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Students can Download Chapter 10 Internal Trade 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 10 Internal Trade

Contents

• Internal Trade – Meaning – Types
• Whole Sale Trade – Meaning – Function – Services to producers and retailers
• Retail Trade – Meaning – functions – services to producers/wholesalers and consumers
• Types of Retail trade – Itinerant Traders – Fixed shop retailers
• Departmental Store – Meaning – Features Advantages – Limitations
• Multiple shop – Meaning – Features – Advantages Limitations
• Mail order business – Meaning – Features Advantages – Limitations
• Consumer co – operative store – Meaning Features – Advantages – Limitations
• Supermarket – Meaning – Features – Advantages Limitations
• Vending Machines – Meaning
• Role of Chamber of Commerce and Industry

Internal Trade:
Buying and selling of goods and services within the boundaries of a nation are called internal trade. Internal trade can be classified into two broad categories viz.

• Wholesale trade
• Retail trade

Wholesale Trade:
Wholesale trade means buying goods in large quantities from the producers and selling them in smaller quantities to the retailers. Wholesalers acts as an important link between manufacturers and retailers.
Functions of Wholesaler:
1. Buying and assembling:
The wholesalers buy goods from different producers and keep them in a central place.

2. Warehousing:
The goods are to be kept in the I warehouses till they are sold to retailers.

3. Grading and packing:
The goods purchased are sorted out on the basis of quality and size. This is called grading. After grading they are packed in attractive packages

4. Pricing:
The wholesaler fix the price of products

5. Transportation:
The wholesalers move the goods from the production centre to the retail shop

6. Risk bearing:
They assume the risk like change in demand, spoilage, theft during transportation etc.

7. Financing:
Wholesalers purchase goods on cash basis from manufacturers and sell them to the retailers on credit basis.

8. Market information:
Wholesalers collect various market information for the benefits of manufacturers so that they can change the products accordingly

Services of Wholesalers to Manufacturers:
1. Facilitating large scale production:
As the wholesalers place bulk orders, the producers are able to undertake production on a large scale and take advantages of economies of scale.

2. Risk bearing:
Wholesaler deals in goods in their own name and bear variety of risks such as the risk of fall in prices, theft, pilferage spoilage, fire etc.

3. Financial assistance:
Wholesalers provide financial assistance to the manufacturers by making cash payment for the purchased goods.

4. Expert advice:
Wholesaler provide various useful information regarding the customer preference, market conditions etc to the manufacturer.

5. Help in marketing function:
As the wholesalers place bulk orders, it relieves the producer from many marketing activities and he can concentrate on production.

6. Storage facilities:
Wholesalers hold the goods in their own warehouses. It reduces the burden of storage of goods by the manufacturers.

7. Facilitate production continuity:
The wholesalers facilitate continuity of production activity throughout the year by purchasing the goods as and when these are produced

Services of Wholesalers to Retailers:
1. Availability of goods:
The wholesalers make the products of various manufacturers readily available to the retailers.

2. Marketing support:
They undertake advertisements and other sales promotional activities to induce customers to purchase the goods.

3. Grant of credit:
The wholesalers generally provide credit facilities to the retailers.

4. Specialised knowledge:
Wholesalers know the pulse of the market. They inform the retailers about the new products, their uses, quality, prices, etc.

5. Risk sharing:
Wholesalers sell goods to retailers in small quantities and thus retailers do not face the risk of storage, pilferage, reduction in prices etc.

Retail Trade:
Buying of goods in large quantities from the wholesalers and selling them in small quantities to the ultimate consumers is known as retail trade. Retailers serve as an important link between the producers and consumers in the distribution of products and services.
Functions of Retailers:

1. A retailer collects different varieties of goods. So he can satisfy different types of customers.
2. A retailer provides market information to wholesalers and manufacturers.
3. A retailer is in close contact with consumers. So he can persuade the consumers to buy the product.
4. Retailers locate their business in residential areas. It helps the consumers to purchase the product easily.
5. The retailers provide credit facilities to the consumers.
6. Retailers provide after sales services to attract consumers.

Services of Retailers to Manufacturers and Wholesalers:
1. Retailers help manufacturers & wholesalers in the distribution of their goods & services to the ultimate consumers.

2. Retailers undertake personal selling efforts and thus, help manufacturers and wholesalers to increase the sale of the products.

3. Retailers help manufacturers and wholesalers to operate production on a large scale by undertaking distribution of goods.

4. As retailers are in constant touch with customers, they can provide various market information such as the tastes, preferences and attitudes etc. of consumers to the producers.

5. Retailers participate in various sales promotional activities conducted by producers and wholesalers.

Services Retailers to Consumers:

1. Retailers provide goods to consumers according to their requirements.
2. Retailers keep large varieties of products of different manufacturers. It enable the customers to select goods according to their choice.
3. Retailers provide important information about the new products to the consumers.
4. Retailers also provide after sales services in the form of home delivery etc. to the customers.
5. Retailers often supply goods on credit to the customers.
6. Retailers keep ready stock of the products needed by the consumers.

Differences between wholesaler and Retailer:

Types of Retail Trade:
Retail trade can be classified into following two categories on the basis whether or not they have a fixed place of business.

• Itinerant Retailers
• Fixed shop Retailers

Itinerant Retailers:
The retailers who do not have a fixed place of business to operate from are called itinerant retailers. They have to move from one place to another along with their goods in search of consumers.
Characteristics of itinerant retailers:

1. They are small traders having limited resources.
2. They generally deal in consumer products of daily use.
3. They emphasize on providing greater customer services.
4. They do not have any fixed place to operate from.

Types of itinerant retailers:
1. Peddlers and hawkers:
They cany the products on a bicycle, a hand cart, a cycle-rickshaw or on their heads, and move from place to place to sell their products at the doorstep of the customers. They generally deal in non-standardised and low- value products such as toys, vegetables, fruits etc.

2. Market traders:
They are the small retailer who open their shops at different places and sell the goods on fixed days such as every Saturday or Tuesday. These trader deals in single line of goods such as toys, readymade garment crockery, etc.

3. Street traders:
These traders display their articles on busy streets, bus stand, railway stations etc. They sell low priced articles like pen, books, magazines, hand kerchiefs etc. They do not change their place of business frequently.

4. Cheap jacks:
They are small retailers who have independent shops of a temporary nature in a business locality. They keep on changing their business from one locality to another but not very frequently. They deal in consumer items such as repair of watches, shoes, buckets etc.

Fixed Shop Retailers:
Retailers who maintain permanent establishment to sell their goods are called fixed shop retailers. Following are the main characteristics of fixed shop retailers.

1. They have greater resources and operate on a relatively large scale.
2. They deal in durable as well as non-durable goods.
3. They provide greater services to the customers such as home delivery, repairs, credit facilities etc.

Types of Fixed Shop Retailers:
The fixed-shop retailers can be classified into two. They are:

• Small shop-keepers
• Large retailers

Fixed Shop Small Retailers:
1. General stores:
These shops carry stock of a variety of products required to satisfy the day-to-day needs of the consumers residing in nearby localities. They deal products of daily use such as grocery items, soft drinks, Stationary etc.

2. Speciality shops:
These retail stores specialise in the sale of a specific line of products. For example, shops selling children’s garments, men’s wear, ladies shoes, toys and gifts, school uniforms etc. The speciality shops are generally located in a central place where a large number of customers can be attracted.

3. Street stall holders:
These shops are generally located at street crossings or in the main street. They usually display their goods on a table or by fixing shelf on the wall. Low priced articles such as pen, cosmetics, magazines etc. are sold in these stalls. .

4. Secondhand goods shop:
These shops deal in secondhand goods such as books, clothes, furniture, automobile etc. People who cannot afford to buy new articles, generally, becomes their customers.

Fixed shop large retailers:
Departmental stores:
A departmental store is a large scale retail shop selling a wide variety of goods in different departments under one and management. Each department deals in separate line of goods like stationary, books, furniture, clothing etc. Consumers can purchase all goods from the departmental store.
Features of a departmental store:

1. It Is a large scale retail organization.
2. A number of retail shops in the same building.
3. It offers a wide variety of products under one roof.
4. It is located at central places of the city
5. The products are arranged in separate departments
6. Sales, control and management are centralized
7. It offers various services and facilities like free home delivery etc

Advantages:

1. Central locations: As these stores are usually located at central places they attract a large number of customers
2. Convenience in buying: By offering large variety of goods under one roof, the departmental stores provide great convenience to customers in buying almost all goods of their requirements at one place.
3. Attractive services: A departmental store aims at providing maximum services to the customers.
4. Economy of large-scale operations: As these stores are organised in a very large-scale, the benefits of large-scale operations are available to them
5. Mutual advertisement:-All the departments are under one roof, so there is economy in advertising
6. Risk distribution: If there is a loss in one department, it may be compensated from the profit of other departments
7. Increased sales: Central location, mutual advertisement etc. will help a departmental store to increase its sales.

Limitations:
1. Lack of personal attention:
Because of the large- scale operations, it is very difficult to provide adequate personal attention to the customers in these stores.

2. Inconvenient location:
As a departmental store is generally situated at a central location, it is not convenient for the consumers who reside away from town.

3. High price:
A departmental store charges high price for the products because of high operating cost.

4. High operating cost:
As these stores give more emphasis on providing services, their operating costs tend to be high.

5. High advertisement cost:
The success and prosperity of a departmental store depends on advertisement. Therefore, it should spent large amount on advertisement.

6. Lack of effective control:
Departmental store works through a large number of departments. It creates so many problems.

7. High risk:
A departmental store keeps a large stock of goods. So changes in fashion, taste, price etc will affect the profitability of the business.

Chain Stores or Multiple Shops:
Multiple shop is a system of branch shops operated under a centralised management and dealing in similar line of goods. Branches are located through out the nation.
Features of multiple shops:

1. It deals in one or two lines of products.
2. All branches are dealing in similar goods
3. It has centralized management and unified system of control
4. It eliminates middlemen.
5. It works on cash and carry principle
6. It has centralized buying and decentralized selling.
7. There is uniformity in operation in all branches.
8. It deals in goods of daily use and durables.

Advantages:

1. It enjoys economies of bulk purchase because the goods for all branches are purchased by head office.
2. There is no risk of bad debts because all sales are on cash basis.
3. The advertisements for all branches are done by the head office. So there is economy in advertisement.
4. Multiple shops are located in towns and cities. They attract a large number of customers.
5. All branches of multiple shops are uniform in style, design and display of goods.
6. All the branches sell quality goods at uniform prices. It creates public confidence.
7. The economy in large scale buying, centralized management, etc. reduce the cost of operations.
8. Products having no demand in one branch can be transferred to another branch. It reduces business risk.
9. Multiple shops enjoy the benefits of quick turn over because of country wide location.

Limitations:

1. The multiple shops deal only in a limited range of products. So consumers have very little choice.
2. They will not provide any credit facilities to consumers.
3. There is lack of personal touch between the company and consumers because branches are managed by salaried managers.
4. Branch manager is only a salaried employee. He has no initiative to increase the profits.
5. As these shops deal in a limited line of goods, fall in demand will affect the business.

Difference between Departmental stores and Multiple shops:

Mail Order Houses/shopping by Post:
Mail order business is a form of retailing where the business transactions are done through post or mail. Under this system orders for goods, delivery of goods and payment is made through VPP (Value Payable Post).

Under this arrangement, the goods are delivered to the customers only on making full payment for the same. There is generally no direct personal contact between the buyers and the sellers in this type of trading.
This type of business is suitable for products that can be:

1. graded and standardised
2. easily transported at low cost
3. have ready demand in the market
4. are available in large quantity throughout the year
5. involve least possible competition in the market
6. can be described through pictures.

Bulky, heavy and perishable products are not suitable for this type of business.
Advantages:

1. It needs only limited capital because there is no need of building and other infra structural facilities.
2. Unnecessary middlemen between the buyers and sellers are eliminated.
3. Since the mail order houses do not extend credit facilities to the customers, there are no chances of any bad debt.
4. Under this system goods are delivered at the doorstep of the customers. This results in great convenience to the customers.
5. There is a wider scope for business.
6. It helps to avoid overstocking of goods as goods are collected only when the orders are received.

Disadvantages:

1. It has to spend a large amount for advertisement.
2. There is no direct personal contact between the buyer and the seller.
3. The buyer qannot inspect the goods personally before purchasing.
4. They are not suitable for heavy and perishable goods.
5. They do not provide credit facilities to customers.
6. There may be delay in getting goods.

Consumer Cooperative Store:
It is a retail store formed by the consumers on the basis of principles of co-operation. Thses stores are owned and managed by consumers. They deal all types of consumer goods.
Features of Consumers Co-operative:

1. It is a voluntary association of consumers.
2. The liability of members is limited.
3. Consumers can purchase quality goods at lowest cost from these stores.
4. Democratic control is exercised.
5. It eliminates middlemen

Advantages:

1. Consumers can purchase quality goods at lowest cost from consumers co-operative store.
2. There is no bad debts as goods are sold on cash basis.
3. Economies of large scale purchasing can be enjoyed.
4. Less advertisement expenses are required.
5. It restricts monopoly and wasteful competition.

Disadvantages:

1. A consumer co-operative store can collect low capital. So they cannot start business on a large scale.
2. The management of a consumer co-operative store is inefficient.
3. It lacks proper warehousing facilities.
4. It will not attract consumers because of no credit facilities.

Supermarkets/Super Bazaar:
Supermarket is a large scale retail organisation selling a wide variety of consumer goods. The important feature of supermarket is the absence of salesman to help consumers in selecting goods. Hence supermarket is also called ‘Self Service Store’.
Features of Super Market:

1. They are located at the centre of a town.
2. They sell goods on cash basis only.
3. They deal wide variety of goods.
4. There is no salesman to help consumers

Advantages:

1. Consumers can purchase everything from supermarket
2. There is no bad debt as sales are on cash basis only.
3. They are located at the centre of a town.
4. It attracts a large number of consumers.
5. Consumers can select goods according to their taste and preferences.
6. Variety of goods is available in a supermarket.

Disadvantages:

1. Large premises at central location is not available easily.
2. It lacks personal advice of salesman
3. They do not provide credit facilities to customers.
4. The employees in a supermarket do not take initiative to increase sales.
5. It requires huge capital investment.
6. It is not suitable for products which require personal selling.
7. There is no personal contact with consumers.

Vending Machines:
They are coin operated machines which are used in selling several products such as milk, soft drinks, chocolates, platform tickets etc in many countries. The latest area in which this concept is getting popular is the cpse of Automated Teller Machines (ATM) in the banking service.

However, the installation cost and expenditure on regular maintenance and repair of these machines are quite high. Moreover, the consumers can neither see the product before buying nor can return the unwanted goods.

Role of Chambers of Commerce and Industry in Promotion of Internal Trade:
Association of business and industrial houses are formed to promote and protect their common interest and goals. They undertake following functions.

1. The chamber of commerce and Industry help in the inter-state movement of goods through various activities.
2. They ensure that imposition of octroi and other local taxes do not affect trade adversely.
3. They also undertake marketing of agro products and related issues.
4. They interact with the Government to make laws relating to weights and measures and protection of brands.
5. They discuss with government to get sound infrastructure so that business activities could be undertaken easily

Students can Download Chapter 9 Small Business 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 9 Small Business

Contents

• Small business – Meaning and Nature – administrative setup – Role of Small Business in India – Problems of Small Business
• Government assistance to small industries- Institutional support
• NABARD
• RSBDC
• NSIC
• SIDBI
• NCEUS
• RWED
• WASME
• SFURTI
• DIC
• Government assistance to small industries – Incentives

Types of Small business units in India:
On the basis of the capital invested, small business units can be categorized into various categories.
They are:
1. Small scale industry:
A small scale industrial undertaking is defined as one in which the investment in fixed assets of plant and machinery does not exceed rupees one crore.

2. Ancillary small industrial unit:
An ancillary industrial unit is one which supplies not less than 50% of its output to another parental unit. It can manufacture parts, components, subassemblies, tools or intermediate products for the parent unit. Also it can do business on its own.

3. Export oriented units:
Export-Oriented unit is one which exports more than 50% of its output and wherein investment in plant and machinery does not exceed rupees one core.

4. Small scale industries owned and managed by women entrepreneurs:
An enterprise promoted by women entrepreneurs is a small scale industrial unit in which she/they individually or jointly have share capital of not less than 51 percent. Such units gets concessions offered by the government, like low interest rates on loans, etc.

5. Tiny Industrial Unit:
A tiny unit is defined as an industrial or business enterprise whose investment in plant and machinery is not more than ₹ 25lakhs.

6. Small scale service and business enterprise:
It is one whose investment in fixed assets of plant and machinery excluding land and building does not exceed ₹ 10 lakhs.

7. Micro business enterprises:
Micro enterprises are those whose investment in plant and machinery does not exceed rupees one lakh.

8. Village industries:
Village Industry has been defined as means any industry located in a rural area which produces any goods, renders any service with or without the use of power and in which fixed capital investment per head does not exceeds ₹ 50,000.

9. Cottage industries:
cottage industries are characterised by certain features

• These are organised by individuals, with private resources;
• Normally use family labour and locally available talent;
• The equipment used is simple
• Capital investment is small
• Produce simple products, normally in their own premises;
• Production of goods using indigenous technology.

Administrative setup for the Small scale, Agro and Rural Industries:
For the promotion and development of small scale industries in India, Government created two ministries, i.e
1. Ministry of Small Scale Industries:
The Ministry of Small Scale Industries designs policies, programmes, and schemes for the promotion and growth of SSIs. The Small Industries Development Organisation (SIDO) is responsible for implementing and monitoring of various policies and programmes formulated.

2. Ministry of Agro and Rural Industries:
Ministry of Agro and Rural Industries is the nodal agency for coordination and development of Village and Khadi industries, tiny and micro enterprises in both urban and rural areas.

Various programs and policies are implemented by the ministry through the Khadi and Village Industries Commission (KVIC), Handicrafts Board, Coir Board, Silk Board etc.

State Governments also provide different promotional and developmental projects and schemes to provide a number of supporting incentives for development and promotion of SSIs. These are executed through the State Directorate of Industries and District Industries Centres (DICs).

Role of small business in India:

1. Small industries are labour intensive and less capital intensive. They generate more number of employment opportunities.
2. The share of product from small industries is 45% of total export from India. So it earn valuable foreign exchange
3. Small scale Industries produces a wide variety of goods e.g. readymade garments, stationery, soaps, leather goods, plastic and rubber goods.
4. The contribution of small industries to the balanced regional development of the country is very significant.
5. Small industries provide ample opportunity for entrepreneurship.
6. It enjoys the advantage of low cost of production because they used local resources in their product.
7. Due to the small size of the organisation, quick and timely decisions can be taken without consulting many people.
8. Small industries are best suited for the products which are designed according to the taste, needs and preferences of the customers.
9. Small industries maintain good personal relations with both customers and employees.

Role of small business in Rural India:

1. Cottage and rural industries provide employment opportunities in the rural areas especially for the traditional artisans and the weaker sections of society.
2. It prevents migration of rural population to urban areas in search of employment.
3. Small business helps to eradicate poverty, income inequalities etc in rural area.
4. Small scale industries are powerful instrument for the accelerated industrial growth and creating productive employment in rural and backward areas.

Problems of small business:
Small businesses are faced with the following problems.

1. Small scale industries find it difficult to get adequate finance from banks and other financial institutions.

2. They are not able to get quality raw materials at reasonable prices.

3. Small business is generally operated by people who may not have all the managerial skills required to run the business.

4. Small business firms cannot afford to pay higher salaries to the employees. So productivity per employee is relatively low and employee turnover is generally high.

5. Small business depends excessively on middlemen for marketing the products. Middlemen exploit them by paying low price and delayed payments.

6. Small business organization uses out dated technology to produce products. So they cannot compete with global enterprises with low quality products.

7. Due to lack of marketing skills or lack of demand, many small business firms have to operate below full capacity.

8. Use of outdated technology results low productivity and uneconomical production.

9. Small-scale units find it very difficult to compete .with the product of multinational companies which are comparatively very cheap and of better quality

10. Other important problems are poor project planning, inefficient management, transportation problems, lack of power, and lack of adequate warehousing, etc.

Government assistance to small Industries and Small Business:
A. Institutional Support:
1. National Bank for Agriculture and Rural Development (NABARD):
NABARD was set up in 1982 to promote integrated rural development.
Functions of NABARD:

• It provides financial support to small Industries, cottage and village industries and agriculture.
• It provides counseling and consultancy services.
• It also organises training and development programme for rural entrepreneurs.

2. The Rural Small Business Development Centre (RSBDC:
It aims at providing management and technical support to current and prospective micro and small entrepreneurs in rural areas. RSBDC has organized several programmes on rural entrepreneurship, skill upgradation workshops, training programmes etc.

3. National small Industries Corporation (NSIC):
This was set up in 1955 to promote, aid and foster the growth of small scale units in India.
Functions of NSIC:

• It supplies imported machines and raw materials to small scale industries on easy hire-purchase schemes.
• It exports the products of small units.
• It provides technology to small scale Industries.
• Helps in up-gradation of technology
• Provides advisory service
• Developing software technology parks and technology transfer centres.

4. Small Industries Development Bank of India (SIDBI):
SIDBI was set up in 1980. SIDBI is the main financial Institution for financing and development of small Business in India.
Functions of SIDBI.

• Helps SSI unit for modernisation and technology upgradation by providing loan
• Meet working capital requirements of SSI.
• Provides means forthe rehabilitation of sick units
• Provides service like leasing, hire purchase, venture capital financing etc.
• Provide equity support to small entrepreneurs.

5. The National Commission for Enterprises in the Unorganised Sector (NCEUS):
The NCEUS was constituted in September, 2004, with the following objectives.

• To recommend measures for improving the productivity of small enterprises
• To generate more employment opportunities in rural areas
• To enhance the competitiveness of the sector in the emerging global environment.
• To link the sector with other institutions in the areas of credit, raw materials, infrastructure, technology upgradation etc.

6. Rural and Women Entrepreneurship Development (RWED):
RWE provides the following services.

• Creating a business environment that encourages initiatives of rural and women entrepreneurs.
• Enhancing the human and institutional capacities required to foster entrepreneurial dynamism and enhance productivity.
• Providing training manuals for women entrepreneurs and training them.
• Rendering all types of advisory services.

7. World Association for Small and Medium Enterprises (WASME):
It is the only International Non-Governmental Organisation of micro, small and medium enterprises based in India, which set up an International Committee for Rural Industrialisation. Its aim is to develop an action plan model for sustained growth of rural enterprises.

There are several schemes to promote the non-farm sector initiated by the Government of India, i.e. IRDP, PMRY, TRYSEM, JRY, Development of Women and Children in Rural Areas (DWCRA) etc.

8. Scheme of Fund for Regeneration of Traditional Industries (SFURTI):
To make the traditional industries more productive and competitive, the Central Government set up a fund with ₹ 100 crores. The main objectives of the scheme are as follows:

• To develop clusters of traditional industries in various parts of the country
• To make traditional industries competitive, profitable and sustainable
• To create employment opportunities in traditional industries.

9. The District Industries Centers (DICs):
DICs were established in May 1978. District Industries Centers is the institution at the district level which provides all the services and support facilities to the entrepreneurs for setting up small and village industries.

Identification of suitable schemes, preparation of feasibility reports, arranging for credit, machinery and equipment, provision of raw materials and other extension services are the main activities undertaken by these centers.

