Monday, 4 December 2023

EVS PROJECT ON BIO DIVERSITY LOSS

 

 

EVS PROJECT ON BIO DIVERSITY LOSS

INTRODUCTION:

Biodiversity loss includes the worldwide extinction of different species, as well as the local reduction or loss of species in a certain habitat, resulting in a loss of biological diversity. The latter phenomenon can be temporary or permanent, depending on whether the environmental degradation that leads to the loss is reversible through ecological restoration/ecological resilience or effectively permanent (e.g. through land loss). The current global extinction (frequently called the sixth mass extinction or Anthropocene extinction), has resulted in a biodiversity crisis being driven by human activities which push beyond the planetary boundaries and so far has proven irreversible.

Even though permanent global species loss is a more dramatic and tragic phenomenon than regional changes in species composition, even minor changes from a healthy stable state can have dramatic influence on the food web and the food chain insofar as reductions in only one species can adversely affect the entire chain (extinction), leading to an overall reduction in biodiversity, possible alternative stable states of an ecosystem notwithstanding. Ecological effects of biodiversity are usually counteracted by its loss. Reduced biodiversity in particular leads to reduced ecosystem services and eventually poses an immediate danger for food security, but also can have more lasting public health consequences for humans International environmental organizations have been campaigning to prevent biodiversity loss for decades, public health officials have integrated it into the One Health approach to public health practice, and increasingly preservation of biodiversity is part of international policy. For example, the UN Convention on Biological Diversity is focused on preventing biodiversity loss and proactive conservation of wild areas.

 

IMPORTANCE:

Biodiversity is linked to cleaner water

Healthy soil is diverse

Biodiversity is vital for healthy plants and animals

Biodiversity means better food security

Biodiversity protects medicinal plants

Less biodiversity means more zoonotic disease

Biodiversity has economic benefits

Biodiversity creates jobs

Biodiversity and extinction are linked

Biodiversity loss can have significant direct human health impacts if ecosystem services are no longer adequate to meet social needs. Indirectly, changes in ecosystem services affect livelihoods, income, local migration and, on occasion, may even cause or exacerbate political conflict additionally, biological diversity of microorganisms, flora and fauna provides extensive benefits for biological, health, and pharmacological sciences. Significant medical and pharmacological discoveries are made through greater understanding of the earth's biodiversity.

Loss in biodiversity may limit discovery of potential treatments for many diseases and health problems. Biodiversity plays a crucial role in human nutrition through its influence on world food production, as it ensures the sustainable productivity of soils and provides the genetic resources for all crops, livestock, and marine species harvested for food. Access to a sufficiency of a nutritious variety of food is a fundamental determinant of health.

 

OBJECTIVE:

1) Conservation of biological diversity leads to conservation of essential ecological diversity to preserve the continuity of food chains.

2) The genetic diversity of plants and animals is preserved.

3) It ensures the sustainable utilization of life support systems on earth.

4) It provides a vast knowledge of potential use to the scientific community.

5) A reservoir of wild animals and plants is preserved, thus enabling them to be introduced, if need be, in the surrounding areas.

6) Biological diversity provides immediate benefits to the society such as recreation and tourism.

7) Biodiversity conservation serves as an insurance policy for the future.

8) It develop national strategies for the conservation and sustainable use of biological resources; establish protected areas, restore degraded ecosystems, control alien species, and establish ex-situ conservation facilities; and sustainable use of biodiversity.

 

METHODOLOGY:

Threats to biodiversity and health.

There is growing concern about the health consequences of biodiversity loss. Biodiversity changes affect ecosystem functioning and significant disruptions of ecosystems can result in life sustaining ecosystem goods and services. Biodiversity loss also means that we are losing, before discovery, many of nature's chemicals and genes, of the kind that have already provided humankind with enormous health benefits.

Nutritional impact of biodiversity.

Biodiversity plays a crucial role in human nutrition through its influence on world food production, as it ensures the sustainable productivity of soils and provides the genetic resources for all crops, livestock, and marine species harvested for food. Access to a sufficiency of a nutritious variety of food is a fundamental determinant of health.

 

Nutrition and biodiversity are linked at many levels: the ecosystem, with food production as an ecosystem service; the species in the ecosystem and the genetic diversity within species. Nutritional composition between foods and among varieties/cultivars/breeds of the same food can differ dramatically, affecting micronutrient availability in the diet. Healthy local diets, with adequate average levels of nutrients intake, necessitates maintenance of high biodiversity levels. 

 

The functions of ecosystems and altering native biodiversity. Such disturbances reduce the abundance of some organisms, cause population growth in others, modify the interactions among organisms, and alter the interactions between organisms and their physical and chemical environments. Patterns of infectious diseases are sensitive to these disturbances. Major processes affecting infectious disease reservoirs and transmission include, deforestation; land-use change; water management e.g. through dam construction, irrigation, uncontrolled urbanization or urban sprawl; resistance to pesticide chemicals used to control certain disease vectors; climate variability and change; migration and international travel and trade; and the accidental or intentional human introduction of pathogens.

 

Climate change, biodiversity and health

Biodiversity provides numerous ecosystem services that are crucial to human well-being at present and in the future.

Climate is an integral part of ecosystem functioning and human health is impacted directly and indirectly by results of climatic conditions upon terrestrial and marine ecosystems.

Marine biodiversity is affected by ocean acidification related to levels of carbon in the atmosphere. Terrestrial biodiversity is influenced by climate variability, such as extreme weather events (i.e. drought, flooding) that directly influence ecosystem health and the productivity and availability of ecosystem goods and services for human use. Longer term changes in climate affect the viability and health of ecosystems, influencing shifts in the distribution of plants, pathogens, animals, and even human settlements.

