Plus One

Students can Download Chapter 8 Gravitation Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 8 Gravitation

Plus One Physics Gravitation One Mark Questions and Answers

Plus One Physics Gravitation Questions And Answers Question 1.
If a satellite of mass m is revolving around the earth with distance rfrom centre, then total energy is

Answer:
(c) \(-\frac{G M m}{2 r}\)
The satellite revolving around the earth has two types of energies.
1. Gravitational potential energy U due to position of the satellite is given by
U = \(-\frac{\mathrm{GMm}}{\mathrm{r}}\).

2. Kinetic energy K due to orbital velocity of satellite is given by

Plus One Physics Chapter Wise Questions And Answers Question 2.
The dimensions of universal gravitational constant are
(a) [M^-2L³T^-2]
(b) [M^-2L²T^-1]
(c) [M^-1L³T^-2]
(d) [ML²T^-1]
Answer:
(c) [M^-1L³T^-2]
According to Newton’s law of gravitation

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 3.
When a radius of earth is reduced by 1% without changing the mass, then change in the acceleration due to gravity will be
(a) increased by 2%
(b) decreased by 1.5%
(c) increased by 1 %
(d) decreased by 1 %
Answer:
(a) increased by 2%

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Question 4.
The kinetic energy of a satellite is 2MJ. What is the total energy of the satellite?
(a) -2M J
(b) -1MJ
(c) -1/2MJ
(d) -4MJ
Answer:
(a) -2MJ
Total energy of satellite – Kinetic energy of satellite.

Plus One Physics Important Questions And Answers Pdf Question 5.
If total energy of this stone is negative, can it escape from the earth’s surface? Justify.
Answer:
It will not escape as the force is attractive.

Gravitation Questions And Answers Question 6.
Why a man can jump higher on the moon than on earth?
Answer:
The value of g on the moon is one sixth of that on earth.

Hss Live Plus One Physics Chapter Wise Questions And Answers Question 7.
The gravitational potential energy of a body of mass ‘m’ is -10⁷J. What energy is required to project the body out of gravitational field of earth?
Answer:
10⁷J

Plus One Physics Chapter Wise Questions And Answers Pdf Question 8.
How much energy is required by a satellite to keep it orbiting? Neglect air resistance.
Answer:
No energy is required because work done by centripetal force is zero.

Gravitation Class 11 Important Questions With Answers Question 9.
What would happen to an artificial satellites, if its orbital velocity is slightly decreased due to some defects in it?
Answer:
It will fall on to earth.

Questions On Gravitation For Class 11 Question 10.
Why a tennis ball bounces higher on hills than in plain?
Answer:
As height increases ‘g’ decreases. The value of g on hills is less than that on the surface of earth. Hence tennis ball bounces higher on hills than in plains.

Gravitation Class 11 Questions And Answers Question 11.
The orbiting velocity of an earth-satellite is 8 km/s. What will be the escape velocity?
Answer:
V_e = \(\sqrt{2}\) V₀
V_e = \(\sqrt{2}\) × 8 = 11.3 kms^-1.

Plus One Physics Gravitation Two Mark Questions and Answers

Gravitation Class 11 Derivations Pdf Question 1.
1. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant

• linear speed
• total energy
• angular momentum

2. What are the consequences if the angular momentum is conserved?
Answer:
1. Highly elliptical orbit:

• No
• Yes
• Yes

2. consequences if the angular momentum:

• Areal velocity of planet becomes constant
• Angular velocity changes according to the position of planet around sun.

Numericals On Gravitation Class 11 Question 2.
Fill in the blanks
R – Radius of earth, g – Acceleration due to gravity h – Height from surface of earth, d – Depth from surface of earth

Answer:
a) g(h) = g(1 – \(\frac{2 h}{R}\))
b) Variation of g with depth
c) g Decreases with depth
d) (Change with radius (polar, equitorial org is maximum on the earth’s surface)

Plus One Physics Gravitation Three Mark Questions and Answers

Gravitation Numericals Class 11 Question 1.
The Earth moving round the Sun in a circular orbit is acted upon by a force.

1. Name the force acting on the Earth.
2. Obtain a mathematical expression for the force.
3. What is the work done by the force?

Answer:

1. Centripetal force
2. F_e = ma
   F_c = m \(\frac{v^{2}}{r}\)
3. zero. (Displacement is always perpendicular to centripetal force).

Question 2.
Escape velocity of objects in a planet depends on the mass and size of the planet.

1. Write down the expression for the escape velocity on the surface of the earth.
2. What is its value on the surface of our planet?
3. Give the reason why the moon does not have an atmosphere around it.

Answer:

1. Escape velocity Ve = \(\sqrt{\frac{2 G M}{R}}\)
2. 11.2km/sec
3. The escape velocity of moon is very low. Hence all the gas molecules can escape from the moon’s surface.

Question 3.
The acceleration due to gravity is measured using ticker timer and is found to be 9.8ms^-2

1. What does the value 9.8 for the acceleration implies?
2. One of your friends argues that the acceleration due to gravity at the centre of earth is infinity. Do you agree with it?
3. Justify your answer with mathematical support.

Answer:
1. An acceleration of 9.8m² implies that velocity change by 9.8ms¹ in every second.

2. No, at centre g = 0

3. At centre d = R
g_d = g × 0 = 0

Plus One Physics Gravitation Four Mark Questions and Answers

Question 1.
A body of mass ‘m’ falls freely under gravity, near the surface of earth.

1. Will the acceleration of the body change if a part of the mass is thrown away from it?
2. What will be the free fall acceleration if it is falling from a height equal to R, the radius of earth?
3. If the mass is taken to the moon’s surface, will the free fall acceleration increase or decrease?

Answer:
1. No. The acceleration of body does not depend on the mass of falling body.

2. We know acceleration due to gravity

3. Decrease.

Question 2.
Geo Stationary satellites are commonly used for communication purpose.

1. Name one geostationary satellite of earth.
2. What are the requirements of a geostationary satellite for its orbital motion.
3. Explain the phenomenon of ‘weightlessness’ in orbiting satellites.
4. Distinguish between gravitational mass and inertial mass.

Answer:

1. INSAT.
2. The period of this satellite must be same as the period of the rotation of earth about its own axis. Its direction of rotation must be the same as that of the earth (from west to east).
3. The weight of satellite is used for providing centripetal force required for rotation. Hence anybody inside the satellite appears to be weightlessness.
4. The mass related with newtons second law of motion is called inertial mass. But the mass related with newtons law of gravitation is called gravitational mass.

Question 3.

1. What is acceleration due to gravity?
2. Does a body have the same weight at the equator and at the poles? Explain.
3. If the value of gravitational constant is 6.6 × 10^-11 Nm²kg^-2 and g = 9.8 m/s², find the mass of the earth. Given, radius is 6.4 × 10⁶m.

Answer:
1. 9.8 m/se²

2. No. The weight of the body depends on the acceleration due to gravity at a place. The magnitude of acceleration due to gravity at equator and poles are different.

3.

Question 4.
Two identical satellites are orbiting in circular orbits around the earth at heights R and 3R respectively from the surface of the earth. The radius of the earth is R.

1. Define orbital velocity.
2. What is its value on the surface of the earth?
3. How do you compare the periods of revolution of these two satellites?

Answer:
1. The minimum velocity required for a satellite to revolve around earth in a stable orbit is called orbital velocity.

2.

3.

Question 5.
The uniform acceleration produced in a freely falling body due to gravitational pull of the earth is known as acceleration due to gravity.

1. What is the value of acceleration due to gravity on the surface of earth.
2. Obtain an expression for acceleration due to gravity at a depth ‘d’ from the surface of earth
3. What is the value of gat the centre of earth?

Answer:
1. 9.8m/s^-2

2.

If we assume the earth as a sphere of radius R with uniform density?
mass of earth = volume × density

Substituting eq(1) in eq(2), we get
g = \(\frac{4}{3}\) πGR_ρ ……(3)
Therefore the acceleration due to gravity at a depth d is given by
g_d = \(\frac{4}{3}\) πG(R – d)ρ ……(4)
eq(4)/eq(3) and solving we get

g_d = g(1 – d/R)
The above equation shows that, when depth increases g decreases.

3. Increase.

Plus One Physics Gravitation Five Mark Questions and Answers

Question 1.
Earth can be treated as a sphere of radius R and mass M. A is a point at a height h above the Earth’s surface and B is another point at a depth h1 below the Earth’s surface. The acceleration due to gravity at Earth’s surface is g.

1. Obtain a formula to evaluate the acceleration due to gravity at A.
2. What is the value of acceleration due to gravity at B?
3. If we move from equator to pole, the value of g.
   • increases
   • decreases
   • remains the same
   • first increases and then decreases
   • first decreases and then increases
4. What is the value of g at the centre of Earth? Explain.

Answer:
1. The acceleration due to gravity on the surface of earth,
g = \(\frac{G m}{R^{2}}\) …..(1)
At a height h, the acceleration due to gravity can be written as,

eq(1)/eq(2) and solving we get
g_h = g(1 – \(\frac{2 h}{R}\))
This equation shows that acceleration due to gravity decreases as height increases. The above equation is valid when h << R.

2.

3. Increase

4. Zero. At centre, the force acting on the body is zero.

Question 2.
Fora particle to leave from earth’s gravitational field it should be projected with a minimum velocity.

1. Name the velocity.
2. Derive an expression for this velocity.
3. What is the magnitude of this velocity when this particle is projected from another planet whose mass and radius is twice that of the earth?

Answer:
1. Escape velocity

2. Expression for escape speed:
Force on a mass m at a distance r from the centre of earth = \(\frac{G M m}{r^{2}}\)
Work done in taking the body to infinity from surface of earth,

This energy is given in the form of K.E. = \(\frac{1}{2}\)mv_e²,
where v_e is the escape speed.

This escape velocity \(\sqrt{2 \mathrm{Rg}}\) is estimated to be 11.2 km/son the earth.

3. Escape velocity of planet,

Question 3.
Moon is the only satellite of earth. Mass of moon is very much lower than that of earth.

1. Is there any difference in gravitational force of moon and earth? Explain.
2. Deduce an equation for gravitational potential energy.
3. A girl argues that even today if we visit moon, we can see the foot steps of Neil Armstrong. What is your opinion?

Answer:
1. Yes. Gravitational force of earth longer than gravitational force of moon.

2. Expression for gravitational P.

Consider the earth as a uniform sphere of radius R and mass M. Considera point A at distance ‘r’ from the centre of earth. P is another point at a distance ‘x’ from O. Q lies at distance dx from P.

By definition, the gravitational potential energy of the body at point A, is the work done in bringing the body of mass ‘m’ from infinity to that point A The gravitational force on the body at the point P.
is given by F = \(\frac{G M m}{x^{2}}\).
If the body is displaced from P to Q Work done, dw = F.dx
= \(\frac{G M m}{x^{2}}\) Therefore, workdone in bringing the body from infinity to the point A,

Since this workdone is stored inside the body as its gravitational potential energy, the gravitational potential energy (U) of a body of mass m at distance rfrom the centre of the earth is given by,

The gravitational potential energy of a body at a point is defined as the amount of workdone in bringing the body from infinity to that point without acceleration.

3. There is no atmosphere on the moon. Hence we can see the foot steps of Neil Armstrong.

Plus One Physics Gravitation NCERT Questions and Answers

Question 1.
Suppose there existed a planent that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer:
T_e = 1 year, T_p = \(\frac{1}{2}\) year, R_p = ?, R_e = 1 AU
Using Kepler’s third law

Question 2.
lo, one of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸m. Show that the mass of the Jupiter is about one-thousandth that of the Sun.
Answer:

Substituting the given data for Jupiter
Mass of Jupiter

Substituting the known data forthe revolution of earth around Sun,

Question 3.
Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete on revolution? Take the diameter of the Milky way to be 10⁵ly.
Answer:

Question 4.
A rocket is fired from the earth towards the Sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 × 10³⁰ kg, Mass of the earth = 6 × 10²⁴kg. Neglect the effect of other planets etc. (Orbital radius = 1.5 × 10¹¹m).
Answer:
Mass of Sun, M = 2 × 10³⁰ kg
Mass of Earth m = 6 × 10²⁴kg.
Distance between Sun and Earth, r= 1.5 × 10¹¹m
Suppose, at the point P, the gravitational force on the rocket due to earth = gravitational force on the rocket due to Sun
Let x be the distance of the point P from the earth

Question 5.
How will you “weigh the Sun’, that is estimate its mass? The mean orbital radius of the earth around the Sun is 1.5 × 10⁸km.
Answer:
For the revolution of earth around sun, the gravitational force between the sun and the earth provides the necessary centripetal force.

Question 6.
A geostationary satellite orbits the earth at a height of nearly 36,000km from the surface of the earth. What is the potentional due to earth’s gravity at the site of this sattelite? (Take the potential energy at infinity to be zero). Mass of the earth = 6 × 10²⁴ kg., radius of earth = 6,400 km.
Answer:
Required potential

Question 7.
A satum year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 10⁸km away from the sun?
Answer:
We know that T² α R³

where subscripts s and e refer to the satum and earth respectively.
Now \(\frac{T_{s}}{T_{e}}\) = 29.5 (given); R_e = 1.50 × 10⁸km

R_s = R_e × [(29.5)²]^1/3 = 1.50 × 10⁸ × (870.25)^1/3km
= 1.43 × 10⁹km = 1.43 × 10¹²m.

Question 8.
A body weighs 63N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Answer:
Let g_h be the acceleration due to gravity at a height equal to half the radius of the earth (h = R/2) and g its value on earth’s surface. Let the body have mass m, we know that

Let W be the weight of body on the surface of earth and W_h the weight of the body at height h.
Then,

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

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

Plus One Economics Chapter 5 Notes Human Capital
Human capital is defined as a human resource where people as a capital asset which yields a stream of economic benefit over their working span of life. Converting human beings through education and training into human resources like teachers, doctors, scientists, etc. can be termed as human capital formation.

Hsslive Plus One Economics Malayalam Notes Sources of Human Capital
Sources of human capital can be classified into five. They are given below:

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

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

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

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

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

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

Sources Of Human Capital Formation Class 12 Human Capital and Human Development
Human capital considers education and health as a means to increase labour productivity. Human development is based on the idea that education and health are integral to human well-being. When people attain certain ability to read and write and acquire ability to lead a long and healthy life, their productivity will go up. Human welfare should be increased through investments in education and health. Education and health improve human welfare. In a welfare state, every individual has the right to get basic right to be literate and lead a healthy life.

Class 12 Human Capital Formation Notes Human Capital Formation in India
We have seen that human capital formation is essential for economic growth and development. Various institutions are operating in India for human capital formation. They are

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

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

Students can Download Chapter 3 Private, Public and Global Enterprises Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 3 Private, Public and Global Enterprises

Plus One Business Studies Chapter 3 Contents

• Private sector and Public sector – Meaning
• Departmental Undertakings – Meaning – Features Advantages & Disadvantages
• Public Corporation – Meaning – Features Advantages & Disadvantages
• Government Company – Meaning – Features Advantages & Disadvantages
• Changing Role of Public Sector – Public Sector Reforms since 1991
• Global Enterprises – Meaning – Features Joint Ventures – Meaning – Benefits

The economy may be classified into two sectors viz., private sector and public sector:
1. Private Sector:
The private sector consists of business owned by individuals or a group of individuals. The various forms of organisation are sole proprietorship, partnership, joint Hindu family, cooperative and company.

2. Public Sector:
The business units owned, managed and controlled by the central, state or local government are termed as public sector enterprises or public enterprises. These are also known as public sector undertakings.

Plus One Business Studies Chapter 3 Notes Characteristics of Public Sector Enterprises:

1. The public enterprises are owned and managed by the central or state government, or by the local authority.
2. The public enterprises get their capital from Government Funds
3. The main objective of public sector enterprises is providing the service or commodity at reasonable prices.
4. Public sector enterprises concentrate on providing public utility services like transport, electricity, telecommunication etc.
5. Public enterprises are are accountable to the public through the Parliament.

Class 11 Business Studies Chapter 3 Notes Forms of organizing Public sector Enterprises

1. Departmental Undertaking
2. Statutory Corporation (public corporation)
3. Government Company

Enterprise Meaning In Malayalam Departmental Undertaking:
This is the oldest and most common form of organization. These are established as departments of the ministry and are financed, managed and controlled by either central govt, or state govt.. They are managed by government employees and work under the control of a minister. eg: Railways, Post & Telegraph, All India Radio, Doordarshan, Defense undertakings etc.

Features

1. The enterprise is financed by annual appropriation from the budget of the Government and all revenue is paid to the treasury.
2. The enterprise is subject to accounting and audit control.
3. It is subject to the direct control of the ministry.
4. Its employees are govt, employees and are recruited and appointed as per govt, rules.
5. They are accountable to the concerned ministry.

Merits:

1. These undertakings facilitate the Parliament to exercise effective control over their operations.
2. These ensure a high degree of public accountability.
3. The revenue earned by the enterprise goes directly to the treasury and hence is a source of income for the Government.
4. Where national security is concerned, this form is most suitable since it is under the direct control and supervision of the concerned Ministry.

Limitations:

1. Departmental undertakings lack flexibility because its policies cannot be changed instantly.
2. The employees are not allowed to take independent decisions, without the approval of the ministry concerned. This leads to delay in decision making.
3. These enterprises are unable to take advantage of business opportunities.
4. There is red tapism in day-to-day operations.
5. There is a lot of political interference through the ministry.
6. These organisations are usually insensitive to consumer needs and do not provide adequate services to them.

Business Studies Class 11 Chapter 3 Question Answers Statutory Corporations ( Public Corporation):
Statutory corporations are public enterprises brought into existence by a Special Act of the Parliament. The Act defines its powers and functions, rules and regulations governing its employees and its relationship with government departments. eg: LIC, IFCI, RBI, SBI, ONGC, UTI, Air India etc.
Features:

1. Statutory corporations are set up under an Act of Parliament and are governed by the provisions of the Act.
2. It is wholly owned by the state.
3. It has a separate legal entity, i.e it can sue and be sued, enter into contract and acquire property in its own name.
4. It is usually independently financed.
5. It is not subject to budget, accounting, and audit laws.
6. The employees of these enterprises are not government or civil servants.

Merits:

• They enjoy independence in their functioning and a high degree of operational flexibility
• It enjoys administrative and financial autonomy
• Since they are autonomous organizations, they can frame their own policies and procedures
• A statutory corporation is a valuable instrument for economic development

Limitations:

• A statutory corporation does not enjoy as much operational flexibility
• Government and political interference has always been there in major decisions
• Where there is dealing with public, rampant corruption exists
• Commercial principles are ignored in the operation of public corporations which leads to inefficiency.

Chapter 3 Business Studies Class 11 Government Company:
A Government Company is established under the Indian Companies Act, 1956. According to the Indian . Companies Act 1956, a government company means any company in which not less than 51 percent of the paid up capital is held by the central Private, Public and Global Enterprises government, or by any state government or partly by central government and partly by one or more state governments.
Features:

1. It is registered under the Companies Act, 1956.
2. It has a separate legal entity. It can sue and be sued and can acquire property in its own name.
3. The management of the company is regulated by the provisions of the Companies Act.
4. Employees are recruited and appointed as per the rules and regulations contained in Memorandum and Articles of association.
5. These companies are exempted from the accounting and audit rules and procedures.
6. The government company obtains its funds from government shareholdings and other private shareholders.

Merits:

1. It has a separate legal entity, apart from the Govt.
2. It enjoys flexibility and autonomy in all management decisions
3. These companies by providing goods and services at reasonable prices are able to control the market.
4. The formation of a Government company is easy as compared to other forms of Government enterprises.
5. It can appoint professional managers on high salaries.