B. Incentives:
The incentives offered by the government to develop backward areas are:

1. Availability of land at concessional rate.
2. Supply power at a concessional rate of 50% or exempt such units from payment in the initial years.
3. Water is supplied on a no-profit, no-loss basis or with 50 percent concession or exemption from water charges for a period of 5 years.
4. All union territories, industries are exempted from sales tax,
5. Most states have abolished octroi.
6. Units located in backward areas get scarce raw materials at concessional rates.
7. Subsidy of 10-15% for building capital assets. Loans are offered at concessional rates.
8. Some states encourage setting up of industrial estates in backward areas.
9. Exemption from paying taxes for 5 or 10 years is given to industries established in backward, hilly and tribal areas.

Students can Download Chapter 9 Mechanical Properties of Solids 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 9 Mechanical Properties of Solids

Summary
Introduction
Elasticity:
The property of a material by virtue of which it regains its original shape on the removal of the deforming force is called elasticity. The deformation caused is known as elastic deformation.

Plasticity:
The property of a material by virtue of which it does not regain its original shape on the removal of the deforming force is called plasticity. The substance which shows this property is called plastic.

Elastic Behavior Of Solids
In a solid, each atoms (or molecule) is surrounded by neighboring atoms (or molecules). These atoms are bonded by inter atomic forces. When a solid is deformed, the atoms are displaced from their equilibrium positions. When the deforming force is removed, the interatomic force bring back atoms into its original position. Thus body regains its original shape and size.

Stress And Strain Stress
It is the restoring force developed per unit area of cross section

S.I unit of stress is Nm² or Pascal (Pa) and its dimensional formula is [ML^-1 T^-2]. The restoring force is equal and opposite to the external applied force.
Strain:
The effect of stress on a body is called strain. Strain is measured as the ratio of the change in dimension produced to the original dimension.

Different types of Stress and Strain:
There are three ways in which a solid may change its dimensions when an external force acts on it. Hence there are three types of stress and corresponding strain.

1. Tensile Stress and Longitudinal Strain:
When the forces acting on the body produces an elongation along its length, then we call the stress as tensile stress (see figure).

When the forces acting on the body produces a compression along its length, then we call the stress as compressive stress. The tensile stress produces a change in the length (∆L). Hence the longitudinal strain can be written as
Longitudinal strain = \(\frac{\text { Change in length }}{\text { Original length }}\)
Longitudinal strain = \(\frac{\Delta \mathrm{L}}{\mathrm{L}}\)
Where ‘L’ is the original length.

2. Tangential (or) Shearing Stress and Shearing Strain:

If two equal and opposite deforming forces are applied parallel to cross-sectional area of the cylinder as shown in figure, there is a relative displacement between the opposite faces of the cylinder. The restoring force per unit area developed due to the applied tangential force is known as tangential or shearing stress.

As a result of applied tangential force, there is a relative displacement between opposite faces of the cylinder (see figure). The strain so produced is known a shearing strain. Shearing strain can be written as,

Where θ is the angular displacement of the cylinder from the vertical (original position of the cylinder). Since θ is very small, tanθ ≈ θ.

3. Hydraulic Stress and Volume Strain:

Consider a solid placed inside the fluid. The force applied by the fluid acts in perpendicular direction at each point of the surface. This leads to decrease in its volume without any change of its geometrical shape.

The internal restoring force per unit area in this case is known as hydraulic stress. The strain produced by a hydraulic pressure is called volume strain. The volume stress can be written as
Volume strain = \(\frac{\text { Change in volume }}{\text { Original volume }}=\frac{\Delta \mathrm{V}}{\mathrm{V}}\)
Note: Hydraulic Stress is equal to the hydraulic pressure.

Hooke’S Law
Hooke’s law:
For small deformations, the stress and strain are proportional to each other, ie: stress ∝ strain. Stress = k × strain Where k is the proportionality constant and is known as modulus of elasticity.

Stress Strain Curve

A typical graph for stress – strain graph for a metal is given in the figure.
Region OA:
In the region between O to A, the curve is linear. Hence in this region Hooke’s law is obeyed. The body regains its original dimensions when the applied force is removed. In this region the material behaves as an elastic body.

Region AB:
In this region stress and strain are not proportional. The material returns to its original dimension when the applied force (stress) is removed. The point B in the curve is known as yield point (also known as elastic limit) and corresponding stress is yield strength of the material.

Region BD:
If the applied force exceeds the yield strength, strain increases rapidly for a small change in the stress. When the applied force (stress) is removed at ‘C’, the body does not regain it’s original dimension, (even when the stress is zero, the strain is not zero).

This deformation is said to be plastic deformation and the material is said to have a permanent set. The point D on the graph is the ultimate tensile strength of the material.

Region DE:
Beyond this point D, additional strain is produced even by a reduced applied force and fracture occurs at the point E. The point E is called fracture point. If the ultimate strength (D) and fracture point (E) are close, the material is said to be brittle. If they are far apart, the material is said to be ductile.

Elastomers:
For rubber like materials, even for a small stress, the strain produced will be very high. Such materials are called elastomers.

There is no plastic flow region in the curves for elastomer materials. The elastic tissues of our blood vessel (aorta) is another example for elastomer.
Note: elastomer materials do not obey Hooke’s law.

Elastic Moduli
According to Hooke’s law

where k is called modulus of elasticity.
Depending on different types of stress and strain, there are 3 types of modulus of elasticity.
Different types of modulus of elasticity

1. Young’s modulus (Y)
2. Shear modulus (G)
3. Bulk modulus (B)

1. Youngs Modulus:
Young’s modulus is defined as the ratio of tensile stress (α) to the longitudinal strain (ε).
ie; Youngs modulus Y = \(\frac{\text { tensile stress }}{\text { longitudinal strain }}\)
Y = \(\frac{\alpha}{\varepsilon}\)
Explanation:
Consider a wire of length L and area of cross section a, fixed at one end and loaded at the free end with a weight mg. Let l be the elongation. Then tensile strain α = \(\frac{F}{a}\)
Longitudinal strain ε = \(\frac{l}{L}\)

2. Determination of youngs modulus of the material of a wire:
Experimental setup:

An experimental arrangement to determine the young’s modulus of a wire is shown in the figure. It consists of two long straight wires A and B. (The wire Ais called reference wire and B is called experimental wire).

A millimeter scale (main scale) is fixed near to the wire A. At bottom of wire A there is a pan to place a weight. The wire B also carries a pan in which known weights can be placed. At bottom of wire B, there is a pointer. A vernier scale is attached to this pointer. This vernier scale of B and main scale of A are placed very close to each other.

Determination of Young’s modulus:
Keep the wires (A and B) as straight by placing sufficient load at pans and note down initial reading. Then the experimental wire (B) is loaded with more weights. Note down the reading again. The difference between initial reading and final reading gives the elongation.

Let rand L be the initial radius and length of the experimental wire, then the area of cross section of the wire is πr². Let µ be the mass that
produced an elongation ∆L in the wire.

3. Shear Modulus:

The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material (represented by G). It is also called the modulus of rigidity.

But from figure tanθ = \(\frac{\Delta x}{L}\) = tanθ ≈ θ
∴ G = \(\frac{F}{A \theta}\) The shearing stress α_s can also be expressed as α_s = Gθ.

4. Bulk modulus (B):
When a body is immersed in a fluid, it undergoes a hydraulic stress (equal in magnitude to the hydraulic pressure.) This leads to the decrease in the volume of the body. This decrease in volume produces a strain is called volume strain.

The ratio of hydraulic stress to the corresponding hydraulic strain is called the bulk modulus. It is denoted by symbol B. ie; Bulk modulus

The minus sign indicates the fact that with an increase in pressure, a decrease in volume • occurs. S.l. unit of bulk modulus is the same as that of pressure ie; Nm^-2 (or) Pa.
Compressibility (k):
The reciprocal of the bulk modulus is called
compressibility, k = \(\frac{1}{B}\)
Note: Bulk modulus of solids > Bulk modulus of liquid > Bulk modulus of gases.

Question 1.
Solids are least compressible but gases are most compressible. Why?
Answer:
The molecules in solids are strongly coupled with neighboring atoms. But molecules in gases are very poorly coupled to their neighbors.

5. Poisson’S Ratio
When a wire is subjected to an extension along its length there will be contraction perpendicular to the length. The strain perpendicular to the applied force is called lateral strain. The strain parallel to the applied force is called longitudinal strain.

Within the elastic limit, lateral strain is directly proportional to the longitudinal strain. The ratio of the lateral strain to the longitudinal strain is called Poisson’s ratio.

If the original diameter of the wire is d and the contraction of the wire under stress is Dd, lateral strain = \(\frac{\Delta \mathrm{d}}{\mathrm{d}}\)
If the original length of the wire is L and the elongation under stress is DL, then

Poisson’s ratio is a ratio of two strains. Hence it is a pure number and has no units or dimensions.
For steel the value is between 0.28 and 0.30.

6. Elastic potential energy in a stretched wire:
When a wire is stretched, work is done against the inter molecular forces. This work is stored in the wire as elastic potential energy.

Consider a wire of length L and area of cross section stretched by a force F. Let l be the elongation. Then stretching force increases uniformly from 0 to F. Therefore the average force

Displacement of free end = l
∴ Workdone in stretching = F_av × l
= \(\frac{1}{2}\)Fl = \(\frac{1}{2}\) × Force × extension
This work is stored as the elastic potential energy of the wire
ie. U = \(\frac{1}{2}\) × Fl
Volume of wire =AL
Workdorie in stretching per unit volume, U = \(\frac{\frac{1}{2} F \ell}{A L}\)

Stress = Y × strain ______(3)
Sub(3) in (2)
U = 1/2 × Y × strain × strain
U = 1/2 × Y × (strain)².

Applications Of Elastic Behavior Of Materials

1. Metallic ropes are used in the cranes for lifting the load because of its high elasticity.
2. While designing bridges and buildings, beams and columns are widely used. Bending of beams and columns under a load depends upon the Young’s modulus of the material used. To reduce the bending of a beam for a given load, material with larger youngs modulus is to be used.
3. The maximum height of the mountain is limited by the elastic properties of the rock which hold the mountain.

Question 2.
Find the radius of the rope that used to lift 10 tons(10000Kg). Elastic limit (maximum stress) of steel is 30 × 10⁷ N/m²
Answer:
The condition to withstand a force of 100 metric ton,

Question 3.
What is meant by buckling?

Answer:
If the load is not properly placed on a beam, the beam will bend side wise as shown in figure. This bending is called buckling.

Question 4.
Why the beams used in construction of bridges have a cross-section of the type I?
Answer:
Consider a solid rectangular beam of length l, breadth b and thickness d. If we place a load Mg at its centre, there will be a depression at the centre.

To reduce the depression at the centre, we have to increase d. But when d is large the beam will buckle unless the load is at the centre. This buckling can be avoided if the beam is given the section of the letter I. Such a beam is called I form girder.
It has following advantages

• It reduces the weight of the beam
• It saves material
• It prevents buckling

Question 5.
Find the height of mountain from the data given below. Maximum shearing stress = 30 × 10⁷ N/m² Density of material of mountain = 3 × 10³ Kgm^-3
Answer:
The force per unit area due to the weight of the mountain (at a depth h) is hrg. The material at the bottom experiences this force. The sides of the mountain are free. Hence the shear component is (hrg) itself.
∴ Shearing stress = hrg
30 × 10⁷ = h × 3 × 10³ × 10
h = 10 km
which is more than the height of Mount Everest.

Students can Download Chapter 8 Sources of Business Finance 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 8 Sources of Business Finance

Contents

• Meaning, Nature and Significance of Business finance
• Equity shares – Features -Merits – Demerits
• Preference shares – Features – Types – Merits – Demerits
• Retained earnings – Features – Merits – Demerits
• Debentures – Features – Types – Merits – Demerits
• Public deposits – Features – Merits – Demerits
• Commercial Banks – Features – Merits – Demerits
• Financial institutions – Features – Merits – Demerits
• Trade credit – Features – Merits – Demerits
• Factoring – Features – Merits – Demerits
• Lease financing – Features – Merits – Demerits
• Commercial paper – Features – Merits – Demerits
• International financing – ADR – GDR – FCCB
• Factors affecting the choice of the source of fund Finance is the life blood of any business. The requirements of funds by business to carry out its various activities are called business finance.

Nature of Business Finance:
1. Fixed capital requirements:
In order to start a business funds are needed to purchase fixed assets like land and building, plant and machinery. This is called fixed capital requirement.

2. Working Capital requirements:
A business needs funds for its day to day operation. This is known as working Capital requirements. Working capital is required for purchase of raw materials, to pay salaries, wages, rent and taxes.

Classification of Sources of Funds:

1. Period Basis:
On the basis of period, the different sources of funds cari divided into 3. They are long-term sources, medium-term sources and short-term sources.
(a) Long Term Sources:
The amount of funds required by a business for more than five years is called long-term finance. Generally this type
of finance is required for the purchase of fixed assets like land and building, plant and machinery furniture etc. It include sources such as shares and debentures, long-term borrowings and loans from financial institutions.

(b) Medium Term Sources:
Where the funds are required for a period of more than one year but less than five years, is called medium-term sources. These sources include borrowings from commercial banks, public deposits, lease Financing and loans from financial institutions. This type of finance is required for modernization, renovation, special promotional programmes etc.

(c) Short Term Sources:
Short-term funds are those which are required for a period not exceeding one year. These sources include Trade credit, loans from commercial banks and commercial papers, etc. Short-term finance is used for financing of current assets such as accounts receivable and inventories.

2. Ownership Basis:
On the basis of ownership, the sources can be classified into ‘owner’s funds’and ‘borrowed funds’.
(a) Owners Fund:
It represent the amount of capital provided by owners and the amount of profit retained in the business. It is a permanent source of capital. Equity shares and retained earnings are the two important sources of ownership capital.

(b) Borrowed Funds:
It refers to funds mobilized from outsiders. It include loans from commercial banks, loans from financial institutions, issue of debentures, public deposits and trade credit.

Such sources provide funds for a specified period, on certain terms and conditions and have to be repaid with interest after the expiry of that period. Borrowed funds are provided on the security of some fixed assets.

3. Source of generation:
Another basis of categorising the sources of funds can be whether the funds are generated from within the organisation or from external sources.
(a) Internal sources:
Internal sources of funds are those that are generated from within the business. eg: ploughing back of profit, disposing of surplus stock etc.

(b) External sources:
External sources of funds are those that are generated from outside the business. eg: issue of debentures, borrowing from commercial banks and financial institutions and accepting public deposits.

SOURCES OF FINANCE:

• Issue of shares- (Equity and Preference Shares)
• Ploughing Back of Profit
• Issue of debentures
• Loan from Commercial bank
• Loan from Financial institutions
• Public deposits
• Lease Financing
• Long Term and Medium Term Sources of Finance
• Short Term Sources of Finance International Sources of Finance.

A business can raise funds from various sources. They are:

Issue of shares:
The capital of a company is divided into smaller units called share. Those who subscribe the shares of a company are known as ‘shareholders’. Two types shares may be issued by a company to raise capital. They are:

1. Equity Shares
2. Preference Shares

1. Equity Shares:
Equity shares represents the ownership capital of a company. They do not enjoy any preferential right in the matter of claim of dividend or repayment of capital. Equity share holders do not get a fixed dividend but are paid on the basis of earnings by the company. They bear the maximum risk. Equity shareholders are the owners of the company. They have right to vote and participate in the management.
Merits:

• Equity shares are suitable for investors who are willing to assume risk for higher returns
• Payment of equity dividend is not compulsory.
• Equity capital serves as permanent capital as it is to be repaid only at the time of liquidation of a company.
• Equity shares do not carry any charge on the assets of the company.
• They have right to vote and participate in the management.
• Equity capital provides credit worthiness to the company

Limitations:

1. Investors who want steady income may not prefer equity shares.
2. The cost of equity shares is generally more as compared to the cost of raising funds through other sources.
3. Issue of additional equity shares dilutes the voting power, and earnings of existing equity shareholders.
4. Issue of Equity shares is time consuming.

2. Preference Share:
The capital raised by issue of preference shares is called preference share capital. The preference shareholders enjoy a preferential right over equity shareholders in two ways:

• The right to get a fixed rate of dividend.
• The right to claim repayment of capital in the event of winding up of the company.

Preference shareholders generally do not enjoy any voting rights. A company can issue different types of preference shares.

Types of Preference Shares:
1. Cumulative and Non-cumulative Preference Share:
in cumulative preference shares, the unpaid dividends are accumulated and carried forward for payment in future years. On the other hand, in non- cumulative preference share, the dividend is not accumulated if it is not paid out of the current year’s profit.

2. Participating and Non-participating Preference Share:
Participating preference shares have a right to share the profit after making payment to the equity shares. The non-participating preference shares do not enjoy such a right.

3. Convertible and Non-convertible Preference Share:
The preference shares which can be converted into equity shares after a specified period of time are known as convertible preference share. Non-convertible shares cannot be converted into equity shares.

4. Redeemable and Irredeemable Preference Share:
Redeemable preference shares are those where the company undertakes to repay it after a specified period. Where the amount of the preference shares is refunded only at the time of liquidation, are known an irredeemable preference shares.
Merits:

1. Preference shares provide reasonably steady income in the form of fixed rate of return and safety of investment.
2. Preference shares are useful for those investors who want fixed rate of return with comparatively low risk
3. It does not affect the control of equity share holders because they have no voting right.
4. Preference shares do not create arty charge on the assets of the company.
5. They have a preferential right of repayment of capital over equity shareholders in the event of liquidation of a company.

Limitations:

1. Preference shareholders have no voting right.
2. The dividend paid is not deductible from profit for income tax.
3. These shares may not attract investors who are expecting higher returns.
4. The rate of dividend on preference shares is generally higher than the rate of interest on debentures.

Differences between Equity shares and Preference shares:

Retained Earnings (Ploughing Back of Profit):
A company generally does not distribute all its earnings amongst the shareholders as dividends. A portion of the net earnings may be retained in the business for use in the future. This is known as retained earnings. It is a source of internal financing or self financing or ‘ploughing back of profits’.
Merits:

1. It is a permanent source of funds available to an organization
2. No costs in the form of interest, dividend or flotation cost.
3. It is more dependable than external sources.
4. There is no commitment to pay dividend.
5. It increases the financial strength and earning capacity of the business.
6. Control over the management of the company remains unaffected.
7. It does not require the security of assets.
8. It may lead to increase in the market price of the equity shares of a company

Limitations:

1. Retained profits cause dissatisfaction among the shareholder because they get low dividend.
2. It attract competition in the market
3. It may attract government regulations.
4. It leads the management to manipulate the value of shares
5. It is uncertain source of fund because it is available only when profits are high.

Debentures:
A debenture is a document issued by a company under its seal to acknowledge its debt. Debenture holders are, therefore, termed as creditors of the corrfpany. Debenture holders are paid a fixed rate of interest.
Types of Debentures:
1. Secured and Unsecured Debentures:
Secured (Mortgaged) debentures are debentures which are secured by a charge on the assets of the company. Unsecured (Simple or naked) debentures do not carry any charge or security on the assets of the company.

2. Registered and Bearer Debentures:
In the case registered debentures, the name, address and other details of the debenture holders are entitled in the books of the company. The debentures which are transferable by mere delivery are called bearer (Unsecured) debentures.

3. Convertible and Non-convertible Debentures:
Convertible debentures are those debentures that can be converted into equity shares afterthe expiry of a specified period. On the other hand, nonconvertible debentures are those which cannot be converted into equity shares.

4. First and Second:
Debentures that are repaid before other debentures are repaid are known as first debentures. The second debentures are those which are paid afterthe first debentures have been paid back.
Merits:

1. It. is preferred by investors who want fixed income at lesser risk
2. Debenture holder do not have voting right
3. Interest on Debentures is a tax deductable expense
4. It does not dilute control of equity shareholders on management
5. Debentures are less costly as compared to cost of preference shares.
6. They guarantee a fixed rate of interest
7. It enables the company to take the advantage of trading on equity.
8. The issue of debentures is suitable when the sales and earnings are relatively stable

Limitations:

1. It is not suitable for companies with unstable future earnings.
2. The company has to mortgage its assets to issue debentures.
3. Debenture holders do not enjoy any voting rights.
4. In case of redeemable debentures, the company has to make provisions for repayment on the specified date, even during periods of financial difficulty.
5. With the new issue of debentures, the company’s capability to further borrow funds reduces.

Differences between Shares and Debentures:

Public Deposits:
The deposits that are raised by organisations directly from the public are known as public deposits. Rates of interest offered on public deposits are usually higher than those allowed by commercial banks.

Companies generally invite public deposits for a period up to three years. This is regulated by the R.B.I. and can not exceed 25% of its paid up share capital and reserves.
Merits:

1. The procedure for obtaining public deposits is simpler than share and Debenture.
2. Cost of public deposits is generally lower than the cost of borrowings from banks
3. Public deposits do not usually create any charge on the assets of the company
4. They do not have voting right therefore the control of the company is not diluted
5. Interest paid on public deposits is tax deductable.

Limitations:

1. New companies generally find it difficult to raise funds through public deposits
2. They are not secured.
3. They are costly as most Of the companies have to offer high interest.
4. It is an unreliable source of finance as the public may not respond when the company needs money

Commercial Banks:
Commercial Banks give loan and advances to business in the form of cash credit, overdraft, term loans, discounting of bills, letter of credit etc. Rate of interest on loan is fixed.
Merits:

1. Commercial Bank provide timely financial assistance to business.
2. Secrecy is maintained about loan taken from a Commercial Banks.
3. This is the easier source of finance as there is no need to issue prospectus and underwriting for raising funds.
4. Loan from a bank is a flexible source of finance.

Limitations:

1. Funds are generally available for short periods
2. Banks may ask for security of assets and personal sureties for sanctioning loan.
3. In some cases, difficult terms and conditions are imposed by banks for the grant of loan

Financial Institutions:
The state and central government have established many financial institutions to provide finance to companies. These institutions aim at promoting the industrial development of a country, these are also called ‘development Bank’.

These are IFCI, ICICI, IDBI and LIC, UTI. This source of financing is considered suitable when large funds for longer duration are required for expansion, reorganisation and modernisation of an enterprise.
Merits:

1. Financial Institution provide long term finance which is not provided by Commercial Bank
2. These institutions provide financial, managerial and technical advice and consultancy to business firms.
3. It increases the goodwill of the borrowing company in the capital market.
4. As repayment of loan can be made in easy installments, it does not prove to be much of a burden on the business.
5. The funds are made available even during periods of depression, when other sources of finance are not available.

Limitations:

1. The procedure for granting loan is time consuming due to rigid criteria and many formalities.
2. Financial Institution place restrictions on the powers of the borrowing company.
3. Financial institutions may have their nominees on the Board of Directors of the borrowing company thereby restricting the autonomy of the company.

Trade Credit:
Trade credit is a short term source of financing. The credit extended by one trader to another for purchasing goods or services is known as trade credit. The terms of trade credit vary from one industry to another and are specified on the invoice. Trade credit facilitates the traders to purchase goods without irpmediate payment.
Merits:

1. Trade credit is a convenient and continuous source of funds
2. It does not create any charge on the assets of the firm.
3. Trade credit may be readily available in case the credit worthiness of the customers is known to the seller
4. Trade credit needs to promote the sales of an organisation.

Limitations:

1. It may induce a firm to indulge in overtrading.
2. Only limited amount of funds can be generated through trade credit;
3. It is generally a costly source of funds as compared to other sources of fund.

Factoring:
Factoring is a method of raising short-term finance for the business in which the business can take advance money from the bank against the amount to be realised from the debtors. By this method, the firm shifts the responsibility of collecting the outstanding amount from the debtors on payment of a specified charge.