OBSERVATION:

The biodiversity losses from disturbances caused by humans tend to be more severe and longer-lasting. Humans (Homo sapiens), their crops, and their food animals take up an increasing share of Earth's land area. Half of the world's habitable land (some 51 million square km [19.7 million square miles]) has been converted to agriculture, and some 77 percent of agricultural land (some 40 million square km [15.4 million square miles]) is used for grazing by cattle, sheep, goats, and other livestock. This massive conversion of forests, wetlands, grasslands, and other terrestrial ecosystems has produced a 60 percent decline (on average) in the number of vertebrates worldwide since 1970, with the greatest losses in vertebrate populations occurring in freshwater habitats (83 percent) and in South and Central America (89 percent). Between 1970 and 2014 the human population grew from about 3.7 billion to 7.3 billion people. By 2018 the biomass of humans and their livestock (0.16 gig ton) greatly outweighed the biomass of wild mammals (0.007 gig ton) and wild birds (0.002 gig ton). Researchers estimate that the current rate of species loss varies between 100 and 10,000 times the background extinction rate (which is roughly one to five species per year when the entire fossil record is considered). In addition, a 2019 report by the Intergovernmental Science-Policy Platform on Biodiversity and Ecosystem Services noted that up to one million plant and animal species are facing extinction due to human activities.

 

ANALYSIS: 

According to the Living Planet Index (LPI) 2020, decline was captured for the 68% of the monitored population (i.e. 20,811 populations of 4,392 species) for the period of 1970-2016.

The LPI "is a measure of the state of the world's biological diversity based on population trends of vertebrate species from terrestrial, freshwater and marine habitats adopted by the Convention of Biological Diversity (CBD) to track the progress" Only one-third of the overall decline (32%) relates to the period since 2000 where the most decline was for the freshwater species (44%) followed by terrestrial species populations (39%) and marine species (8%) (WWF, 2020) (Secretariat of the Convention on Biological Diversity, 2020). However, it is important to keep it in mind that most species are understudied and not discovered yet which also leave room for the larger extinction statistics beyond the existing calculations.

CONCLUSION:

In practice, decisions about natural resources are seldom made due to a single factor, but are based on the cumulative weight of many different considerations.

 The value of wildlife to human subsistence, its economic worth and the ecosystem benefits it provides are all important. Additional critical factors may also be far less tangible, such as the link between a particular species and a faith group, or the emotions it creates in some people. Recognizing the value of wildlife is a vital step in building the motivation needed for effective management and conservation.

 Protected areas can never protect all wildlife, and, in some situations, well- managed lands can be more effective vehicles for wildlife conservation than badly managed or under-resourced protected areas. Forestry managers have a critical role to play by

 

 1) Ensuring that forest management is compatible with the survival of native wildlife;

 

 2) Protecting unmanaged edges and patches within the forest estate;

 

 3) Protecting watercourses;

 

 4) Controlling poaching and the bush meat trade;

 

 5) Halting incursions by invasive species

 

One thing is clear: biodiversity loss cannot be halted if we fail to stabilize the climate, and if we are to stabilize the climate as well as to move into the era of low-carbon living, we must protect the biosphere - the very life-support system of our planet.

 

 

Wednesday, 25 October 2023

HSC PHYSICS CHP 11 MAGNETIC METERIALS

HSC PHYSICS CHP 11 MAGNETIC METERIALS 

MOST  IMPORTANT QUESTIONS 2023/24

1. DEFINE MAGNETIZATION .STATE ITS FORMULA ,SI UNIT & DIMENSIONS

2. WHAT IS MAGNETIZATION

3. DEFINE MAGNETIC INTENSITY AND MAGNETIC SUSCEPTIBLITY 

4. SHOW THAT ORIBITAL MAGNETIC MOMENT OF A REVOLVING ELECTRON IS EVR/2

OR 

5. DERIVE AN EXPRESSION FOR MAGNETIC DIPOLE MOMENT OF REVOLVING ELECTRON

OR 

6. IN AN HYDROGEN ATOM OBTAIN AN EXPRESSION FOR MAGNETIC DIPOLE MOMENT FOR REVOLVING ELECTRON

7. EXPLAIN FERROMAGNETISM ON THE BASIS OF DOMAIN THEORY

8. DISTINGUISH BETWEEN DIAMAGNETIC, FERROMAGNETIC AND PARAMAGNETIC SUBSTANCES.

9. DRAW DIAGRAM SHOWING DIPOLE MOMENTS IN DIMAGNETIC,PARAMAGNETIC AND FERROMAGNETIC SUBSTANCE WHEN EXTERNAL MAGNETIC FIELD IS ABSENT & WHEN IT IS STRONG

10. WHAT IS CURIE TEMPERATURE AND WHAT HAPPENS ABOVE CURIE TEMPERATURE

11.WHAT HAPPENS WHEN A ROD OF DIMAGNETIC METERIAL IS PLACED IN NON UNIFORM MAGNETIC FIELD 

 

