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Maharashtra Board OCM 12th Commerce Solutions Chapter 2 Functions of Management

June 24, 2024June 23, 2024 by Prasanna

Balbharti Maharashtra State Board Organisation of Commerce and Management 12th Textbook Solutions Chapter 2 Functions of Management Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Organisation of Commerce and Management Solutions Chapter 2 Functions of Management

1. (A) Select the correct options and rewrite the sentences

Question 1.
The functions of management start with ……………… function.
(a) organising
(b) planning
(c) co-ordinating
Answer:
planning

Question 2.
The functions of management end with ………………
(a) directing
(b) staffing
(c) controlling
Answer:
controlling

Question 3.
……………. sets out standards for controlling.
(a) Staffing
(b) Planning
(c) Co-ordinating
Answer:
Planning

Question 4.
Organizational function is important for execution of the plans which have been prepared by ……………. management.
(a) top level
(b) middle level
(c) lower level
Answer:
top level

Question 5.
……………… is the function which supports to activate the plans with the help of employees.
(a) Staffing
(b) Directing
(c) Co-ordinating
Answer:
Directing

Question 6.
………………. is the function of execution according to the plan and the organisational structure.
(a) Controlling
(b) Directing
(c) Staffing
Answer:
Directing

Question 7.
………………. arranges the work in such a way that minimum conflicts are raised.
(a) Co-ordinating
(b) Organizing
(c) Controlling
Answer:
Co-ordination.

1. (B) Match the pairs

Question 1.

Group A	Group B
(A) Planning	(1) It Is the process of instructing, guiding, communicating and motivating.
(B) Organizing	(2) It is an integration and synchronization of the efforts of group.
(C) Staffing	(3) Deciding In advance what to do, how to do, when to do and who Is to do it.
(D) Directing	(4) Management is what manager does.
(E) Co-ordlnating	(5) To focus on the role of manager.
	(6) Decides the ways and means to achieve what has been planned.
	(7) It Is the process of comparing the actual performance with the pre-determined standard performance.
	(8) It Is a set of principles.
	(9) It is the process of recruiting, selecting, placing and remunerating.
	(10) To manage is to forecast and plan.

Answer:

Group A	Group B
(A) Planning	(3) Deciding In advance what to do, how to do, when to do and who Is to do it.
(B) Organizing	(6) Decides the ways and means to achieve what has been planned.
(C) Staffing	(9) It is the process of recruiting, selecting, placing and remunerating.
(D) Directing	(1) It Is the process of instructing, guiding, communicating and motivating.
(E) Co-ordlnating	(2) It is an integration and synchronization of the efforts of group.

1. (C) Give one word/phrase/term for the following statements

Question 1.
The right person at the job with right pay.
Answer:
Staffing

Question 2.
A person who shows the correct path as well as guides employees in solving the problems.
Answer:
Director

Question 3.
First function of management.
Answer:
Planning

Question 4.
Last function of management.
Answer:
Controlling

Question 5.
It is an intellectual process of logical thinking and rational decision-making.
Answer:
Planning

Question 6.
The term that is used to denote the structure.
Answer:
Organisation

Question 7.
It is the process of attracting, recruiting, selecting, placing, appraising and remunerating the people.
Answer:
Staffing

Question 8.
The process that leads the employees towards the accomplishment of organisational goals.
Answer:
Directing

Question 9.
It increases the team spirit of work place.
Answer:
Co-ordinating

Question 10.
It is the process of comparing the actual performance with the predetermined standard performance.
Answer:
Controlling.

1. (D) State whether the following statements are True or False

Question 1.
Every function of management is not based on planning.
Answer:
False

Question 2.
Specialization in activities leads to increase in organisational efficiency.
Answer:
True

Question 3.
Qualified, efficient and skilled work force is always an asset of the organization.
Answer:
True

Question 4.
Cooperation is not necessary for smooth flow of organisational activities.
Answer:
False

Question 5.
Co-ordination motivates the employees to take initiative while completing their assigned task.
Answer:
True

Question 6.
Standards are not set for every performance in controlling function.
Answer:
False

1. (E) Find the odd one

Question 1.
Planning, Organizing, Staffing, Writing.
Answer:
Writing

Question 2.
Selecting, Training, Co-ordinating, Placing
Answer:
Co-ordinating.

1. (F) Complete the sentences

Question 1.
The tasks of getting the things done by others is known as ……………..
Answer:
Management

Question 2.
The functions of manager start with …………………
Answer:
Planning

Question 3.
The …………….. function of management initiates action
Answer:
Directing

Question 4.
Recruitments are done under ……………….. function.
Answer:
Staffing

Question 5.
………………. is the fundamental function of management.
Answer:
Planning

Question 6.
………………. integrates departmental activities for achieving common goal of the organisation.
Answer:
Co-ordinating

Question 7.
……………… is the last function of management.
Answer:
Controlling

1. (G) Select the correct option from the bracket

Question 1.
Planning is a detailed programme of (present/ future/past) course of action.
Answer:
future

Question 2.
Directing is a responsibility of (manager/ workers/people) at all levels.
Answer:
manager

Question 3.
Qualified, efficient and skilled workforce is always an (liabilities/assets/expenses) of the organization.
Answer:
assets.

1. (H) Answer in one sentence

Question 1.
What is management?
Answer:
The tasks of getting the things done by others to achieve organisational goal is called management.

Question 2.
What is planning?
Answer:
Planning means deciding in advance what to do when to do, how to do, where to do it and who is to do it.

Question 3.
What is staffing?
Answer:
The process of attracting, recruiting, selecting, placing, appraising, remunerating, developing and retaining the best workforce is called staffing.

Question 4.
What is directing?
Answer:
Directing is the process of instructing, guiding, communicating, inspiring, motivating and supervising the employees to achieve the pre-determined goals of the organisation.

Question 5.
What is controlling ?
Answer:
Controlling is a function of comparing the actual performance with the predetermined standard performance to measure deviation if any, identifying causes of deviation and suggest corrective measures.

1. (I) Correct the underlined word and rewrite the following sentences

Question 1.
Factors of business environment are always fixed.
Answer:
Factors of business environment are always changing.

Question 2.
Staffing is concerned with machines.
Answer:
Staffing is concerned with humans.

Question 3.
Directing is a function of comparing the actual performance with the pre-determined performance.
Answer:
Controlling is a function of comparing the actual performance with the pre-determined performance.

Question 4.
Co-ordination helps to maximise the wastage of resources and controls the cost of work.
Answer:
Co-ordination helps to minimise the wastage of resources and controls the cost of work.

Question 5.
Controlling measures are rigid to some extent.
Answer:
Controlling measures are flexible to some extent.

1. (J) Arrange in proper order

Question 1.
Controlling, Organizing, Planning.
Answer:
Planning, Organising, Controlling.

Question 2.
Directing, Co-ordinating, Staffing.
Answer:
Staffing, Directing, Co-ordinating.

2. Explain the following terms/concepts

Question 1.
Management
Answer:
The task of getting the work done by others to achieve organisational goal is called management. According to L. A. Allen, ‘Management is what manager does. Management is a set of principles which relate to the various functions such as planning, organising, staffing, directing, co-ordinating, controlling, etc. which are helpful in achieving organisational goals.

Question 2.
Planning
Answer:
Planning is the basic function of management. Planning is an intellectual process of logical thinking and rational decision-making. It includes deciding the things to be done in advance. In short, planning is a detailed programme of future course of action. Proper planning and its implementation is key to achieve the objectives of an organisation.

Question 3.
Organising
Answer:
Organising is the process of identifying, bringing the required resources together such as men, money, material, machines and method and arranging them in proper manner to achieve the goals of an organisation. It is prepared by the top level management. Organising function decides the ways and means to achieve what has been planned. Organising is more important in executing the plan.

Question 4.
Staffing
Answer:
Staffing is the function of execution according to plan and organisational structure. It is the process of attracting, recruiting, training, developing, appraising, remunerating, developing and retaining the best workforce. Right person at right job with right pay is the basic principle of staffing. This function is concerned with managing humans and not material.

Question 5.
Directing
Answer:
Directing is the process of instructing, guiding, communicating, inspiring, motivating and supervising the employees to achieve pre-determined goals of an organisation. Director shows correct path as well as guides the employees in solving the problems wherever necessary. Directing is the soul of management function.

3. Study the following case/situation and express your opinion

Question 1.
Mr. Ram, an emerging entrepreneur has designed a structure of his business organization by taking into consideration the required resources such as land, money, machinery, workforce, etc. for his new business. He appointed Mr. Shyam as a manager. Mr. Ram has assigned the responsibilities such as recruitment, selection, training and development and to determine the remuneration of the employees to Mr. Shyam. Mr, Ram. has also appointed Mr. Shubham to supervise the work done by the employees according to the standards given to the employees, Mr. Shubham has also to suggest the remedies to the employees wherever necessary. On this context, find out the management functions performed by
(i) Mr. Ram
(ii) Mr. Shyam
(iii) Mr. Shubham
Answer:
(i) Mr. Ram performs the function of planning and organising. He is an emerging entrepreneur and plans the business structure and organises different resources.

(ii) Mr Shyam is performing the function of staffing as his main duty is to recruit, select, train and develop the employees and to decide their remuneration accordingly.

(iii) Mr. Subham is performing the function of controlling. He compares actual performances of employees with standard performance given. He discovers causes of deviations and suggests remedies to overcome deviations.

Question 2.
In XYZ Company, Mr. Lele gives instructions to the employees working under him, provide guidance and motivates them for their best performance. On the other hand, Mr. Sawed takes effort to harmonize the work done by the employees of different departments while achieving organisational goal. Mr. Desai is looking after the arrangement of required resources the business organization.
Mention the name of employee engaged in following functions :
(i) Organisation
(ii) Direction
(iii) Coordination
Answer:
(i) Mr Desai is engaged in the organising function as he is looking after arrangement of required resources for the business organisation.
(ii) Mr. Lele is engaged in the function of directing as he gives instructions to the employees working under him, provides guidance and motivates them for their best performance.
(iii) Mr. Sayyed is engaged in the function of co-ordination as he takes effort to harmonize the work done by the employees of different departments.

4. Distinguish between

Question 1.
Planning and Organising
Answer:

	Planning	Organising
1. Meaning	Planning is a management function that decides in advance what to do, how to do, when to do, where to do and who is to do it.	Organising refers to the process of putting together various resources and activities of the organisation into a system.
2. Objective	The objective of planning is to set the goals and choose the means to achieve those goals.	The main objective of organising is to identity and bring together all the resources.
3. Area of function	Planning involves setting objectives, identifying alternative courses of actions and selecting the best plan for the organisation.	Organising involves identifying the activities and grouping of relative activities of the organisation.
4. Order	Planning is the first and foremost function of management. It precedes every other function.	In organising function, internal as well as external factors are considered to make arrangement of resources.
5. Nature	Planning is continuous in nature. It is related with those resources which are required for achieving the targets.	Organising is related with all the available resources as they need to be properly arranged.
6. Levels of management	Top management is responsible for preparing planning for the activities of the entire organisation.	Usually, the function of organising is undertaken by the top level management and middle level management.

Question 2.
Organising and Staffing
Answer:

	Organising	Staffing
1. Meaning	Organising refers to the process of putting together various resources and activities of the organisation into a system.	Staffing is a process of recruitment through which competent employees are selected, properly trained, effectively developed and suitably rewarded.
2. Objective	The main objective of organising is to identify and bring together all the resources.	The main objective of the staffing is to obtain the most competent and efficient staff to improve the overall performance.
3. Area of function	Organising involves identifying the activities and grouping of relative activities of the organisation.	Staffing involves selection, recruitment, training, developing, promotion, transfer, etc. of employees.
4. Factors	In organising function, internal as well as external factors are considered to make arrangement of resources.	In staffing function, mostly internal factors such as human factor, finance, work load, etc. are considered.
5. Resources	Organising is related with all the available resources as they need to be properly arranged.	Staffing is related with human resources only.
6. Levels of management	Usually, the function of organising is undertaken by the top level management and middle level management.	Usually, the function of staffing is undertaken by the middle level management.

Question 3.
Staffing and Directing
Answer:

	Staffing	Directing
1. Meaning	Staffing is a process of recruitment through which competent employees are selected, properly placed and trained, effectively developed and suitably rewarded.	Directing means instructing, guiding, inspiring and motivating the subordinate employees so that their efforts result in the achievement of goals.
2. Objective	The main objective of the staffing is to obtain the most competent and efficient people to improve the overall performance.	The main objective of directing is to ensure that the employees at different levels accomplish their tasks according to plans.
3. Area of function	Staffing involves recruitment, selection, training, promotion, development, transfer, etc. of employees.	Directing involves guiding, instructing, inspiring, motivating and communicating with the subordinates.
4. Order	Staffing function follows organising as human resources are required in an organisation.	Directing function follows organising and staffing as direction is needed to guide and inspire the employees.
5. Targets	Targets of staffing include allocation of human resources to achieve better results.	Targets of direction include giving guidance and inspiration to employees to achieve better results.
6. Outcome	Staffing function helps to select right persons for right jobs at right time.	Directing function helps to maintain discipline among the staff.

Question 4.
Directing and Controlling

Question 5.
Co-ordination and Controlling
Answer:

	Co-ordination	Controlling
1. Meaning	Co-ordination refers to the process of developing harmony and integration of different activities to achieve common organisational goals.	Controlling is a managerial function that measures deviation of actual results from the standards set and takes necessary corrective actions.
2. Objective	The main objective of co-ordination is to ensure unify of efforts of the employees and smooth functioning of the organisation.	The main objective of controlling is to ensure that goals or targets must be accomplished as per plan.
3. Area of function	Co-ordination involves efforts of top level, middle level and lower level management.	Controlling involves fixation of standard, measurement of actual performance and finding deviations taking corrective actions to improve performance.
4. Factors	In co-ordinating function only internal factors are considered to create unity of action.	In controlling function internal as well as external factors are taken care for taking corrective actions.
5. Resources	Co-ordination is related with human resources only.	Controlling is related with all the resources as it helps to achieve the given targets.
6. Levels of management	All levels of management are responsible for the co-ordination function to achieve given targets.	Top level management and middle level managements are responsible for controlling of organisational activities.

Question 6.
Planning and Controlling
Answer:

	Planning	Controlling
1. Meaning	Planning is a management function that decides in advance what to do, how to do, when to do, where to do and who is to do it.	Controlling is a managerial function that measures deviation of actual performance from the standards set and takes corrective actions.
2. Objective	The main objective of planning is to set the goals and choose the means to achieve those goals.	The main objective of controlling is to ensure that goals or targets must be accomplished as per plan.
3. Area of function	Planning involves setting objectives, identifying alternative courses of actions and selecting the best plan for the organisation.	Controlling involves fixation of standard, measurement of actual performance and finding deviations taking corrective actions to improve performance.
4. Order	In the sequence of managerial functions, planning is the first and foremost function.	In the sequence of managerial functions, controlling is the last function of the management.
5. Resources	Planning is related with those resources which are required for achieving the targets.	Controlling is related with all the resources as it helps to achieve the given targets.
6. Levels of management	Top management is responsible for preparing planning for the activities of the entire organisation.	Top level management and middle level managements are responsible for controlling of organisational activities.

Question 7.
Organising and Directing

Question 8.
Organising and Co-ordinating
Answer:

	Organising	Co-ordinating
1. Meaning	Organising refers to the process of putting together various resources and activities of the organisation into a system.	Co-ordination refers to the process of developing harmony and integration of different activities to achieve common organisational goals.
2. Objective	The main aim of organising is to identify and bring together all the required resources.	The main aim of co-ordination is to ensure unity of efforts of employees and smooth functioning of the organisation.
3. Area of function	Organising involves identifying the activities and grouping of relative activities of the organisation.	Co-ordination involves deliberate or consistent efforts by the management to create harmony and unity of action.
4. Factors	In organising function, internal as well as external factors are considered to make arrangement of available resources.	In co-ordinating function, only internal factors are considered to create unity of action.
5. Resources	Organising is related with all the resources which are required to be arranged in proper order.	Co-ordination is related with human resources only.
6. Levels of management	Top level management and middle level managements are responsible for the organising of resources to achieve desired objectives.	All levels of management are responsible for the co-ordination to achieve the given targets.

5. Answer in brief

Question 1.
Explain any five points of importance of planning.
Answer:
Importance : The importance of planning is explained as follows:
(1) Helps to set clear objectives : Planning is the process of setting objectives, targets and formulating plans to achieve these objectives. With the help of proper planning, management can analyse the present condition of the organisation and can identify the ways of attaining the desired position in future.

(2) Provides path of action : Planning ensures that the goals or objectives are clearly set. It acts as a guide and provides direction for doing the right things at the right time and in a right way. It helps the employees to understand the organisational goals and what they must do to achieve the same.

(3) Planning improves performance : It helps manager to improve future performances of employees by setting clear objectives and selecting a right course of action. It leads to efficiency in working of the employees. Due to proper planning the employees can work according to guidelines which helps them to improve performance. This results into higher profitability of the organisation.

(4) Minimizes the risk : Planning is the process of looking into the future and anticipating the future changes. By deciding in advance the task to be performed, planning helps to deal with future changes and unforeseen events. Planning helps in anticipation of risk and decide preventive measures accordingly. Though changes or risks cannot be eliminated but proper planning minimizes them.

(5) Planning leads to optimum utilization of resources: Plans are made on the basis of availability of resources with proper allocation for various activities. Proper allocation of resources brings higher efficiency and desired results with minimum wastages.

Question 2.
Explain any five points of importance of organizing.
Answer:
Importance : The importance of organising is explained as follows:
(1) Facilitates administration as well as operation : Organising is the process of identifying, grouping and assigning the activities of administration and proceeding according to its operational activity. Due to proper grouping of the task and employees, there is a reduction in duplication of work results in effective delegation.

(2) Brings specialisation : Organising starts from dividing the total work into smaller units and assigning them to different individuals according to their qualification, capabilities and experience. It leads to increase in overall productivity.

(3) Defines job properly : In the organising function the employees are assigned different jobs according to their qualification, skill and experience and the managers clearly define the details of each job. It clearly spells out what exactly has to be done in every job by each employee.

(4) Clarifies authority and responsibility : The organising function clearly defines authority, power, position of every manager and responsibility, accountability of every employee. This enables proper execution of work and at the same time eliminates confusion, duplication, misunderstanding. It also helps to bring efficiency in working of managers.

(5) Establishes Co-ordination : Organising function helps in establishing co-ordination among different activities of different department. Organising defines clear cut relationship among various positions and ensures mutual co-operation amongst them. Organising helps in co-ordination between different levels of managers of different departments for smooth functioning of an organisation.

Question 3.
Describe any five points of importance of staffing.
Answer:
Importance : The importance of staffing is given as follows:
(1) Effective management function : Staffing is considered an effective managerial function as it deals with human resource. Employees appointed in the organisation through staffing function perform various activities in different areas of the organisation such as production, marketing, finance, etc.

(2) Effective utilization of Human Resources : A well organised staffing department discovers the talented, skilful, experienced and qualified staff. Proper care is taken at every stage of recruitment, selection, placement, etc. It ensures smooth functioning of all the managerial areas of the organisation.

(3) Builds relationship : A sound staffing policy creates a team spirit among the employees. Due to team spirit, a sense of belongingness among the employees is developed. This in turn leads to better communication and co-ordination of managerial efforts in an organisation. A smooth human relation is the key to better flow of co-ordination in an organisation.

(4) Helps Human Resource Development: Skilled and experienced employee is an asset of a business organisation. Staffing function of management is mainly concerned with human factor of production. Efforts are made to utilise the human resources more efficiently.

(5) Helps in effective use of technology and other resources : Staffing function trains employees to use latest technology, capital, material, and method of work more effectively. This brings competitive strength to the organisation. It also helps in improving standard of work and productively in terms of quality and quantity.

(6) Improve efficiency : Regular training and development programmes provide to employees to improve their performance levels. Through proper selection the organisation gets talented and quality employees.

Question 4.
Explain any five points of importance of directing.
Answer:
Importance : The importance of Directing is as follows:
(1) Initiates action : Direction initiates action. It activates employees to put in their best efforts, to achieve the goals. Without effective direction, other managerial functions like planning, organising, staffing, co-ordinating and controlling become ineffective. Managers have to stimulate action by issuing instructions like what to do, how to do, etc. to the subordinates and by supervising them from time to time.

(2) Integrates efforts: At every level of the management there are subordinates under managers. The work assigned to these subordinates is interrelated. The directing function integrates the activities of the subordinates by guidance, supervision and counselling. It results in achievement of organisational goals.

(3) Means of motivation: Objectives of an organisation can be achieved only if the people working in it are properly motivated through monetary and non-monetary incentives. It boosts the morale of employees, contribute their maximum efforts and motivates them to give their best.

(4) Provides stability: Effective direction through supervision, motivation, leadership and communication provides stability and maintains balance in the organisation. This in turn results in the growth of the enterprise at faster rate. For long term survival of the organisation, stability in the organisation are necessary.

(5) Coping up with the changes : Effective direction facilitates changes in the organisation. It enables the enterprise to adopt advance technology, new methods of production, modern techniques of management, etc. It is a direction function which helps the superiors to motivate the subordinates to adapt to the new changes, new challenges, etc. Adapting to the environmental changes is necessary for the growth of the organisation.

Question 5.
Describe any five points of importance of ! coordinating.
Answer:
Importance : The importance of Co-ordinating is explained as follows:
(1) Encourages Team Spirit : Co-ordination is concerned with integrated group efforts. Team work under the direction of the manager encourages the subordinates to work sincerely and give better performance to achieve organisational goals. Co-ordination helps to reduce the conflicts between the employees and departments regarding policies, roles, etc. and increase their team spirit.

(2) Gives Proper Direction : Group or combined efforts of all the employees in an organisation helps to co-ordinate with each other and achieve the desired goals. Thus, combined efforts of all the employees always help an organisation to remove its limitations and achieve organisational objectives. The interdependence of departments gives proper direction to the employees.

(3) Facilitates motivation : In the process of co-ordination the superiors motivate their subordinates by providing them with monetary and other incentives. An effective co-ordination increases efficiency and results in growth and prosperity of the organisation which encourages job security, high income, promotion and incentives.

(4) Optimum utilisation of resources : Proper and effective co-ordination helps to bring together all the resources of the organisation. This in turn helps to make the optimum possible use of available resources to achieve organisational goals. Co-ordination also helps to avoid wastage of resources and control the cost.

(5) Achieve organisational objectives : Proper coordination helps to reduce wastages, delays in completion of targets, departmental disputes, etc. of the organisation to a great extent. This ensures smooth working of the organisation in the process of achievement of objectives.

6. Justify the following statements

Question 1.
Planning is the first function of management.
Answer:
(1) Planning is the basic function of management. Every function of management is based on planning. Planning is an intellectual process of logical thinking and rational decision i making.

(2) Designing i.e. doing a proper planning and implementing it accordingly is the key of achieving the objectives of organisation.

(3) Planning means deciding in advance what to do, when to do, how to do, where to do and who is to do it. Thus, it is a detailed programme of future courses of action.

(4) Planning involves setting objectives, identifying alternative courses of action and selecting the best plan. It focuses on organisation’s objective and develop various course of action to achieve those goals.

Question 2.
Controlling is the last function of management.
Answer:
(1) It is important for am organisation to keep a check on whether things are moving as per plan or not. So controlling function comes as the last but indispensable function of management. The effectiveness of planning can be determined with the function of controlling.

(2) Controlling function helps in comparing the actual performance with the pre-determined standard and performance. It is the process of bringing about conformity of performance with planned action.

(3) Controlling function helps in measuring deviation, if any, identifies the course of deviation and suggests corrective measures. The process of controlling helps in formulation of future plans also.

(4) Controlling helps in checking and measuring performance at all the levels of management, as it compares and finds deviation, analyses the causes of deviation and suggests corrective measures. All planning may fail in the absence of proper controlling measures.

Question 3.
Organizing facilitates administration as well as operation of the organization.
Answer:
(1) Organising function is also called as ‘doing function’ i.e. putting the plan into action. Administration and operation both are doing function as organising is the process of putting together various resources and activities of the organisation into a system.

(2) Organising involves identifying the activities and grouping of relative activities of administration and operational department.

(3) Organising function defines, departmentalizes and assigns activities so that they can be most effectively executed for the smooth flow of administration.

(4) Due to proper grouping of the tasks and the employees, there is increase in production and reduction in wastage. The duplication of work can be avoided and effective delegation becomes possible.

Question 4.
Right person at right job with right pay is the basic principle of staffing.
Answer:
(1) The main function of staffing is to select the right person for the right job with right pay. Selecting the right person for the right job brings efficiency and specialisation in the organisation.

(2) It also bring job satisfaction as adequate remuneration increases morale of the employees. Training and development programmes and job security are the factors which are important in providing job satisfaction.

(3) Proper selection of qualified, efficient and skilful work force is always an asset of the organisation. Proper selection of employees contributes in the higher efficiency and leads to long term positive effects in the organisation.

(4) With proper selection process, right persons for right jobs are placed and regularly appraised on merit basis. The criteria of appraised are duly communicated which brings peace and harmony in the organisation.

Question 5.
Co-ordination between different functions and all levels of management is the essence of organisational success,
Answer:
(1) Co-ordination is an integration of different activities which is essential for their smooth flow. It establishes harmony among all the activities of an organisation in achieving desired goals. Co-ordination will not exist unless efforts are taken at all levels of management.

(2) Co-ordination is the synchronization of the efforts of a group so as to provide unity of action for organisational goals. It is a hidden force which binds all other functions at all levels of management.

(3) In an organisation, a number of persons are working together to achieve a common goal. Their work is closely linked with each other. Co-ordination function brings all the group efforts together and harmonise them carefully.

(4) Co-ordination is orderly arrangement of group efforts to provide unity of action to achieve common goals. Co-operation, team work and higher efficiency level lead to attainment of goals and thus, it is the essence of organisational success.

7. Attempt the following

Question 1.
Explain the importance of planning.
Answer:
Importance : The importance of planning is explained as follows:
(1) Helps to set clear objectives : Planning is the process of setting objectives, targets and formulating plans to achieve these objectives. With the help of proper planning, management can analyse the present condition of the organisation and can identify the ways of attaining the desired position in future.

Question 2.
Describe the importance of organizing.
Answer:
Importance : The importance of organising is explained as follows:
(1) Facilitates administration as well as operation : Organising is the process of identifying, grouping and assigning the activities of administration and proceeding according to its operational activity. Due to proper grouping of the task and employees, there is a reduction in duplication of work results in effective delegation.

Question 3.
Explain the importance of staffing.
Answer:
Importance : The importance of staffing is given as follows:
(1) Effective management function : Staffing is considered an effective managerial function as it deals with human resource. Employees appointed in the organisation through staffing function perform various activities in different areas of the organisation such as production, marketing, finance, etc.

Question 4.
Explain the importance of directing.
Answer:
Importance : The importance of Directing is as follows:
(1) Initiates action : Direction initiates action. It activates employees to put in their best efforts, to achieve the goals. Without effective direction, other managerial functions like planning, organising, staffing, co-ordinating and controlling become ineffective. Managers have to stimulate action by issuing instructions like what to do, how to do, etc. to the subordinates and by supervising them from time to time.

Question 5.
Describe the importance of coordinating.
Answer:
Importance : The importance of Co-ordinating is explained as follows:
(1) Encourages Team Spirit : Co-ordination is concerned with integrated group efforts. Team work under the direction of the manager encourages the subordinates to work sincerely and give better performance to achieve organisational goals. Co-ordination helps to reduce the conflicts between the employees and departments regarding policies, roles, etc. and increase their team spirit.

Question 6.
Explain the importance of controlling.
Answer:
Importance : The importance of controlling function is explained as follows:
(1) Fulfilling goals of organisation : Controlling helps to fulfil and achieve organisational goals. The controlling function ensures that the activities take place according to the plans and if there is any deviation, timely action is taken. When all the activities are conducted successfully, according to plan the organisational goals can be achieved as desired.

(2) Making efficient utilisation of resources : By using various control techniques, managers can keep a close watch over the utilisation of human, physical and financial resources. They can prevent the misuse or wastage of resources and ensure proper utilisation of the same.

(3) Accuracy of standards : Proper and efficient control system help the management to check the standards set are accurate or not. This system also keeps check on the changes taking place in the organisation from time to time. Controlling functions are flexible to some extent. This in turn facilitates the organisation to review the standards by considering such changes.

(4) Motivates Employees : A good control system gives information in advance about the standard performance and discovers efficient and inefficient employees. Efficient employees may be given Financial rewards or incentives to motivate them further. The manager may recommend motivational measures in case it finds that deviations are due to insufficient motivation.

