Plus One Economics Notes Chapter 10 Comparative Development Experience of India with its Neighbours

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Kerala Plus One Economics Notes Chapter 10 Comparative Development Experience of India with its Neighbours

Development Path – A Snapshot View
India, Pakistan, and China have many similarities in their developmental strategies. India, China, and Pakistan became independent and started initiating their developmental strategies at almost the same time. India and Pakistan became independent in 1947 and China in 1949. All three countries adopted the planning strategy for economic growth and development. India’s five-year plan started in 1951, Pakistan’s in 1956 (called medium-term plan), and China’s in 1953. Government and the public sector played a major role in these economies. However, with the introduction of economic reforms in tune with globalisation the role of market was redefined. Till the 1980s, all the three economies had almost similar growth rate and other economic indicators.

China
After the establishment of the People’s Republic of China under one-party rule, all the critical sectors of the economy, enterprises, and lands owned and operated by individuals were brought under government control. The Great Leap Forward (GLF) campaign initiated in 1958 aimed at industrializing the country on a massive scale. People were encouraged to set up industries in their backyards. In rural areas, communes were started. Under the Commune system, people collectively cultivated lands.

In 1958, there were 26,000 communes covering almost all the farm population. GLF campaign met with many problems. A severe drought caused havoc in China killing about 30 million people. When Russia had conflicts with China, it withdrew its professionals who had earlier been sent to China to help in the industrialization process.

In 1965, Mao introduced the Great Proletarian Cultural Revolution (1966-76) under which students and professionals were sent to work and learn from the countryside. The present-day fast industrial growth in China can be traced back to the reforms introduced in 1978.

Pakistan
Pakistan followed the mixed economy model with, co-existence of the public sector and private sector. Pakistan also followed a protectionist policy in international trade. The introduction of the Green Revolution resulted in a rise in the production of food grains. In the 1970s, the nationalization of the capital goods industry took place. In the late 1970s and 1980stherewasashift in the economic policy in favor of de-nationalization and encouragement to the private sector.

Pakistan got substantial financial support from western nations. There was an increase in the number of emigrants and their remittance to their home country. The remittance and Western nation’s support helped the country in stimulating economic growth. In 1988 more reforms were introduced. FDI was encouraged, direct taxes were reduced and many areas of the economy were opened to private and foreign investment.

Demographic Indicators
Demographic indicators of India, China, and Pakistan can be summarised as follows:
Plus One Economics Notes Chapter 10 Comparative Development Experience of India with its Neighbours 1

The table shows the population growth as being highest in Pakistan, followed by India and China. Scholars point out the one-child norm introduced in China in the late 1970s as the major reason for low population growth. They also state that this measure led to a decline in the sex ratio, the proportion of females per 1000 males. However, from the table, you will notice that the sex ratio is low and biased against females in all three countries. Scholars cite son preference prevailing in all these countries as the reason. In recent times, all three countries are adopting various measures to improve the situation.

The One-child norm and the resultant arrest in the growth of the population also have other implications. For instance, after a few decades, in China, there will be more elderly people in proportion to young people. This will force China to take steps to provide social security measures with fewer workers. The fertility rate is also low in China and very high in Pakistan. Urbanization is high in both Pakistan and China with India having 28 percent of its people living in urban areas.

Indicators of Human Development
The HDI (Human Development Index) developed by UNDP (United Nations Development Programme) is an index which has universal acceptance as a good measure of the quality of human life. Since 1990 the UNDP has been publishing information related to HDI. This report ranks countries on the basis of their HDI. The following table presents some of the selected indicators of development.
Plus One Economics Notes Chapter 10 Comparative Development Experience of India with its Neighbours 2
Source: Human Development Report, 2014

Plus One Economics Notes Chapter 9 Environment Sustainable Development

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Kerala Plus One Economics Notes Chapter 9 Environment Sustainable Development

Environment
The environment is defined as the total planetary inheritance and the totality of all resources. It includes all the biotic and abiotic factors that influence each other. While all living elements – the birds, animals and plants, forests, fisheries, etc. are biotic elements and abiotic elements include air, water, land, etc. Rocks and sunlight are all examples of abiotic elements of the environment. A study of the environment then calls for a study of the interrelationship between these biotic and abiotic components of the environment.

Functions of the Environment
The environment performs four vital functions

  • It supplies resources: resources here include both renewable and non-renewable resources.
    Renewable resources are those which can be used without the possibility of the resource becoming depleted or exhausted. That is, a continuous supply of the resource remains available.
  • It assimilates waste
  • It sustains life by providing genetic and bio-diversity
  • It also provides aesthetic services like scenery, etc.

Global Warming and Ozone Depletion
Two important issues faced by our environment is global warming and ozone depletion.

Global warming: Global warming is a gradual increase in the average temperature of the earth’s lower atmosphere as a result of the increase in greenhouse gases since the Industrial Revolution. Much of the recent observed and projected global warming is human-induced. It is caused by man-made increases in carbon dioxide and other greenhouse gases through the burning of fossil fuels and deforestation.

Ozone Depletion: Ozone depletion refers to the phenomenon of reductions in the amount of ozone in the stratosphere The problem of ozone depletion is caused by high levels of chlorine and bromine compounds in the stratosphere The origins of these compounds are chlorofluorocarbons (CFC), used as cooling substances in airconditioners and refrigerators, or as aerosol propellants, and bromo fluorocarbons (halons), used in fire extinguishers. As a result of depletion of the ozone layer, more ultraviolet (UV) radiation comes to Earth and causes damage to living organisms. UV radiation seems responsible for skin cancer in humans; it also lowers production of phytoplankton and thus affects other aquatic organisms. It can also influence the growth of terrestrial plants.

State of India’s Environment
India has abundant natural resources in terms of rich quality of soil, hundreds of rivers and tributaries, lush green forests, plenty of mineral deposits beneath the land surface, vast stretch of the Indian Ocean, ranges of mountains, etc. The black soil of the Deccan Plateau is particularly suitable for the cultivation of cotton, leading to a concentration of textile industries in this region. The Indo-Gangetic plains – spread from the Arabian Sea to the Bay of Bengal – are one of the most fertile, intensively cultivated and densely populated regions in the world.

India’s forests, though unevenly distributed, provide green cover for a majority of its population and natural cover for its wildlife. Large deposits of iron ore, coal and natural gas are found in the country. India alone accounts for nearly 20% of the world’s total iron-ore reserves. Bauxite, copper, chromate, diamonds, gold, lead, lignite, manganese, zinc, uranium, etc. are also available in different parts of the country. However, the developmental activities in India have resulted in pressure on its finite natural resources, besides creating impacts on human health and well-being.

The threat to India’s environment poses a dichotomy -threat of poverty-induced environmental degradation and, at the same time, threat of pollution from affluence and a rapidly growing industrial sector. Air pollution, water contamination, soil erosion, deforestation and wildlife extinction are some of the most pressing environmental concerns of India.
The priority issues identified are :

  • land degradation
  • biodiversity loss
  • air pollution with special reference to vehicular pollution in urban cities
  • management of fresh water and
  • Solid waste management. Land in India suffers from varying degrees and types of degradation stemming mainly from unstable use and inappropriate management practices.

Sustainable Development
The concept of sustainable development was emphasized by the United Nations Conference on Environment and Development (UNCED), which defined it as: ‘Development that meets the need of the present generation without compromising the ability of the future generation to meet their own needs’.

Strategies for sustainable development
Strategies for sustainable development include the following.

  • Use of Non-conventional Sources of Energy: LPG, Gobar Gas in Rural Areas:
  • CNG in Urban Areas
  • Wind Power
  • Solar Power through Photovoltaic Cells
  • Mini-hydel Plants
  • Traditional Knowledge and Practices
  • Bio-composting
  • Bio-pest Control

Plus One Economics Notes Chapter 8 Infrastructure

Students can Download Chapter 8 Infrastructure Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 8 Infrastructure

Infrastructure
Infrastructure means some kinds of permanent installation, which are used over a long period of time for the supply of basic inputs like railway lines, schools, colleges, universities, hospitals, etc. Infrastructural facilities are often referred to as economic and social overheads.

  1. The economic infrastructure consists of energy, transport, communication
  2. The social infrastructure consists of education, health, and housing.

Relevance of Infrastructure
Infrastructure plays important role in economic growth and development. Developed nations have good record of social and economic infrastructure. The contributions of infrastructure are:
• It invites investment which leads to growth.
• It enhances productivity.
• It improves the quality of life of people.

State of Infrastructure in India
Two important infrastructures in India are energy and health. We shall examine their details below:

Energy: Energy is very vital for rapid economic growth. There is a big gap between consumer demand and the supply of electricity in India. Energy is a critical aspect of the development process of a nation. It is, of course, essential for industries. Now it is used on a large scale in agriculture and related areas like the production and transportation of fertilizers, pesticides, and farm equipment. It is required in houses for cooking, household lighting, and heating.

Sources of Energy: There are commercial and non-commercial sources of energy. Commercial sources are coal, petroleum, and electricity as they are bought and sold. Non-commercial sources of energy are firewood, agricultural waste, and dried dung. These are non-commercial as they are found in nature/ forests. While commercial sources of energy are generally exhaustible, non-commercial sources are generally renewable.

Non-conventional Sources of Energy: Both commercial and non-commercial sources of energy are known as conventional sources of energy. There are three other sources of energy which are commonly termed as non-conventional sources – solar energy, wind energy and tidal power.

Power/Electricity: The most visible form of energy, which is often identified with progress in modern civilization, is power, commonly called electricity; it is one of the most critical components of infrastructure that determines the economic development of a country.

Health: Health is an essential element of human resource development. Health is a holistic process related to the overall growth and development of the nation. WHO defines health as Economists judge the health conditions of the people of a country by looking at the following indicators.

  • Infant mortality
  • Maternal mortality
  • Life expectancy
  • Nutritional levels
  • Incidence of communicable and non-communicable diseases
  • Health infrastructures

Health System in India: India’s health infrastructure consists of a three-tier system such as primary, secondary, and tertiary. Primary health care includes health education, health problems, prevention and control of diseases, promotion of nutrition, issues relating to potable water and sanitation, maternal and child health care, immunization against major infectious diseases like Polio, T.B, diphtheria, promotion of mental health and provision of essential drugs, etc.

Primary health care is provided through, sub-centers catering to a population of about 5000, Primary health care centres (PHCs) at block level and community health centres (CHCs) at the district level. The primary health care centres have only limited facilities. When the patient need advanced health care they are referred to secondary or tertiary hospitals.

Secondary care institutions are those which have facilities for clinical investigations like X-ray, clinical laboratory, scanning, etc., specialist doctors like a surgeon, gynecologists, pediatricians, etc. It is mostly available in district headquarters and big towns and cities.

Tertiary health care institutions these are health care institutions at the top of the three-tier system, devoted in health care, health education and research: Medical colleges, super-specialty hospitals and multi-specialty hospitals. All India Institute of Medical Sciences (AIIMS) Delhi, National Institute of Mental Health and Neuro Sciences (NIMHANS) Bangalore, Sree Chithra Institute of Medical Science (SCIM) Trivandrum, etc., are tertiary health care institutions.

Indian Systems of Medicine (ISM): Natural systems of medicine have to be explored and used to support public health. There is a great scope of advancement of medical tourism in India. It includes six systems: Ayurveda, Yoga, Unani, Siddha, Naturopathy and Homeopathy (AYUSH). At present, there are 3,004 ISM hospitals, 23,028 dispensaries and as many as 6,11,431 registered practitioners in India. But little has been done to set up a framework to standardize education orto promote research. ISM has huge potential and can solve a large part of our health care problems because they are effective, safe and inexpensive.