HSC PHYSICS THERMODYNAMICS

HSC PHYSICS THERMODYNAMICS 

MOST IMPORTANT QUESTIONS 2023/24

1. DESCRIBE IN DETAIL REVERSIBLE & IRREVERSIBLE PROCESS

2. DESCRIBE IN DETAIL ISOTHERMAL PROCESS

3. DESCRIBE IN DETAIL ISOBARIC PROCESS 

4. DESCRIBE IN DETAIL ISOCHORIC PROCESS 

5. DESCRIBE IN DETAIL ADIABATIC PROCESS

6. DESCRIBE IN DETAIL CYCLIC PROCESS

7 . WHAT IS FREE EXPANSION

8. DESCRIBE THE HEAT ENGINE CYCLE AND PV DIAGRAM

9. DESCRIBE PV DIAGRAM

10. EXPLAIN ZEROTH LAW OF DYNAMICS 

11. DESCRIBE HEAT ,INTERNAL ENERGY AND WORK 

12. EXPLAIN FIRST LAW OF THERMODYNAMICS 

13. WHAT IS AIR CONDITIONER AND HEAT PUMP 

14. LIMITATIONS OF FIRST LAW OF THERMODYNAMICS

15. EXPLAIN SECOND LAW OF THERMODYNAMICS 

16. SIGNIFICANCE OF REVERSIBILITY IN THERMODYNAMICS

17. EXPLAIN CARNOT CYCLE & CARNOT REFRIGIRATOR 

18. DESCRIBE STERLING CYCLE 

19. WHY SHOULD CARNOT CYCLE HAVE 2 ISOTHERMAL AND 2 ADIABATIC PROCESS

20. WHAT SETS THE LIMITS ON EFFICIENCY OF HEAT ENGINE 

Monday, 7 August 2023

HSC PHYSICS 2024/25 PRACTICAL PAPER PATTERN

 

H.S.C PHYSICS 2024/25 PRACTICAL PAPER PATTERN

TOTAL 30 MARKS

JOURNAL SUBMISSION (5 MARKS)

ORAL/VIVA (5 MARKS)

LONG EXPERIMENT (10 MARKS)

1. DIAGRAM (1 MARKS)

2. SETTING APPARATUS (1 MARKS)

3. FORMULA (1 MARKS)

4. READING /OBSERVATION (3 MARKS)

5. GRAPH/CALCULATION (2 MARKS)

6. RESULT/CONCLUSION (2 MARKS)

SHORT EXPERIMENT (5 MARKS)

1. DIAGRAM (1 MARKS)

2. OBSERVATION (2 MARKS)

3. GRAPH / CALCULATION (1 MARKS)

4. RESULT (1 MARKS)

ACTIVITY (5 MARKS)

1. PERFORM ACTIVITY (3 MARKS)

2. WRITE UP OF ACTIVITY (1 MARKS)

3. RESULT (1 MARKS)

Monday, 31 July 2023

S.S.C 2024 most important questions science 1 c.h.p 1 gravitation

 S.S.C 2024 most important questions 

science 1 c.h.p 1 gravitation

M.C.Q

The minimum velocity of a spacecraft to escape from gravitational force is ………

The value of gravitational acceleration (g) is ………..

The ……………..force is much weaker than other forces in nature.

The value of universal gravitational constant (G) in SI unit is…………

The gravitational force between two bodies is directly proportional to the product of the masses of those bodies and is ……….of the distance between them.

The square of its period of revolution around the sun is directly proportional to the ……….. Of the mean distance of a planet from the sun.

The orbit of a planet revolving around a star is ……

A ……. Is necessary to change the speed as well as the direction of motion of an object.

The value of gravitational acceleration (g) is …………at the equator

The free fall of an object is possible only in…….

The weight of any object on the moon is nearly …………of the weight of the earth

A person weighs 60 N on earth. His weight on the moon will be……

Newton presented the laws of motion, equations of motion and theory of gravitation in his book…….

Laws of planetary motion were discovered by ……………………….

True or false

If the distance between two masses is doubled, the gravitational force between them becomes less than the previous force.

The CGS unit of G is dyne.cm

The value of gravitational acceleration with the centre of the earth is zero.

The value of g is highest at the equator.

The value of G varies from place to place.

As it rises above the earth's surface, its value increases.

The speed of release of an object does not depend on the mass of the object.

Mass is a qualitative measure of the inertia of an object.

Distinguish between

Mass and weight

Give scientific reasons

The weight of the object changes from place to place while the mass is constant

Answer the following

Explain why value of g changes if we go inside the earth

If mass of 2 bodies are m1 and m2 respectively and if distance between them is 3d then force F =

State the 3 planetary laws of KEPLER or of the given figure showing orbit of a planet write the 3 laws related to it

Will the weight of an object on earth will be same as that on mars? Why

What is Free Fall, acceleration due to gravity, escape velocity and centripetal force?

A stone thrown vertically upwards with an initial velocity u reaches a height h before coming down show that the time taken to go up is same as time taken to come down.

Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8T.

Tuesday, 4 July 2023

EVS Project work on conducting energy audit for energy conservation

 

E.V.S Project work on conducting energy audit for energy conservation


Introduction:

An energy audit is an analysis of a facility, indicating how and where

that facility can reduce energy consumption and save energy costs. Its

insight to energy efficiency and conservation can lead to significant

savings on the utility bill of houses, institutions, companies etc. and

suggest the energy conservation techniques. Hence to put more light in

this conservation technique I have selected this topic.

Energy audit refers to examination and verification of energy

consumption in the form of electricity, gas and other forms of fuel energy

used in households, industries, institutes, schools etc. It can be considered

as the first step towards knowing how energy is being used in a given

facility. It indicates the ways in which various forms of energy are being

Used and quantifies energy use according to different functions.

Energy Audit is an integral part of Green auditing. Green Audit is an

unique process of systematic identification, quantification, recording,

reporting and analysis of components of environmental diversity of

various establishments. It allows us to know the uses of available

resources i.e. Energy, Water quality, Built-up Space, Air quality in the

organization. It was initiated in 1970 with the motive of inspecting the

work conducted within the organizations, which can cause risk to

inhabitants and environment.Rapid urbanization and economic development at local, regional and global level has led to several environmental and ecological crisis.

Hence It becomes essential to adopt the system of environmental audit for institutes, organizations etc. It will lead to sustainable development.In the given research I have made an attempt to conduct the energy audit of my house.

Importance of the Project:

Energy conservation means using energy more efficiently and less wastefully. Conservation of energy is an important energy resource because a Unit of energy saved is as good as a unit of energy generated. It helps you to lower energy bills. It helps you to increase the life span of the equipment. Helps in understanding energy consumption pattern of the Institute and house. By identifying energy wastage areas, saving energy and using renewable energy sources, the institute can become more energy Efficient. Thus Energy audit is a small step towards preventing global warming as well as saving the valuable conventional energy resource

Objectives:

To create awareness towards wastage of electrical energy.