There are two methods of factoring-recourse and non-recourse. Under recourse factoring, the client is not protected against the risk of bad debts. Under non recourse factoring, full amount of invoice is paid to the client in the event of the debt becoming bad.
Merits:

1. Obtaining funds through factoring is cheaper than bank credit
2. Factoring provides steady cash inflow so that the company is able to meet its liabilities promptly.
3. It is flexible and ensures cash inflows from credit sales.
4. It does not create any charge on the assets of the firm;
5. The company can concentrate on important areas of business as the responsibility of credit control is shouldered by the factor.
6. Factors may give useful information about the credit standing of customers.

Limitations:

1. This source is expensive
2. The advance finance provided by the factor firm is generally available at a higher interest cost.
3. The factor is a third party to the customer who may not feel comfortable while dealing with it.

Lease Financing:
A lease is a contractual agreement whereby the owner of an asset (lessor) grants the right to use the asset to the other party (lessee). The lessor charges a periodic payment for renting of an asset for some specified period called lease rent.
Merits:

1. It enables the lessee to acquire the asset with a lower investment;
2. Lease rentals paid by the lessee are deductible for computing taxable profits;
3. It provides finance without diluting the ownership or control of business
4. The lease agreement does not affect the debt raising capacity of an enterprise;
5. Simple documentation makes it easierto finance assets.

Limitations:

1. A lease arrangement may impose certain restrictions on the use of assets.
2. The normal business operations may be affected in case the lease is not renewed.
3. The lessee never becomes the owner of the asset.

Commercial Paper(CP):
It is an unsecured promissory note issued by a firm to raise funds for a short period. The maturity period of commercial paper usually ranges from 90 days to 364 days. Being unsecured, only firms having good credit rating can issue the CP and its regulation comes under the purview of the Reserve Bank of India.
Merits:

1. A commercial paper does not contain any restrictive conditions;
2. As it is a freely transferable instrument, it has high liquidity;
3. A commercial paper provides a continuous source of funds.
4. They are cheaper than a bank loan.

Limitations:

1. Only financially sound and highly rated firms can raise money through commercial papers
2. The size of money that can be raised through commercial paper is limited
3. Commercial paper is an impersonal method of financing. Extending the maturity of a CP is not possible.
4. Issue of commercial paper is very closely regulated by the RBI guidelines.

International Financing:
1. Commercial Banks:
Commercial banks all over the world extend foreign currency loans for business purposes. Standard chartered is a major source of foreign currency loan to the Indian industry.

2. International Agencies and Development Banks:
A number of international agencies and development banks provide long and medium term loans and grants to promote the development of economically backward areas in the world. Eg. International Finance Corporation (IFC), EXIM Bank and Asian Development Bank.

3. International Capital Markets:
(a) Global Depository Receipts (GDR’s):
Uhder GDR, shares of the company are first converted into depository receipts by international banks. These depository receipts are denominated in US dollars. Then these depository receipts are offered for sale globally through foreign stock exchanges.

GDR is a negotiable instrument and can be traded freely like any other security. The holder of GDRs are entitled for dividend just like shareholders. But they do not enjoy the voting rights. Many Indian companies like ICICI, Wipro etc. have raised foreign capital through issue of GDRs.
Feature of GDR:

1. GDR can be listed and traded on a stock exchange of any foreign country other than America.
2. It is negotiable instrument.
3. A holder of GDR can convert it into the shares.
4. Holder gets dividends
5. Holder does not have voting rights.
6. Many Indian companies such as Reliance, Wipro and ICICI have issue GDR.

(b) American Depository Receipts (ADR’s):
The depository receipts issued by a company in the USA are known as American Depository Receipts.
Feature of ADR:

1. It can be issued only to American Citizens.
2. It can be listed and traded is American stock exchange.
3. Indian companies such as Infosys, Reliance issued ADR

Differences between ADR and GDR:

Foreign Currency Convertible Bonds (FCCB’s):
Foreign currency convertible bonds are equity linked debt securities that are to be converted into equity or depository receipts after a specific period.

The FCCB’s are issued in a foreign currency and carry a fixed interest rate. These are listed and traded in foreign stock exchange and similar to the debenture.
Factors affecting the choice of the source of funds:
The factors that affect the choice of source of finance are
1. Cost:
The cost of procurement of funds and cost of utilising the funds should be taken into account while deciding about the source of funds that will be used by an organization.

2. Financial strength and stability of operations:
In the choice of source of funds, business should be in a sound financial position so as to be able to repay the principal amount and interest on the borrowed amount. When the earnings of the organisation are not stable, it can issue equity shares to collect the fund.

3. Form of organisation and legal status:
The form of business organisation and status influences the choice of a source for raising money.

4. Purpose and time period:
Business should plan according to the time period for which the funds are required. A short-term need can be met through borrowing funds at low rate of interest through trade credit, commercial paper, etc. For long term finance, sources such as issue of shares and debentures are more appropriate.

5. Risk profile:
Business should evaluate each of the source of finance in terms of the risk involved.

6. Control:
business firm should choose a source keeping in mind the extent to which they are willing to share their control over business.

7. Effect on credit worthiness:
The dependence of business on certain sources may affect its credit worthiness in the market.

8. Flexibility and ease:
Another aspect affecting the choice of a source of finance is the flexibility and ease of obtaining funds.

9. Tax benefits:
Various sources may also be weighed in terms of theirtax benefits. For example, while the dividend on preference shares is not tax deductible, interest paid on debentures and loan is tax deductible.

Students can Download Chapter 8 Gravitation 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 8 Gravitation

Summary
Introduction
Gravitational force is one of the fundamental forces exist in nature. All celestial bodies in the universe are moving under the influence of the gravitational force.

Kepler’S Laws
The three laws of Kepler are Known as

1. Law of orbits
2. Law of areas
3. Law of periods

1. Law of Orbits: All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse.
Ellipse:
An ellipse is the path traced out by a planet around the sun. The closest point ‘P’ is called the perihelion and Athe aphelion. The semimajor axis is half the distance AP.

Drawing an ellipse:

Select two points F₁ and F₂. Take a length of a string and fix its ends at F₁ and F₂ by pins. With the tip of a pencil stretch the string and then draw a closed curve. F₁ and F₂ are called the focii.

2. Law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time, ie. aerial velocity is constant
Proof:

Let the sun be at the origin and let the position and momentum of the planet be denoted by \(\overrightarrow{\mathrm{r}}\) and \(\overrightarrow{\mathrm{p}}\). Let ∆\(\overrightarrow{\mathrm{A}}\) be the area swept out by the planet of mass m in time interval A t.
From the figure:

But we know angular momentum of a planet is constant because the gravitational force is central force. Hence we get \(\frac{\Delta \overline{\mathrm{A}}}{\Delta \mathrm{t}}\) = a constant.
Note:

1. Gravitational force is a central force. If direction of force on the planet is along the vector joining sun and planet, it is called central force.
2. The law of area is the consequence of conservation of angular momentum.

3. Law of periods (Kepler’s third law): The square of the time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet.
T² α a³
Where ‘a’ is semi major axis.

Universal Law Of Gravitation
On the basis of Keplers laws and observations of motion of planets, Newton gave a universal law of force acting between any two particles of matter.

Statement:
Every body in the universe attracts each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Explanation

Consider two masses m₁ and m₂ separated by a distance r. The force of attraction between two particles can be written as

The above equation can be written in vector form as

\(\hat{\mathbf{r}}\) is the unit vector from m₁ to m₂. G is a constant called the gravitational constant. In S.l. Unit G = 6.67 × 10^-11 Nm² × \(\bar{g}^{2}\).
Note:

1. The negative sign of gravitational force shows it is an attractive force.
2. \(\vec{F}_{12}=-\vec{F}_{21}\) ie; force on I^st particle due to II^nd particle = – force on II^nd particle due to I^st particle.

Deduction of Kepler’s third law:
Planet of mass m is orbiting around the sun of mass M in circular orbit of radius r, with constant angular velocity w, then.
Centripetal force = Gravitational force.

The above result holds equally good for elliptical orbit provided we replace r with a. (the semi major axis of the elli pse)

The Gravitational Constant
The value of the gravitational constant G can be determined experimentally. This was first done by English scientist Henry Cavendish in 1798.
Cavendish experiment to determine gravitational constant:

Two equal masses (m) are attached at two ends of the rod of length L. The bar is suspended from a rigid support by a fine wire. When two large lead spheres are brought close to small ones from opposite sides as shown, a deflecting torque is produced.
deflecting torque τ = F × L
= \(\frac{G M m}{d^{2}}\) × L
Where d is the distance between m and M. Due to this deflecting torque a restoring couple is produced on the wire.
For equilibrium,
Deflecting couple = Restoring couple

Where ‘C’ is the restoring couple per unit twist and θ is the angle of rotation. Knowing C, θ, d, M, m and L, we can calculate G. The currently accepted
value is G = 6.67 × 10^-11 Nm²/kg²

Question 1.
is it possible to shield a body from gravitational effect?
Answer:
It is not possible to shield a body from gravitational effect because the gravitational force does not depend on the nature of the intervening medium.

Acceleration Due To Gravity Of The Earth
Acceleration due to gravity is the acceleration experienced by a body falling freely towards the earth.
Relation between acceleration due to gravity and gravitational constant:
Consider a body of mass ‘m’ at a distance ‘r’ from centre of earth. According to Newton law of
Gravitation, F = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\) ______(1)
Force acting on mass ‘m’
F = mg ______(2)
comparing eq(1) and eq(2)
g = \(\frac{G M}{r^{2}}\)
On the surface of earth r = R.
∴ acceleration due to gravity on the surface of earth

The value of g is 9.8 m/s². There is a slight variation for g from place to place depending on the height or depth of the place.

Question 2.
Where is the maximum value of g on the surface of earth. Why?
Answer:
It is on the poles. The distance between the pole and centre of earth is minimum at the poles. Hence
g = \(\frac{G M}{R^{2}}\) is maximum at the poles.

Question 3.
There is a popular statement regarding Convendish: “Cavendish weighed the earth”. Comment on this statement.
Answer:
Acceleration due to gravity g = \(\frac{G M}{R^{2}}\)
∴ M = \(\frac{\mathrm{gR}^{2}}{\mathrm{G}}\)
Knowing g, R, and G, we can find the mass of earth. The value of G is experimentally found out by Covendish. This is the reason for the above statement.

Acceleration Due To Gravity Below And Above The Surface Of Earth
1. Variation of ‘g’ with altitude:
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,
g_h = \(\frac{\mathrm{Gm}}{(\mathrm{R}+\mathrm{h})^{2}}\) _____(2)
eq(1)/eq(2)

This equation shows that acceleration due to gravity decreases as height increases. The above equation is valid when h<<R.

2. Variation of g with depth:

If we assume the earth as a sphere of radius R with uniform density r,
mass of earth = volume × density
M = \(\frac{4}{3}\) πR³ ρ _____(1)
We know acceleration due to gravity on the surface, GM
g = \(\frac{G M}{R^{2}}\) ______(2)
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)

The above equatiqn shows that, when depth increases g decreases.

Gravitational Potential Energy
Gravitational potential energy:
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.
Expression for gravitational P.E:

Consider the earth as a uniform sphere of radius R and mass M. Consider a 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 = \(\frac{\mathrm{GMm}}{\mathrm{x}^{2}}\).
If the body is displaced from P to Q Work done,
dw = F.dx
= \(\frac{\mathrm{GMm}}{\mathrm{x}^{2}}\). 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 r from the centre of the earth is given by,

Question 4.
Gravitational potential energy at a point, U = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)

1. What is meant by negative sign in the above equation.
2. What is the value of U at r = ∞

Answer:

1. Negative sign shows that the potential energy is due to attractive gravitational force. It means, to bring a body from infinity to point P, work has been done by the gravitational field of earth, ie; by attractive force.
2. When r = ∞, the gravitational potential energy becomes zero.

Note: Maximum value of gravitational potential energy is zero.
Gravitational Potential:
The gravitational potential at a point is the amount of workdone in bringing a unit mass from infinity to that point without acceleration.
Explanation
Gravitational potential energy of mass m at a point is v = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)
If we take, m = 1 we get gravitational potential.
∴ Gravitational potential v = \(\frac{-G M}{r}\)
Note: The gravitational potential at infinity is taken to be zero.

Question 5.
What is the difference between gravitational potential energy and gravitational potential?
Answer:
Gravitational potential energy of mass m at a distance ‘r’ from centre of earth.
w = \(\frac{-\mathrm{GMm}}{\mathrm{r}}\)
But gravitational potential at a distance r from the centre of earth, w = \(\frac{-G M}{r}\)
From the equation we can understand that, gravitational potential energy depends on the mass of the body. But gravitational potential at point is independent of the mass of the body.

ESCAPE SPEED
The minimum speed with which a body is projected so that it never returns to the earth is called escape
speed or escape velocity.
Expression for escape speed:
Force on a mass m at a distance r from the centre of earth = \(\frac{\mathrm{GMm}}{\mathrm{r}^{2}}\)
Work done for giving small displacement dr,

Work done in taking the body to infinity from surface of earth,

This energy is given in the form of K.E. = \(\frac{1}{2}\)mv²_e
where v_e is the escape speed

This escape velocity \(\sqrt{2 \mathrm{Rg}}\) estimated to be 11.2 km/s on the earth. But escape velocity of moon is 2.3 km/s.
Note: Escape velocity is independent of mass of the escaping body.

Question 6.
Moon has no atmosphere. Why?
Answer:
Escape velocity on moon is 2.3 km/s. The r.m.s. velocity of the gas molecules is greater than this value. Gas molecules easily escape from the surface of moon. Hence there is no atmosphere on moon.

Earth Satellites
Satellites:
The moon is the natural satellite of the earth. It goes round the earth in about 27.3 days almost in a circular. Orbit of radius 3.84 × 10⁵ km. The gravitational force of attraction between the earth and the moon supplies the necessary centripetal force to keep the moon in its circular orbit.
Artificial Satellites:
A man made satellite is called an artificial satellite.
eg: INSAT, EDUSATetc.
Orbital velocity of a satellite:
Orbital velocity of a satellite is the velocity required for a satellite to revolve round the earth in a fixed orbit.
Expression for orbital velocity:
Consider a satellite of mass m revolving with orbital velocity ‘v’ around the earth at a height ‘h’ from the surface of earth. Let M be the mass of earth and R be radius of earth.

The gravitational force of attraction between earth and satellite.

Centripetal force required forthe satellite

For stable rotation,
Centripetal force = Gravitational force

The above equation shows that orbital velocity decreases as h increases.
Period of satellite:
Period of satellite is time taken by the satellite to revolve once around the planet in a fixed orbit.


Case -1
For a minimum orbit
For a satellite very close to the surface of earth, h can be neglected in comparison to R.
ie; R + h ≈ R

Where T₀ is called period of minimum orbit. If we substitute the value g = 9.8 m/s² and R = 6400 km,
We get

T₀ = 85 minutes.

Energy Of An Orbiting Satellite
The kinetic energy of the satellite in a circular orbit at height h from surface of earth with speed vis k.E
= \(\frac{1}{2}\) mv²

The potential energy at distance (R+h) from the centre of the earth is P.E = \(\frac{-G M m}{(R+h)}\)
∴ Total energy E = k.E. + p.E

Question 7.
What is the physical meaning of -ve energy?
Answer:
If total energy of satellite is negative, it moves in a closed path. This satellite can’t escape from earth’s gravitational field.

Question 8.
What is the condition fortotal energy to escape from earth’s gravitational field.
Answer:
If total energy of satellite is zero or positive, it can escapes to infinity.

Geostationary And Polar Satellites
A geostationary satellite is a satellite which appears to be stationary in the sky. The period of this satellite must be the same as the period of the rotation of the earth about its own axis. Its direction of rotation is from west to east.

Question 9.
Calculate the height of geostationary satellite.
Answer:
If h is the height of satellite, period

Substituting the value of
T = 24 × 60 × 60 sec., and
R= 6400 × 1000 m g = 9.8 m/s²
we get

R + h = 42648.54 km.
h = 42648.54 – R = 42648.54 – 6380
h = 36268.54 km.
Example: INSAT
Polar satellite:

Polar satellite orbits in the north-south direction. The height of polar satellite is 500 to 800 km. Since its time period is around 100 minutes, it crosses any altitude many times a day. The camera fixed on the satellite can view small strips of the earth in one orbit.

Adjacent strips are viewed in the next orbit. So that in effect the whole earth can be viewed strip by strip during the entire day. This type of satellite is used for remote sensing, meteorology and environmental studies.

Weightlessness
A satellite can be considered to be a continuously falling body. Since a freely falling body experiences weightlessness, the satellite and object inside it feel weightlessness.

Students can Download Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Introduction
The element carbon organic chemistry the element carbon has the unique property called catenation due to which it forms covalent bonds with other carbon atoms. It also forms covalent bonds with atoms of other elements like hydrogen, oxygen, nitrogen, sulphur, phosphorus and halogens. The resulting compounds are studied under a separate branch of chemistry called organic chemistry.

General Introduction
In early years of chemistry, compounds were classified into two types. Compounds derived from non-living sources such as rocks, minerals etc. were called ‘inorganic compounds’ and those derived from plants and animals were regarded as ‘organic compounds’. On account of the special nature of organic compounds and their occurrence in living world alone, it was believed that they were produced by a vital force existing in living organisms. This led to the belief that such compounds could not be synthesised in the laboratory. However in 1828, F. Wohler succeeded in preparing urea (an organic compound) from an inorganic material, ammonium cyanate.

Tetravalance Of Carbon

Shapes of Organic Compounds
Carbon (atomic number 6) has the ground state electronic configuration 1 s² 2s²p¹_x2p¹_y2p°_z. Carbon attains noble gas configuration only by sharing electrons with other atoms. Carbon atom, therefore, forms four covalent bonds in all its compounds.

During the formation of bonds (which is an energy releasing process) the two electrons in the 2s orbital get unpaired and one is promoted to the empty 2p_z orbital. This corresponds to the excited state of carbon which has four half-filled orbitals (four valence electrons). The shapes of molecules such as methane (CH₄), ethene (C₂H₄) and ethyne (C₂H₂) are explained in terms of the use of sp³, sp² and sp hybridised orbitals by the carbon atoms in the respective molecules.

Structural Representations Of Organic Compounds

Complete, Condensed and Bond-line Structural Formulas
Structures of organic compounds are represented in several ways. The Lewis structure or dot structure, dash structure, condensed structure and bond line structural formulas are some of the specific types. In Lewis structures bonds are represented by lines and lone pair electrones by dots. For example,

In bond-line structural representation of organic compounds, carbon and hydrogen atoms are not shown and the lines representing carbon-carbon bonds are drawn in a zig-zag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen etc. The terminals denote methyl (-CH₃) groups.

Classification Of Organic Compounds

I. Acyclic or open chain compounds
These compounds are also called as aliphatic compounds and consist of straight or branched chain compounds, for example:

II. Alicyclic or closed chain or ring compounds
Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring (homocyclic). Sometimes atoms other than carbon are also present in the ring (heterocyclic). Some examples of this type of compounds are:
CycloiTexane Tetrahydrofuran These exhibit some of the properties similar to those of aliphatic compounds ane:

Aromatic compounds
Aromatic compounds are special types of compounds. You will learn about these compounds in detail in Unit 13. These include benzene and other related ring compounds (benzenoid). Like alicyclic compounds, aromatic comounds may also have hetero atom in the ring. Such compounds are called heterocyclic aromatic compounds. Some of the examples of various types of aromatic compounds are:

Benzenoid aromatic compounds

Non-benzenoid compound

Heterocyclic aromatic compounds

Organic compounds can also be classified on the basis of functional groups, into families or homolo-gous series.

Homologous series
Homologous series may be defined as a series of similarly constituted organic compound in which the members possess the same functional group and have similar chemical properties and the neighbouring (or consecutive) members differ by – CH₂ unit in their molecular formula,
eg: Alkane family (Cn H_2n+2), alkenes (CnH_2n) alcohols (Cn H_2n+1OH).

Nomenclature Of Organic Compounds
IUPAC names of unbrached saturated hydrocarbons

Nomenclature of branched chain alkanes:
We encounter a number of branched chain alkanes. The rules for naming them are given below.
1. First of all, the longest carbon chain in the molecule is identified.For example,

2. The numbering is done in such a way that the branched carbon atoms get the lowest possible numbers.

and the numbering is not from right to left.

3. The names of alkyl groups attached as a branch are then prefixed to the name of the parent alkane and position of the substituents is indicated by the appropriate numbers. If different alkyl groups are present, they are listed in alphabetical order. Thus, name for the compound shown above is: 6- Ethyl-2- methylnonane.
[Note: the numbers are separated from the groups by hyphens and there is no break between methyl and nonane.]

4. If two or more identical substituent groups are present then the numbers are separated by commas. The names of identical substituents are not repeated, instead prefixes such as di (for 2), tri (for 3), tetra (for 4), Penta (for 5), Hexa (for 6) etc. are used. While writing the name of the substituents in alphabetical order, these prefixes, however, are not considered. Thus, the following compounds are named as:

5. If the two substituents are found in equivalent positions, the lower number is given to the one coming first in the alphabetical listing. Thus, the following compound is 3-Ethyl-6-methyl octane and ‘ not6-Ethyl-3-methyl octane.

6. The branched alkyl groups can be named by following the above mentioned procedures. However, the carbon atom of the branch that attaches to the root alkane is numbered 1 as exemplified below.

Cyclic Compounds
A saturated monocyclic compound is named by prefixing ‘cyclo’ to the corresponding straight chain alkane. If side chains are present, then the rules given above are applied. Names of some cyclic compounds are given below.

Nomenclature Of Organic Compounds Having Functional Group (S)
When a functional group (other than C=C and -C ≡ C) is present in the molecule, it is indicated by adding secondary suffix after the primary suffix. The terminal ‘e’ of the primary suffix is removed before adding secondary suffix (whose name begins with a, i, o, u or y). It is to be noted that some functional group such as alkoxy (-OR), nirto (-NO₂), halogeno etc. are indicated by the prefixes.

For example, CH₃-CH₂-OH Ethane -e+ol = Ethanol (using secondary suffix) CH₃-CH₂-CHO Propane -e+al = Propanal (using secondary suffix) CH₃-CH₂-NO₂ Nitroethane (using prefix)

The systematic name of an organic compound containing functional group can be derived using the following sequence of steps.

The longest carbon chain (parent chain) containing the functional groups is identified. This gives the word root. The name of the compound is then obtained as follows:
Prefixes – word root – primary suffix – secondary suffix The following examples will illustrate the rules

In case of compounds containing more than one similar functional group, the world di, tri etc is added before the secondary suffix which indicates the functional group. In doing so, the last letter ‘e’ of the parent alkane has to be retained. However, the endingne of the parent alkane is dropped in case of compounds having more than one double or triple bond. For example,

If the molecule contains two or more different functional groups, the parent chain must contain maximum possible number of functional groups. The carbon atoms in the parent chain are numbered in such a way that the functional group of higher priority gets the lower number. The priority of various functional groups follows the order.

COOH>-CO-O-CO->-COOR>-COCI>-CONH₂>-CN>- HC=O>CO>-OH>-NH₂>>C=C<-C≡C->-X>-NO₂>R-
The functional group with higher priority is indicated by suitable secondary suffix and the other functional groups are treated as substituents which are specified by suitable prefixes.
For example,

Nomenclature of Substituted Benzene Compounds
For IUPAC nomenclature of substituted benzene compounds, the substituent is placed as prefix to the word benzene as shown in the following examples.

