Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Balbharti Maharashtra State Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders Important Questions and Answers.

Maharashtra State Board 12th Psychology Important Questions Chapter 6 Psychological Disorders

Choose the correct option and complete the following statements.

Question 1.
A persistent but disproportionate and irrational fear of a specific object or situation is called ……………….
(a) anxiety
(b) mania
(c) phobia
Answer:
(c) phobia

Question 2.
A cluster of symptoms is called as ……………….
(a) therapy
(b) maladjustment
(c) syndrome
Answer:
(c) syndrome

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Question 3.
Bipolar disorder is also called ……………… disorder.
(a) manic-depressive
(b) schizophrenia
(c) PTSD
Answer:
(a) manic-depressive

Question 4.
The DSM-5 comprises of …………….. broad categories of mental disorders along with subcategories.
(a) 19
(b) 11
(c) 22
Answer:
(c) 22

Question 5.
Ashok experiences feelings of unhappiness but has no apparent mental illness. According to Illness Wellness Continuum Model, he will be in Quadrant ……………..
(a) 1
(b) 2
(c) 3
Answer:
(b) 2

Question 6.
Mr. X tries to stop his drug usage. He experiences tremors, muscle pains, etc. This indicates, ……………. symptoms.
(a) withdrawal
(b) recovery
(c) negative
Answer:
(a) withdrawal

State whether the following statements are true or false.

Question 1.
Illness and absence of illness are distinct categories.
Answer:
False

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Question 2.
Individual distress and impairment are criteria for psychological disorders.
Answer:
True

Question 3.
Imbalance in neurotransmitters may cause bipolar disorders.
Answer:
True

Question 4.
Extremely traumatic events may lead to the onset of Acute Stress Disorder.
Answer:
True

Question 5.
Most schizophrenics suffer from delusions and hallucinations.
Answer:
True

Question 6.
Generally, negative symptoms of schizophrenia improve much with proper treatment.
Answer:
False

Question 7.
The same therapy can be applied with equal success to all persons who exhibit similar symptoms.
Answer:
False

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Answer the following in one sentence each.

Question 1.
What is the literal meaning of word ‘schizophrenia’.
Answer:
The word ‘schizophrenia’ is derived from the Greek words ‘schizein’ (to split) and ‘phren’ (mind). Thus, the literal meaning of the word ‘schizophrenia’ is split mind.

Question 2.
What is meant by delusion?
Answer:
Delusion refers to false belief held by the individual which cannot be shaken off in spite of clear evidence that is contrary to it.

Question 3.
What are the two states of bipolar disorder?
Answer:
The two states of bipolar disorder are mania state and depression state.

Question 4.
When is the person diagnosed with PTSD?
Answer:
The person is said to be diagnosed with PTSD when the symptoms of Acute Stress Disorder (ASD) continue for more than one month with the same intensity.

Question 5.
What are addictive disorders?
Answer:
Addictive disorders refer to the physical and psychological inability to stop consuming some substance or indulging in some activity although it is harmful.

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Question 6.
What is psychotherapy?
Answer:
Psychotherapy refers to the systematic attempt by a mental health professional to assist a person to overcome some psychological disorders.

Question 7.
What is the suggestible stage of PTSD?
Answer:
Suggestible stage of PTSD refers to a person who may seek guidance from others and may either accept these suggestions unquestioningly or may get extra sensitive.

Explain the following concepts.

Question 1.
Delusions
Answer:
Delusions is a false belief that is strongly held by an individual in spite of the presence of contrary evidence, Delusions are a symptom of mental disorders. Commonly noticed delusions are delusion of grandeur, delusions of persecution, delusion of influence or control, delusion of reference, etc.

Answer the following questions Briefly.

Question 1.
Explain Acute Stress Disorder.
Answer:
If a person (aged 6 years and above) has experienced extremely stressful situations like death of a loved one, serious disease or injury, sexual abuse, natural disasters, etc., then he/she may experience ASD.
The symptoms of ASD include-

  1. emotional numbness and instability
  2. nightmares and sleep disturbances
  3. insomnia, lack of concentration, irritability and guilt feelings
  4. depression

Question 2.
What are the red flags in identifying psychological disorders?
Answer:
Signs that help to identify individuals who require expert (professional) help in mental health are called ‘red flags’.
Symptoms, if present, in any person should satisfy three requirements-

  1. it should be present for a considerably long period of time
  2. symptom has become more severe than before
  3. it has created problems in the person’s life.

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Question 3.
What is Generalized Anxiety Disorder?
Answer:
Anxiety refers to a condition in which a person feels uneasy for a long time for no obvious reasons. The main anxiety disorders include Generalized Anxiety Disorder, Phobia, Panic attacks etc.

In Generalized Anxiety Disorder the person frequently experiences anxiety more intensely so that it starts interfering with the ability to perform daily tasks. Symptoms include irritability, headaches, insomnia, dizziness, breathlessness, etc.

Write short notes on the following.

Question 1.
Identification and treatment of psychological disorders.
Answer:
Signs that help to identify individuals who require expert (professional) help in mental health are called ‘red flags’. Symptoms, if present, in any person should satisfy three requirements

  1. it should be present for a considerably long period of time
  2. symptom has become more severe than before
  3. it has created problems in the person’s life.

The signals indicating a need for professional help are:

  1. inability to concentrate or to sleep well.
  2. severe confusion and memory loss.
  3. intense and uncontrollable negative feelings.
  4. self-neglect or even self-harm.
  5. loss of interest in friends/ family/ activities.
  6. odd statement or strange use of words.

Treatment strategies range form pharmacotherapy (administering drugs) to psychotherapy such as Rational Emotive Behaviour Therapy, Interpersonal Therapy, Group/Family Therapy, etc. Mental health professionals are qualified to decide the plan of treatment.

Maharashtra Board Class 12 Psychology Important Questions Chapter 6 Psychological Disorders

Answer the following questions in 150 – 200 words.

Question 1.
Discuss the red flags indicating the need for psychological help.
Answer:
Signs that help to identify individuals who require expert (professional) help in mental health are called ‘red flags’.
Symptoms, if present, in any person should satisfy three requirements

  1. it should be present for a considerably long period of time
  2. symptom has become more severe than before
  3. it has created problems in the person’s life.

Signals indicating need for professional help-

  1. Inability to concentrate
  2. Inability to sleep well
  3. Various physical complaints
  4. Uncontrollable and frequent bad thoughts
  5. Too intense uncontrollable negative feelings (fear, anger, sadness, worry, etc.)
  6. Severe confusion
  7. Memory loss
  8. Loss of interest in activities and friends
  9. Odd or irrational statements or strange use of words or way of speaking
  10. Self-harm
  11. Signs of self-neglect

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Balbharti Maharashtra State Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues Important Questions and Answers.

Maharashtra State Board 12th Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the following statements by selecting the appropriate option.

Question 1.
The Rio de Janeiro Earth Summit (1992) focused on
(a) Environment and development
(b) Nuclear non-proliferation
(c) International trade
(d) Gender issues
Answer:
(a) Environment and development

Question 2.
The 2002 Earth Summit recognised as the most important goal at all level.
(a) poverty
(b) socio-economic develop
(c) sustainable development
(d) gender issues
Answer:
(c) sustainable development

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 3.
The 2030 Agenda for sustainable Development was adopted in the year
(a) 2015
(b) 2002
(c) 1992
(d) 2019
Answer:
(a) 2015

Question 4.
The Second UN Decade for poverty eradication was in
(a) 1997-2006
(b) 1971-1980
(c) 2008-2017
(d) 1992-2003
Answer:
(c) 2008-2017

Question 5.
The Revolution aimed to make India self-sufficient in food grains.
(a) Import substitution
(b) Green
(c) White
(d) Infrastructural
Answer:
(b) Green

Question 6.
The is a scheme introduced in 1978-79 to create opportunities for self-employment in
the rural sector.
(a) MGNREGA
(b) JRY
(c) IRDP
(d) KVY
Answer:
(c) IRDP

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the following sentences by selecting the appropriate reason.

Question 1.
Green Revolution was introduced in India in the 1960’s to ………………
(a) make India self sufficient in food grains.
(b) generate employment and tackle poverty.
(c) to combine economic growth with social justice.
Answer:
(a) make India self sufficient in food grain

Question 2.
The Department of Women and Child Development was set up as a separate Ministry in 2006 to …………..
(a) facilitate health and education programmes for women.
(b) provided political representation to women
(c) empower women to live with dignity and contribute as equal partners in development.
Answer:
(c) make India self sufficient in food grains.

State the appropriate concept for the given statements.

Question 1.
Introduction of high yielding variety of seeds and irrigation methods.
Answer:
Green Revolution

Question 2.
Think tank of the Government of India, providing both directional and policy inputs.
Answer:
NITI Aayog

Question 3.
The responsible interaction with the environment to avoid depletion or degradation of natural resources.
Answer:
Sustainable development

Question 4.
Deprivation of common necessities that determine the quality of life.
Answer:
Poverty

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 5.
A view that free market economy promotes development.
Answer:
Capitalism

Question 6.
Blueprint for prosperity for people and planet adopted by the UN in 2015.
Answer:
2030 Agenda for Sustainable Development

Question 7.
Agency set up in 1950 to formulate a plan for effective and balanced utilisation of India’s resources.
Answer:
Planning Commission

Question 8.
India’s policy towards poverty and development.
Answer:
Growth with Social Justice and Equity

Question 9.
The process by which oppressed persons gain control over their own lives.
Answer:
Empowerment

Question 10.
Socially constructed characteristics of women and men such as behaviour, norms, roles and relationships.
Answer:
Gender

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Find the odd word.

Question 2.
Pollution, Deforestation, Poverty, Water Scarcity.
Answer:
Poverty (not a direct environment concern)

Question 2.
Brundtland Commission, Rio+20 Summit, Johannesburg Earth Summit, United Nations Development Programme (UNDP).
Answer:
UNDP (not a commission/ summit associated with sustainable development)

Expand the following abbreviations.

(1) UNCED
(2) ECOSOC
(3) UNDP
(4) IRDP
(5) JGSY
(6) MPI
Answer:
(1) UNCED – United Nations Conference on Environment and Development
(2) ECOSOC – Economic and Social Council of the UN.
(3) UNDP – United Nations Development Programme
(4) IRDP – Integrated Rural Development Programme
(5) JGSY – Jawahar Gram Samridhi Yojana
(6) MPI – Multidimensional Poverty Index

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Complete the concept maps.

Question 1.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 1
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 2

Question 2.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 3
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 4

Question 3.
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 5
Answer:
Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991 Humanitarian Issues 6

State whether the following statements are true or false with reason.

Question 1.
Rio Earth Summit (1992) was a landmark in environment issues.
Answer:
This statement is True.

  1. Rio de Janeiro Earth Summit (1992) sought to raise public awareness on the need to integrate environment and development.
  2. It’s objective was to create a partnership between developing and more industrialized nations to ensure a healthy future for the planet.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 2.
The perception of poverty has undergone a change in recent times.
Answer:
This statement is True.

  1. The traditional perception of poverty focused on the deprivation of the basic necessities of life such as food, shelter, etc.
  2. The new perception of poverty as expressed by the ECOSOC, considers poverty as lack of basic capacity to participate effectively in society.
  3. It is perceived as a denial of choices, opportunities and human dignity.

Question 3.
The 2030 Agenda for Sustainable Development (2015) is a significant document.
Answer:
This statement is True.

  1. The 2030 Agenda for Sustainable Development (2015) adopted by the UN, provides a blueprint for prosperity of the people and planet for now and in the future
  2. It mentions seventeen Sustainable Development Goals for action by all countries in a global partnership for e.g., No Poverty, Climate action, Gender equality, etc.

Question 4.
The economic reforms initiated in India after 1991 had far-reaching implications.
Answer:
This statement is True.

  1. After 1991, India introduced the policies of economic liberalisation and privatisation of the public sector and brought an increase in economic and industrial growth rate.
  2. These reforms changed the role of the State to focus more on social development such as in heath care, sanitation, education, etc.

Explain the correlation between the following.

Question 1.
Department of Women and Child Development – Women’s empowerment.
Answer:
The Department of Women and Child Development, Government of India came into existence as separate ministry in 2006. Before this, it was a department under HRD ministry: The Department of Women and Child Development was constituted with the intention of addressing gaps in State action for women and children and to create gender equitable and child-centered legislations, policies and programmes.