To make statistical analysis of electrical energy usage in order to reduce unnecessary consumption.

To sensitize the community about lawful consumption of electrical energy in their houses and work places.

To study about alternatives to cut down the cost as well as reduce the energy loss.

To find the difference in the electricity bills in order to analyses the impact of alternatives used in conserving electricity.

To learn electric energy consumption pattern and its calculation. 

Methodology

I am doing my research project by applying the survey method in which I have used Interview method to do the analysis of my research have selected my own house to conduct the energy audit have compared the electricity bills of 2 months (January & March respectively) in order to find the difference in the electricity consumption. This difference has helped me to find the reduction in electricity consumption due to the use of alternatives. In interview method I have taken interview of two experts, an architecture and an electrical engineer to put light on my research and to gain in-depth knowledge about electricity consumption its conservation. During interview session I have asked 5 questions each to the experts

Name of the Expert & List of questions asked.

Mr. Rajendra Baburao (Electrical Engineer-Reliance)

1. What is the effect of Wattage difference of same appliances on electricity consumption?

2. Can replacement of a part of electrical appliances help to reduce electricity consumption?

3. Is there any user guide available by which we can use electrical energy more wisely?

4. Is there any specific electrical wires available in the market which helps in reducing the electricity consumption?

5. How to calculate electrical energy consumption of appliances?

Observations:




Analysis:



Calculation of electricity consumption was done in following way:

DATA: 1kWh=1unit

Where, W stands for Wattage which means operating power of electrical

Appliances.

Convert Wattage into Watt-hour = Watt/60 mins

Convert Watt-hour into KWh = Watthr/1000

FOR EXAMPLE: CALCULATION OF ELECTRICAL CONSUMPTION BY TUBE LIGHTS.

Wattage of tube light = 40 W

No. of tube lights = 6 K + B + H

Hrs. of Consumption/day = 8 (On an average) {(4+4+8+8+12+12)/6 = 8}

WattHr./day = 40 X 6 X 8 = 1920

KiloWatt Hr. = 1920/1000 = 1.920

Kilowatt Hr. / month = 1.920 X 31 = 59.52 units

Resultant and Conclusion

After understanding the application of Energy audit in energy Conservation, I have taken an initiative to conduct an energy Audit to calculate and compare the electricity usage at my own House for the month of January and March have reduced the usage of appliances with respect to numbers And hours of operation. Based on the results obtained it has been noticed that without Taking much effort I could save 200 units of electricity. I have consciously made an effort for reducing the hours of Operations and number of appliances like Geyser, Fan, and tube light etc. On regular basis. Along with the conservation of energy this initiative has also given me the financial benefit where I could save around 600 rupees.

Monday, 3 July 2023

EVS Projects List

E.V.S  Projects List ( All Projects )


Click on below links to watch & download 馃憞

Visited historic place & environmental issues over there 馃憟( click here )

Project work on global warming 馃憟( click here )

Project work on organic farming 馃憟( click here )

Project work on solar energy  馃憟( click here )

Project work on water pollution 馃憟( click here ) 

Project work on biodiversity 馃憟(click here)

Project work on conservation of natural resources 馃憟(click here)

Project work on airpollution 馃憟(click here)



Wednesday, 28 June 2023

Airpollution

 

Project work on Air Pollution

Introduction:

Air pollution is contamination of the indoor or outdoor environment by any chemical, physical or biological agent that modifies the natural characteristics of the atmosphere. Household combustion devices, motor vehicles, industrial facilities and forest fires are common sources of air pollution.

Pollutants of major public health concern include particulate matter, carbon monoxide, ozone, nitrogen dioxide and sulphur dioxide. Outdoor and indoor air pollution cause respiratory and other diseases and are important sources of morbidity and mortality. WHO data show that almost all of the global population (99%) breathe air that exceeds WHO guideline limits and contains high levels of pollutants, with low- and middle-income countries suffering from the highest exposures.

 

Air quality is closely linked to the earth’s climate and ecosystems globally. Many of the drivers of air pollution (i.e. combustion of fossil fuels) are also sources of greenhouse gas emissions. Policies to reduce air pollution, therefore, offer a win-win strategy for both climate and health, lowering the burden of disease attributable to air pollution, as well as contributing to the near- and long-term mitigation of climate change.

Importance of the Project:

This project on air pollution is important as

Air pollution in countries like India is a serious environmental issue. Of the 30 most polluted cities in the world, 21 were in India in 2019.As per a study based on 2016 data, at least 140 million people in India breathe air that is 10 times or more over the WHO safe limit and 13 of the world's 20 cities with the highest annual levels of air pollution are in India. 51% of the pollution is caused by industrial pollution, 27% by vehicles, 17% by crop burning and 5% by other sources. Air pollution contributes to the premature deaths of 2 million Indians every year. Emissions come from vehicles and industry, whereas in rural areas, much of the pollution stems from biomass burning for cooking and keeping warm. In autumn and spring months, large scale crop residue burning in agriculture fields – a cheaper alternative to mechanical tilling is a major source of smoke, smog and particulate pollution. India has a low per capita emissions of greenhouse gases but the country as a whole is the third largest greenhouse gas producer after China and the United States. A 2019 study on non-smokers has found that Indians have 30% weaker lung function than Europeans.

air pollution effect on climate as

Black carbon, a component of fine particulate matter, is one of the largest contributors to global warming after CO2. Black carbon warms the earth's atmosphere by absorbing sunlight, thereby accelerating the melting of snow and ice.