(5) Ensures order and discipline : An efficient and good control system ensures order and discipline in the organisation. It prevents and reduces unnecessary behaviour on the part of employees. Under this system, regular checking is done by the managers or departmental heads and preventive measures are taken against deviation or indiscipline.

8. Answer the following

Question 1.
Define the term Planning and explain the importance of planning.
Answer:
[A] Meaning : Planning is the fundamental and basic function of management. It is a process of setting goals and choosing the means to achieve these goals. Planning means deciding the future course of action which determines what is to be done, how to do it, when to do it, who is to do it and how results are to be evaluated. It is a detailed programme in which all activities to be performed in future are mentioned keeping in mind the objectives. Thus, it is an intellectual process of logical thinking and rational decision-making.

[B] Importance : The importance of planning is explained as follows:
(1) Helps to set clear objectives : Planning is the process of setting objectives, targets and formulating plans to achieve these objectives. With the help of proper planning, management can analyse the present condition of the organisation and can identify the ways of attaining the desired position in future.

(6) Helps in decision-making : Planning helps the management to achieve to take a rational decision and to select best alternatives by considering all positive and negative outcomes of all the alternatives the decisions are taken after selecting the best suitable alternatives a predefined goals.

(7) Useful is setting the standards for controlling: Planning sets the standards of performance to be achieved and which can be measured with the actual performance for find out about any deviation. Such deviation can be taken care by controlling steps. Thus, planning provides basis for maintaining discipline in an organisation.

(8) Facilitates co-ordination of all activities : Proper planning reduces the overlapping among all activities of business which are closely linked with each other. Planning interrelates such activities of all department work as per overall plan and thus management co-ordination is achieved.

(9) Facilitates other functions : Planning is the primary function of all the functions of management. Every organisational function is set to achieve the organisational goals at the planning stage. Other management functions such as organising, staffing, etc. cannot be undertaken till the plan is ready.

(10) Promotes innovative ideas : Planning is the basic function. The process of decision-making involves promotion of innovative ideas after critical
thinking. It is the most challenging activity for the management as it guides all future activities and actions of an organisation. In the end, these innovative plans result in attainment of the organisation goals.

Question 2.
What is Organising? Explain the importance of organising.
Answer:
[A] Meaning: Organising is the process of putting together various activities, resource and people into a system so that people work together for achieving planned objectives. Organising means arranging everything in an orderly manner. It means making arrangements like money, machinery, materials, man-power and other physical resources to achieve the predefined goals. The synchronization and combination of workforce, physical, financial and information resources in the process of organising.

[B] Importance : The importance of organising is explained as follows:
(1) Facilitates administration as well as operation : Organising is the process of identifying, grouping and assigning the activities of administration and proceeding according to its operational activity. Due to proper grouping of the task and employees, there is a reduction in duplication of work results in effective delegation.

(6) Helps for effective administration : A sound organising structure facilitates in defining the right job to the right individual. Similarly, the functions, duties and role of each and every employee are well defined in the organising function. This facilitates effective administration and ultimately leads to efficient administration.

(7) Helpful for growth and diversification : Smooth and efficient functioning, clearly defined authority and responsibilities and smooth co-ordination leads to the growth of the organisation. Use of appropriate techniques of control brings efficiency and reduces wastages which ultimately leads to higher profitability of the organisation. All this is possible when the structure of the organisation is well defined.

(8) Creates sense of security: Organising function defines and clarifies the jobs, functions and roles, powers and authority assigned to every manager and employee. Clarity in job profile eliminates confusion and gives responsibility. It helps a lot in getting mental satisfaction and develops sense of security.

(9) Scope for innovation : The manager can use his talent, knowledge and experience to take decisions on various matters and problems. For instance, decision to adopt new technique of production in the organisation. Thus, his talent flourishes by adopting new changes in the methods of work.

(10) Optimum utilisation of resources : Organising function lays down the best possible uses of resources for a specific job. Thus, it is possible to use the available resources to their optimum level and thereby avoid wastage as well as their excessive use.

Question 3.
What do you mean by Staffing? Describe the importance of staffing.
Answer:
[A] Meaning : Right person at right job with right pay is the basic principle of staffing, Staffing is the process involved in attracting, identifying, assessing, recruiting, placing, evaluating and directing employees. It is recruitment, selection, development, training and compensation of employees. It is very challenging for the organisation to focus on best utilisation of workforce by using their talents and skills, retaining them and arranging training and t development programme.

[B] Importance : The importance of staffing is given as follows:
(1) Effective management function : Staffing is considered an effective managerial function as it deals with human resource. Employees appointed in the organisation through staffing function perform various activities in different areas of the organisation such as production, marketing, finance, etc.

(7) Long term effect : Sub-functions of staffing, namely, proper selection, training, development, motivation, etc. help to achieve long-term benefits such as increase in productivity and efficiency, loyalty of customers and employees, etc.

(8) Essential contribution : The selection of employees should be based on the ability of the prospective candidates to meet the future challenges. Selection is based on the ability of the prospective employee so that organisation can meet the future challenges wisely. Therefore, in staff selection, the selectors should take into account the contribution of the employees in their future roles.

(9) Provides job satisfaction : A good staffing policy creates job satisfaction in the minds of the employees. For instance, proper placement of the individuals according to their knowledge, experience and aptitude, timely promotions, training etc. give job satisfaction. Fair remuneration and job security are the factors which are important in providing job satisfaction.

(10) Maintains harmony: Proper staffing policy helps to develop good labour relations. The performances of employees are regularly appraised and promotions are made on merits. Due to this, employees develop positive attitude towards the management which, in turn, helps to bring about peace and harmony in the organisation.

Question 4.
Give the meaning of Directing and explain the importance of directing.
Answer:
[A] Meaning : Directing is the process of instructing, guiding, inspiring, motivating and supervising the employees to achieve pre-determined organisational goals. It is a continuous function started at top level and flows through the lower level of an organisation. It is continued through out the tenure of an organisation. A few philosophers called it as ‘Life spark of an Enterprise. Director shows the correct path as well as guides the employees in solving the problems.

[B] Importance : The importance of Directing is as follows:
(1) Initiates action : Direction initiates action. It activates employees to put in their best efforts, to achieve the goals. Without effective direction, other managerial functions like planning, organising, staffing, co-ordinating and controlling become ineffective. Managers have to stimulate action by issuing instructions like what to do, how to do, etc. to the subordinates and by supervising them from time to time.

(6) Efficient utilisation of resources : Constant instructions can be given to the subordinates to make the maximum use of the available resources and to make every possible effort to minimize the wastages of resources. Thus, effective direction helps in optimum use of available resources such as men, materials, money and methods which helps to reduce cost and increase profit.

(7) Creates team spirit : The supervisors through proper direction can guide, lead and motivate their subordinates to co-ordinate the activities. Thus, team spirit is created which helps the employees to perform their activities more efficiently and on time. This results in faster achievement of organisational goals.

(9) Increases efficiency level : Effective direction and guidance results in better performance of the employees. It also enables the managers and other superiors to guide the subordinates as the leader while performing their jobs.

(8) Exploring capabilities of individuals: Every employee in the organisation has certain capabilities and potential. Through proper direction, motivation and encouragement manager can utilise them to their optimum level to achieve organisational goals and increases efficiency of organisation.

(10) Co-operation : Co-operation between different departments from top level to the bottom level and among the people within the department is must. Co-operation is essential for the success and achievement of organisational goals and for smooth flow of all organisational activities.

Question 5.
What is Coordinating? Describe the importance of coordinating.
Answer:
[A] Meaning : Co-ordination as a function of management refers to the task of developing harmony between various sections of departments and between various departments of the organisation. Thus, co-ordination is a hidden force which binds all other functions of the management integrating the group activities to accomplish the organisational goal efficiency. As the work of each one is linked in an organisation is necessary in co-ordination.

[B] Importance : The importance of Co-ordinating is explained as follows:
(1) Encourages Team Spirit : Co-ordination is concerned with integrated group efforts. Team work under the direction of the manager encourages the subordinates to work sincerely and give better performance to achieve organisational goals. Co-ordination helps to reduce the conflicts between the employees and departments regarding policies, roles, etc. and increase their team spirit.

(6) Improve relations : Co-ordination brings develops good relations among the employees working at different levels of management. For instance, marketing department depends upon production department, production department : depends upon purchase department and so on. Proper co-ordination always helps employees to improve and build strong relations among the employees working in different departments.

(7) Leads to Higher Efficiency : Co-ordination facilitates the optimum use of physical and human resources. This leads to higher returns at lesser cost, thereby higher efficiency. Co-ordination ultimately leads to the optimum use of the resources, higher efficiency reduction in cost and reduction in wastages.

(8) Improves goodwill : Higher sales and higher profitability can be achieved due to synchronized efforts. It earns a name and goodwill in the corporate world. This leads to better value of shares in the stock exchange (market).

(9) Unity of direction : Co-ordinating function helps to bring together activities of different departments to achieve common goals and objectives of the organisation. Therefore, co¬ordination is needed to give proper direction to all the departments of the organisation.

(10) Specialisation : In every business organisation all departments are headed by qualified and specialised professionals in their respective field. The specialised knowledge of these departmental heads helps in various managerial decisions. Proper and efficient co-ordination among these professionals helps to achieve organisational goals (targets) as planned by the top management.

Question 6.
Define the term Controlling and explain the importance of controlling.
Answer:
[A] Meaning : Controlling is a function of comparing the actual performance with the predetermined standard performance. It measures deviation, if any, identifies the causes and suggest the corrective measures. It is performed by all levels of management. Controlling is an indispensable function at all levels of management.

[B] Importance : The importance of controlling function is explained as follows:
(1) Fulfilling goals of organisation : Controlling helps to fulfil and achieve organisational goals. The controlling function ensures that the activities take place according to the plans and if there is any deviation, timely action is taken. When all the activities are conducted successfully, according to plan the organisational goals can be achieved as desired.

(6) Facilitates co-ordination : Every manager or superior co-ordinates the activities of subordinates towards the process of controlling. Controlling reveals the weak points where co¬ordination falls short, so that the management can take timely action.

(7) Psychological pressure : Efficient control system puts psychological pressure on the employees to perform better. Their performance is measured and compared with standards set from time to time. All the employees know that their performance will be evaluated and hence they put on their best to perform well.

(8) Ensures Organisational Efficiency and Effectiveness : Efficient and proper control system ensures organisational efficiency and ; effectiveness. The factors of controlling such as motivation for better performance, achievement of co-ordination in the performance and managers’ responsibility ensure that the organisation works i more efficiently.

(9) Build good Corporate image : An efficient controlling system helps to improve overall efficiency and quality of work. As a result organisation achieves its goals according set standards. This in turn helps to build a good corporate image and develops reputation of the business.

(10) Acts as a Guide : Controlling function provides set of standard performance. All levels of managers and employees work according to it. They follow these standards to achieve desired results. The steps taken for controlling an activity guide the management while planning any future activity.

Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current

June 24, 2024June 23, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 10 Magnetic Fields due to Electric Current Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 10 Magnetic Fields due to Electric Current

1. Choose the correct option.

i) A conductor has 3 segments; two straight and of length L each and a semicircular with radius R. It carries a current I. What is the magnetic field B at point P?
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 1
Answer:
(C) \(\frac{\mu_{0}}{4} \frac{I}{R}\)

ii) Figure a, b show two Amperian loops associated with the conductors carrying current I in the sense shown. The \(\oint \vec{B} \cdot d \vec{l}\) in the cases a and b will be, respectively,
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 2
Answer:
(A) -μ₀ I, 0

iii) A proton enters a perpendicular uniform magnetic field B at origin along the positive x axis with a velocity v as shown in the figure. Then it will follow the following path. [The magnetic field is directed into the paper].
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 3
(A) It will continue to move along positive x axis.
(B) It will move along a curved path, bending towards positive x axis.
(C) It will move along a curved path, bending towards negative y axis.
(D) It will move along a sinusoidal path along the positive x axis.
Answer:
(C) It will move along a curved path, bending towards negative y axis.

(iv) A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of 1.3 × 10^-4 T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are
(A) 14 × 10^-4 N, downward.
(B) 20 × 10^-4 N, downward.
(C) 14 × 10^-4 N, upward.
(D) 20 × 10^-4 N, upward.
Answer:
(D) 20 × 10^-4 N, upward.

v) A charged particle is in motion having initial velocity \(\overrightarrow{\mathrm{V}}\) when it enter into a region of uniform magnetic field perpendicular to \(\overrightarrow{\mathrm{V}}\) . Because of the magnetic force the kinetic energy of the particle will
(A) remain uncharged.
(B) get reduced.
(C) increase.
(D) be reduced to zero.
Answer:
(A) remain uncharged.

Question 2.
A piece of straight wire has mass 20 g and length 1m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field B in a horizontal direction. What must be the magnetic of B?
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 4
Answer:
Data: m = 20 g = 2 × 10^-2 kg, l = 1 m, I = 1 A,
g = 9.8 m/s²
To balance the wire, the upward magnetic force must be equal in magnitude to the downward force due to gravity.
∴ F_m = IlB = mg
Therefore, the magnitude of the magnetic field,
B = \(\frac{m g}{I l}=\frac{\left(2 \times 10^{-2}\right)(9.8)}{(1)(1)}\) = 0.196 T

Question 3.
Calculate the value of magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A (Given: µ₀ = 4π × 10^-7 Wb/Am).
Answer:
Data : I = 5A, a = 0.02 m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\) = 10^-7 × \(\frac{2(5)}{2 \times 10^{-2}}\) = 5 × 10^-5 T

Question 4.
An electron is moving with a speed of 3.2 × 10⁶ m/s in a magnetic field of 6.00 × 10^-4 T perpendicular to its path. What will be the radius of the path? What will be frequency and the kinetic energy in keV ? [Given: mass of electron = 9.1 × 10^-31 kg, charge e = 1.6 × 10^-19 C, 1 eV = 1.6 × 10^-19 J]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 8

Question 5.
An alpha particle (the nucleus of helium atom) (with charge +2e) is accelerated and moves in a vacuum tube with kinetic energy = 10.00 MeV.On applying a transverse a uniform magnetic field of 1.851 T, it follows a circular trajectory of radius 24.60 cm. Obtain the mass of the alpha particle. [charge of electron = 1.62 × 10^-19 C]
Answer:
Data: 1 eV = 1.6 × 10^-19 J,
E = 10MeV = 10⁷ × 1.6 × 10^-19 J = 1.6 × 10^-12 J
B = 1.88 T, r = 0.242 m, e = 1.6 × 10^-19 C
Charge of an -partic1e,
q = 2e = 2(1.6 × 10^-19)=3.2 × 10^-19 C
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 9
This gives the mass of the α-particle.
[Note : The value of r has been adjusted to match with the answer. The CODATA (Committee on Data for Science and Technology) accepted value of m_x is approximately 6.6446 × 10^-27 kg.]

Question 6.
Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in apposite directions. Separation between the two wires is 8 mm. The length AB is S = 22 cm. Obtain the direction and magnitude of the magnetic field due to current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. [µ₀ = 4π × 10^-7 T.m/A]
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 5
Answer:
Data: I₁ = I₂ = 10 A, s = 8 mm = 8 × 10^-3 m, l = 0.22 m
By right hand grip rule, the direction of the magnetic field \(\overrightarrow{B_{2}}\) due to the current in wire 2 at AB is into the page and its magnitude is
B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{s}=10^{-7} \times \frac{2(10)}{8 \times 10^{-3}}=\frac{1}{4} \times \mathbf{1 0}^{-3}\) T
The current in segment AB is upwards. Then, by Fleming’s left hand rule, the force on it due to \(\overrightarrow{B_{2}}\) is to the left of the diagram, i.e., away from wire 1, or repulsive. The magnitude of the force is
F_{on 1 by 2} = I₁lB₂ = (10)(0.22) × \(\frac{1}{4}\) × 10^-3
= 5.5 × 10^-4 N

Question 7.
A very long straight wire carries a current 5.2 A. What is the magnitude of the magnetic field at a distance 3.1 cm from the wire? [ µ₀ = 4π × 10^-7 T∙m/A]
Answer:
Data : I = 5.2 A, a = 0.031 m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\) = 10^-7 × \(\frac{2(5.2)}{3.1 \times 10^{-2}}\)
= 3.35 × 10^-5 T

Question 8.
Current of equal magnitude flows through two long parallel wires having separation of 1.35 cm. If the force per unit length on each of the wires in 4.76 × 10^-2 N, what must be I ?
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 14

Question 9.
Magnetic field at a distance 2.4 cm from a long straight wire is 16 µT. What must be current through the wire?
Answer:
Data: a = 2.4 × 10^-2 m, B = 1.6 × 10^-5 T,
\(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
B = \(\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\)
The current through the wire,
I = \(\frac{1}{\mu_{0} / 4 \pi} \frac{a B}{2}=\frac{1}{10^{-7}} \frac{\left(2.4 \times 10^{-2}\right)\left(1.6 \times 10^{-5}\right)}{2}\) = 1.92 A

Question 10.
The magnetic field at the centre of a circular current carrying loop of radius 12.3 cm is 6.4 × 10^-6T. What will be the magnetic moment of the loop?
Answer:
Data: R = 12.3cm = 12.3 × 10^-2 m,
B = 6.4 × 10^-6 T, µ₀ = 4π × 10^-7 T∙m/A
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 17
= 5.955 × 10^-2 J/T (or A∙m²)

Question 11.
A circular loop of radius 9.7 cm carries a current 2.3 A. Obtain the magnitude of the magnetic field (a) at the centre of the loop and (b) at a distance of 9.7 cm from the centre of the loop but on the axis.
Answer:
Data: R = z = 9.7 cm = 9.7 × 10^-2 m, I = 2.3A, N = 1
(a) At the centre of the coil :
The magnitude of the magnetic induction,
B = \(\frac{\mu_{0} N I}{2 R}\)
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 18

Question 12.
A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what be the magnetic field at the centre of the coil?
Answer:
Data: N = 100,R = 8 × 10^-2 m, I = 0.4A,
µ₀ = 4π × ^-7 T∙m/A
B = \(\frac{\mu_{0} N I}{2 R}\)
= \(\frac{\left(4 \pi \times 10^{-7}\right)(100)(0.4)}{2\left(8 \times 10^{-2}\right)}\) = 3.142 × 10^-4 T

Question 13.
For proton acceleration, a cyclotron is used in which a magnetic field of 1.4 Wb/m² is applied. Find the time period for reversing the electric field between the two Ds.
Answer:
Data: B = 1.4 Wb/m², m = 1.67 × 10^-27 kg,
q = 1.6 × 10^-19 C
T = \(\frac{2 \pi m}{q B}\)
t = \(\frac{T}{2}=\frac{\pi m}{q B}=\frac{(3.142)\left(1.67 \times 10^{-27}\right)}{\left(1.6 \times 10^{-19}\right)(1.4)}\)
= 2.342 × 10^-8 s
This is the required time interval.

Question 14.
A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm × 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m². The torsional constant of the spring is 1.5 × 10^-9 Nm/ degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of 30°.
Answer:
Data : N = 50, C = 1.5 × 10^-9 Nm/degree,
A = lb = 5 cm × 3 cm = 15 cm² = 15 × 10^-4 m²,
B = 0.05Wb/m², θ = 30°
NIAB = Cθ
∴ The current through the coil, I = \(\frac{C \theta}{N A B}\)
= \(\frac{1.5 \times 10^{-9} \times 30}{50 \times 15 \times 10^{-4} \times 0.05}=\frac{3 \times 10^{-5}}{5 \times 0 \cdot 5}\)
= 1.2 × 10^-5 A

Question 15.
A solenoid of length π m and 5 cm in diameter has winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.
Answer:
Data: L = 3.142 m, N = 1000, I = 5A,
μ₀ = 4π × 10^-7 T∙m/A
The magnetic induction,
B = μ₀ nI = μ₀(\(\frac{N}{L}\))I
= (4π × 10^-7)(\(\frac{1000}{3.142}\))(5) = \(\frac{20 \times 3.142 \times 10^{-4}}{3.142}\)
= 2 × 10^-3 T

Question 16.
A toroid of narrow radius of 10 cm has 1000 turns of wire. For a magnetic field of 5 × 10^-2 T along its axis, how much current is required to be passed through the wire?
Answer:
Data : CentraI radius, r = 10 cm = 0.1 m, N = 1000,
B = 5 × 10^-2 T, = 4π × 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} N I}{2 \pi r}=\frac{\mu_{0}}{4 \pi} \frac{2 N I}{r}\)
2irr 4ir r
∴ 5 × 10^-2 = 10^-7 × \(\frac{2(1000) I}{0.1}\)
∴ I = \(\frac{50}{2}\) = 25A
This is the required current.

Question 17.
In a cyclotron protons are to be accelerated. Radius of its D is 60 cm. and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated?
(Proton mass = 1.67 × 10^-27 kg, e = 1.60 × 10^-19 C, 1eV = 1.6 × 10^-19 J)
Answer:
Data : R = 0.6 m, f = 10⁷ Hz, m_p = 1.67 × 10^-27 kg,
e = 1.6 × 10^-19C, 1 eV = 1.6 × 10^-19 J
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 10

Question 18.
A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P. Given : \(\frac{\mu_{0} I}{4 \pi R} \sqrt{2}\)
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 6
Answer:
The wire loop is in the form of a circular arc AB of radius R and a straight conductor BCA. The arc AB subtends an angle of Φ = 270° = \(\frac{3 \pi}{2}\) rad at the centre of the loop P. Since PA = PB = R and C is the midpoint of AB, AB = \(\sqrt{2} R\) and AC = CB = \(\frac{\sqrt{2} R}{2}\) = \(\frac{R}{\sqrt{2}}\). Therefore, a = PC = \(\frac{R}{\sqrt{2}}\).

The magnetic inductions at P due to the arc AB and the straight conductor BCA are respectively,
B₁ = \(\frac{\mu_{0}}{4 \pi} \frac{I \phi}{R}\) and B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}\)
Therefore, the net magnetic induction at P is
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 16
This is the required expression.

Question 19.
Two long parallel wires going into the plane of the paper are separated by a distance R, and carry a current I each in the same direction. Show that the magnitude of the magnetic field at a point P equidistant from the wires and subtending angle θ from the plane containing the wires, is B = \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) sin 2θ What is the direction of the magnetic field?
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 15
In above figure, \(\vec{B}_{1}\) and \(\vec{B}_{2}\) are the magnetic fields in the plane of the page due to the currents in wires 1 and 2, respectively. Their directions are given by the right hand grip rule: \(\vec{B}_{1}\) is perpendicular to AP and makes an angle Φ with the horizontal. \(\vec{B}_{2}\) is perpendicular to BP and also makes an angle Φ with the horizontal.
AP = BP = a = \(\frac{R / 2}{\cos \theta}\)
and B₁ = B₂ = \(\frac{\mu_{0}}{4 \pi} \frac{2 I}{a}=\frac{\mu_{0}}{4 \pi} \frac{2 I(2 \cos \theta)}{R}\)
= \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) cos θ
Since the vertical components cancel out, the magnitude of the net magnetic induction at P is
B_net = 2B₁ cos Φ = 2B₁ cos(90° – θ) = 2B₁ sinθ
= 2(\(\frac{\mu_{0}}{\pi} \frac{I}{R}\) cos θ) sin θ = \(\frac{\mu_{0}}{\pi} \frac{I}{R}\) sin 2θ
as required. is in the plane parallel to that of the wires and to the right as shown in the figure.

Question 20.
Figure shows a section of a very long cylindrical wire of diameter a, carrying a current I. The current density which is in the direction of the central axis of the wire varies linearly with radial distance r from the axis according to the relation J = J_or/a. Obtain the magnetic field B inside the wire at a distance r from its centre.
[Answer: B J r
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 7
Answer:
Consider an annular differential element of radius r and width dr. The current through the area dA of this element is
dI = JdA = (J_o \(\frac{r}{a}\))2πrdr = \(\frac{2 \pi J_{0} r^{2} d r}{a}\) …………….. (1)
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 22
To apply the Ampere’s circuital law to the circular path of integration, we note that the wire has perfect cylindrical symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r. Thus,
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 23

Question 21.
In the above problem, what will be the magnetic field B inside the wire at a distance r from its axis, if the current density J is uniform across the cross section of the wire?
Answer:
Figure shows the cross section of a long straight wire of radius a that carries a current I out of the page. Because the current is uniformly distributed over the cross section of the wire, the magnet ic field \(\vec{B}\) due to the current must be cylindrically symmetrical. Thus, along the Amperian loop of radius r(r < a), symmetry suggests that \(\vec{B}\) is tangent to the loop, as shown in the figure.
\(\oint \vec{B} \cdot \overrightarrow{d l}=B \oint d l\) = B(2πr) ……….. (1)
Because the current is uniformly distributed, the current I_encl enclosed by the loop is proportional to the area encircled by the loop; that is,
I_encl = Jπr²
By right-hand rule, the sign of ‘d is positive. Then, by Ampere’s law,
B (2πr) = µ₀ I_encl = µ₀ Jπr² ……………….. (2)
∴ B = \(\frac{\mu_{0} J}{2} r\) ……………… (3)
OR
I_encl = I\(\frac{\pi r^{2}}{\pi a^{2}}\)
By right-hand rule, the sign of I\frac{\pi r^{2}}{\pi a^{2}} is positive. Then, by Ampere’s law,
\(\oint B d l\) = µ₀ I_encl
∴ B(2πr) = µ₀I \(\frac{r^{2}}{a^{2}}\) …………. (4)
∴ B = (\(\frac{\mu_{0} I}{2 \pi a^{2}}\) )r ………… (5)

[Note: Thus, inside the wire, the magnitude B of the magnetic field is proportional to distance r from the centre. At a distance r outside a straight wire, B = (\(\frac{\mu_{0} I}{2 \pi r}\) i.e., B ∝ \(\frac{1}{r}\).]

Theory Exercise

Question 1.
Distinguish between the forces experienced by a moving charge in a uniform electric field and in a uniform magnetic field.
Answer:
A charge q moving with a velocity \(\vec{v}\) through a magnetic field of induction \(\vec{B}\) experiences a magnetic force perpendicular both to \(\vec{B}\) and \(\vec{v}\) . Experimental observations show that the magnitude of the force is proportional to the magnitude of \(\vec{B}\), the speed of the particle, the charge q and the sine of the angle θ between \(\vec{v}\) and \(\vec{B}\). That is, the magnetic force, F_m = qv B sin θ
∴ \(\vec{F}_{\mathrm{m}}=q(\vec{v} \times \vec{B})\)
Therefore, at every instant \(\vec{F}_{\mathrm{m}}\) acts in a direction perpendicular to the plane of \(\vec{v}\) and \(\vec{B}\) .
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 27
If the moving charge is negative, the direction of the force \(\vec{F}_{\mathrm{m}}\) acting on it is opposite to that given by the right-handed screw rule for the cross-product \(\vec{v}\) × \(\vec{B}\).

If the charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle.
The force due to the electric field \(\vec{E}\) is \(\vec{F}_{\mathrm{e}}=q \vec{E}\) .
The total force on a moving charge in electric and magnetic fields is called the Lorentz force :
\(\vec{F}=\vec{F}_{\mathrm{e}}+\vec{F}_{\mathrm{m}}=q(\vec{E}+\vec{v} \times \vec{B})\)
Special cases :
(i) \(\vec{v}\) is parallel or antiparallel to \(\vec{B}\): In this case, F_m = qvB sin 0° = 0. That is, the magnetic force on the charge is zero.
(ii) The charge is stationary {v = 0) : In this case, even if q ≠ 0 and B ≠ 0, F_m = q(0)B sin θ = 0. That is, the magnetic force on a stationary charge is zero.

Question 2.
Under what condition a charge undergoes uniform circular motion in a magnetic field? Describe, with a neat diagram, cyclotron as an application of this principle. Obtain an expression for the frequency of revolution in terms of the specific charge and magnetic field.
Answer:
Suppose a particle of mass m and charge q enters a region of uniform magnetic field of induction \(\vec{B}\). In Fig., \(\vec{B}\) points into the page. The magnetic force \(\vec{F}_{\mathrm{m}}\) on the particle is always perpendicular to the velocity of the particle, \(\vec{v}\) . Assuming the charged particle started moving in a plane perpendicular to \(\vec{B}\), its motion in the magnetic field is a uniform circular motion, with the magnetic force providing the centripetal acceleration.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 28
If the charge moves in a circle of radius R,
F_m = |q|vB = \(\frac{m v^{2}}{R}\)
∴ mv = p = |q|BR …………. (1)
where p = mv is the linear momentum of the particle. Equation (1) is known as the cyclotron formula because it describes the motion of a particle in a cyclotron-the first of the modern particle accelerators.