Plus One Economics Notes Chapter 7 Employment-Growth, Informalisation and Related Issues

Students can Download Chapter 7 Employment-Growth, Informalisation and Related Issues Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 7 Employment-Growth, Informalisation and Related Issues

Workers and Employment
Those activities that contribute to Gross National Product (GNP) or national income are known as economic activities. When farmers work in a field or a labour work in a factory, or doctor works in a hospital they produce goods or services. All those who are engaged in economic activities including self-employed are called workers. The employment situation is diverse and complex in India. It is due to the developing nature of the economy and the socio-economic and demographic factors that influence it.

Participation of People in Employment
The worker-population ratio is an important indicator of the employment situation in an economy.
Worker population-ratio refers to the ratio of workers to the population. It is computed by dividing the number of workers(W) by the total population (P) and express it in terms of percentage (W/P)*100. This ratio is useful in knowing the proportion of the population actively contributing to the production of goods and services in a country. The worker-population ratio in India for 2011 was 39.3%.

Employment in Firms, Factories, and Offices
In the course of the economic development of a country, labour flows from agriculture and other related activities to industry and services. Generally, we divide all economic activities into eight different industrial divisions. They are:

  1. Agriculture
  2. Mining and quarrying
  3. Manufacturing
  4. Electricity, gas and water supply
  5. Construction
  6. Trade
  7. Transport and storage
  8. Services.

For simplicity, all the working persons engaged in these divisions can be clubbed into three major sectors, viz.

  • Primary sector includes (1) and (2)
  • Secondary sector which includes (3), (4), and (5)
  • Service sector which includes divisions (6), (7), and (8).

Unemployment and Types of Unemployment
The unemployment situation in India is highly complex. There are different types of unemployment in our country like open unemployment, disguised unemployment, seasonal unemployment, etc. Unemployment is a situation in which people are willing to work at the prevailing wage rate, but do not get any work.

Open Unemployment: Open unemployment is the situation in which people above a certain age who are able to work and willing to work at the prevailing wage remain unemployed. Open unemployment is involuntary in nature. They are willing to work, but employment opportunities are not available to them. People standing in some selected areas waiting to be recruited as the hired worker is a case of open unemployment.

Disguised Unemployment: When more persons are working in a job than actually required, the situation is termed as disguised unemployment or hidden unemployment. If some workers are withdrawn from work, either total production or productivity falls. This type of unemployment is prominent in Indian agriculture.

Seasonal Unemployment: The type of unemployment caused by a change in seasons is termed as seasonal unemployment. This is normally found in the agricultural sector of India. Agriculture normally provides only seasonal employment and people are employed during the busy sowing and harvesting seasons. Seasonal unemployment could also be found in agro-based industries such as sugar mills, rice mills, cotton-spinning mills, etc.

Plus One Economics Notes Chapter 6 Rural Development

Students can Download Chapter 6 Rural Development Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 6 Rural Development

Rural Development
Rural development is quite a comprehensive term but it essentially means a plan of action for the development of areas which are lagging behind in socio-economic development. It essentially focuses on the action for the development of areas that are lagging behind in the overall development of the village economy. Some of the areas which are challenging and need fresh initiatives for development in India include:

  • Development of human resources including- literacy, more specifically, female literacy, education, and skill development-health, addressing both sanitation and public health
  • Land reforms
  • Development of the productive resources of each locality
  • Infrastructure development like electricity, irrigation, credit, marketing, transport facilities including construction of village roads and feeder roads to nearby highways, facilities for agriculture research and extension, and information dissemination
  • Special measures for alleviation of poverty and bringing about significant improvement in the living conditions of the weaker sections of the population emphasizing access to productive employment opportunities.

Credit and Marketing in Rural Areas
Credit: Growth of rural economy depends primarily on the infusion of capital, from time to time, to realize higher productivity in agriculture and non-agriculture sectors. As the time gestation between crop sowing and realization of income after production is quite long, farmers borrow from various sources to meet their initial investment on seeds, fertilizers, implements and other family expenses of marriage, death, religious ceremonies.

A major change occurred after 1969 when India adopted social banking and a multi-agency approach to adequately meet the needs of rural credit. Later, the National Bank for Agriculture and Rural Development (NABARD) was set up in1982 as an apex body to coordinate the activities of all institutions involved in the rural financing system.

The institutional structure of rural banking today consists of a set of multi-agency institutions, namely, commercial banks, regional rural banks (RRBs), co-operatives and land development banks. Recently, Self-Help Groups (henceforth SHGs) have emerged to fill the gap in the formal credit system because the formal credit delivery mechanism has not only proven inadequate but has also not been fully integrated into the overall rural social and community development. By March end 2003, more than seven lakh SHGs had reportedly been credit linked. Such credit provisions are generally referred to as micro-credit programmes.

Agricultural Market System
Agricultural marketing is a process that involves the assembling, storage, processing, transportation, packaging, grading and distribution of different agricultural commodities across the country. Let us discuss four such measures that were initiated to improve the marketing aspect.

1. The first step was regulation of markets to create orderly and transparent marketing conditions.

2. Second component is provision of physical infrastructure facilities like roads, railways, warehouses, godowns, cold storages and processing units.

3. Co-operative marketing, in realizing fair prices for farmers’ products, is the third aspect of a government initiative.

4. The fourth element is the policy instruments like

  • assurance of minimum support prices (MSP) for 24 agricultural products
  • maintenance of buffer stocks of wheat and rice by Food Corporation of India and
  • distribution of food grains and sugar through PDS.

These instruments are aimed at protecting the income of the farmers and providing food grains at subsidized rate to the poor.

Diversification into Productive Activities
Diversification of farm products has two aspects: The first one relates to the diversification of crop production. The second one relates to the shift of the workforce from agriculture to other allied activities such as livestock, poultry, fishers, etc., and to non-farm sectors like food processing. Diversification of agriculture helps to provide alternative employment opportunities in the non-farm sector and will minimize the risk of depending exclusively on agriculture. These activities related to diversification are given below:

  • Animal husbandry
  • Fisheries
  • Horticulture

Sustainable Development and Organic Farming
In recent years, awareness of the harmful effect of chemical-based fertilizers and pesticides on our health is on a rise. Conventional agriculture relies heavily on chemical fertilizers and toxic pesticides, etc., which enter the food supply, penetrate the water sources, harm the livestock, deplete the soil and devastate natural eco-systems. Efforts in evolving technologies which are eco-friendly are essential for sustainable development and one such technology which is eco-friendly is organic farming.

Benefits of Organic Farming:
1. Organic agriculture offers a means to substitute costlier agricultural inputs (such as HYV seeds, chemical fertilizers, pesticides, etc.) with locally produced organic inputs that are cheaper and thereby generate good returns on investment.

2. Organic agriculture also generates income through international exports as the demand for organically grown crops is on a rise.

3. Studies across countries have shown that organically grown food has more nutritional value than chemical farming thus providing us with healthy foods.

4. Since organic farming requires more labour input than conventional farming, India will find organic farming an attractive proposition.

5. Finally, the produce is pesticide-free and produced in an environmentally sustainable way.

Plus One Botany Notes Chapter 11 Plant Growth and Development

Students can Download Chapter 11 Plant Growth and Development Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 11 Plant Growth and Development

Growth
Growth is defined as an irreversible permanent increase in the size of an organ or its parts of an individual cell.
It is accompanied by metabolic processes (both anabolic and catabolic), that occur at the expense of energy.
Eg: expansion of a leaf.

Plant Growth Generally is Indeterminate
Plant growth is unlimited growth due to the presence of meristems.
Root apical meristem and the shoot apical meristem are responsible for the primary growth of the plants and contribute to the elongation of the plants along their axis.

Role of lateral meristem in plants
In dicotyledonous plants and gymnosperms, the lateral meristems, (vascular cambium and cork-cambium) cause an increase in the girth of the organs. This is known as secondary growth.
Plus One Botany Notes Chapter 11 Plant Growth and Development 1

Growth is Measurable

  • Growth is measured in terms of increase in fresh weight, dry weight, length, area, volume, and cell number.
  • One single maize root apical meristem can give rise to more than 17,500 new cells per hour, cells in a watermelon increase in size by up to 3,50,000 times.
  • In the former, growth is expressed as an increase in cell number.
    latter expresses growth as an increase in the size of the cell.
  • While the growth of a pollen tube is measured in terms of its length, an increase in surface area denotes the growth in a dorsiventral leaf.

Phases of Growth
The period of growth is generally divided into three phases.

  1. Meristematic: The constantly dividing cells, both at the root apex and the shoot apex, represent the meristematic phase of growth.
  2. Elongation: The cells proximal to the meristematic zone represent the phase of elongation.
  3. Maturation: Proximal to the phase of elongation represents the phase of maturation.

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 2

Growth Rates
The increased growth per unit time is termed as growth rate. The growth rate may be

  1. Arithmetic
  2. Geometrical

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 3

On plotting the length of the organ against time, a linear curve is obtained, it is expressed as
Plus One Botany Notes Chapter 11 Plant Growth and Development 4

Lt = L0 + rt
Lt = length at time ‘t’
L0 = length at time ‘zero’.
r = growth rate/elongation per unit time.

Different phases of the Sigmoid curve

  1. lag phase
  2. log or exponential phase
  3. stationary phase

In most systems, the initial growth is slow (lag phase), and it increases rapidly at an exponential rate (log or exponential phase)
In the end, due to the limited nutrient supply, the growth slows down leading to a stationary phase. It is the typical sigmoid or S-curve.
Plus One Botany Notes Chapter 11 Plant Growth and Development 5
The exponential growth can be expressed as
W1 = W0 e rt
W1 = final size (weight, height, number etc.)
W0 = initial size at the beginning of the period
r = growth rate
t = time of growth
e = base of natural logarithms

Quantitative comparisons between the growth of a living system can also be made in two ways:

  • Measurement and the comparison of total growth per unit time is called the absolute growth rate.
  • The growth of the given system per unit time expressed on a common basis.

 

Plus One Botany Notes Chapter 11 Plant Growth and Development 6
In Figure two leaves, A and B, are drawn that are of different sizes but show an absolute increase in area in the given time to give leaves, A1 and B1.

Conditions for Growth
Water, oxygen, and nutrients as very essential elements for growth.
The plant cells grow in size by cell enlargement it requires water.
Turgidity of cells helps in extension growth. Water also provides the medium for enzymatic activities
Oxygen helps in releasing metabolic energy essential for growth activities.
Nutrients (macro and micro essential elements) are required by plants for the synthesis of protoplasm and act as a source of energy.
An optimum temperature range is best suited for plant growth.
Environmental signals such as light and gravity also affect certain phases/stages of growth.

Differentiation, Dedifferentiation, and Redifferentiation
1. The cells derived from root apical and shoot-apical meristems and cambium differentiate and mature to perform specific functions. This is termed as differentiation.
For example, during differentiation, tracheary elements lose their protoplasm and develop a very strong, elastic, lignocellulosic secondary cell wall, to carry water too long distances.

2. The living differentiated cells, that have lost the capacity to divide can regain the capacity of division This phenomenon is termed dedifferentiation.
For example, interfascicular cambium and cork cambium is formed from fully differentiated parenchyma cells.