If benzene ring is disubstituted, the position of substituents is defined by numbering the carbon atoms of the ring such that the substituents are located at the lowest numbers possible. For example, the compound(b) is named as 1, 3-Dibromobenzene and not as 1,5 dibromobenzene. Substituent of the base compound is assigned number 1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order. Some examples are given below.

In the trivial system of nomenclature the terms ortho (o), meta (m) and para (p) are used as prefixes to indicate the relative positions 1,2-;1,3- and Ir-respectively. Thus, 1,3-dibromobenzene (b) is named as m-dibromobenzene (meta is abbreviated as m-) and the other isomers of dibromobenzenel, 2-(a) and 1,4-(c), are named as ortho (or just o-) and para (or just p-)-dibromobenzene, respectively. The substituents appear in the name in alphabetical order.

Isomerism
The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Such compounds are called as isomers.

Structural Isomerism
Compounds having the same molecular formula but different structures (manners in which atoms are linked) are classified as structural isomers. Some typical examples of different types of structural isomerism are given below:

(i) Chain isomerism:
When two or more compounds have similar molecular formula but different carbon skeletons, these are referred to as chain isomers and the phenomenon is termed as chain isomerism.

ii) Position isomerism:
When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers and this phenomenon is termed as position isomerism.

iii) Functional group isomerism :
Two or more compounds having the same molecularformula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism.

iv) Metamerism :
It arises due to different alkyl chains on either side of the functional group in the molecule. For example, C_aH₁₀O represents Methoxypropane (CH₃OC₃H₇) and Ethoxyethane (C₂H₅OC₂H₅).

Fundamental Concepts In Organic Reaction Mechanism
An organic reaction takes place by the attack of a reagent on an organic compound which is designated as a substrate. The steps of an organic reaction showing the breaking and formation of bonds in such substrate leading to the formation of the final product are referred to as its mechanism.

Fission Of A Covalent Bond
A covalent bond can be broken in two different ways,

(i) Homolytic fission :
If a covalent bond breaks in such a way that each atom takes away one electron of the shared pair. It is called homolytic fission or homolysis. The fragments with odd or unpaired electrons formed by homolysis are known as free radicals. For example,

Usually homolysis occurs at high temperature or in presence of high energy radiations. Reactions occurring through homolytic fission are known as free radical reactions (non-polar reactions).

(ii) Heterolytic fission :
When a covalent bond breaks in such a way that both the electrons of the covalent bond are taken away by one of the bonded atoms, the mode of cleavage is called heterolytic cleavage or heterolysis. The products of heterolysis of a covalent bond are positive and negative ions, eg:

Inductive Effect (I Effect)
It is the permanent polarisation of a sigma bond in a molecule by the influence of an adjacent polar bond or group.

For illustration, let us consider a carbon chain in which one terminal carbon atom is joined to a chlorine atom. Since the chlorine atom is more electronegative than carbon, the sigma electrons of the C-Cl bond are displaced towards the chlorine atom. As a result, the chlorine atom acquires a small negative charge and C acquires a small positive charge as shown below. The magnitude of+charge is of the order C₁ > C₂ >C₃.

This type of electron displacement of sigma electrons along a saturated carbon chain due to the presence of an electron-withdrawing group (or electron-donating group) is called Inductive effect:

This effect decreases sharply with increasing dis-tance from the substituent and becomes negligible afterthe third carbon in a chain. Atoms orgroups of atoms that attract or withdraw electrons from a chain are said to have electron withdrawing inductive effect or-l effect, eg:,
-NO₂ >- CN >- COOH >- F >- Cl >- Br >-l

Atoms or groups which push or donate electrons to a carbon chain are said to have electron releasing inductive effect or + l effect. Alkyl groups have +l effect and the order of + l effect of alkyl groups is

Resonance Structure
There are many organic molecules whose behaviour cannot be explained by a single Lewis structure. An example is that of benzene. Its cyclic structure containing alternating C-C single and C=C double bonds shown is inadequate for explaining its characteristic properties.

As per the above representation, benzene should exhibit two different bond lengths, due to C-C single and C=C double bonds. Thus, the structure of benzene cannot be represented adequately by the above structure. Further, benzene can be represented equally well by the energetically identical structures I and ll. The actual structure of benzene cannot be adequately represented by any of these structures, rather it is a hybrid of the two structures (I and II) called resonance structures.

The resonance structures are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability. The energy of actual structure of the molecule (the resonance hybrid) is lower than that of any of the canonical structures. The difference in energy is called the resonance stabilisation energy or simply the resonance energy. The more the number of important contributing structures, the more is the resonance energy. Resonance is particularly important when the contributing structures are equivalent in energy.

Resonance Effect
The resonance effect is defined as ‘the polarity produced in the molecule by the interaction of two π- bonds or between a π-bond and lone pair of electrons present on an adjacent atom’.The effect is transmitted through the chain. There are two types of resonance or mesomeric effect designated as R or M effect.
(i) Positive Resonance Effect (+R effect)
In this effect, the transfer of electrons is away from an atom or substituent group attached to the conjugated system. This electron displacement makes certain positions in the molecule of high electron densities.

(ii) Negative Resonance Effect (- R effect)
This effect is observed when the transfer of electrons is towards the atom or substituent group attached to the conjugated system.The atoms or substituent groups, which represent +R or-R electron displacement effects are as follows:
+R effect: – halogen, -OH, -OR, -OCOR, -NH₂, – NHR,-NR₂,-NHCOR,
– R effect: – COOH, -CHO, >C=O, – CN, -NO₂

Electromeric Effect (E – Effect)
This is temporary effect which involves the complete transfer of n electrons of a multiple bond to one of the bonded atoms in presence of an attacking reagent. However, when the attacking reagent is removed, the polarised molecule shifts back to its original electronic condition.

With transfer of π electrons takes place towards the attacking reagent the effect is called +E effect and when the transfer of π electrons occurs-S away from the attacking reagent the effect is called -E effect.

Hyperconjugation
Hyperconjugation is a general stabilising interaction. It involves delocalisation of σ electrons of C—H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. The σ electrons of C—H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p orbital. Hyperconjugation is a permanent effect.

Methods Of Purification Of Organic Compounds
Sublimation:
On heating, some solid substances change from solid to vapour state without passing through liquid state. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from nonsublimable impurities.

Crystallisation:
It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved . in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. On cooling the solution, pure compound crystallises out and is removed by filtration. The filtrate (mother liquor) contains impurities and small quantity of the compound.

Distillation:
This important method is used to separate (i) volatile liquids from non-volatile impurities and (ii) the liquids having sufficient difference in their boiling points. Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Chloroform (b.p 334 K) and aniline (b.p. 457 K) are easily separated by the technique of distillation. On boiling, the vapours of lower boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapours of higher boiling component form later and the liquid can be collected separately.

Fractional Distillation:
If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask. Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile component. By the time the vapours reach to the top of the fractionating column, these are rich in the more volatile component. The vapours become richer in low boiling component. On reaching the top, the vapours become pure in low boiling component and pass through the condenser and the pure liquid is collected in a receiver.

After a series of successive distillations, the remaining liquid in the distillation flask gets enriched in high boiling component. Each successive condensation and vaporisation unit in the fractionating column is called a theoretical plate. One of the technological applications of fractional distillation is to separate different fractions of crude oil in petroleum industry. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump. Glycerol can be separated from spent-lye in soap industry by using this technique.

Steam Distillation:
This technique is applied to separate substances which are steam volatile and are immiscible with water. In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and collected. The compound is later separated from water using a separating funnelAniline is separated by this technique from aniline-water mixture.

Differential Extraction:
When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution should be immiscible with each other so that they form two distinct layers which can be separated by separatory funnel. The organic solvent is later removed by distillation or by evaporation to get back the compound.

Chromatography:
The name chromatography is based on the Greek word chroma, for colour since the method was first used for the separation of coloured substances found in plants. In this technique, the mixture of substances is applied onto a stationary phase, which may be a solid ora liquid.

A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one another. The moving phase is called the mobile phase. Based on the principle involved, chromatography is classified into different categories. Two of these are:

1. Adsorption chromatography, and
2. Partition chromatography.

1. Adsorption Chromatography:
Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina. When a mobile phase is allowed to move over a stationary phase (adsorbent), the components of the mixture move by varying distances over the stationary phase. Following are two main types of chromatographic techniques based on the principle of differential dsorption.

1. Column chromatography, and
2. Thin layer chromatography.

Column Chromatography
Column chromatography involves separation of a mixture overa column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower soluble in the organic solvent. The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube. An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distances in the column.

Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin layer of an adsorbent (silica gel or alumina) is spread overa glass plate of suitable size. The plate is known as thin layer chromatography plate or chrome plate. The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of the TLC plate. The glass plate is then placed in a closed jar containing the eluant. As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. The relative adsorption of each component of the mixture is expressed in terms of its retardation factor i.e. R_f value
R_f = x/y

Where distance moved by the substance from base line is x and distance moved by the solvent from base line is y. The spots of coloured compounds are visible on TLC plate due to their original colour. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine. Spots of compounds, which adsorb iodine, will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example, amino acids may be detected by spraying the plate with ninhydrin solution.

Partition Chromatography:
Partition chromatography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent ora mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as a chromatogram. The spots of the separated coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed under thin layer chromatography.

Qualitative Analysis Of Organic Compounds
Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin
The elements present in organic compounds are carbon and hydrogen. In addition.to these, they may also contain oxygen, nitrogen, sulphur, halogens and phosphorus.

Detection of Carbon and Hydrogen
Carbon and hydrogen are detected by heating the compound with copper(ll) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue).

Detection of Other Elements
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:

C, N, Sand X come from organic compound. Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

1. Test for Nitrogen
The sodium fusion extract is boiled with iron(ll) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(ll) sulphate and forms sodium hexacyanoferrate(ll). On heating, with concentrated sulphuric acid some iron(ll) ions are oxidised to iron(lll) ions which react with sodium hexacyanoferrate(ll) to produce iron(lll) hexacyanoferrate(ll) (ferric ferrocyanide) which is Prussian blue in colour.

2. Test for Sulphur
1. The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.

2. On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur.

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood-red colour and no Prussian blue since there are no free cyanide ions.

If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give their usual tests.
NaSCN + 2Na → NaCN + Na₂S

3. Test for Halogens
The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.
X^– + Ag⁺ → AgX

X represents a halogen – Cl, Br or I. If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens.

4. Test for Phosphorus
The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

Quantitative Analysis
The percentage composition of elements present in an organic compound is determined by the methods based on the following principles:

Carbon and Hydrogen
Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(ll) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively.
C_xH_y + (x + y /4)O₂ → xCO₂ +(y/2)H₂O

The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series. The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated. Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m₁ and m₂g respectively;

Nitrogen
There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method:
The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.
C_xH_yN_z+(2x + y/2)CuO → x CO₂ + y / 2H₂O +Z / 2N₂ + (2x + y / 2)CU

Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube. Let the mass of organic compound = m g Volume of nitrogen collected = V₁ mL
Room temperature = T₁K
Volume of nitrogen at STP = 7\(\frac{p_{1} v_{1} \times 273}{760 \times T_{1}}\) (LetitbeVmL)

Where p₁ and V₁ are the pressure and volume of nitrogen, p,is different from the atmospheric pressure at which nitrogen gas is collected. The value of pt is obtained by the relation;
p₁ Atmospheric pressure-Aqueous tension 22400 mL N₂ at STP weighs 28 g.
\frac{p_{1} v_{1} \times 273}{760 \times T_{1}}

(ii) Kjeldahl’s method:
The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate.The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution. The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.
taken = V mL
Volume of NaOH of molarity, M. used for titration of excess of H₂SO₄ = V₁ mL
V₁L of NaOH of molarity M= V1 /2 mL of H₂SO₄ 0f molarity M
Volume of H₂SO₄ of molarity M unused= (V-1/1/2) mL (V-1/1/2) mL of H2S04 of molarity M = 2(V-V1/2) mL of NH₃ solution of molarity M.
1000 mL of 1 M NH₃ solution contains 17g NH₃ or 14 g of N
2(V-V1/2) mL of NH₃ solution of molarity M contains:

Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Halogens
Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.
Let the mass of organic compound taken = m g
Mass of AgX formed = m, g
1 mol of AgX contains 1 mol of X
Mass of halogen in m1g of AgX

Sulphur
A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate. Let the mass of organic compound taken = m g and the mass of barium sulphate formed = m₁ g
1 mol of BaSO₄ = 233 g BaSO₄ = 32 g sulphur

Phosphorus
A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH₄)₃PO₄.12MoO₃, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH₄PO₄ by adding magnesia mixture which on ignition yields Mg₂P₂O₇. Let the mass of organic compound taken = m g and mass of ammonium phospho molydate = m1g.
Molar mass of (NH₄)₃PO₄.12MoO₃ = 1877 g.

where, 222 u is the molar mass of Mg₂P₂O₇, m, the mass of organic compound taken, mv the mass of Mg₂P₂O₇ formed and 62, the mass of two phosphorus atoms present in the compound Mg₂P₂O₇.

Oxygen
The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (IJOJ) when carbon monoxide is oxidised to carbon dioxide producing iodine.

The percentage of oxygen can be derived from the amount of carbon dioxide or iodine produced.
Let the mass of organic = mg compund taken
Mass of carbon dioxide = m₁g
44g carbon dioxide = 32 g oxygen

Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1 -3 mg) and displays the values on a screen within a short time.

Students can Download Chapter 7 Formation of a Company 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 7 Formation of a Company

Contents

• Promotion- Functions and Legal position of promoter
• Incorporation-Steps
• Capital subscription-Steps
• Commencement of business- Steps
• Memorandum of Association and its clauses- Articles of Association and its clauses- Prospectus and its clauses
• Differences between Memorandum and Articles of Association

The steps involved in the formation of a company are:

• Promotion
• Incorporation
• Capital subscription
• Commencement of business

Promotion:
Promotion is the first stage in the formation of a company. The identification of business opportunities, analysis of its prospects and initiating steps to form a joint stock company is called promotion. The person who undertakes to form a company is called promoter.
Functions of a Promoter:
1. Identification of business opportunity:
The first and foremost activity of a promoter is to identify a business opportunity.

2. Feasibility studies:
After identifying a business opportunity, the promoters undertake some feasibility studies to determine the viability and profitability of the proposed activity.

• Technical feasibility: To determine whether the raw materials or technology is easily available
• Financial feasibility: To determine the total estimated cost of the project
• Economic feasibility: To determine the I profitability of the proposed project

3. Name approval:
After selecting the name of company the promoters submit an application to the Registrar of companies for its approval. The selected name is not the same or identical to an existing company.

4. Fixing up signatories to the Memorandum of Association:
Promoters have to decide about the members who will be signing the Memorandum of Association of the proposed company.

5. Appointment of professional:
Promoters appoint merchant bankers, auditors etc. to assist them in the preparation of necessary documents.

6. Preparation of necessary documents:
The promoters prepare certain legal documents which are to be submitted to the Registrar of companies. They are

• Memorandum of Association
• Articles of Association,
• Consent of proposed Directors
• Agreement, if any, with proposed managing or whole time director
• Statutory declaration

Position of Promoters
The promoter is neither an agent nor a trustee of the company. The promoter stands in the fiduciary relationship with the company. He should not make any secret profits out of the dealings. Any, such gains are to be disclosed.

The promoter must act honestly, in good faith and in the best interest of the company. The promoter is personally liable for all the preliminary contracts with the other parties before incorporation. The promoter is also liable for any omission of facts or false statements in the prospectus.

Incorporation:
A company comes into existence only when it is registered with the Registrar of Companies. For this purpose the promoter has to take the following steps.
Steps for Incorporation:
(a) Application for incorporation:
Promoters make an application for the incorporation of the company to the Registrar of companies.

(b) Filing of documents:
The following documents must be filed with the Registrar of Companies for incorporation.

1. The Memorandum of Association duly stamped, signed and witnessed
2. Articles of Association duly stamped, signed and witnessed
3. Written consent of the proposed directors
4. Agreement, if any, with proposed managing or whole time director
5. A copy of the Registrar’s letter approving the name of the company.
6. Statutory declaration
7. A notice about the exact address of the registered office.
8. Documentary evidence of payment of registration fees.

The Registrar verifies the entire document submitted. If he is satisfied then he enters the name of the company in his Register. After the registration, the Registrar issues a Certificate called Certificate of Incorporation.

This is called the birth certificate of the company. With effect from November 1, 2000, the Registrar of Companies allots a CIN (Corporate Identity Number) to the Company.

Effect of the Certificate of Incorporation Certificate of Incorporation is the conclusive evidence of the legal existence of the company. A private company can commence its business after receiving Certificate of Incorporation. The certificate of incorporation is the birth certificate of the company.

Capital Subscription:
A public company can raise funds from the public by issuing shares and Debentures. Forthis it has to issue prospectus. The following steps are required for raising funds from the public:

1. A.public company is required to take approval from SEBI. (Securities and Exchange Board of India)
2. File a copy of prospectus or a statement in lieu of prospectus with the Registrar of Companies.
3. Appointment of Bankers, Brokers, Underwriters:
4. Ensure that minimum subscription is received;
5. Application for listing of company’s securities;
6. Refund/adjust excess application money received;
7. Issue allotment letters to successful applicants;
8. File return of allotment with the Registrar of Companies (ROC).

Commencement of Business:
A public company can commence business only after getting certificate of commencement of business from the Registrar. The company must file the following documents to obtain the certificate of commencement of business.

1. Declaration that the minimum subscription has been received in cash to allot shares.
2. A declaration that all directors have taken up and paid for their qualification shares
3. A statutory declaration stating that necessary legal formalities have been complied with has to be filed.

The Registrar shall examine these documents. If these are found satisfactory, a ‘Certificate of Commencement of Business’ will be issued. This certificate is conclusive evidence that the company is entitled to do business.

With the grant of this certificate the formation of a public company is complete and the company can legally start doing business. Documents used in the formation of a company.

Memorandum of Association:
It is the charter or magnacarta of the company. It defines the objects of the company and provides the framework beyond which the company cannot operate. It lays down the relationship of the company with outside world.

Memorandum of Association must be printed, divided into paragraphs, numbered consecutively. The Memorandum of Association must be signed by at least seven persons in case of a public company and by two persons in case of a private company.

Contents of Memorandum of Association:

1. The name clause: Under this clause the name of the company is mentioned. A company can select any name subject to the following restrictions.

1. The proposed name should not be identical with the name of another company
2. A name which can mislead the public
3. In case of a public company the name should end with the word ‘Limited’ and in case of a private company the name should end with the word ‘Private Limited’
4. The name must not directly or indirectly imply any participation of the Central or State Govt.
5. The name must not suggest any connection or patronage of a national hero
6. It should not include the word co operative.

2. Registered office clause:
This clause contains the name of the state, in which the registered office of the company is proposed to be situated. It must be informed to the Registrar within thirty days of the incorporation of the company.

3. Objects clause:
This is the most important clause of the memorandum. It defines the purpose for which the company is formed. A company is not legally entitled to undertake an activity, which is beyond the objects stated in this clause.

4. Liability clause:
It states that the liability of members is limited to the face value of shares held by them or the amount guaranteed to be paid on winding up.

5. Capital clause:
This clause specifies the maximum capital which the company will be authorised to raise through, the issue of shares.

6. Association clause:
In this clause, the signatories to the Memorandum of Association state their intention to be associated with the company and also give their consent to purchase qualification shares.

Articles of Association:
The Articles of Association is the second important document of a company. The Articles define the rights, duties and powers of the officers and the Board of directors. It contains the rules regarding internal management of the company. It shows the relationship between the company and its members.
Contents of Articles of Association:

1. The share capital of the company and its division.
2. Rights of each class of shareholders.
3. Details of contracts made with different parties.
4. Procedure for making allotment of shares.
5. Procedure for issuing share certificate.
6. Procedure for transfer and transmission of shares.
7. Procedure for forfeiture and reissue of shares.
8. Procedure for conducting meetings, voting, proxy and poll
9. Procedure for appointing, removal and remuneration of directors.
10. Procedure for declaration of and payment of dividend.
11. Keeping books of account and audit of the company.
12. Procedure regarding alteration of share capital.
13. Procedure regarding winding up of the company.

Table A:
A public limited company may adopt Table A which is a model set of articles given in the Companies Act. Table A is a document containing rules and regulations for the internal management of a company. If a company adopts Table A, there is no need to prepare separate Articles of Association.

Difference between Memorandum of Association and Articles of Association:

Prospectus:
Prospectus is a document issued by the public companies inviting the public to subscribe for shares or debentures of the company. It contains all information regarding the company’s affairs and its future prospects.

A prospectus must be dated and signed by all the directors. A copy of the prospectus must be filed with Registrar before it is issued to public.
Contents of prospectus:

1. Name and address of the registered office of the company.
2. Main objects of the company.
3. Classes of shares and debentures.
4. Name, address and occupation of the signatories to the memorandum.
5. Details of the borrowing powers of the company.
6. Name, address and occupation of the directors and managing director.
7. Name and address of the promoters.
8. Minimum subscription.
9. Time of opening and closing of subscription.
10. The amount payable on application and allotment of each class of shares.
11. Name of underwriters.
12. Details of preliminary expenses.

Companies which do not want to issue a prospectus may submit a statement in lieu of prospectus to the Registrar of Companies. It is a copy of the prospectus but is not issued to the public.

Statement in lieu of prospectus:
Sometimes a company may not invite public subscription and hence may not issue a prospectus. In such a case the Companies Act provides that at least three days before the first allotment, a statement called Statement in lieu of prospectus must be filled with the Registrar for registration of a company. It is drafted according to the Part 1 of Schedule 3 of the Act.

Minimum Subscription:
Minimum subscription is the minimum amount of shares that must be subscribed by the public. A company can make allotment of shares only after receiving the minimum subscription. Otherwise, the application money received must be returned to the applicants. Minimum subscription is 90% of the total number of shares offered to the public.

Preliminary contract:
During the promotion of the company, promoters enter into certain contracts with third parties on behalf of the company. These are called preliminary contracts. Promoters are personally liable to third parties for these contracts.

Qualification Shares:
According to the Articles of Association, a director must take a certain number of shares in a company to act as a director. These are called Qualification Shares. They have to pay for these shares before the company obtains Certificate of Commencement of Business.

Underwriting:
The process of appointing underwriters to ensure the minimum subscription of capital is known as underwriting. Underwriters undertake to buy the shares if these are not subscribed by the public. For this, they get underwriting commission.

Students can Download Chapter 7 Systems of Particles and Rotational Motion 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 7 Systems of Particles and Rotational Motion

Summary
Introduction
In the earlier chapters we discussed the motion of particle. We applied the results of our study to the motion of bodies of finite size.

A large class of problems with extended bodies can be solved by considering them to be rigid bodies. Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of particles of such a body do not change.

(i) Basically a rigid body can have two types of motion.

1. translational
2. Rotational motion

1. Translational motion:

In pure translational motion, at any instant of time every particle of the body has the same velocity. Explanation
Consider a rectangular block moving down along an inclined plane as shown in the figure: This body is rigid. Hence all the particles have same velocity. Any points like P1 or P2 of the body moves with the same velocity at any instant of time.

2. Rotational motion:
In pure rotational motion at any instant of time every particle of the body have different velocities. Explanation

Consider a cylinder rolling down along inclined plane as shown in figure. Points P1, P2, P3 and P4 have different velocities (shown by arrows) at any instant of time.
Note: The above figure shows a combination motion of both translational and rotational motion.

Centre Of Mass
The centre of mass of a system of particles is the point where all the mass of the system may be assumed to be concentrated.
Position vector of two particle system:

Consider two particles of masses m1 and m2 with position vectors \(\vec{r}_{1}\) and \(\vec{r}_{2}\) respectively with respect to the origin O. Now the position coordinate of the center of mass C is defined as.