Question 3.
What is special about a radial magnetic field ? Why is it useful in a moving coil galvanometer ?
Answer:

Advantage of radial magnetic field in a moving- coil galvanometer:
- As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a ximum depending only on the current in the coil, but not on the position of the coil.
- The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straight forward to calibrate and to read.
Producing radial magnetic field :
- The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
- A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic, field; that is, the magnetic lines of force in the gap are along radii to the central axis.

Question 4.
State Biot-Savert law. Apply it to
(i) infinitely long current carrying conductor and (ii) a point on the axis of a current carrying circular loop.
Answer:
Consider a very short segment of length dl of a wire carrying a current I. The product I \(\overrightarrow{d l}\) is called a current element; the direction of the vector \(\overrightarrow{d l}\) is along the wire in the direction of the current.

Biot-Savart law (Laplace law) : The magnitude of the incremental magnetic induction \(\overrightarrow{d B}\) produced by a current element I \(\overrightarrow{d l}\) at a distance r from it is directly proportional to the magnitude I \(\overrightarrow{d l}\) of the current element, the sine of the angle between the current element Idl and the unit vector r directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both I \(\overrightarrow{d l}\) and \(\hat{r}\) as per the cross product rule.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 24
where \(\hat{\mathrm{r}}=\frac{\vec{r}}{r}\) and the constant µ₀ is the permeability of free space. Equations (1) and (2) are called Biot-Savart law.

The incremental magnetic induction \(\overrightarrow{d B}\) is given by the right-handed screw rule of vector crossproduct \(I \overrightarrow{d l} \times \hat{\mathrm{r}}\). In Fig, the current element I\(\overrightarrow{d l}\) and \(\hat{\mathbf{r}}\) are in the plane of the page, so that \(\overrightarrow{d B}\) points out of the page at point P shown by ⊙; at the point Q, \(\overrightarrow{d B}\) points into the page shown by ⊗.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 25

The magnetic induction \(\vec{B}\) at the point due to the entire wire is, by the principle of superposition, the vector sum of the contributions \(\overrightarrow{d B}\) from all the current elements making up the wire.
From Eq. (2),
\(\vec{B}=\int \overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \int \frac{I \overrightarrow{d l} \times \hat{\mathrm{r}}}{r^{2}}\)
[Notes : (1) The above law is based on experiments by Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841), French physicists. From their observations Laplace deduced the law mathematically. (2) The Biot- Savart law plays a similar role in magnetostatics as Coulomb’s law does in electrostatics.]

Question 5.
State Ampere’s law. Explain how is it useful in different situations.
Answer:
Ampere’s circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to p0 times the net steady current enclosed by the path.
In mathematical form,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀ I …………… (1)
where \(\vec{B}\) is the magnetic induction at any point on the path in vacuum, \(\overrightarrow{d l}\) is the length element of the path, I is the net steady current enclosed and μ₀ is the permeability of free space.

Explanation : Figure shows two wires carrying currents I₁ and I₂ in vacuum. The magnetic induction \(\vec{B}\) at any point is the net effect of these currents.

To find the magnitude B of the magnetic induction :
We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length \(\overrightarrow{d l}\). The direction of dl is the direction along which the loop is traced.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 26

(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.

For each length element of the Amperian loop, \(\vec{B} \cdot \overrightarrow{d l}\) gives the product of the length dl of the element and the component of \(\vec{B}\) parallel to \(\overrightarrow{d l}\) . If θ r is the angle between \(\overrightarrow{d l}\) and \(\vec{B}\) ,
\(\vec{B} \cdot \overrightarrow{d l}\) = (B cos θ) dl
Then, the line integral,
\(\oint \vec{B} \cdot \overrightarrow{d l}=\oint B \cos \theta d l\) …………. (2)
For the case shown in Fig., the net current I through the surface bounded by the loop is
I = I₂ – I₁
∴ \(\oint B \cos \theta d l\) = μ₀ I
= μ₀(I₂ – I₁) …………… (3)
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.

[Note : Ampere’s law in magnetostatics plays the part of Gauss’s law of electrostatics. In particular, for currents with appropriate symmetry, Ampere’s law in integral form offers an efficient way of calculating the magnetic field. Like Gauss’s law, Ampere’s law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral \(\oint \vec{B} \cdot \overrightarrow{d l}\). The current configurations that can be handled by Ampere’s law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]

12th Physics Digest Chapter 10 Magnetic Fields due to Electric Current Intext Questions and Answers

Do you know (Textbook Page No. 230)

Question 1.
You must have noticed high tension power transmission lines, the power lines on the big tall steel towers. Strong magnetic fields are created by these lines. Care has to be taken to reduce the exposure levels to less than 0.5 milligauss (mG).
Answer:
With increasing population, many houses are constructed near high voltage overhead power transmission lines, if not right below them. Large transmission lines configurations with high voltage and current levels generate electric and magnetic fields and raises concerns about their effects on humans located at ground surfaces. With conductors typically 20 m above the ground, the electric field 2 m above the ground is about 0.2 kV / m to I kV / m. In comparison, that due to thunderstorms can reach 20 kV/m. For the same conductors, magnetic field 2 m above the ground is less than 6 μT. In comparison, that due to the Earth is about 40 μT.

Do you know (Textbook Page No. 232)

Question 1.
Magnetic Resonance Imaging (MRI) technique used for medical imaging requires a magnetic field with a strength of 1.5 T and even upto 7 T. Nuclear Magnetic Resonance experiments require a magnetic field upto 14 T. Such high magnetic fields can be produced using superconducting coil electromagnet. On the other hand, Earth’s magnetic field on the surface of the Earth is about 3.6 × 10^-5 T = 0.36 gauss.
Answer:
Magnetic Resonance Imaging (MRI) is a non-invasive imaging technology that produces three dimensional detailed anatomical images. Although MRI does not emit the ionizing radiation that is found in X-ray imaging, it does employ a strong magnetic field, e.g., medical MRIs usually have strengths between 1.5 T and 3 T.

The 21.1 T superconducting magnet at Maglab (Florida, US) is the world’s strongest MRI scanner used for Nuclear Magnetic Resonance (NMR) research. Since its inception in 2004, it has been continually conducting electric current of 284 A by itself. Because it is superconducting, the current runs through some 152 km of wire without resistance, so no outside energy source is needed. However, 2400 litres of liquid helium is cycled to keep the magnet at a superconducting temperature of 1.7 K. Even when not in use this magnet is kept cold; if it warms up to room temperature, it takes at least six weeks to cool it back down to operating temperature. The 45 T Hybrid Magnet of the Lab (which combines a superconducting magnet of 11.5 T with a resistive magnet of 33.5 T) is kept at 1.8 K using 2800 L of liquid helium and 15142 L of cold water.

Question 2.
Let us look at a charged particle which is moving in a circle with a constant speed. This is uniform circular motion that you have studied earlier. Thus, there must be a net force acting on the particle, directed towards the centre of the circle. As the speed is constant, the force also must be constant, always perpendicular to the velocity of the particle at any given instant of time. Such a force is provided by the uniform magnetic field \(\vec{B}\) perpendicular to the plane of the circle along which the charged particle moves.
Answer:
When a charged particle moves in uniform circular motion inside a uniform magnetic field \(\vec{B}\) in a plane perpendicular to \(\vec{B}\), the centripetal force is the magnetic force on the particle. As in any UCM, this magnetic force is constant in magnitude and perpendicular to the velocity of the particle.

Remember this (Textbook Page No. 233)

Question 1.
Field penetrating into the paper is represented as ⊗, while that coming out of the paper is shown by ⊙.
Answer:
In a two-dimensional diagram, a vector pointing perpendicularly into the plane of the diagram is shown by a cross ⊗ while that pointing out of the plane is shown by a dot ⊙ .

Do you know (Textbook Page No. 234)

Question 1.
Particle accelerators are important for a variety of research purposes. Large accelerators are used in particle research. There have been several accelerators in India since 1953. The Department of Atomic Energy (DAE), Govt. of India, had taken initiative in setting up accelerators for research. Apart from ion accelerators, the DAE has developed and commissioned a 2 GeV electron accelerator which is a radiation source for research in science. This accelerator, ‘Synchrotron’, is fully functional at Raja Ramanna Centre for Advanced Technology, Indore. An electron accelerator, Microtron with electron energy 8-10 MeV is functioning at Physics Department, Savitribai Phule Pune University, Pune.
Answer:
Particle accelerators are machines that accelerate charged subatomic particles to high energy for research and applications. They play a major role in the field of basic and applied sciences, in our understanding of nature and the universe. The size and cost of particle accelerators increase with the energy of the particles they produce. Medical Cyclotrons across the country are dedicated for medical isotope productions and for medical sciences. There are many existing and upcoming particle accelerators in India in different parts of country.

(https: / / www.researchgate.net/ publication/ 3209480 83_Existing and upcoming_particle_accelerators_in. India). For cutting-edge high energy particle physics, Indian particle physicists collaborate with those at Large Hadron Collider CERN, Geneva.

Can you recall (Textbook Page No. 238)

Question 1.
How does the coil in a motor rotate by a full rotation? In a motor, we require continuous rotation of the current carrying coil. As the plane of the coil tends to become parallel to the magnetic field \(\vec{B}\), the current in the coil is reversed externally. Referring to Fig. the segment ab occupies the position cd. At this position of rotation, the current is reversed. Instead of from b to a, it flows from a to b, force \(\vec{F}_{\mathrm{m}}\) continues to act in the same direction so that the torque continues to rotate the coil. The reversal of the current is achieved by using a commutator which connects the wires of the power supply to the coil via carbon brush contacts.
Answer:
Electric Motor
From Fig. we see that the torque on a current loop rotates the loop to smaller values of 9 until the torque becomes zero, when the plane of the loop is perpendicular to the magnetic field and θ = 0. If the current in the loop remains in the same direction when the loop turns past this position, the torque will reverse direction and turn the loop in the opposite direction, i.e., anticlockwise. To provide continuous rotation in the same sense, the current in the loop must periodically reverse direction, as shown in Fig.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 11
In an electric motor, the current reversal is achieved externally by brushes and a split-ring commutator.

Use your brain power (Textbook Page No. 242)

Question 1.
Currents in two infinitely long, parallel wires exert forces on each other. Is this consistent with Newton’s third law?
Answer:
Yes, they are equal in magnitude and opposite in direction and act on the contrary parts : \(\vec{F}\)_{on 2 by 1} = \(\vec{F}\)_{on 1 by 2}. Thus, they form action-reaction pair.

Do you know (Textbook Page No. 244)

Question 1.
So far we have used the constant µ₀ everywhere. This means in each such case, we have carried out the evaluation in free space (vacuum). µ₀ is the permeability of free space.
Answer:
Permeability of free space or vacuum, µ₀ = 4π × 10^-7 H/m.
Earlier SI (2006) had fixed this value of µ₀ as exact but revised SI fixes the value of e, requiring µ₀ (and ε₀) to be determined experimentally.

Use your brain power (Textbook Page No. 244)

Question 1.
Using electrostatic analogue, obtain the magnetic field \(\vec{B}\)_equator at a distance d on the perpendicular bisector of a magnetic dipole of magnetic length 2l and moment \(\vec{M}\). For far field, verify that
\(\vec{B}_{\text {equator }}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{-\vec{M}}{\left(d^{2}+l^{2}\right)^{3 / 2}}\)
Answer:
The magnitude of the electric intensity at a point at a distance r from an electric charge q in vacuum is given by
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{|q|}{r^{2}}\)
where ε₀ is the permittivity of free space. This intensity is directed away from the charge, if the charge is positive and towards the charge, if the charge is negative.

A magnetic pole is similar to an electric charge. The N-pole is similar to a positive charge and the S-pole is similar to a negative charge. Like an electric charge, a magnetic pole is assumed to produce a magnetic field in the surrounding region. The magnetic field at any point is denoted by a vector quantity called magnetic induction. Thus, by analogy, the magnitude of the magnetic induction at a point at a distance r from a magnetic pole of strength q_m is given by
B = \(\frac{\mu_{0}}{4 \pi} \frac{q_{\mathrm{m}}}{r^{2}}\)
This induction is directed away from the pole if it is an N-pole (strength + q_m) and towards the pole if it is an S-pole (strength -q_m).

Consider a point P on the equator of a magnetic dipole with pole strengths + q_m and – q_m and of magnetic length 21. Let P be at a distance d from the centre of the dipole,
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 12
The magnetic induction of a bar magnet at an equatorial point

The magnetic induction at P due to the N-pole is directed along NP (away from the N-pole) while that due to the S-pole is along PS (towards the S-pole), each having a magnitude
B_N = B_S = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{q_{\mathrm{m}}}{\left(d^{2}+l^{2}\right)}\)
(∵ NP = SP = \(\sqrt{d^{2}+l^{2}}\))
The inductions due to the two poles are equal in magnitude so that the two, oppositely directed equatorial components, B_N sin θ and B_S sin θ, cancel each other.

Therefore, the resultant induction is in a direction’ parallel to the axis of the magnetic dipole and has direction opposite to that of the magnetic moment of the magnetic dipole. The component of the induction due to the two poles along the axis is
\(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{q_{\mathrm{m}}}{\left(d^{2}+l^{2}\right)}\) cos θ
where θ is the angle shown in the diagram. From the diagram,
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 13
Thus, for a short dipole the induction varies in-versely as the cube of the distance from it.

Question 2.
What is the fundamental difference between an electric dipole and a magnetic dipole?
Answer:
If a magnet is carefully and repeatedly cut, it would expose two new faces with opposite poles such that each piece would still be a magnet. This suggests that magnetic fields are essentially dipolar in character. The most elementary magnetic structure always behaves as a pair of two magnetic poles of opposite types and of equal strengths. Hence, analogous to an electric dipole, we hypothesize that there are positive and negative magnetic charges (or north and south poles) of equal strengths a finite distance apart within a magnet. Also, they are assumed to act as the source of the magnetic field in exactly the same way that electric charges act as the source of electric field. The magnitude of each ‘magnetic charge’ is referred to as its ‘pole strength’ and is equal to q_m = \(\frac{M}{2 l}\), where \(\vec{M}\) is the magnetic dipole moment, pointing from the negative (or south, S) pole to the positive (or north, N) pole.

However, while two types of electric charges exist in nature and have separate existence, isolated magnetic charges, or magnetic monopoles, are not observed. A magnetic pole is not an experimental fact: there are no real poles. To put it in another way, there are no point sources for \(\vec{B}\), as there are for \(\vec{E}\); there exists no magnetic analog to electric charge. Every experimental effort to demonstrate the existence of magnetic charges has failed. Hence, magnetic poles are called fictitious.

The electric field diverges away from a (positive) charge; the magnetic field line curls around a current. Electric field lines originate on positive charges and terminate on negative ones; magnetic field lines do not begin or end anywhere, they typically form closed loops or extend out to infinity.

Do you know (Textbook Page No. 247)

Question 1.
What is an ideal solenoid?
Answer:
A solenoid is a long wire wound in the form of a helix. An ideal solenoid is tightly wound and infinitely long, i.e., its turns are closely spaced and the solenoid is very long compared to its crosssectional radius.

Each turn of a solenoid acts approximately as a circular loop. Suppose the solenoid carries a steady current I. The net magnetic field due to the current in the solenoid is the vector sum of the fields due to the current in all the turns. In the case of a tightly- wound solenoid of finite length, Fig. 10.39, the magnetic field lines are approximately parallel only near the centre of the solenoid, indicating a nearly uniform field there. However, close to the ends, the field lines diverge from one end and converge at the other end. This field distribution is similar to that of a bar magnet. Thus, one end of the solenoid behaves like the north pole of a magnet and the opposite end behaves like the south pole. The field outside is very weak near the midpoint.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 19
For an ideal solenoid, the magnetic field inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid. The field outside is negligible in this case.

Use your brain power (Textbook Page No. 248)

Question 1.
Choosing different Amperean loops, show that out-side an ideal toroid B = 0.
Answer:
From below figure, the inner Amperean loop does not enclose any current while the outer Amperean loop encloses equal number of I_in and I_out. Hence, by Ampere’s law, B = 0 outside an ideal toroid.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 20

Question 2.
What is an ideal toroid?
Answer:
A toroid is a toroidal solenoid. An ideal toroid consists of a long conducting wire wound tightly around a torus, a doughnut-shaped ring, made of a nonconducting material.
Maharashtra Board Class 12 Physics Solutions Chapter 10 Magnetic Fields due to Electric Current 21
In an ideal toroid carrying a steady current, the magnetic field in the interior of the toroid is tangential to any circle concentric with the axis of the toroid and has the same value on this circle (the dashed line in figure). Also, the magnitude of the magnetic induction external to the toroid is negligible.

Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves

June 22, 2024June 21, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 6 Superposition of Waves Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 6 Superposition of Waves

1. Choose the correct option.

i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its
vibrations is …….times the fundamental frequency.
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(D) 5

ii) If two open organ pipes of length 50 cm and 51 cm sounded together produce 7 beats per second, the speed of sound is.
(A) 307 m/s
(B) 327m/s
(C) 350m/s
(D) 357m/s
Answer:
(D) 357m/s

iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.
(A) 0.8
(B) 1.12
(C) 1.25
(D) 1.56
Answer:
(B) 1.12

iv) Which of the following equations represents a wave travelling along the y-axis?
(A) x = A sin(ky – ωt)
(B) y = A sin(kx – ωt)
(C) y = A sin(ky) cos(ωt)
(D) y = A cos(ky)sin(ωt)
Answer:
(A) x = A sin(ky – ωt)

v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string
(A) must be an odd integral multiple of λ/4.
(B) must be an odd integral multiple of λ/2.
(C) must be an odd integral multiple of λ.
(D) must be an even integral multiple ofλ.
Answer:
(A) must be an odd integral multiple of λ/4.

2. Answer in brief.

i) A wave is represented by an equation y = A sin (Bx + Ct). Given that the constants A, B and C are positive, can you tell in which direction the wave is moving?
Answer:
The wave is travelling along the negative x-direction.

ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these, is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?
Answer:
Nodes are the points where the vibrating string can be touched without disturbing its motion.
When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there are no point on the string which can be touched without disturbing its motion.

When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.

When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.

iii) What are harmonics and overtones?
Answer:
A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.

The lowest allowed frequency, n₁, is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being the fundamental or 2n₁, the third harmonic is 3n₁, and so on.

The higher allowed frequencies are called the overtones. Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second overtone, and ‘so on. The relation between overtones and allowed harmonics depends on the system under consideration.

iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the
second harmonic?
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

v) The amplitude of a wave is represented by y = 0.2 sin 4π[\(\frac{t}{0.08}\) – \(\frac{x}{0.8}\)] in SI units. Find
(a) wavelength,
(b) frequency and
(c) amplitude of the wave. [(a) 0.4 m (b) 25 Hz (c) 0.2 m]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 10
y = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Let us compare above equation with the equation of a simple harmonic progressive wave:
y = A sin 2π[\(\frac{t}{T}\) – \(\frac{x}{\lambda}\)] = 0.2 sin 2π[\(\frac{t}{0.04}\) – \(\frac{x}{0.4}\)]
Comparing the quantities on both sides, we get,
A = 0.2 m, T = 0.04 s, λ = 0.4 m
∴ (a) Wavelength (λ) = 0.4 m
(b) Frequency (n) = \(\frac{1}{T}\) = \(\frac{1}{0.04}\) = 25 Hz
(c) Amplitude (A) = 0.2 m

Question 3.
State the characteristics of progressive waves.
Answer:
Characteristics of a progressive wave :

Energy is transmitted from particle to particle without the physical transfer of matter.
The particles of the medium vibrate periodically about their equilibrium positions.
In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency, but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.
A wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.
The velocity of propagation through a medium depends upon the properties of the medium.
Progressive waves are of two types : transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. The progressively changing phase of the successive particles results in the formation of alternate crests and troughs that are periodic in space and time. In an em wave, the electric and magnetic fields oscillate in mutually perpendicular directions, perpendicular to the direction of propagation.
In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave. The progressively changing phase of the successive particles results in the formation of typical alternate regions of compressions and rarefactions that are periodic in space and time. Periodic compressions and rarefactions result in periodic pressure and density variations in the medium. There are no longitudinal em wave.
A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

Question 4.
State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

Stationary waves are produced by the interference of two identical progressive waves travelling in opposite directions, under certain conditions.
The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).
The distance between adjacent nodes (or antinodes) is λ/2.
The distance between successive node and antinode is λ/4.
There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.
A stationary wave does not propagate in any direction and hence does not transport energy through the medium.
In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

[Note : Since the nodes are points where the particles are always at rest, energy cannot be transmitted across a node. The energy of the particles within a loop remains localized, but alternates twice between kinetic and potential energy during each complete vibration. When all the particles are in the mean position, the energy is entirely kinetic. When they are in their extreme positions, the energy is entirely potential.]

Question 5.
Derive an expression for equation of stationary wave on a stretched string.
Answer:
When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.

Consider two simple harmonic progressive waves, of the same amplitude A, wavelength A and frequency n = ω/2π, travelling on a string stretched along the x-axis in opposite directions. They may be represented by
y₁ = A sin (ωt – kx) (along the + x-axis) and … (1)
y₂ = A sin (ωt + kx) (along the – x-axis) …. (2)
where k = 2π/λ is the propagation constant.

By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
y = y₁ + y₂ = A [sin (ωt – kx) + sin (ωt + kx)]
Using the trigonometrical identity,
sin C + sin D = 2 sin \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\),
y = 2A sin ωt cos (- kx)
= 2A sin ωt cos kx [∵ cos(- kx) = cos(kx)]
= 2A cos kx sin ωt … (3)
∴ y = R sin ωt, … (4)
where R = 2A cos kx. … (5)
Equation (4) is the equation of a stationary wave.

Question 6.
Find the amplitude of the resultant wave produced due to interference of two waves given as y₁ = A₁ sin ωt y₂ = A₂ sin (ωt + φ)
Answer:
The amplitude of the resultant wave produced due to the interference of the two waves is
A = \(\sqrt{A_{1}^{2}+2 A_{1} A_{2} \cos \varphi+A_{2}^{2}}\).

Question 7.
State the laws of vibrating strings and explain how they can be verified using a sonometer.
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (n₁) is 1 : 2.

Here, L = 3\(\frac{\lambda}{2}\)
∴ Wavelength, λ = \(\frac{2 L}{3}\) = \(\frac{2 \times 30}{3}\) = 20 cm.

Question 8.
Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.
Answer:
Consider a narrow cylindrical pipe of length l closed at one end. When sound waves are sent down the air column in a cylindrical pipe closed at one end, they are reflected at the closed end with a phase reversal and at the open end without phase reversal. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end.

Taking into account the end correction e at the open end, the resonating length of the air column is L = l + e.
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 20

Let v be the speed of sound in air. In the simplest mode of vibration, there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive anti-node is \(\frac{\lambda}{4}\), where λ is the wavelength of sound. The corresponding wavelength λ and frequency n are
λ = 4L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{4 L}\) = \(\frac{v}{4(l+e)}\) …… (1)
This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.

In the next higher mode of vibration, the first overtone, two nodes and two antinodes are formed. The corresponding wavelength λ₁ and frequency n₁ are
λ₁ = \(\frac{4 L}{3}\) and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{3 v}{4 L}\) = \(\frac{3 v}{4(l+e)}\) = 3n … (2)
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third harmonic.

In the second overtone, three nodes and three antinodes are formed. The corresponding wavelength λ₂ and frequency n₂ are
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 30
which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ,..) is
n_p = (2p + 1)n … (4)
i.e., the pth overtone is the (2p + 1)th harmonic.

Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe closed at one end are n, 3n, 5n, …. That is, only odd harmonics are present as overtones.

Question 9.
Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.
Answer:
Consider a cylindrical pipe of length l open at both the ends. When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.

The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 35
Taking into account the end correction e at each of the open ends, the resonating length of the air column is L = l + 2e.

Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends. The distance between two consecutive antinodes is λ/2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are
λ = 2L and n = \(\frac{v}{\lambda}\) = \(\frac{v}{2 L}\) = \(\frac{v}{2(l+2 e)}\) …. (1)

In the next higher mode, the first overtone, there are two nodes and three antinodes. The corresponding wavelength λ₁ and frequency n₁
λ₁ = L and n₁ = \(\frac{v}{\lambda_{1}}\) = \(\frac{v}{L}\) = \(\frac{v}{(l+2 e)}\) = 2n …. (2)
i.e., twice the fundamental. Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes. The corresponding wavelength λ₂ and frequency n₂ are
λ₂ = \(\frac{2 L}{3}\) and n₂ = \(\frac{v}{\lambda_{2}}\) = \(\frac{3v}{2L}\) = \(\frac{3 v}{2(l+2 e)}\) = 3n …. (3)
or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, …) is
n_p = (p + 1)n … (4)
i.e., the pth overtone is the (p + 1)th harmonic.
Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n, …. That is, all the harmonics are present as overtones.

Question 10.
A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
(b) what will be the smallest distance between two points which are 45º out of phase at an instant of time?
[Ans : π, 8.75 cm ]
Answer:
Data : n = 500 Hz, v = 350 m/s
D = n × λ
∴ λ = \(\frac{350}{500}\) = 0.7 m
(a) in t = 1.0 ms = 0.001 s, the path difference is the distance covered v × t = 350 × 0.001 = 0.35 m
∴ Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.7}\) × 0.35 = π rad

(b) Phase difference = 45° = \(\frac{\pi}{4}\) rad
∴ Path difference = \(\frac{\lambda}{2 \pi}\) × Phase difference
= \(\frac{0.7}{2 \pi}\) × \(\frac{\pi}{4}\) = 0.0875 m

Question 11.
A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency. [Ans : 20 kHz]
Answer:
Data : Distance between two successive nodes =
\(\frac{\lambda}{2}\) = 3.75 × 10^-2 m, v = 1500 m/s
∴ λ = 7.5 × 10^-2m
v = n × λ
∴ n = \(\frac{1500}{7.5 \times 10^{-2}}\) = 20 kHz

Question 12.
Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s) [Ans: ± 0.25, ± 0.75, ± 1.25 and ± 1.75 m from the point at the center]
Answer:
∴ λ = \(\frac{v}{n}\) = \(\frac{330}{330}\) = 1 m
Directly at the cenre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of \(\frac{\lambda}{4}\) from the point at the centre. Other
maxima will be located each \(\frac{\lambda}{2}\) further along.
Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25, ± 1.75 m from the point at the centre.

Question 13.
Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase? [Ans : 1.636 π]
Answer:
Data : v = 330 m/s, n₁ = n₂ = 540 Hz
v = n × λ
∴ λ = \(\frac{330}{540}\) = 0.61 m
Here, the path difference = 3.5 – 3 m = 0.5 m
Phase difference = \(\frac{2 \pi}{\lambda}\) × Path difference
= \(\frac{2 \pi}{0.61}\) × 0.5 = 1.64π rad

Question 14.
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire. [Ans: 1.2 m]
Answer:
Data : m₁ = m₂ = m, L₁ = 60 cm = 0.6 m, T₁ = 1.5 kg = 14.7 N, T₂ = 6 kg = 58.8 N
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 39
The vibrating length of the second wire is 1.2 m.

Question 15.
A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental
frequency. [Ans: 128 Hz]
Answer:
The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let n_C be the fundamental frequency of the closed pipe and n_q, n_q-1, n_q-1 = the frequencies of the q^th, (q + 1)^th and (q + 2)^th consecutive overtones, where q is an integer.

Data : n_q = 640 Hz, n_q-1 = 896 Hz, n_q+2 = 1152 Hz
Since only odd harmonics are present as overtones, n_q = (2q +1) n_C
and n_q+1 = [2(q + 1) + 1] n_C = (2q + 3) n_C
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 40
∴ 14q + 7 = 10q + 15 ∴ 4q = 8 ∴ q = 2
Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.
∴ 5n_C = 640 ∴ b_C = 128Hz

Question 16.
A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length
of tube in the fundamental mode? (speed of the sound = 340 m s^-1). [Ans: 0.5666 m]
Answer:
Data : For the tube open at both the ends, n = 300 Hz and v = 340 m / s Igonoring end correction, the fundamental frequency of the tube is
n = \(\frac{v}{2 L}\) ∴ L = \(\frac{v}{2 n}\) = \(\frac{340}{2 \times 300}\) = 0.566m
The length of the tube open at both the ends is 0.5667 m.