Objectives:

The main objective of this project is to cut down the air pollution

The overall aim of the proposed research is to further understand the link between air pollution levels and chemical composition and to investigate and assess the effects of air pollution on the distribution of related health impacts, socio-economics and welfare in the countries. In pursuing this aim, the following objectives will be investigated:

High spatial resolution emissions: Performing a detailed common country emission inventory Down to 1 km x 1 km resolution, based on emission databases in the individual countries. Official data available will be used as well as preparation and incorporation of additional spatial disaggregation proxies in order to achieve the high spatial resolution emission inventory needed for the air quality modelling. Different methodologies and proxies used in different countries will be analysed and preferable methodologies recommended. Special attention will be given to sources with large contribution to the total emissions in the affected countries, specifically residential wood combustion. The emission inventory will be conducted in WP1 and will be input to WP2.

Integrated modelling: Setting up of a state-of-the-art advanced and integrated air pollution model system from hemispheric scale, European scale, and national scale, for calculation and assessment of high resolution (down to 1 km x 1 km resolution) air pollution levels and human exposure, including assessing the contribution related to different emission sectors and regions. This work is carried out in WP2. The results will be hourly values of air pollutants, on a 1 km x 1 km resolution covering the affected countries over several decades (input to WP3).

Health effects of air pollution: Investigate the potential causal impact of individual chemical air pollutants as well as mixtures of air pollutants on health outcomes. In pursuing this aim, we utilize the unique population-based registers allowing linkage between historical residential address, air pollutants over decades and later health outcomes. By linking the exposure to health outcomes, new exposure-response relationships are determined of health effects for different population Groups (e.g. age, education, ethnicity, gender, lifestyle, and working life vs. retirement conditions) related to air pollution for the individual chemical air pollutants. The objective will be conducted in WP3.

Assessment and quantification of health impacts: An assessment and quantification of overall negative health outcomes of air pollution in terms of premature deaths, hospital admissions, days of reduced activity, respiratory diseases, mental disorders, etc. on high resolution down to 1 km x 1 km in the affected countries for the different population groups, using the integrated model system EVA, based on the impact pathway chain. The EVA system will be further developed utilizing the new exposure-response relationships found in WP3 and will be carried out in WP4.

Methodology

Assessing the effects of air-pollution is a significant problem in the field of modern environmental epidemiology. When modelling these effects it is important that the models must be epidemiologically meaningful and robust (that is, insensitive to variations in the model parameters). The objective of this paper is to propose a methodology for the assessment of the health impact of air pollution. The proposed methodology involves the construction of models for complex dynamic hierarchical systems in environmental epidemiology and their problem-oriented interpretation.

The principal stages of the proposed methodology are:

Creation of a multivariate hierarchical structural model based on system analysis.

Generation of a mathematical formalization for this model.

Development of a statistical model for a particular study case based on the mathematical formalization, using the generalized estimating equations technique and time-series analysis. At this stage, for a dichotomized dependent variable, a special fuzzy algorithm was used. The algorithm employed fuzzy membership functions instead of the binary variable to obtain robust regression models.

Use of the “multi-layered” approach for model interpretation developed by the authors. This approach involved the creation of special functional time-dependent coefficients that reflect the effect of air pollutants at a given time. These coefficients allow an epidemiological meaningful model interpretation. Thus, they can be used for air-pollution health effects assessment.

Observation

Western Europe, eastern North America, and East Asia, the three major industrial regions of the world, are major emission sources of air pollution (AP) and greenhouse gases (GHG). In East Asia, the fossil fuel consumption in 2005 for the three major countries of China, Japan, and Korea was 1,554M, 525M, and 225M tons, respectively (. These three alone use 21.8% of the world’s energy, adding AP and GHG to the atmosphere and causing environmental impacts, which include global warming and climate change .Strikingly, it has been reported that in heavily polluted China, the poor air and water quality affect about 1 million of the 20 million babies born each year, with about 300,000 babies suffering from visible deformities. According to a World Bank study, each year, about 460,000 people in China die prematurely because of air and water pollution.

Air pollution in India is a serious environmental issue. Of the 30 most polluted cities in the world, 21 were in India in 2019. As per a study based on 2016 data, at least 140 million people in India breathe air that is 10 times or more over the WHO safe limit and 13 of the world's 20 cities with the highest annual levels of air pollution are in India 51% of the pollution is caused by industrial pollution, 27% by vehicles, 17% by crop burning and 5% by other sources. Air pollution contributes to the premature deaths of 2 million Indians every year. Emissions come from vehicles and industry, whereas in rural areas, much of the pollution stems from biomass burning for cooking and keeping warm. In autumn and spring months, large scale crop residue burning in agriculture fields – a cheaper alternative to mechanical tilling – is a major source of smoke, smog and particulate pollution.

Analysis:

Here is a sample analysis of air pollution monitoring in Mumbai India

Mumbai (18.97N, 72.82E) with over 10 million population is one of the most populous city in India and ranks fifth among the most populous cities of the world. With considerable industrial activity and vehicular presence, air pollution is a key issue affecting the human health. Continuous air quality monitoring is being done to identify the patterns in key indicators like SOx, NOx and RSPM (respirable suspended particulate matter). We have analysed the pattern in the changes in these species using air quality data at Sion (19.04N, 72.86E) from 2004 to 2014 and Bandra (19.05N, 72.84E) from 2007 to 2014. . In time series analysis of this 10 years data at Sion we are able to get considerable decay in SOx from 30 to 10脗碌g/m3 and in RSPM from 250 to 150脗碌g/m3because industries are displaced from locality, not much change is seen in NOx from 110 to 100脗碌g/m3. Whereas in case of Bandra there is little increase in SOx concentration from 18 to 20脗碌g/m3while decay in concentration of NOx from 65 to 50脗碌g/m3and in RSPM from 140 to 110脗碌g/m3. The paper attempts to analyze the causes for these pattern changes in terms of industrial activity, changes in vehicular pollution norms and other anthropogenic factors.

