to be called as aromatic compound?
(i) It should be planar.
(ii) Obeys Huckel rule i.e. (4n + 2)π electrons.
(iii) Should contain delocalized π – electrons above and below the plane of the molecule.

Question 18.
(a) Which oxide of nitrogen in higher concentration will retard the rate of photosynthesis in plants?
NO2 or Nitrogen dioxide.
(b) Name the compound formed when carbon monoxide binds to haemoglobin.
Carboxyhaemoglobin (Hb + CO → HbCO).

Part – C

III. Answer any FIVE of the following questions carrying THREE marks: ( 5 × 3 = 15 )

Question 19.
Why Beryllium exhibit anomalous behaviour from the rest of the elements in the group.
Answer:
Due to

  1. Smaller size compared to other elements in the group.
  2. High ionisation enthalpy in the ground state.

Question 20.
With the help of hybridization explain the structure of methane.
Answer:
CH4
EC of C = 1s22s22px12py12pz1
The four half filled orbitals overlaps with each other giving 4sp3 hybrid orbitals. These combines with s-orbital of hydrogen along the axis giving 4σ bonds with bond angle 109°28′ and tetraheral geometry.
1st PUC Chemistry Model Question Paper 2 with Answers - 6

Question 21.
(a) What is bond enthalpy? How it is related to the bond order?
Answer:

  1. The amount of energy required to break one mole of bonds of same type to separate them into gaseous atoms.
  2. Bond enthalpy oc Bond order.

(b) Write the resonance structure of CO2.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 7

Question 22.
(a) Give any two differences between bonding and anti-bonding molecular orbitals.
Answer:
BMO :

  1. Formed by addition overlapping ΨAB = ΨA + ΨB
  2. Less energy than ABMO.
  3. More stable.

ABMO :

  1. Formed by the subtraction overlapping of atomic orbitals. of atomic orbitals i.e. ΨAB = ΨAB
  2. More energy than BMO.
  3. Less stable.

(b) What is the dipole moment of BeF2?
Answer:
Dipole moment is 0 (zero)
F ⇌ Be ⇌ F

KSEEB Solutions

Question 23.
Balance the following redox reaction by half reaction method.
MnO4 (aq) + 1 (aq) → MnO2 (s) + I2(s) : Basic medium.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 8

Question 24.
Give the reactions to show amphoteric nature of water.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 9
(ii) Mention any one method of removal of temporary hardness of water.
Answer:
By boiling.

KSEEB Solutions

Question 25.
How caustic soda is commercially prepared from brine by Castner-Kellner cell.
Answer:
Caustic soda by Castner-Kellner cell: NaOH is manufactured by the electrolysis of aqueous solution of NaCl (Brine).
i.e. 2NaCl → 2Na+ + 2Cl
Sodium ions are discharged at mercury cathode.
Sodium deposited at mercury forms sodium amalgam.
Chlorine liberated at anode removed from the cell.
At cathode : 2Na+ + 2e → Na, Na + Hg → Na – Hg
At anode : 2Cl – 2e → Cl2
Na-Hg is treated with water to form NaOH
i.e. Na/Hg + 2H2O → 2NaOH + H2 + Hg.

Question 26.
Give the example of element of group 14
(i) Shows maximum catenation capacity.
Answer:
Carbon

(ii) Used as semiconductor.
Answer:
Silicon

(iii) Which reacts with water.
Answer:
Tin

Part – D

IV. Answer any FIVE of the following questions carry ing FIVE marks: ( 5 x 5 = 25 )

Question 27.
(a) M atch the following :
1st PUC Chemistry Model Question Paper 2 with Answers - 10
Answer:
(i)-(c),
(ii)-(d),
(iii)-(a)

(b) Define molarity. Write the expression to calculate the molarity of the solution for the given mass and volume.
Answer:
It is the number of moles of solute present in 1000 ml solvent or 1dm3.
M = \(\frac{\mathrm{W}}{\mathrm{GMM}} \times \frac{1000}{\mathrm{V}}\)