Question 17.
Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s. [Ans: 660 Hz, 1320 Hz, 1980 Hz]
Answer:
Data : Open pipe, ∠25 cm = 0.25 m, v = 330 m / s
The fundamental frequency of an open pipe ignoring end correction,
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 55
Since all harmonics are present as overtones, the first overtone is, n₁ = 2n₀ = 2 × 660 = 1320 Hz
The second overtone is n₂ = 3n = 3 × 660 = 1980 Hz

Question 18.
A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? (Take velocity of sound to be 330 m/s) [Ans : 27.5 cm, 20.625 cm]
Answer:
Data : Open pipe, n_o = 600 Hz, n_{C, 1} = n_{o, 1} (first overtones)
For an open pipe, the fundamental frequency,
n_o = \(\frac{v}{2 L_{O}}\)
∴ The length of the open pipe is
L₀ = \(\frac{v}{2 n_{O}}\) = \(\frac{330}{2 \times 600}\) = 0.275 m
For the open pipe, the frequency of the first overtone is
2n₀ = 2 × 600 = 1200 Hz
For the pipe closed at one end, the frequency of the first overtone is \(\frac{3 v}{L_{O}}\).
By the data, \(\frac{3 v}{4 L}\) = 1200
∴ L_C = \(\frac{3 \times 330}{4 \times 1200}\) = 0.206 m
The length of the pipe open at both ends is 27.5 cm

Question 19.
A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the free end is an antinode.] [Ans: 6.67 m s^-1]
Answer:
Data : L = 1 m, n = 15 Hz.
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
L = \(\frac{\lambda}{4}\) + 4\(\left(\frac{\lambda}{2}\right)\) = \(\frac{9}{4}\)λ
∴ λ = \(\frac{4 L}{9}\) = \(\frac{4}{9}\) × 1 = \(\frac{4}{9}\) m
v = nλ
∴ Speed of the progressive wave
v = 15 × \(\frac{4}{9}\) = \(\frac{60}{9}\) =6.667m/s

Question 20.
A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones? [Ans: 880 Hz, 1320 Hz]
Answer:
Data: n =440Hz
The first overtone, n₁ = 2n =2 × 400 = 880 Hz
The second overtone, n₁ = 3n = 3 × 400 = 1320 Hz

Question 21.
A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last for k is twice that of the first. Calculate the frequencies of the first and the last for k. [Ans: 28 Hz, 56 Hz]
Answer:
Data : n₈ = 2n₁, beat frequency = 4 Hz
The set of tuning fork is arranged in the increasing order of their frequencies.
∴ n₂ = n₁ + 4
n₃ = n₂ + 4 = n₁ + 2 × 4
n₄ = n₃ + 4 = n₁ + 3 × 4
∴ n₈ = n₇ + 4 = n₁ + 7 × 4 = n₁ + 28
Since n₈ = 2n₁,
2n₁ = n₁ + 28
∴ The frequency of the first fork, n₁ = 28 Hz
∴ The frequency of the last fork,
n₈ = n₁ + 28 = 28 + 28 = 56 Hz

Question 22.
A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How
many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N? [Ans: 12 beats]
Answer:
Data : T₁ =40N, n₁ = 384 Hz, T₂ = 40 – 1.24 = 38.76 N
Maharashtra Board Class 12 Physics Solutions Chapter 6 Superposition of Waves 60
∴ The number of beats produced in two seconds = 2 × 6 = 12

Question 23.
A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire. [Ans: 4.9 × 10^-3 kg/m]
Answer:
Data : L = 0.5 m, T = 5 kg = 5 × 9.8 = 49 N, n = 100 Hz
n = \(\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
∴ Linear density, m = \(\frac{T}{4 L^{2} n^{2}}\)
= \(\frac{49}{4(0.5)^{2}(100)^{2}}\)
= 4.9 × 10^-3 kg/m

Question 24.
The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency
of 160 Hz, where should the person press the string? [Ans : 56 cm from one end]
Answer:
Data : L₁ = 80 cm n₁ = 112 Hz, n₂ = 160 Hz
According to the law of length, n₁L₁ = n₂L₂.
∴ The vibrating length to produce the fundamental frequency of 160 Hz,
L₂ = \(\frac{n_{1} L_{1}}{n_{2}}\) = \(\frac{112(80)}{160}\) = 56 cm

12th Physics Digest Chapter 6 Superposition of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 132)

Question 1.
What is the minimum distance between any two particles of a medium which always have the same speed when a sinusoidal wave travels through the medium ?
Answer:
When a sinusoidal wave travels through a medium the minimum distance between any two particles of the medium which always have the same speed is \(\frac{\lambda}{2}\).
Such particles are opposite in phase, i.e., their instantaneous velocities are opposite in direction.
[Note : The minimum distance between any two particles which have the same velocity is λ]

Do you know? (Textbook Page No. 140)

Question 1.
What happens if a simple pendulum is pulled aside and released ?
Answer:
If a simple pendulum is pulled aside and released, it oscillates freely about its equilibrium position at its natural frequency which is inversely proportional to the square root of its length and directly proportional to the square root of the acceleration of gravity at the place. These oscillations, called as free oscillations, are periodic and tautochronous if the displacement of its bob is small and the dissipative forces can be ignored.

Question 2.
What happens when a guitar string is plucked ?
Answer:
When a guitar string is plucked, two wave pulses of the same amplitude, frequency and phase move out from that point towards the fixed ends of the string where they get reflected. For certain ratios of wavelength to length of the string, these reflected pulses moving towards each other will meet in phase to form standing waves on the string. The vibrations of the string cause the air molecules to oscillate, forming sound waves that radiate away from the string. The frequency of the sound waves is equal to the frequency of the vibrating string. In general, the wavelengths of the sound waves and the waves on the string are different because their speeds in the two mediums are not the same.

Question 3.
Have you noticed vibrations in a drill machine or in a washing machine ? How do they differ from vibrations in the above two cases ?
Answer:
Vibrations in the body of a drill machine or that of a washing machine are forced vibrations induced by the vibrations of the motors of these machines. On the other hand, the oscillations of a simple pendulum or a guitar string are free oscillations, produced when they are disturbed from their equilibrium position and released.

Question 4.
A vibrating tuning fork of certain frequency is held in contact with a tabletop and its vibrations are noticed and then another vibrating tuning fork of different frequency is held on the tabletop. Are the vibrations produced in the tabletop the same for both the tuning forks ? Why ?
Answer:
No. Because the tuning forks have different frequencies, the forced vibrations in the tabletop differ both in frequency and amplitude. The tuning fork whose frequency is closer to a natural frequency of the tabletop induces forced vibrations of a larger amplitude.

Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic Induction

June 22, 2024June 21, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 12 Electromagnetic Induction Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 12 Electromagnetic Induction

1. Choose the correct option.

i) A circular coil of 100 turns with a cross-sectional area (A) of 1 m² is kept with its plane perpendicular to the magnetic field (B) of 1 T. What is the magnetic flux linkage with the coil?
(A) 1 Wb
(B) 100 Wb
(C) 50 Wb
(D) 200 Wb
Answer:
(B) 100 Wb

ii) A conductor rod of length (l) is moving with velocity (v) in a direction normal to a uniform magnetic field (B). What will be the magnitude of induced emf produced between the ends of the moving conductor?
(A) BLv
(B) BLv²
(C) \(\frac{1}{2}\)Blv
(D) \(\frac{2 B l}{\mathrm{v}}\)
Answer:
(A) BLv

iii) Two inductor coils with inductance 10 mH and 20 mH are connected in series. What is the resultant inductance of the combination of the two coils?
(A) 20 mH
(B) 30 mH
(C) 10 mH
(D) \(\frac{20}{3}\) mH
Answer:
(A) 20 mH

iv) A current through a coil of self inductance 10 mH increases from 0 to 1 A in 0.1 s. What is the induced emf in the coil?
(A) 0.1 V
(B) 1 V
(C) 10 V
(D) 0.01 V
Answer:
(A) 0.1 V

v) What is the energy required to build up a current of 1 A in an inductor of 20 mH?
(A) 10 mJ
(B) 20 mJ
(C) 20 J
(D) 10 J
Answer:
(A) 10 mJ

2. Answer in brief.

i) What do you mean by electromagnetic induction? State Faraday’s law of induction.
Answer:
The phenomenon of production of emf in a conductor or circuit by a changing magnetic flux through the circuit is called electromagnetic induction.

Faraday’s laws of electromagnetic induction :
(1) First law ; Whenever there is a change in the magnetic flux associated with a circuit, an emf is induced in the circuit.
(2) Second law : The magnitude of the induced emf is directly proportional to the time rate of change of magnetic flux through the circuit.
[Note : The phenomenon was discovered in 1830 by Joseph Henry (1797-1878), US physicist, and independently in 1832 by Michael Faraday (1791 -1867), British chemchemist and physicist.]

ii) State and explain Lenz’s law in the light of principle of conservation of energy.
Answer:
Lenz’s law : The direction of the induced current is such as to oppose the change that produces it.

The change that induces a current may be (i) the motion of a conductor in a magnetic field or (ii) the change of the magnetic flux through a stationary circuit.

Explanation : Consider Faraday’s magnet-and- coil experiment. If the bar magnet is moved towards the coil with its N-pole facing the coil, as in Fig., the number of magnetic lines of induction (pointing to the left) through the coil increases. The induced current in the coil sets up a magnetic field of its own pointing to the right (as given by Amperes right-hand rule) to oppose the growing flux due to the magnet. Hence, to move the magnet towards the coil against this repulsive flux of the induced current, we must do work. The work done shows up as electric energy in the coil.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 1
When the magnet is withdrawn, with its N-pole still facing the coil, the number of magnetic lines of induction (pointing left) through the coil decreases. The induced current reverses its direction to supplement the decreasing flux with its own, as shown in Fig.. Facing the coil along the magnet, the induced current is in the clockwise sense. The electric energy in the coil comes from the work done to withdraw the magnet, now against an attractive force. Thus, we see that Lenz’s law is a consequence of the law of conservation of energy.

[Note : The above law was discovered by Heinrich Friedrich Emil Lenz (1804-65), Russian physicist.]

iii) What are eddy currents? State applications of eddy currents.
Answer:
Whenever a conductor or a part of it is moved in a magnetic field “cutting” magnetic field lines, or placed in a changing magnetic field, the free electrons in the bulk of the metal start circulating in closed paths equivalent to current-carrying loops. These loop currents resemble eddies in a fluid stream and are hence called eddy or Foucault currents [after Jean Bernard Leon Foucault (1819-68), French physicist, who first detected them].

Applications :
(1) Dead-beat galvanometer : A pivoted moving-coil galvanometer used for measuring current has the coil wound on a light aluminium frame. The rotation of the metal frame in magnetic field produces eddy currents in the frame which opposes the rotation and the coil is brought to rest quickly. This makes the galvanometer dead-beat.

(2) Electric brakes : When a conducting plate is pushed into a magnetic field, or pulled out, very quickly, the interaction between the eddy currents in the moving conductor and the field retards the motion. This property of eddy currents is used as a method of braking in vehicles.

iv) If the copper disc of a pendulum swings between the poles of a magnet, the pendulum comes to rest very quickly. Explain the reason. What happens to the mechanical energy of the pendulum?
Answer:
As the copper disc enters and leaves the magnetic field, the changing magnetic flux through it induces eddy current in the disc. In both cases, Fleming’s right hand rule shows that opposing magnetic force damps the motion. After a few swings, the mechanical energy becomes zero and the motion comes to a stop.

Joule heating due to the eddy current warms up the disc. Thus, the mechanical energy of the pendulum is transformed into thermal energy.

v) Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
Answer:
Assuming that their mutual inductance can be ignored, the equivalent inductance of a parallel combination of two coils is given by
\(\frac{1}{L_{\text {parallel }}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}\) or L_parallel = \(\frac{L_{1} L_{2}}{L_{1}+L_{2}}\)
Hence, the equivalent inductance is less than the inductance of either coil.

Question 3.
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω about an axis perpendicular to the plane of the disc and passing through its centre. The disc is placed in a magnetic field B acting perpendicular to the plane of the disc. Determine the induced emf between the rim and the axis of the disc.
Answer:
Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\) (see the figure in the above Note for reference). \(\vec{B}\) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
\(\frac{d A}{d t}\) = frequency of rotation x dA = fdA
where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.
.’. \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2 πr dr) = ωr dr
The total emf induced between the axle and the rim of the rotating disc is
\(|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}\)
For anticlockwise rotation in \(\vec{B}\) pointing down, the axle is at a higher potential.

Question 4.
A horizontal wire 20 m long extending from east to west is falling with a velocity of 10 m/s normal to the Earth’s magnetic field of 0.5 × 10^-4 T. What is the value of induced emf in the wire?
Answer:
Data : l = 20 m, v = 10 m/s. B = 5 × 10^-5 T
The magnitude of the induced emf,
|e| = Blv = (5 × 10^-5)(20)(10) = 10^-2V = 10 mV

Question 5.
A metal disc is made to spin at 20 revolutions per second about an axis passing through its centre and normal to its plane. The disc has a radius of 30 cm and spins in a uniform magnetic field of 0.20 T, which is parallel to the axis of rotation. Calculate
(a) The area swept out per second by the radius of the disc,
(b) The flux cut per second by a radius of the disc,
(c) The induced emf in the disc.
Answer:
Data: R = 0.3m, f = 20 rps, B = 0.2T
(a) The area swept out per unit time by a given radius = (the frequency of rotations) × (the area swept out per rotation) = f(πr²)
= (20)(3.142 × 0.09) = 5.656 m²

(b) The time rate at which a given radius cuts magnetic flux
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B f(πr²)
= (0.2)(5.656) = 1.131 Wb/s

Question 6.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 10 A in 0.2 s, what is the change of flux linkage with the other coil?
Answer:
Data: M = 1.5 H, I_1i = 0, I_1f = 10A, ∆f = 0.2s
The flux linked per unit turn with the second coil due to current I₁ in the first coil is
Φ₂₁ = MI₁
Therefore, the change in the flux due to change in I₁ is
∆₂₁ =M(∆I₁) = M(I_1f – I_1i) = 1.5 (10 – 0)
= 15 Wb
[Note: The rate of change of flux linkage is M(∆I₁/∆t) = 15/0.2 = 75 Wb/s] .

Question 7.
A long solenoid has 1500 turns/m. A coil C having cross sectional area 25 cm2 and 150 turns (N_c) is wound tightly around the centre of the solenoid. If a current of 3.0A flows through the solenoid, calculate :
(a) the magnetic flux density at the centre of the solenoid,
(b) the flux linkage in the coil C,
(c) the average emf induced in coil C if the direction of the current in the solenoid is reversed in a time of 0.5 s. (µ₀ = 4π × 10^-7 H/m)
Answer:
Data: n = 1.5 × 10³ m , A = 25 × 10^-4 m²,
N_c = 150, I = 3A, ∆t = 0.5s,
µ₀ = 4π × 10^-7 H/m
(a) Magnetic flux density inside the solenoid,
B = u₀ nI = (4π × 10^-7)(1500)(3)
= 5.656 × 10^-3 T = 5.656 mT

(b) Flux per unit turn through the coils of the solenoid, Φ_m = BA
Since the coil C is wound tightly over the solenoid, the flux linkage of C is
N_CΦ_m = N_CBA = (150)(5.656 × 10^-3)(25 × 10^-4)
= 2.121 × 10^-3 Wb = 2.121 mWb

(c) Initial flux through coil C,
Φ_i = N_CΦ_m = 2.121 × 10^-3 Wb
Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is Φ_f = -2.121 × 10^-3 Wb
Therefore, the average emf induced in coil C,
e = \(-\frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}=-\frac{(-2.121-2.121) \times 10^{-3}}{0.5}\)
= 2 × 4.242 × 10^-3 = 8.484 × 10^-3 V = 8.484 mV

Question 8.
A search coil having 2000 turns with area 1.5 cm² is placed in a magnetic field of 0.60T. The coil is moved rapidly out of the field in a time of 0.2 second. Calculate the induced emf across the search coil.
Answer:
Data: N = 2000, A_i = 1.5 × 10^-4 m², A_f = 0,
B = 0.6T, ∆t = 0.2s
Initial flux, NΦ_f = NBA_i = 2000(0.6)(1.5 × 10^-4)
= 0.18 Wb
Final flux, NΦ_f = 0, since the coil is withdrawn out of the field.
Induced emf,e = \(-N \frac{\Delta \Phi_{\mathrm{m}}}{\Delta t}=-N \frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}\)
∴ e = \(-\frac{0-0.18}{0.2}\) = 0.9V

Question 9.
An aircraft of wing span of 50 m flies horizontally in earth’s magnetic field of 6 × 10^-5 T at a speed of 400 m/s. Calculate the emf generated between the tips of the wings of the aircraft.
Answer:
Data : l = 50 m, B = 6 × 10^-5T, v = 400 m/s
The magnitude of the induced emf,
|e| = Blv = (6 × 10^-5)(400)(50) = 1.2V

Question 10.
A stiff semi-circular wire of radius R is rotated in a uniform magnetic field B about an axis passing through its ends. If the frequency of rotation of the wire is f, calculate the amplitude of the alternating emf induced in the wire.
Answer:
In one rotation, the wire traces out a circle of radius R, i.e., an area A = πR².
Therefore, the rate at which the wire traces out the area is
\(\frac{d A}{d t}\) = frequency or rotation × A = fA
If the angle between the uniform magnetic field \(\vec{B}\) and the rotation axis is θ, the magnitude of the induced emf is
|e|= B\(\frac{d A}{d t}\) cosθ = BfA cosθ = Bf(πR²)cosθ
so that the required amplitude is equal to Bf(πR²).

Question 11.
Calculate the value of induced emf between the ends of an axle of a railway carriage 1.75 m long traveling on level ground with a uniform velocity of 50 km per hour. The vertical component of Earth’s magnetic field (B_v) is given to be 5 × 10^-5T.
Answer:
Data: l = 1.75 m, v = 50 km/h = 50 × \(\frac{5}{18}\) m/s.
B_v = 5 × 10^-5 T
The area swept out by the wing per unit time = 1v.
∴ The magnetic flux cut by the wing per unit time
= \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B_v(lv)
=(5 × 10^-5)(1.75)(50 × \(\frac{5}{18}\))= 121.5 × 10^-5
= 1.215 mWb/s
Therefore, the magnitude of the induced emf,
|e| =1.215 mV
[Note: In the northern hemisphere, the vertical com ponent of the Earth’s magnetic induction is downwards. Using Fleming’s right hand rule, the port (left) wing-tip would be positive.]

Question 12.
The value of mutual inductance of two coils is 10 mH. If the current in one of the coil changes from 5A to 1A in 0.2 s, calculate the value of emf induced in the other coil.
Answer:
Data: M = 10 mH = 10^-2 H, I_1i = 5A, I_1f = 1 A,
∆t = 0.2s
The mutually induced emf in coil 2 due to the changing current in coil 1,
e₂₁ = \(-M \frac{\Delta I_{1}}{\Delta t}=-M \frac{I_{1 \mathrm{f}}-I_{1 \mathrm{i}}}{\Delta t}\)
= -(10^-2) \(\left(\frac{1-5}{0.2}\right)\) = 0.2 V

Question 13.
An emf of 96.0 mV is induced in the windings of a coil when the current in a nearby coil is increasing at the rate of 1.20 A/s. What is the mutual inductance (M) of the two coils?
Answer:
Data: |e₂| = 9.6 × 10^-2 V, dI₁/dt = 1.2 A/s
|e₂| = M\(\frac{d I_{1}}{d t}\)
Mutual inductance,
M = \(\frac{\left|e_{2}\right|}{d I_{1} / d t}=\frac{9.6 \times 10^{-2}}{1.2}\)
= 8 × 10^-2 H
= 80 mH

Question 14.
A long solenoid of length l, cross-sectional area A and having N₁ turns (primary coil) has a small coil of N₂ turns (secondary coil) wound about its centre. Determine the Mutual inductance (M) of the two coils.
Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = µ₀nI_s ……………….. (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
Φ_CS = BA = (µ₀nI_s)(πR²) ……………… (2)
Thus, their mutual inductance is
M = \(\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}\) = µ₀πR²nN ……………. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.
replacing n with N₁/l and N with N₂. M = µ₀A = \(\frac{N_{1} N_{2}}{l}\)
[Note: The answer given in the textbook misses out the factor of 1.] .

Question 15.
The primary and secondary coil of a transformer each have an inductance of 200 × 10^-6H. The mutual inductance (M) between the windings is 4 × 10^-6H. What percentage of the flux from one coil reaches the other?
Answer:
Data: L_P = L_S = 2 × 10^-4 H, M = 4 × 10^-6 H
M = K\(\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}\)
The coupling coefficient is
K = \(\frac{M}{\sqrt{L_{\mathrm{P}} L_{\mathrm{S}}}}=\frac{4 \times 10^{-6}}{\sqrt{\left(2 \times 10^{-4}\right)^{2}}}=\frac{4 \times 10^{-6}}{2 \times 10^{-4}}\)
= 2 × 10^-2
Therefore, the percentage of flux of the primary reaching the secondary is
0.02 × 100% = 2%

Question 16.
A toroidal ring, having 100 turns per cm of a thin wire is wound on a nonmagnetic metal rod of length 1 m and diameter 1 cm. If the permeability of bar is equal to that of free space (µ₀), calculate the magnetic field inside the bar (B) when the current (i) circulating through the turns is 1 A. Also determine the self-inductance (L) of the coil.
Answer:
Data : l = 1 m, d = 1 cm, n = 100 cm^-1 = 10⁴ m^-1,
I = 100 A, µ₀ = 4π × 10^-7 H/m
The radius of cross section, r = \(\frac{d}{2}\) = 0.5 cm
= 5 × 10^-3 m
(a) Magnetic field inside the toroid,
B = µ₀nI = (4π × 10^-7)(10⁴)(100)
= 0.4 × 3.142 = 1.257 T

(b) Self inductance of the toroid,
L = µ₀2πRn²A = µ₀n²lA = µ₀n²l(πr²)
= (4π × 10^-7)(10⁴)²(1) [π(5 × 10^-3)²]
= π² × 10^-3 = 9.87 × 10^-3 H = 9.87 mH

Question 17.
A uniform magnetic field B(t), pointing upward fills a circular region of radius, s in horizontal plane. If B is changing with time, find the induced electric field.
[Hint : Part of Maxwell’s equation, applied to a time varying magnetic flux, leads us to the equation \(\oint \vec{E} \cdot \overline{\mathrm{d} l}=\frac{-d \phi_{m}}{d t}\), where \(\vec{E}\) is the electric field induced when the magnetic flux changes at the rate of \(\frac{d \phi_{m}}{d t}\)]
Answer:
The area of the region, A = πs², remains constant while B = B(f) is a function of time. Therefore, the induced emf,
e = \(-\frac{d \Phi_{\mathrm{m}}}{d t}=-\frac{d}{d t}(B A)=-A \frac{d B(t)}{d t}=-\pi s^{2} \frac{d B(t)}{d t}\)
[Note : Emf and electric field are different physical quantities, whose respective SI units are the volt and the volt per metre. The question has accordingly been corrected.]

12th Physics Digest Chapter 12 Electromagnetic induction Intext Questions and Answers

Do you know (Textbook Page No. 274)

Question 1.
If a wire without any current is kept in a magnetic field, then it experiences no force as shown in figure (a). But when the wire is carrying a current into the plane of the paper in the magnetic field, a force will be exerted on the wire towards the left as shown in the figure (b). The field will be strengthened on the right side of the wire where the lines of force are in the same direction as that of the magnetic field and weakened on the left side where the field lines are in opposite direction to that of the applied magnetic field. For a wire carrying a current out of the plane of the paper, the force will act to the right as shown in figure (c).
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 2
Answer:
Force on a current-carrying conductor in a magnetic field, \(\vec{F}=I \vec{L} \times \vec{B}\) (Refer unit 10.5). The field due to acurrent-carrying straight conductor is given by right- hand grip rule. As shown in the figure below, the combined field due to a permanent magnet and a current-carrying conductor force the conductor out of the field. The field is strengthened where the two fields are in the same direction and add constructively while the field is weakened where the two fields are opposite in direction.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 3

Use your brain power (Textbook Page No. 282)

Question 1.
It can be shown that the mutual potential energy of two circuits is W = MI₁I₂. Therefore, the mutual inductance (M) may also be defined as the mutual potential energy (W) of two circuits corresponding to unit current flowing in each circuit.
M = \(\frac{W}{I_{1} I_{2}}\)
M = W[I₁ = I₂ = 1]
Answer:
Mutual inductance of two magnetically linked coils equals the potential energy for unit currents in the coils.
1 H = 1 T∙m²/A (= 1 V∙s/A = 1 Ω ∙ s = 1 J/A²)

Use your brain power (Textbook Page No. 284)

Question 1.
Prove that the inductance of parallel wires of length l in the same circuit is given by L = \(\left(\frac{\mu_{0} l}{\pi}\right)\) ln (d / a), where a is the radius of wire and d is separation between wire axes.
Answer:
If l is the current in each wire, from Ampe’re’s law the magnitude of the magnetic field outside each wire is
B = \(\frac{\mu_{0} I}{2 \pi r}\)
By right hand grip rule, the direction of the magnetic field due to both the wires are in the same direction at the point shown. Hence, by the symmetry of the setup, the total magnetic flux through an area dA = l dr shown is two times that due to one wire.
Maharashtra Board Class 12 Physics Solutions Chapter 12 Electromagnetic induction 4

Do you know (Textbook Page No. 285)

Question 1.
The flux rule is the terminology that Feynman used to refer to the law relating magnetic flux to emf. (RP Feynman, Feynman lectures on Physics, Vol II)
Answer:
Modern applications of Faraday’s law of induction :

Electric generators and motors
Dynamos in vehicles
Transformers
Induction furnaces (industrial), induction cooking stoves (domestic)
Radio communication
Magnetic flow meters and energy meters
Metal detectors at security checks .
Magnetic hard disk and tape, storage and retrieval
Graphics tablets
ATM Credit/debit cards, ATM and point-of-sale (POS) machines
Pacemakers

Faraday’s second law of electromagnetic induction is referred by some as the “flux rule”.

Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials

June 22, 2024June 21, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 11 Magnetic Materials Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 11 Magnetic Materials

1. Choose the correct option.

i) Intensity of magnetic field of the earth at the point inside a hollow iron box is.
(A) less than that outside
(B) more than that outside
(C) same as that outside
(D) zero
Answer:
(D) zero

ii) Soft iron is used to make the core of transformer because of its
(A) low coercivity and low retentivity
(B) low coercivity and high retentivity
(C) high coercivity and high retentivity
(D) high coercivity and low retentivity
Answer:
(A) low coercivity and low retentivity

iii) Which of the following statements is correct for diamagnetic materials?
(A) µ_r < 1
(B) χ is negative and low
(C) χ does not depend on temperature
(D) All of above
Answer:
(D) All of above

iv) A rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves ( each having half of the original length) and one piece is made to oscillate freely. Its period of oscillation is T′, the ratio of T′ / T is.
(A) \(\frac{1}2 \sqrt{2}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{2}\)

v) A magnetising field of 360 Am -1 produces a magnetic flux density (B ) = 0.6 T in a ferromagnetic material. What is its permeability in Tm A^-1 ?
(A) \(\frac{1}{300}\)
(B) 300
(C) \(\frac{1}{600}\)
(D) 600
Answer:
(C) \(\frac{1}{600}\)

2 Answer in brief.

i) Which property of soft iron makes it useful for preparing electromagnet?
Answer:
An electromagnet should become magnetic when a current is passed through its coil but should lose its magnetism once the current is switched off. Hence, the ferromagnetic core (usually iron-based) used for an electromagnet should have high permeability and low retentivity, i.e., it should be magnetically ‘soft’.

ii) What happens to a ferromagnetic material when its temperature increases above curie temperature?
Answer:
A ferromagnetic material is composed of small regions called domains. Within each domain, the atomic magnetic moments of nearest-neighbour atoms interact strongly through exchange interaction, a quantum mechanical phenomenon, and align themselves parallel to each other even in the absence of an external magnetic field. A domain is, therefore, spontaneously magnetized to saturation.

The material retains its domain structure only up to a certain temperature. On heating, the increased thermal agitation works against the spontaneous domain magnetization. Finally, at a certain critical temperature, called the Curie point or Curie temperature, thermal agitation overcomes the exchange forces and keeps the atomic magnetic moments randomly oriented. Thus, above the Curie point, the material becomes paramagnetic. The ferromagnetic to paramagnetic transition is an order to disorder transition. When cooled below the Curie point, the material becomes ferromagnetic again.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 2

iii) What should be retentivity and coercivity of permanent magnet?
Answer:
A permanent magnet should have a large zero-field magnetization and should need a very large reverse field to demagnetize. In other words, it should have a very broad hysteresis loop with high retentivity and very high coercivity.

iv) Discuss the Curie law for paramagnetic material.
Answer:
Curie’s law : The magnetization of a paramagnetic material is directly proportional to the external magnetic field and inversely proportional to the absolute temperature of the material.