Here is sample chart showing sources of air pollution



Resultant and Conclusion:

While the effects of air pollution on materials, vegetation, and animals can be measured, health effects on humans can only be estimated from epidemiological evidence. Most of the evidence comes from occupational exposure to much higher concentrations of pollutants than the general public is exposed to. Moreover, the health effects of smoking and other lifestyle characteristics and exposures confound the observations of air pollutant effects. Ethical considerations preclude deliberate exposure of human subjects to concentrations of pollutants that might produce adverse effects, so evidence from sources other than epidemiology is virtually impossible to obtain. All of the evidence we have suggests that air pollutants threaten human health and well-being to an extent that control of these pollutants is necessary



Wednesday, 21 June 2023

Free Maths Printable Worksheets

 Free Maths Printable Worksheets ( All Grades ) 

Click on below links to download 馃憞馃憞馃憞

Kinrtergarden (KG) click here 馃憟馃憟

Grade 1 ( click here )  馃憟馃憟

Grade 2 ( click here ) 馃憟馃憟

Grade 3 ( click here ) 馃憟馃憟

Grade 4 ( click here ) 馃憟馃憟

Grade 5 ( click here ) 馃憟馃憟

Grade 6 ( click here) 馃憟馃憟

Saturday, 6 May 2023

Read Log & Antilog Table

How To Read Log & Antilog Table 

Watch Click Below 馃憞馃憞


Watch Click Below 馃憞馃憞



Friday, 5 May 2023

12 Tenses in English

 12 Tenses in English 

Watch Click Below 馃憞


Learn all the tense Rule click here 馃憟馃憟馃憟

Wednesday, 26 April 2023

Fast multiplication

Fast multiplication of 2 digit numbers ending 1

Watch click below 

 



Try yourself : 
61*41 ,  81*11 , 51*41  

Tuesday, 14 March 2023

FYBSC IT SEM ( NUMERICAL METHODS QUESTION BANK SOLUTIONS )

 FYBSC IT SEM ( NUMERICAL METHODS QUESTION BANK SOLUTIONS ) 

UNIT 1

1.Let p = 0.54617 and q = 0.54601. Use three significant figures to approximate p − q and determine the absolute and relative errors using  (i) rounding and (ii) chopping 

Solution:

Using three significant figures, we have p = 0.546 and q = 0.546.

(i) Rounding:

p - q = 0.546 - 0.546 = 0.000

Absolute error = |0.000 - 0.00016| = 0.00016

Relative error = |0.00016/0.00016| = 1

(ii) Chopping:

p - q = 0.546 - 0.546 = 0.000

Absolute error = |0.000 - (-0.00017)| = 0.00017

Relative error = |0.00017/0.00017| = 1

Note that the true value of p - q is 0.00016.

2.Find the absolute error and relative error and also determine the number of significant digits in the approximations for the following.

True value =98350 and approximate value =98000

Find the absolute error and relative error and also determine the number of significant digits in the approximations for the following.

solution : 

True value =98350 and approximate value =98000

The absolute error is the difference between the true value and the approximate value: Absolute error = |true value - approximate value| = |98350 - 98000| = 350

The relative error is the absolute error divided by the true value: Relative error = Absolute error / True value = 350 / 98350 = 0.00356 (rounded to 5 decimal places)

To determine the number of significant digits in the approximation, we count the number of digits that are certain plus one uncertain digit. In this case, the approximation has two certain digits (98) and one uncertain digit (000). Therefore, there are three significant digits 

3. Create a hypothetical floating-point number set for a machine that stores information using 7-bit words.

solution : 

For a machine that uses 7-bit words, we could use the following format for a floating-point number:

1 bit for the sign (S)

3 bits for the exponent (E) 

3 bits for the mantissa (M)


4.Determine the absolute and relative errors when approximating p by p when i) p = 0.3000 × 101 and p = 0.3100 × 101 ii) p = 0.3000 × 10−3 and p = 0.3100 × 103

solution : 

To determine the absolute and relative errors when approximating p by p when p = 0.3000 × 10^1 and p = 0.3100 × 10^1, we use the following formulas:

Absolute error = |p - p|

Relative error = |(p - p)/p|

Substituting the values given in the problem, we get:

absolute error = |0.3000 × 10^1 - 0.3100 × 10^1| = |-0.0100 × 10^1| = 0.100

Relative error = |(0.3000 × 10^1 - 0.3100 × 10^1)/(0.3000 × 10^1)| = |-0.0100/0.3000| = 0.0333...

Therefore, the absolute error is 0.100 and the relative error is 0.0333... or approximately 3.33%.

ii) To determine the absolute and relative errors when approximating p by p when p = 0.3000 × 10^-3 and p = 0.3100 × 10^-3, we use the same formulas:

Absolute error = |p - p|

Relative error = |(p - p)/p|

Substituting the values given in the problem, we get:

Absolute error = |0.3000 × 10^-3 - 0.3100 × 10^-3| = |-0.0100 × 10^-3| = 0.000010

Relative error = |(0.3000 × 10^-3 - 0.3100 × 10^-3)/(0.3000 × 10^-3)| = |-0.0100/0.3000| = 0.0333...


UNIT 5

1. Fit a Poisson distribution to the following data and calculate theoretical frequencies.
No. of deaths 0 1 2 3 4  Frequency 122 60 15 2 1

Solution: 
To fit a Poisson distribution, we first need to calculate the mean of the data. The mean of a Poisson distribution is equal to its parameter 位. The formula for the mean of a Poisson distribution is: 位 = 危 (xi * fi) / 危fi
Where xi is the value of the the category (0, 1, 2, 3, 4) and fi is the frequency of that category.
Using the given data, we can calculate the mean as:
位 = (0 * 122 + 1 * 60 + 2 * 15 + 3 * 2 + 4 * 1) / (122 + 60 + 15 + 2 + 1) = 0.534
Now, we can use the Poisson probability mass function to calculate the theoretical frequencies for each category:
P(X = k) = (e^-位 * 位^k) / k!
Where X is the random variable representing the number of deaths, 位 is the mean or parameter of the Poisson distribution, and k is the number of deaths.
Using the mean 位 = 0.534, we can calculate the theoretical frequencies as follows:
P(X = 0) = (e^-0.534 * 0.534^0) / 0! = 0.588
P(X = 1) = (e^-0.534 * 0.534^1) / 1! = 0.313
P(X = 2) = (e^-0.534 * 0.534^2) / 2! = 0.084
P(X = 3) = (e^-0.534 * 0.534^3) / 3! = 0.014
P(X = 4) = (e^-0.534 * 0.534^4) / 4! = 0.002
To calculate the theoretical frequencies, we multiply each probability by the total number of observations
Theoretical frequency of 0 deaths = P(X = 0) * (122 + 60 + 15 + 2 + 1) = 93.97 ≈ 94
Theoretical frequency of 1 death = P(X = 1) * (122 + 60 + 15 + 2 + 1) = 50.32 ≈ 50
Theoretical frequency of 2 deaths = P(X = 2) * (122 + 60 + 15 + 2 + 1) = 13.59 ≈ 14
Theoretical frequency of 3 deaths = P(X = 3) * (122 + 60 + 15 + 2 + 1) = 2.26 ≈ 2
Theoretical frequency of 4 deaths = P(X = 4) * (122 + 60 + 15 + 2 + 1) = 0.31 ≈ 0
Therefore, the theoretical frequencies for a Poisson distribution with 位 = 0.534 are approximately 94, 50, 14, 2, and 0 for 0, 1, 2, 3, and 4 deaths, respectively

2.If 4 jobs arrive on an average per minute. What is the probability of waiting less than or equal to 30 seconds.
Solution:
To solve this problem, we need to first convert the arrival rate from per minute to per second. Since there are 60 seconds in a minute, the arrival rate per second would be:
位 = 4 / 60 = 0.067
Next, we need to use the Poisson distribution to calculate the probability of having 0, 1, 2, 3, or 4 arrivals in the next 30 seconds. The Poisson distribution formula is:
P(X = k) = (位^k / k!) * e^(-位)
where X is the random variable (number of arrivals), 位 is the arrival rate, k is the number of arrivals we are interested in, and e is the mathematical constant e (approximately equal to 2.71828).
Using this formula, we can calculate the probabilities as follows:
P(X = 0) = (0.067^0 / 0!) * e^(-0.067) ≈ 0.934
P(X = 1) = (0.067^1 / 1!) * e^(-0.067) ≈ 0.062
P(X = 2) = (0.067^2 / 2!) * e^(-0.067) ≈ 0.002
P(X = 3) = (0.067^3 / 3!) * e^(-0.067) ≈ 0.00005
P(X = 4) = (0.067^4 / 4!) * e^(-0.067) ≈ 0.000001
To find the probability of waiting less than or equal to 30 seconds, we need to add up the probabilities of having 0, 1, 2, 3, or 4 arrivals in the next 30 seconds:
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ≈ 0.999
Therefore, the probability of waiting less than or equal to 30 seconds is approximately 0.999 or 99.9%.

3.The monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is probability that there will be
i) More than 2 such accidents in the next month
ii) Less than 4 such accidents in the next months
Solution : 
Given that the monthly worldwide average number of airplane crashes of commercial airlines is 2.2, we can assume that the number of such accidents follows a Poisson distribution.
i) Probability of having more than 2 accidents in the next month:
Let X be the number of airplane crashes in the next month. We want to find P(X > 2).
Using the Poisson distribution, we have:
P(X > 2) = 1 - P(X ≤ 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
Where P(X = k) = (e^-位 * 位^k) / k!
位 = 2.2 (monthly worldwide average number of airplane crashes of commercial airlines)
Substituting the values, we get:
P(X > 2) = 1 - [((e^-2.2 * 2.2^0) / 0!) + ((e^-2.2 * 2.2^1) / 1!) + ((e^-2.2 * 2.2^2) / 2!)]
P(X > 2) = 1 - [(0.108 + 0.238 + 0.262)]
P(X > 2) = 0.392
Therefore, the probability of having more than 2 accidents in the next month is 0.392.
ii) Probability of having less than 4 accidents in the next month:
Let X be the number of airplane crashes in the next month. We want to find P(X < 4).
Using the Poisson distribution, we have:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Where P(X = k) = (e^-位 * 位^k) / k!
位 = 2.2 (monthly worldwide average number of airplane crashes of commercial airlines)
Substituting the values, we get:
P(X < 4) = ((e^-2.2 * 2.2^0) / 0!) + ((e^-2.2 * 2.2^1) / 1!) + ((e^-2.2 * 2.2^2) / 2!) + ((e^-2.2 * 2.2^3) / 3!)
P(X < 4) = 0.699
Therefore, the probability of having less than 4 accidents in the next month is 0.699.


4.In a certain examination mean of marks scored by 400 students is 45 and a standard deviation of 15.Assuming the distribution to be normal find
i) Number of students scoring marks between 30 and 60
ii) Number of students who scored marks more than 70
Solution : 
We are given that the distribution of marks is normal, with a mean of 45 and a standard deviation of 15.
Let's use the z-score formula to convert the given range of scores to z-scores:
z = (x - 渭) / 蟽
where x is the score, 渭 is the mean, and 蟽 is the standard deviation.
For the range of scores between 30 and 60:
z1 = (30 - 45) / 15 = -1
z2 = (60 - 45) / 15 = 1
Using a standard normal distribution table or calculator, we can find the proportion of the distribution that falls between z1 and z2:
P(-1 < z < 1) = 0.6827
Therefore, the number of students scoring marks between 30 and 60 is:
0.6827 * 400 ≈ 273
For the number of students who scored marks more than 70:
z = (70 - 45) / 15 = 1.67
Using the standard normal distribution table or calculator, we can find the proportion of the distribution that falls to the right of z:
P(z > 1.67) = 0.0475
Therefore, the number of students who scored marks more than 70 is:
0.0475 * 400 ≈ 19
Hence, the number of students scoring marks between 30 and 60 is approximately 273 and the number of students who scored marks more than 70 is approximately 19.