Limitations:

1. It evades constitutional responsibility as it is not directly answerable to parliament.
2. They are autonomous only in name. Company is operated by the controlling ministry.
3. The law relating to the companies, in general is meaningless for the government companies, as it requires fulfillment of various formalities.

Comparative Features of the various forms of Public Enterprises:

Business Studies Class 11 Chapter 3 Notes Role and importance of Public Sector:
1. Development of infrastructure:
It is the responsibility of the Government to provide infrastructural facilities to the core sector which requires huge capital investment, complex and upgraded technology etc.

2. Regional balance:
The government is responsible for developing all regions and states in a balanced way and removing regional disparities.

3. Economies of scale:
Public sector enterprises are large in size and are, able to avail the advantages of large scale operations.

4. Check over concentration of economic power:
The development of public enterprises prevents concentration of economic power and wealth in the hands of private sector.

5. Employment opportunities:
Public sector enterprises helps to generate a large number of employment opportunities.

6. Import substitution:
It is also necessary for the economic progress of the country that industries which can decrease imports and increase exports are only promoted. Public enterprises also ensure promotion of such industries.

Private Public And Global Enterprises Class 11 Notes Public Sector Reforms:
In the industrial policy 1991, the govt, of India introduced four major reforms in public sector.
They are:
1. Reduction in number of industries reserved for public sector:
The number of industries reseved for public sector is reduced from 17 to 8 in 1991. But in 2001 it is again reduced to 3 industries. These three industries are atomic energy, arms and rail transport.

2. Memorandum of Understanding (MOU):
Under this govt, lays down performance targets for the management and gives greater autonomy to hold the management accountable for the results.

3. Disinvestment:
Disinvestment involves the sale of equity shares to the private sector and public. The objective was to raise resources and encourage wider participation of the general public in the ownership of these enterprises.

4. Restructure and Revival:
All public sector sick units were referred to Board of Industrial and financial Reconstruction (BIFR). Units which were potentially viable were restructured and which could not be revived were closed down by the board.

Objectives of privatizing public sector enterprises:

1. Releasing the large amount of public resources locked up in nonstrategic Public Sector Enterprises (PSEs), so that they may be utilized on other social priority areas such as basic health, family welfare and primary education.
2. Reducing the huge amount of public debt and interest burden;
3. Transferring the commercial risk to the private sector so that the funds are invested in able projects;
4. Freeing these enterprises from government control and introduction of corporate governance; and
5. In many areas where the public sector had a monopoly, for example, telecom sector the consumers have benefitted by more choices, lower prices and better quality of products and services.

Global enterprises (Multi National Companies):
Global enterprises are huge industrial organisations which extend their industrial and marketing operations through a network of their branches in several countries. Their branches are also called Majority Owned Foreign Affiliates (MOFA).

They are characterised by their huge size, large number of products, advanced technology, marketing strategies and network of operations all over the world. Eg. Pepsi, Coca Cola, Cadbury, Sony, Susuki etc.
Features of MNCs:
1. Huge capital resources:
Multinational companies have the ability to raise huge funds from different sources such as equity shares, debentures, bonds etc. They can also borrow from financial institutions and international banks.

2. Foreign collaboration:
Global enterprises usually enter into agreements relating to the sale of technology, production of goods, use of brand name etc. with local firms in the host countries.

3. Advanced technology:
Multinational companies can possess latest and advanced technology so that they can provide quality products.

4. Product innovation:
Multinational companies are able to conduct sophisticated research so that they can develop new products.

5. Marketing strategies:
They use aggressive marketing strategies in order to increase their sales in a short period. Their advertising and sales promotion techniques are normally very effective.

6. International Market:
They operate through a network of subsidiaries, branches and affiliates in host countries. Due to their giant size, they occupy a dominant position in the market.

7. Centralised control:
They have their headquarters in their home country and exercise control over all branches and subsidiaries.

Ch 3 Business Studies Class 11 Notes Joint Ventures:
When two or more independent firms together establish a new enterprise by pooling their capital, technology and expertise, it is known as a joint venture. The risks and rewards of the business are also shared.

The aim of joint venture is business expansion, development of new products or moving into new markets, particularly in another country. Benefits
1. Increased resources and capacity:
Since two or more firms join together to form a joint venture, there is availability of increased capital and other resources, able to face market challenges and take advantage of new opportunities.

2. Access to new markets and distribution networks:
A foreign company gain access to the vast Indian market by entering into a joint venture with Indian Company. They can also take advantage of the established distribution channels.

3. Access to technology:
It provides access to advanced techniques of production which increases efficiency and then helps in reduction in cost and improvement in quality of product.

4. Innovation:
Foreign partners can come up with innovative products because of new ideas and technology.

5. Low cost of production:
Low cost of raw materials, technically qualified workforce, management professionals, excellent manpower etc. helps to reduce cost of production and it results increased productivity.

6. Established brand name:
When one party has well established brands and goodwill, the other party gets its benefits.

Students can Download Chapter 2 Indian Economy 1950-1990 Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 2 Indian Economy 1950-1990

Plus One Economics Chapter 2 Notes The Goals of Five Year Plans
A plan should have some clearly specified goals. The goals of the five-year plans are growth, modernization, self-reliance, and equity. This does not mean that all the plans have given equal importance to all these goals. Let us now learn about the goals of planning in some detail.

1. Growth: It refers to an increase in the country’s capacity to produce the output of goods and services within the country. It implies either a larger stock of productive capital, or a larger size of supporting services like transport and banking, or an increase in the efficiency of productive capital and services. A good indicator of economic growth, in the language of economics, is a steady increase in the Gross Domestic Product (GDP).

2. Modernization: To increase the production of goods and services the producers have to adopt new technology Adoption of new technology is called modernization.

3. Self-reliance: A nation can promote economic growth and modernization by using its own resources or by using resources imported from other nations. The first seven five-year plans gave importance to self-reliance which means avoiding imports of those goods which could be produced in India itself. This policy was considered a necessity in order to reduce our dependence on foreign countries, especially for food. It is understandable that people who were recently freed from foreign domination should give importance to self-reliance.

4. Equity: Now growth, modernization, and self-reliance, by themselves, may not improve the kind of life which people are living. A country can have high growth; the most modem technology developed in the country itself.

Plus One Economics Chapter 2 Agriculture
Land reforms and the green revolution were the two important changes in India’s agricultural sector during the initial periods of independence.

Land Reforms: The purpose of land reforms is twofold. On the one hand, it aims to make rational use of the scarce land resource by enforcing conditions on holding land. Secondly, it aims at the redistribution of land in favour of landless farmers.
Land reform consists of the following measures:

1. Abolition of intermediaries like Zamindars
2. Tenancy reforms, ie. regulation of rent, the security of tenure, etc.
3. Ceiling of landholding
4. Distribution of land among landless by acquiring surplus land from big landlords
5. Consolidation of holding and prevention of subdivision and fragmentation
6. Organization of cooperative farming

The policy decisions and legislative initiatives of the government on land reforms were very progressive. As a result of it, tillers were able to undertake the permanent improvement of their land and this contributed to growth in agriculture. However, the implementation of land reforms was not free from limitations. Some loopholes in the law were conveniently exploited by the landlords and transferred surplus land in the name of their relatives or binamies. But states like Kerala and West Bengal implemented the land reform measures in a relatively successful way.

The Green Revolution: Planners and policymakers of independent India gave top priority to the development of agriculture. In the 1960s the new technology in agriculture was tried in seven districts and was called the Intensive Agricultural Districts Programme (IADP). Later, the High Yielding Varieties Programme (HYVP) “was extended to the entire country, which is popularly known as Green Revolution or modem agricultural technology, seed fertilizer water technology. Norman Borlaug, an American Biologist was known as the father of the green revolution. Prof. M.S. Swaminathan is known as the father of the green revolution in India.
Features of the green revolution are:

1. Use of high yielding variety (HYV) seeds
2. Use of chemical fertilizers and pesticides
3. Use of modem farm implements like power tillers, tractors, water pump sets, etc.
4. Better irrigation facilities
5. Easy availability of credit to farmers at lower interest rates

Plus One Economics Second Chapter Notes Industry and Trade
Economic growth and development depend upon industry and trade. The industry provides employment. In order to promote the industrial sector government introduced industrial policies.

Industrial Policy Resolution 1956 (IPR-1956)
To mould the economy into a socialist pattern, where the state ‘holds the commanding heights’ the Industrial Policy Resolution of 1956 (IPR1956) was adopted. This resolution classified industries into three categories:

1. Industries exclusively owned by the state.
2. Industries in which the private sector could supplement the efforts of the state.
3. Remaining industries were to be in the private sector. But they will be kept under state control through a system of licenses.

Plus One Economics Chapter Notes Small Scale Industries
Small scale industries are crucial for economic development. A small scale industry is defined with the reference to the maximum investment allowed to set up a unit. This limit of investment has changed over a period of time. In 1950 a small-scale industry unit was one which invests a maximum of rupees five lakhs, in 1966 it was raised to rupees 7.5 lakhs, in 1990 it was raised to rupees 60 lakhs and from 2000 it has been fixed at rupees one crore.
Small scale industries have the following advantages:

1. Less capital investment
2. Labour intensive technology – generates more employment opportunities
3. Less dependence on imports
4. Less pollution-environment friendly
5. Rural development

Plus One Economics Notes In Malayalam Pdf Import Substitution
The practice of replacing imports with the products made within the country is known as import substitution. The government allowed a policy of protection in order to protect domestic industries.

HSE Kerala Board Syllabus HSSLive Plus One Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of Plus One Kerala SCERT. Here HSSLive.Guru has given Higher Secondary Kerala Plus One Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus.

Kerala Plus One Chapter Wise Previous Year Questions and Answers

• Plus One Maths Chapter Wise Previous Questions and Answers
• Plus One Physics Chapter Wise Previous Questions and Answers
• Plus One Chemistry Chapter Wise Previous Questions and Answers
• Plus One Botany Chapter Wise Previous Questions and Answers
• Plus One Zoology Chapter Wise Previous Questions and Answers
• Plus One Economics Chapter Wise Previous Questions and Answers
• Plus One Business Studies Chapter Wise Previous Questions and Answers
• Plus One Accountancy Chapter Wise Previous Questions and Answers
• Plus One Computer Science Chapter Wise Previous Questions and Answers
• Plus One Computer Application Chapter Wise Previous Questions and Answers

We hope the given HSE Kerala Board Syllabus HSSLive Plus One Chapter Wise Previous Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus One Chapter Wise Previous Year Important Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus One

Students can Download Chapter 4 Business Services Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 4 Business Services

Plus One Business Studies Chapter 4 Contents

• Meaning – Nature of Service – Differences between Services and Goods – Types of services
• Banks – Types of banks – Commercial bank – Functions of Commercial bank – e-Banking – Benefits of e-banking
• Insurance – Functions of insurance – Principles of insurance
• Types of insurance – Life insurance – Fire insurance – Marine insurance
• Communication services – Postal services Telecom services
• Transportation – Functions of Transportation
• Warehouse – Functions – Types of warehouses Services are those identifiable, essentially intangible activities that provides satisfaction of wants. For example, consulting a doctor for medical help, getting legal opinion from an advocate etc.

Business Studies Class 11 Chapter 4 Notes Pdf Nature of Services:
1. Intangibility:
Services are intangible, i.e., they cannot be touched. They are experiential in nature, eg: Treatment by a doctor.

2. Inconsistency:
Since there is no tangible product, services have to be performed according to the demand and expectations of the different customers, eg: Mobile services/Beauty parlour.

3. Inseparability:
Another important characteristic of services is the simultaneous activity of production and consumption being performed, i.e. They are inseparable, eg: ATM may replace clerk but presence of customer is a must.

4. Less Inventory:
Services cannot be stored for future use.

5. Involvement:
Participation of the customer in the service delivery is a must.

Class 11 Business Studies Chapter 4 Notes Difference between Services and Good:

Business Services Class 11 Explanation Types of service:
1. Business Services:
Business services are those services which are used by business enterprises for the conduct of their activities. For example, banking, insurance etc.

2. Social Services: Social services are generally provided voluntarily in pursuit of certain social goals. Eg. improve the standard of living for weaker sections of society, to provide educational services to their children etc.

3. Personal Services: Personal services are those services which are experienced differently by different customers, eg: tourism, restaurants, etc. Banking.

According to Banking Regulation Act 1949, banking means “accepting, for the purpose of lending and investment of deposits of money from the public, repayable on demand or otherwise and withdrawable by cheques, draft, order or otherwise”. Banks can be classified into the following:

1. Commercial banks
2. Cooperative banks
3. Specialised banks
4. Central bank

1. Commercial Banks:
Commercial Banks are banking institutions that accept deposits and grant short-term loans and advances to their customers. There are two types of commercial banks, public sector and private sector banks.
(a) Public Sector Bank:
Public sector banks are those banks, which are owned and managed by the Government. eg. SBI, PNB, IOB etc. Presently there are 28 public sector banks in India

(b) Private Sector Bank:
Private sector banks are owned and managed by private parties. Eg. HDFC Bank, ICICI Bank, Kotak Mahindra Bank and Jammu and Kashmir Bank etc. ICICI bank is the largest private sector bank in India.

2. Cooperative Banks:
Cooperative Banks are governed by the provisions of State Cooperative Societies Act. They managed on the principles of co-operation, self help and mutual help.

3. Specialised Banks:
Specialised banks are foreign exchange banks, industrial banks, development banks, export-import banks which provide financial aid to industries, joint venture projects and foreign trade.

4. Central Bank:
The Central bank supervises, controls and regulates the activities of all the commercial banks of that country. It also acts as a government banker. It controls and coordinates currency and credit policies of any country. The Reserve Bank of India is the central bank of our country.

Business Services Class 11 Notes Insurance:
Insurance is an agreement between two parties whereby one party undertakes, in exchange for a consideration, to pay the other party an agreed sum of money to compensate the loss, damage or injury caused as a result of some unforeseen events.

• Policy: The agreement or contract entered into by the insured and insurer is known as a ‘policy’.
• Insurer: The firm which insures the risk of loss is known as ‘insurer’.
• Insured: The person whose risk is insured is called ‘insured’.
• Premium: The consideration in return for which the insurer agrees to compensate the insured is known as ‘Premium’.

Functions of Insurance:
1. Providing certainty:
Insurance provides certainty of payment for the risk of loss. The insurer charges premium for providing the certainty

2. Protection:
Insurance provides protection from probable chances of loss.

3. Risk sharing:
Insurance helps in sharing of risk. The premium collected from a large number of people are used for compensating the loss of a few.

4. Assist in capital formation:
The accumulated funds of the insurer received by way of premium are invested in various income generating schemes.

Functions Of Warehousing Class 11 Principles of Insurance:

1. Utmost good faith:
The insured must disclose all material facts about the subject matter to the insured. Otherwise the insurer can cancel the contract. The insurer must disclose all the terms and conditions jn the insurance contract to the insured.

2. Insurable interest:
The insured must have an insurable interest in the subject matter of insurance. Insurable interest means the interest shown by the insured in the continued existence of the subject matter or the financial loss he is subjected to on the happening of an event against which it has been insured.

3. Indemnity:
All insurance contracts, except life insurance are contracts of indemnity. According to the principle of indemnity, in the event of occurrence of loss, the insured will be indemnified to the extent of the actual value of his loss or the sum insured which ever is less. The objective behind this principle is nobody should treat insurance contract as the source of profit.

4. Subrogation:
According to this principle, after the insured is compensated for the loss to the property insured by him the right of ownership of such property passes on to the insurer. This is because the insured should not be allowed to make any profit, by selling the damaged property.

5. Causa proxima:
When the loss is the result of two or more causes, the proximate cause for the loss alone will be considered by the insurance company for admitting the claim.

6. Contribution:
In certain cases, the same subject matter is insured with one or more insurer. In case there is a loss, the insured is eligible to receive a claim only up to the amount of actual loss suffered by him.

7. Mitigation of loss:
This principle states that it is the duty of the insured to take reasonable steps to minimize the loss or damage to the insured property. If reasonable care is not taken then the claim from the insurance company may be rejected.

Types of Insurance:

Business Studies Insurance Notes Life Insurance:
It is a contract whereby the insurer undertakes to pay a certain sum of money either on the death of the insured or on the expiry of a specified period of time in consideration of a certain amount (premium) paid by the insured either in lump sum or in installments.
Elements of Life insurance contract:

1. The life insurance contract must have all the essentials of a valid contract
2. Life insurance is a contract of utmost good faith
3. The insured must have insurable interest in the life assured
4. Life insurance is not a contract of indemnity

Types of life insurance policies:
1. Whole Life Policy:
In this kind of policy, the sum assured becomes payable only on the death of the policy holder.

2. Endowment Life Policy:
Here, the sum assured becomes payable either at the end of the stipulated period or on the premature death of the policyholder which ever is earlier.

3. Joint Life Policy:
Under this policy, the lives of two or more persons are insured jointly. The sum assured becomes payable on the death of any one, to the survivor. Usually, this policy is taken up by husband and wife jointly or by two partners of the firm.

4. Annuity Policy:
Under this policy, the assured sum or policy money is payable after the assured attains a certain age in monthly, quarterly, half yearly or annual installments.

5. Children’s Endowment Policy:
This policy is taken by a person for his/ her children to meet the expenses of their education or marriage. The agreement states that a certain sum will be paid by the insurer when the children attain a particular age.

Business Studies Class 11 Business Services Notes Fire Insurance:
Fire insurance is a contract whereby the insurer, in consideration of premium, undertakes to compensate the insured for the loss or damage suffered due to fire. The premium is payable in single installment.
A claim for loss by fire must satisfy the two following conditions:

• There must be actual loss;
• Fire must be accidental and non intentional.

Elements of Fire insurance contract:

1. The insured must have insurable interest in the subject matter.
2. Fire insurance is a contract of utmost good faith.
3. It is a contract of indemnity.
4. Fire must be the proximate cause of damage or loss.

Meaning Of Business Services Class 11 Marine Insurance:
Marine insurance is an agreement by which the insurance company agrees to indemnify the owner of a ship or cargo against risks which are incidental to marine adventures. eg: collision with other ship, collision of ship with the rocks, fire, storm etc. Marine insurance insures ship hull, cargo and freight.
Elements of Marine insurance contract:

1. Marine insurance is a contract of indemnity
2. It is a contract of utmost good faith
3. Insurable interest must exist at the time of loss
4. The principle of causa proxima will apply to it.

There are three types of Marine insurance. They are:
Ship or hull insurance:
when the owner of a ship is insured against loss on account of perils of the sea, it is known as Ship or Hull insurance.

Cargo insurance:
Marine insurance that covers the risk of loss of cargo is known as Cargo Insurance.

Freight insurance:
The shipping company may seek insurance of the risk of loss of freight. Such a marine insurance is known as freight insurance.

Differences between Life, Fire and Marine insurance:

Business Studies Class 11 Chapter 4 Notes Communication services:
Communication services are helpful to business for establishing links with the outside world viz., suppliers, customers, competitors etc. The main services which help business can be classified into postal and telecom.
1. Postal services:
Indian post and telegraph department provides various postal services across India. Various facilities provided by postal department are:
(a) Financial facilities:
They provide postal banking facilities to the general public and mobilise their savings through the saving schemes like public provident fund (PPF), Kisan Vikas Patra, National Saving Certificate, Recurring Deposit Scheme and Money Order facility.
(i) Mail facilities: Mail services consist of

• Parcel facilities that is transmission of articles from one place to another
• Registration facility to provide security of the transmitted articles
• Insurance facility to provide insurance cover for all risks in the course of transmission by post.