If a paramagnetic material at an absolute temperature T is placed in an external magnetic field of induction \(\), the magnitude of its magnetization
M_z ∝ \(\frac{B_{\text {ext }}}{T}\) ∴ M_z = C\(\frac{B_{\text {ext }}}{T}\)
where the proportionality constant C is called the Curie constant.
[Notes : (1) The above law, discovered experimentally in 1895 by Pierre Curie (1859-1906) French physcist, is true only for values of Bext/ T below about 0.5 tesla per kelvin.
(2) [C] = [M_z ∙ T] / [B_ext] = [L^-1I ∙ ] /[MT^-2I^-1]
= [M^-1L^-1T²I²],
where denotes the dimension of temperature.]

v) Obtain and expression for orbital magnetic moment of an electron rotating about the nucleus in an atom.
Answer:
In the Bohr model of a hydrogen atom, the electron of charge – e performs a uniform circular motion around the positively charged nucleus. Let r, v and T be the orbital radius, speed and period of motion of the electron. Then,
T = \(\frac{2 \pi r}{v}\) …………….. (1)
Therefore, the orbital magnetic moment asso-ciated with this orbital current loop has a magnitude,
I = \(\frac{e}{T}=\frac{e v}{2 \pi r}\) …………… (2)
Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude
M₀ = current × area of the loop
= I(πr²) = \(\frac{e v}{2 \pi r}\) × πr² = \(\frac{1}{2}\) evr ……………… (3)
Multiplying and dividing the right hand side of the above expression by the electron mass m_e,
M₀ = \(\frac{e}{2 m_{\mathrm{e}}}\) (m_evr) = \(\frac{e}{2 m_{\mathrm{e}}}\) L₀ ……………. (4)
where L₀ = m_evr is the magnitude of the orbital angular momentum of the electron. \(\vec{M}_{0}\) is opposite to \(\vec{L}_{0}\).
∴ \(\vec{M}_{0}=-\frac{e}{2 m_{e}} \overrightarrow{L_{0}}\) ……………. (5)
which is the required expression.

According to Bohr’s second postulate of stationary orbits in his theory of hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to n \(\frac{h}{2 \pi}\) , where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 1
L₀ = m_evr = \(\frac{nh}{2 \pi}\) …………… (6)
Substituting for L₀ in Eq. (4),
M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
For n = 1, M₀ = \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\)
The quantity \(\frac{e n h}{4 \pi m_{\mathrm{e}}}\) is a fundamental constant called the Bohr magneton,
µ_B ∙ µ_B = 9.274 × 10^-24 J/T (or A∙m²) = 5.788 × 10^-5 eV/T.
[ Notes : (1) Magnetic dipole moment is conventionally denoted by µ. (2) The magnetic moment of an atom is expressed in terms of Bohr magneton (vµ_B). (3) According to quantum mechanics, an atomic electron also has an intrinsic spin angular momentum and an associated spin magnetic moment of magnitude µ₅. It is this spin magnetic moment that gives rise to magnetism in matter. (4) The total magnetic moment of the atom is the vector sum of its orbital magnetic moment and spin magnetic moment.]

vi) What does the hysteresis loop represents?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

vii) Explain one application of electromagnet.
Answer:
Applications of an electromagnet:

Electromagnets are used in electric bells, loud speakers and circuit breakers.
Large electromagnets are used in junkyard cranes and industrial cranes to lift iron scraps.
Superconducting electromagnets are used in MRI and NMR machines, as well as in particle accelerators of cyclotron family.
Electromagnets are used in data storage devices such as computer hard disks and magnetic tapes.

Question 3.
When a plate of magnetic material of size 10 cm × 0.5 cm × 0.2 cm (length , breadth and thickness respectively) is located in magnetising field of 0.5 × 10⁴ Am^-1 then a magnetic moment of 0.5 A∙m² is induced in it. Find out magnetic induction in plate.
Answer:
Data : l = 10 cm, b = 0.5 cm, h = 0.2 cm,
H = 0.5 × 10⁴ Am^-1, M = 5 A∙m²
The volume of the plate,
V = 10 × 0.5 × 0.2 = 1 cm² = 10^-6 m²
B = μ₀ (H + M_z) = μ₀ (H + \(\frac{M}{V}\))
The magnetic induction in the plate,
∴ B = 4π × 10^-7 (0.5 × 10⁴ + \(\frac{5}{10^{-6}}\))
= 6.290 T

Question 4.
A rod of magnetic material of cross section 0.25 cm² is located in 4000 Am^-1 magnetising field. Magnetic flux passing through the rod is 25 × 10^-6 Wb. Find out (a) relative permeability (b) magnetic susceptibility and (c) magnetisation of the rod.
Answer:
Data: A = 0.25 cm² = 25 × 10^-6 m²,
H = 4000 A∙m^-1, Φ = 25 × 10^-6 Wb
Magnetic induction is
B = \(\frac{\phi}{A}=\frac{25 \times 10^{-6}}{25 \times 10^{-6}}\) = 1 Wb/m²
(a) B = µ₀µ_rK
∴ The relative permeability of the material,
µ_r = \(\frac{B}{\mu_{0} H}=\frac{1}{4 \times 3.142 \times 10^{-7} \times 4000}\)
= \(\frac{10000}{50.272}\) = 198.91 = 199

(b) µ_r = 1 + χ_m
∴ The magnetic susceptibility of the material,
χ_m = µ_r – 1 = 199 – 1 = 198

Question 5.
The work done for rotating a magnet with magnetic dipole momentm, through 90° from its magnetic meridian is n times the work done to rotate it through 60°. Find the value of n.
Answer:
Data : θ₀ = 0°, θ₁ = 90°, θ₂ = 60°, W₁ = nW₂
The work done by an external agent to rotate the magnet from θ₀ to θ is
W = MB (cos θ₀ – cos θ)
∴ W1 = MB(cos θ₀ – cosθ₁)
= MB (cos 0° – cos 90°)
= MB (1 – 0)
= MB

∴ W2 = MB (cos 0°- cos 60°)
= MB(1 – \(\frac{1}{2}\))
= 0.5MB
∴ W₁ = 2W₂ = MB
Given W₁ = nW₂. Therefore n = 2.

Question 6.
An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10^-11 m, with a speed of 2 × 10⁶ ms^-1 Find the resultant orbital magnetic moment and angular momentum of electron. (charge on electron e = 1.6 × 10^-19 C, mass of electron m_e = 9.1 × 10^-31 kg.)
Answer:
Data: r = 5.3 × 10^-11 m, v = 2 × 10⁶ m/s,
e = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg
The orbital magnetic moment of the electron is
M₀ = \(\frac{1}{2}\) evr
= \(\frac{1}{2}\) (1.6 × 10^-19) (2 × 10⁶) (5.3 × 10^-11)
= 8.48 × 10^-24 A∙m²
The angular momentum of the electron is
L₀ = m_evr
=(9.1 × 10^-31) (2 × 10⁶) (5.3 × 10^-11)
= 96.46 × 10^-36 = 9.646 × 10^-35 kg∙m²/s

Question 7.
A paramagnetic gas has 2.0 × 10²⁶ atoms/m with atomic magnetic dipole moment of 1.5 × 10^-23 A m² each. The gas is at 27° C. (a) Find the maximum magnetization intensity of this sample. (b) If the gas in this problem is kept in a uniform magnetic field of 3 T, is it possible to achieve saturation magnetization? Why? (k_B = 1.38 × 10^-23 JK^-1)[Answer: 3.0× 103 A m-1, No]
(Hint: Find the ratio of Thermal energy of atom of a gas ( 3/2 kBT) and maximum potential energy of the atom (mB) and draw your conclusion)
Answer:
Data: \(\frac{N}{V}\) = 2.0 × 10²⁶ atoms/m³,
μ = 1.5 × 10^-23 Am2, T = 27 + 273 = 300 K,
B = 3T, k^B = 1.38 × 10^-23 J/K, 1 eV = 1.6 × 10^-19 J
(a) The maximum magnetization of the material,
M_z = \(\frac{N}{V}\)μ =(2.0 × 10²⁶) (1.5 ×10^-23)
= 3 × 10³ A/m

(b) The maximum orientation energy per atom is
U_m = -μB cos 180° = μB
= (1.5 × 10^-23) (3) = \(\frac{4.5 \times 10^{-23}}{1.6 \times 10^{-19}}\)
= 2.8 × 10^-4 eV

The average thermal energy of each atom,
E = \(\frac{3}{2}\) k_BT
where k_B is the Botzmann constant.
∴ E = 1.5(1.38 × 10^-23)(300)
= 6.21 × 10^-21 J = \(\frac{6.21 \times 10^{-21}}{1.6 \times 10^{-19}}\)
= 3.9 × 10^-2 eV
Since the thermal energy of randomization is about two orders of magnitude greater than the magnetic potential energy of orientation, saturation magnetization will not be achieved at 300 K.

Question 8.
A magnetic needle placed in uniform magnetic field has magnetic moment of 2 × 10^-2 A m², and moment of inertia of 7.2 × 10^-7 kg m². It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field ?
Answer:
Data: M = 2 × 10^-2 A∙m², I = 7.2 × 10^-7 kg∙m²,
T = \(\frac{6}{10}\) = 0.6 S
T = 2π\(\sqrt{\frac{I}{M B}}\)
The magnitude of the magnetic field is
B = \(\frac{4 \pi^{2} I}{M T^{2}}\)
= \(\frac{(4)(3.14)^{2}\left(7.2 \times 10^{-7}\right)}{\left(2 \times 10^{-2}\right)(0.6)^{2}}\)
= 3.943 × 10^-3 T = 3.943 mT

Question 9.
A short bar magnet is placed in an external magnetic field of 700 guass. When its axis makes an angle of 30° with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to most unstable position.
Answer:
Data : B = 700 gauss = 0.07 tesla, θ = 30°,
τ = 0.014 N∙m
τ = MB sin θ
The magnetic moment of the magnet is
M = \(\frac{\tau}{B \sin \theta}=\frac{(0.014)}{(0.07)\left(\sin 30^{\circ}\right)}\) = 0.4 A∙m²
The most stable state of the bar magnet is for θ = 0°.
It is in the most unstable state when θ = 180°. Thus, the work done in moving the bar magnet from 0° to 180° is
W = MB(cos θ₀ – cos θ)
= MB (cos 0° – cos 180°)
= MB [1 – (-1)]
= 2 MB = (2) (0.4) (0.07)
= 0.056 J
This the required work done.

Question 10.
A magnetic needle is suspended freely so that it can rotate freely in the magnetic meridian. In order to keep it in the horizontal position, a weight of 0.1 g is kept on one end of the needle. If the pole strength of this needle is 20 Am , find the value of the vertical component of the earth’s magnetic field. (g = 9.8 m s^-2)
Answer:
Data: M = 0.2g = 2 × 10^-4kg, q_m = 20 A∙m, g =9.8 m/s²

Without the added weight at one end, the needle will dip in the direction of the resultant magnetic field inclined with the horizontal. The torque due to the added weight about the vertical axis through the centre balances the torque of the couple due to the vertical component of the Earth’s magnetic field.
∴ (Mg)\(\left(\frac{L}{2}\right)\) = (q_m B_v) L
The vertical component of the Earth’s magnetic field,
B_v = \(\frac{M g}{2 q_{\mathrm{m}}}=\frac{\left(2 \times 10^{-4}\right)(9.8)}{2(20)}\) = 4.9 × 10^-5 T

Question 11.
The susceptibility of a paramagnetic material is χ at 27° C. At what temperature its susceptibility be \(\frac{\chi}{3}\) ?
Answer:
Data: χ_m1 = χ, T₁ = 27°C = 300 K, χ_m2 = \(\frac{\chi}{3}\)
By Curie’s law,
M_z = C\(\frac{B_{0}}{T}\)
Since M_z = χ_mH = B₀ = μ₀H
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 3
This gives the required temperature.

12th Physics Digest Chapter 11 Magnetic Materials Intext Questions and Answers

Activity (Textbook Page No. 251)

Question 1.
You have already studied in earlier classes that a short bar magnet suspended freely always aligns in North South direction. Now if you try to forcefully move and bring it in the direction along East West and leave it free, you will observe that the magnet starts turning about the axis of suspension. Do you know from where does the torque which is necessary for rotational motion come from? (as studied in rotational dynamics a torque is necessary for rotational motion).
Answer:
Suspend a short bar magnet such that it can rotate freely in a horizontal plane. Let it come to rest along the magnetic meridian. Rotate the magnet through some angle and release it. You will see that the magnet turns about the vertical axis in trying to return back to its equilibrium position along the magnetic meridian. Where does the torque for the rotational motion come from?

Take another bar magnet and bring it near the suspended magnet resting in the magnetic meridian. Observe the interaction between the like and unlike poles of the two magnets facing each other. Does the suspended magnet rotate continuously or rotate through certain angle and remain stable? Note down your observations and conclusions.

Do you know (Textbook Page No. 255)

Effective magneton numbers for iron group ions (No. of Bohr magnetons)

Ion	Electron configuration	Magnetic moment (in terms of /i_B)
Fe³ +	[Ar] 3s²3p⁶3d⁵	5.9
Fe²⁺	[Ar] 3s²3p⁶3d⁶	5.4
Co²⁺	[Ar] 3s²3p⁶3d⁷	4.8
n²⁺	[Ar] 3s²3p⁶3d⁸	3.2

(Courtsey: Introduction to solid state physics by Charles Kittel, pg. 306 )
These magnetic moments are calculated from the experimental value of magnetic susceptibility. In several ions the magnetic moment is due to both orbital and spin angular momenta.
Answer:
In terms of Bohr magneton (µ_B), the effective magnetic moments of some iron group ions are as follows. In several cases, the magnetic moment is due to both orbital and spin angular momenta.

Ion	Configuration	Effective magnetic moment in terms of Bohr magneton (B.M) (Expreimental values)
Fe³ +	3d⁵	5.9
Fe²⁺	3d⁶	5.4
Co²⁺	3d⁷	4.8
n²⁺	3d⁸	3.2

Remember this (Textbook Page No. 256)

Question 1.
Permeability and Permittivity:
Magnetic Permeability is a term analogous to permittivity in electrostatics. It basically tells us about the number of magnetic lines of force that are passing through a given substance when it is kept in an external magnetic field. The number is the indicator of the behaviour of the material in magnetic field. For superconductors χ = – 1. If you substitute in the Eq. (11.18), it is observed that permeability of material µ = 0. This means no magnetic lines will pass through the superconductor.

Magnetic Susceptibility (χ) is the indicator of measure of the response of a given material to the external applied magnetic field. In other words it indicates as to how much magnetization will be produced in a given substance when kept in an external magnetic field. Again it is analogous to electrical susceptibility. This means when the substance is kept in a magnetic field, the atomic dipole moments either align or oppose the external magnetic field. If the atomic dipole moments of the substance are opposing the field, χ is observed to be negative, and if the atomic dipole moments align themselves in the direction of field, χ is observed to be positive. The number of atomic dipole moments of getting aligned in the direction of the applied magnetic field is proportional to χ. It is large for soft iron (χ >1000).
Answer:
Magnetic permeability is analogous to electric permittivity, both indicating the extent to which a material permits a field to pass through or permeate into the material. For a superconductor, χ = -1 which makes µ = 0, so that a superconductor does not allow magnetic field lines to pass through it.

Magnetic susceptibility (χ). analogous to electrical susceptibility, is a measure of the response of a given material to an applied magnetic field. That is, it indicates the extent of the magnetization produced in the material when it is placed in an external magnetic field. χ is positive when the atomic dipole moments align themselves in the direction of the applied field; χ is negative when the atomic dipole moments align antiparallel to the field. χ is large for soft iron (χ > 1000).

Use your brain power (Textbook Page No. 259)

Question 1.
Classify the following atoms as diamagnetic or paramagnetic.
H, O, Zn, Fe, F, Ar, He
(Hint : Write down their electronic configurations)
Is it true that all substances with even number of electrons are diamagnetic?
Answer:

Atoms	Electronic configuration	No. of Electrons	Diamagnetic/Paramagnetic
H	1s¹„	1	Diamagnetic
0	1s²2s²2p⁴	8	Paramagnetic
Zn	1s²2s²2p⁶3s²3p⁶3d¹⁰4s²	30	Diamagnetic
Fe	1s²2s²2p⁶3s²3p⁶4s²3d⁶	26	Neither diamagnetic nor paramagnetic (ferromagnetic)
F	1s²2s²2p⁵	9	Paramagnetic
Ar	1s²2s²2p⁶3s²3p⁶	18	Diamagnetic
He	Is²	2	Diamagnetic

It can be seen that all substances with an even number of electrons are not necessarily diamagnetic.

Do you know (Textbook Page No. 260)

Question 1.
Exchange Interaction: This exchange interaction in stronger than usual dipole-dipole interaction by an order of magnitude. Due to this exchange interaction, all the atomic dipole moments in a domain get aligned with each other. Find out more about the origin of exchange interaction.
Answer:
Exchange Interaction :
Quantum mechanical exchange interaction be-tween two neighbouring spin magnetic moments in a ferromagnetic material arises as a consequence of the overlap between the magnetic orbitals of two adjacent atoms. The exchange interaction in particular for 3d metals is stronger than the dipole-dipole interaction by an order of magnitude. Due to this, all the atomic dipole moments in a domain get aligned with each other and each domain is spontaneously magnetized to saturation. (Quantum mechanics and exchange interaction are beyond the scope of the syllabus.)

Use your brain power (Textbook Page No. 262)

Question 1.
What does the area inside the curve B – H (hysteresis curve) indicate?
Answer:
A magnetic hysteresis loop is a closed curve obtained by plotting the magnetic flux density B of a ferromagnetic material against the corresponding magnetizing field H when the material is taken through a complete magnetizing cycle. The area enclosed by the loop represents the hysteresis loss per unit volume in taking the material through the magnetizing cycle.

Do you know (Textbook Page No. 262)

Question 1.
What is soft magnetic material?
Soft ferromagntic materials can be easily magnetized and demagnetized.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 4
Hysteresis loop for hard and soft ferramagnetic materials.
Answer:
A soft magnetic material, usually iron-based, has high permeability, low retentivity and low coercivity. In other words, it does not have appreciable hysteresis, i.e., its hysteresis loop is very narrow. Such a material magnetizes and demagnetizes more easily, by small external fields.
Maharashtra Board Class 12 Physics Solutions Chapter 11 Magnetic Materials 5

Do you know (Textbook Page No. 263)

Question 1.
There are different types of shielding available like electrical and accoustic shielding apart from magnetic shielding discussed above. Electrical insulator functions as an electrical barrier or shield and comes in a wide array of materials. Normally the electrical wires used in our households are also shielded. In case of audio recording it is necessary to reduce other stray sound which may interfere with the sound to be recorded. So the recording studios are sound insulated using acoustic material.
Answer:
There are different types of shielding, such as electrical, electromagnetic. magnetic, RF (radio fre quency) and acoustic, to shield a given space or sensitive instrument from unwanted fields of each type.

Maharashtra Board Class 11 Sociology Important Questions Chapter 8 Social Change

June 20, 2024June 19, 2024 by Prasanna

Balbharti Maharashtra State Board Class 11 Sociology Important Questions Chapter 8 Social Change Important Questions and Answers.

Maharashtra State Board 11th Sociology Important Questions Chapter 8 Social Change

Choose the correct alternative and complete the statements.

Question 1.
Changes in the ………………… of society have an impact on social relationships.
(direction / structure / nature)
Answer:
structure

Question 2.
Changes that take place in several directions at the same time is known as …………………
(linear / multi-linear / cyclical)
Answer:
multi-linear

Question 3.
Styles of dressing which was popular few generations ago have become popular today is an example of ………………… change.
(linear / cyclical / multi-linear)
Answer:
cyclical

Question 4.
………………… compares society to a biological organism.
(Auguste Comte / Herbert Spencer / Max Weber)
Answer:
Herbert Spencer

Question 5.
Social change is …………………, as it denotes a time sequence.
(temporal / complex / immoral)
Answer:
temporal

Question 6.
Giving up social evils like dowry or early marriage is an example of ………………… change.
(long-term / short-term / neutral)
Answer:
long-term

Question 7.
………………… change happens suddenly.
(Planned / Unplanned / Temporal)
Answer:
Unplanned

Question 8.
………………… factor is also known as geographical or natural factor.
(Cultural / Economic / Physical)
Answer:
Physical

Question 9.
The physical environment has also been adversely affected by human behaviour in the name of …………………
(displacement / conditions / development)
Answer:
development

Question 10.
…………….. factor is also known as demographic factor.
(Biological / Physical / Social)
Answer:
Biological

Question 11.
……………….. is defined as the number of females per thousand males in any given population.
(Demography / Sex ratio / Fertility)
Answer:
Sex ratio

Question 12.
………………… development is affected by size of population.
(Socio-psychological / Socio-economic / Socio-cultural)
Answer:
Socio-economic

Question 13.
………………… showed an interrelation between the teachings of Protestant religion and spread of capitalism in Europe.
(Karl Marx / Auguste Comte / Max Weber)
Answer:
Max Weber

Question 14.
……………….. conflict has resulted in class-conflict.
(Social / Economic / Cultural)
Answer:
Economic

Question 15.
The LPG Policy was adopted by the government in …………………
(1971 / 1991 / 1994)
Answer:
1991

Question 16.
………………… development is an index of the overall progress of society.
(Cultural / Technological / Physical)
Answer:
Technological

Question 17.
………………… is a means to help develop ideas, skills, solve problems and transform people.
(Technology / Knowledge / Education)
Answer:
Education

Question 18.
………………… is a statistical study of human population.
(Migration / Demography / Composition)
Answer:
Demography

Correct the incorrect pair.

Question 1.
(a) Herbert Spencer – Functionalism
(b) Emile Durkheim – Theory of Three Stages
(c) Auguste Comte – Protestant Ethic
(d) Max Weber – Organic Analogy
Answer:
(b) Emile Durkheim – Anomic suicide

Question 2.
(a) YCMOU – University Grant Commission
(b) UGC – Right to Education
(c) RTE – Post Globalization
(d) LPG – Yashwant Chavan Maharashtra Open University
Answer:
(d) LPG – Liberalization, Privatization and Globalization

Question 3.
(a) Max Weber – Cultural Factor
(b) Karl Marx – Economic Factor
(c) Demographic Factor – Technological Factor
(d) The Latur Earthquake – Physical Factor
Answer:
(c) Demographic Factor – Biological Factor

Question 4.
(a) Changes takes place everywhere – Universality .
(b) An endless process – Continuous process
(c) It is amoral – Neutrality
(d) Denotes a time sequence – Chain reaction
Answer:
(d) Denotes a time sequence – Temporal.

Identify the appropriate term from the given options.

(Physical Factor, Educational Factor, Economic Factor, Technological Factor, Demography, Biological Factor, Socio-cultural Factor, Unplanned Factor, Information Age, Planned Change, Migration, Interaction Chain Reaction, Long-term Change.)
Question 1.
Consequences of overpopulation and underpopulation.
Answer:
Biological factor

Question 2.
The growth of large scale industries.
Answer:
Economic factor

Question 3.
Online examination.
Answer:
Technological factor

Question 4.
Cultural diffusion is a source of change.
Answer:
Socio-Cultural factor

Question 5.
A huge increase in school fees, will have an impact on student enrolment
Answer:
Interactional factor

Question 6.
Eradicating strongly embedded customs like dowry, sati system, etc.
Answer:
Long-term change

Question 7.
Definite spaces marked for residences, economic zones, parks etc.
Answer:
Planned change

Question 8.
A statistical study of human population.
Answer:
Demography

Correct the underlined words and complete the sentence.

Question 1.
Emile Durkheim has given us a theory of three stages of human thought.
Answer:
Auguste Comte has given us a theory of three stages of human thought.

Question 2.
Human society is a structure with various parts unrelated.
Answer:
Human society is a structure with various parts interrelated.

Question 3.
Joint families are now undergoing significant changes is an example of functional changes.
Answer:
Joint families are now undergoing significant changes is an example of structural changes.

Question 4.
Social changes take place overnight.
Answer:
Social changes take place over time.

Question 5.
Decrease in school fees may further result in higher ‘drop out’.
Answer:
Increase in school fees may further result in higher ‘drop out’.

Question 6.
Social change is immoral.
Answer:
Social change is amoral.

Question 7.
Social programmes have to be immediately designed and implemented for the natural calamities affected person.
Answer:
Rehabilitation programmes have to be immediately designed and implemented for the natural calamities affected person.

Question 8.
Some purposeful and unplanned changes are promoted by the government and other agencies.
Answer:
Some purposeful and planned changes are promoted by the government and other agencies.

Question 9.
Social change is a result of the interaction of single factors.
Answer:
Social change is a result of the interaction of multiple factors.

Question 10.
Technological factors consist of climatic conditions, bio diversity, natural resources, etc.
Answer:
Physical factors consist of climatic conditions, bio diversity, natural resources, etc.

Question 11.
Social change is to some extent conditioned by religious factors.
Answer:
Social change is to some extent conditioned by physical factors.

Question 12.
The problems of food, housing, unemployment, poverty are problems as well as direct outcomes of changing sex ratio.
Answer:
The problems of food, housing, unemployment, poverty are problems as well as direct outcomes of changing demography.

Question 13.
Economic competition has resulted in class – conflict, increase in capitalism in the society.
Answer:
Economic competition has resulted in class – conflict, increase in materialism in the society.

Question 14.
For Conflict Theorist, culture is considered a basis for change in society.
Answer:
For Conflict Theorist, conflict is considered a basis for change in society.

Question 15.
One of the benchmarks of a so-called civilized society is its extent of cultural development.
Answer:
One of the benchmarks of a so-called civilized society is its extent of technological development.

Question 16.
The government is taking huge efforts to reach to all sections of society through digitalization.
Answer:
The government is taking huge efforts to reach to all sections of society through Right to Education.

Write suitable examples of the given concepts and justify your answer.

Question 1.
Social change.
Answer:
Example : For example, in today’s ‘Information Age’, the role of a teacher in a school is radically different than it was during the early Vedic period. Structural changes always occur in society. For example, with regard to structure in terms of size of family, joint families are now undergoing significant changes. At the same time there are functional changes in the family system. Education was a function of the family previously. Today it has become a specialized function which has been passed on to schools, colleges through a formal means of education.

Social change is a change in the social structure and social relationships of the society. Society change is a continuous, ongoing process. ‘The term’ social change refers to changes that take place in the structure and functioning of social institutions; for example, government education, economy, marriage, family, religion. Social changes also refers to change in performance of social roles of individuals according to changing times.

Question 2.
Long – term change.
Answer:
Example : Giving up social evils like dowry, early marriage or domestic violence. Eradicating strongly embedded customs and practices take decades to get rid of.

Social changes that may take years or decades to produce results are long – term changes. Significant social changes having long-term effects include the industrial revolution, the abolition of slavery etc. social movements play important role in creating awareness and inspiring discontented members of a society to bring about social change.

Question 3.
Unplanned change.
Answer:
Example : Natural calamities such as famine, floods, earthquakes, tsunami, volcanic eruption, etc. are the instances of unplanned changes. Social change which occurs in the natural course is called the unplanned change. The unplanned changes are spontaneous, accidental or the product of sudden decision.

When a natural disaster takes place, there is a loss of human and animal lives as well as property. Rehabilitation programmes have to be immediately designed and implemented for the affected persons. Unplanned change by its very name suggests that it is a type of change that is not planned. It happens suddenly.

Question 4.
Demography.
Answer:
Example : The study of statistics such as births, deaths, income, or the incidence of disease, which illustrate the changing structure of human populations. The problems of food, housing, unemployment, poor health, poverty, low standard of living etc., are problems as well as direct outcomes of changing demography.

Demography is a statistical study of human population. Demography encompasses the study of the size, structure and distribution of the population. It records spatial and temporal changes in population in response to birth, migration, aging and death. The composition of a particular human population is demography.

Question 5.
Direction of social change.
Answer:
Example : Linear Change – Primitive society moving towards a state of industrialism. Multi-linear Change – Cultural diversity.
Cyclical Change – Styles of dressing or hairstyling which were popular few generations ago have become popular today.

Social change is a change in social structure and social relationship of society. Auguste Comte has given us a theory of three stages of human thought. Society change is a continuous process. Sometimes changes proceed from one stage to another in a single direction known as linear change. It may also be a multi-linear change, that is, changes can take place in several different directions at the same time. Then again, the change may be cyclical, that is, human society goes through certain cycle.

Write short notes.

Question 1.
Characteristics of Social Changes.
Answer:
1. Universality : All human societies change. This could include changes in population, beliefs, tools, attire, customs, roles, music, art, architecture etc. Social change is universal as it takes place everywhere. This change is not uniform as it takes place at differing speeds.

2. Continuous speed : There is continuity of change in society. Right from the emergence of human society from the times of nomadic cave dwellers to the present, every aspect of human life and living has changed.