Question 28.
(a) Give any three posulates of Bohr’s model for hydrogen atom.
Answer:

  • Electrons are revolving around nucleus in a closed circular path called orbits or main shells or energy levels.
  • When an electron jumped from higher energy level to lower energy level, the difference of energy emitted as radiation, i.e. E2 – E1 = ∆E= hγ
  • The angular momentum of an electron has discrete values. It is given by the equation
    mvr = \(\frac{n h}{2 \pi}\)

(b) Calculate the mass of a photon with wavelength 5,OA°
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 11

Question 29.
(a) Sketch the shapes of Px and dz2.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 12

(b) Identify the property exhibited by 19K40 and 19Ca40.
Answer:
Isobars (∵ mass number are equal).

(c) Write the orbital (box) type electronic configuration of p4 and d4 according to Hund’s rule of maximum of multiplicity.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 13

KSEEB Solutions

Question 30.
a) Derive ideal gas equation using gas laws.
Answer:
Ideal gas equation:
According to Boyle’s law V ∝ \(\frac{1}{P}\) T at constant T
According to Charle’s law V ∝ at constant P
According to Avogadro’s law V ∝ n at constant T and P
On coming V ∝ \(\frac{1}{P}\) V ∝ \(\frac{1}{P}\) x T x n
or PV=nRT
For ‘n’ moles, PV = RT for 1 mole.
R = gas constant, T = Kelvin Temp, P = Pressure, V = Volume of gas
n = Number of moles of gas.

(b) At 25°C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 10°C and volume of the gas is 640 ml.
Initial conditions
P1 = 760 mm
V1 = 600 mm
T1 = 250C = 298K

Final conditions
P 2= ?
v2 = 640 mL
T2= 10°C + 273k = 283K
1st PUC Chemistry Model Question Paper 2 with Answers - 14

Question 31.
(a) If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourization of 1 mole of water at 1 bar ad 100°C is 41kJ mole-1. Calculate the internal energy change when 1 mole of water is vapourized at 1 bar pressure and 100°C.
H2O (l) → H20 (g)
∆ng = 1 – 0 = 1
W.K.T. LH = ∆U + ∆ngRT
OR ∆U = ∆H – ∆ngRT = 41 x 103 – 1 x 8.314 x 373
= 41000— 3101,122 = 37898.878 J/mol
OR = 37.898878 K/J/mol.

(b) State Hess’s law of constant heat summation.
Answer:
The total amount of heat liberated or absorbed is same whether the reaction takes place in one step or more than one step / (or several steps).

(c) What is the value for standard enthalpy of formation of an element.
Answer:
Zero/0

Question 32.
(a) Calculate ∆G° for conversion of oxygen to ozone, -O2(g) —> O3(g) at 298K. If
Kp for this conversation is 2.47 x 10-29.
Answer:
∆rG° = ? R = 8.314J/k/mol T = 298K Kp = 2.47 x 10-29.
W.K.T. ∆rG° = -2.303RTlogKp = 2.303x 8.314 x 298 x 2.47 x 10-29.
= 163229 J/mol or 163.2 KJ/mol.

(b) What is thermochemical equation? Write the thermochemical equation for the molar combustion of ethanol (Given ∆rH° = -1367 kJ mol-1).

  1. It is a balanced reaction, gives the information about physical state of reactants and product as well as heat liberated or absorbed.
  2. C2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H20(/) ; ∆H° = -1367 KJ/mol

(c) What is the value of ∆G in a spontaneous process?
Answer:
∆G = -ve

KSEEB Solutions

Question 33.
For the equilibrium 2NOCl (g) ⇌ 2NO(g) + Cl2(g) the value of the equilibrium constant Kc is 3.75 x 10-6 at 1069K. Calculate the Kp for the reaction at this temperature?
Answer:
For the reaction 2NOCl (g) ⇌ 2NO(g) + Cl2(g)
∆n = 3 – 2 = 1. ∴ Kp = Kc (RT)∆n
Kp = (3.75 x 10-6 x 0.0831 x 1069) x 1 = 3.75 x 0.0831 x 1069 x 10-6 = 333.13 x 10-6 OR Kp= 3.33 x 10-4

(b) Write any two general characteristics of equilibria involving physical process.
Answer:

  1. It is dynamic in nature.
  2. It can be achieved in a closed vessel.
  3. It depends only on temperature but not on concentration or pressure.

Question 34.
(a) The pH of the blood is 7.4. Calculate the [H+].
pH = 7.4 [H+] = ?
W.K.T [H+] = Antilog (-pH)
[H+]= Antilog 10 (-7.4)
Add-1 to-7 and +1 to-0.4 ∴ [H+] = Antilog \(\overline{8.6}\)
[H+] = 10-8 x Antilog of (+0.6).

(b) Derive the Henderson-Hasselbalch equation for acid buffer.
Answer:
Henderson-Hasselbalch equation:
Consider a weak acid and its salt BA,
1st PUC Chemistry Model Question Paper 2 with Answers - 15

Part – E

V. Answer any TWO of the following questions carrying FIVE marks: ( 2 x 5 = 10 )

Question 35.
(a) For the following bond cleavage, use curved arrows to show electron flow, mention the type of bond cleavage, and reactive intermediate formed.
CH3CH2O + OCH2CH3 → CH3CH2O + OCH2CH3
1st PUC Chemistry Model Question Paper 2 with Answers - 16

  1. Homolytic cleavage
  2. Ethoxide free radicals.

(b) Give the hybridisation and geometry of carbocation.
Answer:
sp2 and planar.

KSEEB Solutions

Question 36.
(a) 0.2033 g of an organic compound on combustion gave 0.3780 g of CO2 and 0.1288 g H2O. In a separate experiment 0.1877 g of the compound on analysis by Dumas method produced 31.7 ml. of nitrogen collected over water at 14°C and 758 mm pressure. Determine the percentage composition of the compound. (Aqueous tension of water at 14°C = 12 mm pressure).
Answer:
Given mass of organic compound = 0.2033 g |
Volume of nitrogen (V1) = 31.7 mL
T1 = 287K P1 = (P – f) = 758 – 12 = 746 mm;
P2 = 760mm U2 = ? T2 = 273K
1st PUC Chemistry Model Question Paper 2 with Answers - 17
To convert the volume of of N2 at STP into mass
22400 mL of nitrogen at STP weighs = 28 g
1st PUC Chemistry Model Question Paper 2 with Answers - 18
1st PUC Chemistry Model Question Paper 2 with Answers - 19
% age of oxygen = 100 – (18.1985 + 50.71 + 7.03) = 24.06%

(b) What are nucleophiles? Give an example.
Answer:
The negatively charged species reacts at nucleus or +ve centre are called nucleophiles.
Example: Cl, Br, I, \(\mathrm{C} \overline{\mathrm{N}}\), NH3, etc.

Question 37.
(a) Explain the mechanism of addition of HBr to propene in the presence of peroxide catalyst.
Answer:
It is a free radical mechanism.
It has the following steps:
(a) Initiation
1st PUC Chemistry Model Question Paper 2 with Answers - 20

(b) Propagation
1st PUC Chemistry Model Question Paper 2 with Answers - 21

(c) Termination
1st PUC Chemistry Model Question Paper 2 with Answers - 22

KSEEB Solutions

(b) Write the structures of cis and trans isomers of But-2-ene.
Answer:
1st PUC Chemistry Model Question Paper 2 with Answers - 23

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7

Students can Download Maths Chapter 10 Ghatankagalu Ex 10.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 1
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 2

KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 3
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 4
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 5
KSEEB Solutions for Class 8 Maths Chapter 10 Ghatankagalu Ex 10.7 6

2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3

Students can Download Basic Maths Exercise 20.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3

Part-A

2nd PUC Basic Maths Indefinite Integrals Ex 20.3 Two Marks Questions and Answers

Question 1.
\(\int \frac{3 x^{2}}{1+x^{3}} d x\)
Answer:
\(\int \frac{3 x^{2}}{1+x^{3}} d x\) = log(1 + x3)+c or using \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x} = log (f(x) + c\)
put 1 + x3 = t
∴ 3x2dx = dt
∴ = ∫\(\frac { 1 }{ t }\)dt = log t + c = log(1 + x3) + c

Question 2.
\(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\)
Answer:
\(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\) = log(2x2 + 3x +5) + c
∴ \(\int \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{d} \mathrm{x}\) = log(f(x)) + C

Question 3.
\(\int \frac{e^{x}-1}{e^{x}-x} d x\)
Answer:
\(\int \frac{e^{x}-1}{e^{x}-x}\) = log(ex – x) + c

KSEEB Solutions

Question 4.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 4
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 5

Question 5.
\(\int \frac{\cos x}{2+\sin x} d x\)
Answer:
\(\int \frac{\cos x}{2+\sin x} d x\) = log(2 + sin x) + c

Question 6.
\(\int \frac{1}{x(2 \log x+5)} d x\)
Answer:
\(\int \frac{1 / x^{d x}}{2 \log x+5}\)
\(\int \frac{d t / 2}{t}=\frac{1}{2} \int \frac{1}{t} d t\)
= \(\frac { 1 }{ 2 }\)log t + c
= \(\frac { 1 }{ 2 }\)log (2 log x + 5) + c
put 2 log x + 5 = t
2. \(\frac { 1 }{ x }\) dx = dt
\(\frac { 1 }{ x }\) dx = \(\frac { dt }{ 2 }\)

Question 7.
\(\int \frac{3 \sin x}{3+4 \cos x} d x\)
Answer:
\(\int \frac{3 \sin x}{3+4 \cos x} d x\)
\(=\int \frac{3 \cdot \frac{d t}{-4}}{t}=\frac{-3}{4} \int \frac{1}{t} d t\)
= \(\frac { -3 }{ 4 }\) log t + c
= \(\frac { -3 }{ 4 }\) log(3 + 4 cos x) + c
put 3 + 4 cos x = t
– 4 sinx dx = dt
sin x dx = \(\frac { dt }{ -4 }\)

KSEEB Solutions

Part-B

2nd PUC Basic Maths Indefinite Integrals Ex 20.3 Five Marks Questions and Answers

Question 1.
\(\int \frac{1}{\sqrt{x}+x} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 1

Question 2.
\(\int \frac{\sin 2 x}{1+\cos ^{2} x} d x\)
Answer:
\(\int \frac{\sin 2 x}{1+\cos ^{2} x} d x\)
= \(\int \frac{-d t}{t}\)
= -log t + c
= -log (1 + cos2x) + c
put 1 + cos2x = t
2 cosx (-sinx) dx = dt
-sin 2x dx = dt
sin 2x dx = -dt

Question 3.
\(\int \frac{e^{2 x}+1}{e^{2 x}-1} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 2

Question 4.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 6
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.3 - 3

KSEEB Solutions

Question 5.
\(\int \frac{\cot x}{3+\log (\sin x)} d x\)
Answer:
\(\int \frac{\cot x}{3+\log (\sin x)} d x\)
\(=\int \frac{d t}{t}\)
= log t + c = log(3 + log(sin x)) + c
put 3 + log(sin x) = t
\(\frac{\cos x}{\sin x} d x=d t\)
cos x dx = dt

Question 6.
\(\int \frac{\csc ^{2} x \cdot \cot x}{4+5 \csc ^{2} x} d x\)
Answer:
\(\int \frac{\csc ^{2} x \cdot \cot x}{4+5 \csc ^{2} x} d x\)
\(=\int \frac{-1 / 10^{\mathrm{dt}}}{t}=\frac{-1}{10} \log t+C\)
= \(\frac { -1 }{ 10 }\)log(4 + 5cosec2x) + c
put 4 + 5 cosec2x = t
– 10 cosec x – cosec x × cot x dx = dt
cosec2x . cot dx = \(\frac { -1 }{ 10 }\)dt

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