(b) Allied Postal Services

1. Greetings Post: Greetings card can be sent through post offices.
2. Media Post: Corporate can advertise their brands through post cards, envelops etc.
3. Speed Post: It allows speedy transmission of articles to people in specified cities.
4. e-bill post: The post offices collect payment of telephone, electricity, and water bills from the consumers.
5. Courier Services: Letters, documents, parcels etc. can be sent through the courier service.

2. Telecom Services:
The various types of telecom services are
1. Cellular mobile services:
Mobile communication device including voice and non-voice messages, data services and PCO services utilising any type of network equipment within their service area.

2. Radio paging services:
It means of transmitting information to persons even when they are mobile.

3. Fixed line services:
It includes voice and non-voice messages and data services to establish linkage for long distance traffic.

4. Cable services:
Linkages and switched services within a licensed area of operation to operate media services which are essentially one way entertainment related services.

5. VSAT services (Very small Aperture Terminal):
It is a Satellite based communication service. It offers government and business agencies a highly flexible and reliable communication solution in both urban and rural areas.

6. DTH services (Direct to Home):
It is a Satellite based media services provided by cellular companies with the help of small dish antenna and a set up box.

Chapter 4 Business Studies Class 11 Notes Transportation:
Transport refers to the activity that facilitates physical movement of goods and individuals from one place to another transportation removes the hindrance of place, i.e., it makes goods available to the consumer from the place of production.
Functions of Transport:

1. Helps in the movement of goods and materials from one place to another
2. Helps in the stabilisation of prices.
3. Helps in the social, economic and cultural development of the country
4. Helps in national and international trade
5. Facilitates large scale production
6. Generates employment opportunities
7. Increases growth of towns and cities
8. Connects all part of the world

Types of Transport:

1. Land Transport
   • Road Transport
   • Rail Transport
   • Pipeline Transport
2. Water Transport
   • Inland water Transport
   • Ocean Transport
3. Air Transport.

Business Studies Class 11 Chapter 4 Question Answers Warehousing:
Warehousing means holding or preserving goods in huge quantities from the time of their purchase or production till their consumption. Warehousing is one of the important auxiliaries to trade. It creates time utility by bridging the time gap between production and consumption of goods.
Functions of Warehousing

1. Warehouse helps in supplying the goods to the customers when it is needed.
2. By maintaining a balance of supply of goods warehousing leads to price stabilization.
3. By keeping the goods in the warehouse, the trader can relieve himself of the responsibility of keeping of goods.
4. The warehouse performs the function of dividing the bulkquantity of goods into smaller quantities.
5. Warehousing helps in the seasonal storage of goods to select businesses.
6. The functions of grading, branding and packing of goods can be done in warehouses.
7. The warehousing receipt can be used as a collateral security for obtaining loans.

Business Services Class 11 Notes Pdf Types of Warehouses:

(a) Private warehouses: Private warehouses are owned by big business concerns or wholesalers for keeping their own products.

(b) Public warehouses:
They are owned by some agencies, offer storage facilities to the public after charging certain fees. The working of public warehouses is subject to some govt, regulations. They are also known as Duty paid warehouses.

(c) Bonded warehouses:
These warehouses are used to keep the imported goods before the payment of import duties. It offers many advantages to the importer, i.e. The importer can releases the goods in part by paying the proportionate amount of duty. The goods can be branded, blended and packed in the warehouse itself.

(d) Government warehouses:
These warehouses are fully owned and managed by the government. For example, Food Corporation of India, State Trading Corporation, and Central Warehousing Corporation.

(e) Co-operative warehouses:
Marketing co-operative societies and agricultural oo operative societies have set up their own warehouses for members of their cooperative society.

Students can Download Chapter 7 Systems of Particles and Rotational Motion Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 7 Systems of Particles and Rotational Motion

Plus One Physics Systems of Particles and Rotational Motion One Mark Questions and Answers

Plus One Physics Chapter 7 Systems of Particles and Rotational Motion Question 1.
The dimension of angular momentum is
(a) M°L¹T^-1
(b) M¹L²T²
(c) M¹L²T^-1
(d) M²L¹T^-2
Answer:
(c) M¹L²T^-1
Angular momentum = Moment of inertia x angular velocity
(Angular momentum) = [M¹L²][T^-1] = [M¹L²T^-1].

Plus One Physics System Of Particles And Rotational Motion Question 2.
If the density of material of a square plate and a circular plate shown in figure is same, the centre of mass of the composite system will be

(a) inside the square plate
(b) inside the circular plate
(c) at the point of contact
(d) outside the system
Answer:
(a) inside the square plate

Plus One Physics Important Questions And Answers Pdf Question 3.
Why spokes are provided in by cycle wheel?
Answer:
This increases moment of inertia even when the mass is small. This ensures uniform speed.

Plus One Physics Chapter Wise Questions And Answers Question 4.
A ballet dancer, an acrobat and an ice skater make use of an important principle in physics. Which is that principle?
Answer:
Conservation of angular momentum.

Physics Class 11 Chapter 7 Important Questions Question 5.
A cat is able to land on her feet after a fall. Which principle of physics is being used by her?
Answer:
Principle of conservation of angular momentum.

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 6.
A body is rotating in steady rate. What is torque acting on the body?
Answer:
Zero. Torque is required only for producing angular acceleration.

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Question 7.
A flywheel is revolving with constant angular velocity. A chip of its rim breaks and flies away. What will be the effect on the angular velocity?
Answer:
The reduction in mass will decrease moments of inertia. Hence angular velocity will be increased in order to conserve angular momentum.

Rotational Motion Class 11 Important Questions Question 8.
Is radius of gyration of a body constant quantity?
Answer:
No. It changes with change in position of the axis of rotation.

System Of Particles And Rotational Motion Class 11 Solutions Question 9.
What is another name for angular momentum?
Answer:
Moment of momentum.

Plus One Physics Systems of Particles and Rotational Motion Two Mark Questions and Answers

Hss Live Plus One Physics Chapter Wise Questions And Answers Question 1.
Moments of inertia of some bodies with axis are given in the table below. Fill in the blanks

Answer:

System Of Particles And Rotational Motion Questions And Answers Question 2.
Match the following:

Answer:
(a) Torque
(b) τ ∆ θ
(c) Rotational motion
(d) \(\frac{1}{2} \frac{L^{2}}{1}\).

System Of Particles And Rotational Motion Important Questions Question 3.
A cat is able to function its feet after a fall, taking the advantage of principle of conservation of angular momentum.

1. Give the law of conservation of a angular momentum.
2. Explain how cat is able to do so

Answer:
1. When there is no external torque, the total angular momentum of a body or a system of bodies are a constant.
τ = \(\frac{d L}{d t}\) (when τ = 0 , we get \(\frac{d L}{d t}\) = 0).
ie L = constant.
But L = Iω
∴ Iω = a constant.

2. When a cat falls, it stretches its body. So that the moment of inertia becomes large. As Iω = constant, the value of angular speed will be decreased due to the increased value of moment of inertia. So cat lands on its feet gently.

System Of Particles And Rotational Motion Class 11 Important Questions Question 4.
If the polar ice cap melts what will happen to the length of the day?
Answer:
For earth, angular momentum is a constant (Lω = constant, ie no torque acts on the earth). When the polar ice cap melts, the water thus formed will flow down to the equatorial region.

The accumulation of water in equatorial line will increase the moment of inertia I of earth. In order to keep the angular momentum as a constant, ω will decrease. The decrease in ‘ω’will increase the length of day.

System Of Particles And Rotational Motion Class 11 Question 5.
A girl has to lean towards right when carrying a bag in her left hand. Why?
Answer:
When a girl carries her bag in her left hand, the centre of gravity of system will shift towards left. In order to bring it in the middle, the girl has to lean towards right.

Question 6.
If the earth loses the atmosphere what will happen to the length of the day?
Answer:
For earth, the angular momentum (L = Iω) is a constant, because there is no torque acting on it. When earth loses the atmosphere, I decreases and ω increases to keep L as constant. Hence length of the day decreases.

Question 7.
A girl standing on a turn table. What happens to the rotation speed, if she stretches her hand?
Answer:
lf a girl rotating with a uniform speed on turn table, it’s angular momentum (L = Iω) will be a constant. When she suddenly stretches her hand, I Increases and ω decreases to keep L as constant.

Question 8.
How does a circus acrobat and a diver take advantage of conservation of angular momentum? Answer:
The diver while leaving the spring board, is throwing himself in a rotating motion. When he brings his hands and legs close, I decrease and ω increases. But before reaching water he will stretch his hands and legs. Hence I increases and ω decreases. So, that he gets a smooth entry into the water.

Plus One Physics Systems of Particles and Rotational Motion Three Mark Questions and Answers

Question 1.
A rigid body consists of ‘n’ particles of mass m1, m2, m3,……The body rotates about an axis with
an angular velocity ω1, ω2, ω3……..

1. Starting from the kinetic energy of a single particle, arrive at an equation for kinetic energy of rotation.
2. Moment of inertia is also called rotational inertia. Why?

Answer:
1. Consider a body rotating about an axis passing through some point O with uniform angular velocity ‘ω’. The body can be considered to be made up of a number of particles of masses m1, m2, m3……etc at distances r1, r2, r3……etc. All the particles will have same angular velocity ω But their linear velocities will be different say v1, v2, v3…….etc.

K.E of 1^st particle = \(\frac{1}{2}\)m₁v₁²
\(\frac{1}{2}\)m₁(r₁ω)²
(∵ v = rω)
K.E of II^nd particle = \(\frac{1}{2}\)m₂(r₂ω)²
∴ K.E of whole body =

But we know moment of inertia,

∴ KE = \(\frac{1}{2}\)Iω²
2. Rotation inertia is measured in terms of moment of inertia. Hence moment of inertia is also called rotational inertia.

Question 2.
The handle of a door is always found at one edge of the door which is located at a maximum possible distance away from hinges.

1. Give reason for it.
2. In which direction will the torque act while the door opens inside the room?
3. If the door handle is fixed at the middle of the door, what difference do you feel in the applied force to open the door.

Answer:
1. Torque τ = r F sin θ
From the above equation it is clear that, we get maximum torque when the handle of a door is located at a maximum possible distance (r) away from hinge.

2. The direction of torque is always along the axis of rotation of door.

3. If the door handle is fixed at middle, more force must be applied to get maximum torque that is required to open the door.

Question 3.
Moment of inertia depends on the mass, axis of rotation and distribution of mass of the body.

1. What are moment of inertia and radius of gyration?
2. How will you distinguish a hard boiled egg from a raw egg by spinning each on the table.

Answer:
1. Moment of inertia I = mr²
Radius of gyration K = \(\sqrt{\frac{I}{m}}\).

2. A raw egg has more monemt of inertia than boiled egg. Hence raw egg spins more time than boiled egg.

Question 4.
Table below given analogy between translational and rotational motions. Match the following.

Answer:

Plus One Physics Systems of Particles and Rotational Motion Four Mark Questions and Answers

Question 1.
1. Show that the total angular momentum of a rotating system remains constant if no torque acts on the system

2. A disc of moment of inertia I₁ is rotating freely with angular speed ω1 and a second non rotating disc with moment of inertia I₂ is dropped on it as shown in the figure. The two then rotate as one unit. Find the angular speed of rotation of the system.
Answer:
1. we know torque τ = \( \frac{d L}{d t}\)
if τ = 0, we get \( \frac{d L}{d t}\) = 0
ie. L = constant.

2. we know if torque acting on the body is zero, its angular momentum will be conserved
ie. I₁ω₁ = I₂ω₂
angular momentum of system, ω₂ = \(\frac{I_{1} \omega_{1}}{\left(I_{1}+\mathrm{I}_{2}\right)}\).

Question 2.
A rigid body can rotate an axis with a constant angular velocity and angular momentum L.

1. What is its moment of inertia about the axis?
2. Obtain a mathematical expression for rotational kinetic energy.
3. If the orientation of the axis of rotation changes, what happens to its moment of inertia

Answer:
1. L = Iω
ie. I = L/ω

2. Consider a rigid body rotating about an axis passing through the point O. Let co be the uniform angular velocity of the body.

The body is imagined to be made up of large number of particles. Consider one such particle of mass ‘m’ at a distance ‘r’ from the axis of rotation.
Linear Velocity of the particle v = rω
K.E of the particle = 1/2mv² = 1/2 mr²ω²
K.E of whole body = Σ1/2mr²ω² = 1/2ω²Σmr²
K.E = 1/2Iω²
Where Σmr² = I, moment of inertia of the body.

3. Moment of inertia will be changed.

Question 3.
A platform diver holds his hands and legs straight and makes loops in air before entering into water.

1. State the principle behind this.
2. What happens when he tries to land in the pool by stretching his arms and legs?
3. In the above situation, rotational kinetic energy is not conserved. Explain.

Answer:
1. Conservation of angular momentum
Statement
Conservation of angular momentum states that, if the total torque acting on a system is zero, its angular momentum will be conserved.

2. Angular velocity decreases.

3. Initial kinetic energy K.E₁ = \(\frac{L^{2}}{2 I_{1}}\) ____(1)
where I₁ is the moment of inertia of diver when he makes loops in air
final kinetic energy K.E₂ = \(\frac{L^{2}}{2 I_{2}}\) _____(2)
where I₂ is the moment of inertia of diver when he stretches his hands
But I₁ < I₂
Hence from eq(1) and eq(2), we get
KE₁ > KE₂
which means that rotational kinetic is not conserved.

Question 4.
Moment of inertia of a thin ring of radius R about an axis passing through any diameter is 1/2MR²

1. What is the radius of gyration of the ring about an axis passing through any diameter.
2. A thin metal ring of radius 0.25m and mass 2kg starts from rest and roll down an inclined plane. If the linear velocity on reaching the foot of the plane is 2m/s, calculate its rotational kinetic energy at that instant.

Answer:
1.

2.

Question 5.

Figures show the two different spinning poses of a ballet dancer.

1. In which spinning pose, the ballet dancer has more angular speed?
2. State the principle used by the ballet dancer to increase his angular speed.
3. “In the above situation, rotational kinetic energy is not conserved” – Justify this statement.

Answer:
1. The pose shown in figure (B).

2. Statement of conservation of angular momentum.

3. I₁ω₁ = I₂ω₂
1/2 I₁²ω₁² = 1/2 I₂²ω₂²
I₁(1/2 I₁ω₁²) = (1/2 I₂ω₂²)I₂, I₁ > I₂
(1/2 I₂ω₂²) > 1/2I₁ω₁²

Plus One Physics Systems of Particles and Rotational Motion Five Mark Questions and Answers

Question 1.
The moment of inertia of a thin ring of radius R about an axis passing through any diameter is \(\frac{1}{2}\)MR²

1. To find the moment of inertia of the same ring about an axis passing through its centre of mass and perpendicular to its plane, which of the following theorem is used and state the theorem.
   • Perpendicular axis theorem.
   • Parallel axis theorem
2. What is the radius of gyration of the ring about an axis passing through its centre of mass and perpendicular to its plane?
3. A thin metal ring has a diameter 0.20 cm and mass 1 kg. Calculate its moment of inertia about an axis passing through any tangent.

Answer:
1. perpendicular axis theorem.

2. Moment of inertia of ring,
I = mr² ____(1)
Moment of inertia of ring in terms of radius of gyration,
I = mk² ____(2)
From eq(1) and eq(2), we get
mk² = mr²
radius of gyration, k = r.

3. I = I₀ + ma²

Question 2.
The earth is moving around the sun in an elliptical orbit and this brings out the seasons.

1. Name the physical quantity conserved in this.
2. As the earth approaches near the sun in its path, it moves faster. State whether this statement is correct or wrong. Why?
3. Determine the duration of the day on earth if it suddenly shrinks to 1/4^th of its present size while the mass of the earth remains unchanged. (Assume earth is a perfect sphere)

Answer:
1. Conservation of angular momentum.

2. This statement is correct.
When earth approaches near the sun, its moment of inertia decreases. To maintain angular momentum as constant, the angular velocity increases.

3. New volume of earth = 1/4 original volume of earth

Total angular momentum before shrinking = Total momentum after shrinking.
ie. I₁ω₁ = I₁ω₂
mr₁²ω₁ = m₂²ω₂ ______(2)
Sub eq(1) in eq(2)

we know period

T₂ = 3 hr
Duration of day T₂ = 3 hr.

Question 3.
1. Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR²/5, where M is the mass of the sphere and R is the radius of the sphere.
2. Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be MR²/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Answer:
1. Applying theorem of parallel axes, moment of inertia of sphere about a tangent

2. Using theorem of perpendicular axis, the moment of inertia of the disc about an axis passing through the centre of the disc

Using theorem of parallel axes, moment of inertia of the disc passing through a point on its edge and the normal to the disc

Question 4.
A solid cylinder of mass 20kg rotates about its axis with angular speed 100rad s^-1. The radius of the cylinder is 0.25m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular mementum of the cylinder about its axis?
Answer:
M = 20kg
angularspeed, co – 100 rad s^-1; R = 0.25m
Moment of inertia of the cylinder about its axis
= \(\frac{1}{2}\)MR² = \(\frac{1}{2}\) × 20 × (0.25)²kgm² = 0.625kgm²
Rotational kinetic energy,
Er = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\) × 0.625 × (100)²J = 3125 J
Angular momentum,
L = Iω = 0.625 × 100 Js = 62.5 Js.

Question 5.
A rope of negligible mass is wound round a hollow cylinder of mass 3kg and radius 40cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope? Assume that there is no slipping.
Answer:
M = 3kg, R = 40cm = 0.4m
F = 30N, α = ?, a = ?
The torque acting on the cylinder,
τ = force × radius of the cylinder = 30N × 0.4m = 12Nm
The moment of inertia of the hollow cylinder about its axis,
I = MR² = 3kg × (0.4)²m² = 0.48kgm²
Now, τ = Iα, where α is angular acceleration
∴ The angular acceleration of the cylinder, r 12
α = \(\frac{\tau}{I}=\frac{12}{0.48}\)rad s^-2 = 25 rad s^-2
The linear acceleration of the rope,
a = Rα = 0.4 × 25m s^-2 = 10 m s^-2.

Question 6.
Explain why friction is necessary to make the disc in roll in the direction indicated.

1. Give the direction of the frictional force at B, and the sense of frictional torque, before perfect rolling begins.
2. What is the force of friction after perfect rolling begins?

Answer:
Friction is necessary for rolling.

1. Frictional force at B opposes velocity of B. So, the frictional force is in the same direction as the arrow. The sense of frictional torque is such as to oppose angular motion, \(\overrightarrow{ω_{0}}\) and \(\overrightarrow{τ_{0}}\) are both normal to the paper, the first into the paper and the second coming out of the paper.
2. Frictional force decreases the velocity of the point of contact B. Perfect rolling ensures when this velocity is zero. Once this is so, the force of friction is zero.

Students can Download Chapter 8 Financial Statements – I & Financial Statements – II Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 8 Financial Statements – I & Financial Statements – II

Plus One Accountancy Financial Statements – I & Financial Statements – II One Mark Questions and Answers

Plus One Accountancy Financial Statement Questions And Answers Question 1.
Closing stock is valued at
(a) Market Price
(b) Cost Price
(c) Market Price or Cost Price whichever is less.
Answer:
(c) Market Price or Cost Price whichever is less.