3. Temporal : Social change is temporal as it denotes a time sequence. Innovations of new things, modifications and renovation of the existing phenomena and the discarding of the old takes time.

4. Interactional chain reaction : The physical, biological, technological, cultural, social economic and other factors may together bring about social change. This is due to mutual interdependence of social phenomenon. Thus, for example, a huge increase in school fees will have an impact on student enrolment.

5. Neutrality : The term social change has no value judgment attached to it. As a phenomenon, it is neither moral nor immoral it is amoral. It is ethically neutral.

6. Short term and long-term change : Some social changes may bring about immediate results while some others may take years or decades to produce results. The purchase of new gadgets for the purpose of entertainment is faster if one has the purchasing capacity when compared to giving up social evils like dowry, early marriage or domestic violence.

7. Planned or unplanned changes : Unplanned changes happens suddenly and is not planned. For example, when a natural disaster takes place, there is a loss of human and animal lives as well as property. Rehabilitation programmes, some purposeful and planned changes promoted by the government are examples of planned change.

Question 2.
Socio cultural factor of social change.
Answer:
Human culture is a process of change. A change in the cultural order is accompanied by a corresponding change in the whole social order. Cultural diffusion is a source of change. Culture includes our values, beliefs, ideas and ideologies, morals, customs and traditions. These are all subject to change and they in turn, cause changes. Ideas and cultural values play a crucial role in social change.

German sociologist, Max Weber gave importance to the cultural factor of social change. He showed an interrelation between the teachings of protestant religion and spread of capitalism in Europe, in his famous book “Protestant Ethic and the spirit of Capitalism”. Also, these are negative consequences of ideologies that promote religious fundamentalism, extremist thinking, superstitious beliefs and practices, mindless values; these leads to untold hardships and human miseries. They become stumbling blocks to change.

Question 3.
Technological factor of social change.
Answer:
One of the benchmarks of a so called civilized society is its extent of technological development. Technological development creates new conditions of life and new conditions for adaptation. Technological development continues to be an index of the overall progress of society. Technological changes have affected social, economic, religious, cultural and political life of human beings.

Opportunities for e-learning, e-library, e-commerce, e-ticketing, online marketing, online examination is possible today, due to technological innovations. We live in digitalized age. Digitalization has helped the government to identify many beneficiaries who can be provided aid.

Question 4.
Educational factor and social changes.
Answer:
Education is a means to help develop ideas and skills, solve problems, transform people. People acquire knowledge, skills, develop, competencies and then use these to seek employment or self¬employment. The purpose of education, its content, its pedagogy is changing. Use of technology within education though e-learning, online education, smart boards, virtual classrooms, national digital library etc have brought about far reaching changes even within the field of education.

Many persons have opportunities to learn due to the efforts of the government. A special effort has been made by the University Grants Commission (UGC) to encourage education for trans¬gender persons. Educations can transform people lives.

Question 5.
Economic factor and social changes.
Answer:
This factor is of unique importance in social change. Stages of economic development in human history are not limited to economic transformation in society. They promote large scale political and social transformations. Economic development affects different institutions. The growth of large-scale industries led to employment opportunities, professionalism, exploitation, trade unionism and so forth. Economic competition has resulted in class-conflict increase in materialism in the society. The class divide continues to this day.

For example, peasant movement, women’s movements, labour movements, student movement, tribal movement etc. for conflict theorists – conflict is considered a basis for change in society. Globalization as a process continues to have a huge impact on Indian society. The LPG policy adopted by the government of India in 1991 has led to far reaching consequences on our political institutions, economy, family, education etc.

Differentiate between.

Question 1.
Physical and biological factors of social change.
Answer:

Physical factor	Biological factor
(i) Physical factor is also known as geographical or natural factor.	(i) Biological factor is also known as demographic factor.
(ii) Physical factors consists of climatic conditions, physical environment, animal life, biodiversity, mineral resources, natural resources, etc.	(ii) The biological factors lies in the biological conditions of social continuity, the perpetuation, growth or decline of a given population, migration or race.
(iii) Social change to some extent conditioned by physical factors.	(iii) Biological factors bring change in population structure.
(iv) Physical factor is modified by man, to meet his needs for example construction, of bridge, dam, highway, road, canal, irrigation etc.	(iv) Population movement from rural to urban area, and other such demographic changes significantly influence the course of social change in a society.

Complete the concept maps.

Identify the significant factor of change for each.
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

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

Question 1.
Social change is a complex phenomenon.
Answer:
This statement is True.

It includes the direction of social change, form of social change, sources of social change, causes of social change, and consequences of social change.
Any alternation, difference, modification that takes place in a human situation through time, can be called social change.
A single factor may trigger a particular change, but it is almost associated with other factors due to interdependence of several factors which makes social change a complex phenomenon.

Question 2.
Cultural diffusion is not a source of change.
Answer:
This statement is False.

Cultural diffusion is a source of change. Culture includes our values, beliefs, ideas and ideologies, morals, customs and traditions.
These are all subject to change and they in turn, cause changes.
Where two cultures meet or clash, social changes are inevitable.

Question 3.
The physical environment has also been adversely affected by human behaviour.
Answer:
This statement is True.

The physical environment has been adversely affected by human behaviour, in the name of ‘development’.
The effects of industrial pollution on the environment and the consequent effects such as global warming, melting of solar caps etc., are examples of the physical environment.
Hence, there is a need for conscious efforts to promote ‘sustainable development for all’. To solve the problem created by physical environment.

Question 4.
Both, overpopulation or under population has its share of consequences.
Answer:
This statement is True.

Both, overpopulation or under-population has its share of consequences.
It leads to regional imbalance, changes in density, skewed sex ratio.
Socio-economic development and access to opportunities is affected by size of population. For example, the growing population of senior citizens, or declining number of youth in some communities creates new challenges for society.

Question 5.
Economic Factor is of unique importance in social change.
Answer:
This statement is True.

Stages of economic development in human history are not limited to economic transformation in society.
They promoted large scale political and social transformations. Amidst these changes, there continues to be a need for a decent livelihood and human welfare.
Economic development affects different institutions like family, education etc. The growth of large-scale industries led to a spurt in cities, employment opportunities, professionalism, exploitation, trade unionism and so forth.

Question 6.
The purpose of education, its content, is changing.
Answer:
This statement is True.

Use of technology within education through e-learning, online education, smart boards, virtual classrooms, National Digital Library etc. have brought about far-reaching changes even within the field of education.
Today many persons have opportunities to learn due to the efforts of the government.
A special effort has been made by the(UGC) to encourage education for transgender persons.

Question 7.
There is no direction to social change.
Answer:
This statement is False.

Sometimes, changes proceed from one stage to another, like in a sequence, and in a single direction.
For example, Auguste Comtes Theory of Three Stages of human thought is an example of linear change.
It may also be multi- linear, that is, changes can take place in several different directions at the same time.
Change may also be cyclical; for example, this is common in the world of fashion. Styles of dressing or which were popular few generations ago have become popular today.

Give your personal response.

Question 1.
The consequences of social change may be constructive or destructive.
Answer:
Every factor of social changes has both constructive or destructive results. Physical factors like natural calamities leads to displacement which affects human life. However, geographic condition may also be favourable for human settlements. Similarly, both over population and under population has its share of consequences. Ideas and cultural values also play a crucial role in social change but some ideologies when promote religious fundamentalism and superstitious beliefs does have negative consequences on society. The growth of large-scale industries has not only increased employment opportunities but also have given rise to exploitation.

Question 2.
The purchase of new gadgets for the purpose of entertainment is faster as compared to giving up social evils like dowry, early marriage or domestic violence.
Answer:
Invention and usage of new gadgets is change in material culture. While social evils like dowry, early marriage or domestic violence is strongly embedded customs and practices and part of non¬material culture. Material culture changes fast as compared to non-material culture which are long-term change.

Question 3.
The social system also becomes dysfunctional at times.
Answer:
Emile Durkheim makes reference to ‘anomic suicide’ where there is a state or normlessness or chaos, which can trigger off suicidal feelings which further leads to instability of society. Similarly, there are many other factors which may disturb the balance of the society. Then human beings have to make conscious efforts to help bring about stability, balance and equilibrium in society.

Answer the following question in detail (about 150 words).

Question 1.
Discuss factors responsible for social change with examples of your own
Answer:
More often than not, social change is a result of the interaction of multiple factors.
1. Physical factors : This factor is also known as geographical or natural factor. Physical factors consist of climatic conditions, physical environment, biodiversity, natural resources etc. Social change is to some extent conditioned by physical factors. Natural calamities such as famine or drought affect human lives. The Latur earthquake in 1993, in Maharashtra had long-term impact in terms of displacement. However, geographical conditions may also be favourable for human settlements. For example, people who live in areas which have plenty of rain, suitable soil conditions, rich in minerals have progressed more rapidly.

2. Biological factor : This factor is also known as demographic factor. Biological factors influence numbers i.e., population, sex composition, birth rate and death rate, fertility rate and the hereditary quality of successive generations factors like size and composition of population produce social change. Socio-economic development and access to opportunities is affected by size of population. For example, the growing population of senior citizens or declining number of youths in some communities creates new challenges for society.

3. Socio-cultural factor : Any change in cultural order is accompanied by a corresponding change in the whole social order. Cultural diffusion is a source of change. Ideas propounded by biologist Charles Darwin, psychoanalyst Sigmund Freud and thinker Karl Marx, for example, in the past century, have had significant impact across the globe. Also, there are negative consequences of ideologies that promote religious fundamentalism, extremist thinking, superstitious beliefs and practices which are stumbling blocks to change.

4. Economic factor : Economic development affects different institutions. The growth of large- scale industries led to a spurt in cities, employment opportunities, professionalism, exploitation, trade unionism and so forth. Economic competition has resulted in class-conflict increase in materialism in society. The ‘conflict approached’ as stated by Karl Marx has had a significant impact on understanding social change and also on movements against injustice in society. For example: peasant movement, labour movement, women’s movement etc. Globalization as an economic process continues to have a huge impact on Indian society.

5. Technological factor : One of the benchmarks of a so-called civilized society is its extent of technological development. Technological development continuous to be an index of the overall progress of society. Our daily life is increasingly loaded with the effects of technology from our homes to our work places, entertainment, voting, banking, e-business, e-governance, etc., opportunities for e-learning, online examination is possible today, due to technological innovations.

6. Educational factor: Education is a means to develop ideas and skills, solve problems, transform people. People do acquire knowledge, skills, develop competencies and then use these to seek employment or self-employment. Use of technology within education through e-learning, online education, smart boards, virtual classrooms, have brought about far reaching changes even within the field of education.

Maharashtra Board Class 11 Sociology Important Questions Chapter 7 Social Stratification

June 20, 2024June 19, 2024 by Prasanna

Balbharti Maharashtra State Board Class 11 Sociology Important Questions Chapter 7 Social Stratification Important Questions and Answers.

Maharashtra State Board 11th Sociology Important Questions Chapter 7 Social Stratification

Choose the correct alternative and complete the statements.

Question 1.
Social ………………. involves the formation of horizontal social divisions.
(stratification / differentiation / hierarchy)
Answer:
differentiation

Question 2.
Social stratification involves ………………. ranking of social strata.
(horizontal / vertical / equal)
Answer:
vertical

Question 3.
Social stratification is a particular form of social ……………….
(equality / inequality / justice)
Answer:
inequality

Question 4.
In ………………. stratification there is no scope for social mobility.
(open / closed / universal)
Answer:
Closed

Question 5.
Caste is the ………………. form of social stratification.
(closed / open / equal)
Answer:
closed

Question 6.
………………. is hereditary and membership is based on birth.
(Class / Caste / Society)
Answer:
caste

Question 7.
Choice of ………………. is not free under caste system.
(prestige, occupation, wealth)
Answer:
Occupation

Question 8.
………………. is essence of caste system.
(Endogamy / Exogamy / Monogamy)
Answer:
Endogamy

Question 9.
A social ………………. is essentially a status group.
(caste / class / division)
Answer:
class

Question 10.
The positions that one inherits at birth is called ………………. status.
(achieved / ascribed / vertical)
Answer:
ascribed

Question 11.
The position that one acquires through effort is called ………………. status.
(ascribed / achieved / prestige)
Answer:
achieved

Question 12.
………………. is associated with prestige.
(Role / Status / Mobility)
Answer:
Status

Question 13.
Higher education gives opportunities for ………………. mobility.
(downward / upward / middle)
Answer:
upward

Question 14.
………………. stratification refers to social ranking, where men typically inhabit higher statuses than women.
(Class / Gender / Caste)
Answer:
Gender

Question 15.
………………. refers to the biological distinction between females and males.
(Gender / Inequality / Sex)
Answer:
Sex

Question 16.
………………. is the belief that one sex is superior that the other.
(Gender / Sexism / Feminism)
Answer:
Sexism

Question 17.
Male dominance is supported by ……………….
(matriarchy / patriarchy / endogamy)
Answer:
patriarchy

Question 18.
The process of ………………. is gendered and creates gender hierarchy.
(stratification / socialization / mobility)
Answer:
socialization

Question 19.
Social ………………. means the movement of position from lower to higher one.
(stratification / movement / equality)
Answer:
mobility

Question 20.
………………. mobility refers to change of residence or job without status change.
(Vertical / Horizontal / Intergenerational)
Answer:
Horizontal

Question 21.
………………. mobility stands for change of social position.
(Horizontal / Vertical / Intergenerational)
Answer:
Vertical

Question 22.
………………. mobility means that one generation changes its social status in contrast to previous generation.
(Vertical / Intergenerational / Intragenerational)
Answer:
Intergenerational

Question 23.
………………. mobility takes place in the lifespan of one generation.
(Vertical / Intergenerational / Intragenerational)
Answer:
Intragenerational

Correct the incorrect pair.

Question 1.
(a) Good connections, networking – Social Capital
(b) Reputation and status – Economic Capital
(c) Knowledge of art – Cultural Capital
(d) Higher education – Upward Mobility
Answer:
(b) Reputation and status – Symbolic Capital

Question 2.
(a) Emotional, Cooperative – Traditional Feminine Traits
(b) Ambitious, Independent – Traditional Masculine Traits
(c) Purity and pollution – Caste
(d) Prestige – Gender Stratification
Answer:
(d) Prestige – Class

Question 3.
(a) Endogamy – Caste
(b) Wealth and income – Class
(c) Sexism – Gender Stratification
(d) Male dominance – Matriarchy
Answer:
(d) Male dominance – Patriarchy

Identify the appropriate term from the given options.

(Open stratification, Prestige, Education, Sex, Vertical Mobility, Horizontal Mobility, Closed Stratification, Lifestyles, Gender Stratification, Sexism, Gender Socialization, Intergenerational Mobility, Intragenerational Mobility.)
Question 1.
The varna and the caste system in India.
Answer:
Close Stratification

Question 2.
Power, property, intelligence, skills etc.
Answer:
Open Stratification

Question 3.
It reflects the speciality in preferences, tastes and values of a class.
Answer:
Lifestyles

Question 4.
It refers to the respect and admiration with which an occupation is regarded by society.
Answer:
Prestige

Question 5.
It gives opportunities for upward mobility.
Answer:
Education

Question 6.
It refers to social ranking where men typically inhabit higher statuses than woman.
Answer:
Gender Stratification

Question 7.
Biological distinction between females and males.
Answer:
Sex

Question 8.
A belief that one sex is superior than the other.
Answer:
Sexism

Question 9.
Boys are given toy cars or ball to play.
Answer:
Gender Stratification

Question 10.
Ascending or descending type of mobility.
Answer:
Vertical Mobility

Question 11.
It refers to change of residence or job without status change.
Answer:
Horizontal Mobility

Question 12.
An individual moves up and occupies a higher social position than his previous social position.
Answer:
Intragenerational Mobility

Correct underlined words and complete the sentence.

Question 1.
Social stratification refers to the existence of structured equalities in a society.
Answer:
Social stratification refers to the existence of structured inequalities in a society.

Question 2.
The pattern of mobility stays same from generation to generation.
Answer:
The pattern of inequality stays same from generation to generation.

Question 3.
Open stratification is one in which individuals do not enjoy the freedom of changing their social strata.
Answer:
Close Stratification is one in which individuals do not enjoy the freedom of changing their social strata.

Question 4.
In closed stratification there is scope for social mobility.
Answer:
In open stratification there is scope for social mobility.

Question 5.
The caste system in modern industrial society is an example of open stratification.
Answer:
The class system in modern industrial society is an example of open stratification.

Question 6.
The Spanish word ‘casta’ means colour.
Answer:
The Spanish word ‘casta’ means breed or race

Question 7.
The membership of caste is determined by effort.
Answer:
The membership of caste is determined by birth.

Question 8.
Hereditary is a scheme, which arranges castes in terms of higher and lower status.
Answer:
Hierarchy is a scheme, which arranges castes in terms of higher and lower status.

Question 9.
In a caste society, there is an equal distribution of privileges and disabilities among its members.
Answer:
In a caste society, there is an unequal distribution of privileges and disabilities among its members.

Question 10.
Every caste insists that its members should marry outside the group.
Answer:
Every caste insists that its members should marry within the group.

Question 11.
Caste is related to status.
Answer:
Class is related to status.

Question 12.
Caste is almost a universal phenomenon.
Answer:
Class is almost a universal phenomenon.

Question 13.
Rank is associated with prestige.
Answer:
Status is associated with prestige.

Question 14.
An open class system is one in which horizontal social mobility is possible.
Answer:
An open class system is one in which vertical social mobility is possible.

Question 15.
Hereditary reflect the specialty in preferences, tastes and values of a class.
Answer:
Life-styles reflect the specialty in preferences, tastes and values of a class.

Question 16.
Bourdieu gives three types of capital to explain hierarchy in society.
Answer:
Bourdieu gives four types of capital to explain hierarchy in society.

Question 17.
A social class is an achievement of people who have same status, rank etc.
Answer:
A social class is an aggregate of people who have same status, rank etc.

Question 18.
One’s class position is directly proportionate to one’s income.
Answer:
One’s class position is not directly proportionate to one’s income.

Question 19.
There is an open reciprocal relationship between social class and education.
Answer:
There is a close reciprocal relationship between social class and education.

Question 20.
Education refers to the respect and admiration with which an occupation is regarded by society.
Answer:
Prestige refers to the respect and admiration with which an occupation is regarded by society.

Question 21.
Race stratification refers to unequal distribution of wealth, power and privilege between the two sexes.
Answer:
Gender stratification refers to unequal distribution of wealth, power and privilege between the two sexes.

Question 22.
Sex refers to the social aspects of differences between male and female.
Answer:
Gender refers to the social aspects of differences between male and female.

Question 23.
Submissive, dependent, timid are traditional notion of masculine traits.
Answer:
Submissive, dependent, timid are traditional notion of feminine traits.

Question 24.
Under horizontal mobility, a person changes one’s occupation but the overall social standing remains the same.
Answer:
Under vertical mobility, a person changes one’s occupation but the overall social standing remains the same.

Question 25.
Intragenerational mobility may be upward or downward.
Answer:
Intergenerational mobility may be upward or downward.

Question 26.
Inequality is social and persists over generation.
Answer:
Stratification is social and persists over generation.

Write suitable examples of given concepts and justify your answer.

Question 1.
Caste as closed system of stratification.
Answer:
Example : Varna System – Hindu society is divided into four varnas – Brahmins, Kshatriya, Vaishyas and Shudras. Occupation social status and lifestyle of every varna is determined by some rules and regulations and it is a group based on birth.

Each position in the caste structure is defined in terms of its ‘purity and pollution’. In a caste stratification system, an individual’s position depends on the status attributes ascribed by birth. Therefore, caste is the closed form of social stratification. Since membership based on birth, morality from one caste to another is impossible. Each caste has its own traditional social status, occupation, customs, rules and regulations. Castes are endogamous group and every caste insist that its members should marry within the group which further makes the caste system rigid.

Question 2.
Gender role socialization.
Answer:
Example : Boys are given toy cars or lego sets or bat and ball to play whereas girls are given household sets, medical sets, dolls etc. It explains why human males and females behave in different ways and learn different social roles.

The concept of gender stratification refers to society’s unequal distribution of wealth, power and privilege between the sexes. The process of socialization is gendered and creates gender hierarchy.

Question 3.
Almost all societies are characterised by sexism.
Answer:
Example : Sexist concepts teaches narratives about traditional gender roles for males and females. Women are considered to be the weaker sex and less capable than man in various areas like business, politics etc., and they are confined to the domestic realm of nurturance.

In most countries throughout the world, societies allocate fewer resources to women than to men. Sexism is the belief that one sex is superior than the other and almost all societies are characterized by sexism. Although, societies have been believing in the superiority of men over women and therefore have been dominating women. The male dominance is supported further by patriarchy, which is a form of social organization in which men dominate, oppress and exploit women.

Question 4.
Social stratification is universal but variable.
Answer:
Example : Influence of caste system on all aspects of the Indian society. Practice of class system, gender roles around the world.
Societies around us are heterogeneous in nature and thus divided into various strata or groups. Most societies exist with social systems of social division and social stratification. Everywhere society is divided into various strata. Hence, social stratification is found everywhere. At the same time the nature of inequality varies. ‘What’ is unequal and ‘how’ unequal, changes within the context of the societies.

Write short notes.

Question 1.
Gender stratification.
Answer:
1. Gender stratification refers to social ranking, where men typically inhabit higher statuses than women. A common general definition of gender stratification refers to the unequal distribution of wealth, power and privilege between the two sexes.

2. Throughout the world, most societies allocate fewer resources to women than men. Almost all societies are characterized by sexism. Sexism is the belief that one sex is superior than the other.

3. Although, societies have been believing in the superiority of men over women and therefore have been dominating women. This male dominance is supported further by patriarchy.

4. The process of socialization is gendered and creates gender hierarchy. E.g., Boys are given toy cars or legs sets or bat and ball to play whereas girls are given household sets, medical sets, dolls, etc.

Question 2.
Types of social stratification.
Answer:
Social stratification is of two types – Closed stratification and Open stratification.
1. Closed stratification is one in which individuals or groups do not enjoy the freedom of changing their social strata. The individual who gets a social strata by birth can never change it in one’s lifetime. In this type of social stratification there is no scope for social mobility. Example – The Varna system and the caste system in India are examples of closed stratification.

2. Open stratification is one in which individuals or groups enjoy the freedom of changing their social strata, i.e., in this type of social stratification there is scope for social mobility. Example – The class system in modern industrial society is an example of open stratification, The criteria of open stratification are power, property, intelligence, skills, etc.

Question 3.
Characteristics of class.
Answer:
1. Wealth and Income: Possession of substantial amounts of wealth is the main characteristics distinguishing the upper class from the other class groups in society. Persons having more wealth and income generally have higher social position and respect in society.

2. Occupation : Occupation is an exceedingly important aspect of social class. It is another determinant of class status. It is a well known fact that some kinds of work are more honourable than others, e.g., doctors, engineers, administrators, professors and lawyers hold a higher position than people who are in labour-intensive professions.

3. Education : There is a close reciprocal relationship between social class and education. Higher education gives opportunities for upward mobility. Thus education, is one of the main levers of social class.

4. Prestige : It refers to the respect and administration with which an occupation is regarded by society. Besides wealth, occupation and education, there are certain other criteria which help a person to gain prestige and subsequent higher social status in the society, e.g., family background, kinship, place of residence, etc.

Differentiate between.

Question 1.
Horizontal Mobility and Vertical Mobility.
Answer:

Horizontal Mobility	Vertical Mobility
(i) It refers to change of residence or job without status change.	(i) It refers to any change in the occupational economic or political status of an individual or a group which leads to change of their position.
(ii) Under this type of social mobility a person changes one’s occupation but the overall social position remains the same.	(ii) It stands for change of social position either upward or downward.
(iii) Certain occupation like doctor, engineer and teacher may enjoy the same status hence when an engineer changes one’s occupation from engineer to teaching engineering, there is a horizontal shift but no change takes place in his social position.	(iii) A person who works hard as a salesman, earns money and starts his own business successfully. In such position there is a clear change in the position of the individual.
(iv) A change in the social hierarchy does not take place.	(iv) Change in the social hierarchy takes place.

Question 2.
Closed stratification and Open Stratification.
Answer:

Closed stratification	Open stratification
(i) Closed stratification is one in which individuals or groups do not enjoy the freedom of changing their social strata.	(i) Open stratification is one in which individuals or groups enjoy the freedom of changing their social strata.
(ii) In this type of social stratification there is no scope for social mobility.	(ii) In this type of social stratification there is scope for social mobility.
(iii) Ascribed status – The individual who gets a social strata by birth can never change it in one’s lifetime.	(iii) Achieved status – Individuals or groups move from one strata to another on the basis of their achievement.
(iv) Example : The varna system and the caste system in India are examples of close stratification.	(iv) Example : The class system in modern industrial society is an example of open stratification.

Explain the following concept with suitable examples.

Question 1.
Intragenerational Mobility
Answer:

This type of mobility takes place in the lifespan of one generation.
Here the individual moves up and occupies a higher social position than previously.

Example : A person may start one’s career as a clerk and after acquiring more education, becomes an IFS Officer.

Question 2.
Horizontal Mobility
Answer:

It refers to change of residence or job without status change.
Under this type of social mobility, a person changes one’s occupation but the overall social standing remains the same.

Example : Certain occupation like doctor, engineer and teacher may enjoy the same status but when an engineer changes one’s occupation from engineer to teaching engineering there is a horizontal shift from one occupational category to another but no change has taken place in the system of social stratification.

Question 3.
Closed stratification
Answer:

Closed stratification is one in which individuals or groups do not enjoy the freedom of changing their social strata.
The individual who gets a social strata by birth can never change it in one’s lifetime.
In this type of social stratification there is no scope for social mobility.
Example : The Varna system and the caste system in India are examples of closed stratification.

Question 4.
Prestige
Answer:

It refers to the respect and admiration with which an occupation is regarded by society.
Prestige is independent of the particular person who occupies a job.
Sociologists have tried to assign prestige rankings to various occupations.
Besides wealth, occupation and education, there are certain other criteria which help a person to gain prestige and subsequent higher social status in the society.
Example : Family background, kinship, place of residence, etc.

Question 5.
Gender and sex
Answer:

Sex refers to the biological distinction between females and males.
In contrast the term gender refers to the social aspects of differences and hierarchies between male and female.
Sex may be male or female whereas gender refers to the social meaning of masculinity and femininity. Gender determines how one should behave in society.

Example : Men are supposed to behave in a masculine manner and certain attributes are assigned to men such as courage, bravery, physical strength. On the other hand, women are assigned with attributes like nurturance, care, love, sacrifice which help them to behave in a feminine manner.

Complete the concept maps.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

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

Question 1.
Occupation is determinant of class status
Answer:
This statement is True.

Occupation is an important aspect of social class and it is another determinant of class status.
It is a well-known fact that some kinds of work are more honourable than others, e.g., doctors, engineers, professors and lawyers hold a higher position than people who are in labour-intensive professions.
Occupation is also one of the best clues to one’s way of life, and therefore to one’s social class membership.

Question 2.
In most countries, societies allocate equal resources to women and men.
Answer:
This statement is False.

In most countries throughout the world, societies allocate fewer resources to women than to men.
Almost all societies are characterized by sexism. Sexism is the belief that one sex is superior that the other.
All through, societies have been believing in the superiority of men over women and therefore have been dominating women which is further supported by patriarchy.

Question 3.
Horizontal mobility can be labelled as ascending or descending type of mobility.
Answer:
This statement is False.

It refers to change of residence or job without status change.
Under this type of social mobility, a person changes one’s occupation but the overall social standing remains the same.
Certain occupations like doctor, engineer, and teacher may enjoy the same status but when an engineer changes one’s occupation from engineer to teaching engineering, there is horizontal shift from one occupational category to another but no actual change takes place in the system of social stratification.

Question 4.
Social stratification is consequential.
Answer:
This statement is True.

Social stratification is consequential as it affects every aspect of life of all individuals.
Some individuals experience positive consequences, while others face negative consequences of the hierarchy in a particular society.
It leads to two important consequences – life chances and life styles.

Question 5.
One’s class position is directly proportionate to one’s income.
Answer:
This statement is False.

One’s class position is not directly proportionate to one’s income.
For example, a car mechanic has a lower social status than a teacher, though the income may be greater than the teacher.
Income is not always the determinant of one’s class position.

Question 6.
Gender and sex are interchangeable term.
Answer:
This statement is False.

Gender and sex are not interchangeable terms. Sex refers to the biological distinction between males and males.
In contrast, the term gender refers to the social aspects of differences and hierarchies between male and female. Gender is result of socialization and a social construct.
Sex may be male or female whereas gender refers to the social meaning of masculinity and femininity. It determines how one should behave in society.