X, Y and Z coordinate of centre of mass of two particle system

Position vector of N particle system:
Consider a system of N particles of mass m₁, m₂…….,mN with position vectors \(\vec{r}_{1}, \vec{r}_{2}, \ldots \ldots \ldots \vec{r}_{N}\) respectively. Then the centre of mass of the system of N particle is defined as

Position vector of CM of continuous distribution of mass:
If body is continuous distribution of mass, we divide the body into n small elements of mass: Dm₁, Dm₂,………..DmN. Let r₁, r₂,……….rN be the position vectors of those small elements respectively.
Then position vector of CM is,

when we take N as bigger ((Dmi) becomes smaller) the summation can be changed into integration.

Motion Of Centre Of Mass
Velocity of centre of mass. The position vector of the centre of mass of N particle system is given

Differentiating the position vector of C.M., we get velocity of CM. ie;

Acceleration of centre of mass:
Acceleration of centre of mass a = \(\frac{\mathrm{d}}{\mathrm{dt}} \overrightarrow{\mathrm{v}}\)

Force acting on the centre of mass:
Force acting on the centre of mass = Total mass at the centre of mass × acceleration of the centre of mass.
ie; \(\overrightarrow{\mathrm{F}}\) = M\(\overrightarrow{\mathrm{a}}\) ______(1)
But F = F_internal + F_extenal
By Newton’s third law, sum of the internal forces is zero.
∴ F = F_extenal
Therefore eq(1) becomes

Therefore the centre of mass moves as if it were a particle of mass equal to the total mass of the system and all the external forces are acting on it.

Linear Momentum Of A System Of Particles
Momentum of centre of mass
Velocity of centre of mass,

The equation means that, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass.
Momentum conservation and centre of mass motion:
statement:
The total momentum of the centre of mass is conserved if no external force acts.
Proof:
According to Newton’s second law, rate of change momentum of centre of mass is directly proportional s to force acting on it.

If the total external force acting on the system is zero; the centre of mass moves with a constant velocity.
Application of the idea of centre of mass:

1. Explosion of a shell inflight

Consider a shell projected into air. If it is not exploded, its path will be a parabola. If it is exploded in air, the centre of mass follows the same parabolic path. Because the forces due to explosion are internal. Internal force can’t change the direction of centre of mass.
Note: External force can change the direction of centre of mass

2. Decay of radio active nuclei at rest:

Consider radioactive decay of radium nucleus moving along a straight line. A radium nucleus disintegrates into a nucleus of radon and an alpha particle. The two particles produced in the decay move in different directions. But the centre of mass (of radon and α particle) moves along a straight line, because the force leading to decay is internal. Internal force can’t change the direction of centre of mass.

Centre of mass frame:
If we observe this decay from the frame of reference in which the centre of mass (of a particle and radon) at rest, the product particle seems to be moving in opposite direction along a straight line.

In centre of mass frame, the motion of product particles become simple.

3. Binary stars:

Consider the trajectories of the two stars of equal mass as shown in figure. If there are no external forces, the centre of mass moves along a straight line as shown in figure.

Centre of mass frame of Binary stars:
If we look the trajectories of S₁ and S₂ from the centre of mass frame, we find that these two stars are moving in a circle as shown in figure.

In this frame of reference, the trajectories of the stars are a combination of

1. uniform motion in a straight line of the centre of mass and
2. circular orbits of the stars about the centre of mass.

Vector Product Of Two Vectors
The vector product of two vectors \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) is


Screw rule:
Rotate a right handed screw from \(\overrightarrow{\mathrm{a}}\) to \(\overrightarrow{\mathrm{b}}\). Then the direction of advance of the. tip of the screw gives the direction of (\(\overrightarrow{\mathrm{a}}\) × \(\overrightarrow{\mathrm{b}}\))

Right hand rule:
Curl the fingers of the right hand from \(\overrightarrow{\mathrm{a}}\) to \(\overrightarrow{\mathrm{b}}\) along the shorter angle. Then the direction in which the thumb points gives the direction of (\(\overrightarrow{\mathrm{a}}\) × \(\overrightarrow{\mathrm{b}}\)).

Properties of cross product:

Note:
If \(\hat{i}, \hat{j}, \hat{\mathbf{k}}\) occur cyclically in the above vector product relation, the vector product is positive. If \(\hat{i}, \hat{j}, \hat{\mathbf{k}}\) do not occur in cyclic order, the vector product is negative.

Question 1.

Answer:

Angular Velocity And Its Relation With Linear Velocity
In earlier chapter, we treated angular velocity as scalar. But angular velocity is a vector quantity. The relation between angular velocity and linear velocity can be written in vector notation as

Direction of angular velocity:

Consider a body moving along the circumference of circle of radius ‘r’with velocity v. Then the direction of angular velocity will be perpendicular to both \(\overrightarrow{\mathbf{V}}\) and \(\overrightarrow{\mathbf{r}}\) (along the axis of rotation).

The figure(1) shows the direction of w, if body rotates in clockwise direction.

The figure(2) shows the direction of w, if body rotates in anticlockwise direction.
1. Angular acceleration:
Rate of change of angular velocity is called angular acceleration.
angular acceleration \(\vec{\alpha}=\frac{\mathrm{d} \vec{\omega}}{\mathrm{dt}}\)
If the axis of rotation is fixed, the direction of ω and hence α is fixed. In this case the vector equation reduces to scalar equation.

Torque And Angular Momentum
Torque and angular momentum are important quantities to discuss the motion of rigid bodies.

1. Moment of force (Torque)
Torque (or the moment) of a force of about a point is the rotating effect of the force about that point.
Explanation

If f is the force acting at a point A, then torque about a point O is given by


Where \(\overrightarrow{\mathbf{r}}\) is the position vector of the point A from the point O. Torque is a vector quantity. The direction of torque is (given by right hand screw rule). Perpendicular to both \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{f}}\).
Unit:
The unit of torque is Nm.
Note:

• If the turning tendency of the force is anticlockwise, then the torque is positive and if it is clockwise, then torque is negative.
• Torque has dimensions ML²T^-2. Its dimensions are the same as those of work or energy. Torque is. vector, while work is scalar.
• Torque plays the same role in rotational motion as force does in translational motion.

Couple:
Two equal and opposite forces, separated by a distance, constitute a couple.

Moment of couple (Torque):
The moment of couple is the product of either of the forces and the perpendicular distance between them.

Question 2.
The door is a rigid body which can rotate about a fixed axis passing through the hinges. What makes the door rotate?
Answer:
A force applied to the hinge line can’t produce any rotation. But a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing motion. The rotational analogue of force is moment of force. It is also referred to as torque.

Question 3.
When you fix a handle on a door where do you fix it?
Answer:
You fix it in such a way that the torque is maximum. For this the lever arm must be maximum. So the handle is fixed as far away from the hinges as possible Note: A couple produces rotation without translation motion.

2. Angular momentum of a particle:
Angular momentum of a particle about a point is defined as the moment of its linear momentum about that point.
Explanation

Considers particle of mass ‘m’ and linear momentum \(\overrightarrow{\mathbf{p}}\) at a position \(\overrightarrow{\mathbf{r}}\) relative to the origin O. The angular momentum \(\overrightarrow{\mathbf{l}}\) of the particle can be written as (with respect to the origin O).

Note:

• The quantity angular momentum is the rotational analogue of linear momentum.
• Direction of \(\overrightarrow{\mathbf{l}}\) is given by right hand screw rule, (angular momentum is perpendicular to both \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{p}}\)).

Relation between angular momentum and torque:
Angular momentum of a particle,

When differentiate on both side, we get


ie The rate of change of angular momentum is the torque applied to it. This is similar to force equal to rate of change of linear momentum.

Torque and angular momentum for a system of particles:
Consider a system having n particles. The total angular momentum of system is the vector sum of angular momentum of individual particles.
∴ total angular momentum of system,

\(\frac{\mathrm{dL}}{\mathrm{dt}}=\tau\) _____(1)
But Στ_i = τ is the total torque acting on the system of particle. Actually torque acting on a system is sum of external torque (τ_ext) and internal torque (τ_int)
ie. τ = τ_ext + τ_int
But τ_int = 0, because internal torque arises due to action – reaction pair. Action reaction pair for a system is zero.
∴ τ = τ_ext ______(2)
substituting eq(2) is eq (1) we get
\(\frac{\mathrm{d} \overrightarrow{\mathrm{L}}}{\mathrm{dt}}=\tau_{\mathrm{ext}}\)
The above equation means that, the time rate of the total angular momentum of a system of particles about a point is equal to the sum of the external torques acting on the system taken about the same point.

Conservation of angular momentum of a system:
For a system of particles,


If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved.
Note: The statement that the total angular momentum is conserved means that each of these three components (Lx, Ly, Lz) is conserved.

Equilibrium Of A Rigid Body
A rigid body is said to be in a mechanical equilibrium, if both it’s linear momentum and angular momentum are not changing with time.

(or)

A body is said to be in mechanical equilibrium, if the body has neither linear acceleration nor angular acceleration. A body moving with constant momentum is said to . be in translational equilibrium. Similarly a body with constant angular momentum is said to be in rotational equilibrium.

Condition for translational equilibrium:
If the total force acting on a rigid body is zero, the body, moves with constant linear momentum (or zero linear acceleration). The body moving with constant linear momentum is said to be in translational equilibrium Undertranslational equilibrium, the total force acting on the rigid body is zero.

Condition for rotational equilibrium:
If the total torque acting on rigid body is zero, the total angular momentum of the body does not change. Then the body is said to be in rotational equilibrium. Mathematical condition for rotational equilibrium is

Principles of moments:

Consider a lever pivoted at some origin (say fulcrum) in mechanical equilibrium. Let \(\overrightarrow{\mathrm{f}}_{1}\) and \(\overrightarrow{\mathrm{f}}_{2}\) he the forces acting at A and B as shown in figure. Let \(\overrightarrow{\mathrm{R}}\) be the reaction of the support at the fulcrum. For translational equilibrium

For rotational equilibrium, sum of moments must be zero.
ie d₁f₁ + -d₂f₂ = 0
d₁f₁ = d₂f2₂ ______(2)
In the case of lever, f₁ is used to lift some weight. Hence f₁ is called load and its distance from the fulcrum d₁ is called load arm. Force f₂ is the effort applied to lift the load, distance d₂ is called effort arm.
Hence d₁f₁ = d₂f₂ can be written as
load arnn × load = effort arm × effort
The above equation expresses the principle of moments for a lever.
The ratio \(\frac{f_{1}}{f_{2}}\) is called mechanical advantage (MA)
Note:
High value of mechanical advantage means that a small effort can be used to lift a large load.
Examples for level

• See – saw
• Beam balance

1. Centre of gravity:
Centre of gravity of a body is that point through which the net gravitational force acts. The centre of gravity of a body may not be the same as its centre of mass. For a body of very large dimensions, the value of acceleration due to gravity is different for its different parts. In this situation the centre of gravity does not coincide with the centre of mass.

If the size of the body is small, the value of acceleration due to gravity is same for its different parts. In this situation, the centre of gravity of the body coincides with the centre of mass. Torque due to gravity.

Moment Of Inertia
A body at rest cannot rotate by itself. A body in uniform rotation cannot stop by itself. This inability of a material body is called rotational inertia. It depends on the mass of the body and axis of rotation. In other words it depends on a quantity called moment of inertia.

a. Moment of inertia of a particle:
Consider a particle of mass ‘m’ capable of rotation about an axis AB.

Let ‘r’ be the perpendicular distance of particle from AB. The moment of inertia of the particle about AB is defined as the product of mass of the particle and square of the distance between the particle and the axis of rotation.

b. Moment of inertia of a rigid body:

Consider a rigid body capable of rotation about an axis AB. Let the body consisting of particles of , masses, m₁, m₂, m₃……….m_n at distances r₁, r₂, r₃………..r_n
Then by definition, moment of inertia of m, about AB = m₁r₁².
M.I of m₂ about AB = m₂r₂²
————————–
M.I. of m_n about AB = m_nr_n²
Therefore total moment of inertia of the body about
I = m₁r₁² + m₂r₂² + ………………m_nr_n²

c. Moment of inertia of a ring about an axis through its centre and perpendicular to its plane:
Consider ring of mass M and radius R. AB the axis of rotation. Let ‘m’ be the mass of small section on the circumference of the ring.

M.I. of ‘m’about AB = mR²
∴ Total moment of inertia about AB,
I = ΣmR²

where Σm = M.

d. Moment of inertia of pair of small masses attached to the two ends of massless rod:
Consider a rigid massless rod of length / with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod. Each mass M/2 is at a distance 1/2 from the axis of rotation.

∴ Total moment of inertia

Radius of gyration:
Radius of gyration of a body is the square root of ratio of moment of inertia and total mass of the body

Question 10.
Why do the concept of radius of gyration introduce?
Answer:
The moment of inertia of a body is given by
I = Σmr²
From the above equation it is clear that, we have to find the product of mass and the square of distances from the axis of rotation for all particles and then sum all these products. But the concept of radius of gyration simplifies the above problem.

In this method we find the centre of mass. Then we find a distance to any axis from this centre of mass, in such a way that the moment of inertia of this point (centre of mass) may be equal to that of I = Σmr². This distance from the axis of rotation to centre of mass is called radius of gyration. Practical utility of moment of inertia.

Question 11.
In a fly wheel (or) wheels of vehicles, most of the mass is concentrated at the rim? Explain why?
Answer:
(i) Flywheel:
The machines (such as steam engine, automobile engine etc) that produce rotational motion have a disc with large moment of inertia.

This disc is called fly wheel. Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents the jerky motions. Thus fly wheel gives a smooth ride forthe passengers on the vehicle.

(ii) The wheels of vehicles:
The moment of inertia of wheels is increased by concentrating most of the mass at the rim of the wheel. If such a wheel gain or loses some K.E of rotation \(\left(\frac{1}{2} I \omega^{2}\right)\) brings a relatively smaller change in its angular speed w(∵ I is large) Hence such a flywheel helps in maintaining uniform rotation.

1. K.E of rotating body:
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.

Theorems Of Perpendicular And Parallelaxes
Theorem of perpendicular axes:
The moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis lying in the plane of the body.

(or)

Moment of inertia of a plane lamina about the z-axis is equal to sum of the moments of inertia about x axis and the y axis, if planer lamina lies in xy plane.
Explanation

Let I_x and I_y be the moments of inertia of the lamina about ox and oy. Let I_z be the moment of inertia of the lamina about an axis perpendicular to the lamina and passing through ‘0’ (about z axis) Then by perpendicular axis theorem.

Question 12.
M.l of disc about one of its diameters.
Answer:
According perpendicular axis theorem, I_z = I_x + I_y
But in case I_x = I_y
I_z = 2 I_x _____(1)
But we know I_z = MR²/2 _____(2)
sub(2) in (1), we get \(\frac{\mathrm{MR}^{2}}{2}\) = 2 I_x
I_x = \(\frac{M R^{2}}{4}\)
The moment of inertia about any diameter is \(\frac{M R^{2}}{4}\).

1. Theorem of parallel axis:
The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

Explanation
Let I be the moment of inertia of a-body about an axisAB. Let I0 be moment of inertia about the axis CD parallel to AB and passing through the centre of gravity (G of the body). Let M be the mass of the body and ‘a’ be the distance between the two axes. Then by parallel axes theorem.
I = I_o + Ma²

Question 13.
What is the moment of inertia of rod of mass M, length l about an axis perpendicular to it through one end.

Answer:
Consider a rod of mass M and length l, rotating about an axis passing through one end. Let d_x be a small element at a distance x from the axis of rotation.
mass of the element dx, dm = \(\frac{M}{l}\)dx
∴ M.I of length small element, dl = \(\frac{M}{l}\) dx x²

Moment of inertia of a ring about a tangent:
Consider a ring of mass M and radius R. If this ring is rotating about tangent, we can use parallel axis theorem.
According to parallel axis theorem,

Kinematics Of Rotational Motion About A Fixed Axis
The quantities ‘q’, w and ‘α’ in rotational motion has corresponding quantities in translational motion (x, ‘v’ and a respectively). For translational motion, we have
v = u + at
v² = u² + 2as
s = ut+ \(\frac{1}{2}\) at²
putting u = w₁, v = w₂ and s = q, we get the equations of motion in rotational motion.
w₂ = w₁ + αt
ω₂² = ω₁² + 2αθ
θ = ω₁t + \(\frac{1}{2}\) αt²

Dynamics Of Rotational Motion About A Fixed Axis
Table 7.2 lists quantities associated with linear motion and their analogues in rotational motion.
a. Work done by a torque:

Consider a particle at P₁. Let r₁ be the position vector at time t = 0. This position vector makes an angle q with x – axis. Let particle be acted by a force \(\overrightarrow{\mathrm{F}}_{1}\). Due to this force the particle subtends an angle dq and reaches at p₁¹.

ds is the linear displacement due to the force \(\overrightarrow{\mathrm{F}}_{1}\) This force makes angle with α₁, the position vector r₁.f₁ is the angle made by \(\overrightarrow{\mathrm{F}}_{1}\) with linear displacement.
Form the triangle the workdone for small displacement ds,

Substituting the eq(2) in (1) we get
dW₁ = τ₁dθ
For rigid body, there are many particles. Hence total work done on it.
dW₁ = (τ₁ + τ₂ +……….)dθ

Where τ is the total torque acting oh the body.

b. Rateofwork done by torque:
Instantaneous power due to torque.
We know dw = τ dθ
dividing both sides by dt, we have \(\frac{d w}{d t}=\tau \frac{d \theta}{d t}\)
P = τω

c. Angular acceleration:
Rate of change of angular velocity is called angular acceleration.
Angular acceleration α = \(\frac{\mathrm{d} \omega}{\mathrm{dt}}\).

d. Relation between torque and angular acceleration:
The rate at which work is done on the body is equal to the rate at which kinetic energy increases.

τω = Iωα
τ = Iα
Note:
Just as force (F = ma) produces acceleration, torque produces angular acceleration in a body.
Newtons second law in rotation about a fixed axis

The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body for rotation about a fixed axis.

Angular Momentum In Case Of Rotation About A Fixed Axis

Consider a rigid body rotating about a given axis with a uniform angular velocity w. Let the body consist of n particles of masses m₁, m₂, m₃…………m_n at perpendicular distances r₁, r₂, r₃………..r_n respectively from the axis of rotation.
If v₁, v₂, v₃…………v_n are the linear velocities of the
respective particles, then
v₁ = r₁w, v₂ = r₂w, v₃ = r₃w……….
The linear momentum of particle of mass m₁ is,
P₁ = m₁v₁
P₁ = m₁r₁w.
The angular momentum of this particle about the given axis,
l₁ = p₁ x r₁
= (m₁ v₁) x r₁
v₁ = r₁w
= m₁r₁²w
Similarly angular momentum of second particle
l₂ = m₂r₂²w
Angular momentum of the about the given axis,

1. Conservation of angular momentum:
Conservation of angular momentum and moment of inertia. When there is no external torque, the total angular momentum of a body or a system of bodies are a constant.

L = constant
But L = Iω
∴ Iω = a constant
Application

Question 14.
If the polar ice cap melts what will happen to the length of the day?
Answer:
For earth, angular momentum is a constant (Lw = constant, ie no torque acts on the earth). When the polar ice cap melts, the water thus formed will flow down to the equtorial 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, w will decrease. The decrease in ‘w’ will increase the length of day.

Question 15.
If the earth loses the atmosphere what will happen to the length of the day?
Answer:
For earth, the angular momentum (L = Iw) is a constant, because there is no torque acting on it. When earth loses the atmosphere, I decreases and w increases to keep L as constant. Hence length of the day decreases.

Question 16.
A girl standing on a turn table. What happens to the rotation speed, if she stretches her hand?
Answer:
If a girl rotating with a uniform speed on turn table, it’s angular momentum (L = Iw) will be a constant. When she suddenly stretches her hand, I increases and w decreases to keep L as constant.

Question 17.
How does a circus acrobat and a divertake 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 w increases. But before reaching water he will stretch his hands and legs. Hence I increases and w decreases. So that he gets a smooth entry into the water.

Rolling Motion

Question 18.
A wheel rolling uniformly along a level road is shown in the figure. The centre is moving with speed (V_CM). Find resultant velocities at P₁, P₂ and P₀.
Answer:
We know that the translational velocity of body is equal to the velocity of centre of mass. The velocity of centre of mass V_CM = Rw. Where R is the radius of wheel.
Velocity at P₁:
The point at P₁ has two velocities.

• Linear velocity (Vl)
• Translational velocity (Vt)

The linear velocity at P₁, V_l = Rw.
Translational velocity at P₁, V_t = Rw
[∵ translational velocities are same for all points on the wheel and its value equal to velocity of centre of mass].
The direction of V_l and V_t are same at P₁.
Hence total velocity at

Velocity at P₂:
Linear velocity at P₂, V_l = rw
[where r is the distance of P₂ from centre of mass]
translational velocity at P₂, V_t = Rw.
∴ Total velocity at

Velocity at P₀:
Linear velocity at P₀, \(\overrightarrow{\mathrm{V}}_{l}\) = Rω Translational at P₀, V_t = Rw.
The direction of V_land V_t are opposite.
∴ Hence total velocity,

= -Rw + Rw = 0
which means that the point P₀ is instantaneously at rest. Hence the friction at P₀ is zero. As a result rolling friction becomes less than the kinetic friction.
Note: The condition that P₀ is instantaneously at rest requires V_CM = Rw. Thus for the disc the condition for rolling without slipping is,
V_CM = RW.

1. Kinetic energy of rolling motion (without slipping):
In this case kinetic energy has two parts,

• due to the linear motion of centre of mass
• due to the rotational motion of the body.

K.E in terms of radius of gyration:
Moment of inertia I = mk²
where K is the radius of gyration of the body and v = Rw.
Substituting I, and V in eq(1), we get

Students can Download Chapter 6 Social Responsibilities of Business and Business Ethics 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 6 Social Responsibilities of Business and Business Ethics

Contents
1. Social Responsibility – Meaning – Arguments for and against social responsibility – Kinds of social responsibility – Social responsibility towards different interest groups

2. Environmental protection – Causes of pollution – Need for pollution control – Role of business in environmental protection

3. Business Ethics – Meaning – Elements of Business Ethics Social responsibility of business refers to its obligation to take those decisions and perform those actions which are desirable in terms of the objectives and values of our society. Social responsibility involves an element of voluntary action on the part of business people for the benefit of society.

Arguments in favour of Social Responsibility:
1. Justification for Existence and Growth:
The prosperity and growth of business is possible only through continuous service to society.

2. Long term Interest of the firm:
A firm can improve its image and builds goodwill in the long run when its highest goal is to serve the society .

3. Avoidance of government regulations:
Business can avoid the problem of government regulations by voluntarily assuming social responsibilities.

4. Maintenance of society:
Law alone can’t help out people with ail the difficulties they face. A socially responsible business can contribute something for social peace and harmony.

5. Availability of resources with business:
Business has valuable financial and human resources which can be effectively used for solving problems of the society.

6. Better environment for doing business:
Social responsibility creates better environment for business operations as it improves quality of life and standard of living of the people.

7. Contribution to social problems:
Some of the social problems have been created by business firms themselves such as pollution, unsafe work places, discrimination etc, Therefore, it is the moral obligation of business to solve such social problems.

Arguments Against Social Responsibility:
1. Violation of profit maximization objective:
According to this argument, business exists only for the maximum profit to its shareholders and do not have responsibility to the society as a whole.

2. Burden on consumers:
Involvement of business in social responsibilities involve a lot of expenditure which will ultimately be borne by the customers.

3. Lack of Social Skills:
The business firms and managers are not expert to tackle the social problems like poverty, over population etc.

4. Lack of public support:
business cannot fulfill social responsibility because of lack of public confidence & cooperation.

Kinds of Social Responsibility:
1. Economic responsibility:
The primary social responsibility of a business is to produce goods and services that society wants and sell them at a profit.

2. Legal responsibility:
Every business has a responsibility to operate within the laws of the land.

3. Ethical responsibility:
This includes the behavior of the firm that is expected by the society but not included in law. eg: Respect religious sentiment and dignity of people while advertising.

4. Discretionary responsibility:
This refers to voluntary obligation that an enterprise assumes. eg: Giving charitable contributions to educational institutions, helping the people in natural calamities etc.

Social Responsibility towards different interest groups:
1. Responsibility towards share holders or owners:

• Provide a fair and regular return on the investment of shareholders.
• Provide regular and accurate information on the financial position of the firm.
• To ensure the safety of their investment.

2. Responsibility Towards the workers:

• Providing fair wages
• Providing good working conditions and welfare amenities.
• Respect democratic rights of workers to form unions.

3. Responsibility toward consumers:

• Supply right quality and quantity of goods and services at reasonable prices.
• Avoding unfair trade practices like adulteration, poor quality, misleading advertisement etc.
• Inform them about new products, its features, uses and other matters relating to the products.
• To handle the customers grievance promptly.

4. Responsibility Towards Government:

• Respect the laws of the country
• Pay taxes regularly and honestly.
• act according to the well accepted values of the society.

5. Responsibility towards community:

• Make employment opportunities
• Protect the environment from pollution.
• To uplift the weaker sections of society

Business & Environmental Protection Causes of Pollution:
Many industrial organisations have been responsible for causing air, water, land and noise pollution.
1. Air Pollution:
Air pollution is mainly due to Carbon monoxide emitted by automobiles and smoke and other chemicals from manufacturing plants. It has created a hole in the ozone layer leading to global warming.

2. Water pollution:
Water becomes polluted primarily from chemical and waste dumping.. It has led to the death of several animals and posed a serious problem to human life.

3. Land Pollution:
Dumping of toxic wastes reduces the quality of land and making it unfit for agriculture or plantation.

4. Noise Pollution:
Noise caused by the running of factories and vehicles create a serious health hazard such as loss of hearing, malfunctioning of the heart and mental disorders.

Need for Pollution Control:
1. Reduction of health hazard:
Pollution control measures can check diseases like cancer, heart attack & lung complications and support a healthy life on earth.

2. Reduced Risk of Liability:
It is a sound business policy to install pollution control devices in its premises to reduce the risk of liability of paying compensation to the affected people.

3. Cost Saving:
An effective pollution control programme is needed to save costs of operating business.

4. Improved Public Image:
A firm that adopts pollution control measures enjoys a good reputation as a socially responsible enterprise.

5. Other social benefits:
Pollution control results in many other benefits like clearer visibility, cleaner buildings, better quality of life, and the availability of natural products in a purer form.

Role of Business in Environmental Protection:

1. A definite commitment by top management to create a work culture for environmental protection
2. Ensuring that commitment of environmental protection is shared throughout the enterprise by all divisions and employees.
3. Developing clear cut policies and programmes for purchasing good quality raw materials, introducing superior technology, using scientific techniques of disposal of waste and developing employee skills for pollution control.
4. Complying with the laws and regulations enacted by the Government for prevention of pollution.
5. Participation in government programs relating to management of hazardous substances, cleaning up of polluted rivers, plantation of trees, and checking deforestation.
6. Periodical assessment of pollution control programmes in terms of costs and benefits with a view to improve them.
7. Arranging educational workshops and training materials to share technical information with everyone involved in pollution control.

Business Ethics:
Ethics is concerned with what is right and what is wrong in human behavior. Business ethics refer to the socially determined moral principles which should govern business activities.

Business ethics is the code of conduct followed and performed by every business. Ethical business behavior improves public image, earn’s public confidence and leads to greater success.
Examples of Business Ethics:

1. Charging fair prices from customers
2. Using fair weights for measurement of commodities
3. Giving fair treatment to workers
4. Earning reasonable profits.
5. Avoiding adulteration, hoarding etc.
6. Using environmentally friendly products

Elements of Business Ethics:
1. Top management commitment:
The Chief Executive Officer and higher level managers must give continuous leadership for developing and upholding the moral values of the organisation.

2. Publication of a Code:
‘Code’ refers to a formal written document of the principles, values and standards that guide a firm’s actions. It may cover the areas of fundamental honesty and adherence to laws, product safety and quality, health and safety in the workplace etc.

3. Establishment of Compliance Mechanism:
A suitable mechanism should be developed to comply with the ethical standards of the enterprise.

4. Employees Involvement:
To make ethical business a reality, employees at all levels must be involved.

5. Measuring Results:
Ethical results must be verified and audited that how far work is being carried according to ethical standards.

Students can Download Chapter 5 Emerging Modes of Business 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 5 Emerging Modes of Business

Contents

• E-business – Meaning and scope of e-business – Differences between traditional and e-business – Benefits and Limitations of e-business
• Online transactions – Steps -e-business risk – Resources required for e-buisness
• Outsourcing – Meaning – features – Benefits and Concerns of outsourcing – Types of outsourcing services

e-Business (Electronic Business):
e-business may be defined as the conduct of industry, trade and commerce using the computer networks. Computer network means internet, e-business versus e-commerce e-business is a wider term which includes e-commerce and other electronically conducted business functions such as production, accounting, finance, personnel etc.

e-commerce covers a firms interactions with its customers and suppliers over the internet, e-business is, therefore, clearly much more than buying and selling over the internet, i.e., e-commerce.

Various constituents of e-business:
1. B2B Commerce:
It is that business activity in which two business units make electronic transaction.
Eg. making enquiries seeking or placing orders, communicating supply of goods, making payments, and so on.

2. B2C Commerce:
When the transaction is between business and consumers, it is called Business to Consumers. It enables a business firm to be in touch with its customers on round the clock basis. It involves consumers placing order on line, electronic payment etc.

3. Intra-B Commerce:
It means interaction and dealings among various departments and persons within the firm. For example, the marketing department may interact regularly with the production department and other departments that help in attaining efficient inventory handling, better cash management, timely and sufficient provision of customer services, and so on.

4. C2C Commerce:
Under it, both the parties involved in electronic transaction are customers. It is required for buying and selling of those goods for which there are no established markets. For example, selling used books and household equipments.

Benefits of e-Business:

1. e-business is relatively easy to start and requires lower capital.
2. Customers can buy goods at any time from any seller located in different parts of the world.
3. Business transactions can .be made easily and speedily.
4. It helps the business units to operate at the national as well as the global level.
5. It helps to reduce clerical and paper work.
6. It helps to eliminate middlemen.
7. Any company can launch its new product in the market through the medium of E-Business.
8. It improves the brand image of the company.

Limitations of e-Business:

1. It lacks personal touch with customers, which makes it unsuitable for medical, legal services etc.
2. The transaction can be finalised quickly, but physical delivery of goods often takes long time and be delayed.
3. For successful implementation of e business, the parties to the transactions have to be familiar with computers.
4. It leads to leakage of confidential information such as credit card details. Also there are problems of virus and hacking.
5. It is difficult to establish identity of the parties .

Differences between Traditional business and e-business:

On line Transactions:
On line transaction means receiving information about goods, placing an order, receiving delivery and making payment through medium of internet.

Buying / Selling Process:

Steps involved in online purchase:
1. Register with the online vendor by filling-up a registration form.

2. Place the order for the items put by customer in his virtual shopping cart, an on-line record of what has been picked up while browsing the Online store.

3. Payment for the purchases through online shopping may be done in a number of ways: i.e Cash on delivery, cheque, net banking transfer, debit/credit card.

Net Banking Transfer:
Modem banks provide to their customers the facility of electronic transfer of funds over the net. In this case, the buyer may transferthe transaction amount to the account of the online vendor who may, then, proceed to arrange for the delivery of goods.

Debit card:
The holder of a debit card can buy goods from approved shops without paying cash against the balance in his bank account. Every purchase reduces bank balance. Debit card is issued to bank account holders only and against the amount deposited with the bank.

Credit cards:
A credit card is an instrument issued by a bank in the name of the customer providing credit up to a specified amount. The person holding a valid credit card uses it for purchasing goods from approved shops without paying cash. The payment is made by the bank to the sellers. The buyers have to pay for the purchase within the credit period.

Security and safety of e- Business:
There are three types Of possible risks as listed below:
(a) Transaction risks:

1. Seller may deny that customer ever placed the order or the customer may deny that he ever placed the order. It is called “Default on Order taking/Giving”.
2. Goods may be delivered at wrong address or wrong goods may be delivered which is referred as “Default on Delivery”.
3. Seller may complaint that he didn’t receive payment while customer may claim that payment was over. This is referred as “Default
   Payment”.

(b) Data storage and transmission risks:

1. VIRUS (Vital Information & Resources Under Siege): Virus can disrupt functioning, damage the data and even may cause complete destruction of the system.
2. Interception: Data maybe intercepted in the course of transmission

(c) Risks of threat to intellectual property and privacy:

1. Once the information is made available over the internet, it moves out of the private domain. So important information may be copied by others.
2. When data furnished goes in the hands of others they may start dumping with lot of advertising & promotional literature into our e-mail box.

Resources Required for Successful e-Business Implementation:
The resources required for the e-Business are:

1. Computer system
2. Internet connection and technically qualified work force
3. A well developed web page
4. Effective telecommunication system
5. A good system for making payment using credit instruments.

Outsourcing or Business Process Outsourcing (BPO):
Outsourcing is a management strategy by which an organisation contracts out its major non-core functions to specialized service providers with a view to benefit from their expertise, efficiency and cost effectiveness, and allow managers to concentrate on their core activities.
Merits of outsourcing:

1. It provides an opportunity to the organisation to concentrate on areas in which it has core competency or strength.
2. It helps better utilisation of its resources as the management can focus its attention on selected activities and attain higher efficiency.
3. It helps the organisation to get an expert and specialised service at competitive prices. It helps in improved service and reduction in costs.
4. It facilitates inter-organisational knowledge sharing and collaborative learning.
5. It enables expansion of business as resources saved from outsourcing can be used for expanding the production capacity and diversified products.

Limitations of outsourcing

1. It reduces confidentiality as outsourcing involves sharing a lot of information with others.
2. It may be opposed by labour unions who feel threatened by possible reduction in their employment.
3. In the name of cost cutting, unlawful activities such as child labour, wage discrimination maybe encouraged in other countries.
4. The organisation hiring others may face the problem of loss of managerial control because it is more difficult to manage outside service providers than managing one’s own employees.
5. It causes unemployment in the home country.

Students can Download Chapter 11 The p Block Elements Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 11 The p Block Elements

Introduction
There are six groups of p-block elements in the periodic table numbering from 13to 18. Boron, carbon, nitrogen, oxygen, fluorine and helium head the groups. Their valence shell electronic configuration is ns² np^1-6(except for He). The inner core of the electronic configuration may, however, differ. The difference in inner core of elements greatly influences their physical properties (such as atomic and ionic radii, ionisation enthalpy, etc.) as well as chemical properties.

In groups 13, 14 and 15, the group oxidation state is the most stable state for lighter elements of the group. However, the oxidation state two units less than the group oxidation state becomes progressively more stable down a group. This is due to the reluctance of ns² electrons to participate in bond formation in the case of heavier elements. This phenomenon is known as inert pair effect. Since p-block contains non-metals (and metalloids), these elements have higher electronegativities and higher ionisation enthalpies. In contrast to metals which form cations, non-metals readily form anions.

The combined effect of size and availability of cf orbitals considerably influences the ability of these elements to form π bonds. The first member of a group differs from the heavier members in its ability to form pπ -pπ multiple bonds to itself ( e.g., C=C, C° C, N° N) and to other second row elements e.g., C=0, C=N, C° N, N=0). This type of π – bonding is not particularly strong for the heavier p-block elements. The heavier elements do form π bonds but this involves d orbitals.

Group 13 Elements: The Boron Family

Electronic Configuration
The outer electronic configuration of these elements is ns² np¹. This difference in electronic structures affects the other properties and consequently the chemistry of all the elements of this group.

Atomic Radii
On moving down the group, atomic radius is expected to increase. However, a deviation can be seen. Atomic radius of Ga is less than that of Al. This can be understood from the variation in the inner core of the electronic configuration. The presence of additional 10 d-electrons offer only poor screening effect for the outer electrons from the increased nuclear charge in gallium. Consequently, the atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).

Ionization Enthalpy
The ionisation enthalpy values as expected from the general trends do not decrease down the group. The decrease from B to Al is associated with increase in size. The observed discontinuity in the ionisation enthalpy values between Al and Ga, and between In and Tl are due to inability of d- and f-electrons, which have low screening effect, to compensate the increase in nuclear charge.

Electronegativity
Down the group, electronegativity first decreases from B to Al and then increases marginally.

Physical Properties
Boron is non-metallic in nature. It is extremely hard and black coloured solid. It exists in many allotropic forms.

Chemical Properties
Oxidation state and trends in chemical reactivity The sum of its first three ionization enthalpies of boron is very high due to its small size. This prevents it to form +3 ions and forces it to form only covalent compounds. But as we move from BtoAl.the sum of the first three ionisation enthalpies of Al considerably decreases, and is, therefore, able to form Al³⁺ ions. The tendency to behave as Lewis acid decreases with the increase in the size down the group. BCl₃ easily accepts a lone pair of electrons from ammonia to form BCl₃.NH₃.

i) Reactivity towards air
Boron has crystalline form which is unreactive. Alu-minium forms a very thin oxide layer on the surface which protects the metal from further attack.

ii) Reactivity towards acids and alkalies
Boron does not react with acids and alkalies even at moderate temperature, but aluminium has amphoteric character.

iii) Reactivity towards halogens
2E(s) + 3X₂(g) → 2EX₃(S) (X = F, Cl, Br, I)

Important Trends And Anomalous Properties Of Boron
The tri-chlorides, bromides and iodides of all these elements being covalent in nature are hydrolysed in water. Species like tetrahedral [M(OH)₄]^– and octahedral [M(H₂O)6]³⁺, except in boron, exist in aqueous medium. The monomeric trihalides, being electron deficient, are strong Lewis acids. Boron trifluoride easily reacts with Lewis bases such as NH₃ to complete octet around boron.
F₃B+: NH₃ → F₃B ← NH₃

It is due to the absence of d orbitals that the maximum covalence of B is 4. Since the d orbitals are available with Al and other elements, the maximum covalence can be expected beyond 4. Most of the other metal halides (e.g., AlCl₃ are dimerised through halogen bridging (e.g., Al₂Cl₆). The metal species ‘ completes its octet by accepting electrons from halogen in these halogen bridged molecules.

Some Important Compounds Of Boron

Borax
It is the most important compound of boron. Formula of the compound is Na₂B₄O₇.10H₂O . In fact it contains the tetranuclear units [B₄O₅(OH)₄]^2- and correct formula; therefore, is Na₂[B₄O₅(OH)₄].8H₂O.

On heating, borax first loses water molecules. On further heating it turns into a transparent liquid, which solidifies into glass like material known as borax bead.

Orthoboric acid
Orthoboric acid, H₃B0₃ is a white crystalline solid, with soapy touch. It is sparingly soluble in water but highly soluble in hot water.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃

Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion:
B(OH)₃ +2HOH → [B(OH)₄]^– + H₃O⁺

Structure of boric acid is given below.

Diborane (B₂H₆)
The simplest boron hydride is diborane (B₂H₆). Diborane can be prepared by treating BF₃ with lithium aluminium hydride in ether. A convenient laboratory method is oxidation of sodium borohydride with iodine.
2NaBH₄ + l₂ → B₂H₆ + 2Nal +H₂

On a commercial scale, diborane is produced by the action of BF₃ on sodium hydride.

Diborane is a colourless toxic gas. It catches fire on exposure to air releasing large amount of energy.
B₂H₆+ 6H₂O → 2B(OH)₃ + 6H₂

Reaction of diborane with NH₃ gives an addition product B₂H₆.2NH₃ which on heating gives borazine (B₃N₃H₃), commonly known as inorganic benzene due to its structural similarity with benzene. Boron forms a series of hydridoborates, the most important being (BH₄)^–.NaBH₄ (sodium borohydride) is a good reducing agent.

Each boron atom in B₂H₆ is sp³ hybridised. The structure contains two types of H- atoms the four-terminal hydrogen atoms and two bridged hydrogen atoms. The four-terminal H atoms and two B atoms lie in the same plane. Above and below this plane lie the bridged H atoms. B-H bonds formed by the terminal hydrogen atoms are normal covalent bonds while the bridge B-H bonds are three centre two-electron bonds. Each B atom forms four bonds even though boron has only three valence electrons. Hence B2H6 is an electron deficient compound.

Group 14 Elements: The Carbon Family
Carbon, silicon, germanium, tin, and lead form the carbon family.
Occurrence:
Carbon is widely distributed in nature in the free and combined states. Graphite, diamond, coal, etc are elemental forms of carbon while in the combined state it occurs as metal carbonates, hydrocarbons and CO₂ in air. Silicon is present in nature as silica and silicates. Ge is found only in traces. Tin occurs as cassiterite (SnO₂) and lead as galena (PbS)

Electronic Configuration
The valence shell electronic configuration of these elements is ns²np². The inner core of the electronic configuration of elements in this group also differs.

Covalent Radius
There is a considerable increase in covalent radius from C to Si, thereafter from Si to Pb a small increase in radius is observed. This is due to the presence of completely filled d and f orbitals in heavier members.

Ionization Enthalpy
The first ionization enthalpy of group 14 members is higher than the corresponding members of group 13. The influence of inner core electrons is visible here also. In general, the ionisation enthalpy decreases down the group.

Electronegativity
Due to small size, the elements of this group are slightly more electronegative than group 13 elements. The electronegativity values for elements from Si to Pb are almost the same.

Physical Properties
All group 14 members are solids. Carbon and silicon are non-metals, germanium is a metalloid, whereas tin and lead are soft metal.

Chemical Properties Oxidation states and trends in chemical reactivity
The group 14 elements have four electrons in outermost shell. The common oxidation states exhibited by these elements are +4 and +2.
Carbon also exhibits negative oxidation states. Since the sum of the first four ionization enthalpies is very high, compounds in +4 oxidation state are generally covalent in nature. In heavier members the tendency to show +2 oxidation state increases in the sequence Ge<Sn (i) Reactivity towards oxygen
All members when heated in oxygen form oxides. There are mainly two types of oxides, monoxide, and dioxide of formula MO and MOs respectively.

(ii) Reactivity towards water

(iii) Reactivity towards halogen
These elements can form halides of formula MX₂, and MX₄ (where X = F, Cl, Br, I). Except carbon, all other members react directly with halogen under suitable condition to make halides.

Hydrolysis can be understood by taking the example of SiCl₄. It undergoes hydrolysis by initially accepting lone pair of electrons from water molecule in d orbitals of Si, finally leading to the formation of Si(OH)₄ as shown below:

Important Trends And Anomalous Behaviour Of Carbon
Carbon differs from rest of the members of its group. It is due to its smaller size, higher electronegativity, higher ionisation enthalpy and unavailability of d orbitals. In carbon, only s and p orbitals are available for bonding and, therefore, it can accommodate only four pairs of electrons around it. This would limit the maximum covalence to four whereas other members can expand their covalence due to the presence of d orbitals.
Carbon has the ability to form pπ – pπ multiple bonds with itself and with other atoms of small size and high electronegativity.

Few examples are: C=C, C° C, C=0, C=S, and C° N. Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation.

Allotropes Of Carbon

Diamond
It has a crystalline lattice. In diamond, each carbon atom undergoes sp³ hybridisation and linked to four other carbon atoms by using hybridised orbitals in tetrahedral fashion. The C-C bond length is 154 pm. In this structure, directional covalent bonds are present throughout the lattice. It is very difficult to break extended covalent bonding and, therefore, diamond is a hardest substance on the earth. It is used as an abrasive for sharpening hard tools.

Graphite
Graphite has layered structure. Layers are held by van der Waals forces and distance between two layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C—C bond length within the layer is 141.5 pm. Each carbon atom in hexagonal ring undergoes sp² hybridisation and makes three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a π bond. The electrons are delocalised over the whole sheet. Electrons are mobile and, therefore, graphite conducts electricity along the sheet. Graphite cleaves easily between the layers and, therefore, it is very soft and slippery. For this reason graphite is used as a dry lubricant in machines running at high temperature, where oil cannot be used as a lubricant.

Fullerenes
Fullerenes are prepared by heating graphite in an electric arc in the presence of helium or argon. The sooty material formed by condensation of the vapours consists of C₆₀ with smaller amounts of C₇₀ and other fullerenes. C₆₀ is named as Buckminster fullerence. The general name fullerence refers to the family of spheroidal carbon-cage molecules. The shape of C₆₀ resembles that of a soccer ball. It contains twelve five-membered rings and twenty 6-membered rings of carbon. The 6-membered rings are fused both to other five and six membered rings. However, the 5-membered rings are fused only to six-membered rings. Both carbon-carbon single (1.435 Å) and double (1.383 Å) bonds are present in this structure. Carbon black, coke and charcoal are impure amorphous forms of graphite or fullerenes. Carbon black is formed by burning hydrocarbon in limited supply of air. Charcoal and coke are obtained by heating wood and coal respectively in the absence of air.

Uses of Carbon
Being good conductor, graphite is used for electrodes in batteries and industrial electrolysis. Crucibles made from graphite are inert to dilute acids and alkalies. Being highly porous, activated charcoal is used in adsorbing poisonous gases. Diamond is a precious stone and used in jewellery.

Some Important Compounds Of Carbon And Silicon
Oxides of Carbon
Two important oxides of carbon are carbon monoxide, CO and carbon dioxide, CO₂.

Carbon Monoxide
Direct oxidation of C in limited supply of oxygen or air yields carbon monoxide.

On commercial scale it is prepared by the passage of steam over hot coke. The mixture of CO and H₂ thus produced is known as water gas or synthesis gas.

When air is used instead of steam, a mixture of CO and N₂ is produced, which is called producer gas.

Water gas and producer gas are very important industrial fuels. Carbon monoxide in water gas or producer gas can undergo further combustion forming carbon dioxide with the liberation of heat. CO arises has the ability to form a complex with haemoglobin, which is about 300 times more stable than the oxygen-haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.

Carbon Dioxide
It is prepared by complete combustion of carbon and carbon-containing fuels in excess of air.

On commercial scale it is obtained by heating limestone. Carbon dioxide, which is normally present to the extent of ~0.03 % by volume in the atmosphere, is removed from it by the process known as photosynthesis. It is the process by which green plants convert atmospheric CO₂ into carbohydrates such as glucose. The overall chemical change can be expressed as:

The increase in combustion of fossil fuels and decomposition of limestone for cement manufacture in recent years seem to increase the CO₂ content of the atmosphere. This may lead to increase in green house effect and thus, raise the temperature of the atmosphere which might have serious consequences. Carbon dioxide can be obtained as a solid in the form of dry ice by allowing the liquified CO₂ to expand rapidly. Dry ice is used as a refrigerant for ice-cream and frozen food.