Financial Statements Without Adjustments Class 11 Solutions Question 2.
Carriage inward is debited to …………… account.
(a) Trading account
(b) Profit and Loss Account
(c) Cash Account
Answer:
(a) Trading Account

Plus One Accountancy Textbook Questions And Answers Question 3.
Balance Sheet is prepared to find out ……………
(a) Capital
(b) Net Profit
(c) Financial position
Answer:
(c) Financial Position

Plus One Accountancy Chapter Wise Questions And Answers Pdf Question 4.
Gross profit is the difference between ……………..
(a) Sales and Purchases
(b) Sales and total cost
(c) Sales and cost of goods sold
Answer:
(c) Sales and cost of goods sold

Trading And Profit And Loss Account Questions Question 5.
Interest on capital is ……………… to the business.
(a) Income
(b) Expenses
(c) Asset
Answer:
(b) Expenses

Plus One Accountancy Chapter Wise Questions And Answers Question 6.
Wages paid before it has become due is shown in the balance sheet as ……………..
(a) Asset
(b) Liability
(c) Expenses
Answer:
(a) Asset

Financial Statement With Adjustment Class 11 Question 7.
Ameer’s trial balance contains the following information.

• Bad debt Rs.400
• Provision for bad debts Rs. 1600

It is desired to maintain a provision for bad debts at Rs. 1500. The amount to be debited to profit and loss a/c is ………….
(a) Rs. 3500
(b) Rs. 100
(c) Rs. 300
Answer:
(c) Rs. 300

Trade Expenses In Final Accounts Question 8.
Profit from Profit and Loss account is transferred to …………… account.
(a) Asset account
(b) Capital account
(c) Liability account
Answer:
(b) Capital Account

Trade Expenses In Profit And Loss Account Question 9.
The Financial statements consist of:
(a) Trial Balance
(b) Profit and Loss account
(c) Balance sheet
(d) Both b and c
Answer:
(d) Both b and c

Find The Odd One Out Rent Salary Insurance Plant And Machinery Question 10.
Choose the correct order of ascertainment of the fol¬lowing profits from the profit and loss account:
(a) Operating profit. Net Profit, Gross profit
(b) Operating profit, Gross Profit, Net Profit
(c) Gross Profit, Operating Profit, Net Profit
(d) Gross Profit, Net Profit, Operating Profit
Answer:
(c) Gross profit, Operating profit, Net Profit

Trading Account Questions And Answers Question 11.
Which of the following are not taken into account at the time of calculation of operating profit?
(a) Normal transactions
(b) Abnormal items
(c) Expenses of a purely financial nature
(d) Both a and c
Answer:
(c) Expenses of a purely financial nature

Trading Profit And Loss Account Questions And Answers Question 12.
If the insurance premium paid Rs. 1,000 and prepaid insurance of Rs. 400. The amount of insurance premium shown in profit and loss a/c will be:
(a) Rs. 1400
(b) Rs. 1000
(c) Rs. 400
(d) Rs. 600
Answer:
(d) Rs. 600

Financial Statement With Adjustment Class 11 Questions Question 13.
When the Manager is entitled to a commission of 10% on profits after charging such commission, it is calculated on profits before charging commission by the formula ………….
(a) 10/90
(b) 10/100
(c) 10/110
(d) 90/100
Answer:
(c) 10/110

Question 14.
Profit and Loss Account is an account prepared to find out ………………
Answer:
Net Profit or Net Loss.

Question 15
Income tax paid for the trader from the business is treated as …………..
Answer:
Drawings.

Question 16.
Adjustments are made in the final accounts to satisfy ………. principle of accounting
Answer:
Matching Principle.

Question 17.
Free samples distributed among customer’s will be credited to ……….. account.
Answer:
Advertisement Account.

Question 18.
…………….. is an expenditure of the revenue nature in the current year, the benefit of which accrues gradually or lasts for more than one accounting year.
Answer:
Deferred Revenue Expenditure

Question 19.
EBIT stands for …………………
Answer:
Earnings Before Interest and Tax

Plus One Accountancy Financial Statements – I & Financial Statements – II Two Mark Questions and Answers

Question 1.
Find the odd one and state the reason.

1. Wages, carriage, Trade expenses, Freight
2. Income tax paid, Salary paid, Rent paid, Wages paid

Answer:

1. Trade expense – It is an indirect expense, all others are direct expenses.
2. Income tax paid – It is a personal expense of the Proprietor, all others are business expenses.

Question 2.
The sales and cost of goods sold by John Brothers are Rs. 40,000 and Rs. 28,000 respectively. What is the amount of his gross profit?
Answer:
Gross profit = Sales – Cost of goods sold
= 40,000 – 28,000 = Rs. 12,000

Question 3.
What are the financial statements?
Answer:
The term Financial Statements generally refers to two statements prepared at the end of an accounting period for an enterprise. These are Trading and Profit and Loss Account, showing profitability of the business operations and Balance sheet, showing the financial position of the enterprises.

Question 4.
A new manufacturer incurred huge expenditure in advertisement. Explain the nature of expenditure.
Answer:
Deferred Revenue Expenditure:
“Expenses incurred today, and the benefit of which accrue gradually in subsequent year is called Deferred Revenue Expenditure. For example Expenses like advertisement, which may be incurred in one lump sum but the benefit of which will be received only within a series of years. The proportionate amount due the current year must be arrived at and debited to the P/L account. The balance must be shown on the assets side of the Balance sheet.

Question 5.
Calculate the value of cost of goods sold.

• Opening stock – 10,000
• Net purchases – 5,000
• Direct expenses – 2,000
• Closing stock – 10,500

Answer:
Cost of goods sold = Opening stock + Net Purchases + Direct Expenses – Closing stock
= (10,000 + 5000 + 2000) – 10500 = Rs. 6,500

Question 6.
What is Operating profit?
Answer:
Operating profit is the profit earned through the normal operations and activities of the business. Operating profit can be calculated as follows:
Operating profit = (Gross profit – Operating expense) + Operating income.

Or

= Net Profit + Non-operating expenses – Non-operating income.
Examples of non-operating expenses are loss on sale of assets, interest paid etc. Non-operating incomes are dividend received, Profit on sale of assets.

Question 7.
Operating profit earned by M/s. Asoka & Sons in 2010-11 was Rs. 17,00,000. Its non-operating incomes were Rs. 1,50,000 and non-operating expenses were Rs.3,75,000. Calculate the amount of net profit earned by the firm.
Answer:
Net profit = Operating profit – Non-operating expenses + Non operating income = 1700000 – 375000 + 150000 Net Profit =Rs. 1475000

Question 8 .
Explain the treatment of goods distributed as sample to customers.
Answer:
In order to increase sales, goods may be distributed among customers free of cost. Such free goods are in the form of samples for test use. It is in the form of advertisement expense. The amount is to be deducted from the purchases in the trading account. The cost of the sample is to be shown as an expenses in the profit and loss account.

Question 9.
To tally a balance sheet, one has to strictly adhere to the basic accounting equation ‘Asset = Liabilities + Captial’. Do you agree with this statement? Substantiate.
Answer:
Every transaction has two aspects which are equal in value but opposite in nature. One of the aspects will form part of liability and the other one asset. Be¬cause of this feature of accounting equation, a balance sheet is always tallied.

Question 10.

• Closing stock – Rs. 30,000
• Sales – Rs. 1,00,000

Gross Profit 20% on sale

• Purchases – Rs. 60,000
• Direct Wages – Rs. 7000

Find out opening stock.
Answer:
Opening stock = Sales + Closing stock – (Purchases + Direct Expense + Gross Profit)
= 1,00,000 + 30,000 – (60,000 + 7,000 + 20,000)
= 1,30,000 – 87,000 = 43,000

Question 11.
Provide provision for bad debt @ 10% on debtors from the particulars given below.

• Good Debts Rs. 10,000
• Doubtful debts Rs. 50,000
• Bad debts Rs. 1000

Answer:
Provision for doubtful debts = Doubtful debts × % of provision for bad debt
= 50000 × 10/100 = Rs. 5000

Question 12.
Payment received from debtors does not result a change in the total assets. Comment.
Answer:
When the payment is received from debtors the asset, cash is increased and also a corresponding decrease happens on the asset, Debtors. Therefore, there won’t be any change in the value of total assets in effect.

Question 13.
What is meant by provision for discount on debtors?
Answer:
This is a discount which is being allowed by an enterprise to its debtors to encourage prompt payments. Discount likely to be allowed to customers in an accounting year can be estimated and provided for by creating a provision for discount on debtors. Here, it is to be remembered that provision for discount made on good debtors which are arrived at by deducting further bad debts and the provision for doubtful debts.

Plus One Accountancy Financial Statements – I & Financial Statements – II Three Mark Questions and Answers

Question 1.
Distinguish between capital and revenue expenditure.
Answer:
1. Capital expenditure increases earning capacity of business whereas revenue expenditure is incurred to maintain the earning capacity.

2. Capital expenditure is incurred to acquire fixed assets for operation of business whereas revenue expenditure is incurred on the day-to-day conduct of business.

3. Revenue Expenditure is recurring expenditure but capital expenditure is non-recurring by nature.

Question 2.
Differentiate between Direct and Indirect Expenses.
Answer:
1. Direct Expenses:
Means all expenses directly connected with the manufacture, purchase of goods and bringing them to the point of sale. Direct expenses include carriage inward, freight inwards, octroi, clearing charges, wages, factory lighting, coal, water, gas, fuel, import duty, cotton waste, royalty on production, heating, dock dues, customs duty etc.

2. Indirect expenses:
Are those expenses which are incurred after the manufacturing of goods. In other words, Indirect expenses are those expenses that are incurred to operate a business as a whole. Indirect expense include – carriage outward, rent, rates and tax, office expenses, selling and distribution expenses.

Question 3.
Mention the difference between a Balance Sheet and a Trial Balance. (Any two points).
Answer:

Question 4.
Calculate closing stock from the following

• Sales – 20,000
• Purchases – 12,300
• Return inwards – 500
• Carriage inwards – 400
• Return outwards – 1,000
• Gross Profit – 8000

Answer:
Trading Account

Question 5.
On 1^st January 2008, a firm had a stock of goods valued at Rs. 10,000. During the year the following transactions took place.

• Sales – 2,00,000
• Purchases – 1,20,000
• Carriage inwards – 500
• Sales returns – 2,000
• Purchase returns – 1,000
• Find out the amount of Gross Profit.

Answer:

1. Gross Profit = Net Sales – Cost of goods sold
2. Net sales = 2,00,000 – 2,000 = 1,98,000
3. Net Purchase = 1,20,000 -1000 = 1,19,000.
4. Cost of goods sold = 10,000 + 1,19,000 + 500 = 1,29,500
5. Gross Profit = 1,98,000 – 1,29,500 = Rs. 68,500

Question 6.
What do you mean by Profit and Loss Account.
Answer:
Profit and Loss a/c is an account prepared to find out the net profit earned or net loss incurred by a business during an accounting period. It is debited with all operating expenses and losses and credited with incomes and profit. This account begins with the Gross Profit or Gross Loss brought down from Trading A/c.

If the total of the credit side of this account is more than the total of the debit side, the difference is net profit. If the total of the debit side exceeds the total of the credit side, the difference is net loss. The amount of net profit or net loss transferred to capital account.

Question 7.
From the following information calculate operating profit.

• Cost of goods sold – Rs. 5,00,000
• Administrative expenses – 25,000
• Selling & Distribution expenses – 35,000
• Net sales – 7,50,000

Answer:

• Operating Profit = Gross Profit – (Operating expenses + Operating income)
• Gross Profit = Net sales – Cos of goods sold = 7,50,000-5,00,000 = 2,50,000
• Operating Expenses = Administration expense + Selling and Distribution exp.
• Operating Profit = 2,50,000 – (25,000+ 35,000) = Rs. 1,90,000

Question 8.
What do you mean by Deferred Revenue Expenditure? Can you present it with a suitable example?
Answer:
“Deferred Revenue Expenditure is an expenditure of the revenue nature in the current year, the benefit of which accrues gradually or lasts for more than one accounting year.”

Examples are advertisements of usually high amounts, cost of shifting business to a more convenient location, etc. The amounts spend on such expenses would be spread over the period for which the benefit arises. Suppose if a concern spends Rs. 5 lakh for advertisement and it is expected that the benefit of its lasts for 5 years, the amount to be treated as current year’s expenditure is only one fifth (1/5) of Rs. 5 lakhs ie. Rs. 1 lakh. The balance Rs. 4 lakhs would be treated as an asset.

Question 9.
State the reasons why the following items appear or not in the Profit and Loss A/c.

1. Bad debts
2. Drawings
3. Provision for bad debts.

Answer:

1. Bad debts is an indirect expenditure and hence it appears in the Profit and Loss A/c.
2. Drawings will reduce the capital a/c. Therefore, it cannot be shown in the P/L a/c.
3. Provision for bad debt is a charge against profit. Hence it will appear in the P/LA/c.

Question 10.
Mr. Narayanan, an accountant of Samay Ltd wrote off Rs. 2000 as bad debts in the year 2004. The total
Sundry debtors for the year 2005 is Rs. 26,000. During this year half of the bad debts written off in the last year were recovered. Give Journal Entry for the recovery of bad debts and also show how it will be dealt in the financial statement for the year 2005.
Answer:

Question 11.
What are adjusting entries? Why are they necessary for preparing final accounts?
Answer:
Entries which are given outside the trial balance are called adjustment entries, to record those entries a proper treatment is required according to the double-entry system. Here it is to be remembered that all the adjustments given outside the Trial Balance are posted at two places.

Adjustment is generally done for those items which are omitted or entered with the wrong amount and/or recorded under wrong heads. The following are reasons for recording or. incorporating these adjustment entries in preparation of final account.

1. Through these adjustment entries, we come to know the actual figure of profit or loss.
2. Because of these adjusting entries, we can assess the true financial position of an organisation based on accrual basis of accounting.
3. These adjustment entries enable us to records the omitted entries and help in rectifying all those errors.
4. These adjusting entries help in providing depreciation and making different provisions, such as Bad Debts and Depreciation.

Question 12.
What is meant by provision for doubtful debts? How are ‘the relevant accounts prepared and what journal entries are recorded in final accounts? How is the amount for provision for doubtful debts calculated?
Answer:
Provision for doubtful debts is a kind of arrangement about the expect bad debts from the debtors. Generally it is provided after deducting the amount of bad debts from the debtors. As provision for doubtful debts is made after preparing the trial balance, to record it we need a kind of adjustment entry in this regard we prepare debtors account and provision for doubtful debts account. For recording bad debts, the following journal entry is passed.

• Profit and Loss A/c Dr
  To Provision for Doubtful Debts A/c (Being provision for doubtful debts is created out of current year profits)

Computation of the Amount of Provision for Doubtful Debts As it is given at the end of a trial balance as an adjustment, little another related adjustment may be there for instance bad debts and discount on debtors. In this case provision for doubtful debt will be created after deducting the figure for bad debts out of the debtor figure.

Plus One Accountancy Financial Statements – I & Financial Statements – II Four Mark Questions and Answers

Question 1.
What is Trading Account? Explain its purpose.
Answer:
The trading account is an account which shows the results of buying and selling of goods or services. It contains summarized form of all the transactions occurring during a trading period. This account is credited with direct incomes and debited with direct expenses.

Trading account is prepared to ascertain the gross result of the business. The gross result of the business is either gross profit or gross loss. If the net sales exceeds cost of goods sold then there is gross profit and if the opposite takes place, there is a gross loss.

Gross profit = Net sales – Cost of goods sold Gross Loss = Cost of goods sold – Net Sales Purposes of a trading account are:

1. To ascertain the gross profit or gross loss.
2. To enable the management to make a comparison of gross profit or gross loss with that of the previous year.
3. To ascertain different ratios such as gross profit ratio, ratio of cost of goods sold to sales etc.

Question2.
From the following details, prepare Trading Account.

• Opening stock – Rs. 12500
• Purchases – Rs. 22,000
• Purchase returns – Rs.2000
• Wages – Rs. 2000
• Carriage inwards – Rs. 500
• Sales – Rs. 57,000
• Sales returns – Rs. 3,000
• Closing stock – Rs. 15,000

Answer:
Trading Account for the year ended ………..

Question 3.
What is a Balance Sheet? Explain the needs for preparing Balance Sheet.
Answer:
A Balance Sheet is a statement prepared to ascertain the true position of assets and liabilities as on a particular date. It is prepared at the end of the accounting period, after the preparation of Trading and Profit and Loss account. It is called Balance sheet, as it is a statement prepared with the balance of accounts left after the preparation of Trading and Profit and Loss account.

It gives clear picture of the financial position of the concern. Accounts of Assets, liabilities and Owner’s equity are shown in the Balance Sheet. Items of liabilities and capital are shown on the left side, known as “liabilities” side and the item of assets are shown on the right-hand side, known as “Assets” side of the balance sheet. Balance sheet is prepared with the following objectives.

1. To ascertain the financial position of the concern.
2. To ascertain the nature of assets and liabilities of the firm.
3. To know about the source and application of funds.
4. To ascertain working capital as on the date of Balance sheet.
5. To ascertain the excess of assets over external liabilities.

Question 4.
What do you mean by Grouping and Marshalling of assets and liabilities?
Answer:
1. Grouping:
The term grouping means putting together items of similar nature under a common heading in the Balance sheet.

2. Marshaling:
Marshaling denotes the order in which the assets and liabilities are shown in the Balance sheet. They are arranged in the following two different ways.

a. In the order of liquidity:
Liquidity means the capacity to raise cash, Under these approach assets are presented in the order of their liquidity. ‘Cash’ being the most liquid item, it is shown as the first item whereas the least liquid item such as ‘Goodwill’ is shown as the last one. The most urgent liability is shown first and the least urgent to pay is shown last.

b. In the order of permanence:
Linder this approach, permanent assets, and liabilities are shown first followed by current assets and liabilities. Joint Stock Companies have to prepare their Balance sheet in the order of permanence. This is just the reverse of the order of liquidity.

Question 5.
Show the treatment in financial statements in respect of the following:

1. Outstanding expenses
2. Managers commission
3. Interest on capital

Answer:
1. Outstanding expenses:
Expenses that have been incurred during the current year, but the payment has not been made is called outstanding expenses. It must be added to respective expense account in the Trading and Profit and Loss account. It will be shown on the ‘liability’ side of the Balance sheet.

2. Managers Commission:
Commission on net profit, at a specific percentage, may be allowed to the manager of a business concern. The commission as a percentage of the net profit may be ‘before’ or ‘after’ charging such commission. In the absence of any special instructions, it is assumed that commission is allowed as a percentage of the net profit before charging such commission.

• If the commission is on the net profit before charging such commission, the formula is
  Profit before commission × \(\frac{\text { rate of commission }}{100}\)
• If the commission is on the net profit after charging such commission the formula is
  Profit × rate/(100 + rate)
• The amount of commission must be debited to Profit and Loss account and it must be shown as a liability in the Balance sheet.

3. Interest on capital:
Sometimes interest is paid on the Proprietor’s capital. Interest is allowed at a certain rate on the capital at the beginning of the year. Such interest is an expense to the business and is debited to Profit and Loss Account. It is shown in the liability side by adding the same to capital.

Plus One Accountancy Financial Statements – I & Financial Statements – II Five Mark Questions and Answers

Question 1.
From the following figures, prepare profit and loss account of M/s. Thomas and Sons for the year ended 31.03.2008.

• Gross Profit – 15000
• Printing charges – 750
• Salaries – 5000
• Carriage outwards – 500
• Interest received – 2000
• Bad debts – 500
• Insurance charges – 400
• Discount allowed 650
• Discount received – 750
• Advertisement – 440

Answer:
Profit and Loss Account of M/s. Thomas & Sons for the year ended 31.03.08

Question 2.
Arrange the following assets and liabilities in the order of liquidity and in the order of permanence.