It aims to achieve the empowerment of women i.e., to live in dignity and to contribute as equal partners in development in an environment free from discrimination, injustice and violence. The Ministry has prepared the National Policy for the Empowerment of Women (2001). The Draft of the National Policy for women (2016) aims at women’s empowerment by following a socially inclusive rights-based approach.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Express your opinion of the following.

Question 1.
There is a need to protect the environment.
Answer:
I agree with this statement.
The environment refers to all living and non-living things that make up our surroundings. Many environmental problems result due to unplanned management, human activities and technological development that interfere with the environment. Protection of environment refers to conservation and preservation of environment by reducing pollution, soil erosion, global warming, deforestation etc.

Main environmental concerns hence are climate change, water scarcity, pollution, loss of biodiversity, etc. Environmental degradation and depletion of resources will result in spread of pests and vector diseases, extinction of species, natural disasters like floods; acid rain, melting of glaciers, etc. Life on this earth cannot exist unless we restore environmental balance.

Question 2.
Poor political representation of women is a major concern.
Answer:
I agree with this statement.
Poor political representation of women is a significant gender concern in India. The first Lok Sabha had only 24 women of total 489 members i.e., 5%, and the 17th Lok Sabha (2019) has 78 women (14%). This is the highest representation of women in Lok Sabha till date. The Women’s Reservation Bill which proposes to reserve 33% of all seats in Lok Sabha and in State Assemblies for women is pending in the Lok Sabha. This bill was first introduced in Parliament in 1996, by the United Front government of Prime Minister H.D. Deve Gowda.

As per the bill, seats will be reserved for women on a rotational basis. The UPA-I government (2004-2009) again introduced the bill in May 2008 and it was passed in May 2010 by the Rajya Sabha. Currently, the bill is still pending in Lok Sabha. However, there exists 33% reservation for women in rural and urban local bodies (as per 73rd and 74th Amendment Acts).

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Answer the following question in 80 to 100 words.

Question 1.
What is the Green Revolution?
Answer:
Green Revolution refers to a large increase in crop production that is achieved by the use of high yielding crop varieties, pesticides, artificial fertilizers, machines and better management of agricultural resources. Norman Borlaug, an agricultural scientist of USA is considered as ‘Father of the Green Revolution’ for which he was awarded the Nobel Peace Prize (1970). In India, M.S. Swaminathan is credited with introducing the Green Revolution in 1965 especially in Punjab, Haryana and Uttar Pradesh.

This was necessitated due to famine and acute food shortages due to low productivity in food production. Green Revolution was mainly in wheat, maize and gram production. During it’s early years, the Green Revolution yielded great economic prosperity leading to significant increases in agricultural output and farmers incomes. However, this did not benefit marginal farmers who could not afford expensive inputs.

Question 2.
What is empowerment? Suggest measures for empowerment of women.
Answer:
Empowerment refers to the process by which oppressed or marginalized sections of the population become stronger, especially in controlling their own life and claiming their own rights. This may happen socially, economically, politically or nationally. The approach to women’s issues has progressed from ‘welfare’ to ‘development’ to ‘empowerment’.

Some of the measures for empowerment of women include-

  1. Political empowerment – giving women better representation in legislative bodies and in the decision making process.
  2. Economic empowerment – such as strengthening women’s access to property inheritance and land rights, skill training, work opportunities, micro-credit, increased right to economic resources and power.
  3. Social empowerment – through literacy/education, training and raising awareness, freedom from domestic violence, access to information, health services and sanitation.

In India, schemes to empower women include Beti Bachao Beti Padhao Yojana (2015), Support to Training and Employment Programme for Women (STEP), Mahila Shakti Kendras (2017) etc.

Question 3.
Describe international efforts to deal with environmental concerns.
Answer:
There have been many international efforts to deal with environmental concerns as well as about the relationship between economic development and environmental degradation.
1. UN Conference on Human Environment (Stockholm, 1972) The purpose was to encourage and provide guidelines for protection of the environment.

2. World Commission on Environment and Development (1983) also called Brundtland commission put forward the concept of sustainable growth.

3. Rio de Janeiro Earth Summit (1992) sought to raise public awareness on the need to integrate environment and development and to create a partnership between developing and more industrialized nations to ensure a healthy future for the planet.

4. Earth Summit at Johannesburg (2002) recognized sustainable development as the most important goal for institutions at the national, regional and international levels

5. UN Conference on sustainable development (Rio-2012) is also called Rio Earth Summit or Rio+20. Environmental sustainability is defined as responsible interaction with the environment to avoid depletion or degradation of natural resources and allow for long¬term environmental quality. It’s goal is to conserve natural resources, to reduce pollution, to develop alternate sources of power, etc.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 4.
Elaborate on the aspects of development in India.
Answer:
The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (Socialist) or role of free market economy (Capitalist). Since the 1990s, the alternative view of development that focuses on both material and non-material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability.

  1. Modernisation of the economy – During the Nehruvian period, the focus was on land reforms, irrigation and large-scale industrialisation to provide employment and increase productivity.
  2. Self – reliance – This included ‘import substitution’ strategy to develop indigenous industry.
  3. Socialist pattern of society – It was based on promotion of the Welfare State, employment generation and importance ,to the public sector. The Planning Commission (March 1950 to 31st December 2014) was created to chalk out Five Year Plans for development.

“Growth with Social Justice and Equity’ has been India’s policy towards poverty and development. Its development strategy has evolved over the years. In the early stages, government played a dominant role through the public sector.

Question 5.
Describe the 2030 Agenda for Sustainable Development adopted in 2015 by the UN.
Answer:
In 2015, the United Nations adopted the ‘2030 Agenda for Sustainable Development. It provides a blueprint for peace and prosperity for people and the planet for now and into the future. The Agenda mentions 17 Sustainable Development Goals (SDGs) for action by all developed and developing countries in a global partnership.

The Sustainable Development Goals are:

  1. No Poverty
  2. Zero hunger
  3. Good health and well being
  4. Quality
  5. Gender equality
  6. Clean water and sanitation
  7. Affordable and clean energy
  8. Decent work and economic growth
  9. Industry, innovation and infrastructure
  10. Reduced inequalities
  11. Sustainable cities and communities
  12. Responsible consumption and production
  13. Climate action
  14. Life below water
  15. Life on land
  16. Peace, justice and strong institutions
  17. Partnership goals

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Answer the following question with reference to the given points.

Question 1.
Explain condition of women in India.
(a) Economic inequality
(b) Trafficking and exploitation
(c) Literacy rate
(d) Political representation
Answer:
All over the world, women have to face serious problems such as discrimination and violence, under representation in economic and political decision-making, inequalities in education, health care, landholdings and workplaces and even in human rights.

Some of the important gender issues:
(a) Economic inequality – Participation of women in the labour market is about 28% only. There is also income inequality due to wide gender wage gaps as well as inequality of opportunities to women e.g., unequal access to education and finance. In many countries, women are either not encouraged to work or employed in the informal sector. Women continue to be underrepresented in high-level, highly paid positions and experience gender discrimination and sexual harassment in the workplace. Women experience high levels of poverty, unemployment and other economic hardships.

(b) Trafficking and exploitation – Women are often exploited and denied basic rights. They are trafficked for purpose of sexual and economic exploitation particularly prostitution, ‘sold’ as brides, subjected to sexual crimes, forced labour, street begging, etc. Trafficking in women means that they are deceived or forced/sold, physically confined, abused with no access to protection or health care.

(c) Literacy rate – low literacy among women is acute in India. As per 2011 census report, female literacy rate is only 65.46% while for males it is 82.14%. The reasons for the low female literacy are-

  1. traditional patriarchal notions that do not consider female education as economically advantageous
  2. Schools in rural areas are not easily accessible and travel may not safe or easy.
  3. Benefits of schemes like R.T.E have not reached many rural females.
  4. Social evils like child marriage, female foeticide, dowry, etc.

(d) Political Representation – Poor political representation of women is a significant gender concern in India. The First Lok Sabha had only 24 women of total 489 members i.e., 5% and the 17th Lok Sabha (2019) has 78 women (14%). This is the highest representation of women in Lok Sabha till date. The Women’s Reservation Bill which proposes to reserve 33% of all seats in Lok Sabha and in State Assemblies for women is pending in the Lok Sabha. However, there exists 33% reservation for women in rural and urban local bodies.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 2.
Describe the following poverty alleviation programmes in India.
(a) Integrated Rural Development Programme (IRDP)
(b) Jawahar Gram Samridhi Yojana (JGSY)
(c) Pradhan Mantri Gramin Awaas Yojana (PMGAY)
(d) Mahatma Gandhi National Rural Employment Guarantee Act. (MGNREGA)
(e) Saansad Adarsh Gram Yojana.
Answer:
The poverty alleviation programmes in India have been designed for both rural areas and urban areas. Most of the programmes are designed to target the rural poverty as prevalence of poverty is high in rural areas.
(a) Integrated Rural Development Programme (IRDP) – It was first introduced in 1978¬79. The main objective of IRDP is to create sustainable opportunities for self-employment in the rural sector.

(b) Jawahar Gram Samridhi Yojana (JGSY) – It is the revised version of the Jawahar Rozgar Yojana (JRY). It was started in 1999. The main aim of this programme was development infrastructure like roads to connect the villages to different areas and other social, education (schools) and infrastructure like hospitals in the rural areas.

(c) Pradhan Mantri Gramin Awaas Yojana – It was started in 1985. This scheme aimed at creating housing for everyone.

(d) Mahatma Gandhi National Rural Employment Guarantee Act. (MGNREGA) – It was launched in 2005. It provides 100 days assured employment every year to every rural household.

(e) Saansad Adarsh Gram Yojana – It was started in 2014. According to this yojana, each Member of Parliament will take the responsibility of developing three villages by 2019. The idea is to make India’s villages to be fully developed with physical and institutional infrastructure.

Question 3.
Discuss the following humanitarian issues.
(a) Environmental degradation
(b) Poverty
(c) Gender issues
Answer:
Humanitarian issues refer to concerns about issues affecting human welfare with the aim of alleviation of sufferings. Examples of humanitarian issues include natural disasters, famine, epidemics, armed conflicts, poverty, inequalities and injustice, environmental problems, etc.

(a) Environmental degradation – Indiscriminate exploitation of natural resources had resulted in environmental depletion and degradation. This results in significant environmental concerns such as deforestation, soil erosion,loss of biodiversity, climate change, endangerment of species, water scarcity, acid rain, pollution, spread of diseases etc. The relationship between economic development and environmental degradation has been considered at various international conferences like Rio Earth Summit, etc.

Environmental degradation has serious consequences for humans, animals, plants as well as the ecological balance. The causes of degradation include resource depletion due to overuse, landfills and illegal dumping, waste production, overpopulation and consumption behaviour, etc.

(b) Poverty – The traditional perception of poverty is where people are unable to provide for their basic necessities of life. It is the deprivation of common necessities that determine the quality of life including food, clothing, shelter and safe drinking water. The alternative view of poverty focuses on both material and non-material aspects i.e., focus on community ties, values and availability of common resources, participatory decision making and political and economic decentralisation.

The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (socialist) or role of free market economy (capitalist). Since the 1990s, the alternate view of development that focuses on both material and non material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability.

(c) Gender issues – There is a distinct link between poverty and unsatisfactory conditions of women. In most parts of the world women are poor due to cultural norms, unequal power equations between men and women. In the 1970s, women empowerment came to be accepted as an important philosophy to bring about gender justice. The UN ‘Decade for Women’ began in 1976 to-

  1. link women’s issues with developmental issues
  2. promote equal rights and opportunities for women across the globe.

Maharashtra Board Class 12 Political Science Important Questions Chapter 3 Key Concepts and Issues Since 1991: Humanitarian Issues

Question 4.
Discuss the vision of the development process in India.
(a) Modernisation of the economy
(b) Self-reliance
(c) Socialist pattern of society
Answer:
The purpose of development is to ensure welfare of the people and is associated with economic growth. The traditional approaches to development focus on the predominant role of the State in promoting economic growth (Socialist) or role of free market economy (Capitalist). Since the 1990s, the alternative view of development that focuses on both material and non material aspects has become acceptable. The focus today is development based on equity, participation, empowerment and sustainability. India’s vision of development had three aspects:

(a) Modernisation of the economy – This implied industrialization to provide employment to a growing labour force and to increase productivity During the Nehruvian period, the focus was on large scale industrialization and factory production. Land reforms and irrigation were looked at as the means to achieve agricultural growth and productivity.