3. Meristems are able to divide and produce cells that once again lose the capacity to divide but mature to perform specific functions. This is called a redifferentiation.
For example, secondary tissues develop from vascular cambium and cork cambium

Development
It is the stage of the life cycle in which germination of the seed to senescence.
Plus One Botany Notes Chapter 11 Plant Growth and Development 7
The plant shows a response to the environment to form different kinds of structures. This ability is called plasticity
Heterophylly is an example of plasticity

Types of Heterophylly
1. The leaves of the juvenile plant are different in shape from those in mature plants.
e.g cotton, coriander, and larkspur.
2. Shapes of submerged leaves are different from those produced in the air.
Eg buttercup.
Plus One Botany Notes Chapter 11 Plant Growth and Development 8
Development is considered as the sum of growth and differentiation.
Development in plants is under the control of intrinsic and extrinsic factors.

A) Intrinsic factors

  1. Intracellular (genetic)
  2. Intercellular factors (chemicals such as plant growth regulators)

B) Extrinsic factors
Light, temperature, water, oxygen, nutrition, etc.

Plant Growth Regulators

Characteristics
The plant growth regulators (plant hormones or phytohormones) include

  1. Indole compounds (indole-3-acetic acid, IAA);
  2. Adenine derivatives (N6-furfurylamino purine, kinetin),
  3. Derivatives of carotenoids (abscisic acid, ABA);
  4. Terpenes (gibberellic acid, GA3) or
  5. Gases (ethylene, C2H4).

The PGRs are divided into two groups based on their functions in a living plant body.
One group of PGRs are involved in growth-promoting activities, e.g., auxins, gibberellins, and cytokinins.
The other group mainly involved in growth-inhibiting activities such as dormancy and abscission. Eg-abscisic acid and ethylene.
Plus One Botany Notes Chapter 11 Plant Growth and Development 9

The Discovery of Plant Growth Regulators
1. Auxin
At first, Charles Darwin and his son Francis Darwin observed that the coleoptiles of canary grass responded to unilateral illumination by growing towards the light source.
After a series of experiments, it was concluded that the tip of the coleoptile was the site of transmittable influence that caused the bending of the entire coleoptile.
Auxin was isolated by F. W. Went from tips of coleoptiles of oat seedlings.

2. Gibberellin
The ‘balance’ (foolish seedling) disease of rice seedlings, was caused by a fungal pathogen Gibberalla fujikuroi.
In this experiment, the uninfected rice seedlings were treated with sterile filtrates of the fungus. It led to the development of the disease. The active substance was gibberellic acid.
It was demonstrated by E. Kurosawa.

3. Cytokinin
The internodal segments of tobacco stems- the callus proliferated in the presence of auxins along with the extracts of vascular tissues, yeast extract, coconut milk or DNA.
Skoog and Miller later identified and crystallized the cytokinesis promoting active substances that they termed kinetin.

4. Abscisic acid(ABA)
During the mid-1960s inhibitory hormones were identified: inhibitor-B, abscission II and dormin.
Later all the three were named abscisic acid (ABA).

5. Ethylene
Ripened oranges that hastened the ripening of stored unripened bananas. Later this volatile substance was identified as ethylene, a gaseous PGR.

Physiological Effects of Plant Growth Regulators
Auxins
Auxins were first isolated from human urine.
They are generally produced by the growing apices of the stems and roots, from where they migrate to the regions of their action.

Two types of auxins

  1. Natural (IAA and indole butyric acid (IBA)
  2. Synthetic. NAA (naphthalene acetic acid) and 2, 4-D (2, 4-dichlorophenoxyacetic)

Plus One Botany Notes Chapter 11 Plant Growth and Development 10

  1. They help to initiate rooting in stem cuttings
  2. Auxins promote flowering e.g. in pineapples.
  3. They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
  4. In most higher plants, the growing apical bud inhibits the growth of the lateral (axillary) buds, a phenomenon called apical dominance.
  5. Removal of shoot tips (decapitation) usually results in the growth of lateral buds Hence it is widely applied in tea plantations, hedge-making, etc.
  6. Auxins also induce parthenocarpy.
  7. They are widely used as herbicides. 2, 4-D is used to kill dicotyledonous weeds, So it is used to prepare weed-free lawns by gardeners.
  8. Auxin controls xylem differentiation and helps in cell division.

Gibberellins
Gibberellic acid (GA3) was one of the first gibberellins to be discovered and remains the most intensively studied form.
GA3 is acidic.

  1. GA3 causes an increase in the length of grapes stalks. .
  2. Gibberellins, cause fruits like apple to elongate and improve its shape
  3. They delay senescence. Hence the fruits are keeping as fresh.
  4. GA3 is used to speed up the malting process in the brewing industry.
  5. Spraying sugarcane crop with gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
  6. Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production.
  7. Gibberellins also promote bolting(internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Cytokinins
Cytokinins were discovered as kinetin (a modified form of adenine, a purine) from the autoclaved herring sperm DNA.
Naturally occurring cytokinin-zeatin was isolated from corn-kernels and coconut milk.
Natural cytokinins are synthesised in regions where rapid cell division occurs, for example, root apices, developing shoot buds, young fruits, etc.

  1. It helps to produce new leaves, chloroplasts in leaves, lateral shoot growth and adventitious shoot formation.
  2. Cytokinins help to overcome apical dominance.
  3. They promote nutrient mo8/+9bilisation which helps in the delay of leaf senescence.

Ethylene
The most widely used compound as a source of ethylene is ethephon.
It is readily absorbed and transported within the plant and releases ethylene slowly.
Ethephon hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits

  1. Ethylene is a gaseous hormone that promotes senescence and ripening fruits.
  2. It promotes horizontal growth of seedlings, swelling of the axis, and apical hook formation in dicot seedlings.
  3. Ethylene promotes senescence and abscission of plant organs, especially of leaves and flowers.
  4. Ethylene is highly effective in fruit ripening. It enhances the respiration rate during the ripening of the fruits (respiratory climactic).
  5. Ethylene breaks seed and bud dormancy and initiates germination in peanut seeds, sprouting of potato tubers.
  6. Ethylene promotes rapid internode/petiole elongation in deepwater rice plants.
  7. Ethylene also promotes root growth and root hair formation, thus helping the plants to increase their absorption surface.
  8. Ethylene is used to initiate flowering and fruit-set in pineapples.
  9. It also induces flowering in mango.
  10. It promotes female flowers in cucumbers

Abscisic acid

  1. It promotes abscission and dormancy.
  2. It acts as an inhibitor of plant metabolism.
  3. ABA inhibits seed germination.
  4. ABA stimulates the closure of stomata and increases the tolerance of plants to various kinds of stresses. Hence it is called the stress hormone.
  5. ABA plays an important role in seed development, maturation, and dormancy.
  6. ABA helps the seeds to withstand desiccation
  7. ABA acts as an antagonist to GA3.

Photoperiodism
It is the phenomenon of relative day and night length for the initiation of flowering.
Plus One Botany Notes Chapter 11 Plant Growth and Development 11

Based on the exposure to photoperiod there are three types of plants

  1. Long day plants: They require exposure to light for a period greater than critical duration (12 hr).
  2. Short-day plants: They require less than critical duration before flowering.
  3. Day-neutral plants: In this type, there is no such correlation between exposure to light duration and induction of flowering response.

Which is the organ of a plant perceives light for photoperiodism?
The site of perception of light/dark duration is the leaves. After receiving the required photoperiod, the hormonal substance migrates from leaves to shoot apices for inducing flowering. The shoot apices become changed into flowering apices prior to flowering.

Vernalisation
It is the phenomenon of exposure of low temperature for the initiation of flowering
Some important food plants, wheat, barley, rye have two kinds of varieties: winter and spring varieties.

Nature of spring and winter varieties
The ‘spring’variety are normally planted in the spring and come to flower and produce grain before the end of the growing season.
Winter varieties, planted in spring fail to flower or produce mature grain within a span of a flowering season.
If they are planted in autumn .they germinate and overwinter come out as small seedlings, resume growth in the spring, and are harvested usually around mid-summer.

Biennials and low-temperature treatment
Biennials are monocarpic plants that normally flower and die in the second season. Biennial plants are subjected to a cold treatment, it stimulates photoperiodic flowering response. Sugarbeet, cabbages, carrots are some of the common biennials.

NCERT Supplementary Syllabus

Seed Germination
The seeds germinate under favourable conditions after the period of dormancy.
After dormancy embryo becomes metabolically active and starts growing. This process is known as seed germination.
The conditions necessary for seed germination are the availability of water and oxygen.

A physiological phenomenon in seed germination
The physical phenomenon associated with seed germination is imbibition. It causes the swelling of seed then rupturing of the seed coat, through which radical emerges out.
It develops into a root system but the shoot system arises from the plumule of another end of the embryonal axis.
The metabolic activities require oxygen for breaking down the food reserves such as polysaccharides, proteins and lipid.
It is converted into soluble materials with the help of enzymes and mobilized to the embryonal axis.
The growth of radical and plumule is due to cell extension, cell division, and several biochemical processes.
The seed also needs a suitable temperature (optimum between 25 to35). The rate of respiration increases rapidly during seed germination.

What is the Viviparous type of germination?
Vivipary is the germination of a seed while it is still attached to the parent plant and is nourished by it. The plants grow in marshy land such as Rhizophora and Sonneratia (halophytes)show this type of germination.
During germination, radical elongates, and the weight of the germinating seed increases. As a result, the seedling separates and fail down vertically into the mud and grow into a new plant.

Seed Dormancy
It is the period of rest or a period of suspended growth due to this

  1. water content, the metabolic activities become extremely low.
  2. the seed coat becomes impermeable to oxygen and moisture and hardens.

The suspension of growth is due to exogenous (environmental conditions) or endogenous control during which metabolic activity of the seed is greatly reduced.

Causes of Dormancy

  1. Impermeable or mechanically resistant seed coats.
  2. Rudimentary or physiologically immature embryos or
  3. Due to the presence of germination inhibitors such as abscisic acid, phenolic acid, short-chain fatty acids, and coumarin.

How can overcome seed dormancy?

  1. Mechanical or chemical scarification of the seed coat (scratching of seed coat or seeds soaked in chemicals to break the dormancy)
  2. Stratification of seeds or changing environmental conditions such as temperature, light, and pressure. Stratification of seeds is subjecting the moist seeds to oxygen for variable periods of low or high temperatures.

Plus One Economics Notes Chapter 5 Human Capital Formation in India

Students can Download Chapter 5 Human Capital Formation in India Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 5 Human Capital Formation in India

Human Capital
Human capital is defined as a human resource where people as a capital asset which yields a stream of economic benefit over their working span of life. Converting human beings through education and training into human resources like teachers, doctors, scientists, etc. can be termed as human capital formation.

Sources of Human Capital
Sources of human capital can be classified into five. They are given below:

1. Investment in Education: Investment in education is considered one of the main sources of human capital. Education increases the efficiency and productivity of the population which will enhance the welfare of people and facilitate growth and economic development.

2. Investment in Health: Good health of a person helps him to produce goods and services to his optimum level. Therefore, spending on health to improve the health status of the population is other way of spending on human capital.

3. Investment in Job Training: On-the-job training under the supervision of skilled professionals increases productivity. Expenditure regarding on the job-training is a source of human capital formation as the return of it is in the form of enhanced labour productivity.

4. Migration: Migration means the movement of people from one place to another. Migration happens due to marriage, education and employment. People migrate in search of jobs that fetch them higher salaries than what may get in their native places. In India, rural-urban migration is very rampant in which rural people migrate to cities in search of better jobs.

5. Investment in Information Acquisition: People spend to obtain information regarding labour market and other markets like education and health. This information is necessary to make decisions regarding investments in human capital for efficient utilization of the acquired human capital stock. So, expenditure incurred for acquiring information relating to labour market is also a source of human capital information.