Answer:

Question 3.
Write the adjustment entries for the following:-
a) Salary outstanding Rs. 5000.
b) Insurance prepaid Rs. 5100.
c) Depreciation of Machinery Rs. 4000
d) The commission received in advance Rs. 1000
e) Interest on drawing Rs. 400
Answer:

Question 4.
What are the closing entries? Give examples.
Answer:
The preparation of trading and profit and loss account requires that the balances of accounts of all concerned items are transferred to it for its compilation. For transferring the balance of all the ledger account to concerned head is done through closing entries.
For examples:
1. Opening stock account, purchase account, wages account, carriage inward account, and direct expense account are closed by transferring to the debit side of the trading and profit and loss account.
The journal entry is:

• Trading A/c Dr
• To opening stock A/c
• To Purchase A/c
• To Wages A/c
• To Carriage inward A/c
• To Direct Expense A/c

2. The purchase return account is closed by transferring its balance to the purchase account. The journal entry is:

• Purchase Return A/c Dr
• To Purchase A/c

3. The sales return account is closed by transferring its balance to the sales account as:

• Sales A/c Dr
• To Sales Return A/c

4. The Sales account is closed by transferring its balance to the credit side of the trading and profit and loss account.
The Journal entry is:

• Sales A/c Dr
• To Trading A/c

Plus One Accountancy Financial Statements – I & Financial Statements – II Six Mark Questions and Answers

Question 1.
The following are the extracts from Trial Balance of a business.

• Sundry Debtors = 40,000
• Bad debts = 4,000
• Provision for bad debts = 5,000

Additional Information:

1. Provide further bad debts Rs. 2000
2. Create 10% provision for bad debts.

Pass Journal entries and show how these items will appear in the final accounts.
Answer:
Journal

Question 2.
Show the treatment of prepaid expenses,
depreciation, closing stock at the time of preparation of final accounts.

1. When given inside the trial balance.
2. When given outside the trial balance.

Answer:
Treatment of prepaid expenses, depreciation and closing stock at the time of preparing the final account.
1. When Given Inside the Trial Balance Prepaid Expenses:
When prepaid expenses are given in the trial balance itself it will be treated as current assets only and will be posted in the Assets Side of the balance sheet. No further adjustment will be required in this case.

Depreciation:
When depreciation is given in the trial balance it will be treated as an expenditure and will be shown in the debit side of the Profit and Loss Account. No further adjustment will be required in this case.

Closing Stock:
When closing stock is given in the trial balance it will purely be treated as assets and will be shown only in the Assets Side of the Balance sheet. No further adjustment will be required in this case.

2. When Given Outside the Trial Balance Prepaid Expenses:
When prepaid expenses are given outside the trial balance it will be treated as an Adjustment and will be posted at two places, first of all, it will be deducted from the concerned expenses in the debit side of Profit and Loss Account and after that, it will be treated as current assets and will be posted in the Assets side of the Balance Sheet.

Depreciation:
When depreciation is given outside the trial balance be treated as an adjustment and will be posted at two places to comply with the rules of the double-entry bookkeeping system. First of all the amount of depreciation will be shown in the debit side of Profit and Loss account as an expenditure and the amount of depreciation will be deducted from the concerned assets in the assets side of Balance Sheet.

Closing Stock:
When closing stock is given outside the trial balance it will purely be treated as an adjustment and will be posted at two places first of all the amount of closing stock will be shown at the credit side of Trading Account and after that it will be shown as an assets in the Assets side of the Balance Sheet.

Plus One Accountancy Financial Statements – I & Financial Statements – II Eight Mark Questions and Answers

Question 1.
The following is the extract takes from the Trial Balance of Vimal.
Trial Balance as on 31.03.2008

Adjustments:

1. Salary outstanding has not been recorded – Rs.8000
2. Prepaid insurance was meant for Proprietor’s son.
3. Write off further Rs. 200 as bad debts and make a provision for doubtful debts @ 5%.
4. Depreciate furniture @ 10%.

You are required to prepare Trading and Profit and Loss Account for the year ended 31.03.2008 and Balance Sheet as on the date.
Answer:
Trading and Profit and Loss a/c for the year ended 31.03.2008

Balance sheet as on 31.03.2008

Question 2.
Prepare Trading and Profit and Loss a/c for the year ended 31.03.05 and Balance sheet as on that date from the following balance.


Answer:
Trading and Profit and Loss A/c of Mr. Babu for the year ended 31.03.05

Balance Sheet of Mr. Babu as on 31.03.05

Question 3.
The following is the Trial Balance of Vineeth as on 30.06.2008.

Additional Information:

1. Closing stock is valued at Rs. 8750.
2. Provide 5% of debtors for bad debts and 2% of debtors and creditors for discount.
3. Provide interest on capital at 10% and charge interest on drawings at 5%.
4. Depreciate delivery van by 20%.
5. Only one-fifth of advertisement is to be treated as expenses of current year.

Prepare Trading and Profit and Loss A/c for the year ended 30.06.2008 and also a Balance Sheet as on that date.
Answer:
Trading and Profit and Loss A/c for the year ended 30.06.08

Balance Sheet as on 30.06.2008

Note: Discount on Debtors
= (20000-1000) × 2/100 = 380
Discount on Creditors = 10500 × 2/100 = 210

Question 4.
Prepare Trading and Profit and Loss Account for the year ended 31.03.2005 and a Balance Sheet as on that date from the following balances.

Adjustments:

1. Stock on 31.03.05 Rs. 72,600.
2. Depreciate Plant and Machinery by 33 1/3%, Furniture by 10% and Freehold property by 5%.
3. Loose tools valued at Rs. 1,760 on 31.3.05.
4. Of the Sundry debtors Rs. 660 are bad and should be written off.
5. Maintain a provision of 5% debtors for doubtful debts.
6. The Manager is entitled to a commission of 10% of the net profits after charging such commission.

Answer:
1. Trading and Profit and Loss A/c for the year ended 31.03.2005


Balance Sheet as on 31.03.2005

Note: Manager’s Commission:
Net Profit before commission = 49280 (111870-62590)
Therefore, Managers commission @ 10% of Net Profit after charging commission = 49280 × 1°/110 = 4480

Question 5.
The following is the Trial Balance of Balu as on 31/12/2011.

Adjustments:

1. Closing stock Rs. 9,500
2. Depreciate plant and machinery at 10% p.a
3. Interest on loan at 12% is due for the whole year.
4. Write off further Rs.400 as bad debts and provision for bad debts is to be made equal to 5% on debtors.
5. Provide 2% for discount on debtors.

Prepare Trading and Profit and Loss A/c for the year ended and Balance sheet as on 31/12/2011.
Answer:
Trading and Profit and Loss Account for the year ended 31/12/2011


Balance sheet as on 31/12/2011

Question 6.
The following is the Trial Balance of Mahesh as on 31/3/11.

Adjustments:

1. Closing stock Rs. 65,000
2. Insurance prepaid Rs. 400
3. Commission earned but not received amounts to Rs. 300
4. Write off 1/5 of leasehold premises and provide a depreciation of 10% on plant and machinery.
5. Debtors include goods of the cost of Rs. 4,000 sent on sale or return basis at an invoice price of Rs. 5,000. The goods are likely to be returned.
6. Transfer 10% of net profit to reserve. Prepare Trading and Profit and Loss A/c and Balance Sheet.

Answer:
Trading and Profit and Loss A/c for the year ended 31/3/2011.


Balance Sheet as on 31/03/2011

Note:
Reserve = 10% of net profit
79,500 -16,800 = 62,700 x 10% = 6,270

Question 7.
From the following balance extracted from the books of M/s. Hariharan Brother, you are required to prepare the trading and profit and loss account and a balance sheet as on December 31, 2005.

Adjustments:

1. Closing stock Rs. 14000
2. Wages outstanding Rs.600, Salaries outstandings Rs. 1,000, Rent outstanding Rs. 200.
3. Fire Insurance premium includes Rs.1200 paid in July 01, 2005, to run for one year from July 01, 2005, to June 30, 2006.
4. Apprenticeship Premium is for three years paid in advance on January 01, 2005.
5. Stationery bill for Rs. 60 remain unpaid.
6. Depreciation on Premises @ 5%, furniture @ 10%, Machinery @ 10%.
7. Interest on loan given accrued for one year @ 7%.
8. Interest on investment @ 5% for half year to December 31, 2005, has accrued.
9. Interest on capital to be allowed at 5% for one year.
10. Interest on drawings to be charged to him ascertained for the year Rs. 160.

Answer:
Books of Hariharan Bros.
Trading and Profit and Loss account for the year ended December 31, 2005


Balance Sheet as at December 31, 2005

Question 8.
Prepare the trading and profit and loss account of M/s.Roni Plastic Ltd. from the following trial balance and a balance sheet as at March 31,2006.


Adjustments:

1. Depreciation on land and building at 5% and Motor vehicle at @ 15%.
2. Interest on loan is @ 5% taken on April 01, 2005
3. Goods costing Rs. 1200 were sent to a customer on sale on return basis for Rs.1400 on March 30, 2006, and has been recorded in the books as actual sales.
4. Salaries amounting to Rs. 1400 and Rates amounting to Rs. 800 are due.
5. The bad debts provision is to be brought up to @ 5% on Sundry debtors.
6. The closing stock was Rs. 13,700.
7. Goods costing Rs. 1,000 were taken away by the proprietor for his personal use but not entry has been made in the books of account.
8. Insurance prepaid Rs. 350.
9. Provide the Manager’s commission at @ 5% on Net profit after charging such commission.

Answer:
Books of Roni’s Plastic Ltd.
Trading and Profit and Loss account for the year ended December 31, 2006


Note: New provision for bad debt = (38200 -1400) 5/100 = 1840
Balance Sheet as on 31.3.2006

Question 9.
From the following balances extracted from the books of Raga Ltd. prepare a trading and profit and loss account for the year ended December 31, 2011, and a balance sheet as on that date.

The additional information is as under

1. The closing stock was valued at the end of the year 20,000.
2. Depreciation on plant and machinery charged @ 5% and land and building @ 10%.
3. Discount on debtors @ 3%.
4. Make a provision @ 5% on debtors for bad debts.
5. Salary outstanding was? 100 and wages prepaid was? 40.
6. The manager is entitled a commission of 5% on net profit after charging such commission.

Answer:
Trading and Profit and Loss Account as on 31^st December 2011

Balance Sheet as on 31^st December 2011

HSSLive.Guru has great pleasure in presenting HSSLive Plus One Malayalam Textbook Questions and Answers, Plus One Malayalam Chapters Summary, Malayalam Chapter Wise Notes, SCERT Malayalam Guide Pdf, Malayalam Study Materials, Plus One Malayalam Handbook, Malayalam Textbook Kerala Syllabus PDF free download according to the latest syllabus prescribed by SCERT under the activity-oriented grading pattern for class 11.

Kerala Plus One Malayalam Textbook Questions and Answers, Notes, Chapters Summary HSSLive

Unit 1 Kinav

• Chapter 1 Sandarsanam
• Chapter 2 Ormayude Njaramb
• Chapter 3 Verukal Nashtapedunnavar
• Chapter 4 Matsyam

Unit 2 Kalca

• Chapter 1 Kayalarikathu
• Chapter 2 Cinemayum Samoohavum Kalavupoya Cyclum
• Chapter 3 Kazhinjupoya Kalaghattavum
• Chapter 4 Kaippad Kelkkunnundo?

Unit 3 Ullariv

• Chapter 1 Kavyakalaye Kurichu Chila Nireekshanangal
• Chapter 2 Oonjalil
• Chapter 3 Anargha Nimisham
• Chapter 4 Lathiyum Vediyundayum

Unit 4 Uravu

• Chapter 1 Peeli Kannukal
• Chapter 2 Anukampa
• Chapter 3 Muhyadheen Mala
• Chapter 4 Vasanavikrithi
• Chapter 5 Samkramanam
• Chapter 6 Shasthrakriya

We hope that HSSLive Plus One Malayalam Study Material will help you improve your quality of learning as per the new grading system and to get an A⁺ grade in Malayalam. Suggestions for further improvement of this material are always appreciated.

Plus One Malayalam Previous Year Question Papers and Answers

HSSLive Plus One

Kerala State Board New Syllabus Plus One Hindi Textbook Answers Unit 3 Chapter 9 आनंद की फूलझडियाँ Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Hindi Textbook Answers Unit 3 Chapter 9 आनंद की फूलझडियाँ

Anand Ki Fuljhadiya Plus One Hindi Chapter 9  प्रश्न 1.
पात्र और घटनाओं का सही मिलान करें।

उत्तर:

आनंद की फुलझड़ियाँ Plus One Hindi Chapter 9 प्रश्न 2.
संक्षेपण करें: “एक बूढ़ा आदमी, जिसके बाल सफ़ेद हो गए थे इसके फल मेरे नाती-पोते खाएँगे।”
उत्तर:

सत्कृत्य:
बूढ़े आदमी से नवजवान ने समझा कि पेड़ पौधे लगाने से प्रकृति सुन्दर हो जाती है और भविष्य में उपयोग में आता है।

Hindi Plus आनंद की फुलझड़ियाँ Chapter 9 प्रश्न 3.
संभ्रांत महिला रेलगाड़ी से कुछ चीजें बाहर फेंकती जा रही है तब सहयात्री और संभ्रांत महिला के बीच का संभावित वार्तालाप तैयार करें।
उत्तर:
संहयात्री : यह आप क्या फेंक रही हैं?
महिला : मैं…….?
सहयात्री : हाँ…. हाँ…….
महिला : तुम देखते नहीं?
सहयात्री : इसलिए तो पूछता हूँ।
महिला : ये तो बीज हैं।
सहयात्री : बीज?
महिला : हाँ… हाँ… फल-फूलों के बीज हैं।
सहयात्री : इनको खिड़की से क्यों फैंकती हैं?
महिला : इनमें कुछ जड़ पकड़ लेंगे।
सहयात्री : तो फिर?
महिला : तब फायदा होगा।
सहयात्री : फायदा? किस प्रकार?
महिला : फूलेंगे, फलेंगे।
सहयात्री : तब?
महिला : मनुष्य के लिए उपयोगी होंगे।
सहयात्री : अरे बापरे! आप तो महान कार्य कर रही हैं।
महिला : यह लो….आप भी फेंकिए।
सहयात्री : हाँ…… हाँ….. दीजिए।

Apoorv Anubhav Class 11 Summary In Hindi Chapter 9 प्रश्न 4.
मान लें, रेलगाड़ी में सफर करनेवाली वृद्ध संभ्रांत महिला की नज़र डिब्बे में चिपके हुए विज्ञापन पर पड़ती है जो रक्तदान के महत्व को रेखांकित करता है। संकेतों के सहारे वह विज्ञापन तैयार करें।

• समभाव
• सहिष्णुता
• मानव-प्रेम
• जीवनदान

उत्तर:

स्वास्थ्य मंत्रालय का विज्ञापन
‘रक्तदान महादान है।’

भाईयो,…… बहनो,…..
रोगावस्था में पीड़ित भाई-बहनों से समभाव रखिए । सहिष्णुता और अनुकंपा रखकर जान बचाने के लिए रक्तदान करके सहायता दीजिए। रक्तदान जीवनदान ही है!! सरकारी रक्तदान केन्द्रों में जाकर खुशी से रक्तदान कीजिए!! आपका रक्त कटेगा नहीं बढ़ेगा!! दूसरों की जान बचेगी।

Plus One 9 Hindi आनंद की फुलझड़ियाँ प्रश्न 5.
‘आज भी वह रक़म अमेरिका में ज़रूरतमंदों के हाथों में घूम रही है’ – मान लें, वह रक़म अपने वर्तमान अनुभवों का आत्मकथा के रूप में ज़िक्र करती है। वह आत्मकथांश लिखें।
उत्तर:

मैं हूँ जरूरतमंदों के सामने

मैं रकम हूँ। मैं साधारण रकम नहीं। मैं एक अपूर्व रकम हूँ। मैं जन्म से अमेरिकी हूँ। अमेरिका के प्रसिडेंट बेंजमिन फ्रैंकलीन के हाथों से मेरा जन्म हुआ। मुझे प्रसिडेंटजी ने एक गरीब विद्यार्थी की सहायता में दिया था। मैंने उसे भाग्यवान् बनाया। विद्यार्थी ने मुझे वापस करने के लिए प्रसिडेंट के पास गया। लेकिन, बेंजमिनजी ने उससे बतायाः “आप इसे अपने ही पास रखिए और जब आपके पास कोई ऐसा ही ज़रूरतमंद आये तो उसे यह दे दीजिए” । उस व्यक्ति ने ऐसा ही किया। आज भी मैं अमेरिका के ज़रूरतमंदों के हाथों में घूम रही हूँ। मैं कितना सौभाग्यवान हूँ! ज़रूरत पड़े, मुझे बुलाईए। मैं ज़रूर आऊँगा।

मेरे इतने जीवनकाल से मैंने समझा कि समाज की पूँजी धनवानों के हाथों में है। पूँजी का समुचित विवरण से समाज का संतुलन होता है। जरूरतमंदों के हाथों में पूँजी का सौगुना मूल्य होगा। मैं रकम, पूँजी समान है। मैं जरूरतमंदों को जीवनदान करता हूँ।

Plus One Hindi आनंद की फूलझडियाँ Important Questions and Answers

प्रश्न 1.
ज़मीन कौन खोद रहा था?
उत्तर:
एक बूढ़ा आदमी।

प्रश्न 2.
बूढ़े आदमी क्या बो रहे हैं?
उत्तर:
आम की गुठलियाँ।

प्रश्न 3.
नौजवान के प्रश्न पर बूढ़े का उत्तर क्या था?
उत्तर:
भविष्य में दूसरों की भलाई के लिए आम की गुठलियाँ बो रहा हूँ।

प्रश्न 4.
पूर्वजों के मनोवृत्ति का फल क्या है?
उत्तर:
वर्तमान के लोगों के लिए भलाई होती है।

प्रश्न 5.
लेखक ने सुंदर स्वभाव की परिभाषा कैसे दी है?
उत्तर:
अशा दूसरों को सुख और आनंद पधुंचानेवाले सात्विक आनंद का स्वभाव।

प्रश्न 6.
हमारे पूर्वजों की इसी मनोवृत्ती का फल है, जो हम जगह-जगह अमराई देखते हैं। कौन-सी मनोवृत्ति?
उत्तर:
अपने स्वार्थी जीवन को त्यागकर दूसरों को सुख और आनंद पहुँचाने के सुंदर स्वभाव की मनोवृत्ति ।

प्रश्न 7.
बूढ़ा आदमी का ‘आम की गुठलियाँ बोना’ घटना का मुख्य आशय क्या है?
उत्तर:
हमें दूसरों को सुख और आनंद पहूँचानेवाले सात्विक आनंद के स्वभाव अपनाना चाहिए।

प्रश्न 8.
मान लें, रेलगाड़ी में सफर करनेवाली वृद्ध संभ्रांत महिला की नज़र डिब्बे में चिपके हुए विज्ञापन पर पड़ती है जो रक्तदान के महत्व को रेखांकित करता है। संकेतों के सहारे वह विज्ञापन तैयार करें।

• समभाव
• सहिष्णुता
• मानव-प्रेम
• जीवनदान

उत्तर:

स्वास्थ्य मंत्रालय का विज्ञापन
‘रक्तदान महादान है।’

भाईयो,…… बहनो,…..
रोगावस्था में पीड़ित भाई-बहनों से समभाव रखिए। सहिष्णुता और अनुकंपा रखकर जान बचाने के लिए | रक्तदान करके सहायता दीजिए। रक्तदान जीवनदान ही है!! सरकारी रक्तदान केन्द्रों में जाकर खुशी से रक्तदान कीजिए!! आपका रक्त कटेगा नहीं बढ़ेगा!! दूसरों की जान बचेगी।

प्रश्न 9.
संभ्रान्त महिला किस उम्मीद से फल और फूलों के बज़ फेंक रही है?
उत्तर:
उनमें से कुछ भी अगर जड़ पकड़ लेगें तो लोगों का इससे कुछ फायदा होगा।