(b) Self-reliance – This included ‘import substitution’ strategy to develop indigenous industry. Aid taken from Soviet Union, France, etc. was utilised mainly in the public sector and for infrastructure.

(c) Socialist pattern of society – The approach of planning for development was based on promotion of the welfare State, employment generation and importance to the public sector. The Planning Commission (March 1950 31st December 2014) was created to chalk out Five Year Plans for development.

“Growth with Social Justice and Equity” has been India’s policy towards poverty and development. Its development strategy has evolved over the years. In the early stages, government played a dominant role through the public sector.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Balbharti Maharashtra State Board Class 12 Psychology Important Questions Chapter 5 Emotions Important Questions and Answers.

Maharashtra State Board 12th Psychology Important Questions Chapter 5 Emotions

Choose the correct option and complete the following statements.

Question 1.
According to the theory, we experience physiological arousal and feelings simultaneously and independently.
(a) James-Lange
(b) Cannon-Bard
(c) Schachter-Singer
Answer:
(b) Cannon-Bard

Question 2.
According to Ekman, there are basic emotions.
(a) 2
(b) 5
(c) 6
Answer:
(c) 6

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 3.
is an emotion that is in response to some threat.
(a) Sadness
(b) Fear
(c) Disgust
Answer:
(b) Fear

Question 4.
Plutchik’s model explains that there are primary emotions.
(a) 8
(b) 6
(c) 12
Answer:
(b) 6

Question 5.
is not a primary/ basic emotion according to Plutchik.
(a) Trust
(b) Anticipation
(c) Jealousy
Answer:
(c) Jealousy

Question 6.
Emotions are controlled by the ……………….. in the brain.
(a) cerebral cortex
(b) limbic system
(c) RAS
Answer:
(b) limbic system

Question 7.
Aniket shows creativity and has meaningful relationships. He has …………. emotional well-being.
(a) high
(b) low
(c) moderate
Answer:
(a) high

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

State whether the following statements are true or false.

Question 1.
“I feel sad because I cry”, This is the premise of the Schachter – Singer theory.
Answer:
False

Question 2.
Pride and guilt are considered as higher cognitive emotions.
Answer:
True

Question 3.
The fight or flight reaction is a reaction to sadness.
Answer:
False

Question 4.
Low serotonin levels are associated with clinical depression.
Answer:
True

Answer the following in one sentence each.

Question 1.
What is the premise of the facial feedback hypothesis?
Answer:
According to the facial feedback hypothesis, our facial expressions provide feedback to our brain about our emotions.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
What are complex emotions?
Answer:
Complex emotions are emotions which result form the combination of basic emotions for e.g., surprise and sadness lead to disappointment.

Question 3.
According to Ekman, which are the basic emotions?
Answer:
According to Ekman, there are six basic (universal) types of emotions, i.e., happiness, sadness, anger, fear, surprise, disgust.

Question 4.
When does a person experience disgust?
Answer:
A person usually experiences disgust as a response to some unwanted stimulus.

Question 5.
According to Plutchik, how do emotions result?
Answer:
According to Plutchik, emotions result due to evolving over a period of time for increasing our chances of survival.

Question 6.
Why is Plutchik’s model important?
Answer:
Plutchik’s model is important from the perspective of emotional literacy, i.e., understanding emotional levels, complexity and change as well as appropriate emotional labelling.

Question 7.
What does emotional well-being mean?
Answer:
Emotional well-being means managing both positive and negative emotions, so that we can lead a healthy and productive life.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 8.
What does anger management mean?
Answer:
Anger management is an intervention programme to prevent anger from turning into a habit or obstacle by creating awareness of and responsibility for our emotions.

Explain the concepts in 25 – 30 words each.

Question 1.
Emotions
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.

Question 2.
Emotional well-being.
Answer:
Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Persons who have high emotional well-being experience benefits such as-

  1. better able to deal with stress
  2. better self-regulation
  3. increased productivity in tasks undertaken
  4. increased creativity and openness to new experiences
  5. life satisfaction due to meaningful activities and relationship.

Question 3.
Emotional Abuse.
Answer:
Emotional abuse is any kind of abuse that is emotional rather than physical in nature. It occurs when one person subjects or exposes another person to intentionally harmful behaviour that may result in anxiety, depression and psychological trauma in the victim.
The types of emotional abuse may be-

  1. verbal abuse such as blaming, insulting, labeling, threatening, swearing, etc.
  2. nonverbal abuse such as ignoring, rejection, bullying, spying, etc.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 4.
Anger Management.
Answer:
Anger management is an intervention programme to prevent anger from turning into a habit or obstacle. It enables the person to create awareness of and responsibility for his/her emotions. Anger management involves two aspects (i) managing one’s own anger (ii) learning to respond effectively to anger in others. The three R’s in anger management are Relax, Reassess and Respond.

Answer the following questions in 35 – 40 words each.

Question 1.
What are the techniques to deal with emotional abuse?
Answer:
Emotional abuse is any kind of abuse that is emotional rather than physical in nature. It occurs when one person subjects or exposes another person to intentionally harmful behaviour that may result in anxiety, depression and psychological trauma for the victim.
The techniques to deal with emotional abuse-

  1. Accept that emotional abuse is not because of you, i.e., don’t justify the actions of the abuser.
  2. Respond assertively to the abuser but seek distance from him/her.
  3. Give yourself time to heal.
  4. Prioritize your self-care, eating right, exercise, etc.
  5. Mobilise support from family and friends. If needed, seek professional help.

Question 2.
What are the benefits of emotional well-being?
Answer:
Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Persons who have high emotional well-being experience benefits such as-

  1. better able to deal with stress
  2. better self-regulation
  3. increased productivity in tasks undertaken
  4. increased creativity and openness to new experiences
  5. life satisfaction due to meaningful activities and relationship.

Write short notes on the following in 50 – 60 words each.

Question 1
Characteristics of emotions
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.
Some characteristics of emotions are-

  1. Emotions may be positive, e.g., joy or negative, e.g., anger.
  2. Emotions may occur for a brief period or may be long lasting.
  3. Emotions may be important for our survival, e.g., fear or for our psychological well-being, e.g., love.
  4. Emotions differ in intensity in expression for e.g., annoyance-anger-rage.
  5. Complex emotions (higher cognitive level emotions) result from the combination of basic emotions, for e.g., surprise and sadness lead to disappointment.
  6. According to Ekman, there are six basic (universal) types of emotions, i.e., happiness, sadness, anger, fear, surprise, disgust.

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
Physiological changes during emotions.
Answer:
Physiological changes refer to automatic reactions that take place in our body in response to some stimulus, for e.g., if you saw a snake, the brain at the cognitive level perceived the stimulus as dangerous. This leads to physiological arousal such as dilated pupils, increased heart rate, increased pulse rate, sweating. At the emotional level you experience fear. The Autonomic Nervous System and Glandular system signal the pituitary gland which activates the adrenal glands to secrete the cortisol hormone. This triggers “the fight or flight” response. At the behavioural level there is an action plan such as running away or calling for help, etc.
Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions 1

Answer the following questions 80 – 100 words each.

Question 1.
Theories of emotions.
Points:
(i) James-Lange theory
(ii) Cannon-Bard theory
(iii) Schachter-Singer theory
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heartrate, thoughts and feelings, i.e. emotional tone and expressive behaviour, i.e., facial expression.
The main theories of emotions are-
(i) James-Lange theory – It was one of the earliest theories to explain emotion. According to William James and Carl Lange, physiological arousal instigates the experience of emotion. It proposes that each specific emotion is attached to a different pattern of physiological arousal. For e.g., we feel sad because we cry. The sequence of events in emotional experience is Emotion Stimulus – Physiological Response Pattern → Affective Experience.

(ii) Cannon-Bard theory – According to Walter Cannon and Philip Bard, we may experience the same physiological arousal but emotions can be different, for e.g., we don’t cry only when we are sad but we also cry when we are angry or extremely happy. We experience physiological arousal and feelings at the same time and independently. For e.g., seeing a man with a gun prompts the feeling of fear as well as a racing heartbeat.

(iii) Schachter and Singer’s Two Factor theory – According to Stanley Schachter and Jerome Singer, emotion is based on two factors, i.e., physiological arousal and cognitive label, for e.g., an environmental stimuli (growling dog) elicits a physiological response (increased heart rate). We cognitively label this response (fear).
Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions 2

Maharashtra Board Class 12 Psychology Important Questions Chapter 5 Emotions

Question 2.
Emotional well-being.
Points:
(i) Meaning
(ii) Aspects
(iii) Benefits
Answer:
The word emotion is derived from the latin word ‘emovere’, which means to stir up or to move. An emotion refers to an involuntary, aroused state of an organism involving physical, cognitive and behavioural components. It is described as a combination of bodily arousal, e.g., increased heart rate, thoughts and feelings, i.e. emotional tone and expressive behaviour i.e. facial expression.

(i) Meaning – Emotional well-being means managing our emotions, both positive and negative ones, so that we can lead a healthy and productive life. Emotional well-being is not easily observable. It can be guaged on the basis of how a person rationally handles a situation involving some stressors or leading to negative emotions. Emotional well-being refers to understanding and managing one’s emotions without getting overwhelmed by negative emotions but instead encouraging positive emotions.

(ii) Aspects – The aspects of emotional well-being are at three levels viz. physical, emotional and social.

  1. Physical level, i.e., well balanced diet, exercise.
  2. Emotional level, i.e., practise mindfulness, raising levels of motivation and optimism.
  3. Social level, i.e., engaging in prosocial behaviour, meaningful relationships.

(iii) Benefits – Persons who have high emotional well-being experience benefits such as-

  1. Coping with stress – It helps to deal with stress using healthy methods such as exercise, social support, etc.
  2. Better self-regulation – It enables the person to label how they feel and accept negative emotions life fear, anger, etc.
  3. Increases productivity in tasks undertaken – The ability to focus is enhanced, the person feels positive and energized.
  4. Increases creativity – The person indulges in divergent thinking, shows curiosity is open to new experiences.
  5. Life satisfaction – The individual is able to have meaningful interactions and relationships, show empathy, altruism and engage in activities like volunteer work.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 1 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 1 Class 7 Maths Solutions

Question 1.
Solve the following:
i. (-16) × (-5)
ii. (72) ÷ (-12)
iii. (-24) × (2)
iv. 125 ÷ 5
v. (-104) ÷ (-13)
vi. 25 × (-4)
Solution:
i. (-16) × (-5) = 80

ii. 72 ÷ (-12) = \(\frac { 72 }{ -12 }\)
= \(\frac{1}{(-1)} \times \frac{72}{12}\)
(-1) × 12
= -6

iii. (-24) × 2 = -48

iv. 125 ÷ 5 = \(\frac { 125 }{ 5 }\)
= 25

v. (-104) ÷ (-13) = \(\frac { -104 }{ -13 }\)
= \(\frac { 104 }{ 13 }\)
= 8

vi. 25 × (-4) = -100

Question 2.
Find the prime factors of the following numbers and find their LCM and HCF:
i. 75,135
ii. 114,76
iii. 153,187
iv. 32,24,48
Solution:
i. 75 = 3 × 25
= 3 × 5 × 5
135 = 3 × 45
= 3 × 3 × 15
= 3 × 3 × 3 × 5
∴ HCF of 75 and 135 = 3 × 5
= 15
LCM of 75 and 135 = 3 × 5 × 5 × 3 × 3
= 675

ii. 114 = 2 × 57
= 2 × 3 × 19
76 = 2 × 38
= 2 × 2 × 19
∴ HCF of 114 and 76 = 2 × 19
= 38
LCM of 114 and 76 = 2 × 19 × 3 × 2
= 228

iii. 153 = 3 × 51
= 3 × 3 × 17
187 = 11 × 17
∴ HCF of 153 and 187 = 17
LCM of 153 and 187 = 17 × 3 × 3 × 11
= 1683

iv. 32 = 2 × 16
= 2 × 2 × 8
= 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2
24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
∴ HCF of 32, 24 and 48 = 2 × 2 × 2
= 8
LCM of 32,24 and 48 = 2 × 2 × 2 × 2 × 2 × 3
= 96

Question 3.
Simplify:
i. \(\frac { 322 }{ 391 }\)
ii. \(\frac { 247 }{ 209 }\)
iii. \(\frac { 117 }{ 156 }\)
Solution:
i. \(\frac { 322 }{ 391 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 1

ii. \(\frac { 247 }{ 209 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 2

iii. \(\frac { 117 }{ 156 }\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 3

Question 4.
i. 784
ii. 225
iii. 1296
iv. 2025
v. 256
Solution:
i. 784
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 4
∴ 784 = 2 × 2 × 2 × 2 × 7 × 7
∴ √784 = 2 × 2 × 7
= 28

ii. 225
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 5
∴ 225 = 3 × 3 × 5 × 5
∴ √225 = 3 × 5
= 15

iii. 1296
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 6
∴ 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
∴ √1296 = 2 × 2 × 3 × 3
= 36

iv. 2025
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 7
∴ 2025 = 3 × 3 × 3 × 3 × 5 × 5
∴ √2025 = 3 × 3 × 5
= 45

v. 256
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 8
∴ 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ √256 = 2 × 2 × 2 × 2
= 16

Question 5.
There are four polling booths for a certain election. The numbers of men and women who cast their vote at each booth is given in the table below. Draw a joint bar graph for this data.