Give your personal response.

Question 1.
Women prohibited by certain gatherings, spaces or positions is a result of gender stratification.
Answer:
Gender stratification leads to discrimination which affects many aspects in the lives of woman. Gender stratification results into gender inequality can be analysed on the bases of prestige, style of life, privileges, and opportunities, association with social groups, income, education, occupation, and power. These highly discriminatory practices are still taking place at an alarming rate, affecting the lives of many women today.

Question 2.
There is a close reciprocal relationship between social class and education. Explain how education is one of the main levers of social class.
Answer:
Education is a means to help develop ideas and skills, generate knowledge and help people to seek employment or self employment which further helps to change one’s social position. Higher education gives opportunities for upward mobility, one’s amount and kind of education affects the class rank he will secure. Thus, education is one of the main levers of a man’s social class.

Question 3.
Membership of the caste is not voluntary or not by selection but is obligatory and accidental. Explain how caste system is the closed form of stratification.
Answer:
In a caste stratification system, an individual’s position depends on the status attributes ascribed by birth. Each position in the caste structure is defined in terms of its ‘purity and pollution’. The individual who gets a social strata by birth can never change it in one’s lifetime. There is no scope for social mobility. Therefore, caste is closed form of social stratification.

Question 4.
Caste discrimination exists in India despite special laws.
Answer:
Despite legislation caste discrimination and caste-based crimes persist throughout India. Scheduled Castes (SC), have been the victims of the Hindu caste system for centuries. They are subjected to numerous forms of caste discrimination.

Answer the following in detail (About 150-200 words).

Question 1.
Discuss caste as a form of social stratification.
Answer:
In traditional Indian society different castes formed a hierarchy of social preference. Each position in the caste structure is defined in terms of its purity and pollution. In a caste stratification system, an individual’s position depends on the status attributes ascribed by birth. Therefore, caste is the closed form of social stratification. Caste is hereditary and thus the status of the person in caste based. Society is not determined by the wealth one owns but it depends on the status of the caste in which one is born.

Each caste has its own traditional social status , occupations, customs, rules and regulations. In a caste society, there is an unequal distribution of privileges and disabilities among its members. Choice of occupation is not free under the caste system as occupations are hereditary. The members of a caste are expected to follow their traditional occupation. Caste are divided into sub-castes and each sub-caste is an endogamous group. Hence, a caste is a rigid social class into which members are born.

Maharashtra Board Class 11 Sociology Important Questions Chapter 6 Socialization

June 20, 2024June 19, 2024 by Prasanna

Balbharti Maharashtra State Board Class 11 Sociology Important Questions Chapter 6 Socialization Important Questions and Answers.

Maharashtra State Board 11th Sociology Important Questions Chapter 6 Socialization

Choose the correct alternative and complete the statements.

Question 1.
Socialization is a ………………….. process.
(lifelong / learning / transitional)
Answer:
Life-long/earning

Question 2.
………………….. is a study of genes and heredity.
(Socialization / Genetics / Biology)
Answer:
Genetics

Question 3.
The process of ………………….. signifies the role of‘nurture’ in human development.
(culture / stratification / socialization)
Answer:
socialization

Question 4.
Socialization transforms the individual self into a ………………….. self.
(political / primary / social)
Answer:
social

Question 5.
………………….. transforms the biological organism into a social being.
(Culture / Art / Socialization)
Answer:
Socialization

Question 6.
………………….. is the first stage of the process of socialization, according to Mead.
(Play Stage / Imitation / Game Stage)
Answer:
Imitation

Question 7.
In ………………….. stage, a child internalises the attitudes of others, significant to him/her.
(imitation / game / play)
Answer:
play

Question 8.
In ………………….. stage, children learn to behave according to the impressions of others.
(imitation / play / game)
Answer:
game

Question 9.
The ………………….. is formed through our interactions with others and our understanding of others responses.
(self / imitation / role)
Answer:
self

Question 10.
The learning which happens in the early years is termed as ………………….. socialization.
(secondary / primary / tertiary)
Answer:
primary

Question 11.
………………….. socialization involves intense cultural learning and takes place in childhood.
(Secondary / Primary / Tertiary)
Answer:
Primary

Question 12.
Family is the ………………….. agency of socialization.
(formal / informal / non-formal)
Answer:
informal

Question 13.
………………….. are friendship groups made up of people of similar age.
(Family / Peer groups / Neighbour)
Answer:
Peer groups

Question 14.
Peer groups use ………………….. sanctions.
(informal / formal / egalitarian)
Answer:
informal

Question 15.
In peer groups, the interactions are ………………….. when compared to family or school.
(dependent / formal / egalitarian)
Answer:
egalitarian

Question 16.
………………….. are the means for delivering impersonal communication directed to a vast audience.
(Family / Peer groups / Mass media)
Answer:
Mass media

Question 17.
One of the significant agents of adult socialization is the …………………..
(school / family / workplace)
Answer:
workplace

Question 18.
………………….. is the process of unlearning and relearning new norms and values.
(Adult socialization / Re-socialization / Socialization)
Answer:
Resocialization

Question 19.
………………….. is credited for having popularized the term ‘total institution’.
(Erving Goffman / Max Weber / George Mead)
Answer:
Erving Goffman

Question 20.
………………….. involves a process of tearing down and rebuilding an individual’s role.
(Adult socialization / Re-socialization / Socialization)
Answer:
Re-socialization

Question 21.
………………….. is often a deliberate and an intense social process.
(Socialization / Adult socialization / Re-socialization)
Answer:
Re-socialization

Question 22.
The term ………………….. Institution refers to new ways of life in controlled and regulated manner.
(Social / Total / Political)
Answer:
Total

Question 23.
…………………. is the second stage of the process of socialization, according to Mead.
(Imitation / Play stage / Game stage)
Answer:
Play stage

Question 24.
…………………. indicates continuous process of learning.
(Socialization / Adult socialization / Re-socialization)
Answer:
Adult socialization

Question 25.
A ………………….. is a ereoeraohicallv localized community within a larger city, town etc.
(neighbourhood / family / school)
Answer:
neighbourhood

Correct the incorrect pair.

Question 1.
(a) Senior citizen group – Peer Group
(b) Cultural festivals, sports activities – Neighbourhood
(c) Professional networks – Informal Groups
(d) Newspaper, television – Mass Media
Answer:
(c) Professional networks – Formal Groups

Question 2.
(a) A Prison sentence – Adult socialization
(b) Role play – Game Stage
(c) A significant other – Play Stage
(d) Lack of understanding – Imitation
Answer:
(a) A Prison sentence – Resocialization

Question 3.
(a) Family, peer group – Primary Socialization
(b) School, professional networks – Secondary Socialization
(c) It enabled media and social media – School
(d) Military camps, prisons – Total Institution
Answer:
(c) It enabled media and social media – Mass Media

Identify the appropriate term from the given options.

(Internet, Peer Group, Childhood, Socialization, Unsocialization, The Self, Imitation, Play Stage, Game Stage, Primary Socialization, Secondary Socialization, Positive Sanctions, School, Technology, Negative Sanctions, Mass Media, Neighbourhood, Workplace, Adult Socialization, Re-socialization.)

Question 1.
Well documented cases of Feral, Mogli, Genies.
Answer:
Unsocialization

Question 2.
It emerges through communication and interactions with others.
Answer:
The Self

Question 3.
A little boy helping his parents clean the floor.
Answer:
Imitation

Question 4.
It extends over the entire life of a person.
Answer:
Secondary Socialization

Question 5.
Grades, prizes, praise.
Answer:
School

Question 6.
It has increased the speed of mass media.
Answer:
Technology

Question 7.
It has wider reach and can spread information in a more democratic manner.
Answer:
Mass Media

Question 8.
It helps a person to broaden his/her horizons in terms of social acceptance and tolerance towards the others.
Answer:
Workplace

Question 9.
It involves complete alteration of the easier life style and thinking.
Answer:
Re-socialization

Correct the underlined words and complete the sentence.

Question 1.
Instincts study the inheritance patterns in traits that are handed down from parents to off-spring.
Answer:
Genetics study the inheritance patterns in traits that are handed down from parents to off¬spring.

Question 2.
Cases like Genies suggest that animal contact is necessary for the basic social and physical development.
Answer:
Cases like Genies suggest that human contact is necessary for the basic social and physical development.

Question 3.
Institution transforms the individual self into a social self.
Answer:
Socialization transforms the individual self into a social self.

Question 4.
George Mead has elaborated on the process of building political self.
Answer:
George Mead has elaborated on the process of building social self.

Question 5.
In the game stage, responses are not organised.
Answer:
In the play stage, responses are not organised.

Question 6.
A child at play stage is likely to be aware of the different responsibilities of people.
Answer:
A child at game stage is likely to be aware of the different responsibilities of people.

Question 7.
A child gets acquainted with values, customs, behavioural norms and manners through secondary socialization.
Answer:
A child gets acquainted with values, customs, behavioural norms and manners through primary socialization.

Question 8.
Planned curriculum refers to things we learn from attending the school such as obedience to authority.
Answer:
Hidden curriculum refers to things we learn from attending the school such as obedience to authority.

Question 9.
Radio has an influence on children from a very young age and affects their cognitive and social development.
Answer:
Television has an influence on children from a very young age and affects their cognitive and social development.

Question 10.
Neighbourhood also sometimes helps in changing the notions of‘right and wrong’.
Answer:
Workplace also sometimes helps in changing the notions of ‘right and wrong’.

Question 11.
Adult socialization is the process of unlearning old norms.
Answer:
Re-socialization is the process of unlearning old norms.

Question 12.
Socialization leads to rehabilitation of one’s behaviour.
Answer:
Re-socialization leads to rehabilitation of one’s behaviour.

Question 13.
Prisons, military camps, private boarding schools, are examples of social institution.
Answer:
Prisons, military camps, private boarding schools, are examples of total institution.

Write suitable examples of the given concepts and justify your answer.

Question 1.
Socialization in family.
Answer:
Example : Children learn basic responsibilities and duties from parents and other family members. The learning involves using sanctions which are mainly informal. The family teaches the child basic discipline and inculcates good habits in his or her mind by using positive or negative sanctions. Positive sanctions involving physical expressions, verbal approval, physical rewards. Negative sanctions ranging from negative remark to physical punishment may be used.

Socialization as a learning process is lifelong. The learning which happens in the early years is termed as primary socialization. A child gets acquainted with values, customs, behavioural norms and manners. Primary socialization takes place in infancy and childhood and involves intense cultural learning. During this phase, family is the main agent of socialization.

Question 2.
Resocialization.
Answer:
Example : The person may be in a jail, hospital, in religious organization, police, army etc. In such institutions there is total break up from the normal social life outside. A prison sentence is a good example. The individual not only has to change and rehabilitate one’s behaviour in order to return to society but must also accommodate the new norms required for living, while in prison. The process of unlearning old norms, roles, values and behavioural patterns and learning new patterns is called re-socialization.

Sometimes an individual is caught in a situation where one has to break away from past experience and internalise different norms and values. Re-socialization can also be defined as a process which subjects an individual to new values, attitudes and skills according to the norms of a particular institution and the person has to completely re-engineer one’s sense of social values and norms.

Question 3.
Total Institution as a form of resocialization.
Answer:
Example : Total institutions are prisons, military camps, mental health facilities and private boarding schools.
The term ‘Total Institution’ was popularised by Erving Goffman. A total institution is usually set apart from the society and are organise by strict rules and norms determined and enforced by a single authority. The fundamental purpose of these institutions is to re-socialise people into changed identities and roles.

Question 4.
Adult socialization at workplace.
Answer:
Example : At the workplace, a person meets people of different age groups, belonging to different social and cultural backgrounds. Through various mechanisms like-on job training, orientation and formal meetings, individuals get familiarized with each other and learns new roles.
Socialization is a life long process. Adult socialization indicates this continuous process of learning. One of the significant agents of adult socialization is the workplace.

Adult individuals spend significant amount of time at the workplace. Socialization through workplace involves acquiring new skills, knowledge and behaviour patterns suitable to the requirements of the job. Social interactions at the workplace, help individuals to broaden their horizons in terms of social acceptance and tolerance towards the others. It also sometimes helps in changing the notions of right and wrong. Workplace thus servers as an important agent of Adult socialization.

Question 5.
Socialization.
Answer:
Example : Cases of Feral, Mogli, Genies, unsocialized children are evident enough to show how human behaviour is largely learnt. The real account of the ‘Wolf-Children of Midnapore’ two small girls reportedly found in a wolf den in 1920. They howled like wolves, preferred raw meat, could not walk upright and lacked many basic human skills due to lack of socialization. Such stories and cases like Genies suggest the human contact is necessary for the basic social and physical development.

Socialization transforms the biological organism into a social being. It is a life long social experience by which human beings are transformed into social beings.
Socialization can be understood as-

A process of learning various forms of behaviour acceptable in a particular culture: Through socialization children learn the ways of their elders and perpetuate the cultural values and social practice.
An ongoing process of continuous learning: The birth of a child gives a new experience of parenting for a couple.
A process through which an individual gradually becomes a member of the society. Socialization facilitates learning the collective way of life.

Write short notes.

Question 1.
Nature versus Nature in Human Development.
Answer:
Genetics suggests that people are born with certain abilities derived from biological imperatives. Instincts or the fixed human traits play a significant role in shaping human behaviour. From this viewpoint, we are born with certain abilities that are integral to our ‘human nature’. The process of socialization on the other hand signifies the role of ‘nurture’ in human development. Sociologists use the term socialization to refer to the lifelong social experience by which human beings are transformed into social beings.

Differentiate between.

Question 1.
Primary Socialization and Secondary Socialization.
Answer:

Primary Socialization	Secondary Socialization
(i) The learning which happens in the early years is termed as primary socialization.	(i) The learning which extends over the entire life of a person is termed as secondary socialization.
(ii) It occurs when a child learns the attitudes, values, customs, behavioural norms and manners as a member of primary group.	(ii) It occurs when an individual learns appropriate behaviour, attitudes, norms as a member of a group.
(iii) Family, peer groups, neighbourhood are the main agents of primary socialization.	(iii) Schooling and education are considered as secondary agencies.
(iv) It is an informal process of socialization.	(iv) It is a formal process of socialization.

Question 2.
Re-socialization and Adult Socialization
Answer:

Re-socialization	Adult Socialization
(i) The process of unlearning old norms, roles and values, behavioural patterns and learning new patterns is called as re-socialization.	(i) Adult socialization is a life long process and indicates continuous process of learning.
(ii) A prison sentence or a total institution is a good example of re-socialization.	(ii) One of the significant agents of adult socialization is the workplace.
(iii) Re-socialization involves a process of tearing down and re-building an individual’s role and socially constructed sense of self.	(iii) Socialization through workplace involves acquiring new skills knowledge and behaviour patterns suitable to the requirements of the job.
(iv) It re-socializes people into changed identities and role.	(iv) It helps in changing the notions of right and wrong.

Explain the following concept with an example.

Question 1.
Significant Others
Answer:
It is a term used by George Herbert Mead to refer to those individuals who are most important in the development of the ‘self.

A significant other is someone whose opinions matter to us and who is in a position to influence our thinking, especially about ourselves.
Example : A significant others can be anyone such as parents, siblings, friends and teachers.

Question 2.
Hidden Curriculum
Answer:
1. Sociologists also discuss about hidden curriculum for conditioning children’s learning.

2. Hidden curriculum refers to things we learn from attending the school such as respect for the system and obedience to authority. It also indicates unquestioned acceptance of the system. Example : While distributing co-curricular tasks, girls and boys are expected to do campus cleaning. Through this, children learn the sexual division of labour. Many schools are making efforts to counter this by allotting similar tasks to boys and girls.

Question 3.
Game Stage
Answer:

According to G. H. Mead, formation of self occurs in three distinct stages.
Game stage is the third stage of the process of socialization.
As a child matures, and as the self gradually develops, the child learns to behave according to the impressions of others.
They learn to understand interactions involving different people with a variety of purposes.
They understand that ‘role play’ in each situation involves following a consistent set of rules and expectations.

Example : A child at this stage is likely to be aware of the different responsibilities of people in a restaurant who together, make for a smooth dining experience.

Complete the concept maps.

Question 1.
Maharashtra Board Class 11 Sociology Important Questions Chapter 6 Socialization 1
Answer:

Question 2.
Maharashtra Board Class 11 Sociology Important Questions Chapter 6 Socialization 3
Answer:

Question 3.
Maharashtra Board Class 11 Sociology Important Questions Chapter 6 Socialization 5
Answer:

Question 4.
Maharashtra Board Class 11 Sociology Important Questions Chapter 6 Socialization 7
Answer:

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

Question 1.
Neighbourhood have negligible influence on shaping social behaviour of the growing child.
Answer:
This statement is False.

Adults in the neighbourhood exert an influence on shaping social behaviour of the growing child. Children often take inputs from people living very closely around.
Neighbourhood community provides the base for an individual to extend social relations and interactions beyond the narrow limits of the home.
Neighbourhood, social networks provide great source of learning for children through the celebration of cultural festivals, organisation of sports activities, taking up of social issues such as environment or traffic.

Question 2.
A total institution is set apart from the society and organized by strict rules and norms.
Answer:
This statement is True.

A total institution is usually set apart from the society by distance, laws, and physical attributes like high walls, barbed wire fences, and locked gates.
They are organised by strict rules and norms determined and enforced by a single authority.
The fundamental purpose of these institutions is to re-socialize people into changed identities and roles, hence it is set apart and organized by strict rules and norms.

Question 3.
The function of all children’s stories is to create a sense in the children of the right and wrong.
Answer:
This statement is True.

The moral of honesty, courage, non-violence, etc., is narrated to children in an effective manner.
For centuries together moral stories are transmitted from one generation to another. It gives them a sense of growing up in their own world and a notion of right or wrong.
Thus, the function of all children’s stories is to create a sense in the children of the right/ wrong, acceptable/non-acceptable, is prevalent in a particular society. Through this process vulnerable infants become self-aware, skilled individuals.

Question 4.
Cases of Feral/unsocialised children denote the significance of socialization.
Answer:
This statement is True.

Cases of Feral like Genie’s suggest that human contact is necessary for the basic social and physical development.
For example, simple skills such as walking upright or using language were missing in Genie. Such examples denote the significance of socialization.
What we think and how we act is taught to us by the larger culture.

Question 5.
Human behaviours are governed only by instinct.
Answer:
This statement is False.

If human behaviours were governed only by instinct, there would be very few differences between societies. Human behaviour would be much the same regardless of place and time.
Different cultures develop different ways of doing things.
What we think and how we act is taught to us by the larger culture that we inherit and share. Thus, human behaviours acquired, of the many diverse ways of living and not governed only by instinct.

Question 6.
Peer groups are considered as important primary agencies of socialization.
Answer:
This statement is True.

They are considered as important primary agencies of socialization because personal interactions with our peers influences our behaviour from how we dress to what we like and what we hate.
In peer groups, the interactions are reasonably egalitarian as there is a greater amount of give and take, when compared to family or school.
We carry the value of friendship with us throughout our lives.

Question 7.
Socialization through family is varied.
Answer:
This statement is True.

Socialization through family is varied because there is no single uniform pattern to do so. A child brought up in a nuclear family will undergo different patterns of socialization than one in an extended family. In nuclear family, parents may be key socializing agents but in the other in the extended family grandparents, an aunt, a cousin may play a significant role.
Patterns of child-rearing vary across families with different caste, class, and ethnic backgrounds
The influence of different family backgrounds can be seen on a child growing up in a poor, marginalized household and a child growing up in an upper caste/upper class family background.

Question 8.
Socialization in family always means unquestioned acceptance of everything that elders say.
Answer:
This statement is False.

Socialization in family does not always means unquestioned acceptance of everything that elders say.
Children can also negotiate, question and develop outlooks contradictory to their elders.
This is more so in the contemporary world in which diverse socializing agencies influence social learning.

Question 9.
Schools project a wider range of values and roles.
Answer:
This statement is True.

Schooling involves learning values and norms at a step higher than those learnt in family.
Skills and values like teamwork, competitive spirit, discipline, conformity to authority are learnt in schools and this helps prepare students for the adult world.
School indicates unquestioned acceptance of the system thus, projecting a wider range of values and roles.

Question 10.
Resocialization involves complete alteration of the easier lifestyles.
Answer:
This statement is True.

Sometimes an individual is caught in a situation where one has to break away from past experience and earlier way of life and internalise radically different norms and values.
Hence, it involves complete alteration of the earlier lifestyle and thinking. The new way of life is not only different but also incompatible with the earlier one.
The basic of re-socialization is to unlearn and then relearn. The individual has to completely re-engineer one’s sense of social values, beliefs and norms.

Give your personal response.

Question 1.
We can come to think of ourselves funny because people laugh at the things we say.
Answer:
‘Significant others’ is a term used by G. H. Mead to refer to those individuals who are most important in the development of the self. A significant other can be anyone such as parents, friends, teachers etc. Their opinions matter to us and influence our thinking especially about ourselves. So, when people laugh at the things we say makes us think of ourselves as funny.

Answer the following question in detail (About 150 words).

Question 1.
Explain the process of socialization with suitable examples and discuss any one agency of socialization you are member of.
Answer:
Process of socialization – Human behaviour and skills have to be taught and learnt. This learning process is called as socialization. According to Mead, formation of self occurs in three distinct stages.
Stage 1 – Imitation : In this stage, children imitate behaviour of adults without understanding it. For example, A little boy might drive his mother to her office by driving his toy car.

Stage 2 – Play stage : A child plays, sometimes as being a mother or a teacher etc. In this stage, responses are not organised. A child internalises the attitudes of others who are significant to her/ his through enacting the roles of others.

Stage 3 – Game stage : As a child matures, and as the self gradually develops, one internalises the expectations of a large number of people . For example, a child at this stage is likely to be aware of the different responsibilities of people in a restaurant .Socialization, in this sense is a process of self-awareness.

I am member of various social groups which are agencies of my socialization like family, peer- group, school, neighbourhood etc.

Family as an agency of socialization I have learnt a range of roles in family-like learning responsibilities and duties from my parents and other family members. Family has played a significant role in developing acceptable behaviour patterns in me thus the process of learning attitudes, norms and behaviour patterns takes place in family.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations

June 20, 2024June 19, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 5 Oscillations Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 5 Oscillations

1. Choose the correct option.

i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(A) \(\frac{\sqrt{3}}{2}\)A
(B) \(\frac{2}{\sqrt{3}}\)
(C) A/2
(D) \(\frac{1}{\sqrt{2}}\)
Answer:
(A) \(\frac{\sqrt{3}}{2}\)A

ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
(A) 36 J
(B) 9 J
(C) 27 J
(D) 18 J
Answer:
(D) 18 J

iii) The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]
(A) \(\frac{1}{6}\) m
(B) 6 m
(C) \(\frac{1}{36}\) m
(D) \(\frac{1}{\sqrt{6}}\) m.
Answer:
(A) \(\frac{1}{6}\) m

iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(A) 1:4
(B) 1:2
(C) 2:1
(D) 4:1
Answer:
(B) 1:2

v) The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(A) The acceleration is maximum at time T.
(B) The force is maximum at time 3T/4.
(C) The velocity is zero at time T/2.
(D) The kinetic energy is equal to total energy at time T/4.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 1
Answer:
(B) The force is maximum at time 3T/4.

2. Answer in brief.

i) Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples : The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.

ii) Using differential equation of linear S.H.M, obtain the expression for
(a) velocity in S.H.M.,
(b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is x = A sin (ωt + α) … (1)
where A is the amplitude, ω is a constant in a particular case and α is the initial phase.
The velocity of the particle is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 10
Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
a = \(\frac{d v}{d t}\) = \(\frac{d}{d t}\) [Aω cos (ωt + α) J at at
∴ a = – ω² A sin (ωt + α)
But from Eq. (1), A sin (ωt + α) = x
∴ a = -ω²x … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.

iii) Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

Consider a simple pendulum of length L₁ – suspended from a rigid support O. When displaced from its initial position of rest through a small angle θ in a vertical plane and released, it performs oscillations between two extremes, B and C, as shown in below figure. At B, the forces on the bob are its weight \(m \vec{g}\) and the tension \(\overrightarrow{F_{1}}\) in the string. Resolve \(m \vec{g}\) into two components : mg cos θ in the direction opposite to that of the tension and mg sin θ perpendicular to the string.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 20
mg cos θ balanced by the tension in the string. mg sin θ restores the bob to the equilibrium position.
Restoring force, F = – mg sin θ
If θ is small and expressed in radian,
sin θ \(\approx\) θ = \(\frac{\text { arc }}{\text { radius }}\) = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{x}{L}\)
∴ F = – mgθ = -mg\(\frac{x}{L}\) …. (1)
Since m, g and L are constant,
F ∝ (-x) …. (2)

Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, a = \(\frac{F}{m}\) = –\(\frac{g}{L}\)x … (3)
Therefore, acceleration per unit displacement
= |\(\frac{a}{x}\)| = \(\frac{g}{L}\) ….. (4)
Period of SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 25
This gives the expression for the period of a simple pendulum.

iv) State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
∴ T ∝\(\sqrt{L}\)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
∴ T ∝ \(\frac{1}{\sqrt{g}}\)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.

v) Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment μ, suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is B_h. The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along B_h. If it is rotated in the horizontal plane by a small
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 30
displacement θ from its rest position (θ = 0), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.

The bar magnet experiences a torque \(\tau\) due to the field B_h. Which tends to restore it to its original orientation parallel to B_h. For small θ, this restoring torque is
\(\tau\) = – μB_h sin θ = – μB_hμ …. (1)

where the minus sign indicates that the torque is opposite in direction to the angular displacement θ. Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.

Question 3.
Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition : Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
I\(\frac{d^{2} \theta}{d t^{2}}\) + cθ = 0 ….. (1)
where I = moment of inertia of the oscillating body, \(\frac{d^{2} \theta}{d t^{2}}\) = angular acceleration of the body when its angular displacement is θ, and c = torsion constant of the suspension wire,
∴ \(\frac{d^{2} \theta}{d t^{2}}\) + \(\frac{c}{I}\)θ = 0
Let \(\frac{c}{I}\) = ω², a constant. Therefore, the angular frequency, ω = \(\sqrt{c / I}\) and the angular acceleration,
a = \(\frac{d^{2} \theta}{d t^{2}}\) = -ω²θ … (2)
The minus sign shows that the α and θ have opposite directions. The period T of angular SHM is
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 80
This is the expression for the period in terms of torque constant. Also, from Eq. (2),

Question 4.
Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 35
The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.
The projection Q of the reference point is described by the y-coordinate,
y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α)
which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is v_y = ωA sin (90° – θ) = ωA cos θ
∴ v_y = ωA cos (ωt + α)
The centripetal acceleration of the reference particle is a_r = ω²A, so that its y-component at time t is a_x = a_r sin ∠OPQ
∴ a_x = – ω²A sin (ωt + α)

Question 5.
Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position
(b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period T = 2π/ω starting from the mean position towards the positive extreme position where ω is the angular frequency. Its displacement from the mean position (x), velocity (v) and acceleration (a) at any instant are
x = A sin ωt = A sin\(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = \(\frac{d x}{d t}\) = ωA cos ωt = ωA cos\(\left(\frac{2 \pi}{T} t\right)\)
a = \(\frac{d v}{d t}\) = -ω² A sin ωt = – ω²A sin\(\left(\frac{2 \pi}{T} t\right)\) as the initial phase α = 0.
Using these expressions, the values of x, v and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 38
Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of π radians between x and a.

Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 40

Consider a particle performing linear SHM with amplitude A and period T = 2π/ω, starting from the positive extreme position, where ω is the angular frequency. Its displacement from the mean position (x), velocity (y) and acceleration (a) at any instant (t) are
x = A cos ωt = A cos \(\left(\frac{2 \pi}{T} t\right)\) (∵ω = \(\frac{2 \pi}{T}\))
v = -ωA sin ωt = – ωA sin \(\left(\frac{2 \pi}{T} t\right)\)
a = -ω²A cos ωt = -ω²A cos \(\left(\frac{2 \pi}{T} t\right)\)
Using these expressions, the values of x, y and a at the end of every quarter of a period, starting from t = 0, are tabulated below.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 41
Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.
There is a phase difference of \(\frac{\pi}{2}\) radians between x and v, and between v and a.
There is a phase difference of n radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π/2 rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at x = ± A) and v is maximum ( = ± ωA) at the mean position (x = 0).