Resonance structures of carbon dioxide

Silicon Dioxide, SiO₂
Quartz, cristobatite and tridymite are some of the crystalline forms of silica, and they are interconvertible at suitable temperature. In Silicon dioxide, each silicon atom is covalently bonded in a tetrahedral manner to four oxygen atoms. Each oxygen atom in turn covalently bonded to another silicon atoms.

Silicones
They are a group of organosilicon polymers, which have (R₂SiO) as a repeating unit. The starting materials for the manufacture of silicones are alkyl or aryl substituted silicon chlorides, RnSiCl_(4-n), where R is alkyl or aryl group.

Silicates
The basic structural unit of silicates if SiO₄^4- tertrahedra. Feldspar, zerolites, mica, asbestose, etc. are examples of silicates. In silicates, either the SiO₄^4- will be present as discrete units or several such units are joined togetherth rough sharing of corner of the tetrahedra using one to four oxygen atoms per silicate unit. Like this, different silicates assume different forms such as chain, ring, sheet or three-dimensional structures. Glass and cement are examples of man-made silicates.

Zeolites
Zeolites are alumino silicates. If a few Si atoms of the three-dimensional network structure of SiO₂ are replaced by Al atoms, the resulting structure is called alumino silicate structure. This structure evidently has negative charge and Na⁺.K⁺ pr Ca²⁺ ions balance the negative charge. Zeolites are used as catalysts in petrochemical industry for cracking of hydrocarbons. ZSM-5 is a type of zeolite used in the conversion of alcohol to gasoline. Zeolites are also used in softening hard water.

Students can Download Chapter 10 The s Block Elements Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 10 The s Block Elements

Introduction
Group 1 of the periodic table consists of the elements: Lithium, Sodium, Potassium, Rubidium, Caesium and Francium. They are collectively known as alkali metals.

Group 2 consists of Beryllium, Mgnesium, Calcium, Strontium, Barium and Radium. These elements except of beryllium are known as the alkaline earth metals. The general electronic configuration of s-block elements is [noble gasjns1 for alkali metals and [noble gas] ns² for alkaline earth metals. The first elements of Group 1 and Group 2 respectively exhibit diagonal similarity, which is commonly referred to as diagonal relationship in the periodic table. The diagonal relationship is due to the similarity in ionic sizes and /or charge/radius ratio of the elements.

Group 1 Elements: Alkali Metals

1) Electronic Configuration:
All the alkali metals have one valence electron, ns¹ outside the noble gas core. The loosely held s-electron readily lose electron to give monovalent M⁺ ions.

2) Atomic And Ionic Radii:
The atomic and ionic radii of alkali metals increase on moving down the group. Hence, ionization enthalpies of the alkali metals are considerably low and decrease down the group.

3) Hydration Enthalpy:
The hydration enthalpies of alkali metal ions decrease with increase in ionic sizes. Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺ Li⁺ has maximum degree of hydration and for this reason lithium salts are mostly hydrated, e.g., LiCl- 2H₂O

Physical Properties
When heat is supplied to alkali metal or its salt the electrons are excited to higher energy levels. As these electrons return to their original level; radiations are emitted which fall in the visible region of electromagnetic spectrum. Thus they appear coloured. Li imparts crimson red colour, K gives violet colour and Na gives golden yellow colour to the flame.

Chemical Properties
The reactivity of these metals increases with their size. They burn vigorously in oxygen forming oxides. Lithium forms monoxide, sodium forms peroxide, the other metals form superoxides. The superoxide O₂ ion is stable only in the presence of large cations such as K, Rb, Cs.
4Li + O₂ → 2LizO(oxide)
2Na + O₂ → Na₂O₂ (peroxide)
M + O₂ → MO₂(superoxide)
(M=K, Rb, Cs)
Because of their high reactivity towards air and water, they are normally kept in kerosene oil.lt may be noted that although lithium has most negative E° value.

They also react with proton donors such as alcohol, gaseous ammonia and alkynes.AII the alkali metal hydrides are ionic solids with high melting points.
2M + H₂ → 2M⁺H^–.

The alkali metals readily react vigorously with halogens to form ionic halides, M⁺X^–. However, lithium halides are somewhat covalent. It is because of the high polarisation capability of lithium-ion. The alkali metals are strong reducing agents, lithium being the most and sodium the least powerful. The alkali metals dissolve in liquid ammonia giving deep blue solutions. The solutions are paramagnetic and on standing slowly liberate hydrogen.

General Characteristics Of The Compounds Of The Alkali Metals

Oxides And Hydroxides
Reactivity of alkali metals with oxygen increases down the group. Lithium, when heated in air, forms the normal oxide (Li₂O) while sodium forms the per-oxide (Na₂O₂). Potassium, Rubidium and caesium form superoxides (MO₂).
4Li + O₂ → 2Li₂O; 2Na+ O₂ → Na₂O₂; K + O₂ → KO₂

The normal oxides dissolve in water to form hydroxides (MOH) which are strong bases. However, LiOH is only slightly soluble in water and it decomposes on heating. The peroxides and superoxides also dis-solve in water to form basic hydroxides. The basic character of alkali metal hydroxides increases down the group.

Halides
Alkali metals react vigorously with halogens to form metal halides of the general formula MX. 2M+X₂ → 2MX X=F, Cl, Br or l and M= alkali metal Reactivity of alkali metal towards halogen increases from Li to Cs. Halides of alkali metals are ionic compounds readily soluble in water. But LiF is almost insoluble due to high lattice energy.

Anomalous Properties Of Lithium
The anomalous behaviour of lithium is due to the :

1. exceptionally small size of its atom and ion, and
2. high polarising power (i.e., charge/ radius ratio).

As a result, there is increased covalent character of lithium compounds which is responsible for their solubility in organic solvents.

Points Of Similarities Between Lithium And Magnesium
The similarity between lithium and magnesium is particularly striking and arises because of their similar sizes: atomic radii, Li = 152 pm, Mg= 160 pm; ionic radii: Li⁺ = 76 pm, Mg²⁺ = 72 pm. The main points of similarity are:

1. Both lithium and magnesium are hander and lighter than other elements in the respective groups.
2. Lithium and magnesium react slowly with water. Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating. Both form a nitride, Li₃N and Mg₃N₂, by direct combination with nitrogen.
3. The oxides, Li₂O and MgO do not combine with excess oxygen to give any superoxide.
4. The carbonates of lithium and magnesium decompose easily on heating to form the oxides and CO₂.

Some Important Compounds Of Sodium Sodium Carbonate (Washing Soda), Na₂CO₃.10H₂O
Sodium carbonate is generally prepared by Solvay Process.
The equations for the complete process may be written as:
2NH₃ + H₂O + CO₂ → (NH₄)₂CO₃
(NH₄)₂CO₃ + H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃ +NaCl → NH₄Cl + NaHCO₃
2NaHCO₃ → Na₂CO₃ +CO₂ +H₂O

In this process, NH₃ is recovered when the solution containing NH₄Cl is treated with Ca(OH)₂. On heating washing soda becomes monohydrate and then completely anhydrous i.e., soda ash.

Sodium Chloride, NaCl
The most abundant source of sodium chloride is seawater. Common salt is generally obtained by evaporation of seawater. Crude sodium chloride, generally obtained by crystallisation of brine solution, contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride as impurities. Calcium chloride, CaCl₂, and magnesium chloride, MgCl₂ are impurities because they are deliquescent (absorb moisture easily from the atmosphere). To obtain pure sodium chloride, the crude salt is dissolved in minimum amount of water and filtered to remove insoluble impurities. The solution is then saturated with hydrogen chloride gas. Crystals of pure sodium chloride separate out. Calcium and magnesium chloride, being more soluble than sodium chloride, remain in solution.

Uses:

• It is used as a common salt or table salt for domestic purpose.
• It is used for the preparation of Na₂O₂, Na0H and Na₂CO₃.

Sodium Hydroxide (Caustic Soda), NaOH
Sodium hydroxide is generally prepared commercially by the electrolysis of sodium chloride in Castner-Kellner cell. A brine solution is electrolysed using a mercury cathode and a carbon anode.

The amalgam is treated with water to give sodium hydroxide and hydrogen gas.
2 Na – amalgam + 2H₂O → 2NaOH + 2Hg +H₂
The sodium hydroxide solution at the surface reacts with the C02 in the atmosphere to form Na₂CO₃.

Uses:
It is used in (i)the manufacture of soap, paper, artificial silk and a number of chemicals,(ii) in petroleum refining, (iii) in the purification of bauxite, (iv) in the textile industries for mercerising cotton fabrics and (v) for the preparation of pure fats and oils.

Biological Importance Of Sodium And Potassium
Sodium ions participate in the transmission of nerve signals. The concentration gradient of Na₊ and K₊ demonstrates that-a discriminatory mechanism called sodium-potassium pump, operates across the cell membranes.

Group 2 Elements: Alkaline Earth Metals
The group 2 elements (except beryllium) are known as alkaline earth metals. The first element beryllium differs from the rest of the members and shows diagonal relationship to aluminium.
1) Electronic Configuration:
These elements have two electrons in the s-orbital of the valence shell. Their general electronic configuration may be represented as [noble gas] ns².

2) Atomic And Ionic Radii:
Within the group, the atomic and ionic radii increase with increase in atomic number due to the increased nuclear charge in these elements. They have low ionisation enthalpy and it decreases down the group with increase in size.

3) Hydration Enthalpy:
Hydration enthalpies of alkaline earth metal ions decrease with increase in ionic size down the group. Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺ The hydration enthalpies of alkaline earth metal ions are larger than those of alkali metal ions.

Physical Properties
Calcium, Strontium and Barium impart characteristic brick red, crimson and apple green colours respectively to the flame. Inflame the electrons are excited to higher energy levels and when they drop back to the ground state, energy is emitted in the form of visible light. The electrons in Be and Mg are too strongly bound to get excited by flame. Hence, these elements do not impart any colour to the flame.

Chemical Properties
The alkaline earth metals are less reactive than the alkali metals. The reactivity of these elements increases on going down the group.
Reactivity towards air and water Beryllium and Magnesium are kinetically inert to oxygen and water because of the formation of an oxide film on their surface. However, powdered beryllium burns brilliantly on ignition in air to give BeO and Be₃N₂.
Reactivity towards halogen
M + X₂ → MX₂ (X = F, Cl, Br, I)

Reactivity towards hydrogen
All the elements except beryllium form their hydrides, MH₂.BeH₂, however, can be prepared by the reaction of BeCl₂ with LiAlH₄.
2BeCl₂ +LiAlH₄ → 2BeH₂ +LiCl + AlCl₃

Reactivity towards acids:
The alkaline earth metals readily react with acids liberating dihydrogen.

General Characteristics Of Compounds Of The Alkaline Earth Metals
i) Oxides and Hydroxides: Alkaline earth metals burn in air or oxygen to form their oxides. (Oxides are also prepared by the thermal decomposition of their carbonates). Be, Mg and Ca form monoxides (MO). The tendency to form peroxide increases as the size of the metal ion increases. Strontium and barium form peroxides (MO₂)
2M + O₂ → MO (M = Be, Mg or Ca)
M+O₂ → MO₂ (M = Sr or Ba)

BeO is amphoteric in character, while the oxides of the rest of the elements in group 2 are basic. The oxides of Ca, Sr and Ba react with water to form their corresponding hydroxides.

The hydroxides of alkaline earth metals are bases except Li(OH)₂ which is amphoteric. The basic strength increases from Mg(OH)₂ to Ba(OH)₂. The solubility and thermal stability of hydroxides increase downward in the group. Be(OH)₂ and Mg(OH)₂ are almost insoluble. Ca(OH)₂ is sparingly soluble, while Sr(OH)₂ and Ba(OH)₂ are increasingly more soluble.

ii) Halides: Group 2 metals directly combine with halogen to form divalent halides of the formula

The s-Block Elements
MX₂ where X is the halogen. The metal halides are also formed by the action of halogen acids on metals, their oxides, carbonates and hydroxides. BeCl₂ is, however, prepared by passing Cl₂ over a hot mixture of BeO and coke.
In contrast to the halides of other alkaline earth metals, beryllium halides are covalent. In the solid-state BeCl₂ has a polymeric chain structure involving Be-CI-Be bridges. The anhydrous halides are hygroscopic and form hydrates such as MgCl₂.6H₂O, CaCl₂.6H₂O etc. Due to this reason, anhydrous calcium chloride is widely used as a dehydrating agent. Fluorides are relatively less soluble due to high lattice energies,

iii) Salts of Oxoacids:
The alkaline earth metals also form salts of oxoacids. Some of these are : Carbonates, Sulphates and Nitrates.

Anomalous Behaviour Of Beryllium
Beryllium differs from the rest element in many of its properties. These are

1. Beryllium has high ionisation enthalpy.
2. Small size of Be atom
3. Be does not exhibit coordination number more than four.
4. The oxides and hydroxides of Be are amphoteric in nature.

Diagonal Relationship Between Beryllium And Aluminium
The ionic radius of Be²⁺ is estimated to be same as that of the Al³⁺ ion. Hence Be resembles Al in some ways. Some of the similarities are:

1. Like AI, Be is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
2. Beryllium hydroxide dissolves in excess of alkali to give a beryllate ion just as aluminium hydroxide gives aluminate ion.
3. The chlorides of both Be and Al have Ch bridged chloride structure in vapour phase. Both the chlorides are soluble in organic solvents and are strong Lewis acids. They are used as Friedel Craft catalysts.
4. Be and Al ions have strong tendency to form complexes, BeF₄^2-, AlF₆^3-.

Some Important Compounds Of Calcium
Important compounds of calcium and their preparations are given below.

Calcium Oxide Or Quick Lime, CaO
It is prepared by the following reaction.
CaCO₃ \(\rightleftharpoons \) Ca0 + CO₂
CO₂ is removed as soon as it is produced to enable the reaction to proceed to completion.
CaO + H₂O → Ca(OH)₂
This process is called slaking of lime. CaO is a basis oxide.

Uses:

• Primary material for manufacturing cement
• It is used in the manufacturing of caustic soda
• Used to purify sugar

Calcium Hydroxide (Slaked Lime), Ca(OH)₂
It is prepared by adding water to CaO. The aqueous solution of Ca(OH)₂ is known as lime water and the suspension of slaked lime is known as milk of lime. When CO₂ is passed through lime water it turns milky due to the formation of CaCO₃
Ca(OH)₂ + CO₂ → CaCO₃ +H₂O

Uses:

• It is used in whitewash due to its disinfectant nature.
• Used in the preparation of bleaching powder.
• Used to purify sugar.

Calcium Carbonate, CaCO₂
It occurs in limestone, chalk, marble etc.
It can be prepared by the following reactions.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
CaCO₃ reacts with dilute acids to liberate carbon dioxide.

Uses:

• It is used as a flux in the extraction of metals.
• It is used as the building material of quick lime.

Calcium Sulphate (Plaster Of Paris), CaSO₄.½H₂O
It is obtained by heating gypsum (CaSO₂.2H₂O)

Above 393K anhydrous calcium sulphate is formed. This is known as ‘dead burnt plaster’

Used:

• It is used in building industry as well as plasters.
• Used to make casts of statues.

Cement
Cement is prepared by combining CaO with other materials such as clay with silica, SiO₂ along with Oxides of Al, iron and magnesium. The average composition of portland cement is:
CaO, 50-60%;
SiO₂, 20-25%;
Al₂O₃, 5-10%;
MgO, 2-3%;
Fe₂O₃, 1-2% and
SO₃, 1-2%.
When limestone and clay are heated we get cement clinker. This clinker is mixed with gypsum to form cement.

Setting of Cement:
When mixed with water the setting of cement takes place to give a hard mass. It is due to the rearrangement and hydration of molecules of constituents. Gypsum is added to slow down the setting process so it gets sufficiently hardened.

Uses:

• Used in construction of building.

Biological Importance Of Magnesium And Calcium
Human body contains about 25g of Mg and 1200g of Ca. Mg is a cofactor in enzymes which use ATP in phosphate transfer process in our body. Photosynthesis in plants takes place in presence of chlorophyll which contains Mg. About 99% of body calcium is found in teeth and bones. Calcium concentration in plasma is regulated at 100mg/litre in presence of hormones such as calcitonin and parathyroid hormone.

Students can Download Chapter 6 Work, Energy and Power 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 6 Work, Energy and Power

Summary
The Scalar Product

The scalar product (or) dot product of any two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) is defined as

Where ‘q’ is the angle between \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\)
Note: The dot product of A and B is a scalar quantity. Geometrical meaning of \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\)
We know \(\overrightarrow{\mathrm{A}}\) . \(\overrightarrow{\mathrm{B}}\) = ABcosθ
= A(Bcosθ)
= B(A cosθ)

Properties

Question 1.

Answer:

Work Energy Theory
Statement:
The change in kinetic energy of a particle is equal to the work done on it by the net force.
Proof:
We know v² = u² + 2as
v² – u² = 2as
Multiplying both sides with m/2; we get

Work
Definition:
The work done by the force is defined as the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Explanation


Consider a constant force \(\overrightarrow{\mathrm{F}}\) acting on an object of mass m. The object undergoes a displacement d in the positive x direction as shown in the figure. The projection of \(\overrightarrow{\mathrm{F}}\) on d is Fcosθ.
Hence work done w = Fcosθ. d

There are three types of workdone

1. positive workdone
2. negative workdone
3. zero workdone

1. Positive workdone:
Work will be positive, if the displacement has a component in the direction of the force. The angle between force and displacement is zero for positive workdone.
When q = 0 w = Fd
Example

• A person carrying a load climbing up a staircase
• A body being pushed along a surface
• A body falling under gravitation force.

2. Negative workdone:
Work will be negative, if the displacement has a component opposite to the force F. The angle between force and displacement lies between 90° and 180°.
Example

• When a person carrying a load on his head climbs down a staircase, (applied force by him on the load is upwards and the displacement is opposite to it)
• When a body slides along a rough surface the displacement is opposite to the frictional force. Therefore the workdone by the frictional force is negative.

3. Zero work done:
Work will be zero, if there is no component along the direction of force. The angle between applied force and displacement is 90°.
Example
1. When a person carrying a load on his head walks along a level road, the displacement is perpendicular to the force and therefore the work done is zero.

2. In uniform circular motion the centripetal force is along the radius and direction of displacement is along the tangent.

Kinetic Energy
Kinetic energy is the energy possessed by the body because of it’s motion. Kinetic energy of a body of mass m and velocity v,

Workdone By Variable Force

Consider a body moving from x_i to x_f under a variable force. The variation of force with position is shown in graph. Consider a small AB = Dx. The force in this interval is nearly a constant.

Hence workdone to move a body from A to B is. Dw = F(x) Dx.F(x)dx gives the area of rectangle ABCD. When we add successive rectangular areas, we get total work as

When we take Dx tends to zero, the summation can be replaced integration.

Work Energy Theory For A Variable Force
Work energy theorem for a variable force can be derived from work energy theorem of constant force. According work energy theorem for constant force, Change in kE = work done
dk = dw
dk = F dx
Integrating from the initial position (x_i) to (x_f) we get

Concept Of Potential Energy
Potential energy of a body is the energy possessed by it because of its position.
Explanation
Considera mass ‘m’on the surface of the earth. If this mass is raised to height ‘h’ against force of gravity,
work done w = Force × displacement
w = mg × h
w = mgh
This work gets stored as gravitational potential energy.
ie; Gravitational energy V = mgh.

1. Relation between gravitational potential and gravitational force:
If we take negative of the derivative of V(h) with respect to height (h), we get

Where F is gravitational force. The above equation shows that gravitational force is the negative derivative of gravitational potential.

2. Relation between kinetic energy and gravitational potential energy:
Considera body of mass ‘m’ at a height ‘h’ from the surface of the earth. The potential energy at height h
pE = mgh ______(1)
If the body is allowed to fall from this height, it attains kinetic energy,
kE = \(\frac{1}{2}\)mv² _______(2)
But velocity at surface can be found from the formula
v² = u² + 2as
v² = 2gh [Since u = 0, a = g, s = h]
Substituting this value in eq(2), we get
kE = \(\frac{1}{2}\) m2gh
kE = mgh
kE = pE [∵ pE = mgh]

Properties of conservative force:

• A force is conservative, if it can be derived from a scalar quantity (ie F = – \(\frac{d V}{d x}\))
• The workdone by the conservative force depends only on the end points.
• The workdone by conservative force in a closed path is zero.

Conservation of mechanical energy for a freely falling body:

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
Potential energy at A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2}\)mv² = 0
(since the body at rest, v = 0).
∴ Total mechanical energy = PE + KE = mgh + 0
= mgh.

Total energy at the point B
The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
v² = u² + 2as
v² = 0 + 2 gx
∴ KE at B, = \(\frac{1}{2}\)mv²
\(\frac{1}{2}\)m2gx
= mgx
P.E. at B, = mg (h – x)
Total mechanical energy = PE + KE
= mg (h – x) + mgx = mgh.

Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2gh
∴ KE at C, = \(\frac{1}{2}\)mv²
\(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE
= 0 + mgh = mgh.

The Potential Energy Of A Spring
Hooks law:
The restoring force developed in the spring is proportional to the displacement x and it is opposite to the displacement,
ie Fα – x

Where k is a constant called the spring constant.
Potential energy stored in a spring:

Consider a massless spring fixed to a rigid support at one end and a body attached to the other end. The body moves on a frictionless surface.

If a body is displaced by a distance dx, The work done for this displacement
dw = Fdx
∴ Total work done to move the body from x = 0 to x


This workdone is stored a potential energy in a spring. Hence potential energy of a spring.

Spring force is a conservative force:
If the spring is displaced from an initial position x_i to x_f and again to x_i;

W = 0
This zero workdone means that spring force is conservative.
Energy of a oscillating spring at any point:

If the block of mass ‘m’ (attached to massless spring) is extended to x_m and released, it will oscillate in between +x_m and -x_m. The total mechanical energy at any point x, (lies between -x_m and +x_m) is

This block mass ‘m’ has maximum velocity at equilibrium equi¬librium position (x = 0). At this position, the potential energy stored in a spring is completely converted in to kinetic energy.

Graphical variation of energy

The Law Of Conservation Of Energy
Statement:
Energy cannot be created or destroyed. It can be transformed from one form to another.

Question 2.
Prove conservation of energy for a freely falling body.
Answer:
Conservation of mechanical energy for a freely falling body:

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
Potential energy at A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2}\)mv² = 0
(since the body at rest, v = 0).
∴ Total mechanical energy = PE + KE = mgh + 0
= mgh.

Total energy at the point B
The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
v² = u² + 2as
v² = 0 + 2 gx
∴ KE at B, = \(\frac{1}{2}\)mv²
\(\frac{1}{2}\)m2gx
= mgx
P.E. at B, = mg (h – x)
Total mechanical energy = PE + KE
= mg (h – x) + mgx = mgh.

Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2gh
∴ KE at C, = \(\frac{1}{2}\)mv²
\(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE
= 0 + mgh = mgh.

Various Form Of Energy
1. Heat:
Heat is a one form of energy, it is the internal energy of molecule.

2. Chemical energy:
Chemical energy arises from the fact that the molecules participating in the chemical reaction have different binding energies.

If the total energy of the reactants is more than the products of the reaction, heat is released and the reaction is said to be exothermic reaction. If the heat is absorbed in chemical reaction it is called endothermic.

3. Electrical energy:
The flow of electrons produce electric current.

4. The equivalence of mass and energy Mass and energy are equivalent and are related by the relation. E = mc², where C, the speed of light in. vacuum.

Question 3.
How much energy will be liberated, when 1 Kg. matter converts in to energy?
Answer:
Energy liberated E = mc²
E = 1 × (3 × 10⁸)²
= 9 × 10¹⁶J.