Polling Booths Navodaya Vidyalaya Vidyaniketan School City High School Eklavya School
Women 500 520 680 800
Men 440 640 760 600

Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 9

Question 6.
Simplify the expressions:
i. 45 ÷ 5 + 120 × 4 – 12
ii. (38 – 8) × 2 ÷ 5 + 13
iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
Solution:
i. 45 ÷ 5 + 120 × 4 – 12
= 9 + 80 – 12
= 89 – 12
= 77

ii. (38 – 8) × 2 ÷ 5 + 13
= 30 × 2 ÷ 5 + 13
= 60 ÷ 5 + 13
= 12 + 13
= 25

iii. \(\frac{5}{3}+\frac{4}{7} \div \frac{32}{21}\)
\(\frac{5}{3}+\frac{4}{7} \times \frac{21}{32}\)
\(\frac{5}{3}+\frac{3}{8}=\frac{40}{24}+\frac{9}{24}\)
\(\frac{49}{24}\)

iv. 3 × {4 [85 + 5 – (15 – 3)] + 2}
= 3 × {4[90 – 5] + 2}
= 3 × {4 × 85 + 2}
= 3 × (340 + 2)
= 3 × 342
= 1026

Question 7.
Solve:
i. \(\frac{5}{12}+\frac{7}{16}\)
ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
Solution:
i. \(\frac{5}{12}+\frac{7}{16}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 10

ii. \(3 \frac{2}{5}-2 \frac{1}{4}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 11

iii. \(\frac{12}{5} \times \frac{(-10)}{3}\)
= 4 × (-2)
= -8

iv. \(4 \frac{3}{8} \div \frac{25}{18}\)
= \(\frac{7}{4} \times \frac{9}{5}\)
= \(\frac { 63 }{ 20 }\)

Question 8.
Construct ∆ABC such that m∠A = 55°, m∠B = and l(AB) = 5.9 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 12

Question 9.
Construct ∆XYZ such that, l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 13

Question 10.
Construct ∆PQR such that, m∠P = 80°, m∠Q = 70°, l(QR) = 5.7 cm.
Ans:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° …. (Sum of the measures of the angles of a triangle is 180°)
∴ 80 + 70 + m∠R = 180
∴ 150 + m∠R = 180
∴ m∠R = 180 – 150
∴ m∠R = 30°
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 14

Question 11.
Construct ∆EFG from the given measures. l(FG) = 5 cm, m∠EFG = 90°, l(EG) = 7 cm.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 15

Question 12.
In ∆LMN, l(LM) = 6.2 cm, m∠LMN = 60°, l(MN) 4 cm. Construct ∆LMN.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 16

Question 13.
Find the measures of the complementary angles of the following angles:
i. 35°
ii. a°
iii. 22°
iv. (40 – x)°
Solution:
i. Let the measure of the complementary
angle be x°.
35 + x = 90
∴35 + x-35 = 90 – 35
….(Subtracting 35 from both sides)
∴x = 55
∴The complementary angle of 35° is 55°.

ii. Let the measure of the complementary angle be x°.
a + x = 90
∴a + x – a = 90 – a
….(Subtracting a from both sides)
∴x = (90 – a)
∴The complementary angle of a° is (90 – a)°.

iii. Let the measure of the complementary angle be x°.
22 + x = 90
∴22 + x – 22 = 90 – 22
….(Subtracting 22 from both sides)
∴x = 68
∴The complementary angle of 22° is 68°.

iv. Let the measure of the complementary angle be a°.
40 – x + a = 90
∴40 – x + a – 40 + x = 90 – 40 + x
….(Subtracting 40 and adding x on both sides)
∴a = (50 + x)
∴The complementary angle of (40 – x)° is (50 + x)°.

Question 14.
Find the measures of the supplements of the following angles:
i. 111°
ii. 47°
iii. 180°
iv. (90 – x)°
Solution:
i. Let the measure of the supplementary
angle be x°.
111 + x = 180
∴ 111 + x – 111 = 180 – 111
…..(Subtracting 111 from both sides)
∴ x = 69
∴ The supplementary angle of 111° is 69°.

ii. Let the measure of the supplementary angle be x°.
47 + x = 180
∴47 + x – 47 = 180 – 47
….(Subtracting 47 from both sides)
∴x = 133
∴The supplementary angle of 47° is 133°.

iii. Let the measure of the supplementary angle be x°.
180 + x = 180
∴180 + x – 180 = 180 – 180
….(Subtracting 180 from both sides)
∴x = 0
∴The supplementary angle of 180° is 0°.

iv. Let the measure of the supplementary angle be a°.
90 – x + a = 180
∴90 – x + a – 90 + x = 180 – 90+ x
….(Subtracting 90 and adding x on both sides)
∴a = 180 – 90 + x
∴a = (90 + x)
∴The supplementary angle of (90 – x)° is (90 + x)°.

Question 15.
Construct the following figures:
i. A pair of adjacent angles
ii. Two supplementary angles which are not adjacent angles.
iii. A pair of adjacent complementary angles.
Solution:
i.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 17

ii.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 18

iii.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 19

Question 16.
In ∆PQR the measures of ∠P and ∠Q are equal and m∠PRQ = 70°, Find the measures of the following angles.
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 20

  1. m∠PRT
  2. m∠P
  3. m∠Q

Solution:
Here, ∠PRQ and ∠PRT are angles in a linear pair.
m∠PRQ + m∠PRT = 180°
∴70 + m∠PRT = 180
∴m∠PRT = 180 – 70
∴m∠PRT = 110°
Now, ∠PRT is the exterior angle of ∆PQR.
∴m∠P + m∠Q = m∠PRT
∴m∠P + m∠P = m∠PRT ….(The measures of ∠P and ∠Q is same)
∴2m∠P = 110
∴m∠P = \(\frac { 110 }{ 2 }\)
∴m∠P = 55°
∴m∠Q =

Question 17.
Simplify
i. 54 × 53
ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Solution:
Simplify
i. 54 × 53
= 54+3
= 57

ii. \(\left(\frac{2}{3}\right)^{6} \div\left(\frac{2}{3}\right)^{9}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 21

iii. \(\left(\frac{7}{2}\right)^{8} \times\left(\frac{7}{2}\right)^{-6}\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 22

iv. \(\left(\frac{4}{5}\right)^{2} \div\left(\frac{5}{4}\right)\)
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 1 23

Question 18.
Find the value:
i. 1716  ÷ 1716
ii. 10-3
iii. (2³)²
iv. 46 × 4-4
Solution:
i. 1716  ÷ 1716
= 170
= 1

ii. 10-3
= \(\frac{1}{10^{3}}\)
= \(\frac{1}{1000}\)

iii. (2³)²
= 23×2
= 26
= 2 × 2 × 2 × 2 × 2 × 2
= 64

iv. 46 × 4-4
= 46+(-4)
= 42
= 4 × 4
= 16

Question 19.
Solve:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
ii. (3x + 2y) (7x – 8y)
iii. (7m – 5n) – (-4n – 11m)
iv. (11m – 12n + 3p) – (9m + 7n – 8p)
Solution:
i. (6a – 5b – 8c) + (15b + 2a – 5c)
= (6a + 2a) + (-5b + 15b) + (-8c – 5c)
= 8a + 10b – 13c

ii. (3x + 2y) (7x – 8y)
= 3x × (7x – 8y) + 2yx (7x – 8y)
= 21x² – 24xy + 14xy – 16y²
= 21x² – 10xy – 16y²

iii. (7m – 5n) – (-4n – 11m)
= 7m – 5n + 4n + 11m
= (7m + 11m) + (-5n + 4n)
= 18m – n

iv. (11m – 12n + 3p) – (9m + 7n – 8p)
= 11m – 12n + 3p – 9m – 7n + 8p
= (11m – 9m) + (-12n – 7n) + (3p + 8p)
= 2m – 19n + 11p

Question 20.
Solve the following equations:
i 4(x + 12) = 8
ii. 3y + 4 = 5y – 6
Solution:
i. 4(x + 12) = 8
∴4x + 48 = 8
∴4x + 48 – 48 = 8 – 48
….(Subtracting 48 from both sides)
∴ 4x = -40
∴ x = \(\frac { -40 }{ 4 }\)
∴ x = -10

ii. 3y + 4 = 5y – 6
∴ 3y + 4 + 6 = 5y – 6 + 6
….(Adding 6 on both sides)
∴ 3y + 10 = 5y
∴ 3y + 10 – 3y = 5y – 3y
….(Subtracting 3y from both sides)
∴ 10 = 2y
∴ 2y = 10
∴ y = \(\frac { 10 }{ 2 }\)
∴ y = 5

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.
Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?
Solution:
Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 2 1
∴ T = 4
∴ Angela had deposited the amount for 4 years.

Question 2.
Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?
Solution:
Let us suppose that 8 men require x days to tar the road.
Number of days required by 10 men to tar the road = 4
The number of men and the number of days required to tar the road are in inverse proportion.
∴ 8 × x = 10 x 4
∴ \(x=\frac{10 \times 4}{8}\)
∴ x = 5
∴ 8 men will require 5 days to tar the road.

Question 3.
Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?
Solution:
Total amount invested = Rs 40,000 + Rs 60,000
= Rs 1,00,000
Profit earned = 30%
∴ Total profit = 30% of 1,00,000
= \(\frac { 30 }{ 100 }\) × 100000
= Rs 30000
Proportion of investment = 40000 : 60000
= 2:3 …. (Dividing by 20000)
Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.
∴ 2x + 3x = 30000
∴ 5x = 30000
∴ x = \(\frac { 30000 }{ 5 }\).
∴ x = 6000
∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000
Mahesh’s profit = 3x = 3 × 6000 = Rs 18000
∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.
The diameter of a circle is 5.6 cm. Find its circumference.
Solution:
Diameter of the circle (d) = 5.6 cm
Circumference = πd
= \(\frac{22}{7} \times 5.6\)
= \(\frac{22}{7} \times \frac{56}{10}\)
= 17.6 cm
∴ The circumference of the circle is 17.6 cm.

Question 5.
Expand:
i. (2a – 3b)²
ii. (10 + y)²
iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)
iv. \(\left(y-\frac{3}{y}\right)^{2}\)
Solution:
i. Here, A = 2a and B = 3b
∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²
…. [(A – B)² = A² – 2AB + B²]
= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y
(10 + y)² = 102 + 2 × 10xy + y²
…. [(a + b)² = a² + 2ab + b²]
= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)
\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)
…. [(a + b)² = a² + 2ab + b²]
\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)
\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)
…. [(a – b)² = a² – 2ab + b²
= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.
Use a formula to multiply:
i. (x – 5)(x + 5)
ii. (2a – 13)(2a + 13)
iii. (4z – 5y)(4z + 5y)
iv. (2t – 5)(2t + 5)
Solution:
i. Here, a = x and b = 5
(x – 5)(x + 5) = (x)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= x² – 25

ii. Here, A = 2a and B = 13
(2a – 13)(2a + 13) = (2a)² – (13)²
…. [(A + B)(A – B) = A² – B²]
= 4a² – 169

iii. Here, a = 4z and b = 5y
(4z – 5y)(4z + 5y) = (4z)² – (5y)²
…. [(a + b)(a – b) = a² – b²]
= 16z² – 25y²

iv. Here, a = 2t and b = 5
(2t – 5)(2t + 5) = (2t)² – (5)²
…. [(a + b)(a – b) = a² – b²]
= 4t² – 25

Question 7.
The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?
Solution:
Diameter of the wheel (d) = 1.05 m
∴ Distance covered in 1 rotation of wheel = Circumference of the wheel
= πd
= \(\frac{22}{7} \times 1.05\)
= 3.3 m
∴ Distance covered in 1000 rotations = 1000 x 3.3 m
= 3300 m
= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]
= 3.3 km
∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.
The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.
Solution:
Length of the rectangular garden = 40 m
Area of the rectangular garden = 1000 sq. m.
∴ length × breadth = 1000
∴ 40 × breadth = 1000
∴ breadth = \(\frac { 1000 }{ 40 }\)
= 25 m
Now, perimeter of the rectangular garden = 2 × (length + breadth)
= 2 (40 + 25)
= 2 × 65
= 130 m
Length of one round of fence = circumference of garden – width of the entrance
= 130 – 4
= 126 m
∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3
= 378 m
∴ Total cost of fencing = Total length of fencing × cost per metre of fencing
= 378 × 250
= 94500
∴ The cost of fencing the garden is Rs 94500.