Human Capital and Economic Growth
Human capital is essential for economic growth. We know that the labour skill of an educated person is more than that of an uneducated person and that the former generates more income than the latter. Economic growth means thd increase in real national income of a country; naturally, the contribution of the educated person to economic growth is more than that of an illiterate person. If a healthy person could provide an uninterrupted labour supply for a longer period of time, then health is also an important factor for economic growth. Thus, both education and health, along with many other factors like on-the-job training, job market information and migration, increase an individual’s income-generating capacity.

Human Capital and Human Development
Human capital considers education and health as a means to increase labour productivity. Human development is based on the idea that education and health are integral to human well-being. When people attain certain ability to read and write and acquire ability to lead a long and healthy life, their productivity will go up. Human welfare should be increased through investments in education and health. Education and health improve human welfare. In a welfare state, every individual has the right to get basic right to be literate and lead a healthy life.

Human Capital Formation in India
We have seen that human capital formation is essential for economic growth and development. Various institutions are operating in India for human capital formation. They are

  • NCERT (National Council of Educational Research and Training) regulates school level education.
  • UGC (University Grants Commission) regulates higher education (colleges, universities and research institutes.
  • AICTE (All India Council of Technical Education) regulates technical education.
  • ICMR (Indian Council for Medical Research) regulates medical education.

Educational Achievements in India
India made significant improvements in the education sector since independence. Our literary rate was 18.3% in 1951. It increased to 74% in 2011. Similarly, the number of students seeking higher education increased in the country. The government also enacted the Right to Education Act in 2010 to ensure free and compulsory education for children.

Plus One Computer Application Chapter Wise Questions Chapter 6 Introduction to Programming

Students can Download Chapter 6 Introduction to Programming Questions and Answers, Plus One Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Application Chapter Wise Questions Chapter 6 Introduction to Programming

Plus One Computer Application Introduction to Programming 1 Mark Questions and Answers

Question 1.
From the following which is ignored by the compiler
a) statement
b) comments
c) loops
d) None of these
Answer:
b) comments

Question 2.
Multi line comment starts with ____ and ends with
a) /’ and’/
b)*/and/*
c) /* and*/
d)’/ and /’
Answer:
c) /* and*/

Question 3.
Single line comment starts with _____.
a) **
b)@@
c) */
d) //
Answer:
d) //

Question 4.
Alvin wants to store the value of π From the following which is correct declaration
a) char pi=3.14157
b) const int pi=3.14157
c) const float pi=3.14157
d) long pi=3.14157
Answer:
c) const float pi=3.14157

Question 5.
To store 70000 which modifier is used with int.
a) long
b) short
c) big
d) none of these
Answer:
a) long

Question 6.
To store 60000 which modifier is used with int.
a) unsigned
b) short
c) big
d) none of these
Answer:
a) unsigned

Question 7.
Consider x++(post fix form). Select the correct definition from the following
a) The operation is performed after the value is . used
b) The operation is performed before the value is used
c) First change then use
d) None of these
Answer:
a) The operation is performed after the value is used

Question 8.
Consider ++x(pre fix form). Select the correct definition from the following
a) The operation is performed after the value is used
b) The operation is performed before the value is used
c) First use then change
d) None of these
Answer:
b) The operation is performed before the value is used

Question 9.
Consider the following
int a=10;
float b=4;
cout<<a/b;
We know that the result is 2.5 a float. What type of conversion is this ?
a) type promotion
b) type casting
c) explicit conversion
d) None of these
Answer:
a) type promotion(implicit conversion)

Question 10.
From the following which has the major priority?
a) ++
b) =
c) ==
d) &&
Answer:
++

Question 11.
From the following which has the major priority?
a) ++
b) =
c) ==
d) &&
Answer:
++

Qn. 11
One of your friend told you that post increment (eg:x++) has more priority than, pre increment (eg: ++x).
State True / False
Answer:
It is true.

Question 12.
Emerin wants to store a constant value. Which key word is used for this?
Answer:
constant.

Question 13.
Suppose x= 5. Then cout<<x++ displays _____.
Answer:
Here post increment first use the value then incremented

Question 14.
Suppose x= 5. Then cout<<++x displays _____.
Answer:
6. Here pre increment first incremented and then use the value.

Question 15.
Pick the odd one out :
a) long
b) short
c) unsigned
d) int
Answer:
d) int. it is fundamental data type and the others are type modifiers

Question 16.
Memory size and sign can be changed using ______ with fundamental data types.
Answer:
Type modifiers

Question 17.
“Its value does not change during execution”. What is it?
Answer:
constant.

Question 18.
“BVM HSS” is called ______.
а) integer constant
b) float constant
c) string constant
d) None of these
Answer:
c) string constant

Question 19.
he address of a variable is called ______.
Answer:
L-value (Location value) of a variable

Question 20.
The content of a variable, is called _____.
Answer:
R-value (Read value) of a variable

Question 21.
Suppose the address of a variable age is 1001 and the content i.e. age = 33. Then what is R-value and L-value?
Answer:
R-value is 33 and L-value is 1001

Question 22.
is it possible to declare a variable as and when a need arise . What kind of declaration is this ?
Answer:
Yes.lt is known as Dynamic declaration

Question 23.
In the following program, some lines are missing. Fill the missing lines and complete it.
#include<iostream.h>
{
4 int num1, num2, sum;
Cout<<“Enter two numbers:”;
…………..
…………
Cout<<“sum of numbers are=”<<sum;
}
Answer:
The correct program is given below.
#include<iostream>
using namespace std;
int main( )
{
int num1,num2,sum;
cout<<“Enter two numbers:”;
cin>>num1>>num2;
sum=num1+num2;
cout<<“Sum of numbers are=”<<sum;
}

Question 24.
The following program finds the sum of three numbers. Modify the program to find the average. . (Average should display fractional part also.)
#include<iostream>
using namespace std;
int main ( )
{
int x, y, z, result;
cout<<“Enter values for x, y, z”;
cin>>x>>y>>z;
result=x+y+z;
cout<<“The answer is =”<<result;
return 0;
}
Answer:
float result
result = (x + y + z) / 3.0;

Question 25.
Some of the C++ statements given below are invalid. ,
i) cin>>m, n;
ii) a + b = c;
iii) void p;
iv) cout<<28;
Identify them from the following alternatives:
a) Statement (i) and (ii) only
b) Statement (Ii) and (iii) only
c) Statement (i), (ii) and (iii) only
d) All the statements
Answer:
c) Statement (i); (ii) and (iii) only

Question 26.
Which one of the following is NOT a valid C++ statement?
a) x = x+ 10;
b) x + = 10;
c)x+10 = x;
d) x=10 + x;
Answer:
c)x+10 = x;

Plus One Computer Application Introduction to Programming 2 Marks Questions and Answers

Question 1.
What do you mean by pre processor directive?
Answer:
A C++ program starts with the pre processor directive i.e., # include, #define, #undef, etc, are such a pre processor directives. By using #include we can link the header files.that are needed to use the functions. By using #define we can define some constants.
Eg: #define x 100.
Here the value of x becomes 100 and cannot be changed in the program.

Plus One Computer Application Introduction to Programming 3 Marks Questions and Answers

Question 1.
We know that a program has a structure. Explain the structure of C++program.
Answer:
A typical C++ program would contain four sections as shown below.
Include files
Function declarations
Function definitions
Main function programs
Eg:
#include<iostream>
using namespace std;
int sum(int x, int y)
{
return (x+y);
}
int main( )
{
cout<<sum(2,3);
}

Question 2.
Write a program to print a message as” Hello, Welcome to C++”.
Answer:
#include<iostream>
using namespace std;
int main( )
{
cout<<” Hello, Welcome to C++”;
}

Question 3.
Following is a sample C++ program. Identify the errors if any in the structure. Correct the program..
#include<iostream>
using namespace std;
int main [ ]
{
*/ It is a simple program */
Cout<< “Welcome to C++”
Cout<< “The End” .
}
Answer:
The multi line comment used in this program is wrong. So the correct code is as follows
#include<iostream>
using namespace std;
int main ( )
{
/* It is a simple program */
cout<< “Welcome to C++”
cout<< “\nThe End”
}

Question 4.
Write a program to read two numbers and find its sum.
Answer:
#include<iostream>
using namespace std;
int main( )
{
int n1,n2,sum;
cout<<“Enter two numbers”;
cin>>n1>>n2;
sum=n1+n2;
cout<<“The sum of “<<n1<<” and “<<n2<<” is “<<sum;
}

Question 5.
Write a program to read three scores and find the average.
Answer:
#include<iostream>
using namespace std;

int main( )
{
ints1,s2,s3;
float avg;
cout<<“Enter the three scores”;
cin>>s1>>s2>>s3;
avg=(s1 +s2+s3)/3.0;
cout<<“The average CE score is “<<avg;
}

Question 6.
Write a program to find the area and perimeter of a circle.
Answer:
#include<iostream>
using namespace std;
int main( )
{
const float pi=3.14157;
float r,area,perimeter;.
cout<<“Enter the radius of a circle”;
cin>>r;
area=pi*r*r;
perimeter= 2*pi*r;
cout<<“Area of the circle is”<<area<<“\nPerimeter of the circle is “<<perimeter;
}

Question 7.
Write a program to find the simple interest.
Answer:
#include<iostream>
using namespace std; ,
int main( )
{
float p,n,r,si;
cout<<“Enter the Principal amount”;
cin>>p;
cout<<“Enter the number of years”;
cin>>n;
cout<<“Enter the rate of interest”;
Cin>>r;
si=p*n*r/100;
cout<<“Simple interest is “<<si;

Question 8.
Write a program to convert temperature from Celsius to Fahrenheit.
Answer:
#include<iostream>
using namespace std;
int main( )
{
float c,f;
cout<<“Enter the Temperature in Celsius:”;
cin>>c;
f=1,8*c+32;
cout<<c<<” Degree Celsius = “<<f<<” Degree Fahrenheit”;
}

Question 9.
Write a program to read weight in grams and convert it into Kilogram.
Answer:
#include<iostream>
using namespace std;
int main( )
{
float gm.kg;
cout<<“Enter the weight in grams:”;
cin>>gm; kg=gm/1000;
cout<<gm<<” grams = “<<kg<<” Kilograms”;
}

Question 10.
Write a program to generate the following table.
Plus One Computer Application Chapter Wise Questions Chapter 6 Introduction to Programming 3
Use a single cout statement for output
Answer:
#include<iostream>
using namespace std;
int main( )
{
cout<<“2013\t100%\n2012\t99.9%\n2011 \t95.5%\n2010\ t90.81 %\n2009\t85%”;
getch( );
}

Question 11.
Write a program to read your height in meter and cm convert it into Feet and inches
Answer:
#include<iostream>
using namespace std;
int main( )
{
int m,cm;
float inch;
int feet;
cout<<“Enter your height in Centimeter:”;
cin>>cm;
cout<<” Your height is “<<cm/100<<” Meters and “<<cm%100<<” cm \n”;
inch=cm/2.54;
feet=inch/12;
inch=(int)inch%12;
cout<<feet<<“feet and”<<inch<<” inch”;
}

Question 12.
Write a program to find the area of a triangle Triangle .
Answer:
#include<iostream>
using namespace std;
int main( )
{
intb,h;
float area;
cout<<“Enter values for b and h”;
cin>>b>>h;
area=0.5*b*h;
cout<<“The area of the triangle is “<<area;
}

Question 13.
Write a program to find simple interest and compound interest.
Answer:
#include<iostream>
using namespace std;
#include<cmath>
int main( )
{
float p,n,r,si,ci;
cout<<“Enter the Principal amount”;
cin>>p;
cout<<“Enter the number of years”;
cin>>n;
cout<<“Enter the rate of interest”;
cin>>r;
si=p*n*r/100;
ci=p*pow((1+r/100),n)-p;
cout<<“Simple interest is “<<si;
cout<<“\nThe compound interest is “<<ci;
}

Question 14.
Write a program to
i) print ASCII for a given digit
ii) print ASCII for backspace.
Answer:
#include<iostream>
using namespace std;
int main( )
{
char ch;
intasc.bak;
cout<<“Enter a digit:”;
cin>>ch;
asc=ch;
cout<<“ASCII for “<<ch<<” is “<<asc;
bak=’\b’;
cout<<“\nASCII for backspace is “<<bak;
}

Question 15.
Write a program to read time in seconds and convert it into hours, minutes and seconds .
Answer:
#include<iostream>
using namespace std;
int main( )
{
long h,m,s;
cout<<“Enter the time in seconds:”;
cin>>s;
h=s/3600;
s=s%3600;
m=s/60.;
s=s%60;
cout<<h<<” hr:”<<m<<” min:”<<s<<” secs”;
}

Question 16.
Consider the following ‘
int a=45.65;
cout<<a;
What is the output of the above. Is it possible to convert a data type to another type? Explain.
Answer:
The output of the code is 45, the floating point number is converted into integer. It is possible to convert a data type into another data type. Type conversions are two types.
1) Implicit type conversion : This is performed by C++ compiler internally. C++ converts all the lower sized data type to the highest sized operand. It is known as type promotion. Data types are arranged lower size to higher size is as follows, unsigned int(2 bytes) ,int(4 bytes),long(4 bytes) , unsigned long(4 bytes), float(4 bytes), double(8 bytes), long double(10 bytes)

2) Explicit type conversion : It is known as type casting. This is done by the programmer. The syntax is given below.
(data type to be converted) expression
Eg. intx=10;
(float)’x; This expression converts the data type of the variable from integer to float.

Question 17.
Match the following numbers and data types in C++ to form the most suitable pairs.

1) 142789 a) Signed
2) 240 b) Double
3) -150 c) Long int
4) 8.4×10-4000 d) Float
5) 0 e) Long double
6) 0.0008 f) Unsigned short
7) -127 g) Short int
8) 2.8×10308 h) Signed char

Answer:

1) 142789 a) Long int
2) 240 b) Short int
3) -150 c) Signed
4) 8.4×10-4000 d) Long double
5) 0 e) Unsigned short
6) 0.0008 f) Float
7) -127 g) Signed char
8) 2.8×10308 h) Double

Question 18.
Determine the data type of the following expression.
If a is an int, b is a float, c is a long int and d is a double
\(\frac{(1-a c)^{25}}{(c-d)}+\frac{(b+a) / c}{(l o n g)(a+d)}\)
Answer:
In type promotion the operands with lower data type will be converted to the highest data type in the expression. So consider the following,
Plus One Computer Application Chapter Wise Questions Chapter 6 Introduction to Programming 2
=double + long
= double (Which is the highest data type)

Question 19.
What is implicit type conversation? Why it is called type promotion?
Answer:
Type conversion : Type conversions are of two types.
1) Implicit type conversion: This is performed by C++ compiler internally. C++ converts all the lower sized data type to the highest sized operand. It is known as type promotion. Data types are arranged lower size to higher size is as follows, unsigned int(2 bytes), int(4 bytes),long (4 bytes) , unsigned long (4 bytes), float(4 bytes), double(8 bytes), long double(10 bytes)

Question 20.
Area of a circle is calculated using the formula πr², where π =3.14 and r is the radius of the circle. Fill in the blanks in the following program which finds the area of a circle.
void main ( )
{
int area,‘rad;
cout<<“Enter the radius”;
cin ………..
area = …………
cout …………
}
Answer:
Plus One Computer Application Chapter Wise Questions Chapter 6 Introduction to Programming 1

Question 21.
With the help of an example, explain type casting in C++ programs?
Answer:
The output of the code is 45, the floating point number is converted into integer. It is possible to convert a data type into another data type. Type conversions are two types.
1) Implicit type conversion : This is performed by C++ compiler internally. C++ converts all the lower sized data type to the highest sized operand. It is known as type promotion. Data types are arranged lower size to higher size is as follows.
unsigned int(2 bytes) ,int(4 bytes),long(4 bytes) , unsigned long(4 bytes), float(4 bytes), double(8 bytes), long double(10 bytes)

2) Explicit type conversion : It is known as type casting. This is done by the programmer. The syntax is given below.
(data type to be converted) expression .
Eg. int x=10;
(float) x; This expression converts the data type of the variable from integer to float.

Question 22.
Comments in a program are ignored by the complier. Then why should we include comments? Write the methods of writing comments in a C++ program.
Answer:
To give tips in between the program comments are used. A comment is not considered as the part of program and cannot be executed. There are 2 types of comments single line and multiline.
Single line comment starts with //(2 slashes) but multi line comment starts with /* and ends with */

Question 23.
if A = 5, B = 4, C = 3, D = 4, then what is the result in X after the following operation?
X=A+B-C*D
OR
Is there any difference between (a) and (b) by considering the following statement?
char Gender;
a) Gender = ‘M’;
b) Gender = “M”;
Answer:
X=A+B-C*D
=5+4-3*4
=5+4-12(Reason :Multiplication has more priority than addition and subtraction).
=9-12
= – 3
OR
a) character constant ‘M’ is stored in the variable Gender.
b)“M” is a string constant hence it cannot be assigned by using an assignment operator. Use string function as follows strcpy(Gender,”Mn);

Plus One Computer Application Introduction to Programming 5 Marks Questions and Answers

Question 1.
Two pairs C++ expressions are given below.
i) a=10 a==10
ii) b=a++, b=++a
a) How do they differ?
b) What will be the effect of the expression ?
Answer:
i) = is an assignment operator that assigns a value 10 to the LHS (Left Hand Side)variable a But == is equality operator that checks whether the LHS and RHS are equal or not. If it is equal it returns a true value otherwise false .
ii) In a++ , ++ is a post(means after the operand) increment operator and in ++a, ++ is a pre(means before the operand) increment operator. They are entirely different.
Post increment
Here first use the value of ‘a’ and then change the value of ‘a’.
Eg: if a=10 then b=a++. After this statement b=10 and a=11
Pre increment
Here first change the value of a and then use the value of a.
Eg: if a=10 then b=++a. After this statement b=11 anda=11

Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators

Students can Download Chapter 5 Data Types and Operators Questions and Answers, Plus One Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Application Chapter Wise Questions  Chapter 5 Data Types and Operators

Plus One Computer Application Data Types and Operators 1 Mark Questions and Answers

Question 1.
________ is the main activity carried out in computers
Answer:
Data processing

Question 2.
The data used in computers are different. To differentiate the nature and size of data _____is used.
Answer:
Data types

Question 3.
Classify the following data types.
int
array
function
char
pointer
void
float
double
structure
Answer:
Fundamental data types
int
float
double
void
char
Derived data types
array
function
pointer
structure

Question 4.
Sheela wants to store her age. From the following which is the exact data type.
a) void
b) char
c) int
d) double
Answer:
c) int

Question 5.
Integer data type uses Integer data type ;
a) 5
b) 2
c) 3
d)4
Answer:
d)4

Question 6.
char data type uses ______ bytes of memory
a) 1
b) 3
c) 7
d) 8
Answer:
a)1

Question 7.
From the following which data type uses 4 bytes of memory
a) float
b) short
c) char
d) double
Answer:
a) float

Question 8.
Expand ASCII _______.
Answer:
American Standard Code for Information Inter-change

Question 9.
Ramu wants to store the value of From the fol-lowing which is correct declaration
a) char pi=3.14157
b) int pi=3.14157
c) float pi=3.14157
d) long pi=3.14157
Answer:
c) float pi=3.14157

Question 10.
From the following which is not true , to give a variable name.
a) Starting letter must be an alphabet
b) contains digits
c) Cannot be a key word
d) special characters can be used
Answer:
d) special characters can be used

Question 11.
Pick a valid variable name from the following
a) 9a
b) float
c)age
d) date of birth
Answer:
c)age

Question 12.
To perform a unary operation how many number of operands needed?
a) 2
b)3
c) 1
d) None of these
Answer:
c) 1 (Unary means one)

Question 13.
To perform a binary operation how many number of operands needed?
a) 2
b) 3
c) 1
d) None of these.
Answer:
a) 2 (binary means two)

Question 14.
To perform a ternary operation how many number of operands needed?
a) 2
b) 3
c) 1
d) None of these.
Answer:
b) 3 (eg: ternary means three)

Question 15.
In C++ 13% 26 =
a) 26
b) 13
c) 0
d) None of these
Answer:
% is a mod operator i.e, it gives the remainder. Here the remainder is 13.

Question 16.
In C++ 41/2 =
a) 20.5
b) 20
c) 1
d) None of these
Answer:
b) 20. (The actual result is 20.5 but both 41 and 2 are integers so .5 must be truncated)

Question 17.
++ is a ______ operator
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
a) Unary.

Question 18.
Conditional operator is ______ operator
a) Unary
b) Binary
c) Ternary
d)None of these
Answer:
c) Ternary

Question 19.
% is a operator ______.
a) Unary
b) Binary
c) Ternary
d) None of these
Answer:
b) Binary

Question 20.
State true or false:
a) Multiplication, division, modulus have equal priority.
b) Logical and (&&) has less priority than logical or ( )
Answer:
a) True
b) False

Question 21.
_____ is composed of operators and operands.
a) expression
b) Key words
Identifier
d) Punctuators
Answer:
a) expression

Question 22.
Supply value to a variable at the time of declaration is known as _____.
Answer:
Initialisation

Question 23.
From the following which is initialisation
a) int k;
b)int k=100;
c) int k[10];
d) None of these
Answer:
b) int k=100;

Question 24.
State True / False
In an expression all the operands having lower size are converted(promoted) to the data type of the highest sized operand.
Answer:
True

Question 25.
Classify the following as arithmetic / Logical expression
a) x+y*z
b) xz
c) x/y
d) x>89 || y<80
Answer:
a) and c) are Arithmetic
b) and d) are Logical

Question 26.
Suppose x=5 and y=2 then what will be cout<<(float) x/y
Answer:
2.5 The integer x is converted to float hence the result.

Question 27.
Consider the following.
a = 10;
a =100;
Then a =
a) a = 100
b) a = 50
c)a = 10
d) a = 20
Answer:
a) a = 100. This short hand means a = a*10

Question 28.
Consider the following.
a=10 ;
a+=10;
Then a=
a) a= 30
b) a= 50
c)a=10
d) a=20
Answer:
d) a=20. This short hand means a=a+10

Question 29.
Pick the odd one out
a) structure
b) Array
c) Pointer
d) int
Answer:
d) int, it is fundamental data type the others are derived data types

Question 30.
From the following select not a character of C++ language
a) A
b) 9
c)\
d)@
Answer:
d)@

Question 31.
Consider the following
float x=25.56; cout<<(int)x; Here the data type of the variable is converted. What type of conversion is this?
a) type promotion
b) type” casting
c) implicit conversion
d) None of these
Answer:
b) type casting (explicit conversion)

Question 32.
Identify the error in the following C++ statement and correct it.
Answer:
The maximum number that can store in short type is less than 32767. So to store 68000 we have to use long data type.

Question 33.
Consider the following statements in C++ if(mark>=18)
cout<<“Passed”; ’
else ,
cout<<“Failed”;
Suggest an operator in C++ using which the same , output can be produced.
Answer:
Conditional operator (?:)

Plus One Computer Application Data Types and Operators 2 Marks Questions and Answers

Question 1.
Analyses the following .statements and write True or False. Justify
i) There* is an Operator in C++ having no special character in it
ii) An operator cannot have more than 2 operands
iii) Comma operator has the lowest precedence
iv) All logical operators are binary in nature
v) It is not possible to assign the constant 5 to 10 different variables using a single C++ expression
vi) In type promotion the operands with lower data , type will be converted to the highest data type in expression.
Answer:
i) True (size of operator)
ii) False( conditional operator can have 3 operands
iii) True
iv) False
v) False(Multiple assignment is possible,
eg: a=b=c=…..= 5
vi) True

Question 2.
Consider the following declaration.
const int bp;
bp = 100;
Is it valid? Explain it?
Answer:
This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration js as follows, const int bp=100;

Question 3.
Consider the following statements in C++
1) cout<<41/2;
2) cout<<41/2.0; Are this two statements give same result? Explain?
Answer:
This two statements do not give same results. The first statement 41/2 gives 20 instead of 20.5. The reason is 41 and 2 are integers. If two operands are integers the result must be integer, the real part must be truncated. To get floating result either one of the operand must be float. So the second statement gives 20.5. The reason is 41 is integer but 2.0 is a float.

Question 4.
If mark = 70 then what will be the value of variable result in the following result = mark > 50 ? ’P’: ‘F’;
Answer:
The syntax of the conditional operator is given below Condition? Value if true: Value if false;
Here the conditional operator first checks the condition i.e.,70>50 it is true. So ‘P’ is assigned to the variable result.
So the result is ‘P’

Question 5.
Is it possible to initialise a variable at the time of execution. What kind of initialisation is this? Give an example
Answer:
Yes it is possible. This is known as Dynamic initialisation.. The example is given below
Eg: int a=10,b=5;
int c=a*b;
here the variable c is declared and initialised with the value 10*5.

Question 6.
Boolean data type is used to store True / False in C++. Is it true? Is there any data type Called Boolean in C++?
Answer:
No there is no data type for storing boolean value true / false.
But in C++ non -zero (either negative or positive) is treated as true and zero is treated as false.

Question 7.
Consider the following
n=-15;
if (n)
cout<<“Hello”;
else
cout<<“hai”;
What will be the output of the above code?
Answer:
The Output is Hello, because n = – 15 a non zero number and it is treated as true hence the result.

Question 8.
Is it possible to declare a variable in between the program as and when the need arise.? Then what is it?
Answer:
Yes it is possible to declare a variable in between the program as and when the need arise. It is known as dynamic initialisation.
Eg. int x=10,y=20;
…………..;
…………
int z=x*y

Question 9.
char ch;
cout<<“Enter a character”; cin>>ch;
Consider the above code, a user gives 9 to the variable ‘ch’. Is there any problem? Is it valid?
Answer:
There is no problem and it is valid since 9 is a character. Any symbol from the key board is treated as a character.

Question 10.
“With the same size we can change the sign and range of data”. Comment on this statement.
Answer:
With the help of type modifiers we can change the sign and range of data with same size. The important modifiers are signed, unsigned, long and short.

Question 11.
Write short notes about C++ short hands?
Answer:
x=x + 10 can be represented as x+=10, It is called short hands in C++. It is faster.
This is used with all the arithmetic operators as follows.
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 1

Question 12.
What is the role of ‘const’ modifier? ‘
Answer:
This ‘const’ key word is used to declare a constant, Eg. const int bp=100;
By this the variable bp is treated as constant and cannot be possible to change its value during execution.

Question 13.
Specify the most appropriate data type for handling the following data.
i) Rollno. of a student.
ii) Name of an employee.
iii) Price of an article.
iv) Marks of 12 subjects
Answer:
i) short Rollno;
ii) char name[20];
iii) float price;
iv) short marks[12j;

Question 14.
Write C++ statement for the following.
a) The result obtained when 5 is divided by 2.
b) The. remainder obtained when 5 is divided by 2.
Answer:
a) 5/2
b) 5%2

Question 15.
Predict the output.justify
int k = 5,
b = 0;
b = k++ + ++k;
cout<<b;
Answer:
Output is 12. In this statement first it take the value of k in 5 then increment it K++. So first operand for + is 5. Then it becomes 6. Then ++k makes it 7. This is the second operand. Hence the result is 12.

Question 16.
Predict the output.
a) int sum = 10, ctr = 5;
sum = sum + ctr –;
pout<< sum;
b) int sum = 10, ctr = 5;
sum = sum + ++ctr; cout<<sum;
Answer:
a)15
b) 16

Question 17.
Predict the output.
int a; .
float b; .
a = 5;
cout << sizeof(a + b/2);
Answer:
Output is 4. Result will be the memory size of floating point number

Question 18.
Predict the output.
int a, b, c;
a = 5; b = 2;
c = a/b;
cout<<c; Answer: Output is 2. Both operands are integers. So the result will be an integer.

Question 19.
Explain cascading of i/o operations
Answer:
The multiple use of input or output operators in a single statement is called cascading of i/o operators.
Eg: To fake three numbers by using one statement is as follows cin>>x>>y>>z;
To print three numbers by using one statement is
as follows
cout<<x<<y<<z;

Question 20.
Trace out and correct the errors in the following code fragments
i) cout<<“Mark=”45;
ii) cin <<“Hellow World!”; iii) cout>>”X + Y;
iv) Cout<<‘Good'<<‘Moming’
Answer:
i) cout<<“Mark=45”;
ii) cout <<“Hellow World!”;
iii) cout< iv) Cout<<“Good Morning”;

Question 21.
Raju wants to add value 1 to the variable ‘p’ and store the new value in ‘p’ itself. Write four different statements in C++ to do the task.
Answer:
1) P=P+1; 2) p++;(post increment) 3) ++p; (pre increment) 4) p+=1; (short hand in C++)

Question 22.
Read the following code: charstr[30]; cin>>str;
cout<<str; . If we give the input “Green Computing”, we get the . output “Green”. Why is it so? How can you correct that?
Answer:
The input statement cin>> cannot read the space. It reads the text up to the space, i.e. the delimiter is space. To read the text up to the enter key gets( ) or getline( ) is used

Question 23.
Match the following :
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 2
Answer:
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 3

Question 24.
Write a C++ expression to calculate the value of the following equation.
\(x=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\)
Answer:
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 13

Question 25.
A student wants to insert his name and school address in the C++ program that he has written. But this should not affect the compilation or execution of the program. How is it possible? Give an example.
Answer:
He can use comments to write this information. In C++ comments are used to write information such as programmer’s name, address, objective of the codes etc. in between the actual codes. This is not the part of the programme.
There are two types of comments.
i) Single line (//) and ii) Multi line (/* and */)
i) Single line (//) : Comment is used to make a single line as a comment. It starts with //.
Eg : //programme starts here.
ii) Multi line (/* and */): To make multiple lines as a comment. It starts with /* and ends with */.
Eg : !* this programme is used to find sum of two numbers.*/

Question 26.
Consider the following C++ statements :
char word (20]
cin>>word;
coUt<<word; gets(word); puts(word); If the string entered is “HAPPY NEW YEAR”, predict the output and justify your answer.
Answer:
cin>>word;
cout<<word;
It displays “HAPPY” because cin takes characters upto the space. That is space is the delimiter for cin. The string after space is truncated. To resolve this use gets ( ) function. Because gets ( ) function reads character upto the enter key.
Hence gets (word);
puts (word);
Displays “HAPPY NEW YEAR”

Question 27.
Write the difference between x = 5 and x ==5 in C++.
Answer:
x = 5 means the value 5 of the RHs is assigned to the LHS variable x . Here = is the assignment operator. But x ==5, ==this is the relational (comparison) operator. Here it checks whether the value of RHS is equal to the value of LHS and this expression returns a boolean value as a result. It is the equality operation.

Question 28.
a) What is the output of the following program?
# include
void main ( )
{
int a;
a = 5+3*5;
cout << a;
}
b) How do 9, ‘9’ and “9” differ in C++ program?
Answer:
Here multiplication operation has more priority than addition.
hence a = 5 + 15 = 20
b) Here
9 is an interger
‘9’ is an character
“9” is a string

Question 29.
Read the following C++ program and predict the output by explaining the operations performed.
#include
void main ( )
{
int a=5, b=3;
cout<<a++ /–b;
cout<<a/ (float) b; . . }
Answer:
Here a = 5 and b = 3 a++ / – – b = 5/2 = 2 That is a++ uses the value 5 and next it changes its value to 6 So a/(float) b = 6/(float)2 = 6/2.0 = 3 So the output, is 2 and 3

Question 30.
What is preprocessor directive statement? Explain. with an example.
Answer:
A C++ program starts with the pre processor directive i.e., # include, #define, #undef, etc, are such a pre processor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
Eg. #define x 100. Here the value of x becomes 100 and cannot be changed in the program. No semicolon is needed.

Question 31.
The following C++ code segment is a part of a program written by Smitha to find the average of 3 numbers. int a, b, c; float avg; cin>>a>>b>>c;
avg=(a+b+c)/3; ‘
cout<<avg; , What will be the output if she inputs 1, 4 and 5? How can you correct it?
Answer:
=(1 +4+5)/3 =10/3 =3.3333 Instead of this 3.3333 the output will be 3. This is because if both operands are integers an integer division will be occurred, that is the fractional part will be truncated. To get the correct out put do as follows
case 1: int a,b,c; is replaced by float a,b,c;
OR
case 2: Replace (a+b+c)/3 by (a+b+c)/3.0;
OR
case 3:Type casting. Replace avg=(a+b+c)/3; by avg=(float)(a+b+c)/3;

Plus One Computer Application Data Types and Operators 3 Marks Questions and Answers

Question 1.
In a panchayath or municipality all the houses have a house number, house name and members. Similar situation is in the case of memory.Explain Answer: The named memory locations are called variable. A variable has three important things 1) variable name : A variable should have a name 2) Memory address : Each and every byte of memory has an address, it is also called location (L) value 3) Content: The value stored in a variable is called content. It is also called Read(R) value.

Question 2.
Briefly explain constants
Answer:
A constant or a literal is a data item its value doe not change during execution. The keyword const is used to declare a constant. Its declaration is as follows , const data type variable name=value; eg.const int bp=100; const float pi=3.14157; const char ch=’a’; const ehar[]=”Alvis”; .
1) Integer literals : Whole numbers withbut fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal. Eg. For decimal 100,150,etc For octal 0100,0240, etc For hexadecimal 0x100, 0x1 A,etc
2) Float literals : A number with fractional parts and its value does not change during execution is called floating point literals. Eg. 3.14157,79.78, etc
3) Character literal: A valid C++ character enclosed in single quotes, its value does not change during execution. Eg. ‘m’, ‘f’, etc
4) String literal One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’) . Eg. “Mary’s”,”India”,etc

Question 3.
Consider the following statements int a=10,x=20; float b=49000.34,y=56.78; i) a=b; ii) y=x; Is there any problem for the above statements? What do you mean by type compatibility?
Answer:
Assignment operator is used to assign the value of RHS to LHS. Following are the two chances

1) The size of RHS is less than LHS. So there is no problem and RHS data type is promoted to LHS. Here it is compatible.
2) The size of RHS is higher than LHS. Here comes the problem sometimes LHS cannot possible to assign RHS. There may be a chance of wrong answer. Here it is not compatible.
Here
i) a=b; There is an error since the size of . LHS is 2 but the size of RHS is 4.
ii) y=x; There is no problem because the size of LHS is 4 and RHS is 2.

Question 4.
A company has decided to give incentives to their salesman as per the sales. The criteria is given below. If the total sales exceeds 10,000 the incentive is 10% .
i) If the total sales >=5,000 and total sales <10,000, the incentive is 6 % ii) If the total sales >=1,000 and total sales <5,000, the incentive is 3 %
Write a C++ program to solve the above prob-lem and print the incentive after accepting the total sales of a salesman. The program code should not make use of ‘if statement. )
Answer:
#include
using namespace std;
int main( )
{
float sales,incentive;
cout<<“enter the sales”; cin>>sales;
incentive=(sales>10000 ? salesMO: (sales > =5000 ? sales * .06 : (sales >= 1000 ? sales * .03: 0)>>;
cout<<“\nThe incentive is ” << incentive;
}

Question 5.
A C++ program code is given below to find the value of X using the expression
\(x=\frac{a^{2}+b^{2}}{2 a}\)
where a and b are variables
#include
using namespace std;
intmain( )
{
int a;b;
float x ’
cout<<“Enter the values of a and b; cin>a>b;
x=a*a+b*b/2*a;
cout>>x;
}

Predict the type of errors during compilation, execution and verification of the output. Also write the output of two sets of input values
i) a=4 b=8
ii) a=0 b=2
Answer:
This program contains sortie errors and the correct program is as follows.
#include
using namespace std;
int main( )
{
int a,b; . . .
float x;
cout<<“Enter the values of a and b”; cin>>a>>b;
x=(a*a+b*b)/(2*a);
cout<<x;
}
The output is as follows
i) a=4 and b= 8 then the output is 10
a=0 and b= 2 then the output is an error divide by zero error(run time error)

Question 6.
A list of data items are given below
45,8.432, M, 0.124,8 , 0, 8.1X 1031, 1010, a, 0.00025, 9.2 X1012O, 0471 ,-846, 342.123E03
a) Categorise the given data under proper headings of fundamental data types in C++
b) Explain the specific features of each data type. Also mention any other fundamental data type for which sample data is not given.
Answer:
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 4
b) i) int data type:- It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory.i.e.- 232 . numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve ) So a total of 232 numbers. We can store a number in between -231 to + 231-1.

ii) char data type :- Any symbol from the key board, eg. ‘A’ , ‘?’, ‘9’, It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

iii) float data type:- It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory. Eg. 67.89, 89.9 E-15.

iv) double data type:- It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

v) void data type :- void means nothing. It is used to represent a function returns nothing.

Question 7.
Write valid reasons after reading the following statements in C++ arid comment on their correctness by give reasons.
i) char num=66;
char num =’B’;
ii) 35 and 35L are different
iii) The number 14,016 and OxE are one and the same
iv) Char data type is often said to be an integer type.
v) To store the value 4.15 float data type is preferred over double
Answer:
i) The ASCII number of B is 66. So it is equivalent.
ii) 35 is of integer type but 35L is Long
iii) The decimal number 14 is represented in octal is 016 and in hexadecimal is OxE.
iv) Internally char data type stores ASCII numbers.
v) To store the value 4.15 float data type is better because float requires only 4 bytes while double needs 8 bytes hence we can save the memory.

Question 8.
Suggest most suitable derived data types in C++ for storing the following data items or statements :
rived data type
i) Age of 50 students in a class
ii) Address of a memory variable’
iii) A set of instructions to find out the factorial of a number
iv) An alternate name of a previously defined variable
v) Price of 100 products in a consumer store
vi) Name of a student
Answer:
i) Integer array of size 50
ii) Pointer variable
iii) Function
iv) Reference
v) Float array of size 100
vi) Character array

Question 9.
Considering the following C++ statements.
Fill up the blanks
i) If p=5 and q=3 then q%p is
ii) If E1 is true and E2 is False then E1 && E2 will be
iii) If k=8, ++k <= 8 will be = ______.
iv) If x=2 then (10* ++x) % 7 will be ______.
v) If t=8 and m=(n=3,t-n), the value of m will be ______.
vi) If i=12 the value i after execution of the expression i+=i– + –i will be _____.
Answer:
i) 3
ii) False
iii) False(++k makes k=9. So 9<=8 is false) ‘
iv) 2(++x becomes 3 ,so 10 * 3 =30%7 =2)
v) 5( here m=(n=3,8-3)=(n=3,5), so m=5, The maximum value will take)
vi) Herei=12
i + = i– + -i
here post decrement has more priority than pre decrement. .
So i– will be evaluated first. Here first uses the value then change so it uses the value 12 and i becomes 11
i + =12 + –i
now i =11.
Here the value of i will be changed and used so i~ becomes 10.
i — = 12 + 10
= 22
So i =22+10
i =32
So the result is 32.

Question 10.
The Maths teacher gives the following problem to Riya and Raju.
X=5 + 3*6. Riya got x=48 and Raju got x=23. Who is right and why it is happened ? Write down the operator precedence in detail?
Answer:
Here the answer is x=23. It is because of precedence of operators. The order of precedence of operators are given below.
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 5
Here multiplication has more priority than addition.

Question 11.
b) Explain the data types in C++
Answer:
Fundamental data types: It is also called built in data type. They are int, char, float, double and void
i) int data type: It is-used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory.i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 2 numbers. We can store a number in between -231 to + 2311.

ii) char data type: Any symbol from the key board,
eg. ‘A’,’?’, ‘9’ It consumes one byte(8 bits) of memory. If is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code,48.

iii) float data type: It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory. Eg. 67.89, 89.9 E-15.

iv) double data type: It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

v) void data type: void means nothing. It is used to represent a function returns nothing.
User defined Data types : C++ allows programmers to define their own data type. They are Structure(struct), enumeration (enum), union, class, etc.
Derived data types : The data types derived from fundamental data types are called Derived data types. They are Arrays, pointers, functions, etc

Question 12.
Predict the output of the following C++ statements:
int a = -5, b = 3, c = 4;
C+ = a+++ — b; .
cout<<a<<b<<c;
Answer:
a = -4, b = 2 and c = 1

Question 13.
Match the following
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 6
Answer:
1. (vi) *
2. (v) &&
3. (ii) >=
4. (iii) >>
5. (i) ++
6. (iv) ?:

Question 14.
Write any five unary operators of C++. Why are they called so?
Answer:
A unary operator is an operator that need only one operand to perform the operation. The five unary operators of C++ are given below.
Unary +, Unary -, ++, – – and ! (not)

Question 15.
Write C++ examples for the following:
a) Declaration statement
b) Assignment statement
c) Type casting
Answer:
a) int age;
b) age= 16;
c) avg=(float)a+b+c/3;

Plus One Computer Application Data Types and Operators 5 Marks Questions and Answers

Question 1.
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 7
Consider the above data, we know that there are different types of data are used in the computer. Explain different data types used in C++.
Answer:
i) int data type: It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory.i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 2311.

ii) char data type: Any symbol from the key board,
eg; ’A’.,’?’, ‘9’,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

iii) float data type: It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bit?) of memory.
Eg. 67.89,89.9 E-15.

iv) double data type: It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

v) void data type : void means nothing. It is used to represent a function returns nothing.

Question 2.
Define an operator and explain operator in detail.
Answer:
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators
1) lnput(>>) and output(<<) operators are used to perform input and output operation. Eg. cin>>n;
cout<<n;
2) Arithmetic operators: It is a binary operator. It is used to perform addition(+), subtractionf-), division (/), multiplication (*) and modulus (%- gives the remainder) operations.
Eg. If x=10 and y=3 then
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 8

x/y = 3, because both operands are integer, “to get the floating point result one of the operand must be float.
3) Relational operator: It is also a binary opera- tor. It is used to perform comparison or relational operation between two values and it gives either true(1) px false(O). The operators are <,<=,>,>=,== (equallty)and !=(not equal to)
Eg. If x-10 and y=3 then
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 9
4) Logical operators : Here AND(&&), OR(||) are binary operators and NOT(!) is a unary operator. It is used to combine relational operation? and it gives either true(1) or false(O).
If x=True and y=False then
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 10
Both operands must be true to get a true value in the case of AND(&&) operation If x=True and y=False then
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 11
Either one of the operands must be true to get a true value in the case of OR(||) operation If x=True and y=False then
Plus One Computer Application Chapter Wise Questions Chapter 5 Data Types and Operators 12
5) Conditional operator: It is a ternary operator hence it needs three operands. The operator is?:. Syntax: expression ? value if true : value if false. First evaluates the expression if it is true the second part wilt be executed otherwise the third part will be executed.
Eg. If x=10 and y=3 then x>y ? cout< Here the output is 10.

6) sizeof( ): This operator is used to find the size used by each data type.
Eg. sizeof(int) gives 2.

7) Increment and decrement operator:  These are unary operators.

a) Increment operator (++): It is used to increment the value of a variable by one i.e., x++ is equivalent to x=x+1;
b) Decrement operator (–) : It is used to decrement the value of a variable by one i.e., x-is equivalent to x = x-1.

8) Assignment operator (=) : lt is used to assign the value of a right side to the left side variable.eg. x=5; Here the value 5 is assigned, to the variable x.

Plus One Computer Application Chapter Wise Questions Chapter 4 Getting Started with C++

Students can Download Chapter 4 Getting Started with C++ Questions and Answers, Plus One Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Application Chapter Wise Questions Chapter 4 Getting Started with C++

Plus One Computer Application Getting Started with C++ 1 Mark Questions and Answers

Question 1.
IDE means _______.
Answer:
Integrated Development Environment

Question 2.
We know that C++ is a high level language. From the following which statement is true.
a) C++ contains English like statements.
b) C++ contains mnemonics
c) C++ contains only 0 and 1
d) None of these
Answer:
a) C++ contains English like statements.

Question 3.
C++ is a language ______.
a) High level
b) Low level
c) Middle level
d) None of these
Answer:
a) High level

Question 4.
C++ was developed at ______.
a) AT & T Bell Laboratory
b) Sanjose Laboratory
c) Kansas University Lab
d) None of these
Answer:
a) At & T Bell Laboratory

Question 5.
C ++ is a successor of ______ language.
a) C#
b) C c
c) Java
d) None of these
Answer:
b) C c

Question 6.
The most adopted and popular approach to write programs is ______.
Answer:
Structured programming

Question 7.
From the following which uses OOP concept _____.
a) C
b) C++
c) Pascal
d) Fortran
Answer:
b) C++

Question 8.
______ is the smallest individual unit
Answer:
Token

Question 9.
Pick the odd one out:
a) float
b) void
c) break
d) Alvis
Answer:
d) Alvis, the others are keywords.

Question 10.
Reserved words for the compiler is _______.
a) Literals
b) Identifier
c) Key words
d) None of these
Answer:
c) Key words

Question 11.
Pick an identifier from the following
a) auto
b) age
c) float
d) double
Answer:
b) age

Question 12.
Pick the invalid identifier
a) name
b) Date of birth
c) age
d) joining time
Answer:
b) Date of birth, because it contains space.

Question 13.
Pick the octal integer from the following
a) 217
b) 0x217
c) 0217
d) None of these
Answer:
c) 0217, an octal integer precedes

Question 14.
Pick the hexa decimal integer from the following
a) 217
b) 0x217
c)0217
d) None of these
Answer:
b) 0x217, an hexa decimal integer precedes Ox

Question 15.
From the following pick a character constant
a)’A’
b)’ALL’
c)’AIM’
d) None of these
Answer:
a) ‘A’, a character enclosed between single quote

Question 16.
Non graphic symbol can be represented by using ______.
Answer:
Escape Sequence

Question 17.
Manish wants to write a program to produce a beep sound. Which escape sequence is used to get an alert (sound)
a) \a
b) \d
c) \s
d)None of these
Answer:
a)\a

Question 18.
Ajo wants to print a matter in a new line. Which escape sequence is used for this?
a) \a
b) \n
c) \s
d)None of these
Answer:
b)\n

Question 19.
To represent null character ____ is used.
a) \n
b) \0
c) \f
d) \s
Answer:
b)\0

Question 20.
State True/ False a string is automatically appended by a null-character.
Answer:
True

Question 21.
From the following pick a string constant .
a) ‘a’
b) “abc”
c) ‘abc’
d) None of these
Answer:
“abc”, a character constant must be enclosed between double quotes.

Question 22.
C++ was developed by ______.
a) Bjarne stroustrup
b) James Gosling
c) Pascal
d) None of these
Answer:
a) Bjarne stroustrup

Question 23.
From the following which is not a character constant.
а) ‘c’
b) ‘e’
c) ‘d’
d) “c”
Answer:
d) “c”, It is a string constant the others are character constant.

Question 24.
From the following which is a valid declaration.
а) int 91;
b) int x;
c) int 9x;
d) int “x”;
Answer:
b) intx

Question 25.
Symbols used to perform an operation is called _____.
a) Operand
b) Operator
c) Variable
d) None of these
Answer:
b) Operator

Question 26.
Consider the following
C = A + B. Here A and B are called ______.
a) Operand
b) Operator
c) Variable
d) None of these
Answer:
b) Operand

Question 27.
The execution of a program starts at ______ function.
Answer:
main( )

Question 28.
The execution of a program ends with ______ function.
Answer:
main( )

Question 29.
______ is used to write single line comment.
a) //
b) P
c) */
d) None of these
Answer:
a) //

Question 30.
const k=100 means
a) const float k=100
b) const double k=100
c) const int k=100
d)const char k=100
Answer:
c) const int k=100

Question 31.
Qn. 31 ,
Each and every statement in C++ must be end with ______.
a) Semi colon
b) Colon
c) full stop
d) None of these
Answer:
a) Semi colon

Question 32.
From the following select the input operator.
a)>>
b)<< c) >
d) <
Answer:
a)>>

Question 33.
From the following select the output operator.
a)>>
b)<<
c) >
d) <
Answer:
b)<<

Question 34.
From the following which is known as string terminator.
a) ‘\0’
b) ‘\a’
c) ‘\s’
d) ‘\t’
Answer:
a) ‘\0’

Question 35.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated. From the following which file is a must to run the program .
a) sum.exe
b) sum.obj
c) sum.vbp
d) sum.htm
Answer:
a) sum.exe

Question 36.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated namely sum.obj and sum.exe. From this which file is not needed to run the program.
Answer:
sum.obj is not needed and can be deleted.

Question 37.
From the following which is ignored by the compiler.
а) statement
b) comments
c) loops
d) None of these
Answer:
b) comments

Question 38.
To write a C++ program, from the following which statement is a must _____.
a) sum( )
b) main( )
c) #include
d) #include
Answer:
b) main( ).
A C++ program must contains at least one main( ) function.

Question 39.
State True / False .
Comment statements are ignored by the compiler.
Answer:
True .

Question 40
More-than one input / output operator in a single statement is called ________.
Answer:
Cascading of I/O operator

Question 41.
Is 0x85B a valid integer constant in C++? If yes why ?
Answer:
Yes. It is a hexa decimal number

Plus One Computer Application Getting Started with C++ 2 Marks Questions and Answers

Question 1.
Mr. Dixon declared a variable as follows int 9age. Is it a valid identifier. If not briefly explain the rules for naming an identifier.
Answer:
It is not a valid identifier because it violates the rule
1. The rules for naming an identifier is as follows.
1) It must be start with a letter(alphabet)
2) Under score can be considered as a letter
3) White spaces and-special characters cannot be used.
4) Key words cannot be considered as an identi-fier

Question 2.
How many bytes used to store ‘\a’.
Answer:
To store ‘\a’ one byte is used because it is an escape sequence. An escape sequence is treated as one character. To store one character one byte is used.

Question 3.
How many bytes used to store “\abc”.
Answer:
A string is automatically appended by a null character.
Here one byte for \a (escape sequence).
One byte for character b.
One byte for character c.
And one byte for null character.
So a total of 4 bytes needed to store this string.

Question 4.
How many bytes used to store “abc”.
Answer:
A string is automatically appended by a null character.
Here one byte for a.
One byte for character b.
One byte for character c.
And one byte for null character.
So a total of 4 bytes needed to store this string.

Question 5.
Consider the following code
{
cout<<“welcome to C++”;
}
After you compile this program there is an error called prototype error. Why it is happened? Explain
Answer:
Here we used the output operator cout<<. It is used to display a message “welcome to C++” to use this operator the corresponding header file must be included. We didn’t included the header file hence the error. .

Question 6.
In C++ the size of the string “book” is 5 and that of “book\n” is 6. Check the validity of the above statement. Justify your answer.
Answer:
A string is automatically added by a null character). The null character is treated as one character. So the size of string “book” is 5. Similarly a null character (\0) is also added to “book\n”. \n and \0 is treated as single characters. Hence the size of the string “book\n” is 6.

Question 7.
Pick the odd man out. Justify
TOTSAL, TOT_SAL, totsal5, Tot5_sal, SALTOT, tot.sal,
Answer:
tot.sal. Because it contains a special character dot(,). An identifier cannot contain a special character. So it is not an identifier. The remaining satisfies the rules of naming identifier. So they are valid identifier.

Question 8.
Write a C++ statement to print the following sentence. Justify.
“\ is a special character”
Answer:
cout<<“\\ is a special character” \\ is treated as an escape sequence.

Question 9.
A student type a C++ program and saves it in his personal folder as Sample.cpp. After getting the output of the program, he checks the folder and finds three files namely Sample.cpp, Sample.obj and Sample.exe. Write the reasons for the generation of the two files in the folder.
Answer:
After the compilation of the program sample.cpp, the operating system creates two files if there is no error. The files are one object file (sample.obj) and one executable file(sample.exe). Now the source file(sample.cpp) and object file(sample.obj) are hot needed and can be deleted. To run the program sample.exe is only needed.

Question 10. Mention the purpose of tokens in C++. Write names of any four tokens in C++.
Answer:
Token : It is the smallest individual units similar to a word in English orMalayalam language. C++ has 5 tokens.
1) Keywords
2) Identifier
3) Literals (Constants)
4) Punctuators
5) Operators

Question 11. The following are some invalid identifiers. Specify its reason.
a) Sum of digits
b) 1 year
c) First jan
d) For
Answer:
a) Sum of digits —> space not allowed hence it is invalid
b) 1 year —> First character must be an alphabet hence it is invalid
c) First.jan —> special characters such as dot (.) not allowed hence it is invalid.
d) For —> It is valid. That is it is not the key word for

Question 12.
Some of the literals in C++ are given below. How do they differ? (5, ‘5’, 5.0, “5”)
Answer:
5 – integer literal
‘5’ – Character literal
5.0- floating point literal
“5”- string literal

Question 13.
Identify the invalid literals from the following and write reason for each:
a) 2E3.5
b) “9”
c) ‘hello’
d) 55450
Answer:
a) 2 C 3.5 – The mantissa part (3.5) will not be a floating point number. Hence it is invalid
c) ‘hello’ -> It is a string hence it must be enclosed in double quotes instead of single quotes. It is invalid.

Question 14.
Which one of the following is a user-defined name?
a) Key-word
b) Identifier
c) Escape sequences
d) All of these
Answer:
b) Identifier

Question 15.
Identify whether the following are valid identifiers or not? If not give the reason.
a) Break
b) Simple.interest
Answer:
a) Break – It is valid( break is the keyword, not Break);
b) Simple.interest – It is not valid, because dot(.) is used.

Question 16.
Identify the invalid literals from the following and write a reason for each:
a) 2E3.5
b) “9”
c) ‘hello’
d) 55450
Answer:
a) Invalid, because exponent part should not be a floating point number
b) valid

Plus One Computer Application Getting Started with C++ 3 Marks Questions and Answers

Question 1.
Rose wants to print as follows
\n is used for New Line
Write down the C++ statement for the same.
Answer:
#include
using namespace std;
int main( )
{
cout<<“\\n is used for New Line”;
}

Question 2.
Alvis wants to give some space using escape sequence as follows
Welcome to C++
Write down the C++ statement for the same .
Answer:
#include
using namespace std;
int main( )
{
cout<<“Welcome to \t C++”;
}

Question 3.
We know that the value of pi=3.14157, a constant (literal). What is a constant? Explain it?
Answer:
A constant or a literal is a data item its value doe not change during execution.
1) Integer literals Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
Eg. For decimal 100,150,etc
For octal 0100,(5240, etc
For hexadecimal 0x100, 0x1A,etc
2) Float literals A number with fractional parts and its value does not change during execution is called floating point literals.
Eg. 3,14157,79.78,etc
3) Character literal: A valid C++ character enclosed in single

Question 4.
Write a program to print the message “TOBACCO CAUSES CANCER” on screen,
Answer:
#include using namespace std; int main( )
{
cout<<” TOBACCO CAUSES CANCER”;
}

Question 5.
You are supplied with a list of tokens in C++ pro-gram, Classify and Categorise them under proper headings.
Explain each category with its features. tot_mark, age, M5,. …, break,( ), int, _pay, ; , cin
Answer:
Plus One Computer Application Chapter Wise Questions Chapter 4 Getting Started with C++ 1

Question 6.
Write a program to print the message “SMOKING IS INJURIOUS TO HEALTH” on screen. “SMOKING IS INJURIOUS TO HEALTH”
Answer:
#include
using namespace std;
int main( )
{
cout<<“SMOKING IS INJURIOUS TO HEALTH”;
}

Plus One Computer Application Getting Started with C++ 5 Marks Questions and Answers

Question 1.
Consider the following code
The new line character is \n.
The output of the following code does not contain the \n. Why it is happened? Explain.
Answer:
\n is a character constant and it is also known as escape sequence. This is used to represent the non graphic symbols such as carriage return key(enter key), tab key, back space, space bar etc. It consists of a back slash symbol and one more characters.
Plus One Computer Application Chapter Wise Questions Chapter 4 Getting Started with C++ 2

Question 2.
You are about to study the fundamentals of C++ programming Language. Do a comparative study of the basics of the new language with that of a formal language like English or Malayalam to familiarize C++?. Provide sufficient explanations for the compared items in C++ Language,
Answer:
Character set: To study a language first we have to familiarize the character set. For example to study English language first we have to study the alphabets. Similarly here the character set includes letters(A to Z & a to z), digits(0 to 9), special characters+,-,?,*,/, ) white spaces(non printable)
etc..
Token: It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens
1) Keywords
Eg: float is used to declare variable to store numbers with decimal point. We can’t use this for any other purpose

2) identifier: These are user defined words. Eg: variable name, function name, class name, object name etc…

3) Literals (Constants): Its value does not change during execution
Eg: In maths π = 3.14157 and boiling point of water is 100.

4) Punctuators: In English or Malayalam language punctuation mark are used to increase the read-ability but here it is used to separate the tokens. Eg:{,},(,)……….

5) Operators: These are symbols used to perform an operation(Arithmetic, relational, logical, etc…)
These are reserved words for the compiler. We can’t use for any other purposes.