प्रश्न 10.
निम्नलिखित गद्यांश पढ़ें और प्रश्नों का उत्तर लिखें।

एक वृद्ध संभ्रांत महिला रेलगाड़ी से सफ़र कर रही थी। वे खिड़की के पास बैठकर, बीच-बीच में अपनी मुट्ठी से कुछ चीज़ बाहर फेंकती जा रही थीं। एक सहयात्री ने, जो यह देख रहा था, पूछा, “यह आप क्या कर रही हैं?” उस महिला ने जवाब दिया, “ये सुंदर फलों और फूलों के बीज़ हैं। मैं इन्हें इस उम्मीद से फेंक रही हूँ कि इनमें से कुछ भी अगर जड़ पकड़ लेंगे तो लोगों का इससे कुछ फायदा होगा । पता नहीं इस रास्ते से फिर गुजरूँ या न गुजरूँ, इसलिए क्यों न मैं इस संधि का उपयोग क लूँ?”

i. रेलगाड़ी से कौन यात्रा कर रहीं थी?
उत्तर:
एक वृद्ध संभ्रात महिला।

ii. वे क्या कर रही थीं?
उत्तर:
अपनी मुट्ठी से सुंदर फलों और फूलों के बीज़ बाहर फेंकती जा रही थीं।

iii. सहयात्री ने क्या पूछा?
उत्तर:
सहयात्री ने पूछा : यह आप क्या कर रही हैं?

iv. गद्यांश का संक्षेपण करें।
उत्तर:
एक वृद्ध संभ्रात महिला रेलगाड़ी से सफर करते समय सुंदर फलों और फूलों के बीज़ बाहर फेंकती जा रही थीं। उनका उद्धेश्य था कि उनमें से कुछ भी अगर जड़ पकड़ लेंगे तो, लोगों का उससे कुछ फायदा होगा।

v. संक्षेपण केलिए उचित शीर्षक दें।
उत्तर:
जन-सेवा।

प्रश्न 11.
हमें किसके परे रहना चाहिए?
उत्तर:
घर भौतिकवाद और भोग-विलास की हाय-हाय से परे ।

प्रश्न 12.
किस प्रकार के लोगों को देखकर मानव जाति के भविष्य पर श्रद्धा और विश्वास कर सकते हैं?
उत्तर:
जमाने के अंधकार में भी आनंद की फुलझड़ियों से प्रकाश फैलाते रहनेवालों को देखकर ।

प्रश्न 13.
बेंचमिन फ्रांक्लिन ने विद्यार्थी की मदद कैसे की?
उत्तर:
उन्होंने विद्यार्थी को बीस डॉलर देकर मदद की।

प्रश्न 14.
कुछ दिनों के बाद विद्यार्थी डॉलर लौटाने आए तो फ्रांक्लिन ने क्या कहा?
उत्तर:
उन्होंने कहा : “मुझे याद तो नहीं है कि मैंने यह रक़म आपको कब दी। लेकिन खैर, आप इसे अपने ही पास रखिए और जब आपके पास कोई ऐसा ही जरूरतमंद आए तो उसे यह दे दीजिए।”

प्रश्न 15.
बीस डॉलर किनके हाथों में घूम रही है?
उत्तर:
जरूरतमंदों के हाथों में।

प्रश्न 16.
ज़रूरतमंद कौन-कौन हो सकता है?
उत्तर:
सहायता मिलने के लिए व्याकुल रहनेवाले सभी लोग ज़रूरतमंद होते हैं।

प्रश्न 17.
हमारा जीवन मुसीबतों से भरा पड़ा है । कैसे?
उत्तर:
अब लड़ाई, गरीबी, महंगाई और गुलामी से।

प्रश्न 18.
दुनिया रहने लायक कैसे बनी?
उत्तर:
निस्वार्थ और आदर्श-प्रिय लोगों की उपस्थिति से।

प्रश्न 19.
पैसा वापस देने आया छात्र औह बेंजमिन फ्रैंकलीन के बीच का वार्तालाप तैयार करें?
उत्तर:
छात्र : नमस्कार जी!
फ्रैंकलीन : नमस्कार।
छात्र : आप मुझे जानते हैं?
फ्रैंकलीन : याद नहीं, लगता है कि देखा है।
छात्र : मैंने आप से कुछ डॉलर माँग लिया था।
फ्रैंकलीन : कब?
छात्र : कुछ साल पहले मेरी पढ़ाई केलिए। अब मेरी पढ़ाई खतम हुई। वह डॉलर वापस देने आया हूँ।
फ्रैंकलीन : यह डॉलर मुझे नहीं, किसी ज़रूरतमंद को दो।
छात्र : आप का मन कितना अच्छा है!
फ्रैंकलीन : तुम से कोई ज़रूरतमंद व्यक्ति माँगे है तो उसे यह डॉलर दो। जाओ।
छात्र : ठीक है, धन्यवाद ।

प्रश्न 20.
किसके लिए छीना-झपटी होती थी?
उत्तर:
टिकट लेने के लिए।

प्रश्न 21.
टिकट बाबू किन बातों को सुना-अनसुना करके अपना काम करते रहे?
उत्तर:
टिकट लेने के लिए छीना-झपटी करनेवालों के परिहासों और धमकियों को।

प्रश्न 22.
टिकट बाबू की परेशानी का कारण क्या था?
उत्तर:
टिकट लेने के लिए बड़ी भीड़ थी। वहाँ पर छीना-झपटी होती थी। छोटी-सी खिड़की से टिकट के लिए घुसानेवाले अनेक हाथों को टिकट देने केलिए टिकटबाबू अकेला था।

प्रश्न 23.
टिकट बाबू पर इन शब्दों का अजीब असर पड़ा । क्यों?
उत्तर:
लेखक ने टिकट बाबू की परेशानी समझकर बड़ी सहानुभूति से व्यवहार करके मुसाफिरों को शांत किया।

प्रश्न 24.
टिकट बाबू को नई ताकत कैसे मिली?
उत्तर:
लेखक की सहानुभूति के शब्दों से टिकट बाबू को नयी … ताकत मिली।

प्रश्न 25.
टिकट बाबू का हृदय कब मोम-सा हो गया?
उत्तर:
टिकट बाबू के प्रति लेखक सहानुभूति दर्शाने पर।

प्रश्न 26.
टिकट बाबू से संबंधित घटना का ज़िक्र करते हुए लेखक अपने मित्र को पत्र लिखता है। वह पत्र तैयार करें।
उत्तर:

स्थान,
तारीख,

प्रिय मित्र रामू,

तुम कैसे हो? ठीक हो न? यहाँ पर मैं ठीक ही हूँ।
रामू, कुछ दिन पहले अपने काम से मुंबई जाने के लिए मैं टिकट लेने गया। वहाँ पर बड़ी छीना-झपटी होती थी। टिकट बाबू बड़ी परेशानी में था। लोग हल्ला करते थे, टिकट बाबू का परिहास करते थे। कुछ लोग उनको धमकी भी करता था। लेकिन टिकट बाबू इन बातों को सुना- अनसुना करके आपना काम करते रहे। मैं पास ही खड़ा था। मैंने बड़ी सहानुभूति से टिकट बाबू के बारे में यात्रियों से बात की। मेरी सहानुभूति देखकर टिकटबाबू को बड़ा आश्वास मिल गया। उन्होंने जल्दी मेरा टिकट दे दिया और एक नए उत्साह से अन्य यात्रियों को वे टिकट देने लगे।

प्रिय मित्र, दुसरों की कठिनाइयों को समझकर हमें व्यवहार करना चाहिए। हमारे कर्म और वचन से दूसरों को आश्वास मिला चाहिए।
यहाँ पर तुम कब आवोगो? तुम्हारी प्रतीक्षा में मित्र,

(हस्ताक्षर)
शेवड़े।

पताः
नाम

प्रश्न 27.
क्लर्क का तमाम दिन कैसे बीतता है?
उत्तर:
बैंक के रूखे आँकड़ों से माथापच्ची करते -करते क्लर्क का तमाम दिन बीतता है।

प्रश्न 28.
क्लर्क के किस गुण का सम्मान किया गया?
उत्तर:
अच्छी हस्तलिपि का।

प्रश्न 29.
क्लर्क का चेहरा प्रसन्नता से क्यों खिल उठा?
उत्तर:
लेखक के थोड़े-से शब्दों द्वारा क्लर्क के जीवन में किंचितमात्र सुख पहुँचने पर।

प्रश्न 30.
लेखक को क्यों प्रसन्नता महसूस हुई?
उत्तर:
प्रय लेखक को थोड़े-से शब्दों द्वारा बैंक के क्लर्क के जीवन में थोड़ा सुख पहूँचते देखकर लेखक को प्रसन्नता महसूस हुई।

प्रश्न 31.
बैंक का क्लर्क अपनी हस्तलिपि की तारीफ सुनने पर बहुत खुश हुआ। घर आने पर पत्नी उसकी खुशी का कारण जानना चाहती है-दोनो के बीच का संभावित वार्तालप लिखें।
उत्तर:
पत्नी : आज आप बहुत खुश है ….
क्लर्क : हाँ-हाँ….
पत्नी : कारण क्या है?
क्लर्क : एक कारण है।
पत्नी : मुझे भी बताओ….
क्लर्क : तुम जानना चाहती हो?
पत्नी : क्यों नहीं?
क्लर्क : तुम अनंत गोपाल शेवड़े को…
पत्नी : ओहो… सुप्रसिद्ध लेखक?
क्लर्क : जानती हो उन्हें?
पत्नी : सुनी तो है।
क्लर्क : आज उन्होंने मेरे बैंक में…
पत्नी : बैंक में?
क्लर्क : आये थे।
पत्नी : तो ?
क्लर्क : उन्होंने मेरी हस्तलिपि की …..
पत्नी : प्रशंसा की?
क्लर्क : हाँ…हाँ…
पत्नी : बड़ी बात है।
क्लर्क : हाँ…हाँ…

प्रश्न 32.
बैंक क्लर्क ने अपनी डायरी में क्या लिखा होगा? वह डायरी तैयार करें।
उत्तर:
तारीख
आनंद नगर :
आज का दिन बहुत अच्छा है। आज मुझे मिली प्रशंसा एक पंखुडी के समान मुझे अब भी सहलाती है। वह आदमी कितना अच्छा है! आज पहली बार बैंक के बोरिंग समय में कुछ राहत मिली। उस आदमी ने खाता खोलने के लिए आया था। मैं उनका नाम और पता लिखते समय उन्होंने मेरी हस्तलिपि पर ध्यान दिया और कहा ‘ब्यूटिफुल’! मेरा मन खुशी से भरा। सब लोगों ने मेरी हस्तलिपि देखी थी, पर किसीने भी मुझ से कुछ नहीं कहा था । अब मैं गर्व का अनुभव करने लगा। बड़ी खुशी से मैंने उनसे बातें कीं। पत्नी से भी यह बात कही। यह दिन में कभी नहीं भूलूंगा।

प्रश्न 33.
आनंद की फुलझड़ियाँ निबन्ध में लेखकने रेल विभाग के एक टिकट बाबू से करुणा प्रकट की थी। उस दिन के टिकट बाबू की डायरी तैयार करें।
उत्तर:

तारीख

आनंद शहर :
आज का दिन अच्छा दिन था। एक आदमी ने मुझे आज ठीक समझा है। यह बात मेरे मन में खुशी भरती है। यात्रियों की भीड़, उनकी परेशानियाँ, धमकियाँ आदि – आदि ने मुझे बहुत परेशानियाँ देती थीं। टिकट कौंटर में हर दिन अकेला रह गया हूँ। मेरी कठिनाइयों पर किसी ने ध्यान नहीं किया था। पर आज एक सज्जन ने मुझपर ध्यान देकर मेरी मदद की। टिकट लेने केलिए खड़े लोगों से मेरी परेशानियों के बारे में बताने की कृपा उन्होंने की। यह एक अजीब बात थी। उनकी सहानुभूति देखते वक्त मेरा हृदय मोम जैसा बन गया। वे शब्द मेरे मन में सांत्वना देने लायक थे। उस घटना के बाद मैं शांत भाव से टिकट देने में समर्थ हुआ।

वे कौन होंगे? जाते वक्त उन्होंने कहा कि फिर मिलें। ज़रूर उनसे मिलना चाहिए। ऐसे सज्जनों से परिचय पा लेना कितनी अच्छी बात है! आज का दिन मैं कभी नहीं भूलूंगा।

प्रश्न 34.
निम्नलिखित अर्थों के समानार्थी मुहावरों को लेख से छाँटकर लिखें।
(नष्ट होना, विपत्ति के दिनों के बाद सुख का दिन आना, सहन करना, हिम्मत करना, किसी के अच्छे काम की न्यायदृष्टि से प्रशंसा करना)
उत्तर:
रसातल में जाना = नष्ट होना
दिन फिरना = विपत्ति के दिनों के बाद सुख का दिन आना
ताना कसना = सहन करना
दिल कडा करना = हिम्मत करना
दाद देना = किसी के अच्छे काम की न्यायदृष्टि से प्रशंसा करना

प्रश्न 35.
सूचनाः यह गद्यांश पढ़िए और नीचे दिए प्रश्नों के उत्तर लिखिए:

नाविक भोलाराम रेलगाड़ी के वातानुकूलित डिब्बे में बैठ रहे थे। उनके पास काफ़ी रुपये थे। एक लड़की भी उस डिब्बे में आ बैठी। उसने नाविक से बातें शुरू की। उसने अपनी गरीबी का जिक्र किया तो नाविक ने पूछा कि इतनी गरीबी में भी वातानुकूलित डिब्बे में तुम क्यों यात्रा कर रही हो? तब लड़की ने कहा कि उसकी शादी तय हो चुकी है। और अपने ससुरालवालों को प्रभावित. करने केलिए वह इस डिब्बे में यात्रा कर रही है। फिर उस लड़की ने नाविक से कुछ रुपया माँग लिया तो उसने देने से इनकार कर दिया। तब लड़की ने उसको धमकी देकर कहा कि मुझे बीस हज़ार रुपये दें दो, नहीं तो मैं तुम्हारे ऊपर झूठे इल्ज़ाम लगाउँगी। गाड़ी रुकी तो लड़की ने पुलिस से कहा कि नाविक ने अपने हाथ से मेरा मुँह बंद कर लिया और दूसरे हाथ से मुझे खींचकर मेरी इज्जत लूटने की कोशिश की है। लेकिन जाँच करने पर पुलिस को मालूम हुआ कि नाविक के दोनों हाथ कटे हुए हैं। झूठे इल्ज़ाम लगाने के अपराध में लड़की पकड़ी गयी।
(इल्ज़ाम – आरोप, इज्जत लूटना – अपमानित करना)

i) इस गद्यांश से कौन सा सन्देश मिलता है?
उत्तर:
हमें कभी भी कपट न होना चाहिए।

ii) इस गद्यांश का संक्षेपण करें और उचित शीर्षण लिखे।
उत्तर:
कपटता :
एक लड़की ने भोलाराम नामक एक बिना हाथवाला नाविक के साथ रेलगाड़ी में यात्रा करते समय धोखा देने की कोशिश की। पुलिस आकर लड़की को पकड़ा।

प्रश्न 36.
हिंदी भारत की राजभाषा एवं राष्ट्रभाषा है- ‘सीखें हिंदी, सिखाएँ हिंदी’ – इस विषय पर निबंध लिखिए।
उत्तर:

सीखें हिंदी, सिखाएँ हिंदी

भारत में अनेक भाषाएँ हैं। भाषाओं को उपभाषाएँ और प्रादेशिक भाषाएँ भी हैं। लेकिन भारत के अधिकांश लोगों से बोलनेवाली भाषा हिंदी है। इसलिए हिंदी को भारत की संपर्क भाषा के रूप में माना जाता है। हिंदी भारत की राजभाषा एवं राष्ट्रभाषा है। हिंदी एक सरल भाषा है। हमें यह जानने से खुशी होगी कि संसार में सबसे ज्यादा बोलनेवाली तीसरी भाषा हिंदी है। आज हिंदी अन्तर्देशीय भाषा के रूप में प्रचलित होती जाती है।

भारत विविधता का देश है। लेकिन हिंदी एकता की कड़ी है। भारत की संस्कृति हिंदी से जुड़ी रहती है। हिंदी समृद्ध साहित्य से भी संपन्न है।

हिंदी का प्रचार करना प्रत्येक भारतीय का दायित्व है। हिंदी के प्रचार से भारत में एकता बढ़ेगी। इससे यह .मतलब नहीं है कि अन्य प्रादेशिक भाषाएँ महत्वपूर्ण नहीं है और वे तिरस्कृत हो जायें। प्रादेशिक भाषाओं का भी संरक्षण होना चाहिए। हिंदी के प्रचार से भारत की अखंडता सदा सुरक्षित रखें।

आनंद की फूलझडियाँ Previous Years Questions & Answers

प्रश्न 1.
निम्नलिखित सहायक बिंदु के आधार पर वार्तालाप तैयार कीजिए।
संभ्रान्त महिला रेलगाड़ी से कुछ चीजें बाहर फेंकती जा रही थी। तब सहयात्री और संभ्रान्त महिला के बीच का संभावित वार्तालाप तैयार कीजिए। सहायक बिंदुः

• रेलगाड़ी से चीजें बाहर फेंकना
• सहयात्री द्वारा पूछा जाना
• निःस्वार्थ सेवा
• दुनिया में रहने लायक

उत्तर:
सहयात्री : यह आप क्या फेंक रही हैं?
महिला : मैं……?
सहयात्री : हाँ…. हाँ…….
महिला : तुम देखते नहीं?
सहयात्री : इसलिए तो पूछता हूँ।
महिला : ये तो बीज हैं।
सहयात्री : बीज?
महिला : हाँ… हाँ… फल-फूलों के बीज हैं।
सहयात्री : इनको खिड़की से क्यों फैंकती हैं?
महिला : इनमें कुछ जड़ पकड़ लेंगे।
सहयात्री : तो फिर?
महिला : तब फायदा होगा।
सहयात्री : फायदा? किस प्रकार?
महिला : फूलेंगे, फलेंगे।
सहयात्री : तब?
महिला : मनुष्य के लिए उपयोगी होंगे।
सहयात्री : अरे बापरे! आप तो महान कार्य कर रही हैं।
महिला : यह लो….आप भी फेंकिए।
सहयात्री : हाँ…… हाँ….. दीजिए।

प्रश्न 2.
मान लीजिए, आनंद की फूलझड़ियाँ इस निबंध का लेखक आत्मकथा लिखता है। आत्मकथा में टिकट बाबू के प्रसंग का उल्लेख है। निम्नलिखित सहायक बिंदु के आधार पर वह आत्मकथांश तैयार कीजिए।
सहायक बिंदुः

• लेखक का मुंबई जाना।
• टिकट काऊंटर के पास भीड़ लगना ।
• टिकट बाबू का परेशान होना ।
• लेखक द्वारा टिकट बाबू के प्रति सहानुभूति प्रकट करना।

उत्तर:
आत्मकथा  :
कुछ साल पहले की बात है। मुझे जल्द ही मुंबई पहूँचना था। मैं टिकट लेने केलिए टिकट काउंटर पहूँचा । लड़ाई के कारण गाड़ियों की संख्या कम थी। इसलिए काउंटर के पास बहुत भीड़ लगी हुई थी। टिकट बाबू परेशानी से टिकट बनाते थे। लेकिन भीड़ ज्यादा होने के कारण उसे ठीक तरह से सभी लोगों को टिकट बन नहीं पा रहे थे। लोगों की ओर से कई प्रकार के बुरे टिप्पणियाँ उन पर हो रहे थे। इसका असर उपनर और बुरी तरह से हो रहे थे। बाबू कितने ही ईमानदारी से काम करें लोग उनपर शक से बात करते थे। यह सब देखकर मुझे बहुत दुख हुआ। वह पूरी ताकत से काम कर रहे थे, फिर भी लोग उनपर गालियाँ दे रहे थे। मैं वहाँ के लोगों को समझाया कि बाबू बहुत कोशिश कर रहे हैं और उसे कुछ समय दीजिए। मैं बाबु से ऐसे बातें किया कि उसे कुछ आश्वास मिला। मेरा सहानुभूति का असर उनपर हुआ। वह फिर पूरी कोशिश की और मुझे धन्यवाद भी अदा की। मुझे पूरा यकीन था कि अच्छे वाक्यों का अच्छा असर हो जायेगा।

आनंद की फूलझडियाँ Summary in Malayalam

आनंद की फूलझडियाँ शब्दार्थ

Students can Download Chapter 3 Liberalisation, Privatisation and Globalisation – An Appraisal Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 3 Liberalisation, Privatisation and Globalisation – An Appraisal

Background of the economic reforms
India introduced economic reforms in 1991. It was due to several reasons. Important among them are:

• Policies such as MRTP and FEMA prevented large scale domestic and foreign investments.
• Reserving certain sectors exclusively for the public sector prevented private investment less attractive for such sectors.
• Gulf war and subsequent events created a severe foreign exchange crisis in our country.
• Import bill of petroleum products increased alarmingly leading to BoP deficit.
• Political instability.

Plus One Economics Chapter 3 Notes Liberalisation
Liberalization implies liberating trade from unwanted government controls and restrictions. Indian economy prior to the nineties was following a restrictive policy and excessive government interferences in all economic activities. This interference created the license-permit-raj as indicated earlier. This has led to extensive corruption, red-tapism, undue delay, and inefficiency. Most of the policies such as the licensing system, FERA, MRTP hindered economic growth, and industrialisation. The aim of the liberalization policy was very comprehensive, promoting economic growth by reducing factors hindering it and makes the economy very competitive at international standards.

Liberalisation policies included reforms in the following sectors.

• Industrial sector reforms
• Financial sector reforms
• Tax reforms
• Foreign exchange reforms

Plus One Economics Chapter 3 Notes Pdf Privatisation
Privatisation refers to any process that reduces the participation of the state/public sector in the economic activities of a country. In other words, the conversion of ownership or management of a government-owned enterprise into a private enterprise is known as privatization or denationalization. India started privatization as part of the Structural Adjustment Programme (SAP). The process of privatisation can take place either by the withdrawal of government ownership and management of public sector companies or by the outright sale of public sector companies (disinvestment).

Economics Notes Class 11 Kerala Syllabus Aims of disinvestment:

1. Better performance of public sector units (PSUs) through better management techniques
2. Enforcing financial discipline and improving financial performance
3. Enhancing the ability of companies to raise financial resources from the market
4. Raising revenue of the government from sale of equity
5. A strong impetus to the flow of FOI (Foreign Direct Investment)

Plus One Economics Notes Chapter 3 Globalisation
Globalisation is a complex phenomenon. The term globalisation indicates the opening up of domestic economy for the world market, or integration of an economy with global economy. It involves creation of network and activities transcending economic, social and geographical boundaries. It attempts to establish links in such a way that the happening in India can be influenced by events happening miles away. Integration of economies is possible through interlinking domestic market with world market through foreign trade. Therefore, it is treated as a very complex phenomenon.

Plus One Economics Notes Pdf Outsourcing
Outsourcing is an important feature of globalisation. It is practice where a company hires regular service from external sources mostly from other countries which previously provided internally or within the country.

Plus One Economics Malayalam Notes World Trade Organisation (WTO)
WTO was founded in 1995 replacing GATT. GATT was established in 1948. Following are the aims of WTO.

• Provides equal opportunities to all participating nations in international trade.
• To ensure optimum utilization of world resources and protect the environment.
• Remove of tariffs (tax) and non-tariffs (quota). This leads to the removal of restrictions on trade thereby facilitating free-entry and free exit of goods
• To encourage multi-lateral trade (more than two nations) rather than bilateral trade (two countries).
• Extension of a trade by including trade in services like banking, insurance communication.
• To include Trade-Related Intellectual Property Rights (TRIPs), commonly known as Patent Rights and Trade-Related Investment Measures (TRIMs) within the span of international trade.

Students can Download Chapter 3 Motion in a Straight Line Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 3 Motion in a Straight Line

Plus One Physics Chapter 3 Notes Pdf Summary
Motion In A Straight Line
In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity, acceleration and relative velocity. We also develop a set of simple equations called Kinematic equations.

• Motion: Motion is change in position of an object with time.
• Rectilinear motion: The motion along a straight line is called rectilinear motion.
• Point object: If the distance travelled by the body is very large compared with its size, the size of the body may be neglected. The body under such a condition may be taken as a point object. The point object can be represented by a point.

Example:

• The length of bus may be neglected compared with the length of the road it is running.
• The size of planet is ignored compared with the size of the orbit in which it is moving.

Position, Path Length And Displacement
1. Reference point, Frame of reference:
In order to specify position of object, we take reference point and a set of axes. Consider a rectangular coordinate system consisting of three mutually perpendicular axes, labelled x, y, and z axes. The point of intersection of these three axes is called origin (O) and serves as the reference point.

The coordinates (x, y, z) of an object describe the position of the object. To measure time, we place a clock in this coordinate system. This coordinate system along with a clock is called a frame of reference.

Straight-line motion in coordinate system

To describe the motion along a straight line we can choose x-axis. The position of a carat different time are given in the figure 3.1. The position to the right of 0 is taken as positive and to the left of 0 as negative. The position coordinates of point P and Q are +360m +240m. The position coordinate of R is-120m.

2. Path Length (Distance):
The total length of the path travelled by an object is called path length.
Explanation:
Consider a car moving along straight line. The positions of car at different time are given in the x-axis. (See figure 3.1)
Case-1:
The car moves from 0 to P. In this case the distance moved by car is OP = +360.
Case-2:
The car moves from 0 to P and then moves back from P to Q.
In this case, the distance travelled is OP + PQ = +360 + (+120) = +480m.

3. Displacement:
The distance between initial point and final point is called displacement.

OR

The change of position of the particle in a particular direction is called displacement.
Explanation:
Consider a car moving along a straight line. The positions of car at different time is given in the x-axis.
See figure (3.1)
Let us take two cases
Case-1:
The car moves from 0 to P, in this case displacement = (360 – 0) = 360
Case-2:
The car moves from 0 to P and moves back from P to Q.
In this case,
Displacement = 240m
Let x₁ and x₂ be the positions of an object at time t₁ and t₂. Then displacement in time Dt = (t₂ – t₁) can be written as Dx = x₂ – x₁
If x₁ < x₂, Dx is positive and if x₂ < x₁, Dx is negative.
Note: The magnitude of displacement may or may not be equal to the path length traversed by an object.

4. Position Time Graph:
Motion of an object can be represented by a position-time graph.
Position time graph for a stationary object:
For a stationary object, the position does not change with time. Hence the position time graph will be a straight line parallel to time axis.

Position time graph in a uniform motion:
Uniform motion:
A body is said to be uniform motion, if it undergoes equal displacements in equal intervals of time. In uniform motion velocity is constant The figure below shows the positiontime graph of such a motion.

Plus One Physics Chapter 3 Notes Question 1.
The position-time of a car is given below. Analyze the graph and explain the motion of car.

Answer:
The car starts from rest a time t=0s from the origin 0 and picks up speed till t=10s. After 10 sec, the car moves with uniform speed till t=18 sec. Then the brakes are applied and the car stops at t = 20s and x = 296m.

Class 11 Physics Chapter 3 Notes Question 2.
Draw the position-time for an object

1. moving with positive velocity
2. moving with negative velocity.

Answer:
1.

2.

Average Velocity And Average Speed
1. Average Velocity:
The average velocity of a particle is the ratio of the total displacement to the time interval.

Explanation:
To explain average velocity, consider a position time graph of a body given below.

Let x₁ be the position of body at a time t₁ and x₂ be the position at t₂.
The average velocity during the time interval Dt = (t₂ – t₁)

where Dx = x₂ – x₁, and Dt = t₂ – t₁,
\(\overline{\mathbf{v}}\) is the average velocity.

Motion In A Straight Line Class 11 Notes Pdf Question 3.
Find the slope of position time graph given below of uniform motion and explain the result.

Answer:

Slope of displacement time graph gives average velocity.

Class 11th Physics Chapter 3 Notes Question 4.
Displacement time graph of a car is given below.

1. Find the average velocity during the time interval 5 to 7 sec.
2. Find the average velocity by taking slope in the interval 5 to 7 sec.

Answer:
1.

2. Slope, tan q

In this case, slope and average velocity are equal in the same interval.

2. Average Speed:
Average speed of a particle is the ratio of the total distance to total time taken.

Physics Chapter 3 Class 11 Notes Question 5.
A car is moving along a straight line. Say OP in figure. It moves from 0 to P in 18s and returns from P to Q in 6s. What are the average velocity and average speed of the car in going?

1. From 0 to P? and
2. from 0 to P and back to Q. (See Figure 3.1)

Answer:
1. Average velocity

Average speed

In this case the average speed is equal to the magnitude of the average velocity.

2. In this case
Average velocity

Average speed

In this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction. So that the distance is greater than displacement.
Note: In general, the velocity is always less than or equal to speed.

Instantaneous Velocity And Speed
Nonuniform Motion:
A body is said to be nonuniform motion, if it undergoes unequal displacements in equal intervals of time.

OR

A body moving with varying velocity is called nonuniform motion.

1. Instantaneous Velocity:
Chapter 3 Physics Class 11 Notes Question 6.
Why the concept of instantaneous velocity is introduced?
Answer:
In nonuniform motion the average velocity tells us how fast the object has been moving over a given interval. But it does not tell us how it moves at different instants during that interval. For this we define instantaneous velocity. The velocity at an instant is called instantaneous velocity.
Explanation:
Position-time of a body moving along a straight line is given below.

Let us find average velocity in the interval 2 sec (3s to 5s), centered at t = 4 sec. In this case, the slope of line P₁P₂ give the value of average velocity, ie. Slope of P₁P₂,

Decrease the value of Dt from 2.to 1 sec. (ie. 3.5 to 4.5 sec). Then line P₁P₂ becomes Q₁Q₂. Then the slope of gives average velocity overthe interval 3.5 sec to 4.5sec.
ie. slope of Q₁Q₂

In the limit Dt ® 0, gives the instantaneous velocity at t = 4sec and its value is nearly 3.84m/s.

Motion In A Straight Line Class 11 Notes Question 7.
When average velocity of a body becomes instantaneous velocity?
Answer:
In the limit, Dt goes to zero, the average velocity becomes instantaneous velocity.

But lim \(\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{d x}{d t}\)
\Instantaneous velocity,

Here \(\frac{d x}{d t}\) is the differential coefficient of x with respect to time. It is the rate of change of position with respect to time at an instant.

Class 11 Physics Notes Chapter 3 Question 8.
The table given below gives the value of \(\frac{\Delta x}{\Delta t}\) for Dt equal to 2s, 1s, 0.55, 0.1s and 0.01s centered at t = 4 sec. (See figure given above). What conclusions can be made from this table?
Answer:
The value of average velocity \(\left(\frac{\Delta x}{\Delta t}\right)\) becomes instantaneous velocity (3.8 m/s), in the limit of Dt goes to zero, (ie Dt is infinitesimally small).

Class 11 Physics Chapter 3 Notes Pdf Download Question 9.
The position of an object moving along x-axis is given by x = a + bt² where a = 8.5m, b = 2.5 m/s² and t is measured in seconds

1. What is the velocity at t = 0s and t = 2s.
2. What is the average velocity between t = 2s and t = 4s?

Answer:
1.

when t = 0
we get v = 2 × 2.5 × 0
v = 0
when t = 2sec
v = 2 × 2.5 × 2 v = 10m/s.

2. The average velocity

Note: If a body is moving with constant velocity, the average velocity is the same as instantaneous velocity at all instants.

2. Instantaneous Speed:
The speed at an instant is called instantaneous speed.
Note:

• The average speed over a finite interval of time is greater or equal to the magnitude of the average velocity.
• Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant.

Motion In Straight Line Notes Pdf Acceleration
1. Average Acceleration:
Average acceleration of a particle is ratio of the change in velocity to the time interval.

Explanation
Consider a body moving along a straight line. Let v₁ and v₂ be the instantaneous velocities at time t₁ and t₂ respectively.


where Dv = change in velocity, Dt = Time interval

2. Instantaneous Acceleration:
Acceleration at any instant is called instantaneous acceleration.
Explanation
In the limit Dt ® 0, (Dt goes to zero) the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Instantaneous acceleration is the rate of change of velocity with respect to time.

3. Uniform Acceleration:
A body is said to be in uniform acceleration, if velocity changes equally in equal intervals of time.

Motion In A Straight Line Class 11 Pdf Question 10.
The velocities of two bodies A and B are given in the tables. From this table, find which body is moving with uniform acceleration. Explain.
Body A

Body B

Answer:
The body A is moving with uniform acceleration be-cause the velocity of body increases at the rate of 2 m/s².
The body B is moving with constant velocity. Hence this motion is called uniform motion.

4. Velocity-Time Graph For Uniformly Accelerated Motion:

An example for velocity-time of a uniformly accelerated motion is given in the above figure.
Let v_t1 and v_t2 be the velocities at instants t₁ and t₂respectively.
The slope of graph in the interval (t₂ – t₁) can be written as,

∴ tan q = acceleration
Thus the slope of the velocity-time gives the acceleration of the particle.

Motion In A Straight Line Notes Pdf Question 11.
Velocity-time of a body is given below. From this graph draw corresponding acceleration time graph.

Answer:
The slope of velocity-time graph increases in the interval (0 – 10) sec which means that acceleration of the body increases in this interval.

Velocity is constant in the interval (10 – 18) sec. Hence ’ the slope is zero which means that acceleration is zero in this range.

The slope in the interval (18 – 20) sec is constant and negative. Hence acceleration in this is a negative value. The acceleration – time graph for the above motion is given below.

Motion In Straight Line Class 11 Notes Question 12.
The position-time graph of a car is given below.

1. Draw corresponding velocity-time graph. Explain the reason for your answer.
2. From velocity-time graph draw acceleration-time graph and identify the regions of

• positive acceleration
• Negative acceleration
• zero acceleration.

Answer:
1. In the time interval (0 – t₁) sec, the slope of x – t graph increases which means that velocity is increasing in this time interval.

In the time interval (t₁ – t₂) sec, slope is constant. Hence velocity remains constant in this time interval.

In the time interval (t₂ – t₃) sec, the slope is decreasing and finally becomes zero. Which means that velocity decreases to zero.

2. Slope is constant throughout the interval (0 – t₁) sec which means that acceleration constant.

In the interval (t₁ – t₂) sec, slope is zero. Which means that acceleration is zero in this region.

Slope is constant (but negative) in the interval (t₂ – t₃)sec. Hence acceleration is constant and negative in this time interval.

Motion In A Straight Line Class 11 Notes Pdf Download Question 13.
Find the region of

1. positive acceleration
2. zero acceleration
3. negative acceleration from the above x-t graph

Answer:

1. Region OA – Positive acceleration
2. Region AB – zero acceleration
3. Region BC – Negative acceleration

Straight Line Class 11 Notes Question 14.
Match the following.

Answer:
1) – d, 2) – c, 3) – b, 4) – a.

5. Area Under Velocity-Time Graph:
Area under velocity-time graph represents the displacement over a given time interval.
Explanation
Consider a body moving with constant velocity v. Its velocity-time graph is given below.

The area of the rectangle has height v and bast t. Therefore,

Note: The acceleration and velocity of a body cannot change values abruptly at an instant. Changes are always continuous.

Kinematic Equations For Uniformly Accelerated Motion
For uniformly accelerated motion, we can derive some simple equations.

1. Velocity-time relation
2. Position-time relation
3. Position-velocity relation

These equations are called kinematic equations for uniformly accelerated motion.
1. Velocity-Time Relation:

Consider a body moving along a straight line with uniform acceleration ‘a’. Let ‘u’ be initial velocity and ‘v ‘ be the final velocity at time t.
We know acceleration a = \(\frac{\text { Change in velocity }}{\text { Time interval }}\)
a = \(\frac{v-u}{t}\)
at = v – u

2. Position-Time Relation:

Consider a body moving along a straight line with uniform acceleration a. Let ‘u’ be initial velocity and ‘v’ be the final velocity. ‘S’ is the displacement travelled by the body during the time interval ‘t‘.
Displacement of the body during the time interval t,
S = average velocity × time
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at ____(2)
Substitute eq.(2) in eq.(1), we get

3. Position-Velocity Relation:
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at
\(\frac{v-u}{a}\) = t _____(2)
Substitute eq.(2) in eq.(1)

2as = v² – u²
v² – u² = 2as

Free-fall:
An object released (near the surface of earth) is accelerated towards the earth. If air resistance is neglected, the object is said to be in free fall. The acceleration due to gravity near the surface of earth is 9.8 m/s².
Note: Free-fall is a case of motion with uniform acceleration.

Question 15.
A body is allowed to fall freely. Draw the following graph.

1. Acceleration-time
2. Velocity-time
3. Position-time

Answer:
1.

2.

3.

Stopping distance of vehicles:
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.

Question 16.
Derive an expression for stopping distance of a vehicle in terms of initial velocity (u) and retardation (a).
Answer:
Let the distance travelled by the vehicle before it stops be ‘s’.
Then we can find ‘s’ using the formula
v² = u² + 2as
0 = u² + -2as

3.7 Relative Velocity
Suppose the distance between two bodies changes with time in magnitude, or in direction or in both. Then each body is said to have a velocity relative to the other.

For example, consider two cars A and B moving in the same direction with equal velocities. To a person in A, the car B would appear to be rest.

Hence the velocity of B relative to A is zero.
ie. V_BA = 0
Similarly, the velocity of A with respect to B is zero.
or V_AB = 0
Let A be moving with a velocity V_A and B be moving with a greater velocity V_B in the same direction. Then the person in A feels that the car B is moving away from him with a velocity V_BA. The velocity of B relative to A

For an observer in B, the car A is going back with a velocity. The velocity of A relative to B
V_AB = -(V_B – V_B).

Question 17.
The position-time graph of two bodies A and B (at different situations) are given in the following graphs. Find the relative velocities of the following graph.

Answer:
a) The slope of Aand B are equal. Hence velocity of A and B are equal. So velocity of A with respect to B, V_AB = 0

b) The body A and B meet at t = 3sec

Velocity of A w.r. to B, V_AB = V_A – V_B
= 20-10 = 10 m/s Velocity of B w.r. to A, V_BA = V_B – V_A
= 10 – 20 = -10 m/s

c) The body A and B meet at t = 1 sec.
The velocity of body in the interval t = 1 sec,

Velocity of A w. r. to B,
V_AB = V_A – V_B
= 20 – 10 = 30 m/s
Similarly velocity of B w.r. to A,
V_BA = V_B – V_A
= 10 – +20 = -30 m/s
The magnitude of V_BA or V_AB (=30 m/s) is greater than the magnitude of velocity A or that of B.

Students can Download Chapter 1 Sets Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 1 Sets

Plus One Maths Sets Three Mark Questions and Answers

Plus One Maths Chapter Wise Previous Questions And Answers Question 1.
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical A, 50 to chemical B and 30 to both chemical A and B, Find the number of individuals exposed to

1. Chemical A but not chemical B. (1)
2. Chemical B but not.chemical C. (1)
3. Chemical A or chemical B. (1)

Answer:
1. Given; n(U) = 200; n(A) = 120;
n(B) = 50; n(A∩B) = 30
n (Chemical A but not chemical B)
= n(A ∩ B’) = n(A) – n(A ∩ B) = 120 – 30 = 90

2. n (Chemical B but not chemical A)
= n(A’ ∩ B) = n(B) – n(A ∩ B) = 50 – 30 = 20

3. n (Chemical A or chemical B)
= n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 120 + 50 – 30 = 140.

Plus One Maths Sets Previous Questions And Answers Question 2.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer:
Let A – Apple juice; O – Orange juice be the sets.
Given; n(U) = 400; n(A) = 100;
n(O) = 150; n(A ∩ O) = 75
n (neither apple juice nor orange juice)
= n(A’ ∩ O’) = n((A ∪ O)’)
= n(U) – n(A ∪ O)
= 400 – [n(A) + n(O) – n(A ∩ O)]
= 400 – [100 + 150 – 75] = 400 – 175 = 225.

Plus One Maths Sets Questions And Answers Pdf Download Question 3.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speaks at least one of these two languages?
Answer:
Let F – French; S – Spanish be the sets.
Given; n(F) = 50; n(S) = 20;w(F ∩ S) = 10
n (speaks at least one of these two languages)
= n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
= 50 + 20 – 10 = 60.

Plus One Maths Textbook Questions And Answers Question 4.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis, how many like tennis only and not cricket? How many like tennis?
Answer:
Let C – Cricket; T – Tennis be the sets.
Given;
n(C’ ∪ T) = 65; n(C) = 40; n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) – 10 ⇒ n(T) = 35
n (tennis only and not cricket)
= n(T ∪ C’) = n(T) – n(T ∩ C) = 35 – 10 = 25.

Plus One Maths Chapter Wise Questions And Answers Pdf Question 5.
Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find

1. n(B) (1)
2. n(B – A) (1)
3. n(A – B) (1)

Answer:

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
   ⇒ 42 = 20 + n(B) – 4 ⇒ n(B) = 26
2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22
3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

Plus One Maths Chapter Wise Questions And Answers Question 6.
A = {x: x is a natural number less than 8}

1. Write in roster form. (1)
2. Write a subset of A containing all even numbers in A. (1)
3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]? (1)

Answer:

1. A = {1, 2, 3, 4, 5, 6, 7}
2. {2, 4, 6} or {2, 4, 6, 7}
3. 10. (since other are powers of 2.)

Plus One Maths Sets Four Mark Questions and Answers

Plus One Maths First Chapter Questions And Answers Question 1.
Observe the Venn diagram.

1. Write in roster form. (1)
2. Verify that (A – B) ∪ (A ∩ B) = A (2)
3. Find (A ∩ B)’ (1)

Answer:

1. A = {1, 3, 4, 8} ; B = {2, 3, 5}
2. A – B = {1, 4, 8}; A ∩ B = {3}
   ⇒ (A – B) ∪ (A ∩ B) = {1, 3, 4, 8}
   Hence; (A – B) ∪ (A ∩ B) = A
3. (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 8, 9}

Plus One Maths Sets Practice Problems Questions and Answers

Plus One Maths Chapter Wise Questions Question 1.
Write the following sets in roster form.

1. A = {x: x xis an integer and -3 < x < 7}
2. B = {x: x ∈ N; x ≤ 6}
3. C = {x : x is a vowel in English alphabet}
4. D = {x : x is a two-digit natural number such that the sum of its digits is 8}
5. E = {x: x ∈ Z; \(-\frac{1}{2}<x<\frac{9}{2}\)}

Answer:

1. A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
2. B = {1, 2, 3, 4, 5, 6}
3. C = {a, e, i, o, u}
4. D = {17, 71, 26, 62, 35, 53}
5. E = {1, 2, 3, 4}

Plus One Maths Text Book Questions And Answers Question 2.
Write the following sets in Set builder form.

1. A = {3, 6, 9, 12}
2. B = {2, 4, 8, 16, 32}
3. C = \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}\)
4. D = {5, 25, 125, 625}
5. E = {2, 4, 6……..} (1 score each)

Answer:

1. A = {x: x = 3n, n ∈ N, n< 4 }
2. B = {x: x = 2ⁿ; n ∈ N; n < 5 }
3. C = {x: x = \(\frac{n}{n+1}\); n ∈ N, n ≤ 6}
4. D = {x: x = 5ⁿ; n ∈ N, n ≤ 4 }
5. E = {x: x is an even number}

Plus One Maths Chapter Wise Questions And Answers Pdf Hsslive Question 3.
Write the following in interval form.

1. {x: x∈ R, -4 < x ≤ 6}
2. {x: x∈ R, 0 ≤ x < 7 }
3. {x: x∈ R, 3 ≤ x ≤ 4 } (1 score each)

Answer:

1. (-4, 6]
2. [0, 7)
3. [3, 4]

Plus One Maths Sets Questions And Answers Pdf Question 4.
Write the following in set builder form.

1. [0, 10]
2. [-2, 7)
3. (3, 4)

Answer:

1. {x: x∈ R, 0 ≤ x ≤ 10 }
2. { x: x ∈ R, -2 ≤ x < 7 }
3. {x: x ∈ R, 3 < x < 4 }

Plus One Maths Questions And Answers Question 5.
Find Set A, B and Universal set U (1 score each)

Answer:
A = {e, f, d}; B = {a, b, c, d} and U = {a, b, c, d, e, f, g, h}

Hsslive Maths Textbook Answers Plus One Question 6.
Write all subset of the following

1. {1, 2}
2. {a, b, c}
3. Φ (1 score each)

Answer:

1. Φ, {1}, {2}, {1, 2}
2. Φ, {a}, {b}, {c} ,{a, b}, {a, c}, {b, c}, {a, b, c}
3. Φ

Plus One Mathematics Questions And Answers Question 7.
Let A = {1, 2, {3, 4}, s, d, θ} , Which of the following statements are true/false and why?

1. 3 ∈ A
2. {1, {3, 4}} ∈ A
3. {1, 2, 3} ⊂ A
4. Φ ∈ A
5. 1 ⊂ A (1 score each)

Answer:

1. False
2. True
3. False
4. False
5. False

Important Questions For Class 11 Maths Sets Question 8.
If A = {1, 2, 4, 6, 7, 8}; B = {2, 5, 7, 9, 10} and C = {4 , 5, 9, 10}. Find

1. A ∪ B
2. B ∪ C
3. A ∪ C
4. A ∩ B
5. B ∩ C
6. A∪ B ∪ C
7. A ∩ B ∩ C
8. (A ∩ B) ∪ (C ∩ A) (1 score each)

Answer:

1. A ∪ B = {1, 2, 4, 5, 6, 7, 8, 9, 10}
2. B ∪ C = {2, 4, 5, 7, 9, 10}
3. A ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
4. A ∩ B = {2, 7}
5. B ∩ C = {5, 9, 10}
6. A ∪ B ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
7. A ∩ B ∩ C = Φ
8. (A ∩ B) ∪ (C ∩ A) = {2, 7} u {4} = {2, 4, 7}

Plus One Maths Important Questions And Answers Question 9.
If A = {2, 4, 6, 7, 8, 12}; B = {2, 7, 9, 10} and C = {5, 9, 10, 12}. Find

1. A – B
2. B – C
3. A – C
4. B – A
5. C – A
6. (A ∪ B) – C
7. A – {B ∩ C)
8. (A ∩ B) – (C ∩ A) (1 score each)

Answer:

1. A – B = {4, 6, 8, 12}
2. B – C = {2, 7}
3. A – C = {2, 4, 6, 7, 8}
4. B – A = {9, 10}
5. C – A = {5, 9, 10}
6. (A ∪ B) – C = {2, 4, 6, 7, 8, 9, 10, 12} – {5, 9, 10, 12} = {2, 4, 6, 7, 8}.
7. A – (B ∩ C) = {2, 4, 6, 7, 8, 12} – {9, 10} = {2, 4, 6, 7, 8, 12}
8. (A ∩ B) – (C ∩ A) = {2, 7} – {12} = {2, 7}

Plus One Maths Previous Questions And Answers Question 10.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; A = {2, 7, 9, 10} B = {5, 9, 10, 12} and C = {1, 4, 5, 7, 11}. Find

1. A’
2. B’
3. A’ – C
4. (B – A)’
5. B’ ∩ C’
6. (A ∪ B)’
7. A’ ∩ B’ (1 score each)

Answer:

1. A’ = { 1, 3, 4, 5, 6, 8, 11, 12}
2. B’ = { 1, 2, 3, 4, 6, 7, 8, 11}
3. A’ – C = {1, 3, 4, 5, 6, 8, 11, 12} – {2, 3, 6, 8, 9, 10, 12} = {1, 4, 5, 11}
4. (B – A)’ = {5, 12}’ = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11}
5. B’ ∩ C’ = {1, 2, 3, 4, 6, 7, 8, 11} – {2, 3, 6, 8, 9, 10, 12} = {2, 3, 6, 8}
6. (A ∪ B)’ = {2, 5, 7, 9, 10, 12}’ = {1, 3, 4, 6, 8, 11}
7. A’ ∩ B’ = {1, 3, 4, 5, 6, 8, 11, 12} ∩ {1, 2, 3, 4, 6, 7, 8, 11} = {1, 3, 4, 6, 8, 11}

Plus One Maths Text Book Answers Question 11.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have?
Answer:
Given; n(X ∪ Y) = 50; n(X) = 28; n(Y) = 32
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 50 = 28 + 32 – n(X ∩ Y)
⇒ n(X ∩ Y) = 60 – 50 = 10

Students can Download Chapter 4 Motion in a Plane Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 4 Motion in a Plane

Summary
Motion In A Plane Class 11 Notes Pdf Introduction
In this chapter, we will study, about vector, its ’ addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.

Plus One Physics Chapter 4 Notes Scalars And Vectors

a. Scalars:
A quantity which has only magnitude and no direction is called a scalar quantity.
Eg: length; volume, mass, time, work etc.

b. Vectors:
(i) The need for vectors:
In one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.

(ii) Vector:
A physical quantity which has both magnitude and direction is called a vector quantity.
Eg: Displacement, Velocity, Acceleration, Force, momentum.

1. Position and Displacement Vectors:
Position vector:
Consider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.

A vector representing the position of an object P with respect to an origin O is called position vector \(\overrightarrow{\mathrm{OP}}\) of the object. This position vector may be represented
by an arrow with tail at O and head at P.

The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\(\overrightarrow{\mathrm{OP}}\)|.

Displacement vector:

Consider the motion of an object in a plane. Let P be the position of a moving object at a time t and p¹ that at a later time t¹. \(\overrightarrow{\mathrm{OP}}\) and \(\overrightarrow{\mathrm{OP}^{1}}\) are the position vectors at time t and t¹ respectively. So the vector \(\overrightarrow{\mathrm{PP}^{1}}\) is called displacement vector corresponding to the motion in the time interval (t – t¹).

2. Equality of vectors:
Two vectors are said to be equal if they have the same magnitude and direction.

The above figure shows two vectors \(\vec{A}\) and \(\vec{B}\) having the same magnitude and direction.
∴ \(\vec{A}\) = \(\vec{B}\).

Motion In A Plane Class 11 Notes Hsslive Question 1.
Observe the following figures (a) and (b) and find which pair does represents equal vectors?

Answer:
Figure a represent that A and B are equal vectors. Two vectors A¹ and B¹ are unequal, because they were in different directions.

Multiplication Of Vectors By Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number I gives a vector whose magnitude is changed by the factor λ.

The direction λ\(\vec{A}\) is the same as that of \(\vec{A}\).
Examples:

A vector \(\vec{A}\) and the resultant vector after multiplying \(\vec{A}\) by a positive number 2.

A vector A and resultant vector after multiplying it by a negative number-1 and -1.5.

Addition And Subtraction Of Vectors – Graphical Method
Vectors representing physical quantities of the same dimensions can be added or subtracted. The sum of two or more vectors is known as their resultant.

1. When two vectors are acting in the same direction:

2. When two vectors act in opposite direction:
In this case, the angle between the vectors is 180°.

The resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.

3. When two vectors are inclined to each other:
The sum of two vectors inclined at an angle q can be obtained either by

• the law of triangle of vectors
• the parallelogram law of vectors

(i) Triangle method:
This law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

(ii) Parallelogram law of vector addition:
This law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

To find \(\vec{A}\) + \(\vec{B}\), we bring theirtails to a common origin Q as shown below.


The diagonal of parallelogram OQSP, gives the resultantof (\(\vec{R}\) = \(\vec{A}\) + \(\vec{B}\)) of two vectors \(\vec{A}\) and \(\vec{B}\).
Note: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.

4. Substraction of vectors:

To substract \(\vec{B}\) from \(\vec{A}\), reverse the direction of \(\vec{B}\).

Then add –\(\vec{B}\) with \(\vec{A}\) using parallelogram law or tri¬angle law.

The resultant of \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{R}\).
Null vector or zero vector:
A vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \(\vec{O}\). Since the magnitude is zero, we don’t have to specify its direction.
Properties of null vector:

Class 11 Physics Motion In A Plane Notes Pdf Question 2.
Explain a zero vector using an example.
Answer:
Suppose that an object which is at P at time t, moves to p¹ and then comes back to P. In this case displacement is a null vector.

Resolution Of Vectors Unit Vectors
A vector divided by its magnitude is called unit vector along the direction of that vector. A unit vector in the direction of \(\vec{A}\) is written as \(\hat{A}\).

Orthogonal unit vectors:

In the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) respectively and are known as orthogonal unit vectors.
For unit vectors

Resolution of vector into rectangular components:
The components of a vector in two mutually perpendicular directions are called its rectangular components.
Explanation

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in x-y plane as shown in figure. To resolve \(\overrightarrow{\mathrm{A}}\), draw lines from the head of \(\overrightarrow{\mathrm{A}}\) perpendicularto the coordinate axes as shown below.

The quantities A_x and A_y are called x and y components of the vector \(\overrightarrow{\mathrm{A}}\). Hence the vector \(\overrightarrow{\mathrm{A}}\) can be written in terms of rectangular components as

Magnitude of \(\overrightarrow{\mathrm{A}}\):

From the figure, the magnitude of \(\overrightarrow{\mathrm{A}}\) can be written as,

Vectors Physics Class 11 Notes Pdf Question 3.
A vector \(\overrightarrow{\mathrm{A}}\) in xyz plane is given below. A_x, A_y and A_z are the perpendicular components in x,y and z directions respectively.

1. Write \(\overrightarrow{\mathrm{A}}\) in terms of rectangular components.
2. Write the magnitude of \(\overrightarrow{\mathrm{A}}\).

Answer:

The magnitude of vector \(\overrightarrow{\mathrm{A}}\) is

Vector Addition – Analytical Method
The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use analytical method to add vectors.
Explanation

The vectors obey commutative and associative laws. Hence

Projectile Motion Class 11 Notes Question 4.
Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{A}}\) in terms of their magnitudes and angle between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle q. Using the parallelogram method of
vectors, the resultant vector \(\overrightarrow{\mathrm{R}}\) can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosq and SN = Bsinq
Substituting these values in eq.(1), we get
OS² = (OP + Bcosq)² + (Bsinq)²
But OS = R and OP = A
R² = (A + Bcosq)² + B²sin²q
= A² + 2ABcosq + B²cos²q + B²sin²q
R² = A² + 2 ABcosq + B²

The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \overrightarrow{\mathrm{A}}. From the right angled triangle OSN,

But SN = Bsinq PN = Bcosq

Motion In A Plane

1. Position vector and displacement vector Position vector:

Consider a small body located at P with reference to the origin O. The position vector of the point ‘P’

Displacement vector


where Dx = x¹ – x¹, Dy = y¹ – y
Velocity:
If Dt is the time taken to reach from P to P¹
The average velocity, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\Delta r}}{\Delta \mathrm{t}}\) ____(3)
Substitute eq.(2) in eq.(3), we get

The direction of average velocity is the same as that of \(\overrightarrow{\Delta r}\).
The instantaneous velocity can be written as

Acceleration:
If the velocity of an object changes from \(\overrightarrow{\mathrm{v}} \text { to } \overrightarrow{\mathrm{v}^{1}}\) in time Dt, then its average acceleration is given by

Instantaneous acceleration:
The acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Motion In A Plane With Constant Acceleration
Consider an object moving in xy plane with constant acceleration ‘a’. Let \(\vec{u}\) be the initial velocity at t=0 and \(\vec{v}\) be the final velocity at time t.
Then by definition acceleration

In terms of components
v_x = u_x + a_xt
v_y = u_y + a_yt
Displacement in a plane
If \(\overrightarrow{\mathrm{r}_{0}}\) and \(\vec{r}\) be position vectors of particle at t = 0 and time t respectively, then
displacement = \(\vec{r}-\vec{r}_{0}\) _______(1)
For uniformly accelerated motion, displacement,

In terms of components
x = x₀ + u_xt + 1/2 a_xt²
y = y₀ + u_yt + 1/2 a_yt²
The eq.(2) shows that, the above motion in xy plane can be treated as two separate one dimensional motions along two perpendicular directions.

Relative Velocity In Two Dimensions
Consider two bodies A and B moving along a plane with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\). Then velocity of A relative to that of B is,

Similarly velocity of B relative to that of A

Projectile Motion
Projectile:
A body is projected into air and is allowed to move under the influence of gravity is called projectile.

Consider a body which is projected into air with a velocity u at an angle q. The initial velocity ‘u’ can be divided into two components ucosq along horizontal direction and using along vertical direction.

1. Time of flight:
The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinq, we get
0 = usinqt – 1/2gt²
1/2 gt² = usinqt

2. Vertical height:
Vertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v² = u² + 2as
When we substitute v=0, a = -g, s = H and u = usinq, we get
0 = (usinq)² + 2 – g × H
2gH = u²sin²q

3. Horizontal Range:
If we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

The eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q₀ = 45°.
The maximum horizontal range

Equation for path of projectile

Motion In A Plane Class 11 Pdf Question 5.
What is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
Answer:
S = ut+ 1/2at²
y = usinqt – 1/2gt²
But we know horizontal displacement, x = ucosq × t

In this equation g, q and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

Uniform Circular Motion
The motion of an object along the circumference of a circle is called circular motion.
Uniform circular motion:
When an object follows a circular path at a constant speed, the motion is called uniform circular motion.
Period:
The time taken by the object to complete one full revolution is called the period.
Frequency:
The number of revolutions completed per second is called the frequency u of the circular motion.
If the period of a circular motion isT, its frequency

Angular Displacement (Dq):
The angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
Angular velocity:
The rate of change of angular displacement is called the angular velocity.

If T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
Therefore in one second it sweeps out an angle \(\frac{2 \pi}{T}\).
∴ Angular velocity of the object

Expression for velocity and acceleration in uniform circular motion:

The direction of velocity is in the direction of tangent at that point. The change in velocity vectors \((\overrightarrow{\Delta v})\) is obtained by triangle law of vector as shown in figure (b).

a. Speed and angular speed in uniform circular motion:
Let the Dq be the angle constructed by the body during the time interval ∆t. The angular velocity can be written as

If the distance travelled by the object during the time Dt is Dr (ie. PP¹ = Ds) then speed

But Ds = RDq
where R = \(|\vec{r}|=|\overrightarrow{r^{\prime}}|\)
Substituting Dr = RDq in eq.(1)
we get

b. Acceleration in uniform circular motion:

The direction of this acceleration should be in the direction of \(\overrightarrow{\Delta V}\). The fig(b) shows that \(\overrightarrow{\Delta V}\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration.

The force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as
F = ma_c

But ω = \(\frac{V}{R}\). Hence we get K