Question 9.
From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.
Solution:
Maharashtra Board Class 7 Maths Solutions Miscellaneous Problems Set 2 2
In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20
∴ According to Pythagoras’ theorem,
∴ l(AC)² = l(BC)² + l(AB)²
∴ l(AC)² = 21² + 20²
∴ l(AC)² = 441 + 400
∴ l(AC)² = 841
∴ l(AC)² = 29²
∴ l(AC) = 29
Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)
= 20 + 21 + 29
= 70
∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.
If the edge of a cube is 8 cm long, find its total surface area.
Solution: ,
Total surface area of the cube = 6 × (side)²
= 6 × (8)²
= 6 × 64
= 384 sq. cm
The total surface area of the cube is 384 sq.cm.

Question 11.
Factorize: 365y4z3 – 146y2z4
Solution:
= 365y4z3 – 146y2z4
= 73 (5y4z3 – 2y2z4)
= 73y2 (5y2z3 – 2z4)
= 73y2z3(5y2 – 2z)

Practice Set 1 Class 7 Answers Chapter 1 Geometrical Constructions Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 1 Solutions Maharashtra Board

Std 7 Maths Practice Set 1 Solutions Answers

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 1
Line AB is the perpendicular bisector of seg PQ.

ii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 2
Line UV is the perpendicular bisector of seg ST.

iii.
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 3
Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 4

ii. 55°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 5

iii. 90°
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 6

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 7
The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 8
The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

  1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
  2. Is l(PM) = l(SM)?

Solution:
Maharashtra Board Class 7 Maths Solutions Chapter 1 Geometrical Constructions Practice Set 1 9

  1. Here, m∠CMS = 90°
  2. Also, l(PM) = l(SM) = 2cm
    ∴ line CD is the perpendicular bisector of seg PS.

Std 7 Maths Digest

Practice Set 24 Class 7 Answers Chapter 5 Operations on Rational Numbers Maharashtra Board

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 24 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 24 Solutions Maharashtra Board

Std 7 Maths Practice Set 24 Solutions Answers

Question 1.
Write the following rational numbers in decimal form.
i. \(\frac { 13 }{ 4 }\)
ii. \(\frac { -7 }{ 8 }\)
iii. \(7\frac { 3 }{ 5 }\)
iv. \(\frac { 5 }{ 12 }\)
v. \(\frac { 22 }{ 7 }\)
vi. \(\frac { 4 }{ 3 }\)
vii. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 13 }{ 4 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 1

ii. \(\frac { -7 }{ 8 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 2

iii. \(7\frac { 3 }{ 5 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 3

iv. \(\frac { 5 }{ 12 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 4

v. \(\frac { 22 }{ 7 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 5

vi. \(\frac { 4 }{ 3 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 6

vii. \(\frac { 7 }{ 9 }\)
Maharashtra Board Class 7 Maths Solutions Chapter 5 Operations on Rational Numbers Practice Set 24 7

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 24 Intext Questions and Activities

Question 1.
Without using division, can we tell from the denominator of a fraction, whether the decimal form of the fraction will be a terminating decimal? Find out. (Textbook pg. no. 40)
Solution:
If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.
Consider the fractions \(\frac { 17 }{ 20 }\) and \(\frac { 19 }{ 6 }\)
Now, 20 = 2 x 2 x 5, and 6 = 2 x 3
∴ \(\frac { 17 }{ 20 }\) is terminating decimal form while \(\frac { 19 }{ 6 }\) is recurring decimal form.

Practice Set 4 Class 6 Answers Maths Chapter 3 Integers Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 4 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 4 Solutions Maharashtra Board

Std 6 Maths Practice Set 4 Solutions Answers

Question 1.
Classify the following numbers as positive numbers and negative numbers.
-5, +4, -2, 7, +26, -49, -37, 19, -25, +8, 5, -4, -12, 27
Solution:

Positive Numbers +4, 7, +26, 19, +8, 5, 27
Negative Numbers -5, -2, -49, -37, -25, -4, -12

Question 2.
Given below are the temperatures in some cities. Write them using the proper signs.

Place Shimla Leh Delhi Nagpur
Temperature 7 °C below 0° 12 °C below 0° 22 °C above 0° 31 °C above 0°

Solution:

Place Shimla Leh Delhi Nagpur
Temperature with proper sign -7 °C -12 °C +22 °C +31 °C

Question 3.
Write the numbers in the following examples using the proper signs.

  1. A submarine is at a depth of 512 meters below sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas, is 8848 meters.
  3. A kite is flying at a distance of 120 meters from the ground.
  4. The tunnel is at a depth of 2 meters under the ground.

Solution:

  1. A submarine is at a depth of -512 meters from sea level.
  2. The height of Mt Everest, the highest peak in the Himalayas is +8848 meters.
  3. A kite is flying at a distance of +120 meters from the ground.
  4. The tunnel is at a depth of -2 meters from the ground.

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 4 Intext Questions and Activities

Question 1.
Take warm water in one beaker, some crushed ice in another and a mixture of salt and crushed ice in a third beaker. Ask your teacher for help in measuring the temperature of the substance in each of the beakers using a thermometer. Note the temperatures. (Textbook pg. no. 13)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 1
Solution:
( Students should attempt this activity on their own)

Question 2.
Look at the picture of the kulfi man. Why do you think he keeps the kulfi moulds in a mixture of salt and ice? (Textbook pg. no. 14)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 2
Solution:
Kulfi man keeps the kulfi moulds in a mixture of salt and ice because such a mixture helps in keeping the kulfi cool for a longer period of time. The kulfi kept in the said mixture relatively takes more time to melt. This mixture is Considered ideal as it has the temperature of -4°C as against the temperature of ice i.e. 0°C.

Question 3.
My class, i.e. Std. VI, is a part of my school. My school is in my town. My town is a part of a taluka. In the same way, the taluka is a part of a district, and the district is a part of Maharashtra State. In the same way, what can you say about these groups of numbers? Textbook pg. no. 15)
Maharashtra Board Class 6 Maths Solutions Chapter 3 Integers Practice Set 4 3
Solution:
By observing the above given groups of numbers, we can infer that natural numbers are a part of whole numbers. In turn, whole numbers are a part of integers.

Std 6 Maths Digest

Maharashtra Board 8th Class Maths Miscellaneous Exercise 2 Solutions

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Miscellaneous Exercise 2 8th Std Maths Answers Solutions.

Miscellaneous Exercise 2 8th Std Maths Answers

Question 1.
Questions and their alternative answers are given. Choose the correct alternative answer.
i. Find the circumference of a circle whose area is 1386 cm²? [Chapter 15]
(A) 132 cm²
(B) 132 cm
(C) 42 cm
(D) 21 cm²
Solution:
(B) 132 cm

Hint:
i. Area of the circle = πr²
1386 = \(\frac { 22 }{ 7 }\) x r²
r² = 1386 x \(\frac { 7 }{ 22 }\)
= 63 x 7
= 441
r = √441 … [Taking square root of both sides]
= 21 cm
Circumference of the circle = 2πr
= 2 x \(\frac { 22 }{ 7 }\) x 21
= 132 cm

ii. The side of a cube is 4 m. If it is doubled, how many times will be the volume of the new cube, as compared with the original cube? [Chapter 16]
(A) Two times
(B) Three times
(C) Four times
(D) Eight times
Solution:
(D) Eight times

Hint:
ii. Original volume = (4)³ = 64 cu.m
New side = 8 m
∴ New volume = (8)² = 512 cu.m
Now, \(\frac{\text { new volume }}{\text { original volume }}=\frac{512}{64}\) = 8
original volume 64
∴ volume of new cube will increase 8 times as compared to the volume of original cube.

Question 2.
Pranalee was practicing for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows. [Chapter 11]
18, 17, 17, 16,15, 16, 15, 14,16, 15, 15, 17, 15, 16,15, 17, 16, 15, 14,15
Find the mean of the time taken for running.
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 1
∴ The mean of the time taken for running 100 m race is 15.7 seconds.

Question 3.
∆DEF and ∆LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles in the correspondence. [Chapter 13]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 2
∆EDF ≅ ∆LMN
∴side ED ≅ side LM
side DF ≅ side MN
side EF ≅ side LN
∠E ≅∠L
∠D ≅∠M
∠F ≅∠N

Question 4.
The cost of a machine is Rs 2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years. [Chapter 14]
Solution:
Here, P = Cost of the machine = Rs 2,50,000
R = Rate of depreciation = 4%
N = 3 Years
A = Depreciated price of the machine
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 3
∴The cost of the machine after three years will be Rs 2,21,184.

Question 5.
In ☐ABCD, side AB || side DC, seg AE ⊥ seg DC. If l(AB) = 9 cm, l(AE) = 10 cm, A(☐ABCD) = 115 cm² , find l(DC). [Chapter 15]
Solution:
Given, side AB || side DC.
∴ ☐ABCD is a trapezium.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 4
Given, l(AB) = 9 cm, l(AE) = 10 cm,
A(☐ABCD) = 115 cm²
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC) x l(AE)]
∴ 115 = \(\frac { 1 }{ 2 }\) x [9 + l(DC)] x 10
∴ \(\frac { 115 \times 2 }{ 10 }\) = 9 + l(DC)
∴ 23 = 9 + l(DC)
∴ l(DC) = 23 – 9
∴ l(DC) = 14cm

Question 6.
The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre?
[π = \(\frac { 22 }{ 7 }\)] [Chapter 16]
Solution:
Given: For cylindrical tank:
diameter (d) = 1.75 m, height (h) = 3.2 m
To Find: Capacity of tank in litre
diameter (d) = 1.75 m
= 1.75 x 100
….[∵ 1 m = 100cm]
= 175 cm
∴ radius (r) = \(=\frac{\mathrm{d}}{2}=\frac{175}{2}\) cm
h = 3.2 cm
= 3.2 x 100
= 320 cm
Capacity of tank = Volume of the cylindrical tank
= πr²h
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 5
∴ The capacity of the tank is 7700 litre.

Question 7.
The length of a chord of a circle is 16.8 cm, radius is 9.1 cm. Find its distance from the centre. [Chapter 17]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 6
Let CD be the chord of the Circle with centre O.
Draw seg OP ⊥ chord CD
∴l(PD) = \(\frac { 1 }{ 2 }\) l(CD)
…[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(PD) = \(\frac { 1 }{ 2 }\) x 16.8 …[l(CD) = 16.8cm]
∴l(PD) = 8.4 cm …(i)
∴In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² …..[Pythagoras theorem]
∴(9.1)² = [l(OP)]² + (8.4)² … [From (i) and l(OD) = 9.1 cm]
∴(9.1)² – (8.4)² = [l(OP)]²
∴(9.1 + 8.4) (9.1 – 8.4) = [l(OP)]²
…[∵ a² – b² = (a + b) (a – b)]
∴17.5 x (0.7) = [l(OP)]²
∴12.25 = [l(OP)]²
i.e., [l(OP)]² = 12.25
∴l(OP) = √12.25
…[Taking square root of both sides]
∴l(OP) = 3.5 cm
∴The distance of the chord from the centre is 3.5 cm.

Question 8.
The following tables shows the number of male and female workers, under employment guarantee scheme, in villages A, B, C and D.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110

i. Show the information by a sub-divided bar-diagram.
ii. Show the information by a percentage bar diagram. [Chapter 11]
Solution:
i.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110
Total 375 400 300 250

Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 7

ii.