5. Nuclear energy:
Nuclear energy is obtained from the sun. In this case four light hydrogen nuclei fuse to form a helium nucleus, whose mass is less than the sum of the masses of the reactants. This mass difference (called the mass defect on) is the source of energy.

Power
Power is defined as the time rate at which work is done.

Expression for power in terms of F and V:
The work done (dw) by a force F for a displacement dr is


Where \(\overrightarrow{\mathrm{V}}\) is the instantaneous velocity when the force is \(\overrightarrow{\mathrm{F}}\).
Unit:
Unit of power is watt. 1 watt = 1J/S.
There is another unit of power, namely the horse power (hp)
1 hp = 746w
Kilowatt hour kwh:
Kilowatt hour (kwh) is the unit of energy used to mea-sure electrical energy. One kilowatt hour is the energy consumed in one hour at the rate of 1000 watts/ second.
1 kwh = 1000 watts × 60 × 60 seconds
= 3.6 × 106ws
1 kwh = 3.6 × 106J
Note : kwh is a unit of energy and not of power.

Collisions
There are two types of collisions.

1. Elastic collision
2. Inelastic collision

1. Elastic collision:
Elastic collision is one in which both momentum and kinetic energy are conserved.
Eg:

• collision between molecules and atoms
• collision between subatomic particles.

Characteristics of elastic collision:

• Momentum is conserved
• Total energy is conserved
• K. E. is conserved
• Forces involved during collision are conservative forces

2. Inelastic collision:
Inelastic collision is one in which the momentum is conserved, but KE is not conserved.
Example.

• Mud thrown on a wall
• Any collision between macroscopic bodies in every day life.

Characteristics of inelastic collision:

• Momentum is conserved
• Total energy is conserved
• K.E. is not conserved
• Forces involved are not conservative
• Part or whole of the KE is converted into other forms of energy like heat, sound, light etc.

2. Collisions in one Dimension:
If the initial velocities and final velocities of both the bodies are along the straight line, then it is called one dimensional motion.

Consider two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ in the same direction and in the same line. If u₁ > u₂ they will collide. After collision let v₁ and v₂ be their velocities.
By conservation of linear momentum.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ______(1)
m₁u₁ – m₂u₂ = m₁v₁ – m₂v₂ ______(2)
This is an elastic collision, hence K.E. is conserved.

To find v₁ and v₂:


Discussion
Case -1 Mass of two bodies are equal
(i.e. m₁ = m₂ = m). Substitute these values in (7) and (8), we have

ie. bodies exchange their velocities.

Case – 2 (If u₂ =0 and m₂ >> m₁ ie; m₁ – m₂ ≈ -m₂, m₁ + m₂ ≈ -m₂)

The second body remains at rest while the first body rebounds with the same velocity.
Collisions in Two Dimensions:

Consider two bodies of masses m₁ and m₂ moving with velocities u₁ and u₂ along parallel lines. If u₁ > u₂ they will collide. Let v₁ and v₂ be their velocities after collision along directions θ₁ and θ₂. v₁ and v₂ can be resolved in to v₁ cosθ₁, v₂cosθ₂ parallel to x axis and v₁ sinθ₁ and v₂sinθ₂ parallel to y axis.
By conservation of momentum parallel to X-axis,

m₁u₁ + m₂u₂ = m₁v₁ cosθ₁ + m₂v₂ cosθ₂
By conservation of momentum parallel to y-axis.
m₁v₁sinθ₁ + m₂v₂ sinθ₂ = 0 + 0 = 0
By conservation of energy

Students can Download Chapter 9 Hydrogen Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Notes Chapter 9 Hydrogen

Introduction
Hydrogen has the simplest atomic structure all the elements around us in nature. It consists of only one proton and one electron.

Position Of Hydrogen In The Periodic Table
Hydrogen is the first element in the periodic table. Hydrogen has electronic configuration 1 s1. On one hand, its electronic configuration is similar to the outer electronic configuration (ns¹) of alkali metals. On the other hand, it is short by one electron to the corresponding noble gas configuration, helium (1s²). It has resemblace to both alkali metals and halogens.

Dihydrogen, H₂

Isotopes Of Hydrogen
There are three isotopes of hydrogen with mass numbers 1,2 and 3. They are called protium, deuterium and tritium respectively. Their natural abundances . are in the ratio l:1.56 × 10^-2: 1 × 10^-17 respectively.

1. Protium (ordinary hydrogen)(\(_{ 1 }^{ 1 }{ H }\)): It is the most abundant isotope of hydrogen. Its nucleus contains one proton and no neutron.
2. Deuterium (heavy hydrogen, \(_{ 1 }^{ 2 }{ H }\) or D): Heavy hydrogen is prepared from heavy water (D₂O) which is obtained by electrolysis of ordinary water.
3. Tritium has 2 neutrons in the nucleus.

Preparation Of Dihydrogen, H₂
It is usually prepared by the following reactions:

3. Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yields hydrogen.

The mixture of CO and H₂ is called water gas. As this mixture of CO and H₂ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or ‘syngas’. Nowadays ‘syngas’ is produced from sewage, sawdust, scrap wood, newspapers etc. The process of producing ‘syngas’ from coal is called ‘coal gasification’.

This reaction is called water-gas shift reaction.

Properties Of Dihydrogen

Physical Properties
Dihydrogen is a colourless, odourless, tasteless, combustible gas. It is lighter than air and insoluble in water.

Chemical Properties
Dihydrogen is not particularly reactive because of its high bond dissociation enthalpy. However, hydrogen forms compounds with almost all elements at high temperature or in presence of catalysts.
Reaction with halogens:
H₂ (g) + X₂(g) → 2HX(g) (X = F, Cl, Br, l)
Reaction with dioxygen:

Uses Of Hydrogen

1. Hydrogen is used in the manufacture of ammonia by Haber process, water gas, fertilisers etc.
2. It is used in the hydrogenation of vegetable oils and as a reducing agent.
3. It is used in the production of methanol and synthetic petrol.
4. Liquid hydrogen is used in as rocket fuel along with liquid oxygen.
5. It is used in oxy-hydrogen torch used for welding.

Hydrides
Hydrogen can form binary compounds with almost all elements. These are known as hydrides.
The hydrides are classified into three categories:

1. Ionic or saline or salt like hydrides
2. Covalent or molecular hydrides
3. Metallic or non-stoichiometric hydrides

Ionic Or Saline Hydrides
These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. However, significant covalent character is found in the lighter metal hydrides such as LiH, BeH₂ and MgH₂.

Covalent Or Molecular Hydride
Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are CH₄, NH₃, H₂O and HF. For convenience hydrogen compounds of nonmetals have also been considered as hydrides. Molecular hydrides are further classified according to the relative numbers of electrons and bonds in their Lewis structure into :

1. electron-deficient,
2. electron-precise,and
3. electron-rich hydrides.

Group13 elements form electron deficient compounds. They act as Lewis acids i.e., electron acceptors. eg.B₂H₆ Group 14 elements form electron precise compounds. They have required number of electrons. eg.CH₄. Electron-rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. (NH₃ has 1 – lone pair, H₂O – 2 and HF -3 lone pairs).They will behave as Lewis bases.

Metallic Or Non-Stoichiometric (Or Interstitial) Hydrides
These are formed by many d-block and f-block elements. However, the metals of group 7, 8 and 9 do not form hydride. Even from group 6, only chromium forms CrH. These hydrides conduct heat and electricity though not as efficiently as their parent metals do. Unlike saline hydrides, they are almost always nonstoichiometric, being deficient in hydrogen. For example, LaH_2.87 & YbH_2.55

Water
Water is a colourless tasteless liquid. A major part of all living organisms is made up of water.The unusual properties of water is due to the presence of extensive hydrogen bonding between water molecules.

Structure Of Water
In the gas phase water is a bent molecule with a bond angle of 104.5°, and O-H bond length of 95.7 pm

(a) The bent structure of water;
(b) the water molecule as a dipole

Structure Of Ice
The crystalline form of water is ice. At atmospheric pressure, ice crystallises in the hexagonal form, but at very low temperatures it condenses to cubic form. Hydrogen bonding gives ice a rather open type structure with wide holes. These holes can hold some other molecules of appropriate size interstitially. Density of ice is less than that of water. Therefore, an ice cube floats on water. In winter season ice formed on the surface of a lake provides thermal insulation which ensures the survival of the aquatic life.

Chemical Properties of Water
1) Amphoteric Nature:
It has the ability to act as an acid as well as a base i.e., it behaves as an amphoteric substance. In the Bronsted sense it acts as an acid with NH₃ and a base with H₂S.

2) Redox Reactions Involving Water
Water can be reduced and oxidised:
2H₂O(l) + 2Na(s) → 2NaOH(aq) + H₂(g): reduction Water is oxidised to O₂ during photosynthesis
6CO₂(g) +12H₂O(I) → C₆H₁₂O₆ (aq) + 6H₂O(I) + 6O₂(g)

3) Hydrolysis Reaction:
Due to high dielectric constant, it has a very strong hydrating tendency.
P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq)

4) Hydrates Formation:
From aqueous solutions, many salts can be crystallised as hydrated salts. Such an association of water is of different types viz.,
i) Coordinated water e.g.,
[Cr(H₂O)₆]₃+3Cl^–
ii) Interstitial water.g., BaCl₂.2H₂O
iii) hydrogen-bonded water.g.,
[Cu(H₂O)₄]²⁺ SO₄^2-.H₂O in CuSO₄.5H₂O

Hard And Soft Water
Water which produces lather with soap solution readily is called soft water. For example, rainwater, distilled water etc. Water which does not produce lather with soap solution readily is called hard water, eg: Sea water, water from certain rivers.

Hardness of water is due to the presence of bicarbonates, chlorides and sulphates of calcium and magnesium. The calcium and magnesium ions present in hard water form insoluble salts with soap and prevent the formation of lather.

Temporary Hardness
Temporary hardness is due to the presence of mag-nesium and calcium hydrogencarbonates.
It can be removed by boiling.

Permanent Hardness
It is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. Permanent hardness is not removed by boiling. It can be removed by the following methods:
i) Treatment with washing soda (sodium carbonate):
Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.
MCl₂ → MCO₃ ↓ 2NaCl (M=Mg, Ca)
MSO₄ + Na₂CO₃ → MCO₃ ↓ +NaSO₄

ii) Calgon’s method:
Sodium hexametaphosphate (Na₆P₆O₁₈), commercially called ‘calgon’, when added to hard water, the following reactions take place.
Na₆P₆O₁₈ → Na^{+ + Na₄P₆O₁₈2- (M=Mg, Ca)
M2+ + Na₄P₆O₁₈2- → [Na₂MP₆O₁₈]2- + 2Na+}

iii) Ion-exchange method:
This method is also called zeolite/perm utit process. Hydrated sodium aluminium silicate iszeolite/permutit.Forthe sake of simplicity, sodium aluminium silicate (NaAlSiO₄) can be written as NaZ.
2NaZ(s) + M²⁺(aq) → MZ₂(s) + 2Na⁺(aq) (M=Mg, Ca)
MZ₂ (S) + 2NaCl(aq) → 2NaZ(s) + MCl₂(aq)

iv) Synthetic resins method:
Nowadays hard . water is softened by using synthetic cation exchangers. This method is more efficient than zeolite process.Ion exchange resin (RSO₃H) is changed to RNa by treating it with NaCI. Here R is resin anion.
2RNa(s) + M²⁺(aq) → R₂M(s) + 2Na⁺(aq)

The resin exchanges Na+ ions with Ca²⁺ and Mg²⁺ ions present in hard waterto make the water soft.

HYDROGEN PEROXIDE (H₂O₂)
It can be prepared by the following methods.

Structure
Hydrogen peroxide has a non-planar structure.

Chemical Properties
i) Oxidising action in acidic medium
PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)

ii) Reducing action in acidic medium
HOCl + H₂O → H₃O⁺ +Cl^– + O₂

iii) Oxidising action in basic medium
Mn²⁺ +H₂O₂ → Mn⁴⁺ + 2OH^–

iv) Reducing action in basic medium
2MnO₄^– + 3H₂O₂ → MnO₂ + O₂ + 2H₂O + OH^–

Uses

1. As a bleaching agent for textiles, wood and paper pulp
2. In the manufacture of chemicals such as sodium perborate, epoxides etc.
3. A dilute solution of H₂O₂ is used as a disinfectant. This solution is used as an antiseptic for wounds, teeth and ears under the name perhydrol.
4. iv) It is used in pollution control treatment of domestic and industrial effluents.

Heavy water. D₂O
It is extensively used as a moderator in nuclear reactors and in exchange reactions for the study of reaction mechanisms. It can be prepared by exhaustive electrolysis of water or as a by-product in some fertilizer industries.lt is used for the preparation of other deuterium compounds.

Dihydrogen As A Fuel
Due to extensive use, our reserves of fossil fuels are fast depleting. A prospective alternative in this regard is what is known as hydrogen economy. The major idea behind hydrogen economy is the storage and transportation of energy in the form of gaseous and liquid hydrogen. Hydrogen can replace fossil fuels in automobiles, and coal or coke in industrial processes involving reduction. Hydrogen fuel can release more energy per unit weight of the fuel than our conventional fuels. Hydrogen oxygen fuel cells can be used for generating power in automobiles. Liquid hydrogen has already been used as rocket fuel along with liquid oxygen.

The technology involves the production of bulk quantities of hydrogen and its storage in liquid form in vacuum insulated cryogenic tanks. Transport of liquid hydrogen by road or rail, or through pipelines is feasible. Certain metal alloys can be used as smaller storage units for hydrogen.

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

Kerala Plus One Chemistry Notes Chapter 8 Redox Reactions

Introduction
The reaction which involve both oxidation and reduction reactions is called Redox reaction.

Classical Idea Of Redox Reactions Oxidation And Reduction Reactions
“Oxidation” is defined as the addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance. Examples of oxidation:

1. Addition of oxygen 2Mg + O2 → 2MgO
2. Removal of hydrogen 2H₂S + O₂ → 2S + 2H₂O
3. Addition of electronegative element Mg + Cl₂ → MgCl₂

The term reduction been broadened these days to include removal of oxygen/electronegative element from a substance or addition of hydrogen /electropositive element.

1. Removal of electronegative element FeCl₃ + H₂ → 2FeCl₂ + 2HCl
2. Removal of Oxygen (2H₂O → 2Hg + O₂)
3. Addition of Hydrogen (H₂ + Cl₂ → 2HCl)

Redox Reactions In Terms Of Electron Transfer Reactions
According to electronic concept, the processes which involves loss of electrons are called oxidation reactions. Similarly, processes which involve gain of electrons are called reduction reactions.
The atom which reduced, act as oxidising agent and the atom which oxidised act as reducing agent.For example;
2Na(s) + Cl₂(g) → 2Na⁺Cl^–(s) or 2NaCl(s)
Here Na is oxidised and Cl is redused.

Competitive Electron Transfer Reaction
Place a strip of metallic zinc in an aqueous solution of copper nitrate. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn²⁺ ions among the products can easily be judged when the blue colour of the solution due to Cu2+ has disappeared. The reaction is,
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) zinc is oxidised, releasing electrons, something must be reduced, accepting the electrons lost by zinc. Copper ion is reduced by gaining electrons from the zinc.

Oxidation Number
Oxidation number of an element may be defined as the charge which an atom of the element has or appears so have when present in the combined state in a compound.

1. Electrons shared between two like atoms are divided equally between the sharing atoms.
2. Electrons shared between two unlike atoms are counted with the more electronegative atom. Atoms can assume positive, zero or negative values of oxidation numbers depending on their state of combination. Oxidation number can be a fraction in some cases.

The rules for calculation of oxidation number are:
1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H2 has the oxidation number zero.

2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na+ ion has an oxidation number of +1, Mg²⁺ion, +2, Fe³⁺ ion, +3, Cl^– ion, -1, O^2- ion, -2; and so on. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.

3. The oxidation number of oxygen in most compounds is-2. However, we come across two kinds of exceptions here.in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of—1, in superoxides (e.g., KO₂, RbO₂) each oxygen atom is assigned an oxidation number of -(½). The second exception appears rarely, i.e. when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.

4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). For example, in LiH, NaH, and CaH₂, its oxidation number is —1.

5. In all its compounds, fluorine has an oxidation number of-1. Other halogens (Cl, Br, and I) also have an oxidation number of-1, when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.

6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO₃)^2- must equal -2. A term that is often used interchangeably with the oxidation number is the oxidation state. Oxidation state of a metal is a compound is sometimes represented by Stock notation. According to this, the oxidation number is written as Roman numeral in parenthesis after the symbol of the metal in the molecular formula. e.g.,Fe(ll)0, Sn(IV), Cl₄,Mn(IV)O₂.

Problem
Using Stock notation, represent the following compounds HAUCl₄, Ti₂O, FeO, Fe₂O₃, Cul, CuO, MnO and MnO₂.

Solution
By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
HAuCl₄ → Au has 3
Tl₂O → Tl has 1
FeO → Fe has 2
Fe₂O₃ → Fe has 3
Cul → Cu has 1
CuO → Cu has 2
MnO → Mn has 2
MnO₂ → Mn has 4

Therefore, these compounds may be represented as
HAU(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)l, Cu(II)O, Mn(II)O, Mn(IV)O₂.

In terms of oxidation number, oxidation may be defined as a chemical change in which there occurs an increase in the oxidation number of an atom or atoms. Reduction may be defined as a chemical change in which there occurs a decrease in the oxidation number of an atom or atoms. Thus, a redox reaction may be defined as a reaction in which the oxidation number of atoms undergoes a change.

Types Of Redox Reactions
1. Combination Reactions:
A combination reaction may be denoted in the manner
A + B → C

2. Decomposition Reaction:
Decomposition reactions are the opposite of combination reactions.
For example, 2H₂O → 2H₂ + O₂

3. Displacement Reaction:
In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as:
X +YZ → XZ + Y
Displacement reactions fit into two categories:
metal displacement and non-metal displacement.

a) Metal displacement:
A metal in a compound can be displaced by another metal in the uncombined state.
CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq)

b) Non-metal displacement:
The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement.
2Na(s) + 2H₂O(I) → 2NaOH(aq) + H₂(g)

The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table.

Note:
fluorine is the strongest oxidising agent; there is no way to convert F^– ions to F₂ by chemical means. The only way to achieve F₂ from F^– is to oxidise electrolytically,

4. Disproportionation Reactions:
In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.

Balancing Of Redox Reactions
There are two ways to balance a redox equation.
They are oxidation number method and Half Reaction Method.

a) Oxidation Number Method
The various steps involved in this method are:

1. Write the skeletal equation and assign oxidation numbers to each element. Identify the elements undergoing change in oxidation number.
2. Find out the increase or decrease of oxidation number per atom. Multiply the increase or decrease of oxidation number with number of atoms undergoing the change.
3. Multiply the formulae of the oxidising agent and the reducing agent by suitable integers so as to equalize the total increase or decrease in oxidation number as determined in the above step.
4. Balance the equation with respect to all atoms other the term reduction has than oxygen and hydrogen.
5. Balance oxygen atoms by adding equal number of H₂O molecules to the side deficient in oxygen atoms.
6. For reaction taking place in acidic medium, add H⁺ ions to the side of deficient in hydrogen atoms.
7. For reaction taking place in basic medium, add H₂O molecules to the side deficient in hydrogen atoms and simultaneously add equal number of OH ions on the other side of the equation.

Problem
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation forthe reaction.
Solution:
The skeletal ionic equation is:
MnO₄^–(aq) + Br^–(aq) → MnO₂(s) + BrO₃^–(aq)

Assign oxidation numbers for Mn and Br

this indicates that permanganate ion is the oxidant and bromide ion is the reductant.

Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.

As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH^– ions on the right to make ionic charges equal.
2MnO₄^–(aq) + Br^–(aq) → 2MnO₂(s) + BrO₃^–(aq) + 2OH^–(aq)

Finally, count the hydrogen atoms and add appropri- ‘ ate number of water molecules (i.e. one H20 molecule) on the left side to achieve balanced redox change.
2MnO₄^–(aq) + Br^–(aq) → 2MnO₂(s) + Br0₃^–(aq) + 2OH^–(aq)

b) Half Reaction Method
This method involves identifying the oxidation and reduction reactions in the given skeletal equation and then splitting the reaction accordingly as two half reactions. Each half reaction is then balanced systematically in various steps as outlined below.

Step 1.
Write the skeletal equation and identify the oxidant and reductant.

Step 2.
Write the half reactions for oxidation and reduction separately.

Step 3.
Balance the half reaction with respect to atoms that undergo change in oxidation number. Add electron to whichever side is necessary, to make up for difference in ON.

Step 4.
Balance O-atoms by adding proper number of H₂O molecules to the side deficient in oxygen atoms.

Step 5.
For ionic equations in acid medium, add sufficient H⁺ ions to the side deficient in hydrogen. If the reaction occurs in basic medium, add sufficient H₂O molecules to the side deficient in H atoms to balance H atoms and equal number of hydroxyl ions on the opposite side.

Step 6.
Equalise the number of electrons lost or gained by multiplying the half reaction with suitable integer and add the half reactions to get the final balanced equation.

Problem
Permanganate (VII) ion, MnO₄^– in basic solution oxidises iodide ion, l^– to produce molecular iodine (l₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:

Redox Reactions As The Basis For Titrations
In redox systems, the titration method can be adopted to determine the strength of a reductant/ oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below:
1. In one situation, the reagent itself is intensely coloured, e.g., permanganate ion, MnO₄^–. Here MnO₄^– – acts as the self indicator. The visible endpoint, in this case, is achieved after the last of the reductant (Fe²⁺ or C₂O₄^2-) is oxidised and the first lasting tinge of pink colour appears at MnO₄^– concentration as low as 10^-6 mol dm^-3 (10^-6 mol L^-1), This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.

2. If there is no dramatic auto-colour change (as with Mn04 – titration), there are indicators which are oxidised immediately after the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr2072-, which is not a self-indicator, but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signalling the endpoint.

Redox Reactions And Electrode Pro-Cesses
When zinc rod is dipped in copper sulphate solution, zinc gets oxidised to Zn²⁺ while Cu²⁺ ions are reduced to Cu due to direct transfer of electrons. However, if a zinc rod dipped in ZnSO₄ solution taken in a breaker is connected externally by a conducting wire to a copper rod placed in CuSO₄ solution in another beaker, electrons are transferred indirectly from Zn to Cu. Now, each beaker contains both the oxidised and reduced form of the same substance ‘ called a redox coupe. In this experiment the redox couples developed are Zn²⁺/Zn and Cu²⁺/Cu When the solutions in the two beakers (called electrodes) are joined by a salt bridge (a U-tube containing a solution of KCl, solidified in presence of agar-agar), electrons flow from Zn to Cu while current flows in the reverse direction. The salt bridge provides electrical continuity between the solutions without allowing them to mix with each other. The flow of current is due to a potential difference between Cu and Zn electrodes (or half cells). This experimental set up gives an electrochemical cell or galvanic cell.

The potential of an electrode is a measure of its ability to lose (oxidation) or gain (reduction) electrons. When the concentrations of solutions in the half cells are unity and the temperature is 298 K, the potential of each electrode is known as Standard Electrode Potential (E°). By convention, E° of hydrogen electrode is zero volts and the potential of other electrodes will be a measure of the relative tendency of the active species to be in oxidised/reduced form. A negative E°shows that the redox couple is a stronger reducing agent than H⁺/H₂ couple.

A positive E° shows that the redox couple is a weaker reducing agent than H⁺/H₂ couples. The values of standard reduction potentials of various electrodes are given in the increasing order in an electrochemical series (electromotive series)