(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of 2π rad. There is a phase difference of π rad between v and a. a is minimum (equal to zero) at the mean position (x = 0) and a is maximum ( = \(\mp\)ω²A) at the extreme positions (x = ±A).
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 43

Question 6.
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting r on the particle is F = – kx, where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
v = ω\(\sqrt{A^{2}-x^{2}}\)
where ω = \(\sqrt{k / m} .\) The kinetic energy (KE) of the particle is
KE = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) mω² (A² – x²)
= \(\frac{1}{2}\)k(A² – x²) … (1)
If the phase of the particle at an instant t is θ = ωt + α, where α is initial phase, its velocity at that instant is
v = ωA cos (ωt + α)
and its KE at that instant is
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)mω²A² cos²(ωt + α) ….. (2)
Therefore, the KE varies with time as cos² θ.

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibirum.

Suppose the particle in below figure is displaced from P₁ to P₂, through an infinitesimal distance dx against the restoring force F as shown.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 45
The corresponding work done by the external agent will be dW = ( – F)dx = kx dx. This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
∴ PE = \(\int\)dW = \(\int_{0}^{x}\) kx dx = \(\frac{1}{2}\) kx² … (3)
The displacement of the particle at an instant t being
x = A sin (wt + α)
its PE at that instant is
PE = \(\frac{1}{2}\)kx² = \(\frac{1}{2}\)kA² sin²(ωt + α) … (4)
Therefore, the PE varies with time as sin²θ.

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (2), total energy is E = PE + KE
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)k(A² – x²)
= \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)kA² – \(\frac{1}{2}\)kx²
∴ E = \(\frac{1}{2}\)kA² = \(\frac{1}{2}\)mω²A² … (5)
As m is constant, ω and A are constants of the motion, the total energy of the particle remains constant (or its conserved).

Question 7.
Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.

In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.

The period of a simple pendulum at a given place is
T = 2π\(\sqrt{\frac{L}{g}}\)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

Question 8.
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data : v = \(\frac{1}{2}\)v_max, 2A = 10 cm
∴ A = 5 cm
v = ω\(\sqrt{A^{2}-x^{2}}\) and v_max = ωA
Since v = \(\frac{1}{2}\)v_max,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 50
This gives the required displacement.

Question 9.
In SI units, the differential equation of an S.H.M. is \(\frac{d^{2} x}{d t^{2}}\) = -36x. Find its frequency and period. Find its frequency and period.
[Ans: 0.9548 Hz, 1.047 s]
Answer:
\(\frac{d^{2} x}{d t^{2}}\) = -36x
Comparing this equation with the general equation,
\(\frac{d^{2} x}{d t^{2}}\) = -ω²x
We get, ω² = 36 ∴ ω = 6 rad/s
ω = 2πf
∴ The frequency, f = \(\frac{\omega}{2 \pi}\) = \(\frac{6}{2(3.142)}\) = \(\frac{6}{6.284}\) = 0.9548 Hz
and the period, T = \(\frac{1}{f}\) = \(\frac{1}{0.9548}\) = 1.047 s

Question 10.
A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \(\frac{1}{30}\)s after it has crossed the mean position. [Ans: 34.2 m/s²]
Answer:
Data : A = 4 cm = 4 × 10^-2 m, f = 5Hz, t = \(\frac{1}{30}\)s
ω = 2πf = 2π (5) = 10π rad/s
Therefore, the magnitude of the acceleration,
|a| = ω²x = ω²A sin ωt
= (10π)² (4 × 10²)
= 10π² sin \(\frac{\pi}{3}\) = 10 (9.872)(0.866) = 34.20 m/s²

Question 11.
Potential energy of a particle performing linear S.H.M is 0.1 π² x² joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data : PE = 0.1 π² x² J, m = 20 g = 2 × 10^-2 kg
PE = \(\frac{1}{2}\)mω²x² = \(\frac{1}{2}\)m (4π²f²)x²
∴ \(\frac{1}{2}\)m(4π²f²)x² = 0.1 π² x²
∴ 2mf² = 0.1 ∴ f² = \(\frac{1}{20\left(2 \times 10^{-2}\right)}\) = 2.5
∴ The frequency of SHM is f = \(\sqrt{2.5}\) = 1.581 Hz

Question 12.
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 cm/s]
Answer:
Data : m = 2 kg, E = 40 J
The speed of the body while crossing the centre of the path (mean position) is v_max and the total energy is entirely kinetic energy.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 60

Question 13.
A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data : T = 4 s, x = A/3
The displacement of a particle starting into SHM from the mean position is x = A sin ωt = A sin \(\frac{2 \pi}{T}\) t
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 61
∴ the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.

Question 14.
A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 × 10^-2 N]
Answer:
Data : L = 100 cm, m = 50 g = 5 × 10^-2 kg, x = 3 cm, g = 9.8 m/s²
Restoring force, F = mg sin θ = mgθ
= (5 × 10^-2)(9.8)\(\left(\frac{3}{100}\right)\)
= 1.47 × 10^-2 N

Question 15.
Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75
m/s² to 9.80 m/s². [Ans: Decreases by 5.07 mm]
Answer:
Data : g₁ = 9.75 m/s², g₂ = 9.8 m/s²
Length of a seconds pendulum, L = \(\frac{g}{\pi^{2}}\)
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 62
∴ The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m.

Question 16.
At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data : A = 8 cm, KE = 3 PE
KE = \(\frac{1}{2}\) (A² – x²) and PE = \(\frac{1}{2}\)kx²
Given, KE = 3PE.
∴ \(\frac{1}{2}\)k(A² – x²) = 3\(\left(\frac{1}{2} k x^{2}\right)\)
∴ A² – x² = 3x² ∴ 4x² = A²
∴ the required displacement is
x = ±\(\frac{A}{2}\) = ±\(\frac{8}{2}\) = ± 4 cm

Question 17.
A particle performing linear S.H.M. of period 2π seconds about the mean position O is observed to have a speed of \(b \sqrt{3}\) m/s, when at a distance b (metre) from O. If the particle is moving away from O at that instant, find the
time required by the particle, to travel a further distance b. [Ans: π/3 s]
Answer:
Data : T = 2πs, v = b\(\sqrt{3}\) m/s at x = b
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 63
∴ Assuming the particle starts from the mean position, its displacement is given by
x = A sin ωt = 2b sin t
If the particle is at x = b at t = t₁,
b = 2b sint₁ ∴ t₁ = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)s
Also, with period T = 2πs, on travelling a further distance b the particle will reach the positive extremity at time t₂ = \(\frac{\pi}{2}\)s.
∴ The time taken to travel a further distance b from x = b is t₂ – t₁ = \(\frac{\pi}{2}\) – \(\frac{\pi}{6}\) = \(\frac{\pi}{3}\)s.

Question 18.
The period of oscillation of a body of mass m₁ suspended from a light spring is T. When a body of mass m₂ is tied to the first body and the system is made to oscillate, the period is 2T. Compare the masses m₁ and m₂ [Ans: 1/3]
Answer:
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 64
This gives the required ratio of the masses.

Question 19.
The displacement of an oscillating particle is given by x = asinωt + bcosωt where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = \(\sqrt{a^{2}+b^{2}}\)
Answer:
x = asinωt + bcosωt
Let a = A cos φ and b = A sin φ, so that
A² = a² + b² and tan φ = \(\frac{b}{a}\)
∴ x = A cos φ sin ωt + A sin φ cos ωt
∴ x = A sin (ωt + φ)
which is the equation of a linear SHM with amplitude A = \(\sqrt{a^{2}+b^{2}}\) and phase constant φ = tan^-1 \(\frac{b}{a}\), as required.

Question 20.
Two parallel S.H.M.s represented by x₁ = 5sin (4πt + \(\frac{\pi}{3}\)) cm and x₂ = 3sin(4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54° 23′]
Answer:
Data: x₁ = 5 sin (4πt + \(\frac{\pi}{3}\)) = A₁ sin(ωt + α),
x₂ = 3 sin (4πt + \(\frac{\pi}{4}\)) = A₂ sin(ωt + β)
∴ A₁ = 5 cm, A₂ = 3 cm, α = \(\frac{\pi}{3}\) rad, β = \(\frac{\pi}{4}\) rad
(i) Resultant amplitude,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 65

(ii) Epoch of the resultant SHM,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 66

Question 21.
A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60° and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (π³ ≈ 31)
[Ans: 0.04133 N m]
Answer:
Data: R = 10cm = 0.1 m, M = 0.2 kg, θ_m = 60° = \(\frac{\pi}{3}\) rad, T = 1 s, π³ ≈ 31
The Ml of the disc about the rotation axis (perperdicular through its centre) is
I = \(\frac{1}{2}\)MR² = (0.2)(0.1)² = 10^-3 kg.m²
The period of torsional oscillation, T = 2π\(\sqrt{\frac{I}{c}}\)
∴ The torsion constant, c = 4πr²\(\frac{I}{T^{2}}\)
The magnitude of the maximum restoring torque,
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 67

Question 22.
Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of 1.6 × 10^-5 Wb/m². The magnet has moment of inertia 3 × 10^-6 kgm² and magnetic moment 3 A m². [Ans:38.19 osc/min.]
Answer:
Data : B = 1.6 × 10^-5 T, I = 3 × 10^-6kg/m²,
µ = 3 A.m²
The period of oscillation, T = 2π \(\sqrt{\frac{I}{\mu B_{\mathrm{h}}}}\)
∴ The frequency of oscillation is
f = \(\frac{1}{2 \pi}\)\(\sqrt{\frac{\mu B}{I}}\)
∴ The number of oscillations per minute
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 68
= 38.19 per minute

Question 23.
A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (g ≈ π² ≈ 10 m s^-2)
[Ans: n_max = 1/s, E/m = 1.25 J/kg]
Answer:
Data : A = 0.25 m, g = π² = 10 m/s²
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is a_max = g, downwards.
Maharashtra Board Class 12 Physics Solutions Chapter 5 Oscillations 69
This gives the required frequency of the piston.

12th Physics Digest Chapter 5 Oscillations of Waves Intext Questions and Answers

Can you tell? (Textbook Page No. 112)

Question 1.
Why is the term angular frequency (ω) used here for a linear motion ?
Answer:
A linear SHM is the projection of a UCM on a diameter of the circle. The angular speed co of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.

Can you tell? (Textbook Page No. 114)

Question 1.
State at which point during an oscillation the oscillator has zero velocity but positive acceleration ?
Answer:
At the left extreme, i.e., x = – A, so that a = – ω²x = – ω²(- A) = ω²A = a_max

Question 2.
During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa ?
Answer:
Velocity v is positive (to the right) while displacement x is negative when the particle in SHM is moving from the left extreme towards the mean position. Velocity v is negative (to the left) while displacement x is positive when the particle in SHM is moving from the right extreme towards the mean position.

Can you tell? (Textbook page 76)

Question 1.
To start a pendulum swinging, usually you pull it slightly to one side and release. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.

Question 2.
Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy,
and this cycle then repeats.

Question 3.
Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.

Can you tell? (Textbook Page No. 109)

Question 1.
Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.

Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits

June 20, 2024June 19, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 13 AC Circuits Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 13 AC Circuits

1. Choose the correct option.

i) If the RMS current in a 50 Hz AC circuit is 5A, the value of the current \(\frac{1}{300}\) seconds after its value becomes zero is
(A) 5\(\sqrt {2}\) A
(B) 5\(\sqrt{\frac{3}{2}}\) A
(C) \(\frac{5}{6}\) A
(D) \(\frac{5}{\sqrt{2}}\) A
Answer:
(B) 5\(\sqrt{\frac{3}{2}}\) A

ii) A resistor of 500 Ω and an inductance of 0.5 H are in series with an AC source which is given by V = 100 \(\sqrt {2}\) sin (1000t). The power factor of the combination
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\frac{1}{\sqrt{3}}\)
(C) 0.5
(D) 0.6
Answer:
(A) \(\frac{1}{\sqrt{2}}\)

iii) In a circuit L, C & R are connected in series with an alternating voltage of frequency f. the current leads the voltage by 450. The value of C is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 1
Answer:
(B) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)

iv) In an AC circuit, e and i are given by e = 150 sin (150t) V and i = 150 sin (150 t + \(\frac{\pi}{3}\)) A. the power dissipated in the circuit is
(A) 106W
(B) 150W
(C) 5625W
(D) Zero
Answer:
(C) 5625W

v) In a series LCR circuit the phase difference between the voltage and the current is 45°. Then the power factor will be
(A) 0.607
(B) 0.707
(C) 0.808
(D) 1
Answer:
(B) 0.707

2. Answer in brief.

i) An electric lamp is connected in series with a capacitor and an AC source is glowing with a certain brightness. How does the brightness of the lamp change on increasing the capacitance ?
Answer:
Impedance, Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), where R is the resistance of the lamp, w is the angular frequency of AC and C is the capacitance of the capacitor connected in series with the AC source and the lamp. When C is increased, \(\) decreases. Hence, Z increases.
Power factor, cos Φ = \(\frac{R}{Z}\)
As Z increases, the power factor decreases.
Now, the average power over one cycle,
P_av = v_rms i_rms cos Φ
= V_rms \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) cos Φ
= \(\frac{V_{\mathrm{rms}}^{2}}{\mathrm{Z}} \cos \phi\)
∴ P_av decreases as Z increases and cos Φ decreases.
As the current through the lamp \(\left(\frac{V_{\mathrm{rms}}}{Z}\right)\) decreases, the brightness of the lamp will decrease when C is increased.

ii) The total impedance of a circuit decreases when a capacitor is added in series with L and R. Explain why ?
Answer:
For an LR circuit, the impedance,
Z_LR = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\), where X_L is the reactance of the inductor.
When a capacitor of capacitance C is added in series with L and R, the impedance,
Z_LCR = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) because in the case of an inductor the current lags behind the voltage by a phase angle of \(\frac{\pi}{2}\) rad while in the case of a capacitor the current leads the voltage by a phase angle of \(\frac{\pi}{2}\) rad. The decrease in net reactance decreases the total impedance (Z_LCR < Z_LR).

iii) For very high frequency AC supply, a capacitor behaves like a pure conductor. Why ?
Answer:
The reactance of a capacitor is X_C = \(\frac{1}{2 \pi f C}\), where f is the frequency of the AC supply and C is the capacitance of the capacitor. For very high frequency, f, X_C is very small. Hence, for very high frequency AC supply, a capacitor behaves like a pure conductor.

iv) What is wattless current ?
Answer:
The current that does not lead to energy consumption, hence zero power consumption, is called wattless current.
In the case of a purely inductive circuit or a purely capacitive circuit, average power consumed over a complete cycle is zero and hence the corresponding alternating current in the circuit is called wattless current.
[Note : In this case, the power factor is zero.]

v) What is the natural frequency of L C circuit ? What is the reactance of this circuit at this frequency
Answer:
The natural frequency of LC circuit is \(\frac{1}{2 \pi \sqrt{L C}}\) ,
where L is the inductance and C is the capacitance. The reactance of this circuit at this frequency is
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 18

Question 3.
In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. Calculate P₁ / P₂ .
Answer:
For a series LR circuit, power factor,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 17

Question 4.
When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) – cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 5.
Prove that an ideal capacitor in an AC circuit does not dissipate power
Answer:
In an AC circuit containing only an ideal inductor, the current i lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad. Here, for e = e₀ sin ωt, we have, i = i₀ sin(ωt – \(\frac{\pi}{2}\))
Instantaneous power, P = ei
= (e₀ sin ωt) [i₀ (sin ωt cos \(\frac{\pi}{2}\) + cos ωt sin \(\frac{\pi}{2}\))]
= – e₀i₀ sin ωt cos ωt as cos \(\frac{\pi}{2}\) = 0 and sin \(\frac{\pi}{2}\) = 1.
Average power over one cycle, P_av
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).
∴ P_av = 0, i.e., the circuit does not dissipate power.

Question 6.
(a) An emf e = e₀ sin ωt applied to a series L – C – R circuit derives a current I = I₀ sinωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
Answer:
(a) Instantaneous power,
P = ei
= (e₀ sin ωt) [i₀ (sin ωt ± Φ)]
= e₀i₀ sin ωt(sin ωt cos Φ ± cos ωt sin Φ)
= e₀i₀ sin² ωt ± e₀i₀ sin Φ sin ωt cos ωt
Average power over one cycle,
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 15

= e_rms i_rms cos Φ = e_rms i_rms (\(\frac{R}{Z}\)), where the impedance Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\).

(b) P_av = e_rms i_rms cos Φ
The factor cos Φ is called as power factor. For circuits used for transporting electric power, a low power factor means the power available on transportation is much less than e_rms i_rms It means there is significant loss of power during transportation.

Question 7.
A device Y is connected across an AC source of emf e = e₀ sinωt. The current through Y is given as i = i₀ sin(ωt + π/2)
a) Identify the device Y and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for Y.
c) How does the reactance of the device Y vary with the frequency of the AC ? Show graphically
d) Draw the phasor diagram for the device Y.
Answer:
(a) The device Y is a capacitor. Its reactance is X_c = \(\frac{1}{\omega C}\),
where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.

(b)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 6

(c) X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). Thus X_C ∝ \(\frac{1}{f}\), where f is the frequency of AC. Suppose C = \(\left(\frac{1000}{2 \pi}\right)\) pF
For f= 100 Hz, X_C = 1 × 10⁷Ω = 10MΩ;
for f = 200 Hz, X_C = 5 MΩ;
for f = 300 Hz, X_C = \(\frac{10}{3}\) MΩ;
for f = 400 Hz, X_C = 2.5 MΩ
for f = 500 Hz, X_C = 2 MΩ and so on
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 8

(d)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 9
The phasor representing the peak emf (e₀) makes an angle (ωt) in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i₀) is turned 90° anticlockwise with respect to the phasor representing emf e₀. The projections of these phasors on the vertical axis give instantaneous values of e and i.

Question 8.
Derive an expression for the impedance of an LCR circuit connected to an AC power supply.
Answer:
Figure shows an inductor of inductance L, capacitor of capacitance C, resistor of resistance R, key K and source (power supply) of alternating emf (e) connected to form a closed series circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 10
We assume the inductor, capacitor and resistor to be ideal. As these are connected in series, at any instant, they carry the same current i = i₀ sin ωt. The voltage across the resistor, e_R = Ri, is in phase with the current. The voltage across the inductor, e_L = X_Li, leads the current by \(\frac{\pi}{2}\) rad and that across the capacitor, e_C = X_Ci, lags behind the current by \(\frac{\pi}{2}\) rad. This is shown in the phasor diagram.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 11
is the effective resistance of the circuit. It is called the impedance.

Question 9.
Compare resistance and reactance.
Answer:
(1) Resistance is opposition to flow of charges (current) and appears in a DC circuit as well as in an AC circuit.
The term reactance appears only in an AC circuit. It occurs when an inductor and/or a capacitor is used.

(2) In a purely resistive circuit, current and voltage are always in phase.
When reactance is not zero, there is nonzero phase difference between current and voltage.

(3) Resistance does not depend on the frequency of AC.
Reactance depends on the frequency of AC. In case of an inductor, reactance increases linearly with frequency. In case of a capacitor, reactance decreases as frequency of AC increases; it is inversely proportional to frequency.

(4) Resistance gives rise to production of Joule heat in a component.
In a circuit with pure reactance, there is no production of heat.

Question 10.
Show that in an AC circuit containing a pure inductor, the voltage is ahead of current by π/2 in phase.
Answer:
Figure 13.8 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a key K and a pure inductor of inductance L to form a closed circuit.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 2
On closing the key K, an emf is induced in the inductor as the magnetic flux linked with it changes with time. This emf opposes the applied emf and according to the laws of electromagnetic induction by Faraday and Lenz, we have,
e’ = -L\(\frac{d i}{d t}\) ………………. (1)
where e’ is the induced emf and i is the current through the inductor. To maintain the current; e and e’ must be equal in magnitude and opposite in direction.

According to Kirchhoff’s voltage law, as the resistance of the inductor is assumed to be zero, we
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 3
where C is the constant of integration. C must be time independent and have the dimension of current. As e oscillates about zero, i also oscillates about zero and hence there cannot be any time independent component of current.
∴ C = 0. ∴ i = –\(\frac{e_{0}}{\omega L}\)cos ωt = – \(\frac{e_{0}}{\omega L}\)sin(\(\frac{\pi}{2}\) – ωt)
∴ i = \(\frac{e_{0}}{\omega L}\) sin(ωt – \(\frac{\pi}{2}\)) ……………. (3)
as sin (-θ) = – sin θ
From Eq. (3), i_peak = i₀ = \(\frac{e_{0}}{\omega L}\)
∴ i = i₀ sin(ωt – \(\frac{\pi}{2}\)) ………………. (4)
Comparison of this equation with e = e₀ sin ωt shows that e leads i by \(\frac{\pi}{2}\) rad, i.e., the voltage is ahead of current by \(\frac{\pi}{2}\) rad in phase.

Question 11.
An AC source generating a voltage e = e₀ sinωt is connected to a capacitor of capacitance C. Find the expression for the current i flowing through it. Plot a graph of e and i versus ωt.
Answer:
Figure 13.12 shows an AC source, generating a voltage e = e₀ sin ωt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle, the
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 4
capacitor is alternately charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V = \(\frac{q}{C}\) ∴ q = CV. q and V are functions of time, with V = e = e₀ sin ωt. The instantaneous current in the circuit is i = \(\frac{d q}{d t}=\frac{d}{d t}\)(CV) = C \(\frac{d v}{d t}\) = C \(\frac{d}{d t}\) (e₀ sin ωt) = ωC e₀ cos ωt
∴ i = \(\frac{e_{0}}{(1 / \omega C)} \sin \left(\omega t+\frac{\pi}{2}\right)=i_{0} \sin \left(\omega t+\frac{\pi}{2}\right)\)
where i₀ = \(\frac{e_{0}}{(1 / \omega C)}\) is the peak value of the current.
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 5
Table gives the values of e and i for different values of cot and Fig shows graphs of e and i versus ωt. i leads e by phase angle of \(\frac{\pi}{2}\) rad.

Question 12.
If the effective current in a 50 cycle AC circuit is 5 A, what is the peak value of current? What is the current 1/600 sec.after if was zero ?
Answer:
Data : f = 50 Hz, i_rms = 5 A, t = \(\frac{1}{600}\) s
The peak value of the current,
i₀ = i_rms\(\sqrt {2}\) = (5)(1.414) = 7.07 A
= i₀sin (2πft)
= 7.07 sin [2π(5o) (\(\frac{1}{600}\))]
= 7.07 sin (\(\frac{\pi}{6}\)) = (7.07)(0.5)
= 3.535 A
This is the required current.

Question 13.
A light bulb is rated 100W for 220 V AC supply of 50 Hz. Calculate (a) resistance of the bulb. (b) the rms current through the bulb.
Answer:
Data: Power (V_rms i_rms) = 100 W, V_rms = 220V,
f = 50 Hz
The rms current through the bulb,
i_rms = \(\frac{\text { power }}{V_{\mathrm{rms}}}=\frac{100}{220}\) = 0.4545 A
The resistance of the bulb,
R = \(\frac{V_{\mathrm{rms}}}{i_{\mathrm{rms}}}=\frac{220}{(100 / 220)}\) = (22) (22) = 484 Ω

Question 14.
A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak)
in the circuit. If the frequency is doubled, what will happen to the capacitive reactance and the current.
Answer:
Data : C = 15 µF = 15 × 10-6 F, V_rms = 220V, f = 50 Hz,
The capacitive reactance = \(\frac{1}{2 \pi f C}\)
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 12
If the frequency is doubled, the capacitive reactance will be halved and the current will be doubled.

Question 15.
An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor (π = 3.142)
Answer:
Data : L = 2H, i₀ = 0.25 A, f = 60 Hz, π = 3.142
ωL = 2πfL = 2(3.142)(60)(2) = 754.1 Ω
The effective potential difference across the inductor = ωLi_rms = ωL \(\frac{i_{0}}{\sqrt{2}}\)
= \(\frac{(754.1)(0.25)}{1.414}\) = 133.3 V

Question 16.
Alternating emf of e = 220 sin 100 πt is applied to a circuit containing an inductance of (1/π) henry. Write an equation for instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?
Answer:
Data: e = 220 sin 100 πt, L = (\(\frac{1}{\pi}\))H
Comparing e = 220 sin 100 πt with
e = e₀ sin ωt, we get
ω = 100 π ∴ ωL = (100 π) (\(\frac{1}{\pi}\)) = 100 Ω
∴ The instantaneous current through the circuit
= i = \(\frac{e_{0}}{\omega L}\) sin(100 πt – \(\frac{\pi}{2}\))
= \(\frac{220}{100}\) sin (100 πt – \(\frac{\pi}{2}\)) = 2.2 sin (100 πt – \(\frac{\pi}{2}\)) in ampere [assuming that e is in volt.]
i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{2.2}{1.414}\) = 1.556 A is the reading of the AC galvanometer connected in the circuit.

Question 17.
A 25 µF capacitor, a 0.10 H inductor and a 25Ω resistor are connected in series with an AC source whose emf is given by e = 310 sin 314 t (volt). What is the frequency, reactance, impedance, current and phase angle of the circuit?
Answer:
Data: C = 25 µF = 25 × 10^-6F, L = 0.10H, R = 25 Ω ,
e = 310 sin (314 t) [volt]
Comparing e = 310 sin (314 t) with
e = e₀ sin (2πft), we get,
the frequency of the alternating emf as
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 13
cos Φ = \(\frac{R}{Z}=\frac{25}{99.2}\) = 0.2520
∴ The phase angle, Φ = cos^-1(0.2520) = 75.40° = 1.316 rad

Question 18.
A capacitor of 100 µF, a coil of resistance 50Ω and an inductance 0.5 H are connected in series with a 110 V-50Hz source. Calculate the rms value of current in the circuit.
Answer:
Data : C = 100 µF = 100 × 10^-6 F = 10^-4 F,
R = 50 Ω, L = 0.5H, f = 50 Hz, V_rms = 110 V
∴ ωL = 2πfL = 2 (3.142)(50)(0.5) = 157.1 Ω
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 14
2500 + 15700 = 18200 Ω²
∴ Impedance, Z = \(\sqrt {18200}\) Ω = 134.9 Ω
The rms value of the current in the circuit,
i_rms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{110}{134.9} \mathrm{~A}\)
= 0.8154 A

Question 19.
Find the capacity of a capacitor which when put in series with a 10Ω resistor makes the power factor equal to 0.5. Assume an 80V-100Hz AC supply.
Answer:
Data : R = 10 Ω, power factor = 0.5, f = 100 Hz
Power factor = \(\frac{1}{2 \pi f C R}\)
∴ 0.5 = \(\frac{1}{2(3.142)(100) C(10)}\)
∴ C = \(\frac{1}{3.142 \times 10^{3}}\)
= \(\frac{10 \times 10^{-4}}{3.142}\)
= 3.182 × 10^-4 F
This is the capacity of the capacitor.

Question 20.
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
Answer:
Data : f = 50 Hz, i = \(\frac{i_{0}}{\sqrt{2}}\) ∴ \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
i = i₀ sinωt
∴ sinωt = \(\frac{i}{i_{0}}=\frac{1}{\sqrt{2}}\)
∴ ωt = \(\frac{\pi}{4}\) rad
∴ 2πft = \(\frac{\pi}{4}\)
∴ t = \(\frac{1}{8 f}=\frac{1}{8(50)}=\frac{1}{400}\)
= \(\frac{1000 \times 10^{-3}}{400}\) = 2.5 × 10^-3 s
This is the required time.

Question 21.
Calculate the value of capacitance in picofarad, which will make 101.4 micro henry inductance to oscillate with frequency of one megahertz.
Answer:
Data : f_r = 10⁶ Hz, L = 101.4 × 10^-6 H
Maharashtra Board Class 12 Physics Solutions Chapter 13 AC Circuits 19
= \(\frac{10000 \times 10^{-10}}{4(3.142)^{2}(101.4)}\) = 2.497 × 10^-10 F
= 249.7 × 10^-12 F = 249.7 picofarad
This is the value of the capacity.

Question 22.
A 10 µF capacitor is charged to a 25 volt of potential. The battery is disconnected and a pure 100 m H coil is connected across the capacitor so that LC oscillations are set up. Calculate the maximum current in the coil.
Answer:
Data: C = 10 µF = 10 × 10^-6F = 10^-5F,
L = 100mH = 100 × 10^-3 H = 10^-1 H, V = 25V
For reference, see the solved example (8) above.
\(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴i² = \(\frac{C}{L} V^{2}=\frac{10^{-5}}{10^{-1}}(25)^{2}\)
∴i = 25 × 10^-2 A = 0.25 A
This is the maximum current in the coil.

Question 23.
A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.
Answer:
Data: C = 100 µF = 100 × 10^-6 F = 10^-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= \(\frac{1}{2}\)CV²
The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li²
Here, \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)Li²
∴ L = C\(\frac{V^{2}}{i^{2}}\)
∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10^-4 × 10²
= 10^-2H
This is the value of the inductance.