Villages A B C D
No. of females 150 240 90 140
No. of males 225 160 210 110
Total 375 400 300 250
Percentage of females 40% 60% 30% 56%
Percentage of males 60% 40% 70% 44%

Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 8

Question 9.
Solve the following equations.
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
iii. 5(1 – 2x) = 9(1 -x)
[Chapter 12]
Solution:
i. 17 (x + 4) + 8 (x + 6) = 11 (x + 5) + 15 (x + 3)
∴ 17x + 68 + 8x + 48 = 11x + 55 + 15x + 45
∴ 17x + 8x + 68 + 48 = 11x + 15x + 55 + 45
∴ 25x + 116 = 26x + 100
∴ 25x + 116 – 116 = 26x + 100 – 116
… [Subtracting 116 from both the sides]
∴ 25x = 26x – 16
∴ 25x – 26x = 26x – 16 – 26x
… [Subtracting 26x from both the sides]
∴ -x = -16
∴ \(\frac{-x}{-1}=\frac{-16}{-1}\)
∴ x = 16

ii. \(\frac{3 y}{2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{3 y \times 2}{2 \times 2}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4}+\frac{y+4}{4}=5-\frac{y-2}{4}\)
∴ \(\frac{6 y}{4} \times 4+\frac{y+4}{4} \times 4=5 \times 4-\frac{y-2}{4} \times 4\)
……[Multiplying both the sides by 4]
∴ 6y + y + 4 = 20 – (y – 2)
∴ 7y + 4 = 20 – y + 2
∴ 7y + 4 = 22 – y
∴ 7y + 4 – 4 = 22 – y – 4
…..[Subtracting 4 from both the sides]
∴ 7y = 18 – y
∴ 7y + y = 18 – y + y
…[Adding y on both the sides]
∴ 8y = 18
∴ \(\frac{8 y}{8}=\frac{18}{8}\) … [Dividing both the sides by 8]
∴ \(y=\frac { 9 }{ 4 }\)

iii. 5(1 – 2x) = 9(1 – x)
∴ 5 – 10x = 9 – 9x
∴ 5 – 10x – 5 = 9 – 9x – 5
….[Subtracting 5 from both the sides]
∴ -10x = 4 – 9x
∴ -10x + 9x = 4 – 9x + 9x
… [Adding 9x on both the sides]
∴ -x = 4
∴ -x x (- 1) = 4 x (- 1)
… [Multiplying both the sides by – 1]
∴ x = – 4

Question 10.
Complete the activity according to the given steps.
i. Draw rhombus ABCD. Draw diagonal AC.
ii. Show the congruent parts in the figure by identical marks.
iii. State by which, test and in which correspondence ∆ADC and ∆ABC are congruent.
iv. Give reason to show ∠DCA ≅ ∠BCA, and ∠DAC ≅ ∠BAC
v. State which property of a rhombus is revealed from the above steps. [Chapter 13]
Solution:
a.
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 9
b. In ∆ADC and ∆ABC,
side AD ≅ side AB …..[Sides of a rhombus]
side DC ≅ side BC …..[Sides of a rhombus]
side AC ≅ side AC … [Common side]
∆ADC ≅ ∆ABC … [By SSS test]
∠DCA ≅ ∠BCA …[Corresponding angles of congruent triangles]
∠DAC ≅ ∠BAC …[Corresponding angles of congruent triangles]
From the above steps, property of rhombus revealed is ‘diagonal of a rhombus bisect the opposite angles’.

Question 11.
The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. l(PQ) = 170 m,
l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m, l(PR) = 260 m.
Find the area of the field in hectare (1 hectare = 10,000 sq.m). [Chapter 15]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 10
Area of the field = A(∆PQR) + A(∆PSR)
In ∆PQR, a = 170 m, b = 250 m, c = 260 m
Semiperimeter of ∆PQR = s
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 11
Area of the field = A(∆PQR) + A(∆PSR)
= 20400 + 12000
= 32400 sq.m
= \(\frac { 32400 }{ 10000 }\)
…[1 hectare = 10,000 sq.m]
= 3.24 hectares
∴ The area of the field is 3.24 hectares.

Question 12.
In a library, 50% of total number of books is of Marathi. The books of English are \(\frac { 1 }{ 3 }\) of Marathi books. The books on Mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library? [Chapter 12]
Solution:
Let the total number of books in the library be x
50% of total number of books is of Marathi.
Number of Marathi books = 50% of x
= \(\frac { 50 }{ 100 }x\)
= \(\frac { x }{ 2 }\)
The books of English are \(\frac { 1 }{ 3 }\) of Marathi books.
Number of books of English = \(\frac{1}{3} \times \frac{x}{2}\)
= \(\frac { x }{ 6 }\)
The books on Mathematics are 25% of the English books.
Number of books of Mathematics
= 25% of \(\frac { x }{ 6 }\)
= \(\frac{25}{100} \times \frac{x}{6}\)
= \(\frac { x }{ 24 }\)
Since, there are 560 books of other subjects, the total number of books in the library are
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 12
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 13
∴ 24x – 17x = 17x + 13440 – 17x
∴ 7x = 13440
∴ \(\frac{7 x}{7}=\frac{13440}{7}\)
∴ x = 1920
∴ The total number of books in the library are 1920.

Question 13.
Divide the polynomial (6x³ + 11x² – 10x – 7) by the binomial (2x + 1). Write the quotient and the remainder. [Chapter 10]
Solution:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 14
∴ Quotient = 3x² + 4x – 7,
remainder = 0
Explanation:
Maharashtra Board Class 8 Maths Solutions Miscellaneous Exercise 2 15

Maharashtra Board Class 8 Maths Solutions

Work and Energy Class 9 Science Chapter 2 Questions And Answers Maharashtra Board

Class 9 Science Chapter 2

Balbharti Maharashtra State Board Class 9 Science Solutions Chapter 2 Work and Energy Notes, Textbook Exercise Important Questions and Answers.

Std 9 Science Chapter 2 Work and Energy Question Answer Maharashtra Board

Class 9 Science Chapter 2 Work and Energy Question Answer Maharashtra Board

1. Write detailed answers?

a. Explain the difference between potential energy and kinetic energy.
Answer:

Kinetic Energy Potential Energy
(i) Kinetic energy is the energy possessed by the body due to its motion. (i) Potential energy is the energy possessed by the body because of its shape or position.
(ii) K.E = 1/2 mv2 (ii) P.E = mgh
(iii) e.g., flowing water, such as when falling from a waterfall. (iii) e.g., water at the top of a waterfall, before the drop.

b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Answer:
Suppose a stationary object of mass ‘m’ moves because of an applied force. Let ‘u’ be its initial velocity (here u = 0). Let the applied force be ‘F’. This generates an acceleration a in the object, and after time T, the velocity of the object becomes equal to ‘v’. The displacement during this time is s. The work done on the object is
W = F x s ……………….. (1)
Using Newton’s 2nd law of motion,
F = ma ……………….. (2)
Using Newton’s 2nd equation of motion
\(s=u t+\frac{1}{2} a t^{2}\)
However, as initial velocity is zero, u = 0
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 1

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Answer:
Let us look at the kinetic and potential energies of an object of mass (m), falling freely from height (h), when the object is at different heights.

As shown in the figure, the point A is at a height (h) from the ground. Let the point B be at a distance V, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B and C.

(1) Let the velocity of the object be vB when it reaches point B, having fallen through a distance x.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 2
(2) When the object is stationary at A, its initial velocity is u = 0
∴ K.E = 1/2 mass x velocity2
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 3

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

(3) Let the velocity of the object be vc when it reaches the ground, near point C.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 4
From equations (i) and (iii) we see that the total potential energy of the object at its initial position is the same as the kinetic energy at the ground.

d. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.
Answer:
When an object is displaced by displacement ‘s’ and by applying force ‘F’ at an ’angle’ 30°. work done can be given as
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 25

e. If an object has 0 momenta, does it have kinetic energy? Explain your answer.
Answer:

  • No, it does not have kinetic energy if it does not have momentum.
  • Momentum is the product of mass and velocity. If it is zero, it implies that v = 0 (since mass can never be zero).
  • Now K.E = ~ mv2, So if v = 0 then K.E also will be zero.
  • Thus, if an object has no momentum then it cannot possess kinetic energy.

f. Why is the work done on an object moving with uniform circular motion zero?
Answer:

  • In uniform circular motion, the force acting on an object is along the radius of the circle.
  • Its displacement is along the tangent to the circle. Thus, they are perpendicular to each other.
    Hence θ = 90° and cos 90 = θ
    ∴ W = Fs cos θ = 0

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

2. Choose one or more correct alternatives.

a. For work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed

b. Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy

c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction

d. Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time

e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force

3. Rewrite the following sentences using a proper alternative.

a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer:
(iii) sleeping on the ground

b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer:
(iii) increases

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer:
(ii) will not change

d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer:
(iii) initial velocity of the object

4. Study the following activity and answer the questions.

1. Take two aluminium channels of different lengths.
2. Place the lower ends of the channels on the floor and hold their upper ends at the same height.
3. Now take two balls of the same size and weight and release them from the top end of the channels. They will roll down and cover the same distance.

Questions
1. At the moment of releasing the balls, which energy do the balls have?
2. As the balls roll down which energy is converted into which other form of energy?
3. Why do the balls cover the same distance on rolling down?
4. What is the form of the eventual total energy of the balls?
5. Which law related to energy does the above activity demonstrate? Explain.
Answer:
1. At the moment of releasing the ball they possess Potential energy as they are at a height above the ground.
2. As the balls roll down, the Potential energy is converted into Kinetic energy since they are now in motion.
3. Since they have been released from the same height, they will cover the same distance.
4. The eventual form of the total energy of the balls is “Mechanical Energy” i.e, a combination of Potential energy and Kinetic energy
5. The above activity demonstrates the “Law of Conservation of Energy”

5. Solve the following examples.

a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Answer:
Given:
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 min = 60 s
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Mass of water (m)= ?
Formula:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 5
Water lifted by the pump is 1224.5 kg

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

b. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Answer:
Given: Initial height (h1) = 10m
Let Initial (P.E1) = 100
Final (P.E2) = 100 – 40
= 60

To Find:
Final height (h2) = ?
Formula:
P.E. = mgh
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 6
The ball will rebound by 6 m.

d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg? (Ans. : 131250 J)
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 23
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 24
Work done to increase the velocity = 131250 J

e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Answer:
Given:
Force (F) = 10 N
θ = 0°, (Since force and displacement are in same direction)
Displacement (s) = 30 cm = 30/100 m
To Find:
Work (W) = ?
Formula:
W = Fs cos θ
Solution:
W = Fs cos θ
Solution:
The work done by Ravi is 3J
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 7
Numericals For Practice

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Class 9 Science Chapter 1 Laws of Motion Intext Questions and Answers

Question 1.
What are different types of forces? Give examples.
Answer:
Forces are of two types.

  • Contact force e.g.: Mechanical force, frictional force, muscular force
  • Non-contact force e.g.: gravitational force, magnetic force, electrostatic force

Question 2.
Monashee wants to displace a wooden block from point A to point B along the surface of a table as shown. She has used force F for the purpose.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 22
(a) Has all the energy she spent been used to produce an acceleration in the block?
(b) Which forces have been overcome using that energy?
Answer:
(a) Only part of the energy applied by Minakshee is used in accelerating the block.
(b) Force of friction has been overcome using the energy.

Question 3.
Mention the type of energy used in the following examples.
(i) Stretched rubber string.
(ii) Fast-moving car.
(iii) The whistling of a cooker due to steam.
(iv) A fan running on electricity.
(v) Drawing out pieces of iron from garbage, using a magnet.
(vi) Breaking of a glass window pane because of a loud noise.
(vii) The drackers exploded in Diwali.
Answer:
(i) Potential energy
(ii) Kinetic energy
(iii) Sound energy
(iv) Electrical energy
(v) Magnetic energy
(vi) Sound energy
(vii) Sound energy, light energy and heat energy

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 4.
Study the pictures given below and answer the questions:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 26
(a) In which of the pictures above has work been done?
(b) From scientific point of view, when do we say that no work was done?
Answer:
(a) Girl studying : No work done
Boy playing with ball: Work is done
Girl watching T.V.: No work done Person lifting sack of grains : Work is done
(b) No work is said to be done when force is applied but there is no displacement.

Question 5.
Make two pendulums of the same length with the help of thread and two nuts. Tie another thread in the horizontal position.

Tie the two pendulums to the horizontal thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and observe. What do you see?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 27
Answer:
You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other.