5.The length of life in years X of heavily used terminal in a student computer laboratory is exponentially distributed with mean =0.5 years. Write the PDF of the distribution and find the probability that the life of a computer terminal is at least 2 years
Solution
The PDF (Probability Density Function) of an exponential distribution with mean 渭 is given by:
f(x) = (1/渭) * exp(-x/渭), for x ≥ 0
where 渭 = 0.5 years is the mean life of a computer terminal.
So, the PDF of the distribution of the length of life in years X of heavily used terminal in a student computer laboratory is:
f(x) = (1/0.5) * exp(-x/0.5) = 2 * exp(-2x), for x ≥ 0
To find the probability that the life of a computer terminal is at least 2 years, we need to calculate:
P(X ≥ 2) = ∫2∞ f(x) dx
= ∫2∞ (2 * exp(-2x)) dx
= -exp(-2x) | 2∞
= exp(-4)
≈ 0.0183
Therefore, the probability that the life of a computer terminal is at least 2 years is approximately 0.0183 or 1.83%.

6.For the following probability distribution. Obtain i) P(X>2) ii) V(X)
X -2 -1 0 1 2 3    ,    P(X) 0.1 0.2 0.2 0.3 0.15 0.05
Solution : 
To obtain P(X>2), we need to add up the probabilities of all the values of X that are greater than 2:
P(X>2) = P(X=3) + P(X=2) = 0.15 + 0.05 = 0.2
Therefore, the probability of X being greater than 2 is 0.2.
ii) To obtain V(X), we need to use the formula:
V(X) = E(X^2) - [E(X)]^2
where E(X) is the expected value of X and E(X^2) is the expected value of X^2.
First, we calculate E(X):
E(X) = (-2)(0.1) + (-1)(0.2) + (0)(0.2) + (1)(0.3) + (2)(0.15) + (3)(0.05) = 0.4
Next, we calculate E(X^2):
E(X^2) = (-2)^2(0.1) + (-1)^2(0.2) + (0)^2(0.2) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.05) = 1.7
Now we can substitute these values into the formula for V(X):
V(X) = E(X^2) - [E(X)]^2 = 1.7 - (0.4)^2 = 1.54
Therefore, the variance of X is 1.54.

7.In a factory 40% workers are skilled. A group of 5 workers is randomly selected from this factory. Find probability that a group consists of.
i) All skilled workers ii) At least 3 skilled workers.
Solution: 
i) To find the probability that all 5 workers are skilled, we need to find the probability of selecting 5 skilled workers from the total number of workers who are skilled.
Probability of selecting one skilled worker from the total number of workers = 0.4
Probability of selecting 5 skilled workers out of 5 workers = (0.4)^5
Therefore, the probability that all 5 workers are skilled = (0.4)^5 = 0.01024

ii) To find the probability that at least 3 workers are skilled, we need to find the probability of selecting 3 or 4 or 5 skilled workers from the total number of workers.

Probability of selecting three skilled workers out of 5 workers = (0.4)^3 x (0.6)^2 x 10 = 0.3456 where 10 is the number of ways of selecting 3 workers out of 5 workers.
Probability of selecting four skilled workers out of 5 workers = (0.4)^4 x (0.6)^1 x 5 = 0.0768
where 5 is the number of ways of selecting 4 workers out of 5 workers.
Probability of selecting five skilled workers out of 5 workers = (0.4)^5 x (0.6)^0 x 1 = 0.01024 where 1 is the number of ways of selecting 5 workers out of 5 workers.
Therefore, the probability that at least 3 workers are skilled = 0.3456 + 0.0768 + 0.01024 = 0.43264.
Hence, the probability that a group of 5 workers consists of all skilled workers is 0.01024 and the probability that a group of 5 workers consists of at least 3 skilled workers is 0.43264.


8.Fit a Binomial distribution to the following data
X 0 1 2 3 4 5
F 10 10 30 25 15 10
To fit a Binomial distribution, we need to know the number of trials and the probability of success for each trial. In this case, we don't have that information directly, but we can estimate it from the data.
Let's start by finding the total number of trials. We can do this by adding up the frequencies: n = 10 + 10 + 30 + 25 + 15 + 10 = 100
Next, we need to estimate the probability of success for each trial. Recall that the Binomial distribution represents the number of successes in a fixed number of trials, where each trial has the same probability of success. So, we need to think about what the "success" would be in this case. One way to approach this is to consider the distribution as a count of successes out of 5. For example, if we assume that the data represents the number of successes out of 5 trials, then we can estimate the probability of success as the mean number of successes divided by 5:
p = (110 + 230 + 325 + 415 + 510) / (5100) = 2.45/5 = 0.49
With these values, we can now fit a Binomial distribution to the data. The Binomial distribution with parameters n and p is given by:
P(X = k) = (n choose k) p^k (1-p)^(n-k)
where "n choose k" is the binomial coefficient.
Using our estimated values for n and p, we can compute the probabilities for each value of k:
k P(X = k)
0 0.00001024
1 0.00144900
2 0.02008114
3 0.13566406
4 0.35604104 
5 0.30875352
These probabilities represent the expected frequencies of each value of k, under the assumption that the data follows a Binomial distribution. We can compare these expected frequencies to the observed frequencies to see how well the distribution fits the data.