Question 6.
Ajay and Atul have been asked to determine the potential energy of a ball of mass m kept on a table as shown in the figure. What answers will they get? Will they be different? What do you conclude from this?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 28
Answer:

  • According to Ajay P.E1 = mgh1 and according to Atul P.E2 = mgh2.
  • Yes, the answer will be different as the two heights are different.
  • Potential energy is relative.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 7.
Discuss the directions of force and of displacement in each of the following cases.
(i) Pushing a stalled vehicle.
(ii) Catching the ball which your friend has thrown towards you.
(iii) Tying a stone to one end of a string and swinging it round and round by the other end of the string.
(iv) Walking up and down a staircase; climbing a tree.
(v) Stopping a moving car by applying brakes.
Answer:
(i) Force and displacement are in the same direction.
(ii) Force and displacement are in the opposite direction.
(iii) Force and displacement are perpendicular to each other.
(iv) Force and displacement are in the opposite direction.
(v) Force and displacement are in the opposite direction.

Question 8.
(A) An arrow is released from a stretched bow.
(B) Water kept at a high flows through a pipe into the tap below.
(C) A compressed spring is released.
(a) Which words describe the state of the object in the above examples?
(b) Where did the energy required to cause the motion of the objects come from?
(c) If the obj ects were not brought in those states, would they have moved?
Answer:
(a) Words such as stretched bow, water kept at a height and compressed spring describe the state of the objects.
(b) The energy required for the objects came from its specific state or motion in the form of potential energy.
(c) No, if the objects were not brought in those states, they would have not moved.

Question 9.
Study the activity and answer the following questions.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 29
(a) Figure A – Why does the cup get pulled?
(b) Figure B – What is the relation between the displacement of the cup and the force applied through the ruler?
(c) In Figure C-Why doesn’t the cup get displaced?
(d) What is the type of work done in figures A, B and C?
(e) In the three actions above, what is the relationship between the applied force and the displacement?
Answer:
(a) The cup gets pulled as the force of the nut and the displacement of the cup is in the same direction.
(b) The displacement of the cup and the force applied through the ruler is in the opposite direction.
(c) Tire cup does not get displaced as two equal forces are working in opposite directions.
(d) The work done in figure A is positive, figure B is negative and in figure C is zero.
(e) In figure A the applied force and the displacement is in the same direction, in figure B the applied force and the displacement is in the opposite direction and in figure C the applied force and displacement is perpendicular to each other.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 10.
From the following activities find out whether work is positive, negative or zero. Give reasons for your answers.
(a) A boy is swimming in a pond.
(b) A coolie is standing with a load on his head.
(c) Stopping a moving car by applying brakes.
(d) Catching the ball which you friend has thrown towards you.
Answer:
(a) A boy is swimming in a pond: The work done is positive because the direction of applied force and displacement are the same.
(b) A coolie is standing with a load on his head: The work done is zero because the applied force does not cause any displacement.
(c) Stopping a moving car by applying brakes: The work done is negative because the fore applied by the brakes acts in a direction opposite to the direction of motion of car.
(d) Catching the ball which you friend has thrown towards you : Negative work because the force required to stop the ball, acts opposite to the displacement of the ball.

Question 11.
(a) Can your father climb stairs as fast as you can?
(b) Will you fill the overhead water tank with the help of a bucket or an electrical motor?
(c) Suppose Raj ashree, Yash and Ranjeet have to reach the top of a small hill. Raj ashree went by car. Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last? (Think before you answer.
Answer:
(a) No, father takes more time to climb stairs.
(b) Overhead water tank can be filled with the help of one electric motor rather than filling it with bucket.
(c) Raj ashree will reach first, followed by Yash and Ranjeet will reach last because car moves faster than a cycle and a person walking.

Class 9 Science Chapter 1 Laws of Motion Additional Important Questions and Answers

1. Choose and write the correct option:

Question 1.
Forces are of …………………… types.
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Example of Contact force is ………………….. .
(a) Gravitational Force
(b) Magnetic Force
(c) Electrostatic Force
(d) Muscular Force
Answer:
(d) Muscular Force

Question 3.
Example of Non-contact force is ………………….. .
(a) Mechanical Force
(b) Frictional Force
(c) Muscular Force
(d) Electrostatic Force
Answer:
(d) Electrostatic force

Question 4.
Work is said to be done on a body when a …………………… is applied on object causes displacement of the object.
(a) Direction
(b) Area
(c) Volume
(d) Force
Answer:
(d) force

Question 5.
W = ………………. .
(a) mgh
(b) mdh
(c) mv2
(d) mfe
Answer:
(a) mgh

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 6.
The energy stored in the dry cell is in of ………………. energy.
(a) Light
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(b) chemical

Question 7.
The work done is zero if there is no ……………… .
(a) Direction
(b) Displacement
(c) Mass
(d) Angle
Answer:
(b) displacement

Question 8.
Flowing water has ………………. energy.
(a) Potential
(b) Chemical
(c) Solar
(d) Kinetic
Answer:
(d) kinetic

Question 9.
By stretching the rubber strings of a we store ………………. energy in it.
(a) Potential
(b) Chemical
(c) Electric
(d) Kinetic
Answer:
(a) potential

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 10.
………………. is the unit of force.
(a) Both B and C
(b) Newton
(c) Dyne
(d) Volts
Answer:
(a) Both B and C

Question 11.
For a freely falling body, kinetic energy is ………………. at the ground level.
(a) Maximum
(b) Minimum
(c) Neutral
(d) Reversed
Answer:
(a) Maximum

Question 12.
Energy can neither be ………………. nor ……………… .
(a) Destroyed
(b) Created
(c) Saved
(d) Both A and B
Answer:
(d) Both A and B

Question 13.
Work and …………………… have the same unit.
(a) Energy
(b) Electricity
(c) Force
(d) Both B and C
Answer:
(a) Energy

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 14.
S.I. unit of energy is ………………….. .
(a) Joule
(b) Ergs
(c) m/s2
(d) Both A and B
Answer:
(a) Joule

Question 15.
Work is the product of ………………….. .
(a) force and distance
(b) displacement and velocity
(c) kinetic and potential energy
(d) force and displacement
Answer:
(d) force and displacement

Question 16.
S.I. unit of work is ………………….. .
(a) dyne
(b) newton-meter or erg
(c) N/m2 or joule
(d) newton-meter or joule
Answer:
(d) newton-meter or joule

Question 17.
…………………… is the capacity to do work.
(a) Energy
(b) Force
(c) Power
(d) Momentum
Answer:
(a) Energy

Question 18.
Kinetic energy of a body (KE) = ………………….. .
(a) mv2
(b) 1/2 mv2
(c) mgh
(d) Fs
Answer:
(b) 1/2 mv2

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 19.
Potential energy of a body is given by (P.E.) = ………………….. .
(a) Fs
(b) mgh
(c) ma
(d) mv2
Answer:
(b) mgh

Question 20.
1 hp = ………………….. .
(a) 476 watts
(b) 746 watts
(c) 674 watts
(d) 764 watts
Answer:
(b) 746 watts

Question 21.
…………………… is the commercial unit of power.
(a) kilowatt second
(b) dyne
(c) kilowatt
(d) erg
Answer:
(c) kilowatt

Question 22.
1 kWh = …………………… joules.
(a) 3.6 x 103
(b) 3.6 x 106
(c) 6.3 x 106
(d) 6.3 x 103
Answer:
(b) 3.6 x 106

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Based on Practicals

Question 23.
The work done by a force is said to be …………………… when the applied force does not produce displacement.
(a) positive
(b) negative
(c) zero
(d) none of these
Answer:
(c) zero

Question 24.
When some unstable atoms break up, they release a tremendous amount of …………………… energy.
(a) chemical
(b) potential
(c) nuclear
(d) mechanical
Answer:
(c) nuclear.

Name the following:

Question 1.
Unit of energy used for commercial purpose.
Answer:
Kilowatt-hour kW h is the unit of energy used for commercial purpose.

Question 2.
Unit used in industry to measure power.
Answer:
Horse power (hp) is the unit used in industry to express power.

Question 3.
SI unit of energy.
Answer:
SI unit of energy is Joule (J).

Question 4.
Two types of mechanical energy.
Answer:
Potential energy and kinetic energy are the two types of mechanical energy.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
An example where force acting on an object does not do any work.
Answer:
In a simple pendulum, the gravitational force acting on the bob does not do any work as there is no displacement in the direction of force.

Question 6.
The relationship between 1 joule and 1 erg.
Answer:
1 joule = 107 erg.

Question 7.
Various forms of energy
Answer:
The various forms of energy are mechanical, heat, light, sound, electro-magnetic, chemical, nuclear and solar.

State whether the following statements are true or false:

(1) The potential energy of a body of mass 1 kg kept at height 1 m is 1 J.
(2) Water stored at some height has potential energy.
(3) Unit of power is joule.
(4) Mechanical energy can be converted into electrical energy.
(5) Work is a vector quantity.
(6) Power is a scalar quantity.
(7) The kilowatt hour is the unit of energy.
(8) The CGS unit of energy is dyne.
(9) The SI unit of work is newton.
(10) Kinetic energy has formula – mv2
Answer:
(1) False
(2) True
(3) False
(4) True
(5) False
(6) True
(7) True
(8) False
(9) False
(10) True

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Find the odd man out.

Question 1.
Work, Energy, Power, Force.
Answer:
Force.

Question 2.
A stretched spring, A body placed in at some height, A bullet fired from gun.
Answer:
A bullet fired from gun.

Question 3.
A stretched spring, A rock rolling downhill, A bullet fired from gun.
Answer:
A stretched spring.

Write the formula of the following.

Question 1.
Kinetic energy
Answer:
\(\frac{1}{2}\)mv2

Question 2.
Potential energy
Answer:
mgh

Question 3.
Work
Answer:
Fs or Fs cosθ

Question 4.
Force
Answer:
ma

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
Power
Answer:
\(\frac{w}{1}\)

One line answer.

Question 1.
(i) When is work done said to be zero?
Answer:
Work done is zero when force acting on the body and its displacement are perpendicular to each other.

(ii) Which quantities are measured in ergs?
Answer:
Work and energy are measured in ergs.

(iii) What is the relationship between newton, meter and joule?
Answer:
1 joule = 1 newton x 1 meter

(iv) What is energy?
Answer:
The ability of a body to do work is called energy.

(v) Give 4 examples of energy
Answer:
Solar, wind, mechanical and heat.

(vi) Which device converts electrical energy into heat?
Answer:
Electric water heater (Geyser) converts electrical energy into heat.

(vii) What is the relationship between second, horsepower and joule?
Answer:
1 horse power = \(\frac{746 \text { joules }}{1 \text { second }\)

Question 2.
Find whether work is positive, negative or zero.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

(a) Person moving along circle from A to B.
Answer:
Work done is positive as direction of applied force and displacement are the same.

(b) Person completing one circle and returns to position A.
Answer:
Work done is zero because there is no displacement for the person.

(c) Person pushing a car in the forward direction.
Ans,
Work done is positive as the motion of car is in the direction of the applied force.

(d) A car coming downhill even after pushing it in the opposite uphill direction.
Ans,
Work done is negative as the motion of car is in opposite direction of the applied force.

(e) Motion of the clock pendulum.
Answer:
work done is zero as there is no displacement of the pendulum and it comes back to its original position.

Give Scientific reasons:

Question 1.
A moving ball hits a stationary ball and displaces it.
Answer:

  • The moving ball has certain energy.
  • When it hits the stationary ball, the energy is transferred to the stationary ball, because of which it moves.
  • Hence, a moving ball hits a stationary ball and displaces it.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Flowing water from some height can rotate turbine.
Answer:

  • Flowing water has certain energy.
  • When it hits the turbine, energy is transferred to the turbine, because of which it rotates.
  • Hence, flowing water from some height can rotate a turbine.

Question 3.
A stretched rubber band when released regains its original length.
Answer:

  • When we stretch a rubber band we give energy to it.
  • This energy is stored in it.
  • Hence, when we release it, it regains its original length.

Question 4.
Wind can move the blades of a windmill.
Answer:

  • Wind has certain energy.
  • When it hits the windmill energy is transferred to the windmill because of which it moves.
  • Hence, wind can move the blades of a wind mill.

Question 5.
An exploding firecracker lights up as well as makes a sound.
Answer:

  • The exploding firecracker converts the chemical energy stored in it into light and sound respectively.
  • Here, energy is converted from one type to another.
  • Hence, an exploding firecracker lights as well as makes a sound.

Question 6.
Work done on an artificial satellite by gravity is zero while moving around the earth.
Answer:

  • When the artificial satellite moves around the earth in a circular orbit, gravitation force acts on it.
  • The gravitational force acting on the satellite and its displacement are perpendicular to each other. i.e. 0 = 90°
  • For 0 = 90°, work done is zero. [ v cos 90 = 0)
  • Hence, work done on an artificial satellite by gravity is zero while moving around the earth.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Difference between :

Question 1.
Work and Power:
Answer:

Work Power
(i) Work is the product of force and displacement.
(ii) Work is given by the formula : W = Fs
(iii) MKS unit – joule, CGS unit-erg
(i) Power is the rate of doing work.
(ii) Power is given by the formula : \(\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}\)
(iii) MKS unit – joule/sec, CGS unit – erg/sec

Question 2.
Work and Energy:
Answer:

Work Energy
(i) It is the product of the magnitude of the force acting on the body and the displacement of the body in the direction of the force.
(ii) It is the effect of energy.
(i) It is the capacity to do work.
(ii) It is the cause of work.

Solve the following:

Type – A

Formula:
W = Fs cosθ
If force and displacement are in same direction, then θ = 0°, and cos θ = 1
If force and displacement are in opposite direction, then θ = 180°, and cos θ = -1
If force and displacement are perpendiculars, then θ = 90°, and cos θ = 0

Question 1.
Pravin has applied a force of 100 N on an object, at an angle of 60° to the horizontal. The object gets displaced in the horizontal direction and 400 J work is done. What is the displacement of the object? (cos 600 =12)
To Find:
Displacement (s) = ?
Formula:
W = Fs cos θ
Solution:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 8
The object will be displaced through 8 m.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
A force of 50 N acts on an object and displaces it by 2 m. If the force acts at an angle of 60° to the direction of its displacement, find the work done.
Answer:
50 J

Question 3.
Raj applied a force of 20 N and moved a book 40 cm in the direction of the force. How much was the work done by Raj?
Answer:
8J

Type -B

Formula:
1) W = K.E = 1/2 mv2
2) W = P.E = mgh
• W = P.E, W = K.E
1 km/hr =
\(\frac{1000}{3600} \mathrm{~m} / \mathrm{s}=\frac{5}{18} \mathrm{~m} / \mathrm{s}\)

Question 4.
A stone having a mass of 250 gm is falling from a height. How much kinetic energy does it have at the moment when its velocity is 2 m/s?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 9
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 10
The kinetic energy of the stone is 0.5 J

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 5.
500 kg water is stored in the overhead tank of a 10 m high building. Calculate the amount of potential energy stored in the water.
Answer:
Given:
Mass (m) = 500 kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Potential energy (P.E) = ?
Formula:
P.E = mgh
Solution:
P.E = mgh
= 500 x 9.8 x 10
= 500 x 98
= 49000J
The P.E of the stored water is 49000 J

Question 6.
Calculate the work done to take an object of mass 20 kg to a height of 10 m. (g = 9.8 m/s2)
Answer:
Given:
Mass (m) = 20 kg
Acceleration due to gravity (g) = -9.8 m/s2
Displacement (s) = (h) = 10 m.
To Find:
Work done (W) = ?
Formula:
(i) W = P.E = mgh
Solution:
W = mgh
= 20 x (-9.8) x 10
= -1960J
The work done to take an object of mass 20 kg to a height of 10 m is -1960 J.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 7.
A body of 0.5 kg thrown upwards reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement.
Answer:
Given:
Mass (m) = 0.5 kg
Acceleration due to gravity (g) = -9.8 m/s2
Displacement (s) = 5 m.
To Find:
Work done (W) = ?
Formula:
W = P.E = mgh
Solution:
W = mgh
= 0.5 x (-9.8) x 5
= -24.5 J
The work done by the force of gravity is -24.5 joule.

Question 8.
1 kg mass has a kinetic energy of 2 joule. Calculate its velocity.
Answer:
Given:
Mass (m) = 1 kg
Kinetic Energy (K.E) = 2 J
To Find:
Velocity (v) = ?
Formula:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 11
The velocity is 2 m/s

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 9.
A rocket of mass 100 tonnes is propelled with a vertical velocity 1 km/s. Calculate kinetic energy.
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 12
The kinetic energy of the rocket is 5 x 1010 J

Type – C

Formula:
\(\text { 1) Power }=\frac{\text { work }}{\text { time }}=\frac{\text { mgh }}{t}\)
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 13
Power should be expressed in kW
Time should be expressed in hours
1 k Wh = 1 unit

Question 10.
Swaralee takes 20 s to carry a bag weighing 20 kg to a height of 5 m. How much power has she used?
Given:
Mass (m) = 20 kg
Height (h) = 5 m
Time (t) = 20s
Acceleration due to gravity (g) = 9.8 m/s2
To Find:
Power (P) = ?
Formula:
\(\mathrm{P}=\frac{\mathrm{mgh}}{\mathrm{t}}\)
Solution:
\(\begin{aligned}
P &=\frac{m g h}{t} \\
&=20 \times 9.8 \times \frac{5}{20} \\
&=9.8 \times 5
\end{aligned}\)
= 49 W
Power used by Swaralee is 49 W

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Write notes on the following:

Question 1.
Derive the expression for potential energy.
Answer:
(i) To carry an object of mass ‘m’ to a height ‘h’ above the earth’s surface, a force equal to ‘mg’ has to be used against the direction of the gravitational force.

(ii) The amount of work done can be calculated as follows:
Work = force x displacement
∴ W = mg x h
∴ W = mgh

(iii) The amount of potential energy stored in the object because of its displacement.
PE = mgh (W = P.E)

(iv) Displacement to height h causes energy equal to mgh to be stored in the object.

Question 2.
When can you say that the work done is either positive, negative or zero?
Answer:

  • When the force and the displacement are in the same direction, the work done by the force is positive.
  • When the force and displacement are in the opposite directions, the work done by the force is negative.
  • When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other, the work done by the force is zero.

Question 3.
Explain the relation between, the commercial and SI unit of energy.
Answer:
The commercial unit of energy is a kilowatt-hour (kWh) while the SI unit of energy is the joule. Their relation is
1 kWh = 1kW x 1hr
= 1000 Wx 3600 s
= 3600000J
(Watt x Sec = Joule)
1 kWh = 3.6 x 106 J.

Question 4.
How is work calculated if the direction of force and the displacement are inclined to each other?

Answer:
If the direction of force and the displacement are inclined to each other then, we must convert the applied force into the force acting along the direction of displacement.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

If θ is angle between force and displacement, then force (F1) in direction of displacement is
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 14

Complete the flow chart.

Question 1.
Transformation of energy
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 15
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 16

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
Transformation of energy
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 17
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 18

Write effects of the following with examples.

Question 1.
Force
Answer:

  • A force can move a stationary object. The force of engine makes a stationery car to move.
  • A force can stop a moving object. The force of brakes can stop a moving car.
  • A force can change the speed of a moving object. When a hockey player hits a moving ball, the speed of ball increases.
  • A force can change the direction of a moving object. In the game of carrom ,when we take a rebound then the direction of striker changes because the edge of the carrom board exerts a force on the strike.
  • A force can change the shape and size of an object. The shape of kneaded wet clay changes when a potter converts it into pots of different shapes and sizes because the p otter applies force on the kneaded wet clay.

Give two examples in each of the following cases:

Question 1.
Potential energy
Answer:

  • Water stored in a dam
  • A compressed spring

Question 2.
Kinetic energy
Answer:

  • Water flowing
  • Bullet fired from a gun

Question 3.
Chemical energy
Answer:

  • Chemical in cell
  • Explosive mixture of a bomb

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 4.
Zero work done
Answer:

  • A stone tied to a string and whirled in a circular path
  • Motion of the earth and other planets moving around the sun

Question 5.
Negative work done
Answer:

  • A cyclist applies brakes to his bicycle, but the bicycle still covers some distance.
  • When a body is made to slide on a rough surface, the work done by the frictional force.

Question 6.
Positive work done
Answer:
(i) A boy moving from the ground floor to the first floor.
(ii) A fruit falling down from the tree.
= 0.5 hr x 30 days
= 15 hrs
To Find:
Energy consumed = ?
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 19
The units of energy consumed in the month of April by the iron is 18 units.

Question 7.
A 25 W electric bulb is used for 10 hours every day. How much electricity does it consume each day?
Answer:
Given:
Power (P) = 25 W
25/1000 kW
Time (E) = 10 hrs
To Find:
Electric energy consumed = ?
Formula:
Electric energy consumed = power x time
Solutions:
Electric energy consumed = power x time
= 25/1000 x 10
= 0.25 kWh
The electric bulb consumes 0.25 kWh of electricity each day.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 8.
If a TV of rating 100W is operated for 6 hrs per day, find the amount of energy consumed in any leap year?
Answer:
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 20
= 2196 hrs.
To Find:
Electric energy consumed
Formula:
Electric energy consumed = power x time
Solution:
Electric energy consumed = power x time
= 0.1 x 2196
= 219.6 kWh
The amount of energy consumed is 219.6 kWh

Complete the paragraph.

Question 1.
………….. is the measure of energy transfer when a force (F) moves an object through a ………….. (d). So when ………….. is done, energy has been transferred from one energy store to another, and so: energy transferred = ………….. done. Energy transferred and work done are both measured in ………….. (J)
Answer:
Work is the measure of energy transfer when a force (F) moves an object through a distance (d). So when work is done, energy has been transferred from one energy store to another, and so: energy transferred = work done. Energy transferred and work done are both measured in joules (J).

Question 2.
………….. energy and ………….. done are the same thing as much as ………….. energy and work done are the same thing. Potential energy is a state of the system, a way of ………….. energy as of virtue of its configuration or motion, while ………….. done in most cases is a way of channeling this energy from one body to another.
Answer:
Potential energy and work done are the same thing as much as kinetic energy and work done are the same thing. Potential energy is a state of the system, a way of storing energy as of virtue of its configuration or motion, while work done in most cases is a way of channeling this energy from one body to another.

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 3.
In physics, ………….. is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the ………….. equal to one ………….. per second.

Power is a ………….. quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more ………….. is needed when the work is done in a shorter amount of time.
Answer:
In physics, power is the rate of doing work or, i.e., the amount of energy transferred or converted per unit time. In the International System of Units, the unit of power is the watt. equal to one joule per second.

Power is a scalar quantity that requires both a change in the physical system and a specified time interval in which the change occurs. But more power is needed when the work is done in a shorter amount of time.

Activity-based questions

Answer in detail:

Question 1.
State the expression for work done when displacement and force makes an angle θ OR State the expression for work done when force is applied making an angle θ with the horizontal force.
Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy 21
Answer:
Let ‘F’ be the applied force and Fj be its component in the direction of displacement. Let ’S’ be the displacement.

The amount of work done is given by W = F1s ……………………………………… (1)
The force ‘F’ is applied in the direction of the string.

Let ‘θ’ be the angle that the string makes with the horizontal. We can determine the component ‘F1‘, of this force F, which acts in the horizontal direction by means of trigonometry.
\(\begin{aligned}
\cos \theta=\frac{\text { base }}{\text { hypotenuse }} \\
\therefore \quad \cos \theta=\frac{\mathrm{F}_{1}}{\mathrm{~F}} \\
\therefore \quad \mathrm{F}_{1}=\mathrm{F} & \cos \theta
\end{aligned}\)
Substituting the value of F1 in equation 1
Thus, the work done by F1 is
W cos θ s
∴ W = Fscosθ

Maharashtra Board Class 9 Science Solutions Chapter 2 Work and Energy

Question 2.
When a body is dropped on the ground from some height its P.E is converted into K.E but when it strikes the ground and it stops, what happens to the K.E?
Answer:
When a body is dropped on the ground, its K.E appears in the form of:

  • Heat (collision between the body and the ground).
  • Sound (collision of the body with the ground).
  • The potential energy of change in state of the body and the ground.
  • Kinetic energy is also utilized to do work i.e., the ball bounces to a certain height and moves to a certain distance vertically and horizontally till Kinetic energy becomes zero.
  • The process in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy is called the dissipation of energy.

Question 3.
Explain the statement “Potential Energy is relative”.
Answer:

  • The potential energy of an object is determined and calculated according to a height of the object with respect to the observer.
  • So, the person staying on 6th floor more potential energy than those staying on the 3rd floor.
  • But, the person on the 6th floor will have lesser potential energy than on the 8th floor. Hence potential energy is relative.

9th Std Science Questions And Answers: