1st PUC

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Karnataka 1st PUC Accountancy Question Bank Chapter 1 Introduction to Accounting

1st PUC Accountancy Introduction to Accounting One Mark Questions and Answers

Question 1.
What is Book-keeping?
Answer:
Book-keeping is the art and science of recording in the books of account.
The monetary aspect of commerical and financial transactions.

Question 2.
Define Accounting.
Answer:
“The art of recording classifying and summarising in a significant manner and in terms of ‘ money transactions event which are, in part atleast, of a financial character and interpreting the results there of “American certificed public accountants”.

Question 3.
Write any two features of Accounting.
Answer:
Features of accounting are :

1. It is a process of recording business transactions.
2. Accounting is grouping the transactions according to their nature of heads.

Question 4.
Define Accountancy.
Answer:
According to Eric Kohler “Accountancy is the theory and practices of Accounting.

Question 5.
Mention the branches of accounting.
Answer:
Branches of accounting are :

1. Financial accounting.
2. Cost accounting.
3. Management accounting.

Question 6.
Mention the objectives of Accountancy.
Answer:

• To maintain the record of financial transactions of a business accurately.
• To ascertain the profit or loss made by business.
• To present the true and fair view of financial position.
• To know the amount due to creditors individually.

Question 7.
What are transactions?
Answer:
Business transactions means, any activity, dealing or event which has value measurable in terms of money related to business.

Question 8.
What is cash a transactions?
Answer:
Any business transaction which involves immediate payment or receipt of cash called cash transactions.

Question 9.
What is credit transactions?
Answer:
Any business transaction which involves postpone of payment or receipt to a future date called credit transactions.

Question 10.
What is capital?
Answer:
Capital represents the owner’s claim or share in the assets of the business. Amount invested by owner of business called capital.

Question 11.
Write the meaning of drawings.
Answer:
The amount of cash or any asset withdrawn by the owner of the business for his personal use or domestic use we called as drawings.

Question 12.
What are Assets?
Answer:
Assets are the properties or resources which are owned by the business entity.
Ex : Machinery, stock, goodwell, etc.

Question 13.
Who is a debtor?
Answer:
Debtor is a person who owes any amount to business. In other words, who purchase goods from business on credit basis is called debtors.

Question 14.
Who is a creditor?
Answer:
Creditor is a person to whom any sum of money is owed by business, other words the person who give benefits to business and amount payable, such person called creditors.

Question 15.
What are goods?
Answer:
The term goods includes all commodities, articles or products which are purchased for the purpose of re-sale.

Question 16.
What is stock?
Answer:
The goods purchased for sale, remain unsold called goods. It is a asset for the business.

Question 17.
Give the meaning of profit.
Answer:
It is an excess amount of revenues over the related cost or expenses, [profit = Revenue – expenditure].

Question 18.
Write the meaning of income and gain.
Answer:
Income: It refer to an amount received for sale of goods and service or for use of any rights belonging to business.
Gain: Increase in the value of assets or resources of business called gains.

Question 19.
What is Discount?
Answer:
Discount: Reducing the value of sales called discounting. Discounts are 2 types, Trade Discount and Cash Discount.

Question 20.
What is Vocher?
Answer:
Voucher: is the document which helps in reseeding business transactions.

Question 21.
Define Book-keeping.
Answer:
“Book keeping is the art of recording business transaction in a systematic manner”-Rosen.

Question 22.
Write any two features of Book-keeping.
Answer:
Features of book-keeping are :

1. It is the recording of only business transactions.
2. It is the recording of business transactions in terms of money. .
3. It is very systamatic and principled manner.

Question 23.
Question Write two objects of book-keeping.
Answer:
Objects of book of book-keeping are :

1. To have a permanent record of all business transactions.
2. To ascertain the net result of the business
3. To know the exact reasons for net profit or loss.
4. To know the progress of business from year to year
5. To minimise errors and frauds.

Question 24.
Differentiate between Book-keeping and Accounting.
Answer:

Book-keeping	Accounting
It is only a recording of business transaction.	It is a recording of, analysing, summarising of business transactions.
Book-keeping just maintaining business information.	Book-keeping is accounting, analysing and interprets the information.

Question 25.
Who is an accountant?
Answer:
An officer who is entrasted with the accounting function of the organisation called accountant.

Question 26.
Mention the two classification of Book-keeping.
Answer:
The two system of book-keeping are:

1. Single entry system of book-keeping.
2. Double-entry system of book-keeping.

Question 27.
Write the two advantages of single entry system.
Answer:

1. It is a simple method of recording transactions.
2. It is less costly when compared to double entry system.

Question 28.
Write any two disadvantages of single entry system.
Answer:

1. This system is an incomplete system of book-keeping.
2. This system is not supporting to prepare trial balance.

Question 29.
What is double entry system?
Answer:
The system of making two sides in the books of each contracting party for recording a transactions completely called double entry system.

Question 30.
Write the advantages of double entry system.
Answer:

• It provides a complete or full records of all transaction .
• It is a systematic and scientific manner of recording business transactions.

Question 31.
Write the disadvantages of double entry system.
Answer:

• Under this method number of books of accounts have to be maintained.
• It consumes more time and money.

Question 32.
Write the different types of business transactions.
Answer:
The business transactions can be classified as:

1. Cash transactions
2. Credit transactions
3. Barter transactions
4. Non-cash transactions

Question 33.
What is Entity?
Answer:
Entity means an area of economic interest of a particular industry or group of industries. Seperate books of accounts are kept for each entity.

Question 34.
What is intangible assets? Give examples.
Answer:
The assets which we cannot see and touch called intangible assets, example: Goodwell, Trade mark, patents, copy rights.

Question 35.
What is Liabilities?
Answer:
Liabilities are debts owed by the business entity to outsiders.
Example : Creditors, Bills payable, bank over draft, etc.

Question 36.
Write the meaning of solvent.
Answer:
Solvent is a person whose assets are equal or more than that of his liability.

Question 37.
Who is insolvent?
Answer:
Insolvent is a person whose assets are not sufficient to make payment of his liabilities infull.

Question 38.
What is purchases?
Answer:
Any articles, commodities or products bought for resale called purchses.

Question 39.
Give the meaning of entry.
Answer:
Entry: It means recording of business transactions in the books of Journal or subsidiary books.

Question 40.
What is Folio?
Answer:
Folio: It means the page number of books of accounts, it helps for referring the entry.

Question 41.
Write the meaning of carried down and Brought down.
Answer:
Carried down: It is the process of taking a balance of an account to the next period at the time of its closing. The short term is c/d.
Brought down : It is the process of bringing down the closing balance of the previous period to current year in the same account. It indicates opening balance. The short term is b/d.

Question 42.
Write the meaning of carry forward and brought forward.
Answer:
Carried forward: It is the process of taking the closing amount at the foot of the page of joumal/ledger etc. The short form is c/f.
Brought forward: It is the process of bringing forward the amount of previous page at the top of next page. The short term form is b/f.

Question 43.
Give the meaning of expenses and loss.
Answer:
Expenses: These are the amount spent for purchasing assets or material which is necessary for business.
Loss: Reduction in the value of assets or resources without any benefit called losses.

Question 44.
What is expenditure?
Answer:
Expenditure means a payment of cash or incurring a liability for acquiring assets, goods or service.

Question 45.
What is-Revenue?
Answer:
Revenue is the amount that adds to the capital. It represents cash generated by sale of goods or service offered.

Question 46.
What is accounting year?
Answer:
The accounts of a year are kept in a single set of books which contains 12 months, called accounting year. Generally it starts from 1st April and ends in 31st march of every year.

Question 47.
What is Accounting cycles?
Answer:
It refers to the flow of accounting data, in the course of accounting during the period of accounting.

Question 48.
What is accounting?
Answer:
Accounting is an art of recording, classifying, measuring and summarizing interms of money, of the business transaction.

Question 49.
Accounts is an art as well as.
Answer:
Science.

Question 50.
Mention one of the objective of accounting.
Answer:
One of the objective of accounting is
(a) Providing accounting information to its users.

Question 51.
Match the following.
(a) Internal users of accounting : Investors, govt. etc.
(b) External users of accounting : Management, share holders.
Answer:
(a) Internal users of accounting : Management, share holders.
(b) External users of accounting : Investors, govt etc.

Question 52.
Give examples for external users of accounting information.
Answer:
External accounting users are : Investors, suppliers / creditors government, customers etc.

Question 53.
Accounting information should be comparable. Do you agree?
Answer:
Yes. This statement is agreeable.

Question 54.
Accounting information should be comparable. Give reasons.
Answer:
Accounting information is always comparable reasons are :
(a) It helps in planning for future.
(b) It helps to compare different business organisation.

Question 55.
Shares is the example for…
Answer:
Revenue or income.

Question 56.
Commission received is a example for
(a) Revenue
(b) Cost
(c) Expenses
(d) Production
Answer:
(a) Revenue.

Question 57.
The primary use of accounting standards is
Answer:
Helps to maintain books in international market requirements.

Question 58.
Mention one feature of accounting.
Answer:
Feature of accounting is; Transactions are recorded in-terms of money.

Question 59.
What is profit?
Answer:
Profit is the excess of revenue over the expenses of a given period.

Question 60.
Give the meaning of gain.
Answer:
Gain refers to a revenue which not generated through regulate business activities.

Question 61.
Give examples for expenses.
Answer:
Rent, wages, salaries are examples for expenses.

Question 62.
Give one example for revenue.
Answer:
Sales, interest received, rent received are for examples.

Question 63.
Mention the different types of assets.
Answer:
The different types of assets are : Fixed assets and current assets.

Question 64.
Mention the different types of fixed assets.
Answer:
The different types of fixed assets are:

1. Tangible fixed assets
2. Intangible fixed assets.

Question 65.
Give one example for fixed assets.
Answer:
Fixed assets examples are : Land and building, plants and machinery furniture, vesicles etc.

Question 66.
Give one example for tangible assets.
Answer:
Examples for tangible for tangible assets.

Question 67.
Give one example for Intangible assets.
Answer:
Examples for intangible assets are : goodwill, patents, copyrights.

Question 68.
Give examples for current assets.
Answer:
Example for current assets are : Cash, stock, debtors, short term investment etc.

Question 69.
What is Fixed Assets?
Answer:
Fixed Assets are assets held on a longterm basis. Such as land Building Machinery etc. These assets are used for the normal operations of the business.

Question 70.
What is revenue?
Answer:
These are the amount of the business earned by selling its product (or) services to customer called revenue.

Question 71.
Give the meaning for examples.
Answer:
Costs incurred by a business in the process of earning revenue are called as Expenses. Example:- Depreciation, Rent, Wages, Salaries etc.

Question 72.
What is Capital?
Answer:
Amount invested by the owner to the business is known as capital.
Balance sheet / Equation = Capital = Assets – Liabilities.

1st PUC Accountancy Introduction to Accounting Two Marks Questions and Answers

Question 1.
Define Accounting.
Answer:
According to American Institute of Certified Public Accountants “Accounting is an art of recording, classifying & summarising in a significant manner & in-terms of money, transactions & events which are, in part at least of a financial character & interpreting the results thereof’.

Question 2.
What is end product of financial accounting?
Answer:
Balance sheet is the end product of financial accounting. It show the true financial positions of a business concern, that provides required informations like assets & liabilities of a business firm.

Question 3.
Enumerate main objectives of accounting.
Answer:
The main objectives of accounting are

1. Maintenance of Records of Business transaction.
2. Calculation of profit and Loss.
3. Depiction of financial position.
4. Providing Accounting information to its users.

Question 4.
Who are the users of Accounting information.
Answer:

• Internal users:- Management, who needs timely information for planning, controlling & decision making.
• External users:- Investors, Government, customers, competitors etc. obtain necessary information & rely on financial statement.

Question 5.
State the nature of accounting information required by long term lenders.
Answer:
Accounting information required by Longterm lenders are credit worthiness of the company & its ability to repay loans with Interest.

Question 6.
Who are External users of accounting information?
Answer:
External users of accounting informations are

a. Investors.
b. Suppliers & creditors
c. Customers
d. Government
e. Common man or society .
f. Lenders & financial institution.

Question 7.
Enumerate information needs to Management.
Answer:
Management needs timely information on cost of sales, profitability etc for planning, controlling & decision making.

Question 8.
Give any three Examples of revenues.
Answer:
Examples for revenues are ;
(a) Sales
(b) Commission received
(c) Interest received
(d) Dividend received.

Question 9.
Distinguish between Debtors & Creditors
Answer:
Debtors Creditors

Debtors	Creditors
(i) Debtors are persons and other entities who owe to an enterprise an amount for buying Goods & services on credit.	(i) Creditors are persons and other entities who have to be paid by an Enterprise goods & services on credit.
(ii) The total amount standing against such Persons (or) Entities on closing Date, is Show in the balance sheet as sundry Debtors on Assets side.	(ii) Total amount standing to the favour of such persons on closing Date, is shown in the Balance as sundry creditors on liability side.

Question 10.
Distinguish between profit & gain.
Answer:

Profit	Gain
(i) Profit is the Excess of revenue over the expenses of a given period, usually a year.	(i) Gain refers to a revenue which not generated through routine or regular business activities.
(ii) Profits increases the investment office owners	(ii) Gain increases the profit of an enterprises

Question 11.
Accounting information should be comparable. Do you agree with this statement. Give two reasons.
Answer:
Yes this Statement is agreeable.
The reasons are as follows:-
a. It helps in planning for the future.
b. It also useful in the areas of decision making in an organization.

Question 12.
If accounting information is not clearly presented, which of the qualitative characteristics of accounting is violated?
Answer:
If accounting information is not clearly presented, then the qualitative characteristics like comparability, reliability & understandability are violated.

Question 1.3.
The Role of accounting has changed over the period of time? Do you agree? Give reasons.
Answer:
The role of accounting is over changing. While in earlier times, accounting was merely concerned with recording the financial events i.e. record keeping activity. However, now a days, accounting is done with the rationale of not only maintaining records, but also providing information to various accounting users area.

Question 14.
With Example, Explain each of the following accounting term.
(a) Fixed Assets
(b) Revenue
(c) Expenses
(d) Short term liability
(e) Capital
Answer:
(a) Fixed Assets Fixed Assets are assets held on a longterm basis. Such as land Building Machinery etc. These assets are used for the normal operations of the business.
(b) Revenue These are the amount of the business earned by selling its product (or) services to customer called revenue.
Example Sales, Commission received, Rent received etc.
(c) Expenses:- costs incurred by a business in the process of earning revenue are called as . Expenses.
Example:- Depreciation, Rent, Wages, Salaries etc.
(d) Short term liability:- Short term liability are obligations that are payable within a period of one year.
Examples:- Creditors, Bills payable, etc.
(e) Capital:- Amount invested by the owner to the business is known as capital.
Balance sheet / Equation = Capital = Assets – Liabilities.
Example:- if, on a given date, the total Assets of a business are ?6,000 & the total liabilities of business are ?20,000 the excess of the total assets over total liabilities of the business (60,000 -20,000) ?40,000 will be owners capital.

Question 15.
Define revenues & expenses.
Answer:
According to American Accounting Association “Revenue is the monetary Expression of the aggregate of products (or) services transferred by the Enterprises to its customers during a period of time.”
According to Robert. N. Anthony “Expenses are the costs incurred in connection with the earnings of revenue”.

Question 16.
What is the primary reason for the business students & others to familiarize themselves with the accounting discipline?
Answer:
The reason for why business students & others should familiarize themselves with the accounting discipline are given below:-

• It helps in learning the various aspects of accounting.
• It helps in learning how to maintain books of accounts.
• It helps in learning how to summarise accounting information.

Question 17.
Give the meaning of expenses and loss.
Answer:
Expenses: These are the amount spent for purchasing assets or material which is necessary for business.
Loss: Reduction in the value of assets or resources without any benefit called losses.

Question 18.
What is expenditure?
Answer:
Expenditure means a payment of cash or incurring a liability for acquiring assets, goods or service.

Question 19.
What is Revenue?
Answer:
Revenue is the amount that adds to the capital. It represents cash generated by sale of goods or service offered.

1st PUC Accountancy Introduction to Accounting Six Marks Questions and Answers

Question 1:
Explain the factors, which necessitated systematic accounting?
Answer:
The factors that necessitated systematic accounting are given below:

a. Only financial transactions are recorded: Those events that are financial in nature are only recorded in the books of accounts. For example, salary of an employee is recorded in the books but not recorded educational qualification.

b. Transactions are recorded in monetary terms: Only those transactions which can be expressed in monetaiy terms are recorded in the books. For example, if a business has two buildings and four machines, then their monetary values is recorded in the books,
i. e. two buildings costing ? 20,000, four machines costing ? 8,00,000. Thus the total value of assets is ? 8,20,000.

c. Art of recording: Transactions are recorded in the order of their occurrence (Date, wire).

d. Classification of Transaction: Business transactions of similar nature are classified and posted under their respective accounts. For example, all the transactions relating to machinery will be posted in the Machinery Account.

e. Summarising of data: All business transactions are summarized in the form of Trial Balance, Trading Account, Profit and Loss Account and Balance Sheet that provides necessary information to various users.

f. Analysing and interpreting data – Systematic accounting records enable users to analyse and interpret the accounting data in a proper and appropriate manner. These accounting data and information are presented in the form of graphs, statements, charts, that leads to easy communication and understand ability by various users. Moreover, this facilitates in decision making and future predictions.

Question 2.
Describe the brief History of accounting.
Answer:
a. The history of accounting can be traced long back in civilization. Around 4000 B.C., in Babylonia and Egypt, payment of wages and taxes were recorded on clay tablets. As history claims that Egyptians kept the record of gold and valuables deposits and withdrawal from the treasuries. These records were reported on daily basis by the incharge of treasuries to the wazir, who used to forward the monthly reports to the king. Babylonia and Egypt used this method to rectify and remove errors, frauds and inefficiency from the records. Around 2000 B.C., China used sophisticated form of accounting.

In Greece, accounting was used to maintain total receipts and total payments and to balance government accounts.
In Rome, around 700 B.C., receipts and payments were recorded in daybook and were posted in the ledger at the end of the month.

b. In India, around twenty three centuries ago, Kautilya wrote the book Arthshastra, which describes how accounting records have to be maintained.

c. In 1494, Luca Pacioli wrote the book ‘Summa’ de Arithmetica Geometria Proportioniet Proportionality. In this, he explained the term debit and credit, which are used in accounting till date.

Question 3.
Explain the development and role of accounting.
Answer:
Development of accounting
In ancient times, around 4000 B.C., accounting was used for recording wages and salaries, deposits and withdrawals of valuable goods (such as gold and silver) from the treasures of the king. Afterwards, it was used to record the receipts and payments and balancing of government financial transactions.

During 1500 A. D., accounting was used by business firms for recording transactions related to business.
In 1800 A.D., accounting was used to record transactions and also to provide information to various users of financial data.

Role of accounting.
While in the earlier times accounting was merely concerned with recording the financial events (i.e. record-keeping activity); however, now-a-days, accounting is done with the rationale of not only maintaining records, but also providing an information system that provides important and relevant information to various accounting users. .

a. Substitute of memory : As, it is beyond human capabilities to remember each and every business transaction, so accounting plays an important role in recording these transactions in the book of accounts.
b. Assistance to management: Management uses accounting information for short term and long term planning of business activities and to control various costs and budgets.
c. Comparative study : In order to ascertain the performance of the business, accounting enables comparison of current year’s profit with that of previous years (intra-firm comparison) and also with other firms in the same business (inter-firm comparison).
d. Evidence in court: It acts as evidence that can be used or presented in the court, if any discrepancy arises in the future.

Question 4.
Define accounting and state its objectives.
Answer:
In 1970, American Institute of Certified Public Accountants changed the definition and stated, “The function of accounting is to provide quantitative information, primarily financial in nature, about economic entities, that is intended to be useful in making economic decisions.”

Objectives of Accounting:

a. Recording business transactions systematically: It is necessary to maintain systematic
records of every business transaction, as it is beyond human capacities to remember such large number of transactions. Skipping the record of any one of the transactions may lead to erroneous and faulty results. ,

b. Determining profit earned or loss incurred: In order to determine the net result at the end of an accounting period, we need to calculate profit or loss. For this purpose trading and profit and loss account are prepared. It gives information regarding how much of goods have, been purchased and sold, expenses incurred and amount earned during a year.

c. Ascertaining financial position of the firm: Ascertaining profit earned or loss incurred is not enough proprietor also interested in knowing the financial position of his/her firm, i.e. the value of the assets, amount of liabilities owed, net increase or decrease in his/her capital. This purpose is served by preparing the balance sheet that facilitates in ascertaining the true financial position of the business.

d. Assisting management: Systematic accounting helps the management in effective decision making, efficient control on cash management policies, preparing budget and forecasting, etc.

e. Assessing the progress of the business: Accounting helps in assessing the progress of business from year to year, as accounting facilitates the comparison both inter-firm as well as intra-firm.

f. Detecting and preventing frauds and errors: It is necessary to detect and prevent fraud and errors, mismanagement and wastage of the finance. Systematic recording helps in the easy detection and rectification of frauds, errors and inefficiencies, if any.

g. Communicating accounting information to various users: The important step in the accounting process is to communicate financial and accounting information to various users including both internal and external users like owners, management, government, labour, tax authorities, etc. This assists the users to understand and interpret the accounting data in a meaningful.

Question 5.
Describe the informational needs of external users.
Answer:
There are various external users of accounting who need accounting information for decision making, investment planning and to assess the financial position of the business. The various external users are given below.

a. Banks and other financial institutions: Banks provide finance in form of loans and advances to various businesses. Thus, they need information regarding liquidity, creditworthiness, solvency and profitability to advance loans.

b. Creditors: These are those individuals and organisations to whom a business owes money on account of credit purchases of goods and receiving services; hence, the creditors require information about credit worthiness of the business.

c. Investors and potential investors: They invest or plan to invest in the business. Hence, in order to assess the viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

d. Tax authorities: They need information about sales, revenues, profit and taxable income in order to determine the levy various types of tax on the business.

e. Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policies measures and to address various economic problems like employment, poverty etc.

f. Researcher: Various research institutes like NGOs and other independent research institutions like CRISIL, stock exchanges, etc. undertake various research projects and the accounting information facilitates their research work.

g. Consumer: Every business tries to build up reputation in the eyes of consumers, which can be created by the supply of better quality products and post-sale services at reasonable and affordable prices. Business that has transparent financial records, assists the customers to know the correct cost of production and accordingly assess the degree of reasonability of the price charged by the business for its products and thus helps in repo building of the business.

h. Public: Public is keenly interested to know the proportion of the profit that the business spends on various public welfare schemes; for example, charitable hospitals, funding schools, etc. This information is also revealed by the profit and loss account and balance sheet of the business.

Question 6.
What do you mean by an asset? Explain the different types of assets.
Answer:
Any valuable thing that has monetary value, which is owned by a business, is its asset. In other words, assets are the monetary values of the properties or the legal rights that are owned by the business organisations.
The different types of assets are ;

a. Fixed Assets: These are those assets that are hold for the long term and increase the profit earning capacity and productive capacity of the business. These assets are not meant for sale, for example, land, building machinery etc.
b. Current Assets : Assets that can be easily converted into cash or cash equivalents are termed as current assets. These are required to run day to day’ business activities; for example, cash, debtors, stock, etc.
c. Tangible Assets : Assets that have physical existence, i.e., which can be seen and touched, are tangible assets; for example, car, furniture, buildings, etc.
d. Intangible Assets : Assets that cannot be seen or touched, i.e. those assets that do not have physical existence, are intangible assets; for example, goodwill, patents, trade mark, etc.
e. Liquid Assets : Assets that are kept either in cash or cash equivalents are regarded as liquid assets. These can be converted into cash in a very short period of time; for example, cash, bank, bills receivable, etc.
f. Fictitious Assets : These are the heavy revenue expenditures, the benefit of whose can be derived in more than one year. They represent loss or expense that are written off over a period of time.

Question 7.
Write the meaning of gain and profit. Distinguish between gain and profit.
Answer:
Profits: Excess of revenue over expense is known as profit. It is normally categorized into ’ gross profit or net profit. It increases the owner’s capital as it is added to the capital at the end of each accounting period.
For example: Goods costing ₹ 1,000 is sold at ₹ 1,200 then the sale proceeds of ₹ 1,200 is the revenue and 1,000 is the expense to generate this revenue. Hence, accounting profit of ₹ 200 (i.e. ₹ 1,200 – ₹ 1,000) is the difference between the revenue and expense that is earned by the business.

Gain: It arises from irregular activities or non-recurring transactions. In other words, a gain is a result of transactions that are incidental to the business, other than operating transactions. For example: an old machinery of book value ₹ 2,000 is sold at ₹ 2,500. Hence, the gain is ₹ 5,00 (i.e. ₹ 25,00 – ₹ 2,000). Here, the sale of the old machinery is an irregular activity; so, the difference is termed as gain. Thus, in other words the only difference between profit and gain is that profit is the excess of revenue over expense and gain arises from other than operating transactions.

	Profit	Gain
1.	Profit is derived from regular business regular business activity	Gain is derived from investment over on period of time not falling under regular business activity.
2.	It is return on capital employed.	It is return on investment
3.	Profit is the summation of total income less total expenses	Gain is the process received from the sale of fixed or financial assets.

Question 8.
Explain the qualitative characteristics of accounting information.
Answer:
The following are the qualitative characteristics of accounting information:

a. Reliability: It means that the user can rely on the accounting information. All accounting information is verifiable and can be verified from the source document (voucher), viz. cash memos, bills, etc. Hence, the available information should be free from any errors and unbiased.

b. Relevance: It means that essential and appropriate information should be easily and timely available and any irrelevant information should be avoided. The users of accounting information need relevant information for decision making, planning and’ predicting the future conditions.

c. Understandability: Accounting information should be presented in such a way that
every user is able to interpret the information without any difficulty in a meaningful and appropriate manner. ,

d. Comparability- It is the most important quality of accounting information. Comparability means accounting information of a current year can be comparable with that of the previous years. Comparability enables intra-firm and inter-firm comparison. This assists in assessing the outcomes of various policies and programmes adopted indifferent time horizons by the same or different businesses.

Question 9.
Describe the role of accounting in the modern world.
Answer:
The role of accounting has been changing over the period of time. In the modem world, the role of accounting is not only limited to record financial transactions but also to provide a basic framework for various decision making, providing relevant information to various users and assists in both short run and long run planning.

The role of accounting in the modem world is given below:

Assisting management: Management uses accounting information for short term and long term planning of business activities, to predict the future conditions, prepare budgets and various control measures.

Comparative study: In the modem world, accounting information helps us to know the performance of the business by comparing current year’s profit with that of the previous years and also with other firms in the same industry. .

Substitute of memory: In the modem world, every business incurs large number of transactions and it is beyond human capability to memorise each and every transaction. Hence, it is very necessary to record transactions in the books of accounts.

Information to end user: Accounting plays an important-role in recording, summarizing and providing relevant and reliable information to its users, in form of financial data that helps in decision making.

1st PUC Accountancy Introduction to Accounting Additional Questions and Answers

Question 1.
Write the advantages and disadvantages of book-keeping.
Answer:

	Book-keeping Advantages	Book-keeping Disadvantages
1.	It provide full information about all expenses and loss.	It gives only monetary transaction information.
2.	It is return on capital employed.	It is return on investment
3.	It helps to know the true financial position of business.	This recordings do not give exact information and timely information.
4.	It helps to know the progress of business from year to year.	Final accounts prepared under book-keeping do not provide timely information.
5.	It helps to know from whose money due and to whom due.	With the help of only book-keeping management can’t take decision and correct action cannot be possible.
6.	It keep control over the properties and activities of business.	Book-keeping depends on personal judgment.

Question 2.
Explain the features of accounting.
Answer:
The features / Natures of accounting are :

• Recording of business transaction as and when they occur in single book is the first feature of accounting.
• Classify or grouping of entries according to their nature in to appropriate heads of account.
• Accounting also involves summarising the transactions, classified in the ledger and financial position presentation.
• Accounting also analyse and interpret the recorded transactions.

Question 3.
Differentiate book-keeping and accounting. .
Answer:

	Book-keeping	Accounting
1.	It is an art of recording money transaction in a-set of books.	It is a process of designing the system of records of books of a/cs.
2.	It is a mechanical and routine Work.	It requires specialised knowledge and creative ability.
3.	Book-keeper is a clerk.	Accountant is a professionalist.
4.	Book-keeping creates a data base.	It process the finacial data and finds results and conclusion.
5.	It helps to know from whose money due and to whom due.	With the help of only book-keeping management can’t take decision and correct action cannot be possible.

Question 4.
Accounting is an art as well as science. Discuss.
Answer:Accounting is an art: art means application of knowledge to produce the desired result, accounting ivolves the designing of information system, record filing system, standardisation of forms and statements etc. The working of an accounting needs creative and active skills . and involvement of all staff.

Accounting is also science: A ‘Science’ means a systematic body of knowledge relating to universal phenomenon. It consists of concept theories, rules, and techniques developed by the process of rational thinking and spirit of enquiry, accounting is a social science..
So accounting is a applied science as well as art of recording a business transactions.

Question 5.
Write the objectives or advantages of accounting.
Answer:
The objectives or advantages of accounting are:

• To maintain the record of financial transactions of a business accurately.
• To ascertain the profit or loss made by business.
• To present the true and fair view of financial position.
• To know the amount due to creditors individually.
• To know the amount due from debtors to the business.
• To compute the tax liability of the concern.
• To supply the required financial information to the mangement it helps for decision making.

Question 6.
Explain the role or growth of accounting.
Answer:
The origin and growth of accounting can be summarised as follows:

• Accounting is said to be very old, as old as money.
• The present accounting is the result of constant innovations to the requirements of business activities.
• Accounting has developed to meet the emerging needs and requirements of fast developing society.
• “Arthashastra” the book written by Kautilya is the base or foundation for accounting.
• The present system of accounting based on the double entry system founded by pacilio in 1494 at Italy.
• Futher various publications were made and an important publications was that of edward jones in 1975 who invovated the concept of ‘Two Column journal’.

Question 7.
State the different terms in Accounting.
Answer:
Drawings: The amount of cash or any asset withdrawn by the owner of the business for his personal use or domestic use we called as drawings.

• Assets: are the properties or resources which are owned by the business entity.
  Ex: Machinery, stock, goodwell, etc.
  Liabilities are debts owed by the business entity to outsiders.
  Example: Creditors, Bills payable, bank over draft, etc.
• Debtor: is a person who owes any amount to business. In other words, who purchase goods from business on credit basis is called debtors.
• Creditor: is a person to whom any sum of money is owed by business. Other words the person who give benefits to business and amount payable, such person called creditors.
• Goods: The term goods includes all commodities, articles or products which are purchased for the purpose of re-sale.
• Purchases: Any articles, commodities or products bought for resale called purchases.
• Sales: Any goods purchase by customer called sales mean sale of goods and not the assets. Stock: The goods purchased for sale, remain unsold called goods. It is a asset for the business.
• Profit: The accounts of a year are kept in a single set of books which contains 12 months, called accounting year. Generally it starts from 1st April and ends in 31st march of every year.
  • Expenditure means a payment of cash or incurring a liability for acquiring assets, goods or service.
  • Revenue is the amount that adds to the capital. It represents cash generated by sale of goods or service offered.
• Discount: Reducing the value of sales called discounting. Discounts are 2 types, Trade Discount and Cash Discount.
• Voucher: is the document which, helps in reseeding business transactions.
• Income: It refer to an amount received for sale of goods and service or for use of any rights belonging to business.
• Gain: Increase in the value of assets or resources of business called gains.
  Capital represents the owner’s claim or share in the assets of the business. Amount invested by owner of business called capital.
• Entity: means an area of economic interest of a particular industry or group of industries. Separate books of accounts are kept for each entity

You can Download Chapter 13 Computerised Accounting System Questions and Answers, Notes, 1st PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Accountancy Question Bank Chapter 13 Computerised Accounting System

1st PUC Accountancy Computerised Accounting System One Mark Questions and Answers

Question 1.
What is computerised accounting system?
Answer:
It is an accounting information system that processes the financial transaction and events an per GAAP.

Question 2.
Expand GAAP
Answer:
Generally accepted accounting principles = GAAP.

Question 3.
Mention the elements computer system?
Answer:
Computer system elements are Hardware, Software people, procedure, data and connectivity.

Question 4.
What is Hardware?
Answer:
Hardware of computer includes – physical components like, keyboard, mouse, monitor, CPU etc. These are electronic and electromechanical components.

Question 5.
What is software or program of computer?
Answer:
A set of program, which is used to work with hardware is called its software.

Question 6.
What is firmware?
Answer:
A coded set of instructions stored in the form of circuits is called firmware.

Question 7.
Mention the different types of software.
Answer:
The different types of software are ;

1. Operating system
2. Utility program
3. Application software
4. Language processors
5. System software
6. Connectivity software.

Question 8.
What is procedure as a element of computer system?
Answer:
A procedure means a series of operations in a certain order or manner to achieve desire results.

Question 9.
Mention the different types of procedure in computer system.
Answer:
The different types of computer system procedures are:

1. HardWare – oriented
2. SoftWare – oriented
3. Internal procedure.

Question 10.
List out the capabilities of computer system.
Answer:
Capabilities of computer system are:

• Speed
• Accuracy
• Reliability
• Storage

Question 11.
List out any two limitations of computer system.
Answer:
The demerits or limitations of computer system are;

• Lack of common sense
• Lack of decision making
• Compulsory instructor.

Question 12.
Mention two features of computerised accounting system.
Answer:
The two features of computerised accounting system are :

1. Online input and storage of accounting data.
2. Grouping of ledger accounts is done in the beginning itself.
3. Quick reporting to management and various types of reports can be produce.
4. Invoices like purchase and sales can be prepared.

Question 13.
What is accounting information system?
Answer:
Accounting information system, identifies, collects, process and communicates economic information about an entity to a wide variety of users.

Question 14.
What is accounting report?
Answer:
Information supplied to meet a particular need of management is called report. It is helpful for management decision making.

Question 15.
List out the requisite of good accounting report.
Answer:
An accounting report must fulfill;

• Relevance
• Timeliness
• Accuracy
• Summarisation

1st PUC Accountancy Computerised Accounting System Six Marks Questions and Answers

Question 1.
State the four basic requirements of a Database Applications.
Answer:
The following are the four basic requirements of a Database Application:
a. Front-end Interface- It acts as an interactive connecting link between the user and the database oriented software through which the user communicates or interacts to the backend database.

b. Back-end Database- It is the data storage system that is hidden from the users. It responds to the requirement of the users to the extent the user is authorised to access.

c. Data Processing- It is a sequence of actions that are taken to transform the input data into useful information for taking various decisions.

d. Reporting System- It is an integrated set of objects that includes all the relevant information that constitutes a report.

Name the various categories of Accounting Package.
The Accounting Packages are classified into the following categories:
a. Ready-to-use or Readymade Software
b. Customised Software
c. Tailored or Tailor-made Software

List the various advantages of Computerised Accounting Systems.
The mentioned below are the various advantages of Computerised Accounting Systems:
a. Speed
b. Accuracy
c. Reliability
d. Up-to-Date Information
e. Real Time User Interface
f. Automated Document Production
g. Scalability
h. Legibility

Give two examples each of the organisations where ‘ready-to-use’, ‘customised’, and ‘tailored’ accounting packages respectively suitable to perform the accounting activity.

‘Ready-to-use’ accounting packages are basically used by the small-sized enterprises. For example, grocery stores, medical stores, etc.

On the other hand, ‘Customised’ accounting packages are basically used by the medium and large business. For example, shopping malls, hospitals, etc.

Whereas, ‘Tailored’ accounting packages are basically used by the geographically scattered businesses. For example, MNC’s, Communication Industries, etc.

Question 2.
Distinguish between ‘ready-to-use’ and ‘tailored’ accounting software.
Answer:

Basis of Difference	Ready-to-Use Accounting Software	Tailored Accounting Software
1. Nature of Business	This software is used in small and conventional businesses.	This software is used in large and typical businesses.
2. Adaptability	Its adaptability is very high.	Its adaptability is very specific and cannot be used by every business houses.
3. Linkage to other Information System	Its interface with the other information system is limited	Its interface with the other information system is unlimited
4. Number of Users	It has limited users.	It has huge number of users.
5. Installation and Maintenance Costs	The installation and maintenance cost is low.	The installation and maintenance cost is comparatively higher

Question 3.
Define a Computerised Accounting System. Distinguish between a Manual and Computerised Accounting Systems.
Answer:
Computerised Accounting Systems is based on the concept of database. It is an accounting information system that processes the financial transactions and events in accordance to the Generally Accepted Accounting Principles (GAAP) to produce reports as per the requirements of the users.

Manual Accounting	Computerised Accounting
1. As the data a recorded under manual accounting are visible.	As the data stored in computers are not visible.
2. The rail of events under manual accounting can be easily established	The rail of transactions (or) events cannot be established easily under computersied accounting.
3. The data recorded in manual accounting are not subjected to the risk of manipulation	The data recorded in computerised ac­counting system are subject to the risk of manipulation.
4. In manual accounting, accounting data cannot be adjusted to produce various special statement and reports	Under a computerised accounting system the accounting data can be easily adjusted to generate various special statement and reports.
5. The cost of preparing statement and report under manual accounting is high.	The cost of preparing statements and re­ports under computerised accounting is low
6. The reports prepared under manual accounting are mostly board and general purpose reports.	Under computerised accounting a number of special reports can be prepared to meet the needs of different mangers.

Question 4.
Discuss the advantages of Computerised Accounting Systems over the Manual Accounting Systems.
Answer:
The following are the various advantages of the Computerised Accounting Systems over the Manual Accounting Systems:
a. Speed: The speed of a computer is very high and takes very less time in performing various difficult operations. The accounting data is processed comparatively faster through the Computerised Accounting Systems than it can be done through the manual efforts.

b. Accuracy: In Computerised Accounting Systems, the possibility of errors is minimised or reduced as the primary accounting data is entered only once for preparing various accounting reports and for subsequent usage and processes.

c. Reliability: As the Computerised Accounting Systems is well-equipped in performing repetitive operations, so it is comparatively more reliable to perform the operations than the manual system. Also, the Computerised Accounting Systems overcome the limitation of Manual Accounting Systems such as tiredness, boredom or fatigue, etc., thereby enhances the degree of reliability.

d. Up-to-Date Information: In the Computerised Accounting Systems, whenever the new accounting data is entered and stored, the existing accounting records automatically gets updated.

e. Real-Time User Interface- Most of the automated accounting systems are interlinked through a network of computers. The availability of information to various users at the same time on the real-time basis is facilitated under computerised system of accounting. This is very difficult to avail such facility under manual system as this call for availability of multiple copies of the accounting records that can be accessed by many users at the same time.

f. Automated Document Production: Under Computerised Accounting Systems, the accounting reports such as, Cash Book, Trial balance, Statement of Accounts, etc. is very easy to obtain. This is because most of the computerised systems have standardised and user-defined format of accounting reports that are generated automatically.

g. Scalability: The computerised systems of accounting are highly scalable as there requirement of additional manpower is mainly confined to data entry for recording and storing the additional vouchers in the computers. Thus, the additional cost of processing additional transactions is meagre as compared to the cost associated with hiring new accountants to handle additional transactions.

h. legibility: In Computerised Accounting Systems, the accounting records are typed and presented in standard fonts. The various characters especially numbers, alphabets, graphics, etc. are more clear and can be read without any difficulty and ambiguity.

Question 5.
Describe the various types of accounting software along with their advantages and limitations.
Answer:
The various types of accounting software are:
Ready-to-use Software- This type of software is readily available in the market with prescribed and standard features. This accounting software is basically used by the small size business enterprises, where the number of transactions is not so large. The cost of its installation and maintenance is also low. It has limited number of users, its adaptability is very high as it is relatively easier to learn and operate. It does not have a wide scope to link it with other information systems.

Advantages of Ready-made Accounting Software
a. This software is easily available in the market.
b. It is less expensive, as it comes with basic and standard features.
c. It involves a lesser need for training.
d. It is less sophisticated.
e. Its adaptability is very high as it is relatively easier to learn and operate.
f. It is suitable for small-size business enterprises.

Disadvantages of Ready-made Accounting Software
a. It has limited number of users.
b. It is not suitable for medium and large business organisation, where the number of transactions is very large.
c. It fails to cater the specific needs of the users, d. It suffers from the low level of data secrecy.
e. It does not have a wide scope to link it with other information systems.

Customised Software: Customised software is the software that has standardised features to meet the special requirements of the users. It provides the scope of changing the features of accounting software. The functions of this software can be programmed as per the needs and requirements of the users. This type of software best suits the needs of medium and large businesses. Its cost of installation and maintenance is comparatively higher. It can be easily linked to the other information systems.

Tailored Software: Tailored or Tailor-made accounting software is the software that is developed as per the specifications and requirements of the users. This accounting software is generally used in the large business organisations with multi-user and geographically scattered locations. It is designed to meet the specific needs of the users and form an integral part of the organisational MIS. It has infinite number of users.

Advantages of Customised and Tailor-made Accounting Software
a. This software is suitable for medium and large business organisation.
b. It caters the specific requirements and needs of the users.
c. It can be modified as per the needs of the organisation.
d. It has high level of security’ and minimises the loss and unauthorised access of data.
e. It cannot be easily imitated or duplicated in the market due to difference in the needs and requirements of different users.
f. It does not involve high cost of training as the training can be imparted within the ‘ organisation by the experienced personnel.
g. It can be easily linked to the other information systems, h. It forms an integral part of the organisational MIS.

Disadvantages of Customised and Tailor-made Accounting Software
a. It involves high cost of installation and maintenance.
b. Developing customised software is a time-consuming process and involves high cost of development.
c. Maintenance of this software is difficult as there exists limited availability of knowledge to the developers.
d. It lacks standard training module.

Question 6.
‘Accounting software is an integral part of the Computerised Accounting Systems’ Explain. Briefly list the generic considerations before sourcing accounting software.
Answer:
The accounting software does form an integral part of the Computerised
Accounting Systems. The accounting software should be selected after considering the level of skill and proficiency of the accounting professionals. This is one of the important aspects that should be taken care of before introducing Computerised Accounting Systems, as the accounting professionals are responsible for accounting and the not computers.

The following are some of the important points that should be taken into consideration before introducing accounting software in an organisation.
1. Flexibility: This is the most important factor that should be considered before sourcing accounting software. The accounting software should be flexible in form of data entry, retrieval of data and generating design of reports. The software should be able to run on different computers having different operating systems and having different configurations. It should provide some flexibility among its users.

2. Cost of Installation and Maintenance: The selection of accounting software largely depends upon its cost to the organisation. The cost of accounting software includes cost of installing the related components and hardware, maintenance and alteration costs, cost of training the staff and cost involved in recovering data in case of data failure. An organisation needs to evaluate the benefits of the software against its costs.

3. Size of Organisation The size of an organisation also determines the selection of accounting software. The small-sized organisations, where the volume of business transactions is not so large, usually opt for simple and single user oriented software. On the other hand, large scale organisations, where the volume of business transactions is very large choose the latest and sophisticated software for meeting the multi-user requirements.

4. Training Needs- Another factor that affects the choice of software is the training needs.There is some accounting software that requires comparatively lesser training and is more user-friendly. While, there are some other complicated software that requires continuous and thorough training.

5. Level of Secrecy: The level of expected security is one of the important factors that an organisation bears in mind before sourcing accounting software. Software should be able to prevent the unauthorised access and manipulation of data. It should have in built features of security.

6. Utilities/MIS Reports- Another factor which helps in determining the software selection isthe MIS reports and the extent to which they are used in the organisation.

7. Vendor Reputation and Capability- The selection of software is also affected by the capability and competence of the vendor. It depends upon the reputation of the vendor in the market, the user-reviews of the similar software, the extent of post-sales support services from the vendors, etc.

‘Computerised Accounting Systems are best form of accounting system’. Do you agree? Comment.
Yes, we agree with this statement that ‘Computerised Accounting Systems are best form of accounting system’. It becomes very easier to work with Computerised

Accounting Systems leading to reduction in the accounting errors. Moreover, the computerised accounting reports are highly reliable, thereby enhances the overall efficiency.

Due to the following positive aspects, the Computerised Accounting Systems certainly enjoy an edge over the Manual Accounting Systems.
a. Speed: The speed of a computer is very high and takes very less time in performing various difficult operations. The accounting data is processed Comparatively faster through the Computerised Accounting Systems than it can be done through the manual efforts.

b. Accuracy: In Computerised Accounting Systems, the possibility of errors is minimised or reduced as the primary accounting data is entered only once for preparing various accounting reports and for subsequent usage and processes.

c. Reliability: As the Computerised Accounting Systems is well-equipped in performing repetitive operations, so it is comparatively more reliable to perform the operations than the manual system. Also, the Computerised Accounting Systems overcome the limitation of Manual Accounting Systems such as tiredness, boredom or fatigue, etc., thereby enhances the degree of reliability.

d. Up-to-Date Information: In the Computerised Accounting Systems, whenever the new accounting data is entered and stored, the existing accounting records automatically gets updated.

e. Real-Time User Interface- Most of the automated accounting systems are interlinked through a network of computers. The availability of information to various users at the same time on the real-time basis is facilitated under computerised system of accounting.

f. Automated Document Production: Under Computerised Accounting Systems, the accounting reports such as, Cash Book, Trial balance, Statement of Accounts, etc. is very easy to obtain. This is because most of the computerised systems have standardised and user-defined format of accounting reports that are generated automatically.

g. Scalability: The computerised systems of accounting are highly scalable as the requirement of additional manpower is mainly confined to data entry for recording and storing the additional vouchers in the computers. Thus, the additional cost of processing additional transactions is meagre as compared to the cost associated with hiring new accountants to handle additional transactions.

h. Legibility: In Computerised Accounting Systems, the accounting records are typed and presented in standard fonts. The various characters especially numbers, alphabets, graphics, etc. are more clear and can be read without any difficulty and ambiguity.

You can Download Chapter 6 Shishu Makkaligolida Madeva Questions and Answers Pdf, Notes, Summary, 1st PUC Kannada Textbook Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Kannada Textbook Answers Sahitya Sanchalana Chapter 6 Shishu Makkaligolida Madeva

Shishu Makkaligolida Madeva Questions and Answers, Notes, Summary

You can Download Chapter 8 Redox Reactions Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 8 Redox Reactions

1st PUC Chemistry Redox Reactions One Mark Questions and Answers

Question 1.
Define oxidation in terms of electronic concept.
Answer:
Oxidation is a process in which loss of electrons take place.

Question 2.
Define by reduction in terms of electronic concept.
Answer:
Reduction is a process in which gain of electrons takes place.

Question 3.
What are redox reactions ? Give an example.
Answer:
Redox reaction is reaction in which oxidation and reduction take place
simultaneously, e.g. Zn(s)+ Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s):

Question 4.
What is oxidation number of alkali metals in their compounds?
Answer:
+1.

Question 5.
Zn(s)+Cu²⁺ → Zn²⁺ + Cu(s) is this reaction redox reaction ? If yes, name the
∴ oxidizing agent as well as reducing agent.
Answer:
Yes, Cu²⁺ is oxidising agent whereas Zn is reducing agent.

Question 6.
Which of the following does not conduct electric current and why?
Molten NaCl, Solid Pb, AgNO₃ solution and methanol.
Answer:
Methanol will not conduct electric current because it does not ionise. .

Question 7.
Define cathode and anode.
Answer:
Cathode is electrode towards which cations are attracted. Anode is electrode which : attracts anions.

Question 8.
What is oxidation state of Cr in CrO⁵ and why?
Answer:
Cr has +6 oxidation state because it has 6 valence electrons, therefore can form 6 covalent bonds.

Question 9.

which is the strongest oxidising agent out of them ?
Answer:
Fe³⁺ is strongest oxidising agent because it has standard reducing potential. (i.e., +ve value) value.

Question 10.
What is meant by oxidation potential of an electrode ?
Answer:
Oxidation potential measures the tendency of an element or anion to lose the electrons.

Question 11.
What is relationship between standard oxidation potential and standard reduction potential?
Answer:
Both are equal in magnitude but opposite in sign.

Question 12.
Identify the oxidant and reductant in the following reactions :
(i) Zn(s) + \(\frac { 1 }{ 2 }\) O₂(g) → ZnO(s)
(ii) Zn(s) + 2H⁺ (aq) → Zn²⁺(aq) + H₂(g)
Answer:
(i) Zn is reducing agent (reductant) and O₂ is oxidizing agent (oxidant)
(ii) Zn is reducing agent (reductant) whereas H⁺ is oxidizing agent (oxidant).

Question 13.
How will you define (i) Oxidant and (ii) Reductant in terms of oxidation no.?
Answer:
(i) Oxidant: Which decreases its own oxidation no. and increases the ox. no. of the other in a reaction.
(ii) Reductant : Which increases its own ox. no. and decreases that of other in a chemical reaction.

Question 14.
What do you mean by disproportionation reaction ?
Answer:
A disproportionation reaction is an oxidation-reduction process in which the same substance is oxidized and reduced e.g., 2Cu⁺ → Cu + Cu²⁺

Question 15.
Determine the oxidation number of C in the following : C₂H₆, C₄H₁₀, CO, CO₂ and HCO₃^–
Answer:
O.N. of C = -3 in C₂H₆ ; -2.5 in C₄H₁₀ ; +2 in CO ; +4 in CO₂ and +4 in HCO₃^–

Question 16.
Determine the oxidation number of O in the followiing : Na₂O₂ and CH₃COOH.
Answer:
O.N. of+1 in Na₂O₂ and -2 in CH₃COOH

Question 17.
Find out the oxidation number of sulphur in the following species : and HSO₄^–.
Answer:
O.N. of S = + 6 in HSO₄^–.

Question 18.
Determine the oxidation number of all the atoms in the following well known oxidants KCI04.
Answer:
K = +1, Cl = +7, O = -2

Question 19.
Determine the change in the oxidation number of S in H₂S and SO₂ in the following industrial reaction : 2H₂S(g) + SO₂(g) → 3S(g) + 2H₂O(g)
Answer:
O.N. of S changes from -2 in H₂S and +4 in SO₂ to zero in elemental sulphur.

Question 20.
What is oxidised and what is reduced in the reaction ?
H₂S + Cl₂ → 2HCl + S
Answer:
H₂S is oxidised to S and Cl₂ is reduced to HCl.

Question 21.
In the reaction SnCl₂ + 2HgCl₂ → SnCl₄ + Hg₂Cl₂
which is oxidant and which is reductant ?
Answer:
SnCl₂ is a reductant and HgCl₂ is an oxidant.

Question 22.
Identify the strongest and weakest reducing agents from the following list: Zn, Cu, Ag, Na, Sn.
Answer:
Strongest reducing agent in Zn and weakest reducing agent in Ag.

Question 23.
Can oxidation number of an atom in a chemical species he fractional ? Illustrate by an example.
Answer:
Yes, it can be fractional. For example oxidation no. of Pb in Pb₃O₄ = + \(\frac { 8 }{ 3 }\)

1st PUC Chemistry Redox Reactions Two Marks Questions and Answers

Question 1.
Calculate the oxidation number of underlined atoms in the following compounds and ions: CH₄, Sb₂O₅, C₆H₁₂O₆
Answer:
CH₄
X + 4 = 0;
X = -4

Sb₂O₅
2x – 10= 0;
2x = 10
x = +5

C₆H₁₂O₆
6x + 12 – 12 = 0
x = 0

Question 2.
Identify the oxidant and reductant in the following chemical reaction:
2I^– (aq) + Cl₂ (g) → 2Cl^– (aq) + I₂(s)
Answer:
2I^– (aq) + Cl₂ (g) → 2Cl^– (aq) + I₂(s)
Oxidant is Cl₂ (g) reductant is I^– (aq)

Question 3.
The standard reduction potentials of Al and Ni are -1.66 V and -0.28 V, respectively. Is Al a stronger or weaker reducing agent than Ni ? Explain.
Answer:
‘Al’ is stronger reducing agent than Ni because it has lower reduction potential. Al³⁺ is more stable than Ni²⁺ ion.

Question 4.
The standard reduction potentials of Zn²⁺, Mg²⁺ and Na⁺ are -0.76 V, -2.37 V and -2.71 Y respectively, which of the following is the strongest oxidizing agent?
Answer:
Zn²⁺ is strongest oxidising agent. It can gain electrons easily. It has highest standard reduction potential.

Question 5.
Which of the following is best reducing agent and why ? Li, Cu, Br₂, F₂, H₂, K.
Answer:
Lithium (Li) is best reducing agent because it has lowest standard reduction potential, i.e., Li⁺ is most stable.

Question 6.
Calculate the oxidation number of (i) C in CH₂Cl₂ and (iii) Pb in Pb₃O₄.
Answer:
(i) C in CH₂Cl₂. Let the oxidation number of C in CH₂Cl₂ be x. Writing the oxidation number of each atom above its symbol.

x + 2(+1) + 2(-1) = 0 ,x = 0

Question 7.
Balance the equation by ion electron method.
Fe(OH)₂(s) + H₂O₂ → Fe(OH)₃ (s) + H₂O is basic solution.
Answer:

Question 8.
Balance the equation by ion electron method.
Al(s) + NO₃ → Al(OH)₄^– + NH₃ in basic solution.
Answer:

Question 9.
Balance the equation by ion electron method.
PbO₂ + Cl^– → ClO^–+ [Pb(OH)₃]^– in basic solution.
Answer:
Here 2e^– + PbO₂ → [Pb(OH)₃]^–
2H₂O + 2e^–PbO2₂ → [Pb(OH)₃] + OH^– …(I)
2OH + Cl^– → ClO^– + 2e^–+H₂O …(ii) Add(i)&(ii)
PbO₂ + OH^– + Cl^– + H₂O → [Pb(OH)₃ ]^– + ClO^–

Question 10.
Balance the equation by ion electron method.
CrO₄^– + H₂O₂ →CrO₄^2- in basic solution.
Answer:
[2OH^– + CrO^–₃ → CrO₄^2- + e^– + H₂O] x 2 …(i)
2e^–+ H₂O₂ → 2OH^– …(ii) (Add(i) & (ii))
2OH^–+2CrO₃^– +H₂O₂ → 2CrO₄^2-+ 2H₂O

Question 11.
Starting with the correctly balanced half reactions write the overall net ionic reactions. Chloride ion is form oxidized to Cl₂ by MnO₄^– in acid
Answer:

Question 12.
Starting with the correctly balanced half reactions write the overall net ionic reactions for HNO₂ reduces MnO₄^– in acid solution.
Answer:

Question 13.
Starting with the correctly balanced half reactions write the overall net ionic reactions for HNO₂ oxidizes I^– to I₂ in acid solution.
Answer:

Question 14.
Starting with the correctly balanced half reactions write the overall net ionic reactions. ClO₃^– oxidizes Mn²⁺ to MnO₂(s) in acid solution.
Answer:

Question 15.
Consider the following cell notation :
Al(s)|Al³⁺(aq)|Ni²⁺(aq)|Ni(s), here
Which substances act as anode ? Which of them acts as cathode ? Write the net ionic equation for the cell reaction.
Answer:
At acts as anode, Ni acts as cathode.

Question 16.
Describe the standard half cell that is used in electrochemistry to measure standard potentials. Write an equation for its half reaction and show its standard potential.
Answer:
Standard hydrogen elctrode is used to measure standard electrode potential.
It consists of H₂(g) filled at 1 atm, 298 K in a sealed tube having Pt rod for metallic contact, coated with platinum black which acts as catalyst. It is dipped in 1 M solution of HCl. The reaction 2H⁺ (aq) + 2e^– → H₂(g) \(\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{\circ}=0\)

Question 17.
Calculate the E°_cell using these electrodes whose half reactions are:
Fe(OH)₂(s) + 2e^– → Fe(s) + 2OH^–(aq)  E° = -0.88V
NiO₂(s) + 2H₂O(l) + 2e^– → Ni(OH)₂(s) + 2OH^–(aq)  E° = +0.49V
Answer:

Question 18.
Calculate E°_cell for the reaction Cl₂(g) + 2l^– → I₂(s) + 2Cl^–(aq) with the help of these half cell reactions:
Cl₂ (g) + 2e^– → 2Cl^–(aq) ; E° = +1.36V
I₂(g) + 2e^– → 2I^–(aq) ;E° = +0.54V
Answer:

Question 19.
(a) Give one use of heavy water in nuclear reactor.
(b) Write down balanced chemical equations of the reaction of cone. Nitric acid with (i) Copper and (ii) Iodine.
Answer:
(a) Heavy water is used as moderate and coolant in nuclear reactor.
(b) (i) Cu(s) + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Question 20.
Balance the following ionic reaction with the help of oxidation number method : MnO₄^– + I^– → MnO₂ + IO₃^– (alkaline medium)
Answer:
MnO₄^– + I^– → MnO₂ + IO₃^– (alkaline medium)
H₂O + 2MnO₄^– + I^– → 2MnO₂ + IO₃^– + 2OH^– is balanced equation.

Question 21.
Complete the following equations :
(a) PbS(s) + H₂O₂(aq) → ………………………… (b) MnO₄^–(aq) + H₂O₂ →………………….
Answer:
PbS + 4H₂O₂ → PbSO₄ +4H₂O
2MnO₄^– + 3H₂O₂ → 2MnO₂ + 3O₂ + 2H₂O + 2OH^–

Question 22.
Write the cell reactions for Zn|Zn²⁺(lM)||cd²⁺(lM)|Cd
E°Zn²⁺ /Zn = -0.76V, E°Cd²⁺ /Cd = 0.40V
Answer:

Question 23.
Zinc Silver oxide cell is used in heating aids or utensils and electric watches:
Zn → Zn²⁺ + 2e^–; E° = +0.76V
Ag₂O + H₂O + 2e^– → 2Ag + 2OH^–; E° = +0.80V
Which is oxidized and reduced?
Answer:
Zn is oxidised, Ag₂O is reduced.

Question 24.
The standard emf value of some elements are listed below at 298 K. Which of the following two electrodes should be combined to form a cell having highest emf ? Identify the cathode and anode and write the cell reaction. Also mention the direction of flow of electrons in the external and internal circuit.

Answer:
For maximum emf; anode should have minimum reduction potential and cathode should have maximum reduction potential. Hence A²⁺ / A couple should be anode and C²⁺/C couple acts as cathode.
Cell reaction A + C²⁺ → A²⁺ + C
Direction of flow of electrons is from A to C in external and from C to A internal circuit. Direction of flow of current is opposite to the direction of flow of electrons.

Question 25.
Silver jewellery turns greyish black due to the formation of Ag₂S layer on it. Can this tarnish be removed by dipping tarnished jewllery in an aluminium vessel containing inert electrolytic solution. The standard electrode potential for the half reactions are given below :
Ag₂S + 2e^– → 2Ag + S^2-; E° =-0.71V ;and Al³⁺ +3e^– →Al; E° = -1.66V
Answer:
Tarnish (black colour) will be removed if the following reaction occur.
Al + Ag₂S → A1³⁺ +2Ag + S² ; here E° for this reaction = -0.71-(-1.66) = 0.95 V
EMF is positive. ∆G will be negative this reaction will occur and tarnish will be removed.

Question 26.
Justify that the reaction : 2Na(s) + H₂(g) → 2NaH is a redox reaction.
Answer:
In this reaction, Na is oxidised to Na⁺ and hydrogen is reduced from H2 and H- ion. Therefore it is a redox reaction.
2Na⁰ + H2⁰ → 2Na⁺ H^-1 (s)

Question 27.
Out of silver and aluminium vessel which one will be more suitable to store 1 M HCl solution and why ?
Answer:
Since reduction potential of silver is more than that of hydrogen (E°H⁺ /H₂Pt = 0), silver vessel will be suitable to store 1 M HCl. On the other hand E° of Al³⁺ /Al is less than that of hydrogen (E°H⁺ /H₂Pt) so that hydrogen will be liberated is stored in aluminium vessel.

Question 28.
What is the oxidation number of Cr in (i) K₂CrO₄ (ii) Ni and Ni(CO)₄
Answer:
Let the Ox. no. of Cr in K₂CrO₄ be x.
2(+1) + x + 4(-2) = 0
+2 + X – 8 = 0
x = +6
Ox. no. of Cr in K₂CrO₄ = +6 A
Ox. state of Ni in Ni(CO)₄ is x + 4(0) = 0;x – 0
Ox. state of Fe in Fe(CO)₅ is x + 5(0) = 0 or x = 0

Question 29.
Can Fe³⁺ oxidize Br^– to Br₂ at 1 m concentrations ?
Answer:
E°(Fe³⁺/ Fe²⁺) is lower than that of E°(Br / B^–). Therefore Fe²⁺ reduce Br₂ but Br^– cannot reduce Fe³⁺ . Thus Fe³⁺ cannot oxidize Br^– to Br₂.

Question 30.
What is the oxidation state of S in (i) H₂SO₃ (ii) P₄ (ii) (iii) PH₃ (iv) H₃PO₄
Answer:
(i) H₂SO₃ :2(+1) + x + 3(-2) = 0
x = +4
Ox. state of is IV in H₂SO₃
Answer:
(i) Ox. no. of P in P₄ = 0
(iii) Ox. no. of P in H₃PO₄ .
3(+1) + x + 4(-2) = 0
+3 + x – 8 = 0
x = +5
Ox. no. of P in H₃PO₄ = +5

Question 31.
An iron rod is immersed in a solution containing NiSO₄ and ZnSO₄. When the concentration of each salt is 1M, predict giving reasons which of the following reactions is likely to proceed ? (i) Iron reduces Zn²⁺ ions, (ii) Iron reduces Ni²⁺ ions.
Answer:
Given : E°(Zn²⁺ /Zn) = 0.76V, E°(Fe²⁺ /Fe) = 0.44V and E°(Ni²⁺ /Ni) = 0.25V
Answer:
(i) The reduction potential of iron is more than that of zinc. Therefore, iron will be reduced. In other words Zn²⁺ will not be reduced.
(ii) The reduction potential of Ni²⁺ is more than that of Iron. Therefore, Ni²⁺ will be reduced by iron.

Question 32.
Which of the following redox reaction is oxidation and which is reduction?
(i) Zn → Zn²⁺ + 2e^–
(ii) Cl2 + 2e^– → 2Cl^–
(iii) Fe → Fe²⁺ + 2e^–
(iv) Sn⁴⁺ + 2e^– → Sn²⁺
Answer:
(i) Oxidation
(ii) Reduction
(iii) Oxidation
(iv) Reduction

Question 33.
Arrange the molecules, NH₃, NO₃, HN₃, NO₂^–, N₂O₄ and N₂H₄ in the decreasing order of the oxidation states of nitrogen.
Answer:

Question 34.
Determine the change in oxidation number of S in H₂S and SO₂ in the following industrial process: 2H₂S + SO₂ → 3S + 2H₂O
Answer:
Let us write the oxidation numbers of atoms of all the reactants and products taking part in the reaction:

O.N. of S in H₂S increases by 2 O.N. of S in SO₂ decreases by 4

Question 35.
Can the reaction Cr₃O₇^2- + H₂O ⇌ CrO₄^2- +2H⁺ be regarded as a redox reaction ?
Answer:
Ox. no. of Cr in Cr₂O₇^2- ⇌ +6 Ox. no. of Cr in Cr₂O₇^2-= +6
Since the ox. no. of Cr has neither increased nor decreased in the above reaction, therefore this reaction cannot be regarded as a redox reaction.

Question 36.
Balance the oxidation reduction reaction FeS₂ + O₂ → Fe₂O₃ + SO₂
Answer:

Question 37.
Which of the two ClO₂^– or ClO₄^– show disproportionation reaction and why ?
Answer:
The oxidation state of Cl in ClO₂ is +3 and in ClO₄ is +7. Chlorine is present in highest oxidation state of +7 in ClO₄^– and it cannot increase its oxidation state. Hence ClO₄^– is not disproportionate.
The disproportion reaction of ClO₂^– is

Question 38.
Which type of electrolytic are used in salt bridge ?
Answer:
Only those electrolytes for which cations and anions have nearly the same ionic mobilities (i.e. distance travelled by an ion per second under a potential gradient of one voltmeter) are used as electrolytes in the salt bridge. Thus KCl, KNO₃, K₄SO₄, and NH₄NO₃ are preferred over NaCl, NaNO₃ and Na₂SO₄.

Question 39.
Mention the cation and anion which have highest ionic mobility.
Answer:
Among cations, H⁺ ion has the highest ionic mobility and among anions, OH^– has the highest ionic mobility.

Question 40.
The standard electrode potential corresponding to the reaction Au³⁺(aq) + 3e^– → Au(s) is 1.5 V. Predict if gold can be dissolved in 1M HCl solution and on passing hydrogen gas through gold salt solution, metallic gold will be precipitated or not.
Answer:
Consider the half reaction, 2H⁺(aq) + 2e^– → H₂(g); E° = 0.0V
Au³⁺(aq) + 3e^– → Au(s); E° = 1.50V
Since E° (1.50V) for (Au³⁺ / Au) is higher than that H+ / – \(\frac { 1 }{ 2 }\)H₂ (0.0V), therefore, Au³⁺
can be more easily reduced than H⁺ ions Au³⁺ can be reduced to Au metal by H₂ butr H⁺ cannot oxidize metallic gold to Au³⁺ ions. In other words, metallic gold does not dissolve in 1M HCl instead H₂ gas can reduce gold salt to metallic gold.

Question 41.
What is the oxidation number of S in (i) Na₂S₄O₆, (ii) S₂Cl₂?
Answer:
(i)

In b & c, s has oxidation number is zero. In a & d s has oxidation number +5
(ii)

Question 42.
What is the name of the reaction ?
2CH₃CH₂CH₂SH → cCH₃CH₂CH₂ – S – S – CH₂CH₂CH₃
Whether condensation, oxidation, reduction or polymerization?
Answer:
This is an example of oxidation reaction since two H-atoms have been removed.

Question 43.
Both Cr₂O₇^2-(aq) and MnO₄^– (aq) can be used to titrate Fe²⁺(aq). If a given titration, 24.50 cm3 of 0.1 M Cr₂O₇^2- were used, then what volume of 0.1 M MnO₄^– solution would have been used for the same titration ?
Answer:
Suppose V₂cm³ of M₂Fe²⁺ is titrated against 24.50 cm3 0.1M Cr₂O₇^2-  and V₁cm³ of 0.1M MnO₄^– solution, then

1st PUC Chemistry Redox Reactions Three/Four Marks Questions and Answers

Question 1.
Balance MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
Answer:
Solution x
Step – 1: Assign oxidation number:

Step – 2 : Write the oxidation number changes.
Change in oxidation number of Mn + 4 to +2 = 2
Change in oxidation number of Cl – 1 to 0 = +1

Step – 3 : Cross multiply the numbers. Co-efficient 2 is multiplied to HCl and 1 is multiplied to MnO₂.
MnO₂ + 2HCl → MnCl₂+H₂O + Cl₂

Step – 4 : Check the oxidation number, and 2 molecules of HC1 on left hand side to balance the chlorine atoms of MnCl₂. In order to balance oxygen and hydrogen atom 2 molecule of H₂O has to be added on the right hand side.
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Question 2.
Balance Fe²⁺+ Cr₂O₇^2- → Fe³⁺ + Cr³⁺ is acid media.
Answer:
Solution:
Step -1: (LEO = Loss of Electron Oxidation)

(GER = Gain of Electron Reduction)

Step-2: Fe²⁺ → Fe³⁺ … (1) (2 → 3 = 1 unit ON) ,
Cr₂O₇^2- → 2Cr³⁺ … (2) (+6 →+3 = 3 units 2 sets => 6 units ON)

Step – 3 : Cross multiplication i.e., (1) x 6 and (2) x 1
(1) x 6 ⇒ 6Fe²⁺ → 6Fe³⁺ … (3)
(2) x 1 ⇒ Cr₂O₇^-2 → 2Cr³⁺ …(4)

Step – 4 : Add (3) + (4) ⇒ 6Fe²⁺ + Cr₂O₇^2- → 6Fe³⁺ + 2Cr³⁺

Step – 5 : Balance Oxygen and then Hydrogen
6Fe²⁺ + Cr₂O₇^2- + 14H⁺ → 6Fe³⁺ + 2Cr³⁺+ 7H₂O

Question 3.
Evaluate the equivalent weight of I2 from the following equation.
I₂ + 2S₂O₃ → S₄O₆^2-+2I^–
Answer:

Question 4.
Find equivalent weight of KMnO₄ from

Answer:
Balance the given equation
2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]

Method – 1: Stoichiometry
We know equivalent weight of Oxygen = 8
Atomic weight of Oxygen = 16
∴ 2 equivalent of oxygen = 1 Atomic weight of oxygen
Here 5[O] = 10 equivalent of oxygen

Method – 2 : (Oxidation Number)
2 KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]
+7 ← 5 Unit GER Reduction → +3
KMnO₄ acts as oxidising agent

Question 5.
Evaluate the equivalent weight of oxidizing and reducing agents in the
following reaction (4M) K₂Cr₂0₇ +FeS0₄ >Cr³⁺ +Fe³⁺
Answer:
Assigning oxidation number,

Question 6.
The half cell reaction with their oxidation potentials are
Pb(s) → Pb²⁺(aq) + 2e^–; E^o_cell = +0.13V
Ag(s) → Ag(aq) + e^–; E^o_cell = -0.80V
Write the cell reaction and calculate its EMF.
Answer:
Rewrite the two equations in the reduction form thus.
Pb²⁺ (aq) + 2eE^– → Pb(s) ; E° = -0.13V .. (i)
Ag⁺(aq) + e^– → Ag(s) ; E° = +0.80V …(ii)
To obtain the equation for the cell reaction, multiply Eq. (ii) with 2 and subtract Eq. (i) from it, we have,
Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s) ; E^o_cell =+0.80-(-0.13) =+0.93V

Question 7.
Justify that the reaction ; 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g) is a redox
reaction. Identify the species oxidized, reduced, which acts as oxidation and which acts as reductant.
Answer:
Writing the oxidation number of each atom above its symbol we have,

Here, in the reaction, the oxidation number of copper decreases from +1 in Cu₂O or Cu₂S to 0 in copper metal, therefore, copper is reduced and is .oxidizing agent or oxidants. Further, the oxidation number of S increases from -2 in Cu₂S to +4 in SO₂. Therefore, sulphur is oxidized. Thus, it is reducing agent of reductant.

Question 8.
Using stock notation, represent the following compounds : HAuCl₂ , TI₂O, FeO, Fe₂O₃, Cul, CuO, MnO and MnO₂.
Answer:
By applying the various rules for determining the oxidation number of different atoms in a compound, the oxidation number of each metallic element in the given compounds is as follows:
In HAuCl₂ , Au has +3; in TI₂0, TI has +1; in FeO, Fe has +2; in Fe₂O₂ , Fe has +3 ; . in Cul, Cu has +1 : in CuO, Cu has +2 ; in MnO, Mn has +2 while in Mn₂O , Mn has +4 oxidation state. Therefore, these compounds may be represent as :
HAu(II)Cl₄, Tl₂(1)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, and Mn(IV)O₂.

Question 9.
Balance the equation, Mg(aq) + HNO₃(aq) → Mg(NO₃)₂(aq) + N₂O(g) + H₂O(1)
Answer:
Find out the elements which undergo a change in oxidation number (O.N.)
O.N. increases by 2 per Mg atom

Here, O.N. of Mg increases from 0 in Mg metal to +2 in Mg(NO₃)₂ and that of N decreases from +5 in HNO₃ to +1 in N₂O.

Question 10.
Split the following redox reactions into the oxidation and reduction half reactions.
(a) 2K(s) + Cl₂(g) →2KCl(s) (b) 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺+(aq) + 3Cu(s)
Answer:

Question 11.
Predict the oxidation state of (a) C in C₃O₂, (b) Cr in Br₃O₃, (c) S in S₃O₃^2-
Answer:
(a) The structure of C₃O₂ (carbon -2 +2 0 +2 -2 suboxide) isO = C = C = C = 0 each of the two terminal oxygen atoms have an oxidation state of -2 and the carbons to which they are atteached have an oxidation state of +2. (Since when ever a covalent bond is formed between similar atoms, each of the atoms is given an oxidation state of zero), therefore, the oxidation state of central carbon atom in C₃O₂ is zero.

Thus, the two terminal carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state. Therefore, the average oxidation state of carbon in

(b) Tribromooctaoxide (Br₃O₈)

We can easily find out that the terminal bromine atoms are present in +6 oxidation state while the middle one has an oxidation state of +4. However, the average oxidation number of Br in Br308 turns out to be

(c) Tetrathionate ion

Since each of the two terminal sulphur atoms is connected to two oxygen atoms by a double bond and one oxygen atom by a single bond, therefore, the oxidation state of each of these terminal sulphur atoms is +5.

Since two central sulphur atoms are linked to each other by a single bond and each sulphur is further attached to similar species on either side, the electron pair forming the S – S bond remain in the centre and hence each of the two central sulphur atoms has an oxidation state of zero.
However, the average oxidation state of the sulphur atoms is

Question 12.
Mention which is reduced and oxidized in the following reaction :
(a) 4Na(s) + O₂(g) → 2Na₂0(s)
(b) 2Na(s) + Cl₂(g) → 2NaCl(s) ,
(c) 2Na(s) + S(s) → Na₂S(s)
Answer:

Question 13.
Identify the oxidant and reductant in the following reactions :
(a) 10H⁺ (aq) + 4Zn(s) + NO₃^– (aq) → 4Zn²⁺(aq) + NH⁺₄ +3H₂O(l)
(b) I₂(g) + H₂S(g) → 2IH(g) + S(s)
Ans:
(a) Writing the O.N. of all the atoms above their symbols, we have

O.N. of Zn changes from zero in Zn to +2 in Zn²⁺ and, therefore, it is oxidized and hence Zn acts as a reductant.
The O.N. of N decreases from +5 in NO₃^– to -3 in NO₄⁺ and, therefore, it is reduced and hence NO₃^– _{acts as the oxidant.}

(b) Writing the ON of all the atoms above their symbols, we have

The ON of I₂ decreases from zero in I2 to -1 in HI, therefore, I₂ is reduced and hence it acts as on oxidant.
The O.N. of S increases from -2 in H₂S to zero in S, therefore, H₂S is oxidized and hence it acts as the reductant.

Question 14.
0.2g of a sample of H₂O₂ reduced 20 ml of 0.1M KMnO₄ solution in acidic medium. What is the percentage purity of H₂O₂ ?
Answer:
No of moles of KMnO₄ present in 20ml of 0.1M
KMnCh solution = \(\frac{20}{1000}\) x 0.1 = 2 x 10^-3

The balanced equation for the redox reactions is :
2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂ …(1)
From the equation, 2 moles of KMnO₄ = 5 moles of H₂O₂.
2 x 10^-3 moles of KMnO₄ will react with H₂O₂ = \(\frac{5}{2}\) x 2 x 10^-3 = 5 x 10^-3 moles
Molecular wt. of H₂O₂ = 34
Amount of H2O2 actually present = 34 x 5 x 10^-3 = 0.17g .
Percentage purity of in 20 ml is =  \(\frac{0.17}{0.20}\) x l00 = 85

Question 15.
16.6g of pure potassium iodide was dissolved in water and the solution was . made upto one litre. V cm³ of this solution was acidified with 20 cm³ of 2M HCl. The resulting solution required 10 cm3 of decinormal KlO₃ for complete oxidation of I- ions to ICl. Find out the value of V.
Answer:
The chemical equation for the redox reactions is :

Question 16.
Which of the following species, do not show disproportionation reaction and why ? CIO^–, CIO₂^–, CIO₃^– and CIO₄^– Also write reactions for each of the species that disproportionates.
Answer:
The oxidation state of Cl in all the given species are :

Out of above species, CIO₄^– does not undergo disproportionation since in this oxoanion chlorine is already present in the highest oxidation state of +7 and hence cannot be further oxidized.

All the remaining oxoanions have oxidation state lower than the highest (i.e., +7) and higher than the lowest (i.e. -1) therefore, all these species undergo disproportionation reactions as shown below.

Question 17.
How does CU₂O act as both oxidant and reductant ? Explain with proper reactions showing the change of oxidation numbers in each example.
Answer:
Cu₂O undergoes disproportionation to form Cu²⁺ and Cu.
2Cu²⁺(aq) >Cu²⁺(aq) + Cu(s)
Thus, Cu₂O acts both as an oxidant as well as a reductant.
(i)

and Cu₂O acts as a reductant and reduces O₂ to O^2-

(ii) When heated with Cu₂S it oxidizes S^2- to SO₂ and hence Cu₂O acts as an oxidants

Question 18.
Determine the volume of \(\frac { M }{ 8 }\) KMnO₄ solution required to react completely with 25.0cm3 \(\frac { M }{ 4 }M/4\) FeSO₄ solution in acidic medium.
Answer:
The balanced ionic equation for the reaction is
MnO₄^–+ 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
From the balanced equation, it is evident that 1 mole of KMnO₄ = 5 moles of FeSO₄.
Applying molarity equation to the balanced redox equation,

Thus, the volume of \(\frac { M }{ 4 }M/4\) KMnO₄ solution required =10.0 ml.

Question 19.
Calculate the concentration of hypo (Na₂S₂O₃.5H₂O) solution in gdm^-3 if 10.0ml of this solution decolourised 15mL of M/40 iodine solution.
Answer:
The balanced equation for the redox reaction is 2S₂O₃^2- + I₂ → 2I^– + S₄O₆ ^2-
From the balanced equation, it is evident that 2 moles of Na₂S₂C>O₃ ≡ 1 mole of I₂.

Question 20.
Calculate the volume of 0.05M KMnO₄ solution required to oxidize completely 2.70g of oxalic acid (H₂C₂O₄) in acidic medium.
Answer:
Balanced equation for the redox reaction is :
2KMnO₄ + 5(COOH)₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O
No of moles oxalic acid = 2.70/90 = 0.03 mole
From the balanced equation, 5 moles of (COOH)₂ = 2 mole KMn0₄ .
Then 0.03 mole (COOH)₂ = pppp— x 0.03 = 0.012 moles of KM_nO₄.
Now, 0.05 mole of KMnO₄ is present in solution given = 1000 cm³.
0.012 mole of KMnO₄ is present in solution = \(\frac { 2 }{ 5 }\) = 240 cm³

Question 21.
Write the rules for assigning oxidation number.
Answer:

• All atoms in the elemental or molecular state have ‘zero’ oxidation state.
• Elements of group I and II in the periodic table always have ‘+1’ and ‘+2’ oxidation sstates respectively.
• Hydrogen ‘+1’ oxidation state in all its compounds except metal hydrides where it is ‘-1’
• Oxygen has been assigned an oxidation number of ‘-2’ in all its compounds except peroxides and oxygen fluoride. In peroxide it is ‘-1’ and in oxygen fluoride it is ‘+2’.
• Halogens generally have ‘-1’ oxidation state. Except fluorine, other halogens may have positive oxidation states in their oxides and inter halogen compounds (Fluorine always has ‘-1’ state).
• The algebraic sum of oxidation numbers of all the elements in a compound is zero and that in an ion is equal to the net charge on the ion.

Question 22.
Balance MnO₄^– + Fe²⁺ → Fe³⁺ + Mn²⁺ in acidic medium by ion electron method.
Answer:
MnO₄^– + Fe²⁺ → Fe³⁺ +Mn²⁺
[Fe²⁺ → Fe³⁺] x 5 ……………(i)
8H⁺ + MnO₄^– >Mn²⁺ + 4H₂O …(ii)
… (ii) Adding (i) and (ii), we get
5Fe²⁺ + 8H⁺ + MnO₄^– → Mn²⁺ + 4H₂O + 5Fe³⁺

Question 23.
Find the oxidation state of sulphur in the following compounds :
Answer:

Question 24.
Find the oxidation number of element underlined in the following species :

Answer:

Question 25.
Calculate the oxidation number of (i) N in NO₃^– ; (ii) P in H₃P₂O₇^– CO₃^2-.
Answer:
(i) N in NO₃^– , let the oxidation number of N in NO₃^– be x. Writing the oxidation number of each atom above its symbol. \(\begin{array}{cc}{\mathbf{x}} & {-2} \\ {\mathbf{N}} & {\mathbf{O}_{3}}\end{array}\)
x + 3(-2) = -1
x – 6 = -1, x = +5
(ii) P in H₃P₂O₇^– , let the oxidation number of P in H₃P₂O₇^– be x. Writing the oxidation
number of each atom above its symbol.
\(\begin{array}{ll}{\text { th } x} & {-2} \\ {\text { H }_{3} P_{2}} & {\text { o_{ } }}\end{array}\)
H₃P₂O₇^–= 3(+1) + 2 (x) = -1 +7(-2) or 2x – 11 = -1; x = + 5
Thus, the oxidation number of P in H₃P₂O₇^– is +5.

Question 26.
What is the oxidation number of metals (i) [Fe(CN)₆ ]^4- and (ii) MnO₄^– ?
Answer:
(i) Fe in [Fe(CN)₆ ]^4-. Let the oxidation number of Fe in [Fe(CN)₆ ]^4- bexand CN be -1 i.e..,

∴  Sum of oxidation numbers of all the atoms in
[Fe(CN)₆ ]^4- =x + 6(-l) = -4 = x-6 = -4; x = +2
Thus, the oxidation number of Fe in [Fe(CN)₆ ]^4- f is +2.

(ii) Mn in MnO₄^–. Let the oxidation number of Mn in MnO₄^– be x. Writing oxidation number of each atom above its symbol, we get.

Thus, the oxidation number of Mn in MnO₄^– is +7
Predict the oxidation number of the element underlined NaCl, MgSO₄, AlF₃, SiCl₄, PF₃, P₄O₁₀, SF₆, SO₃, HClO₄, Cl₂O₇.
Answer:

Question 27.
What is the maximum and minimum oxidation numbers of N, S and Cl?
Answer:
(i) The highest oxidation number (O.N) of N is +5 since it has five electrons in the valence shell (3s² 2p³ ) and its minimum O.N. is -3 since it can be accept three more electrons to acquire the nearest inert gas (Ne) configuration.

(ii) Similarly, the highest O.N. of S is +6 since it has six electrons in the valence shell (3s² 2p⁴) and its minimum O.N. is -2 since it needs two more electrons to acquire the nearest inert gas (Ar) configuration.

(iii) Likewise the maximum O.N. of Cl is +7 since it has seven electrons in the valance shell (3s² 2p⁵ ) and its minimum O.N. is -1 since it needs only one more electron to acquire the nearest (Ar) Noble gas configuration.

Question 28.
Nitric acid acts only as an oxidising agent while nitrous acid acts both as an oxidising as well as a reducing agent. Why ?
Ans:
(i) Nitric Acid HNO₃ : Here Oxidation number of N is HNO₃ = +5 ; But
Maximum oxidation number of N = +5 ; Minimum oxidation number of N = -3 Since the oxidation number of N in HNO₃ is maximum (+5), therefore it can only decrease by accepting electrons. Hence HNO₃ acts only as an oxidizing agent.

(ii) Nitrous Acid HNO₂ : Here Oxidation number of N in HNO₂ = +3
Maximum oxidation number of N = +5 ; Minimum oxidation number of N = -3 Thus, the oxidation number of N can either increase by losing electrons or can decrease by accepting electrons. Therefore, HNO₂ acts both as an oxidizing as well as a reducing agent.

Question 29.
Write correctly the balanced equations for the following redox reactions using half reaction:
(i) H₂S + Fe³⁺ → Fe²⁺ + S + H⁺
(ii) I^–+IO₃^– + H⁺ → I₂+H₂O
(iii) I^– + O₂ + H₂O → I₂ + OH^–
(iv) Cu(s) + Au⁺ → Au(s) + Cu²⁺
Answer:

Question 30.
Consider the following reactions that produce electricity in the galvanic cell:
(i) 2Fe³⁺(aq) + 2Cl^–(aq) → 2Fe²⁺ (aq) + Cl₂(g)
(ii) 2Cr(s) + 3Cu²⁺ → 3Cu(s) + 2Cr³⁺
Write the anode and cathode reactions for the galvanic cell. Write cell in the usual notation.
Cathode
Answer:

Question 31.
Balance the following reaction :
(i) Cu + NO₃^– → NO₂ + Cu²⁺
Write correctly balanced half cell reactions and overall equations for the following equations by ion electron method:
Answer:

Question 32.
NO₃^– + Bi(s) → Bi³ + NO₂ in acid solution by ion electron method.
Answer:

Question 33.
Cr₂O₇^2- + C₂H₄O^– → C₂H₄O₂ + Cr³⁺ in acid solution by ion electron method.
Answer:

Question 34.
MnO₄ + H₂C₂O₄ → Mn²⁺ + CO₂ in acid solution by ion electron method.
Answer:

Question 35.
A cell is prepared by dipping a zinc rod in 1M ZnSO₄ solution and a lead rod in 1 M Pb(NO₃)₂ solution. The standard electrode a potentials for Pb⁺²/Pb and Zn⁺² /Zn electrode are -0.126 V and -0.763 V respectively. _
(a) How will you represent the cell ?
(b) Write the half cell reactions and the overall cell reaction.
Answer:

Question 36.
Copper dissolves in dilute HNO₃ but not in dilute HCl. Why ? Explain.

Answer:
Electrode potential of Cu²⁺/Cu is higher than H⁺/H₂ and so H⁺ ions cannot oxidize Cu to Cu²⁺ ions and therefore copper does not dissolve in dilute HCl. On the other hand, the electrode potential of NO₃^– ion i.e., NO₃^–/NO electrode is higher than that of copper electrode and hence it can oxidize Cu to Cu²⁺ ion; so copper dissolve in dilute HNO₃.
Thus copper dissolved in dilute HNO₃ due to oxidation of Cu by NO₃^– ions and not by H⁺ions.

1st PUC Chemistry Redox Reactions Five Marks Questions and Answers

Question 1.
Write the following redox reactions using half equations.
(i) Zn(s) + PbCl₂ (aq) → Pb(s) + ZnCl₂(aq)
(ii) 2Fe³⁺ (aq) + 2e^– (aq) →  I₂(s) + 2Fe²⁺ (aq)
(iii) 2Na(s) + Cl₂ (g) →  2NaCl(s)
(iv) Mg(s) + Cl₂(s) → MgCl₂(s)
(v) Zn(s) + 2H⁺ (aq) →  Zn²⁺ (aq) + H₂ (g)
In each of the reactions given above, mention
(i) Which reactant is oxidized ? To what ?
(ii) Which reactant is the oxidizer ?
(iii) Which reactant is reduced ? To what ?
(iv) Which reactant is the reducer ?
Answer:
(i) Zn → Zn²⁺ + 2e^– (oxidation), Pb²⁺ → 2e^– +Pb (reduction) Zn is oxidized to Zn²⁺, Pb²⁺ is reduced to Pb; Pb²⁺ is the oxidizer and Zn is the reducer.

(ii) 2Fe³⁺ + 2e^– → 2Fe²⁺ (reduction), 2l^– → I₂ +2e^– (oxidation)
Fe³⁺ is reduced to Fe²⁺, I^– is oxidized to I₂; I^– is the reducer and Fe³⁺ is the oxidizer.

(iii) 3Na → 2Na⁺ +2e^– (oxidation), Cl₂ + 2e^– > 2Cl^– (reduction)
Na is oxidized to Na⁺ and Cl₂ is reduced to Cl^–; Na is the reducer and Cl₂ is the oxidizer.

(iv) Mg → Mg²⁺ + 2e^– (oxidation), Cl₂ + 2e^– → 2Cl (reduction)
Mg is oxidized to Mg while Cl2 is reduced to Cl ; Mg is the reducer and Cl2 is the oxidizer.

(v) Zn → Zn²⁺ +2e^– (oxidation), 2H⁺ + 2e^– >H₂ (reduction)
Zn is oxidized to Zn²⁺ while H⁺ is reduced to H₂;Zn is the reducer and H⁺ is the oxidizer.

Question 2.
Write the reaction at anode and cathode and the net cell reaction in the following cells. Which electrode would be +ve terminal in each cell ?

Answer:

Question 3.
Mention which element undergoes disproportionate reaction in alkaline media.

Answer:

Question 4.
Calculate emf of the following cell:

Answer:

Question 5.
In the reaction given below^ identify the species undergoing oxidation and reduction:
(i) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
(ii) CH₂ = CH₂(g) + H₂(g) → H₃C – CH₃(g)
(iii) H₂S(g) + O₂(g) → 2S(s) + 2H₂O(1)

Answer:
(i) CH₄ is oxidized to CO₂ while O₂ is reduced to H₂O.
(ii) H₂S is oxidised to S while O₂ is reduced to H₂O.
(iii) CH₂ = CH₂ is reduced to CH₃ – CH₃ while H₂ is oxidized to CH₃ – CH₃.
(iv) Hg²⁺ is reduced to Hg while O^2- is oxidized to O₂.

Question 6.
Using electron-transfer concept, identify the oxidation and reduction in the following redox reactions.
(a) Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
(b) 2[Fe(CN)₆]^4- (aq) + H₂O₂(aq) + 2H⁺(aq) → 2[Fe(CN)₆]^3- (aq) + 2H₂O(aq)
(c) 2[Fe(CN)₆]^3- (aq) + 2OH^–(aq) + H₂O₂(aq) → 2[Fe(CN)₆]^4-(aq) + 2H₂O(l)
(d) BrO₃^–(aq) + F₂(g) + 2OH^–(aq) → BrO₄^–(aq) + 2F^–(aq) + H₂O(l)
(e) 2NaClO₃(aq) + l₂(aq) → 2NaIO₂(aq) + Cl₂(g)
Answer:
Oxidants : (a) H⁺ (b) H₂O₂, (c) [Fe(CN)₆]^3-, (d) Fe, (e) l₂.
Reductants : (a) Zn, (b) [Fe(CN)₆]^4- , (c) H₂O2, (d) BrO₃^– , (e)NaClO₃

Question 7.
Why do the following reactions proceed differently ?
Pb₃O₄ + 8HCl → 3PbCl₂ +Cl₂ + 4H₂O and
Pb₃O₄ + 4HNO₃   → 2Pb(NO₃)₂ + PbO₂ + 4H₂O
Answer:
Pb₃O₄ is actually a stoichiometric mixture of 2 moles of PbO and one mole of PbO₂, i.e., 2PbO. PbO₂. In PbO₂, lead is present in + 4 oxidation state, whereas in PbO, lead is present in +2 oxidation state. Since +2 oxidation state or Pb is more stable, therefore, PbO₂ acts as an oxidant (oxidizing agent) and hence oxidises OF ions of HCl into Cl₂. Further more, PbO is a basic oxide and hence reacts with HCl to form PbCl₂ and H₂O. Thus the reaction of Pb₂O₄ with HCl can be split into two , reactions, namely, acid-base and redox reaction as shown below :

Since HNO₃ is an oxidizing agent, it does not react with PbO₂ which is also an oxidizing agent. Therefore, no redox reaction occurs between PbO₂ and HNO₃. However, acid-base reaction between PbO and HNO₃ occurs as follows :

2PbO + 4HNO₃ → 2Pb(NO₃ )₂ + 2H₂O

The overall reaction of PbO₃ with HNO₃ can then be written as
2PbO + PbO₂ + 4HNO₃ → 2Pb(NO₃ )₂ + PbO₂ + 2H₂O or
Pb₃O₄+ 4HNO₃ → 2Pb(NO₃ )₂ + PbO₂ + 2H₂O
Thus, it is the passive nature of PbO₂ against HNO₃that makes the reaction of Pb₃O₄ with HNO₃ different from that with HCl.

Question 8.
Explain applications of Redox Reactions.
Answer:
1. Extraction of metals : By using a suitable reducing agent, metal oxides can be ‘ reduced to metals. For example, Fe₂O₃ is reduced to iron in the blast furnace using coke as the reducing agent.

Fe₂O₃(s) + 3C(S) → 2Fe(s) + 3CO(g)

Simialrly, Al₂O₃ is reduced to aluminium by cathodic reduction in an electrolytic cell. Other metals such as lithium, sodium, potassium, magnesium, calcium, etc., are also obtained commercially by electrolytic methods.

2. Electrochemical cells or batteries : Electrochemical cells or batteries based on redox reactions are widely used in our day today life to run a number of small and big gadgets and equipments. For example storage cells are used to supply all the electrical needs of our cars, trucks, buses, trains aeroplanes, etc. Similarly, electrical energy needed in the space capsule is obtained by the reaction of hydrogen and oxygen in fuel cells which are electro chemical cells using oxygen and hydrogen electrodes.

3. Photosynthesis : Green plants convert carbon dioxide and water into carbohydrates in presence of sunlight. This reaction is called Photosynthesis and is sensitized by chlorophyll.

During this reaction, CO₂ is reduced to carbohydrates while water is oxidized to oxygen. The energy needed for the reaction is provided by sunlight.

This reaction is a source of food for plants and animals. It also maintain a constant supply of 21% of O₂ by volume in the atmosphere needed for combustion of fuels and breathing of a living creatures in the world.

4. Supply of energy : The energy required for our daily needs is obtained by oxidation of fuels. For example, oxidation of fuels such as wood, gas, kerosene, petrol, etc., produces a large amount of energy which we need for various purposes in our daily life.
Fuels (wood, petrol, kerosene, gas) + O₂ → CO₂ + H₂O + other products + Energy

Human body also needs energy for proper functioning. This is obtained by the oxidation of glucose in our body to CO₂ and water.

5. Production of chemicals : Many chemicals of our daily needs such as caustic soda, chlorine, fluorine, etc, are produced by electrolysis which is based on redox reaction.

6. Quantitative analysis : Redox reactions are very useful in quantitative analysis by redox titrations. These titrations involve the reactions between oxidizing and reducing agents and help in estimating the amount of unknown substances in solutions.

Question 9.
Balance the following equations in acidic medium by oxidation number

Answer:

Question 10.
Balance the following equation by ION-Electron method :
Zn(s) + NO₄⁺ → Zn²⁺ (aq) + NH₄⁺ (aq) + H₂O(l)
Answer:

Question 11.
Split the following redox reaction in the oxidation and reduction half reactions:
(a) 2K(s) + Cl₂(g) >2KCl(s)
(b) 2Al(s) + 3Cu²⁺(aq) >2Al³⁺(aq) + 3Cu(s)
Answer:
(a) Let us write the oxidation numbers of the atoms of all the reactants and products taking part in reaction.

(b) Let us write the oxidation numbers of the atoms of all the reactants and products taking part in the reaction.
Increase in Oxidation Number

Question 12.
Balance the following reaction by ION-ELECTRON method Cl₂(g) + OH^– → Cl^– (aq) + CIO₃^– (aq) + H₂O (in basic medium)
Answer:
Cl₂(g) + OH^– → Cl^– (aq) + CIO₃^– (aq) + H₂O

Cl₂(g) + 6H₂O(l) → 2ClO₃^–(aq) (… Go for balancing oxygen is a abasic media odd off on RHS )
Cl₂(g) + 6H₂O(l) + 12OH^– → 2ClO₃^–(aq) + 12H₂O(l)
Cl₂(g) + 12OH^– (aq) → 2ClO₃^–(aq) + 6H₂O(l) (now balance e-)
Cl₂(g) I +12OH^– (aq) → 2ClO₃^–(aq) + 6H₂O(l) + 10e^– . ..(1)
Reduction half-reaction
Cl₂(g) → Cl^–(aq)
Cl₂(g) → 2Cl^–(aq)
Cl₂(g) + 3e → 2Cl^–(aq) …(2)
Multiply equation no. (2) by 5 to equalize electrons
Add both the half-reactions

Question 13.
Write the over all net ionic reaction for the following change if the half reactions are correctly balanced: Cl^– ion is oxidized to Cl₂ by MnO₂ in acidic solution: Chloride ion is oxidized to Cl₂ by MnO₄^–(in acid solution)
Answer:
The skeletal equation is

Where ox. no. of chlorine changes from -1 in Cl ion to O in Cl₂

Question 12.
A cell is prepared by dipping a chromium rod in 1 M Cr₂(SO₄)₃ solution and an ion rod in 1 M FeSO₄ solution. The standard reduction potentials of chromium and iron electrodes are -0.75V and -0.45 V respectively.
(a) What will be the cell reaction ?
(b) What will be the standard EMF of the cell ?
(c) Which electrode will act as anode ?
(d) Which electrode will act as cathode ?
Answer:
The two half cell reduction equations me :

Fe²⁺(aq) + 2e^– → Fe(s);E° =-0.45 V …(i)
Cr³⁺(aq) + 3e^– → Cr(s);E° = -0.75V …(ii)

Since Cr³⁺ / Cr electrode has lower reduction potential therefore it acts as the anode while Fe²⁺ /Fe electrode with higher electrode potential acts as the cathode.

To equalize the number of electrons, multiply Eq. (i) by 3 and Eq. (ii) by 2. But do not multiply their E° values. Thus,

3Fe²⁺ (aq) + 6e^– → 3Fe(s); E° = -0.45V .. .(iii)
2Cr³⁺(aq) + 6e^– → 2Cr(s) ; E° =-0.75V …(iv)

To obtain equation for the cell reaction, subtract Eq. (iv) from Eq. (iii), we have,
2Cr(s) + 3Fe²⁺(aq) → 2Cr³⁺(aq) + 3Fe(s)
E°_cell = -0.45-(-0.75V) = +0.30V Thus, the EMF of the cell = + 0.30V

Question 13.
Write the net ionic equation for the reaction of potassium dichromate (VI), K₂Cr₂O₇ with sodium sulphate, Na₂SO₃, in acid solution to give chromium (III) ion and sulphate ion.
Answer:
Step-1: Write the skeleton equation for the given reaction.

Step – 2 : Find out the elements which undergo a change in oxidation number (O.N.) O.N. decreases by 3 per Cr. atom

Here, O.N. of Cr decreases from +6 in Cr₂O₇^2- to +3 in Cr³⁺ while that of S increases from +4 SO₃^-2 to +6 in SO₄^2-
Step – 3 : Balance increase / decrease in O.N. since the total increase in O.N. is 2 and decrease is 6, therefore, multiply SO₃^2- on L.H.S. and SO₄^2- on R.H.S. of Eq. (ii) by 3. Combining steps 2 and 3, we have,

Step – 7 : Balance O atoms by adding H₂O molecules.

Step – 7 : Balance H atoms by adding H⁺

Thus, Eq. (v) represents the correct balanced equation.

Question 14.
Permanganate ion react with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balance chemical equation for the reaction.
Answer:
Step – 1: Write the skeletal equation. The skeletal equation for the given reaction is :

Step – 2 :
Find out the elements which undergo a change in oxidation number (O.N.) O.N. increases by 6 per Br atom

Here, O.N. of Br inceases from – 1 Br^– to +5 in BrO₃^–, therefore, Br^– acts as reductant. Further, O.N. of Mn decreases from +7 in MnO₄^– to +4 in MnO₂, therefore, MnO₄^– acts as oxidant.

Step – 3 : Balance O atoms by adding H₂O molecules.

Step – 4 s Balance H atoms by adding H₂O and OH^– ions since the reaction occurs in basic medium, therefore, add 2H₂O to L.H.S. and 2OH^– to R.H.S. of Eq. (iv), we have,

Question 15.
Write the half reactions for the following redox reactions.
(a) Fe²⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(aq)
(b) Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
(c) Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s)
Answer:

Question 16.
Balance the equation, by ion electron method.
AS₂S₃(S) + NO₃^–(aq) + H⁺ (aq) → AsO₄^3-(aq) + S(s) + NO(g) + H₂(l)
Answer:
Step 1 : To identify the atoms whose oxidation numbers have undergone a change. Writing the oxidation number of each atom above its symbol, we have

Step 2:

Step 3 : To balance the oxidation half Eq. (i)
(a) Balance all the atoms other than H and O. Multiply AsO₄^3- by 2 and S by 3 on RHS of Eq. (i) we have,

(c) Balance charge by adding H+ ions. The total charge on RHS of Eq. (iv) is -16 and zero the LHS therefore, add 16H+ to RHS of Eq. (iv) we have,

(d) Balance O atoms by adding H₂O molecules.

The H-atoms get automatically balanced. Thus Eq. (vi) represents the balanced oxidation half equation.

Step 4 : To balance the reduction half Eq. (ii)
(a) Balance oxidation number by adding electrons.

(b) Balance charge by adding H+ ions.

(c) Balance H+ by adding H20

Thus, Eq. (ix) represents the balanced reduction half equation Eq. (ix) by 10 and Eq,
(vi) by 3 and (ix) 10 + (vi) x 3

Step 5 :

Question 17.
Permanganate (VII) ion, in basic solution oxidizes iodide ion 1- to produce molecular iodine (I₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Answer:
Step 1 s Write pie skeletal equation for the given reaction.

Step 2 : Write the O.N. of all the elements above their respective symbols.

Step 4 : To balance oxidation half Eq. … (ii)
(a) Balance all atoms 2I^–(aq) → I₂(s) … (iv)
(b) Balance O.N. by adding electrons.
2I^–(aq) → I₂(s) + 2e^– …(v)
Charge on either side of Eq(v) is balanced. Thus, Eq. (v) represents the balanced oxidation half equation.

Step 5 : Balance the reduction half equation (iii)
Balance O.N. by adding electrons.

Step 6: To balance the electrons lost in Eq. (v) and gained in Eq. (viii) Equation (v) x 3 + Equation (viii) x 2 we have,

This represents are final balanced redox equation.

Question 18.
In passing chlorine gas through a concentrated solution of alkali, we get chloride and chlorate ions. Obtain balanced chemical equation for this reaction by ion -electron method.
Answer:
Step 1: Write the skeletal equation for the given reaction.

Step 2 : Write the ON of all the elements above their respective symbols O.N. of Cl increases by 5 per Cl atom.

O.N. of Cl decreases by 1 per Cl atom Total increase = 2 x 5 = 10 Total decrease = 2 x -1 = -2

Step 3 : Find out the oxidant and the reductant and split the skeletal Eq (i) into two half reactions.

Step 4 : To balance the reduction half equation (ii). (a) Balance all atoms.
Cl₂(aq) →2Cl^–(aq) … (iv)
(b) Balance oxidation number by adding electrons.
Cl₂(g) + 2e^– →2Cl^–(aq) … (v)
represents the balanced reduction half reactions.

Step 5 : To balance the oxidation half equation (iii)
(a) Balance all atoms Cl₂(g) »2ClO₃^–
(b) Balance ON by adding electrons.
Cl₂(g) → 2ClO₃^–(aq) + 10e^– … (vi)
(c) Balance charge by adding OH” ions
Cl₂(g) + 120H^–(aq) → 2ClO₃^–(aq) + 10e^– … (ix)
(d) Balance O atoms, the RHS of Eq. (viii) contains six O atoms but on the LHS there are 12. Therefore, add 5H₂O to the RHS, we have

Eq (ix) represents the balanced oxidation half equation.

Step 6 : (v) x 5 + (ix) we have

This represents the final balanced redox equation.

Question 19.
Suggest a scheme of classification of the following redox reactions.
(a) N₂(g) + O₂(g) →2N(Xg)
Ob) 2Pb(NO₃)₂(s) → 2PbO(s) + 2NO₂(g) + 1/2 O₂(g)
(c) NaH(s) + H₂O(1)→  4NaOH(aq) + H₂(g)
(d) 2NO₂(g) + 20H^–(aq) → NO₂^–(aq) + NO₃^– (aq) + H₂O(l)
Answer:
(a)
In this reaction, the compound nitric oxide is formed by combination of elemental substances like nitrogen and oxygen. Since the oxidation number of nitrogen increases from 0 to +2 and of oxygen decreases from 0 to -2 therefore, it is a combination redox reaction.

(b)

In this reaction, lead nitrate decomposes to form three products, viz, lead oxide, nitrogen dioxide and oxygen. Since the oxidation number of nitrogen decrease from +5 in lead nitrate to +4 in NO2 in O2, therefore, it is a decomposition redox reaction.

(c)

In this reaction, hydrogen of water has been displaced by hydride ion to form dihydrogen gas. Therefore, it is a displacement reaction. Since in this reaction, the oxidation number of hydrogen increases from -1 in hydride ion to zero in dihydrogen gas and that of hydrogen decreases from +1 in water to zero in dihydrogen, therefore, it is a displacement redox reaction.

(d)

This is a disproportionation reaction since here the oxidation state of nitrogen decreases from +4 in NO₂ to +3 in NO₂^– ion, as well as increases from +4 in NO₂ to +5 in NO₃ ion.

Question 20.
Calculate the oxidation number of the element underline in each of the following cases,
(a) Al in Al₂O₃
(b) P in P₂O₅
(c) S in Na₂S₂O₃
(d) Cl in NaClO₃
(e) Mn in KM_nO₄
(f) HA₄Cl₄
(g) Na₂PdCl₄
(h) K₄MO₂Cl₈
(i) ClO^–
(j) Si0₃2^–
(k) H₂PO₄^–
(1) Cr₂O₇2^–
(m) H₂PO₄^–
Answer:
Solution : The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

b) P₂O₅ : 2 x (oxidation no. of P) + 5(-2) = 0
oxidation of P = pp = +5

c) Na₂S₂O₃ : 2(+l) + 2(oxidation no. of S) + 3(-2) = 0
2(oxidation no. of S) = 4

d) NaClO₃ : 1(+1) + (oxidation no. of Cl) +3 (-2) = 0
oxidation no. of Cl = +6 – 1 = +5

e) KMnO₄ : 1(+1) + (oxidation no. of Mn) + 4(-2) = 0
+1 + (oxidation no. of Mn) – 8 =0
∴ oxidation no. of Mn = + x7
Answer:
Solution : The rule used here is that the algebraic sum of the oxidation numbers of all the atoms in an ion is equal to the net charge on the ion.

f) HA₄Cl₄ 1(ON of H) + 1C (ON of A4) + (ON of Cl) = 0
(+1) + x + 4(-l) = 0 => l + x-4 = 0 => x = +3

g) Na₂PdCl₂ = 2(+l) + x + 4(-l) = 0 => 2+x-5=0 => x = +2

h) x be ON to be find
K₄MO₂Cl₈ = 4(+l) + (x)2 + 8(-l) = 0
4 + 2x – 8 = 0 => 2x – 4 = 0 => x = +2

Question 21.
What is oxidation number of sulphur in the following molecules / ions (n) sinH₂S sin SO₄^2- sinH₂SO₄
Answer:
(i) The oxidation number of Hydrogen is +1, and s = x than

(ii) SO₄^2- let x be ON of s ; Oxidation Number of oxygen is -2

(iii) H₂SO₄ let x, +1, -2 be on of S, H and O

Question 22.
Find the oxidation number of the underlined atoms.
(a) CCl₄
(b) CH₄
(C) ClO₃^–
(d) BrF₃
(e) Na₂ B₄O₇
(f) HClO₃
(g) HPO₂ ^–
(h) Fe₃ O₄
Answer:

You can Download Chapter 3 Motion in a Straight Line Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 3 Motion in a Straight Line

1st PUC Physics Motion in a Straight Line TextBook Questions and Answers

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. a railway carriage moving without jerks between two stations.
2. a monkey sitting on top of a man cycling smoothly on a circular track.
3. a spinning cricket ball that turns sharply on hitting the ground.
4. a tumbling beaker that has slipped off the edge of a table.

Answer:

1. The railway carriage moving without jerks between two stations, so the distance between two stations is considered to be large as compared to the size of the train. Therefore the train is considered as a point object.
2. As the distance covered by the monkey is large in a reasonable time, so the monkey is considered a point object.
3. A point object does not have a spinning motion. Therefore ball can not be considered a point object.
4. As the beaker is tumbling and then slips off, so the distance covered by it is not large in a reasonable time. Therefore it is not treated as a point object.

Question 2.
The position-time (x-i) graphs for two children A and B returning from their school! O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;

1. (A/B) lives closer to the school than (B/A)
2. (A/B) starts from the school earlier than (B/A)
3. (A/B) walks faster than (B/A)
4. A and B reach home at the (same/ different) time.
5. (A/B) overtakes (B/A) on the road (once/twice).

Answer:
1. The distance between P and O (the school) is less than the distance between Q and O. (from the graph)
⇒ A lives are closer to the school than B.
2. From the graph, we see that the line for A starts before the line for B.
⇒ A starts from school earlier than B.
3. Velocity = Slope of the x-t graph. Clearly, the slope for B > slope for A.
⇒ B walks faster than A
4. From the graph, we see that both lines end at the same time.
⇒ A and B reach home at the same time.
5. From the graph, we see that the lines intersect once. Also, B walks faster.
⇒ B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks at a speed of 5 km h^-1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Time for the woman to reach her office = \(\frac{2.5 \mathrm{km}}{5 \mathrm{km} \mathrm{h}^{-1}}\) = 30 min.
∴ She reaches her office at 9.30 am.
Time for the woman to return home = \(\frac{2.5 \mathrm{km}}{25 \mathrm{km} \mathrm{h}^{-1}}\) = 0.1 h = 6 min.
∴ She reaches her home at 5.06 pm

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:

From the above table, see that every 8 s the drunkard moves 2 m forward.

By looking at the shape of the graph, we can see that the graph for the entire motion will be like this:-

From the graph, we see that the drunkard will fall into the pit at 137 s.
Analytically:-
The drunkard covers 2 m every 8 s.
∴ He will cover 8 m in 32 s. When he takes 5 steps forwards, he will fall into the pit.
∴ Total time = 32 + 5 = 37 s.

Question 5.
A jet airplane travelling at the speed of 500 km h^-1 ejects its products of combustion at the speed of 1500 km h^-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Velocity of jet =υ_J = + 500 km h^-1 (away from observer)
Relative velocity of ejection with respect to jet = υ_ej = – 1500 kmh^-1.
If υ_e is the velocity of ejected products, then =V_e-V_j
=>  v_e =v_e. + vi = – 1500 + (500) = – 1000 km h^-1
∴ v_eg = v_e-v_g = – 1000 km h^_1  (u_g=0)

Question 6.
A car moving along a straight highway with a speed of 126 km h^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Answer: v₀ = 126 km h^-1
= 126 × \(\frac{5}{18}\) m s^-1
= 35 m s^-1
Stopping distance = (x – x₀) = 200 m
Since the car is stopped , v = 0.
v₂ = v₀² + 2 a (x – x₀)
0 = 35² + 2a (200)
⇒ a = – 3.06 m s^-2
Retardation = 3.06 m s^-2
v = v₀ + at
⇒ 0 = 35 – 3.06 t
⇒ t = 11.4 s
Time for car to stop = 11.4 s.

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B Just brushes past the driver of A, what was the original distance between them?
Answer:
Initially,

Relative Acceleration a = 1 ms^-2
v₀ =72 km h^-1 = 72 × \(\frac{5}{18}\) m s^-1
= 20 ms^-1
Relative displacement = x
Relative velocity initially = V₀ – V₀ = 0
X = 1/2 at²
= 1/2 × 1 × 50² = 1250 m
Original distance between the trains = 1250 m.

Question 8.
On a two-lane road, car A Is travelling at a speed of 36 km h^-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

V_A= 36 km h^-1 = 10 m s^-1
VB = Vc = 54 km h^-1 = 15 m s^-1
Velocity Of B with respect to A
= V_BA = V_B – V_A
= 15m -10 = 5 m s^-1
Velocity of C with respect to A .
V_CA = V_C – V_A
= 15 – ( -10) = 25 m s^-1
S = V_CA t
1000 m = 25 × t
⇒ t = 40 s
C will take 40 s to cover AC. In that time, B must cover AB to overtake and avoid collisions.
s = V_BA t + 1/2 at²
1000 = 5 × 40 + 1/2 × a × 40²
⇒ a = 1 ms^-2
Acceleration of B = 1 ms^-2

Question 9.
Two towns A and B are connected by regular bus service with a bus leaving In either direction every T minutes. A man cycling with a speed of 20 km h^-1 In the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let the speed of the bus be v km h^-1
Distance covered by one bus before the next one leaves the same town is vT For buses going in the same direction as the cyclist, frequency of the bus is 18 min i.e.,
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 18 min
\(\frac{v T}{v-20}\) = 18 ……………… (1)
Similarly, for buses going in the opposite direction, frequency of the bus = 6 min
\(\frac{v T}{\text { relative velocity of the bus w.r.t. the cyclist }}\) = 6 min
\(\frac{v T}{v+20}\) = 6 ……………… (2)
\((1) \div(2)\)
⇒ \(\frac{v+20}{v-20}\)
Substitute for v in (1)
\(\frac{40 \times T}{40-20}\) =18 ⇒ T = 9 min.

Question 10.
A player throws a ball upwards with an initial speed of 29.4 ms^-1.

1. What is the direction of acceleration during the upward motion of the ball?
2. What are the velocity and acceleration of the ball at the highest point of its motion?
3. Choose the X = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of X-axis, and give the signs of position, velocity, and acceleration of the ball during its upward, and downward motion.
4. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s^-2 and neglect air resistance).

Answer:
1. The direction of acceleration is vertically downwards (towards the Earth)
2. At the highest point, velocity = 0 m s^-1 acceleration = g = 9.8 m s^-2 vertically downwards.
3. During upward motion:

• position – positive
• velocity – negative
• acceleration – positive

During downward motion:

• position – positive
• velocity – positive
• acceleration – positive

4. v = 0 at the highest point
v₀ = 29.4 m s^-1
g = -9.8 m s^-2
v² = v₀² + 2 g h
0 = 29.4² = 2 × (-9.8) × 4
⇒ h = 44.1m
Height of ascension = 44.1 m
v = v₀ + g t
0 = 29.4 – 9.8 t
t = 3 s
Time of ascension = Time of descension
∴ Time for the ball to return = 6 s

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity,
3. with constant speed must have zero acceleration,
4. with a positive value of acceleration must be speeding up.

Answer:

1. True. A particle thrown upward has zero speed at the highest point but a = g = 9.8 ms^_2  in a vertically downward direction.
2. False. Because the magnitude of velocity = speed. When speed is zero, velocity may not be non-zero.
3. True. Because the particle rebounds instantly with the same speed, it means it has infinite acceleration which is not possible.
4. False. It can be true only when the chosen positive direction is along the direction of motion.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Velocity of the ball when it hits the ground = v
v₀ = 0, h = 90 m, g = 9.8m s^-2
v² = v₀² + 2 g h
v² = 0² + 2 × 9.8 × 90
⇒ v = 42 ms^-1
Time for descent = t
v = v₀ + g t
42 = 0 + 9.8 t
t = 4.3 s
During the rebound, the initial speed of the ball = 9/10 of v
= 9/10 × 42
= 37.8 ms^-1
Total time for ascent and descent = t₁
Net displacement = 0
0 = 37.8 t – 1/2 gt²
37.8 t = 1/2 × 9.8 × t² ⇒ t = 7.7 s
Time at which maximum height is achieved = 4.3 + 1/2 7.7
= 8.15 s

Question 13.
Explain clearly, with examples, the distinction between:
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;

(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show In both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Answer:
(a) Consider a person who leaves home for work in the morning and returns home in the evening.
The magnitude of the displacement of the person during the interval is zero but the total length of path covered = 2 × distance between the home and the workplace.
Magnitude of displacement is the magnitude of the shortest length between the two points. The second quantity is clearly always greater than or equal to the first quantity. The equality holds if the body has not moved at all (path length = magnitude of displacement = 0)

(b) Average velocity = \(\frac{\Delta x}{\Delta t}\). If Δ X is zero (like in the example explained in (a)) then average velocity = 0.
Average speed = \(\frac{\text { Total path length }}{\text { Time taken }}\)
2 × \(\frac{\text { Distance between the home and the work place }}{\text { Time taken to traverse this distance }}\)
= Clearly average speed ≥ magnitude of average velocity. Equality exists when body does not undergo any motion.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the

1. magnitude of average velocity, and
2. average speed of the man over the interval of time

(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 tp 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:
v₁ = 5 km h^-1, v₂ = 7.5 km h^-1, d = 2.5 km
Time for the man to reach the market
= t₁ = d/v₁= 2.5/5 = 0.5 h = 30 min
Time for the man to return from the market
= t₂= d/v₂ = 2.5/7.5 = 1/3 h= 20 min

1. To find the average velocity,
Average velocity = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min,
  Displacement = 2.5 km
  time = 30 minutes
  ∴ Average velocity = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• interval between 0 and 50 minutes,
  Displacement = 0
  ∴ Average velocity = 0
• During between 0 to 40 minutes,
  Displacement = 2.5 km – 7.5 kmh^-1 × 10 min = 1.25 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{1.25 \mathrm{km}}{40 \mathrm{min}}\) = 1.875 km h^-1
  

2. To find the average speed:
Average speed = \(\frac{\text { Total path length }}{\text { Total time taken }}\)

• During the interval between 0 to 30 min
  Path length = 2.5 km
  time = 30 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}}{30 \mathrm{min}}\) = 5 km h^-1
• During the interval between 0 and 50 minutes,
  Path length = 2.5 km + 2.5 km
  Time taken = 50 minutes
  ∴ Average speed = \(\frac{2.5 \mathrm{km}+2.5 \mathrm{km}}{50 \mathrm{min}}\) = 6 km h^-1
• During 0 to 40 minutes
  Path length = 5 kmh^-1 × 30 min + 7.5 kmh^-1 × 10 min
  2.5 km + 1.25 km = 3.75 km
  time = 40 minutes
  ∴ Average velocity = \(\frac{3.75 \mathrm{km}}{40 \mathrm{min}}\) = 5.625 km h^-1

Question 15.
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
When we consider arbitrarily small intervals of time, the magnitude of displacement is always equal to the magnitude of distance.

Question 16.
Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent the one-dimensional motion of a particle.

Answer:
None of the four graphs can represent the one-dimensional motion of the particle.
(a) A particle cannot have two different positions at the same time.
(b) A particle cannot have two values of velocity at the same time.
(c) Speed is always positive.
(d) Total path length of a particle can never decrease with time.

Question 17.
Figure 3.21 shows the X- t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
No, it is wrong to make such statements about the trajectory of the particle because an x-t graph does not show the particle’s trajectory.
Context:
Consider a body dropped from the top of a tower at time t = 0. If the vertically downward direction is chosen as the positive direction, then the body’s x -t graph would resemble the one given in the question.

Question 18.
A police van moving on a highway with a speed of 30 km h^-1 fires a bullet at a thief’s car speeding away In the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 m s^-1, with what speed does the bullet hit the thief’s car?
Answer:
Speed of the police van = v_p = 30 km h^-1 = 25/3 ms^-1.
Speed of the thief’s car is v_t = 192 km h^-1 = 160/3 ms^-1
Speed of the bullet = v_b = 150 ms^-1
Speed of the bullet with respect to the police car = v_b + v_p = 150 + 25/3 = 475/3 msv^-1
Speed with which the bullet hits the thief’s car = Speed of the bullet with respect to the thief’s car
= 475/3 – v_t
= 475/3 – 160/3
=105 ms^-1

Question 19.
Suggest a suitable physical situation for each of the following graphs (Fig 3.22):


Answer:
(a) The x-t graph shows that initially x = 0, attains a certain value of x, again x becomes zero, and then x increases in the opposite direction till it settles at a constant x (i.e., comes to rest). Therefore, it may represent a physical situation such as a ball (initially at rest) on being hit moves with constant speed, rebounds from the wall with less rebound speed, and then moves to the opposite wall and then stops.

(b) From the υ-t graph, it follows that velocity changes sign again and again with the passage of time and every time losing some speed. Therefore, it may represent a physical situation such as a ball falling freely (after throwing up), on striking the ground rebounds with reduced speed after each hit against the ground.

(c) The a-t graph shows that the body gets accelerated for a short duration only. Therefore, it may represent a physical situation such as a ball moving with uniform speed is hit with a bat for a very small time interval.

Question 20.
Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t= 0.3 s, 1.2 s, -1.2 s.

Answer:
Let the maximum amplitude of the sine wave be A, where A is positive. We can see that the particle obeys the
equation x = – A sin \(\left(\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right)\)
Where T = 2 s = period of the sine wave
∴ position = x = – A sin (π t)
velocity – v = \(\frac{dx}{dt}\) = – A_π cos (π t)
acceleration = a = \(\frac{dv}{dt}\) = A_π² sin(π t)
At t = 0.3 s
x = – A sin (0.3_π) = negative
v = – A cos (0.3_π) = negative
a = A _π² sin (0.3_π) = positive
Since sin(0.3_π) > 0 and cos(0.3_π) > 0
At t = 1.2 s
x = – A sin (1.2_π) = – A sin (1.2_π) = positive
v = – A _π cos (1.2_π) = – A cos (1.2_π) = positive
a = A_π² sin (1.2_π) = negative
since sin(1.2_π)<0 and cos(1.2_π) < 0
At t = – 1.2 s
x = – A sin (- 1.2_π) = A sin (1.2_π)= negative
v = – A it cos(- 1.2_π) = – A_π cos (1.2_π) = positive.
a = A_π² sin (- 1.2_π) = – A_π 2sin(1.2_π) = positive
Since sin(-θ) = – sinθt cos(-θ) = cosθ sin(1.2_π) <0 and cos(1.2_π) < 0

Question 21.
Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each Interval.

Answer:
The average speed of particle = slope of x-t graph. The slope of x -t is maximum at t = 3, and minimum at t = 2. Therefore, the average speed of a particle is greatest at t = 3 add least at t = 2. o > 0 in 1 and 2, and υ< 0 in 3.

Question 22.
Figure 3.25 gives a speed-time graph of a particle In motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion gives the signs of v and an in the three Intervals. What are the accelerations at points A, B, C, and D?

Answer:
The magnitude of the slope of the speed-time graph is greatest for interval 2. Hence average acceleration is greatest in magnitude in interval 2. The value of speeds in interval 3 is the highest. Hence the average speed in interval 3 is the highest. Sign of a = sign of the slope of the speed-time graph, a > 0 for 1 and 3 a < 0 for 2. At points, A, B, C, and D, the tangent to the curve will be parallel to the time axis. Hence slope at the points is zero,
∴ a = 0 at points A, B, C, and D.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:
Initial velocity u = 0.
Acceleration a = 1 ms^-2.
Acceleration time = 10 s
Let s_n be the total displacement (or distance in this case) in time ‘n’ seconds.
s_n = ut + 1/2 at²
s_n = 0(n) + 1/2 a(n²) = 1/2 an² for n ≤ 10
Distance covered in ‘n’^th second d_n = s_n – s_n – 1
= 1/2 an² – 1/2 a(n-1)²
= an – 1/2 a (for n ≤ 10)
= (n – 1/2) m
At the end of 10 seconds, velocity acquired v = u + at = 0 + a (10)
10 a = 10 ms^-1
Distance covered in the ‘n’^th second = 10m for n >10
∴ d_n = {(n-1/2)m n ≤ 10,
10m n > 10
The plot is a straight line inclined to the time axis for uniformly accelerated motion. (It is a straight line parallel to the time axis for uniform motion).

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^-1. How much time does the bail take to return to his hands? if the lift starts moving up with a uniform speed of 5 m s^-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Answer:
Initial velocity n = 49 ms^-1
Acceleration g = -9.8 ms²
Displacement s = 0
s = ut + 1/2 at²
0 = 49 t + 1/2 (-9.8) t²
⇒ t = 0 or t = 10 s
t = 0 represents initial time. The required ‘t’ here is t = 10 s. In the case where the lift also moves, the relative velocity of the ball with respect to the boy remains the same, i.e, 49 ms^-1. The relative displacement is also 0. Hence the time required in this case is also 10 s.

Question 25.
On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the

1. speed of the child running in the direction of motion of the belt?.
2. speed of the child running opposite to the direction of motion of the belt?
3. time taken by the child in (1) and (2)?

Which of the answers alter If motion is viewed by one of the parents?

Answer:
1. Speed of the child running in the direction of motion of the belt = Speed of the child with respect to the belt + Speed of the belt with respect to the ground
= 9 + 4 = 13 km h^-1.
2. Speed of the child running opposite to the direction of motion of the belt = speed of the child with respect to the belt speed of the belt with respect to the ground
= 9 – 4 = 5 km h^-1.
3. Time taken by the child in both cases is the same because the speed of the child with respect to the belt does not change.
Speed v = 9 km h^-1 = 2.5 ms^-1
Distance d = 50 m
Time t = \(\frac{\text { Distance }}{\text { Speed }}\) = \(\frac{50}{2.5}\) = 20 s
The answer to (a) and (b) when the motion is viewed by one of the parents is 9 km h^-1. This is because the parents are also on the belt and are moving with respect to the ground. They only see the motion of the child with respect to the belt. In (c), the answer remains unaltered because the speed of the child with respect to the belt does not depend on the speed of the parents.

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with Initial speeds of 15 m s^-1 and 30 m s^-1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take
g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:
Initial velocity of stone 1 is u₁ = 15 ms^-1
Initial velocity of stone 2 = u₂ = 30 ms^-1
Acceleration = g = -10ms^-2 (vertically upwards direction is chosen as positive)
x₁ = 200 + u₁t + 1/2 at²
= 200 + 15 t – 5 t²
x₂ = 200 + u₂t + 1/2 at²
= 200 + 30 t – 5 t².
Also x₁ = 0 for t = 8 s and
x₂ = 0 for t = 10s.
∴ For 0 ≤ t ≤ 8 s
x₂ – x₁ = 15 t
At t = 8 s,
x₂ – x₁ = 120 m
For 8 < t; t ≤ 10 s
x₂ – x₁ = 200 + 30 t – 5 t² – 0
= 200 + 30 t – 5 t².
For t > 10 s,
x₂ – x₁ = 0
∴ The given graph is correct and because it matches with the equations obtained.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown In Fig. 3.28. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s.
(b) t = 2 s to 6 s.

What is the average speed of the particle over the intervals in (a) and (b)?.
Answer:
(a) Distance covered = Area under the speed – time graph
= 1/2 × 10 × 12 = 60 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{60 m}{10 s}\) = 6 ms^-1

(b) For 0 ≤ t ≤ 5 s,
Speed v = \(\frac{12}{5}\) t ms^-1
For 5 < t ≤ 10 s,
Speed v = 12 – \(\frac{12}{5}\) (t – 5) ms^-1
∴ Speed at t = 2 s is 4.8 ms^-1 and speed at t = 6 s is 9.6 ms^-1
Distance covered = Area of trapezium ABEF + Area of trapezium BCDE

= 1/2 (4.8 + 12) × 3 + 1/2 (12 + 9.6) × 1
= 36 m
Average speed = \(\frac{\text { Distance covered }}{\text { Time taken }}\)
= \(\frac{36}{6-2}\) = 9 ms^-1

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂

1. x(t₂) = x(t₁ + v (t₁) (t₂ – t₁ + (1/2) a (t₂– t₁)²
2. v(t₂) = v(t₁) + a(t₂ – t₁)
3. v_average = (x(t₂) – x(t₁)/(t₂-t₁)
4. average = (v(t₂ – v(t₁))/(t₂ – t₁)
5. x(t₂) = x(t₁) + v_average (t₂ – t₁) + (1/2) a_average (t₂ – t₁)²
6. x(t₂) – x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.

Answer:

1. Wrong, because it is not known whether the acceleration ‘a’ is constant.
2. Wrong, because it is not known whether the acceleration ‘a’ is constant.
3. Correct. By definition.
4. Correct. By definition
5. Wrong. The formula should contain v₁ (t) instead of v_average.
6. Correct By definition.

1st PUC Physics Motion in a Straight Line One Mark Questions and Answers

Question 1.
When do you say that a body is in motion?
Answer:
A body is said to be in motion when it changes its position with respect to time and surroundings.

Question 2.
What is a particle?
Answer:
A particle is a geometrical mass point.

Question 3.
Define displacement.
Answer:
The change of position of a body in a particular direction is called displacement.

Question 4.
Can the displacement of a moving body be zero?
Answer:
Yes

Question 5.
Define speed.
Answer:
The rate of change of position of a body in any direction is known as speed.

Question 6.
Define velocity.
Answer:
The rate of change of position of a body in a particular direction is called velocity or the Rate of displacement of a body is called velocity.

Question 7.
Define uniform velocity.
Answer:
If a body covers equal distances in equal intervals of time in a given direction, however small the intervals maybe then its velocity is said to be uniform.

Question 8.
Define variable velocity.
Answer:
If the velocity of a body either changes in magnitude or in direction or both, then its velocity is said to be variable.

Question 9.
Define average velocity.
Answer:
The average velocity of a body is defined as the ratio of its total displacement to the total time.

Question 10.
Define acceleration.
Answer:
Rate of change of velocity of a body is called acceleration.

Question 11.
Define uniform acceleration.
Answer:
If the velocity of a body changes by an equal amount in equal interval of time, however, small the interval maybe then its acceleration is said to be uniform.

Question 12.
What is a position-time graph?
Answer:
A graph drawn by taking time along the x-axis and displacement along the y-axis is called a position-time graph.

Question 13.
What does the slope of the position-time graph indicate?
Answer:
The slope of the position-time graph indicates velocity.

Question 14.
Draw the position-time graph of a particle which is at rest.
Answer:

Question 15.
Draw the position-time graph of an object starting from rest and moving with uniform velocity.
Answer:

Question 16.
What is a velocity-time graph?
Answer:
A graph drawn by taking time along x-axis and velocity along the y-axis is called a velocity-time graph.

Question 17.
What does the slope of the velocity-time graph indicate?
Answer:
The slope of the velocity-time graph indicates the acceleration.

Question 18.
What does the area under the velocity-time graph indicate?
Answer:
Area under the velocity-time graph indicates the displacement.

Question 19.
An object is moving with uniform velocity. What Is its acceleration?
Answer:
Zero

Question 20.
How do you determine the instantaneous velocity of a particle from the position-time graph?
Answer:
Instantaneous velocity at any point is given by the slope of the tangent drawn to the position-time graph at that point.

Question 21.
Draw the velocity-time graph of a particle moving with

1. uniform velocity
2. variable velocity.

Answer:
1)

2)

Question 22.
Draw the velocity-time graph of an object starting from rest and moving with uniform acceleration.
Answer:

Question 23.
Draw the velocity-time graph of an object moving with constant retardation.
Answer:

Question 24.
Draw the position time of the particle which is initially at rest and moving with uniform acceleration.
Answer:

Question 25.
What is the acceleration time graph?
Answer:
A graph drawn by taking time along x-axis and acceleration along the y-axis is called a velocity-time graph.

Question 26.
Write the expression for the distance travelled by a body during the nth second of its motion.
Answer:
Distance travelled by a body during the n^th second of motion is given by
s_n = u + a(n – 1/2)
where u is the initial velocity and a is the uniform acceleration.

Question 27.
Write the expression for the acceleration in terms of distance travelled in two consecutive intervals of time.
Answer:
Acceleration is given by,
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
where s₁ and s₂ are the distances travelled in two consecutive intervals of time ‘t’ seconds each.

Question 28.
What is relative velocity?
Answer:
The velocity of a particle in motion relative to another particle is called relative velocity.

Question 29.
When is the relative velocity of two bodies maximum?
Answer:
Relative velocity of two bodies is maximum when they are moving in opposite directions.

Question 30.
What does the area of a-t graph indicate?
Answer:
Area of a-t graph represents the change in velocity of the body in a given time interval.

Question 31.
Draw the a-t graph of a body moving with uniform acceleration.
Answer:

Question 32.
Why is it not necessary for a body following another, to stop, to avoid collision?
Answer:
If the relative velocity is zero, no collision will occur.

Question 33.
If in case df a motion, displacement is directly proportional to the square of the tune elapsed, what do you think about its acceleration, le, constant or variable? Explain why?
Answer:
x ∝ t²
⇒ x = kt² Where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{d}{d t}\) (2 k t)
= 2k = constant
∴ Acceleration is constant.

Question 34.
Why does the earth Impact the same acceleration to all bodies?
Answer:
Since acceleration is the force on unit mass, the acceleration due to gravity is constant.

Question 35.
What will be the nature of the velocity-time graph for a uniform motion?
Answer:
The v -t graph will be a line parallel to the time axis.

Question 36.
The position coordinate of a moving particle is given by x = 6 + 18t + 9t2(x is in metres and t is in seconds). What is its velocity at t s 2s?
Answer:
x = 6 + 18 t + 9t²
v= \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = 18 + 18 t
v at t = 2 s = 18 + 18 (2)
= 54 ms^-1

Question 37.
A ball is thrown straight up. What is its velocity and acceleration at the top?
Answer:
At the top, v = 0, a = g.

1st PUC Physics Motion in a Straight Line Two Mark Questions and Answers

Question 1.
Distinguish between speed and velocity.
Answer:

1. Speed is the rate of change of position of a body in any direction while velocity is the rate of change of position of a body in a particular direction.
2. Speed is a scalar quantity whereas velocity is a vector quantity.

Question 2.
What is a velocity-time graph? What Is Its importance?
Answer:
A graph drawn by taking time along the x-axis and velocity along the y-axis is called a velocity-time graph. The velocity-time graph can be used

1. To find the nature of the motion of a body.
2. To determine the velocity of a body at any instant.
3. To calculate the displacement of a body in a given time.

Question 3.
If V₁ and v₂ are the velocities of two bodies then what is their relative velocity when they move in the

1. same and
2. opposite direction?

Answer:

1. When the bodies move in the same direction, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ – v₂
2. When the bodies move in the opposite directions, the relative velocity of 1^st body w.r.t 2^nd is v₁₂ = v₁ + v₂.

Question 4.
A car moving with a uniform velocity of 54 kmph is brought to rest in travelling a distance of 5m What is the retardation produced by brake?
Answer:
Initial velocity of the car.,
u = 54km/hr
= \(\frac{54 \times 1000}{3600}\) = 15 ms^-1
Final velocity of the car, v = 0.
Let a be the retardation of the car. Then using the relation.
v² = u² + 2as
0² = 15² + 2a × 5
a = \( -\frac{15 \times 15}{2 \times 5}\) = \(\frac{225}{10}\) = -22.5 ms^-2.

Question 5.
An automobile moving with a uniform velocity of 500ms^-1 is brought to rest in travelling a distance of 5m. What is the acceleration produced by the brakes?
Answer:
Initial velocity of the car, u = 500ms^-1
Final velocity of the car v = 0
Let a be the acceleration of the car.
Then using the relation.
v² = u² + 2aS
0² = 500² + 2a × 5
a = \( -\frac{500^{2}}{10}\) = – 25 × 10³mS^-2.

Question 6.
The equation of motion of a body is given by S = 2t + 2t², where S is in meter and time in second. What is the acceleration of the body?
Answer:
S = 2t + 2t². This is of the standard form S = ut + 1/2 at²
By comparison, 1/2 a = 2 or a = 4m/s².

Question 7.
Draw a velocity-time graph for a particle in the following situations:

1. Starts from rest and moves with uniform acceleration
2. Moving with uniform retardation.

Answer:
1. Body moving the uniform acceleration.

2. Body moving with uniform retardation.

Question 8.
Define velocity and acceleration.
Answer:
Velocity is defined as the rate of displacement. Acceleration is known as the rate of change of velocity.

Question 9.
Two parallel nail tracks run North-South. Train A moves due north with a speed of 54 km h^-1 and train B moves due south with a speed of 90 km h^-1. What is the relative speed of B with respect to A in ms^-1?
Answer:
V_A = 54 km h^-1
V_B = – 90 km h^-1
(north direction is chosen as positive)
V_BA = V_B – V_A
= – 90 – 54 = -144 km h^-1
= – 40 ms^-1

Question 10.
Differentiate between the average speed and the instantaneous speed of an object.
Answer:
The distance covered in unit time is called average speed.
V_avg = \(\frac{x\left(t_{2}\right)-x\left(t_{1}\right)}{t_{2}-t_{1}}\)
The speed at any instant of time is called instantaneous speed.
V_inst= \(\Delta \mathrm{t} \rightarrow 0 \frac{\Delta \mathrm{x}}{\Delta \mathrm{t}}\)

Question 11.
The velocity-time graph of a moving object is shown below. What is the acceleration of the object? Also, draw a displacement time graph for the motion of the object.

Answer:
Acceleration = Zero (because velocity is constant) Displacement time graph is shown below:

Question 12.
Two straight lines are drawn on the same displacement time graph make angles 30° and 60° with the time axis respectively in the figure. Which line represents greater velocity? What is the ratio of the velocity of line A to that of line B?

Answer:
Velocity = Slope of x -t graph
V_A = tan 30° = \(1_{\sqrt{3}}\)
V_B = tan 60° = Clearly, V_B > V_A.
\(\frac{V_{A}}{V_{B}}\) = \(\frac{1_{\sqrt{3}}}{\sqrt{3}}\) = \(\frac{1}{3}\)

1st PUC Physics Motion in a Straight Line Three Mark Questions and Answers

Question 1.
The distance x travelled by a body In a straight line is directly proportional to t². Decide on the type of motion associated. If x ∝ t³ what change will you observe?
Answer:
x ∝ t² = kt² where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (2 k t )
= 2 k
∴ Acceleration is constant / uniform.
If x ∝ t³ where k is a constant
a = \(\frac{d}{d t}\left(\frac{d x}{d t}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\) (3 k t²)
= 6 k t
∴ Acceleration is non-uniform

Question 2.
Draw the following graph for an object under free-fall:

1. Variation of acceleration with respect to time.
2. Variation of velocity with respect to time
3. Variation of distance with respect to time.

Answer:
1.

2.

Line 2 appears only when the body bounces.
3.

Question 3.
A body that starts from rest accelerates uniformly along a straight line at the rate of 10 ms^-2 for 5 seconds. It moves for 2 seconds with a uniform velocity of 50 ms^-1. Then it retards uniformly and comes to rest in 3s. Draw the velocity-time graph of the body and find the total distance travelled by the body.
Answer:

Total distance travelled = Area under the v – t graph
= 1/2 × 50 × (10 + 2)
= 300 m

Question 4.
The displacement (in metre) of a particle moving along x-axis given by x = 18 t + 5t². Calculate:-

1. the instantaneous velocity at t = 2s
2. average velocity between ta 2 s and 3 s
3. instantaneous acceleration

Answer:
x = 18 t + 5 t²
v(t) = \(\frac{d x}{d t}\) = 18 + 10 t,
a(t) = \(\frac{d v}{d t}\) = 10 ms^-2
1. v (2) = 18 + 10 (2) = 38 ms^-1
2. v (2) = 38 ms^-1

v(3) = 18 + 10 (3) = 48 ms^-1
Since acceleration is constant,
V_avg \(\frac{v(3)+v(2)}{2}\) = \(\frac{48+38}{2}\) = 43 ms^-1.

3. Instantaneous acceleration a = 10 ms^-2.

1st PUC Physics Motion in a Straight Line FourFive Mark Questions and Answers

Question 1.
Derive an expression for the velocity of the particle after time ‘t’ or Derive v = u + at using a v-t graph.
Answer:
Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its initial velocity and v be its velocity after a time t. The velocity-time graph AB of this particle is shown in the figure.

Acceleration of the particle is given by the slope of the v-t graph.
From the graph,
slope of the line AB = \(\frac{B C}{A C}=\frac{B C}{O D}=\frac{B D-C D}{O D}\)
But, BD = v, the final velocity,
OA = u, the initial velocity,
OD = t, the time.
∴ Slope of the graph AB is,
a = \(\frac{B D-C D}{O D}\)
= \(\frac{v-u}{t}\)
Thus a = \(\frac{v-u}{t}\) or v = u + at.

Question 2.
What is the v-t graph? Derive an expression for distance covered by the particle in time ‘t’ or Derive the equation s = ut + 1/2 at² using v-t graph.
Answer:
A graph drawn by taking time along the x-axis and velocity along the y-axis is called a velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 3.
Derive an expression for the velocity of the particle after covering distance ‘s’ OR Derive the equation v² = u² + 2as using the velocity-time graph.
Answer:
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or
2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1),
2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 4.
Derive an expression for distance travelled during the n^th second of motion.
OR
Derive s_n= u + a(n – 1/2)
Answer:
Consider a particle moving with uniform acceleration ‘a’.Let u be the initial velocity of the particle.
Distance travelled during n^th second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n-1) + 1/2 a(n-1)²]
= un + 1/2 an² – [ un-u + 1/2 a(n²-2n +1)]
= un + 1/2 an² – [ un-u + 1/2 an²-an + 1/2 a]
= un + 1/2 an² – un+u – 1/2 an²+an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)

Question 5.
Derive an expression for acceleration in terms of distance travelled in two successive equal interval of time.
OR
Derive the equation a = \(\frac{s_{2}-s_{1}}{t^{2}}\)
Answer:
Consider a particle moving with uniform acceleration a. Let it start from a point with initial velocity ‘u’ and covers the distances s₁ and s₂ in two successive equal interval of time ‘t’ each. Distance covered in the first interval
s₁ = ut + 1/2 at² ………. (1)
Distance covered at the end of time 2t,
s₁ + s₂ = u(2t) + 1/2 a (2t)²
s₁ + s₂ = 2ut + 2at²……………. (2)
(s₁ + s₂ ) – 2s₁ = (2ut + 2at²) – (2ut + at²)
s₂ – s₁ = at²
a = \(\frac{s_{2}-s_{1}}{t^{2}}\)

Question 6.
Define uniform velocity obtain v² – u² = 2as from the v – t graph, where the symbols have the usual meaning.
Answer:
If a body undergoes equal displacements in equal intervals of time, however small interval of time may be then velocity is said to be uniform velocity.
particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t= 0 and ‘v’ be the velocity after ‘t’ seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelles by the particle is given by,
s = area under v-t graph
= area of the trapezium OACD
= 1/2 (OA+CD) × OD
But OA = u, CD = v and OD = t
∴ s = 1/2 (u + v) t or 2s = (u+v) t ………(1)
But, a = slope of the graph = \(\frac{v-u}{t}\)
∴ t = \(\frac{v-u}{a}\)
On substituting in (1), 2s = \((u+v)\left(\frac{v-u}{a}\right)\)
i.e., v² – u² = 2as
or v² = u² + 2as.

Question 7.
A particle is moving with uniform acceleration covers a distance of 45 m in 6^th second and 75 m in 12^th second during its motion. Calculate the displacement of the particle after 20s?
Answer:
S₆ = 45m, S₁₂=75m
using S_n = u + a (n – 1/2 )
i.e., S₆  = u + a or 45 = u + 5.5a ……… (1)
S₁₂= u + a(12 – 1/2) or 75 = u + a(11.5) …….. (2)
Eq(2) – (1) gives 6 a = 30
Or a = 30/6 = 5m/s²
From (1), 45 = u + 5 × 5.5
u = 17.5m/s.
Distance travelled in t = 20s,
S = ut+ 1/2at2
= 17.5 × 20 + 1/2 × 5 × 20²
=1350m.

Question 8.
Draw a velocity-time graph of uniformly accelerated motion in one dimension. From the velocity-time graph of uniformly accelerated motion deduce the equations of motion in distance and time.
Answer:
A graph drawn by taking time along the x-axis and velocity along the y-axis is called a velocity-time graph. Consider a particle moving along a straight line with a constant acceleration ‘a’. Let u be its velocity at time t = 0 and v be the velocity after t seconds. The velocity-time graph AC of this particle is shown in the figure.

Distance travelled by the particle in t seconds is given by,
s = area under v-t graph
= area of the triangle ABC + area of the rectangle ABDO
= 1/2 BC × AB + OA × OD From the graph,
BC = CD – BD = v – u,
AB = OD = t and OA = u
∴ s=1/2 ( v-u) t + ut …….. (1)
But, a = Slope of the graph = \(\frac{v-u}{t}\)
∴ v – u = at
On substituting in (1) we have,
s = 1/2 at.t + ut
s = ut + 1/2 at².

Question 9.
Derive an equation for the distance covered by a uniformly accelerated body in the n^th second of its motion. A body travels half its total path in the last second of its fall from rest. Calculate the time of its fall.
Answer:
Consider a particle moving with a uniform acceleration ‘a’.
Let u be the initial velocity of the particle.
Distance travelled during nth second = Distance travelled in ‘n’ seconds – Distance travelled in (n-1) seconds
s_n = un + 1/2 an² – [ u(n – 1) + 1/2 a(n – 1)²]
= un + 1/2 an² – [ un – u + 1/2 a(n² – 2n + 1)]
= un + 1/2 an² – [ un – u + 1/2 an² – an + 1/2 a]
= un + 1/2 an² – un + u – 1/2 an² + an – 1/2 a
= u + an – 1/2 a
s_n =u + a(n – 1/2)
s_n = 4 + a (n – 1/2)
Given u = 0, a = g
s = 1/2 g t² = 1/2 g n²
By the given condition 1/2 s = s_n
⇒ \(\frac{\mathrm{g} n^{2}}{4}\) = g(n – 1/2)
⇒ n² – 4n + 2 = 0
⇒ n = \(\frac{4 \pm \sqrt{16-8}}{2}\)
⇒ n = 2 ± \(\sqrt{2}\)
∴ Time of the body’s fall = (2 ± \(\sqrt{2}\)) s

1st PUC Physics Motion in a Straight Line Numerical Problems Questions and Answers

Question 1.
A man runs from his home to office at a speed of 2 ms^-1 on a straight road and returns back to home at a speed of 4 ms^-1. Find

1. Average speed
2. average velocity.

Solution:
1. To find the average speed. Let ‘s’ is the distance between the home and the office.
∴ Time taken to reach the office
t₁ \(\frac{\text { distance }}{\text { velocity }}\) = \(\frac{s}{2}\)
Similarly, time taken to walk back to the home from the office is t₂ = \(\frac{s}{4}\)
∴ Total time taken = t₁ + ₂ = \(\frac{s}{2}\) + \(\frac{s}{4}\) = \(\frac{3 s}{4}\)
Total distance moved = s + s = 2s
Hence. the average speed \(=\frac{\text { toal distance moved }}{\text { total time taken }}\)
= 2.67 ms^-1
2. To find the average velocity Since the man comes back to home, the final and initial position is same.
∴ Total displacement is zero.
Hence average velocity
\(=\frac{\text { toal displacement }}{\text { total time }}\) = 0

Question 2.
A car moving along a straight high-way with a speed of 35 m/s is brought to stop within a distance of 200 m. What Is the retardation, and how long does it take for the car to stop? Solution:
The initial speed of the car, u = 35 ms^-1
The final speed of the car, v = 0
Let a be the retardation of the car.
Then using the relation,
v² = u² + 2as, we have 0 = (35)² + 2a × 200
a = – \(\frac{35 \times 35}{2 \times 200}\) = – 3.06 ms^-2
Let t be the time taken by the car to come to stop. Then,
v = u + at ;
0 = 35 – 3.06 × t or
t = \(\frac{35}{3.06}\) = 11.43 s.

Question 3.
A car starts from rest and accelerates from rest uniformly for 10 s to a velocity of 36 kmhr^-1. It then runs at a constant velocity and is finally brought to rest in 50 m with uniform retardation. If the total distance covered by the car is 500 m, find

1. acceleration and
2. retardation.

Solution:
1. Initial velocity of the car u = 0;
Final velocity v = 36 kmhr^-1
= \(\frac{36 \times 1000}{3600}\) ms^-1
= 10 ms^-1
time taken t=10 s
∴ acceleration a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) = \(\frac{\mathrm{10}-\mathrm{0}}{\mathrm{10}}\) 1ms^-2

2. To find the retardation
During the motion with retardation, Initial velocity of the car u = 10 ms^-1
Final velocity v = 0, distance travelled, s = 50 m
Using the relation v² = u² + 2as,
0 = 10² – 2a × 50
a = \(-\frac{100}{2 \times 50}\) = -1 ms^-2

Question 4.
A body travelling with a uniform acceleration travels a distance of 100m in the first 5 second and 200m in the next 5 second, calculate the initial velocity and acceleration of the body.
Solution:
If s₁ and s₂ are the distances travelled by a body in two successive intervals of t seconds, then the acceleration is given
by, a = \(\frac{s_{2}-s_{1}}{t^{2}}=\frac{200-100}{5^{2}}=\frac{100}{25}\) = 4ms^-2
we have, s = ut + 1/2 at²
Here, s = s₁ = 100m, a = 4ms^-2, t = 5 s
100 = u × 5 + 1/2 × 4 × 25 OR
5u = 50 ∴u= 10ms^-1.

Question 5.
A body moving with uniform acceleration along a straight line covers 60m in the 5^th second and 80m in the 8^th second of its motion. Calculate the initial velocity and uniform acceleration.
Solution:
The distance travelled during the n^th second is given by,
s_n = u + a/2(2n – 1), n = 5 and s_n = 60
∴ 60 = u+ a/2 (2 × 5 – 1)
60 = u + 9a/2 ………… (1)
Similarly, for n = 8, s_n =80
∴ 80 = u+ a/2 (2 × 8 – 1)
80 = u + 15a/2 …………….. (2)
(2) – (1), gives
20 = 6a/2 = 3a
∴ a = 20/3 ms^-2
Using this in equation (1), we get
60 = u + \(\frac{9 \times 20}{2 \times 3}\)
u = 60 – 30 = 30 ms^-1.

Question 6.
A stone is thrown vertically upwards with an initial velocity of 19.6 m^-1. After how long will the stone strike the ground? Take g = 9.8 ms^-2.
Solution:
Initial velocity u = +19.6 ms^-1. The stone after reaching the highest point comes back to the initial point
∴ The displacement = 0.
a = g = – 9.8 ms^-2
Let t be the time taken to reach the ground
From s = ut + 1/2at², we have
0=19.6t – 1/2 × 9.8 × t²
4.9t² – 19.6t = 0
Dividing throughout by 4.9t, we have t = 4s. Hence, the stone reaches the ground after 4s.

Question 7.
A stone is thrown vertically upwards with a velocity of 10ms^-1 from the top of a tower 40m tall and it finally falls to the ground,

1. Find the time taken by the stone to reach the ground
2. After how long will it pass through the point of projection
3. Calculate the velocity when it strikes the ground. Take g = 10 ms^-2

Solution:
Let the stone be thrown upward with a velocity u = 10ms^-1 from the tower AB of height 40 m. Let C be the highest point reached.

1. Let t be the total time taken by the stone to move from B to C and back to the ground.
∴ The displacement = BA = – 40 m (directed downward since displacement is measured from the initial to the final position)
a = -10 ms^-2.
From s = ut + 1/2at², we have
– 40= 10 × t – \(\frac{10 t^{2}}{2}\)
– 40 = 10t – 5t²
– 8 = 2t -t²
t² – 2t – 8 = 0
(t – 4) (t + 2) = 0.
OR
t = + 4s or – 2s
As time cannot be negative total time taken to reach the ground = 4s.

2. For the motion from B to C and back to B,
displacement = 0
u = 10ms^-1, a = g = -10ms^-2
Let t₁ be the time taken.
From s = ut + 1/2at², we have
0= 10t₁ – \(\frac{10 t_{1}^{2}}{2}\) or 10₁² – 5t₁² = 0
∴ t₁ = 2s.
Hence, the stone reaches the point of projection after 2 seconds.

3. Let ‘v’ be the velocity of the stone when it strikes the ground. For the motion from B to C and back to A, we have,
u = 10 ms^-1, a = g = -10 ms^-2, t = 4s
From, v = u + at, we have
v = 10 – 10 × 4 = – 30 ms^-1.
The -ve sign indicates that it is directed downward.

Question 8.
A stone is projected vertically upwards from the ground with a velocity of 49 ms^-1. At the same time, another stone is dropped from a height 98m to fall freely along the same path as the first. Find where and when the two stones meet each other.
Solution:

Let P be the stone dropped from B at a height 98m from the ground A. Q is the other stone thrown up vertically upwards with a velocity 49ms^-1 at the same instant. Let the two stones meet at C after a time t seconds.
For the stone P, we have
u = 0, a = + g, s = + h (=BC), t = t
From s = ut + 1/2 at², we have,
+h = 0 × t + 1/2 gt² or h = 1/2 gt² ………. (1)
For the stone Q,
u = 49ms^-1,
a = -g,
s = +AC = +(98 – h),
(98 – h) = 49t – 1/2 gt²
Using eqn. (1) in (2), we have,
(98 – h) = 49t – h or 49t =98
∴ t = 2s.
Using this in (1), we have
h = BC = 1/2 × 9.8 × 4 = 19.6 m
∴ AC =96 – 19.6 = 76.4 m
∴ Two stones meet at a height of 76.4 m from the ground, after 2 seconds.

Question 9.
A train is moving southwards with a speed of 30ms^-1. A man is running on the roof of the train with speed of 5ms^-1 with respect to the train. Find the velocity of the man as observed by the person on the ground if the man is running

1. southwards
2. northwards.

Solution:
If v_A and v_B are the velocity of the two bodies moving along the same direction with respect to the ground, relative velocity of A with respect to B is given by,
V_AB = V_A – V_B …………. (1)
Here, the velocity of the man with respect to the train,
v_AB = 5ms^-1.
velocity of the train with respect to the ground is,
v_B = 30 ms^-1
∴ velocity of the man with respect to the ground,
v_A = v_B + v_AB = 30 + 5 = 35 ms^-1 if the man moves southwards.
On the other hand, if the man moves northwards,
v_AB = – 5 ms^-1
∴ V_A = V_B – V_AB
= 25 ms^-1.

Question 10.
A police van moving on a highway with a speed of 10ms^-1 fires a bullet at a smuggler’s car speeding away in the same direction with a speed of 30ms^-1. If the muzzle velocity of the bullet is 140ms^-1, with what speed the bullet will hit the smuggler’s car?
Solution:
Speed of the police car v_p = 10 ms^-1.
Muzzle velocity of the bullet v_M=140 ms^-1
Net velocity of the bullet,
v_A = v_p + v_M = 10 + 140 = 150 ms^-1
(∵ gun is mounted on the van)
speed of the smuggler’s car v_B = 30 ms^-1
∴ Velocity of the bullet relative to smuggler’s car,
v_AB = V_A – V_B
= 150 – 30 = 120 ms^-1.

Question 11.
Two trains A and B are moving on two parallel tracks with a uniform speed of 20ms^-1 in the same direction with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms^-2. After 50 seconds, they are moving together, find the original distance of separation.
Solution:
Initially, both A and B are moving with the same velocity in the same direction,
∴ The initial relative velocity .
u_AB = u_A – u_B = 0
when t = 50 s,
relative acceleration = a_AB = 1 ms^-2
If s is the distance moved, then
From s = ut + 1/2at²
S = u_AB × t + 1/2 a_ABt²
S = 0 × t + 1/2 × 1 × 50²
= 2500/2 = 1250 m.
Hence, the Initial distance of separation is 1250m.

Question 12.
A body is starting from rest and is subjected to a uniform acceleration of 10ms^-2 determine.

1. Velocity of the body at the end of 5s
2. Displacement of the body In the first 3s
3. velocity of the body after a displacement of 20m.
4. Displacement of the body In the 4^th second of its motion (Tumkur 05)

Solutions:
1. Initial velocity, u = 0
acceleration, a = 10ms^-2
time, t = 5s,
v=?
From, the Equation, v = u + at we have,
v= 0 + 10 × 5
v= 50 ms^-1

2. Initial velocity, u = 0
acceleration a = 10 ms^-2
time, t = 36s
Distance travelled, s =?
From the equation s = ut+ 1/2 at² we have
= 0 + 1/2 × 10 × 9
s = 45 m

3. Initial velocity u= 0,
acceleration a = 10ms^-2
Displacement s = 20m
velocity v=?
From the equation v²= u² + 2as ,
= 0² + 2 × 10 × 20
= 400
v = \(\sqrt{400}\) = 20ms^-1

4. Initial velocity u=0
Acceleration a = 10ms^-1
n = 4 sec
Displacement S₄ = ?
From the Equation, S_n = u+ a/2 (2n-1)
S₄=0 + 10/2 (2 × 4 – 1) = 0 + 5(7)
S₄ = 35m.

Question 13.
A stone dropped from the top of a building travels 24.5 m in the last second of its fall. Find the height of the tower, (g = 9.8 m/s2)
Answer:
Distance travelled in the last second of its fall is S_n = 24.5m
Also, S_n = ut + a (n – 1/2)
But u = 0, a = g
∴ S_n = 0 + g (n – 1/2)

24.5 = 9.8n – 4.9
n =\(\frac{24.5+4.9}{9.8}\) = 3 seconds.
we know that S = ut + 1/2 at²
∴ Height of the tower;
h = 0 × 3+ 1/2 × 9.8(3)²
(∵ s = h, u = 0, a = g = 9.8m/s)
∴ h = 0 + 4.9(3)²
= 44.1m.

Question 14.
A body moving along a straight line with uniform acceleration covers 23m and 35m respectively In the 5^th and 8^th second of its motion calculate the distance travelled by the body in

1. 10^th second
2. complete 10 second and what is its velocity at the end of 6 seconds.

Answer:
S₅ = 23m, S₈ = 35m
W.k.that s_n = u + a(n – 1/2)
i.e., S₅ = u + a(5 – 1/2) or 23 = u + 4.5 a….(1)
S₈ = u + a(8 – 1/2) or 35 = u + a(7.5)……….(2)
Eq(2) – (1) gives 3a = 12
or a = 12/3 = 4 m/s²
From (1), 23 = u + 4.5 × 4
u = 5 m/s
1. Distance traveled in 10^th second
S₁₀=u + a(n – 1/2) = 5 + 4(10 – 1/2) = 43m

2. Distance traveled in 10s is
S= ut + 1/2at² = 5 × 10 + 1/2 × 4 × 10² = 250m
Velocity at the end of 6 seconds is v = u + at = 5 + 4 × 6 =29 m/s.

Question 15.
A point object is thrown vertically upwards at such a speed that it returns the thrown after 6 seconds. With what speed was it thrown up and how high did it rise? Plot speed-time graph for the object and use it to find the distance travelled by it in the last second of its journey.
Answer:
Time of rise t₁ = time of fall = t₂
t = 6 s
v = u – g t (upward direction is chosen as positive)
v = – u
– u=u-gt
⇒ – 2u = – 9.8 × 6
⇒ u = 29.4 ms^-1
v² = u² – 2 gh
At the top, v = 0
⇒ h = \(\frac{u^{2}}{2 g}\)
= \(\frac{(29.4)^{2}}{2 \times 9.8}\)
h = 44.1 m
Initial speed = 29.4 ms^-1
Height the object attained = 44.1 m

Distance covered in last second = area under the speed-time graph.
= 1/2 × (6 – 5) × (19.6 + 29.4)
= 24.5 m

Question 16.
A race car is moving on a straight road with a speed of 180 km h^-1. If the driver stops the car in 25 s by applying the brakes, calculate the distance covered by the car during the time brakes are applied. Assume acceleration of the car is uniform throughout the retarding motion.
Answer:
Initial speed μ = 180 km h^-1
μ = 50 ms^-1
t = 25 s
v = u – at
v = 0
∴ 0 = 50 – a × 25
⇒ a = 2 ms^-2
v² = u² – 2as
⇒ s = μ²/2 a
\(\frac{50^{2}}{2 \times 2}\)
∴ Stopping distance = 625 m

Question 17.
A body covers 12 m in the 2^nd second and 20 m in the 4^th second. Find what distance the body will cover in the 4 seconds after the 5^th second.
Answer:
s_n= u + a/2 (2n -1)
s₂ = 12
⇒ 12 = u+ a/2 (2 × 2 – 1)
⇒ 12 = u + 3 a/2 ………. (1)
s₄ = 20
⇒ 20 = u+ a/2 (2 × 4 – 1)
⇒ 20 = u + 7a/2 ………. (2)
Subtract equation (1) from (2)
s = 4a/2
⇒ a = 4 ms^-2
12 = u + 3/2 × 4
⇒ u = 6 ms^-1
Distance covered in the 4 seconds after 5^th second = S₉ – S₅
= u (9) + 1/2 a (9)² – [u(5) + 1/2 a (5)² ]
= 4u + a/2 (81 – 25)
= 4 × 6 + 4/2 × 56 = 136 m.

Question 18.
A ball is thrown vertically upwards with a velocity of 20 ms^-1 from the top of a building of height 25 m from the ground.

1. How high will the ball reach?
2. How long will the ball tale to reach the ground ? (g = 10 ms^-2)

Answer:
u = 20 ms^-1
h = 25 m
g = 10ms^-2
1. v² = u² – 2gh₁
At the top, v = 0
⇒ h₁ = \(\frac{\mathrm{u}^{2}}{2 \mathrm{g}}\)
= \(\frac{20^{2}}{2 \times 10}\)
= 20 m
∴ Height the ball reaches = 20 + 25 = 45 m from the ground.
2. – h = u t – 1/2 g t²
– 25 = 20 t – 1/2 × 10 t²
⇒ 5 t² – 20 t – 25 = 0
⇒ t² – 4 t – 5 = 0
⇒ (t – 5 ) (t + 1 ) = 0
= t = 5, -1
∴ t = 5 s (Since t cannot be negative) Total time to reach the ground = 5 s

1st PUC Physics Motion in a Straight Line Hard Questions and Answers

Question 1.
The speed of a train increases at a constant rate from zero to v and then remains constant for an interval, and finally decreases to zero at a constant rate p. If L be the total distance covered prove that the total time taken is \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\infty}+\frac{1}{\beta}\right)\)
Answer:
During time t₁, the train accelerates uniformly from 0 to V.
v = α t₁
Distance covered d₁ = 1/2 α t₁²
= 1/2 (∝ t₁) t₁
d₁ = 1/2 v t₁
⇒ \(t_{1}=\frac{2 d_{1}}{v} \alpha\)
During time t₂, the train travels at a uniform speed of v
Distance covered d₂ = vt₂
⇒ t₂ = \(\frac{\mathrm{d}_{2}}{\mathrm{v}}\)
During time t₃, the train decelerates uniformly from v to 0.
0 = v – β t₃
⇒ v = β t₃
Distance covered d₃ = vt₃ – 1/2 β t₃²
= vt₃ – 1/2 (β t₃) t₃
d₃ = vt₃ – 1/2 v t₃
⇒ d₃ = 1/2 v t₃
⇒ t₃ = \(\frac{2 \mathrm{d}_{3}}{\mathrm{v}}\)
Total time T = t₁ + t₂ + t₃

Using v² = u² + 2as
d₁ = \(\frac{v^{2}}{2 \alpha}\) and d₃ = \(\frac{v^{2}}{2 \beta}\)
∴ T = \(\frac{L}{v}+\frac{v}{2 \alpha}+\frac{v}{2 \beta}\)
T = \(\frac{L}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
Hence proved.

Question 2.
In a car race, car A takes ‘t’ seconds less than car B and passes the finish line with a velocity V’ more than that of car B. Of the cars start from rest and travel with constant acceleration a₁ and a₂ respectively, show that
V = \(t \sqrt{a_{1} a_{2}}\).
Answer:
Let car A take t_A seconds to finish the race.
Car B takes t_B seconds to finish the race. Let the final velocities of the cars be v_A and v_B
By the given conditions,
t_B – t_A = t
v_A = a₁ t_A
v_B = a₂ t_B
v_A – v_B = v
Since the distance d covered is the same,

Now, v_A = a₁ t_A
⇒ v_B + v = a₁ (t_B – t)
⇒ a₂ t_B + v = a₁ (t_B – t)
⇒ v = (a₁ – a₂) t_B – a₁ t
Substitute for t from (1),

Substitute for t_B from (2),

Hence proved.

Question 3.
An object moves with a deceleration \(\alpha \sqrt{\mathbf{v}}\) in a straight line (a is a constant) At time t = 0, the velocity is v₀. What is the distance it traverses before coming to rest? What will be the total time taken?
Answer:
Deceleration = \(\alpha \sqrt{\mathbf{v}}\)
⇒ Acceleration a = – \(\alpha \sqrt{\mathbf{v}}\)
\(\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}\) = – αv^1/2
⇒ \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}^{1 / 2}}\) = – αdt
Integrating,

2v^1/2 = – αt + c
At t = 0, v = v₀
⇒ 2v₀^1/2 = c
∴ 2v^1/2 = – αt + 2v₀^1/2 …………… (1)
When t = T, v = 0
∴ 2(0)^1/2 = – αT + 2v₀^1/2
⇒ T = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
∴ Total time taken = \(\frac{2 \sqrt{v_{0}}}{\alpha}\)
From (1), we get

You can Download Chapter 14 Oscillations Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 14 Oscillations

1st PUC Physics Oscillations Textbook Questions and Answers

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(b) and (c)

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

1. the rotation of earth about its axis.
2. motion of an oscillating mercury column in a U-tube.
3. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.
4. general vibrations of a poly-atomic molecule about Its equilibrium position.

Answer:

1. Periodic but not SHM (Simple Harmonic Motion)
2. SHM
3. SHM
4. Periodic but not SHM

Question 3.
The figure below depicts four x-t plots for the linear motion of a particle. Which of the piots represent periodic motion? What is the period of motion (in case of periodic motion)?


Answer:
(b) and (d) are in periodic motion with period of 2 sec.

Question 4.
Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion?
Give period for each case of periodic motion (ω is any positive constant):
(a) sin ω t – cos ω t
(b) sin³ ω t
(c) 3 cos( π/4 – 2 ω t)
(d) cos ω t + cos 3 ω t + cos 5 ω t
(e) exp (- ω² t²)
(f) 1 + ω t+ ω² t²
Answer:
(a) sin ω t – cos ω t = \(\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \omega t-\frac{1}{\sqrt{2}} \cos \omega t\right)\)
= \(\sqrt{2} \sin (\omega t-\pi / 4)\) ; SHM , period = \(\frac{2 \pi}{\omega}\)
(b) sin³ ω t = \(\frac{1}{4}\) (3 sin ωt – sin 3ωt) both terms are in SHM hence sin³ ωt is periodic.
period = \(\frac{2 \pi}{\omega}\)
(c) 3 cos (π/4 – 2 ω t) S H M with period = \(\frac{\pi}{\omega}\)
(d) cos ω t + cos 3 ω t + cos 5 ω t Superposition of 3 periodic motion,period = \(\frac{2 \pi}{\omega}\)
(e) exp (- ω² t²) non periodic motion
(f) 1 + ω t+ ω² t² non periodic motion

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration, and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:

Question 6.
Which of the following relationships between the acceleration α and the displacement x of a particle involve simple harmonic motion?

1. α = 0.7x
2. α = -200x²
3. α = -10x
4. α = 100x³

Answer:
α = -10 x (∵ acceleration is proportional and opposite to displacement)

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ω t + Φ). If the initial (t = 0) position of the particle is 1 cm and Its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ω t + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Given: x(t) = A cos(ω t + Φ) ……. (1)
at t = 0, x(0) = 1cm, ω = π s^-1 ,
substituting in (1) we get
1 cm = A cos (0 × π + Φ)
⇒ A cos Φ =1     …… (2)
Differentiate equation (1) w.r.t. t
⇒ \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) – A ω sin (ω t + Φ)    …… (3)
at t = 0, v = ω cm /s, ω = π s^-1
Substituting in 3
⇒ ω = – A ωsin Φ
⇒ 1 = – Asin Φ     ….. (4)
Divide equation (4) by (2)
⇒ 1 = \(\frac{-\sin \phi}{\cos \phi}\)
⇒ tan Φ = -1
⇒ Φ = \(\frac{-\pi}{4}\)
Φ = initial phase.
Substitute Φ value in equation (2) we get
1 = A cos \(\left(\frac{-\pi}{4}\right)\)
⇒ A = \(\sqrt{2} \mathrm{cm}\)
If x (t) = B sin (ω t + α)        ……. (5)
at t = 0,  x = 1 cm, ω = π s^-1
Substituting in (5) we get
1 = B sin(α)                 …… (6)
Differentiate (5) w.r.t. ‘ t’
⇒ v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\)= B ω cos (ω t + α) dt
at t = 0, v = ω = π
⇒ 1 = B cos(α)           …… (7)
Divide(7) by (6)
we get tan Φ = 1
⇒ Φ = π/4 is the initial phase
Substitute Φ in equation (6) we get
1 = B sin (π/4) ⇒ B = \(\sqrt{2} \mathrm{cm}\)

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Maximum length Y_max = 0.2m
Maximum weight M_max = 50 kg
We know that,
F = Mg = ky, at M = M_max, Y = Y_max

⇒ m = 22.36 kg
Weight of the body = 22.36 × 9.8 = 219.1 N

Question 9.
A spring having a spring constant. 1200 N m^-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass Is then pulled sideways to a distance of 2.0 cm and released.

Determine

1. the frequency of oscillations,
2. maximum acceleration of the mass, and
3. the maximum speed of the mass.

Answer:
Given:
K = 1200 Nm^-1, m = 3 kg, a = 2 cm
1.  Frequency of Oscillation
We know that T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}=2 \pi \sqrt{\frac{3}{1200}}\).
⇒ T = 0.3144 s
⇒ f = \(\frac{1}{\mathrm{T}}\) = 3.18/s

2. Maximum acceleration of mass (α_max) maximum acceleration is obtained when y = a
⇒ m α_max = Ka
⇒ α_max = \(\frac{\mathrm{Ka}}{\mathrm{m}}=\frac{1200 \times 0.02}{3}\) = 8 ms^-2

3. Maximum speed of the mass
V_max = a ω = \(a \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}\)
V_max = 0.4 ms^-1

Question 10.
In Exercise 14.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
1. at the mean position,
2. at the maximum stretched position, and
3. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the Initial phase?
Answer:
Given a = 2 cm k = 1200Nm^-1 m = 3 kg
⇒ \(\omega=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=\sqrt{\frac{1200}{3}}=20 \mathrm{s}^{-1}\)
let equation of S H M ⇒ x = A sin ω t
1. time is measured from mean position.
x = A sin ω t
⇒ x = 2 sin 20 t

2.  at maximum structured position phase angle is π/2
⇒ x = a sin (ω t + π/2)
⇒ x = 2 cos (20t)

3. At maximum compressed position phase angle is 3 π/2
x= a sin (ω t + 3 π/2)
⇒ x = – a cos ω t

Question 11.
The figure below correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anticlockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:

(a) Let A be any point on the circle shown in fig (a). Draw AM ⊥ to x-axis. Point M refers to x – projection of the radius vector.
Now, ∠POA = θ = ωt, T = 2s OA = 3cm ⇒ ∠OAM =θ=ωt (∵ Alternate angles)

⇒ x = 3 sin \(\frac{2 \pi}{\mathrm{T}}\)t(cm)
⇒ x = – 3 sin π t (cm)

(b) let A be any point on the circles of fig. (b). From A draw AM ⊥ to x-axis
Now ∠MOA= θ = ωt

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the Initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anti-clockwise In every case: (x Is In cm and t is in s).

1. x = – 2 sin (3t + π/3)
2. x = cos(π/6 – t)
3. x = 3 sin (2 π t + π/4)
4. x = 2 cos π t

Answer:
1.


2.  x = cos(π/6 – t)
radius r = 1 cm
at t = 0, x = cos (π/6) = \(\frac{\sqrt{3}}{2} \mathrm{cm}\),
Φ = – π/6 = – 30°
ω = 1 rad / s

3.

4. x = 2 cos π t
r = 2 cm
ω = π rad / s
at t = 0 , x = 2 cm

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

1. What is the maximum extension of the spring in the two cases?
2. If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer:
1. Consider figure (a)
let y be the extension produced in the spring F = ky
consider fig (b) each mass acts as if it is fixed w.r.t the other
⇒ F = ky  ⇒ y = F/k

2. consider fig (a)
F = – ky
ma = – ky   a = \(\frac{-\mathrm{k}}{\mathrm{m}} \mathrm{y}\)
⇒ ω² = \(\frac{\mathrm{k}}{\mathrm{m}}\)
Therefore, \(\mathrm{T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)
Consider fig (b)
let us assume (1) as centre of system and 2 springs each of length 1/2 attached to two masses. So k’ is the spring factor of each spring.
k’ = 2k

⇒ \(\mathrm{T}=\frac{2 \pi}{\omega}\)
\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}\) is the period of oscillation in the case of (b).

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/ min, what is its maximum speed?
Answer:
Stroke = 1 m
Amplitude = \(\frac{\text { stroke }}{2}\) = 1/2 m
ω = 200 rad / min
V_max = ωA
= 200 × 1/2
= 100 m / min
V_max = 1.67 m/s

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s^-2)
Answer:
g_m =1.7 ms^-2  g_e = 9.8 ms^-2  T_e = 3.5 s

Question 16.
Answer the following questions :

1. Time period of a particle in SHM depends on the force constant k and mass m of the particle: T =\(2 \pi \sqrt{\frac{m}{k}}\) . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
2. The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than \(2 \pi \sqrt{\frac{1}{g}}\) . Think of a qualitative argument to appreciate this result.
3. A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
4. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Answer:

1. In case of a simple pendulum, k is directly proportional to m. Hence ratio of m / k is a constant. Hence time period doesn’t depend on mass.

2. Restoring force that brings body of pendulum back to its mean position F = – mg sin θ
Form small θ, sin θ ≈ θ = \(\frac{\mathrm{y}}{1}\)

If approximation for sin θ = θ is not taken into account then Time period T > \(2 \pi \sqrt{\frac{1}{\mathrm{g}}}\) This happens when θ is not small.

3. Wristwatch work on the principle of spring action. Hence acceleration due to gravity plays no role in the functioning of the wristwatch hence it gives correct time during free fall.

4. During free-fall acceleration, due to gravity is zero hence the pendulum will not vibrate (i.e.: frequency of oscillation is zero).

Question 17.
A simple pendulum of length I and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations In a radial direction about its equilibrium position, what will be its time period?
Answer:
Body of the pendulum is under the action of two accelerations, acceleration due to gravity ‘g’ and centripetal acceleration \(\alpha=\frac{v^{2}}{R}\)
effective acceleration = \(\alpha^{\prime}=\sqrt{\alpha^{2}+g^{2}}\)

Question 18.
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density \(\rho_{\mathrm{t}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{1} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Initially in equilibrium,
weight of cork = weight of water displaced

When cork is pushed in, then the restoring force acting on it is :
f = – weight of the portion dipped after pushing

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic, motion.
Answer:
Restoring force acting on the liquid when suction pump is removed is,
f = – mg  ⇒ f= -(A × 2y)ρ × g
where, A = Cross – section area of U tube
ρ = density of liquid
⇒ f = – 2Aρgy
acceleration produced in liquid column
a = f/mass of liquid

hence it is a SHM.

Question 20.
An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal.

Answer:
Volume = V
mass of ball = m
cross-sectional Area = A
initial pressure on either side of the ball = atmospheric pressure = P
Let the charge in volume of air when ball is pressed be ∆V
∆V = Ay (y = displacement)
Now, Bulk modulus of elasticity


The above equation is of the form,

hence it is in SHM.

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of Its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of

1. the spring constant k and
2. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Answer:
1. Total mass = 3000 kg
mass supported by each wheel = 750 kg
y = 0.15 m
We know that,
mg = kg

2.

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the
same period.
Answer:
Let the particle executing SHM starts from mean position.
Displacement can be given by
x = Asin ω t
Velocity v = A ω cosωt
Kinetic energy over one complete cycle

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillators is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J= – α θ, where j is the restoring couple and θ the angle of twist).
Answer:
We know that,

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is

1. 5 cm
2. 3 cm
3. 0 cm.

Answer:
Given,
r = 5 cm
T = 0.2 s
\(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{0.2}=10 \pi \mathrm{rad} / \mathrm{s}\)
acceleration A = – ω²y
velocity V = \(\omega \sqrt{r^{2}-y^{2}}\)
1. y = 5 cm
A = – (10π)² × 0.05
⇒ A = 0.493 m / s²
V = \(10 \pi \sqrt{(0.05)^{2}-(0.05)^{2}}=0\)

2. y = 3 cm
A = (10π)² × 0.03
⇒ A = – 0.296 m / s²
V= \(10 \pi \sqrt{(0.05)^{2}-(0.03)^{2}}\)
V = 0.4 πm/s

3. y = 0 cm
A = 0
V= \(10 \pi \sqrt{(0.05)^{2}-0}\)
= 0.5 πm/s

Question 25.
A mass attached to a spring is freeto oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x₀ and v₀.
[Hint : Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]
Answer:
Let x = a cos (ωt + θ)
∴ v = \(\frac{d x}{d t}\) = – a ω sin (ωt + θ)
When t = 0, x = x₀ and \(\frac{d x}{d t}\) = – v₀
⇒ x_o = a cos θ ….. (1) and
– v₀ = – a ω sin θ
⇒ a sin θ = \(\frac{\mathrm{v}_{0}}{\omega}\) …..(2)
Squarring and adding (1) and (2),
a² = \(=x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}} \Rightarrow a=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}\)

1st PUC Physics Oscillations One Mark Questions and Answers

Question 1.
Mention the relation between period and frequency of periodic motion.
Answer:
Frequency f = \(\frac{1}{T}\)

Question 2.
A particle executes simple harmonic motion. At what point on its path is the acceleration maximum?
Answer:
Acceleration is maximum at a point of maximum displacement from the mean position.

Question 3.
At what position the KE of an oscillating simple pendulum Is maximum?
Answer:
Kinetic energy is maximum at the mean position.

Question 4.
What is the condition for motion of a particle to be SHM?
Answer:
Acceleration should be proportional to displacement and always directed towards mean position.

Question 5.
What is Oscillation?
Answer:
To and for motion in the same path is called Oscillation.

Question 6.
How will the period of a simple pendulum change when its length is doubled?
Answer:
T_new = \(\sqrt{2}\) T_old

Question 7.
Will a pendulum’s time period increases or decreases when taken to the top of the mountain?
Answer:
Increases, g decreases as high altitude and \(\mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{g}}}\)

Question 8.
Two simple pendulum of equal length cross each other at mean position. What is their phase difference?
Answer:
180° (π radians)

Question 9.
How many times in one vibration, KE and PE become maximum?
Answer:
Two.

Question 10.
When is the tension maximum in the spring of a simple pendulum?
Answer:
Mean position.

Question 11.
A spring of spring constant k Is cut Into two equal parts. What is the spring constant of each part?
Answer:
2 K

Question 12.
What Is the phase difference between the displacement and velocity in a SHM?
Answer:
90° (π/2 radians)

Question 13.
State force law for a SHM.
Answer:
Force, F = mω²x
m = mass
ω = angular speed
x = displacement

Question 14.
A pendulum is making one oscilla¬tion in every two seconds. What is the frequency of oscillation?
Answer:
f = \(\frac{1}{T}=\frac{1}{2} s^{-1}\)

Question 15.
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
Frequency is 0 because the acceleration zero during free fall.

Question 16.
A simple pendulum Is inside a spacecraft. What should be its time period vibration?
Answer:
Pendulum does not oscillate.

Question 17.
What are isochronous vibrations?
Answer:
When the time period is independent of amplitude such an oscillation is called isochronous.

1st PUC Physics Oscillations Two Marks Questions and Answers

Question 1.
Define simple harmonic motion (SHM). Give an example.
Answer:
A particle is said to have SHM if the acceleration of the particle is directly proportional to its displacement from the mean position and directed towards the mean position.
E.g.:

1. Vertical oscillations of a loaded spring.
2. Oscillations of bob of a simple pendulum.
3. Vibrations of string of musical instruments.
4. Motion of air particles during the propagation of sound waves.
5. Vibration of a tuning fork.

Question 2.
Mention expression for velocity and acceleration of a particle executing SHM.
Answer:
Velocity of a particle executing SHM is v = \(\omega \sqrt{A^{2}-y^{2}}\)
Acceleration of a particle executing SHM is a = – ω²y
y is the displacement of a particle from its mean position in t seconds. A is the amplitude and ω is the angular frequency.

Question 3.
Mention expression for K.E, P.E and total energy of a particle executing SHM.
Answer:
Potential energy of particle at any instant t second is E_p = 1/2 m ω² y².
Kinetic energy of particle at any instant t second is, E_K = 1/2 m ω² (A² – y²).
Total energy of particle at any instant is constant E = E_K + E_p = 1/2 m ω² A².
A is the amplitude and w is the angular frequency and m is mass of the particle y is the displacement of a particle from its mean position in t seconds.

Question 4.
The acceleration of a particle executing SHM Is 20ms^-1 at a distance of 5 m from the mean position. Calculate its time period and frequency.
Answer:
a = 20m/s², y = 5m.
We know that a = | ω² y |
⇒ \(\omega=\sqrt{\frac{a}{y}}=\sqrt{\frac{20}{5}}\)= 2 rad/s or 2πf = 2
Frequency f = \(\frac{2}{2 \times 3.14}\) = 0.3185 Hz
Period T = \(\frac{1}{f}=\frac{1}{0.3185}\) = 3.14s

Question 5.
Differentiate between forced oscillations and resonance.
Answer:
1. Forced Oscillations:
A body oscillates with the help of external periodic force with a frequency different from natural frequency of body.

2. Resonance:
A body oscillating with its natural frequency with the help of external periodic force whose frequency is equal to natural frequency of body.

Question 6.
Two linear, simple harmonic motion of equal amplitudes and frequencies ω and 2ω are impressed on a particle along the axis of X and Y respectively. If the initial phase difference is π/2, then find the resultant path followed by particle
Answer:
Let
X = Asinωt …… (1)
Y= Acos2ωt …….(2)
(1) and (2) represent SHM with equal amplitude and phase difference π/2 with frequencies ω and 2ω.

Question 7.
Derive an expression for K.E and P.E of a particle SHM.
Answer:
We know that, Potential energy
U = 1/2 kx²
For a particle in SHM, k = mω² and x = Asin ωt
⇒ U = 1/2 mω² A² sin² ωt
Kinetic energy, k= 1/2 mv²
now, v = ωy = ωA cos ωt
⇒ K= 1/2 mω² A²cos²ωt

Question 8.
The amplitude of a oscillating simple pendulum Is doubled. What will be its effect on

1. periodic time
2. total energy
3. maximum velocity

Answer:

1. Periodic time does not change. T is independent of amplitude.
2. Total energy, T.E = 1/2 mω² A²
   A is doubled ⇒ T.E_new = 4 TE_old
3. Maximum velocity, V_max = ωA
   A is doubled ⇒ V_max is doubled.

Question 9.
The frequency of oscillations of a mass m suspended by a spring is V₁. If the length of spring is cut to one half, the same mass oscillates with frequency V₂. Calculate V₂/V₁.
Answer:
V₁ = \(\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\)
where K is the spring constant and m is the mass
V₂ = \(\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}\)
∴ \(\frac{V_{2}}{V_{1}}=\sqrt{2}\)

Question 10.
Discuss some important characteristics of wave motion.
Answer:

1. Wave transports energy.
2. particles are not transported.
3. Elasticity and inertia determine the motion of particle distribution.

Question 11.
Differentiate between free oscillations and forced oscillations with the help of examples.
Answer:
Consider a pendulum in free space (without air) it oscillates freely. This is free oscillation. Consider another pendulum in a viscous liquid it oscillates only if there is external force this is forced oscillation.

Question 12.
List any two characteristics of SHM.
Answer:

1. SHM is always directed towards mean position.
2. Acceleration is directly proportioned to displacement but opposite in direction.

1st PUC Physics Oscillations Three Marks Questions and Answers

Question 1.
Define the terms

1. amplitude
2. period
3. frequency
4. phase related with a particle executes SHM.

Answer:
1. Amplitude:
Maximum displacement of the particle from the mean position is called amplitude.

2. Period :
Time taken by the particle to complete one oscillation is called time period. (T)

3. Frequency :
The number of oscillations completed by the particle in one second is called frequency (f)

4. Phase:
Phase of a particle is defined as the fraction of the time period that has elapsed since the particle last passed through its mean position in the positive direction.

Question 2.
For an oscillating pendulum, establish the relation \(\frac{d^{2} \theta}{d t^{2}}=-\omega^{2} \theta\) where \(\omega=\sqrt{\frac{g}{1}} \theta\) = small angular displacement.
Answer:

restoring force F_r = – mg sin θ
torque on pendulum \(\tau=\mathrm{I} \alpha=\mathrm{ml}^{2} \alpha\)
restoring torque = torque on pendulum

We know that = \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=\alpha\) and \(\omega=\sqrt{\frac{g}{1}}\)
⇒ \(\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-\omega^{2} \theta\)

Question 3.
A body of mass 1 kg is suspended from a weightless spring having force constant 600 Nm^-1. Another body of mass 0.5 kg moves vertically upwards lift the suspended body with a velocity of 3 ms^-1 and tets embedded in it. Find the frequency of oscillation and amplitude of motion.
Answer:
Total mass = (1 + 0.5)kg = 1.
k = 600 nm^-1
frequency of oscillations

Let V₁ be velocity of mass after collision
⇒ m₁ V₂ = (m₁ + m_2,) V₁
⇒ 0.5 × 3 = 1.5 V₁
⇒ V₁ = 1 m/s
According to law of conservation of energy
(PE)_max = (KE)_max
\(\frac{1}{2} m v_{1}^{2}=1/2^{K A^{2}}\)
\(\frac{1}{2} 1.5 \times 1^{2}=\frac{1}{2} 600 \mathrm{A}^{2}\)
⇒ A = 5 cm

Question 4.
The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.
Answer:

Let the rikshaw of mass M be at any point P in the dip of radius R. Let O be the centre of this circular path. This case is similar to that of a simple pendulum and assume that e is small.
restoring force
F = -mg sinθ
F = – mg θ
displacement of rickshaw = R θ

Question 5.
Find an expression for body vibrating in SHM.
Answer:
Let the SHM equation be
x = Asin ω t

Potential energy is given by
= 1/2 kx² = 1/2 m ω ² k²
= 1/2 mω² A² sin² ω t
Kinetic energy = 1/2 mv²
= 1/2 m ω ² A² cos² ω t
Total energy    = PE + KE
= 1/2 mω² A² sin² ω t + 1/2 mω² A² cos² ω t
= 1/2 mω ² A²

Question 6.
Show that when a particle is moving in SHM its velocity at a distance \(\frac{\sqrt{3}}{2}\) its amplitude from the central position is half its velocity in central position.
Answer:
For a particle in SHM
\(\mathrm{V}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}\)

Question 7.
A body oscillates with SHM according to the equation.
x(t) = 5cos (2π t + π /4) Where x is in meters and t is in seconds calculate,

1. displacement att = 0
2. Angular frequency
3. V_max

Answer:
1. displacement at t = 0
x(0) = 5cos(2π 0 + π/4)
⇒ x = \(\frac{5}{\sqrt{2}} m\)

2. Angular frequency
ω = 2π radian/s

3. Vmax = aω = 2π × 5 = 10π m/s

Question 8.
What is spring factor? Find its value in case of two springs connected in

1. series
2. Parallel

Answer:
Spring factor (k):
Force acting for unit extension produced is called spring factor.
1. When two springs connected in series,

force experienced by both springs is same.
Total extension (x) is sum of individual extension.
⇒ x = x₁ + x₂
\(\frac{F}{-K_{e q}}=\frac{F}{-K_{1}}+\frac{F}{-K_{2}}\) ⇒ 1/K_eq = 1/K₁ + 1/K₂
K_eq = \(\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{1}+\mathrm{K}_{2}}\)

2. When two springs are in parallel,

Net force experienced is the sum of force experienced in both springs. But the extension will be the same.
F_eq = F₁ + F₂
K_eqx =K₁ X + K₂X
⇒ K_eq = K₁ + K₂

Question 9.
Explain the relation in phase between displacement, velocity, and acceleration in SHM, graphically as well as theoretically.
Answer:
Let displacement x = Asinωt
Velocity v= ω Acosω t
acceleration a = – ω² Asin ω t = ω² Asin(ω t + π )
Displacement and velocity have phase difference of π/2 radians.
v and α has \(\frac{\pi}{2}\) radians of phase difference. α ans x has π radians of phase difference.
Assume ω >1

Question 10.
For a particle in SHM, the displacement x of the particle at a function of time t is given as x = Asin(2πt) …..(1) where x is in centimeters and t is in seconds. Let the time taken by the particle to travel from x=0 to x = A/2 beT₁ and the time taken to travel from x = A/2 to x = A be T₂. Find T₁/T₂
Answer:
ω =2π = \(\frac{2 \pi}{\mathrm{T}}\) ⇒ T= 1 s
at t = 0 , x = 0
now if x = A/2

now in a SHM time taken from 0 to A is \(\frac{T}{4}\)
⇒ time taken for x = A/2 to A is \(\frac{T}{4}\) -T/12

Question 11.
What is SHM? Show that in SHM acceleration is directly proportional to Its displacement at a given instant.
Answer:
Simple Harmonic Motion is a type of motion in which displacement is always directed towards mean position and acceleration is directly proportional to displacement and opposite in direction.
Let a SHM be represented by
x = Asin ωt
⇒ \(\frac{\mathrm{d} x}{\mathrm{dt}}\) = A ω cos ωt

acceleration is proportional to displacement and opposite in direction.

Question 12.
A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretched 7cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?
Answer:
Mass added m = 0.02
Length stretched x= 7 cm = 0.07 m

Time period T = \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)

Question 13.
A particle executes SHM of time period 10 sec. The displacement of particle at any instant is x = 10 sinωt(cm) Find

1. Velocity of body 2 s after it passes through mean position.
2. Acceleration of body 2 s after it passes mean position.

Answer:
Given x = 10 sin ωt T = 10 sec
A = 10 cm   ω = \(\frac{2 \pi}{10}\) rad/s
Let at t = 0 body be at the mean position.
Now at t = 2
1.  V = Aω cos ωt
= 10ω cos(ω2)cm/s
⇒ \( V=10 \times \frac{2 \pi}{10} \cos \left(\frac{2 \pi}{10} 2\right)\)
⇒ V = 1.94 cm/s

2. Acceleration
a = – Aω² sinωt
\(= – 10 \times \frac{4 \pi^{2}}{100} \sin \left(\frac{4 \pi}{10}\right)\)
⇒ a = – 3.75 cm/s²

Question 14.
What is a simple pendulum? Show that the motion of the pendulum is SHM and hence deduce an expression for the time period of pendulum. Also, define Second’s pendulum.
Answer:
Simple pendulum is a point mass body suspended by a weightless thread or string from a rigid support about which it is free to oscillate.

Now restoring force acting on body can be given by
F = – mgsinθ (from figure)
If  θ is small
F= – mgθ    (∵ sin θ ≈ θ)
Let P be any point on the path of pendulum and makes ∠OMP = θ which is small and arc OP be x (displacement) then
\(\dot{\theta}=\frac{\mathrm{OP}}{1}=\frac{\mathrm{x}}{1}\)
⇒ F = \(\frac{-m g x}{1}\) ……(1)
⇒ force ∝ displacement and opposite in direction hence an SHM
Now this of form F = -kx ……(2)
⇒ k = mg/l (from (1) and (2)

Question 15.
A body of mass ‘m’ suspended from a spring executes SHM. Calculate the ratio of kinetic energy and potential energy of the body when it is at half the amplitude far from the mean position.
Answer:
KE of a mass suspend from a spring
KE = mω²(a² – y²)
Potential energy of a mass suspend from a spring
PE = 1/2 mω² y²
\(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right) 2}{\mathrm{y}^{2}}\)
at y = a/2
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\left(\mathrm{a}^{2}-\mathrm{a}^{2} / 4\right)^{2}}{\mathrm{a}^{2} / 4}\)
⇒ \(\frac{\mathrm{KE}}{\mathrm{PE}}=3\)

Question 16.
If x = a cos ωt + b sin ωt, show that it represents SHM.
Answer:
We have,
x = a cos ωt + b sin ωt
Now, \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – aω sinωt + bω cosωt
\(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – aω² cos ωt + bω² sin ωt
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω² (a cos ωt+ b sin ωt)
⇒ \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = – ω²x
⇒ α = – ω²x
Hence an SHM

Question 17.
Find the expression for the total energy of a particle executing SHM.
Answer:
For a SHM PE = 1/2 kx²
= 1/2 mω² x² = 1/2 mω A² sin² ωt
KE= 1/2 mv²
v = \(\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}\) = – Aω cos ωt
⇒ KE = 1/2 mA²ω² cos² ωt
Total energy = KE + PE = 1/2 mω² A² cos² ωt + 1/2 mω² sin² ωt
Total energy = 1/2 mω² A²

1st PUC Physics Oscillations Five Marks Questions and Answers

Question 1.

1. Find the total energy of particle executing SHM?
2. Show graphically the variation of PE and KE with time In SHM.
3. What is the frequency of these energies w.r.t the frequency of the particle executing SHM?

Answer:
1. PE at any instant
PE = 1/2 kx²
= 1/2 mω² A² sin² ωt
(∵ k = mω² and x = A sin ωt)
⇒ PE = 1/2 mω² A² sin² ωt
KE at any instant KE = 1/2 mv²
KE = 1/2 mA² ω² cos² ωt
Total energy = KE + PE
⇒ Total energy = 1/2 mω² A²sin² ωt + 1/2 mω² A² cos² ωt
⇒ Total energy = 1/2 mω² A²

2. Graph

3. Frequency = \(\frac{2}{\text { period }}=\frac{2}{\mathrm{T}}\)

1st PUC Physics Oscillations Numerical Problems Questions and Answers

Question 1.
A spring compressed by 0.1m develops a restoring force of 10N. A body of mass 4 kg is placed on it. Deduce the

1. force constant of the spring
2. depression of the spring under the weight of the body
3. period of oscillation, if the body is disturbed.

Answer:
Given
Restoring force, F = 10N
Mass of the body, m = 4kg
displacement, x =0.1m

1. \(\mathrm{k}=\frac{\text { force }}{\text { displacement }}=\frac{10}{0.1}\) = 100 Nm^-1
2. Depression due to weight \(=\frac{\text { force }}{\mathbf{k}}=\frac{4 \times 10}{100}\) = 0.4 m
3. Period of oscillation T = \(2 \pi \sqrt{\frac{m}{k}}\) \(=2 \pi \sqrt{\frac{4}{100}}\) = 0.4 πs
   T = 0.4 πs

Question 2.
A vertical U- tube of uniform crosssection contains water up to a height to 20cm. Calculate the time period of the oscillation of water when it is disturbed.
Answer:
The length of liquid column
L = 2 × 20cm = 40 cm
Time period of oscillation
\(=2 \pi \sqrt{\frac{L}{2 g}}=2 \pi \sqrt{\frac{40}{2 \times 9.80}}\) = 0.9 s

Question 3.
A cylindrical piece of cork of base area ‘A’ and height ‘h’ floats in a liquid of density \(\rho_{\mathrm{e}}\). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period of \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h} \rho}{\rho_{\mathrm{e}} \mathrm{g}}}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let x be the depression created.
Excess upthrust caused is, U_e= g\(\rho_{\mathrm{e}}\) Ax
Restoring force = U = gA\(\rho_{\mathrm{e}}\) x
mass = m = Ahρ
Now F = mα;  α = acceleration
Ahρα = -gA\(\rho_{\mathrm{e}}\) x

Question 4.
If this earth were a homogenous sphere and a straight hole bored in it through its centre. Show that if a body were dropped into the hole it would execute a SHM. Also, find its time period.
Answer:
Let mass of body dropped = m
mass of earth = M
radius of earth = R
mg = \(=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}\) ⇒  g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
⇒  \(\mathrm{g}=\frac{\mathrm{G}}{\mathrm{R}^{2}} \frac{4}{3} \pi \mathrm{R}^{2} \rho=\frac{4 \pi \mathrm{GR} \rho}{3}\)
where ρ is mean density of earth

Let R be the depth of the body fallen from centre then g at the point is
\(\mathrm{g}^{\prime}=\frac{\mathrm{Gm}^{\prime}}{(\mathrm{R}-\mathrm{d})^{2}}\)
where m’ is mass of sphere of radius (R – d)

Hence acceleration is proportional to displacement and in opposite direction go on SHM.

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Karnataka 1st PUC Chemistry Question Bank Chapter 6 Thermodynamics

1st PUC Chemistry Thermodynamics One Mark Questions and Answers

Question 1.
What is thermodynamics ?
Answer:
Thermodynamics is a branch of chemistry that deals with the study of interconversion of with other forms energy during physical and chemical changes.

Question 2.
What you mean by enthalpy?
Answer:
The total heat content of a system is called enthalpy.

Question 3.
Write the relationship between ∆H and ∆E.
Answer:
∆H = ∆E + RT∆n, ∆H = Enthalpy change, ∆E = Internal energy change
∆n = Number of gaseous product number of gaseous reactant or number of moles.

Question 4.
Define (Heat of reaction) Enthalpy of reaction.
Answer:
Heat of a reaction is the change in enthalpy produced when the number of moles of the reactants as represented in the balanced chemical equation have completely reacted with each other.

Question 5.
Heat (enthalpy) of formation.
Answer:
Heat of formation of a compound is the change in enthalpy produced when one mole of the compound is formed from its elements denoted as ∆H_f.

Question 6.
What is Heat Capacity ?
Answer:
Het capacity of a system is defined as quantity of heat required to raise temperature by 1°.

Question 7.
What is specific heat capacity ?
Answer:
It is defined as the quantity of heat required to raise temperature of 1 gram of a substance by 1° C or 1K

Question 8.
What is the relation between C_P and C_v for an ideal gas.
Answer:
C_P– C_v = R

Question 9.
What is entropy ?
Answer:
A measure of degree of disorder of a system.

Question 10.
What is the unit of entropy ?
Answer:
J / K (Joule per Kelvin)

Question 11.
What is Gibbs-Helmholtz equation ?
Answer:
∆G° = ∆H° – T∆S°.

Question 12.
Give an expression for the work done in a reversible isothermal expansion of an ideal gas.
Answer:
w = -2.303nRT log \(\frac{V_{2}}{V_{1}}\)

Question 13.
Explain : Heat of formation a acetylene is + 54 kcal.
Answer:
When 1 mole of gaseous acetalylene is formed from its elements, carbon (s) and hydrogen (g), 54 kcal of heat is absorbed.

Question 14.
White phosphorous is less stable than red phosphorous. Mention whether the process of conversion of white phosphorous to red phosphorous is exothermic or endothermic reaction.
Answer:
Exothermic reaction.

Question 15.
What is the value of the standard enthalpy of formation of an element ?
Answer:
Zero

Question 16.
The standard enthalpies of formation of Hydrogen fluoride and Hydrogen chloride are -268.6 kJ and -92.3 kJ respectively. Which is between the two is stable?
Answer:
Hydrogen fluoride is more stable.

Question 17.
Give an example for a reaction in which AH = AE
Answer:
C(s) + O₂(g) → CO₂(g)

Question 18.
What is the meaning of the statement the enthalpy of formation of PCl₃ is -373 kJ ?
Answer:
When 1 mole of PCl₅ is formed from its elements, P(s) and Cl₂(g), 373 kJ of heat is liberated.

Question 19.
Define ‘standard enthalpy of formation’.
Answer::
Standard enthalpy of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states under standard conditions i.e., at 298 K and 101.3 kPa pressure.

Question 20.
Define enthalpy of combustion.
Answer:
Enthalpy of combustion Of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.

Question 21.
Write the thermochemical equation for the combustion of glucose.
Answer:
C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(l) ΔH = -xkJ

Question 22.
The enthalpy of combustion of graphite and diamond are -393.5 kJ and -395 kJ respectively. Which of them is more stable ?
Answer:
Graphite.

Question 23.
Explain : Heat of combustion of methane is -890 kJ.
Answer:
Heat of combustion of methane is -890 kJ means when one mole of methane (CH4)
i. e., 16 grams of methane is completely burnt in oxygen, 890 kJ of heat is liberated.

Question 24.
Write the thermochemical equation for the reaction involving the burning of hydrogen gas in excess of air forming water and liberating heat.
Answer:
2H₂(g) + O₂(g) → 2H₂O(l); ΔH = -xkJ

Question 25.
The enthalpy of combustion of ethyl alcohol is – 1360 kJ / mol. Write the thermochemical equation of the reaction.
Ans:
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = -1360 kJ / mole.

Question 26.
Define heat of enthalpy of a soultion.
Answer:
Heat of a solution is the change in enthalpy produced when one mole of the solute is dissolved in excess of a solvent so that further dilution does not produce any heat exchange.

Question 27.
Give one difference between isolated system and closed system.
Answer:
Isolated system can neigher exchange matter nor energy with the surroundings. Closed system can exchange energy but not matter with the surroundings.

Question 28.
Which of the following is an intensive property : Surface tension, mass, volume, enthalpy, density ?
Answer:
Surface tension and density are intensive properties.

Question 29.
What happens to the internal energy of the system if:
(a) Work is done on the system ? (b) Work is done by the system ?
Answer:
(a) If work is done on the system, internal energy will increase, (b) If work is done by the system, internal energy will decrease.

Question 30.
For the reaction, N₂(g) + 3H₂(g) → 2NH₃(g) predict whether the work is done on the system or by the system.
Answer:
Volume is decreasing therefore, work is done on the system.

Question 31.
What is the limitation of first law of thermodynamics ?
Answer:
It cannot tell us the direction of the process.

Question 32.
Which of the following is an extensive property ?
(a) Volume, (b) Surface tension, (c) Viscosity, (d) Density
Answer:
(a) Volume is an extensive property.

Question 33.
Write the relation between standard free energy change and equilibrium constant Kp for a reversible reaction.
Answer:
Relationship between standard free energy change ∆G° and the equilibrium constant ∆G° = -2.303 RT log K_p, where R is the gas constant = 8.314 JK^-1. T is the temperature in Kelvin.

1st PUC Chemistry Thermodynamics Two Marks Questions and Answers

Question 1.
What is exothermic reaction ? Give example.
Answer:
The reaction in which heat is evolved is called exothermic reaction.
Example : C_(s) + O_2(g) → CO_2(g) ∆H = -393.5 kJ

Question 2.
What is endothermic reaction ? Give example.
Answer:
The reaction in which heat is absorbed is called endothermic reaction.
Example : N_2(g) + O_2(g) → 2NO_(g) ∆H = +180.8kJ

Question 3.
Mention different types of process.
Answer:

1. Isothermal process
2. Adiabatic process
3. Isoehoric process
4. Isobasic process
5. Reversible process
6. Irriversible process
7. Cyclic process

Question 4.
What is open system ? Give example.
Answer:
An open system is one in which there is an exchange of both matter and energy with its surroundings.
Example : Water in an open beaker.

Question 5.
What is Extensive propery ? Give example.
Answer:
It is a property of which depends on the amount of the substance present in the system.
Example : Mass, Volume, Energy

Question 6.
Write any two difference between Isothermal and Adiabatic process.
Answer:

Question 7.
Mention the factors affecting enthalpy of reaction.
Answer:

1. Physical state of reactant and product.
2. Amount of reactant and product.
3. Temperature of the reaction.,
4. Allotropic ferrous.,
5. Condition of constant volume or constant pressure.,
6. Reaction stoichiometry.

Question 8.
State and illustrate Hess’s law.
Answer:
Hess’s law : Whether a chemical reaction takes place in a single step or in several steps, the total change in enthalpy remains the same.
Consider the formation of CO₂ from C.

Question 9.
Explain the meaning of thermochemical equation for the reaction.
2H₂(g) + O₂(g) → 2H₂O(l) +136 kcal.
Answer:
When two moles i.e., 4 grams of gaseous H₂ reacts with 1 mole (32g) of gaseous O₂ to give 2 moles of liquid water i.e., 36 gram of liquid water, 136 kcal of heat is liberated.

Question 10.
Explain heat of neutralisation with an example.
Answer:
Heat of neutralisation is the change in enthalpy produced when one gram equivalent weight of an acid is neutralised by one gram equivalent weight of a base in dilute solution.
Example : HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O;= ΔH – 57.3 kJ

Question 11.
With example explain the term ‘Heat of transition’.
Answer:
Heat of transition is the charge in enthalpy produced when one mole of one allotropic form of an element is converted into another allotropic form.
Example : C(diamond) > C(graphite); ΔH = -X₁ kJ When 1 mole of diamond is coverted into graphite x kJ of heat is liberated, -x kJ is the heat of transition in this case.

Question 12.
Define heat (enthalpy) of transition.
Answer:
It is enthalpy change when one mole of an element is changed from one equilibrium to another.

Question 13.
What do you understand by the statements :
(i) enthalpy of formation of nitric oxide is +90.7 kJ.
(ii) enthalpy of combustion of carbon disulphide is =1065.2 kJ
Answer:
(i) When 1 mole of NO(g) is formed from its elements N₂(g). and O₂(g), 90.7 kJ of heat is absorbed.
(ii) When 1 mole of CS₂ (l) is completely burnt in O₂,1065.2 kJ of heat is liberated.

Question 14.
Enthalpy of neutralisation of NH₄OH with HC is 51.45 kJ. Calculate the enthalpy of ionization of ammonium hydroxide.
Answer:
Enthalpy of ionzators of NH₄OH = -51-41 – (57.3) = 5.8500 J

Question 15.
Explain: “The enthalpy of formation of nitric oxide is +90.4 kJ”.
Answer:
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g) → ANO_(g)ΔH = +90.4kJ When 1 mole of nitric oxide is formed from its elements 90.4 kJ of heat is absorbed. (Actually ΔH\+90.4kJ Error in the question)

Question 16.
Distinguish between exothermic and endothermic reactions.
Answer:

Question 17.
State and explain the first law of thermochemistry.
Answer:
If enthalpy of formation of a compound is -x kJ, enthalpy of its decomposition intoits elements is + xkJ.
Example : C(s) + O₂(g) → CO₂; ΔH = -xkJ According to the law (Lavoisier & Laplae law) CO₂ (g) → C(s) + O₂ (g); ΔH = +x kJ

Question 18.
Which of the following has highest heat of combustion out of the following and why ?
(a) C₂H₆ (b) C₂H₄ (c) C₂H₂ (d) CH₄
Answer:
C₂H_6(g) will have highest heat of combustion because it has highest molecular weight
and second highest calorific value.
ΔH_c = calorific value of kJ/g × mol. wt. = 52 × 30 = 1560 kJ/mol.

Question 19.
What do you understand by state functions ? Neither q nor w is a state function but q + w is a state function ? Explain.
Answer:
State function is a property whose value depends only uopon the state of system an is independent of the path, q and w are not state functions because they depend upon path. But q + w = AE which is a state function. AE does not depend upon path.

Question 20.
Acetic acid and hydrochloric acid react With KOH solution. The enthalpy of neutralization of acetic acid is -55.8 kJ mol^-1 while that of hydrochloric acid is-57.3 kJ mol^-1. Why?
Answer:
It is because HCl is strong acid, it ionises completely in aqueous solution whereas acetic acid is weak acid which does not ionise completely and some energy is used in its complete ionisation.

Question 21.
Taking a specific example show that ΔS_total is a criterion for spontaneity of a change.
Answer:
ΔG = -TΔS_total
If ΔS_total is +ve, ΔG will be -ve, reaction will be spontaneous.
If ΔS_total is -ve, ΔG will be +ve, reaction will be non-spontaneous,
ΔS_total = ΔS_syst + ΔS_surroundings
For example in case of freezing of water at 273 K, TAStotai is +0.08 J/K/mol whereas at 274 K it is -0.080 J/K/mol. Since the ΔS total is +ve at 273 K, freezing of water will take place. On the other hand at 274 K, the total entropy change is negative, the freezing will not occur.

Question 22.
What is meant by entropy driven reaction ? How can a reaction with positive changes of enthalpy and entropy be made entropy driven ?
Answer:
Those reactions which are endothermic and entropy is increasing such that TΔS > ΔH are entropy driven reactions. If ΔS total is +ve, the reaction will be spontaneous. If ΔS is +ve and is small, but T is large, the reaction will be spontaneous.
A reaction which is endothermic, i.e., ΔH is +ve and ΔS is +ve, T must be large.
ΔG = ΔH – TΔS ; ΔG will be -ve if TΔS > ΔH, i.e., the reaction will be spontaneous at higher temperature because entropy will increase with increase in temperature which will make this reaction spontaneous.

Question 23.
Justify the following statements:
(a) An exothermic reaction is always thermodynamically spontaneous.
(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.
Answer:
(a) It is false. Exothermic process are not always spontaneous. If ΔS = -ve and TΔS > ΔH, the process will be non spontaneous even if it is exothermic.
(b) The entropy of vapour is more than that of liquid therefore entropy increases during vaposisation.

Question 24.
Evaporation of water is an endothermic process but spontaneous. Explain.
Answer:
It is because entropy increases -during this process because water vapour have more entropy than liquid water. ΔG become -ve because TΔS > ΔH.

Question 25.
What is meant by toral entropy change of a process ? Assuming the thermodynamic relationship ΔG = ΔH – TΔS, derive the relationship ΔG = TΔS_total for a system.
Answer:
Total entropy change of a process if entropy change taking place in universe, i.e.
ΔS_system + ΔS_surrounding = ΔS_Total
ΔS_Total = ΔS_sys – \(\frac{q}{T}\)= ΔS_Sys – \(\frac{\Delta \mathrm{H}}{\mathrm{T}}\)
TΔS_Total = TAS_Sys – ΔH [∵ ΔG = ΔH – TΔS]
TΔS_Total = -ΔG
ΔG = -TΔS_Total

Question 26.
At a certain temperature ‘T’ endothermic reaction A → B proceeds virtually to end. Determine the sign of ∆S for the reaction A → B and ∆G for the reverse reaction B →A.
Answer:
∆S – +ve ∆H = +ve, ∆G = -ve the reaction proceeds virtually to end in reverse reaction B → A, ∆S = -ve, ∆G = +ve

Question 27.
Starting with the thermodynamics relationship G = H – TS, derive the following relationship ∆G = -T∆S_Total
Answer:
At initial state of the system G_i = H_i – T S_i
At final state of the system G_f = H_f – TS_f
∴ Change is free energy ∆G = G_f – G_i = (H_f – H_i) – T (S_t – S_i)
∆G = ∆H – T∆S

Question 28.
Explain the help of example, the difference between bond dissociation energy and bond energy.
Answer:
Bond disoociation energy is energy required to break 1 mole of bonds e.g.
H – H(g)) → 2H(g) ∆H = 436 kJ mol^-1
Bond energy is energy released when 1 mole of bonds are formed example
2H →H₂(g) ∆H = -436kJ mol^-1
In diatomic molecule both bond dissociation energy and bond energy are equal in magnitude but opposite in sign.

Question 29.
What is meant by the free energy of a system ? What will be the direction of chemical reaction when (i) ∆G = 0 (ii) ∆G > 0 (iii) ∆G < 0 Answer: Free energy is defined as energy which can be converted into useful work (i) ∆G = 0, the reaction will be in equilibrium (ii) ∆G > 0, the reaction will not take place
(iii) ∆G < 0, the reaction will be spontaneous

Question 30.
Why most of the exothermic process (reactions) spontaneous ?
Answer:
∆G = ∆H – ∆H – T∆S; For exothermic reactions ∆H is -ve. For a spontanec s process AG is to be -ve.
Thus decrease in enthalpy (-∆H) contributes significantly to the driving force (to make AG negative).

Question 31.
The enthalpy of combustion of sulphur is 297 kJ. Write the thermochemical equation for combustion of sulphur. What is the value of Δ_fH of SO₂ ?
Ans wer:
S(s) + O₂(g) → SO₂(g); ΔH = -297kJ and Δ_fH of SO₂ = -297 kJ mol^-1.

Question 32.
What is the most important condition for a process to be reversible in . thermodynmics ?
Answer:
The process should be carried out infinitesimally slowly of the driving force should be infinitesimally greater than the opposing force.

Question 33.
Why absolute value of enthalpy cannot be determined ?
Answer:
As H = E = +PV
Absolute value of E – the interned energy cannot be determined as it depends upon various factors whose value cannot be determined.
∴ Absolute value of H cannot be determined.

Question 34.
What are the applications of Hess’s Law of constant heat summation ?
Answer:

• It helps to calculate the enthalpies of formation of those compounds which cannot be determined experiementally.
• It helps in determine the enthalpy of allotropic transformation like C (graphite) → C (diamond)
• It helps to calculate the enthalpy of hydration.

Question 35.
Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vaccum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion ?
Answer:
We have q = – w : p_ext (10 – 2) = 0 × 8 = 0
No. work done; No heat, is absorbed.

Question 36.
Define (i) Molar enthalpy of fusion (ii) Molar enthalpy of vaporization ?
Answer:
(i) The enthalpy change that accompanies melting of one mole of a solid substance at its melting point is called molar enthalpy of fusion.
(ii) Amount of heat required to vaporize one mole of a liquid at constant temperature and understand pressure (1 bar) is called molar enthalpy of vapourisaton.

Question 37.
Define : (a) Enthalpy of atomization (b) Lattice enthalpy.
Answer:
Enthalpy of atomization : It is defined as the enthalpy change accompyning the breaking of one mole of a substance completely into its atoms in the gas phase.
H₂(g) → 2H(g); ∆_aH° = 435.0kJ mol^-1
Lattice enthalpy ; Lattice of an comp; the enthalpy change which occure when one mole of an ionic compound dissociates into its gaseous ionic state.
NaCl(s) → Na⁺ (g) + Cl^– (g); ∆H = +788 ks / mol

Question 38.
Explain Laplace- Lavoisier law.
Answer:
The quantity of heat that must be supplied to decompose a compound into its elements is equal and opposite to the heat evolved when the same compound is formed from its elements.
C(s) + O_2(g) → CO_2(g) ∆H = -393.5 kJml^-1
CO_2(g) → C_(s) + O_2(g) ∆H = +393.5 kJml^-1

Question 39.
If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ moh1. What is the enthalpy change for the C (graphite) → C (diamond) ?
Answer:
C(diamond) + O₂(g) → CO₂(g) ∆H =-395.4 kJmol^-1 …(1)
C(graphite) + O₂(g) → CO₂(g) ∆H = -393.6kJmol^-1 … (2)
C(graphite) → C(diamond ) substracting (1) from (2), we get
C(graphite) → C(diamond) ∆H = -393.6 kJ – (-395.4 kJ), ∆H = +1.8 kJ mol^-1

Question 40.
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(g) Δ_fH = -242kJ mol^-1. Bond energy of H₂ and O₂ are 436 mol^-1 and 500 mol^-1 respectively. What is bond energy of O – H bond ?
Answer:
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(g); Δ_fH° = -242kJ mol^-1
H = Bond energy of reactions – Bond energy of products
ΔH = B_H-H + \(\frac { 1 }{ 2 } \)B_{O – O} + B_{O – H}  ; 242kJ = 436 + \(\frac { 1 }{ 2 }\) (500) – 2B_{O – H} ⇔ -928kJ = -2B_{O – H}; B_{O – H} = \(\frac { 928 }{ 2 }\) = 464kJ mol^-1

Question 41.
What is spontaneous process ? Give example.
Answer:
Process which takes place itself, without any external aid under the given condition is called spontaneous process.
Example : Flow of heat from higher to lower temperature.

Question 42.
Define non-spontaneous process.
Answer:
Process which does not takes place itself or on its own but with the help of external aid under the given condition is called non-spontaneous process.
Example : Flow of heat energy from lower to higher temperatures.

1st PUC Chemistry Thermodynamics Three Marks Questions and Answers

Question 1.
Classify the following processes as reversible or irreversible :
(i) Dissolution of sodium chloride
(ii) Evaporation of water at 373 K and 1 atm. pressure
(iii) Mixing of two gases by diffusion
(iv) Melting of ice without rise in temperature
(b) When an ideal gas expands is vaccum, there is neither absorption nor revalutions of what ? why ?
Answer:
(a)(i) Irreversible, (ii) Irreversible, (iii) Irreversible, (iv) Reversible
(b) It is because no work is done,
i.e., w = 0 w = -p_ext × ∆V = 0 × ∆V = 0
In ∆U = q + w
q = 0 because gas chamber is insulated ∆U = 0 + 0 = 0

Question 2.
Justify the following statements :
(a) Reaction with ∆G° < 0 always have an equilibrium constant greater thanl.
(b) Many thermodynamically feasible reaction do not occur under ordinary conditions.
(c) At low temperatures enthalpy change dominates the AG expression and at high temperature it is the entropy which dominates the value of AG.
Answer:
(a) ∆G° = -2.303RT logK If K > 1, ∆G° will be less than zero because products formed are more than that of reactants, i.e., process is spontaneous in forward direction.
(b) It is because heat energy is required to overcome activation energy.
(c) ∆G = ∆H – T∆S, At low temperatures ∆H > T∆S whereas at high temperature T∆S >∆H
∴ ∆G decreases, i.e., becomes negative.

Question 3.
If for the reaction PbO₂ → PbO, ΔG° < 0
And for the reaction SnO₂ → SnO, ΔG° > 0
What are the most probable oxidation states of Pb and Sn ?
Answer:
Oxidation states of the elements in the following compounds are

Since PbO₂ → PbO change is accompanied by decrease in free energy it is a spontaneous change hence Pb in +2 state is more stable than in +4 state.

Question 4.
(i) What is the value Δ_rH° of the following reaction :
H⁺(aq) + OH^–(aq) →H₂O(l).
(ii) Give an example in which enthalpy change is equal to internal energy ?
Answer:
(i) Δ_rH° = -57.1kJmol^-1
(ii) H₂(g) + I₂(g) → 2HI(g) here ΔH = ΔU

Question 5.
Consider the reaction. A + B → C + D
(i) If the reaction is endothermic and spontaneous in the direction indicated, comment on the sign of ΔG and ΔS.
(ii) If the reaction is exothermic and spontaneous in the direction indicated, can you comment on the sign of G and S?
(iii) If the reactin is exothermic and spontaneous only in the direction opposite to the indicated, comment on the sign of AG and AS for the direction indicated in the equation.
Answer:
(i) ΔG = -ve, ΔS = +ve
(ii) ΔG = -ve, ΔS = -ve (if ΔH > TΔS)
(iii) ΔG = ΔS = -ve

Question 6.
Predict in which of the following entropy increases / decreases.
i) Temperature of a crystalline solid is raised from 0 K to 115 K.
ii) 2NaHCO₃ → Na₂CO₃(s) +CO₂ (g) + H₂O(g)
iii) H₂(g) → 2H(g
Answer:
(i) When temperature is raised, disorder in molecules increases and therefore entropy increases.
(iii) Reactant is a solid and hence has low entropy. Among the products there are two gases and one solid so products represent a condition of higher entropy.
(iv) Here 2 moles of H atoms have higher entropy than one mole of hydrogen molecule.

Question 7.
Which of the following is / are exothermic and which are endothermic ?
(i) Ca(g) → Ca²⁺(g) + 2e^–
(ii) O(g) + e^– → O^– (g)
(iii) N^2-(g) + e^– → N^3-(g)
Answer:
(i) Endothermic (ionisation energy is required)
(ii) Exothermic (first electron affinity – energy is released)
(iii) Endothermic (higher electron affinities energies are required)

1st PUC Chemistry Thermodynamics Numerical Problems and Answers

Question 1.
Calculate the heat of reaction of the following reaction :

C₆H₁₂O₆(s) + 6O_2(g) → 6CO_2(g)+ 6H₂O(g); ΔH = ?
C(graphite) + O₂(g) → CO₂(g); DH = -395.0 kJ … (1)
H₂ (g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l); DH = -269.5 kJ … (2)
6C(graphit.e) + 6H₂(g) + O₂(g) → C₆H₁₂O₆(s); DH = -1169.8 kJ … (3)
Answer:
Multiplying equation (1) and (2) each by 6 reversing (3), we get,
6C_(graphite) + 6O_2(g) → 6CO_2(g); ΔH = -2370 kJ … (4)
6H_2(g) + 3O_2(g) → 6H₂O_(l); ΔH = -1616.4 kJ … (5)
C₆H₁₂O_6(s) → 6C_(graphite) + 6H_2(g) + 3O_2(g) ; ΔH = +1169-8 kJ …. (6)
Adding (4), (5) and (6), C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(g) ;
AH(C₆H₁₂O₆) = -2370.0 – 1616.4 + 1169,8 = -2816.6 kJ

Question 2.
Calculate the heat of reaction of the following reaction :
CO₂(g) + H₂(g) → CO(g) + H₂O(g)
Given that the Δ_fH°CO(g) =-110.5kJ, Δ_fH°CO₂(g) =-393.8kJ,
Δ_fH°H₂O(g) = -241.8kJ respectively.
Answer:
The required equation is CO₂(g) + H₂(g) → CO(g) + H₂O(g)
ΔH – Σ ΔH_f(products) — Σ ΔH_f(reactants)
= ΔH_f²CO(g) + ΔH_f²H₂O(g) – ΔH_f²CO_2(g) – ΔH_f²H_2(g)
= -110.5 kJ – 241.8 kJ – (-398.3 kJ) – 0 = -352.3 kJ + 393.8 kJ = 41.5 kJ.

Question 3.
1 m³ of C₂H₄ at STP is burnt in oxygen, according to the thermochemical
reaction: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l); ΔH = -1410 kJ mol^-1
Assuming 70% efficiency, determine how much of useful heat is evolved in the reaction.
Answer:
22.4 L of C₂H₄ at STP produces 1410 kJ of energy.

Question 4.
With the help of thermochemical equation, calculate Δ_fH° at 298 K for the following reactions:
C(graphite) + O₂(g) → CO₂(g) ; Δ_fH° =-393.5kJ/mol
H₂ (g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l) ; Δ_fH° = -285.8kJ/mol
CO₂ (g) + 2H₂O(l) → CH₄ (g) + 2O₂(g) ; Δ_fH° = +890.3kJ/mol
Answer:
C(graphite) + O₂(g) → CO₂(g) ; Δ_fH° = -393.5kJ/mol …(1)
H₂(g) + \(\frac { 1 }{ 2 }\)O₂(g) → H₂O(l) ; Δ_fH° = -285.8kJ/mol …(2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) Δ_fH° = +890.3kJ/mol …(3)
Here we want one mole of C(graphite) as reactant, so we write down equation (1) as such, we want two moles of H₂(g) as reactant, so we multiply equation (2) by 2, we want

C(graphite) + O₂ (g) → CO₂(g) ; Δ_fH° = -393.5kJ / mol …(1)
2H₂(g) + O₂(g) → 2H₂O(1) ; Δ_fH° = 2(-285.8kJ/mol) …(2)
CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ; Δ_fH° =+890.3kJ/mol . … (3)
Adding we obtain: .
C(graphite) + 2H₂(g) → CH₄(g); Δ_fH° =-74.8kJ/mol

Question 5.
The heat of combustion of C₂H₆ is -368.4 kcal. Calculate heat of combustion of C₂H₄, heat of combustion of H₂ is 68.32 kcal mol^-1.
C₂H₄(g) + H₂(g) → C₂H₆(g) ; ΔH = -37.1 kcl
Answer:
C₂H₄(g) + H₂(g) → C₂H₆(g); ΔH = -37.1 k cal.
ΔH_cC₂H₆ = -368.4 k cal, ΔH_cC₂H₂ = ?; ΔH_cH_2(g) = -68.32 k cal.

ΔH  = ΣΔH_c(reactanta) – ΣΔH_c(products)
ΔH = ΔH_cC₂H₄+ΔH_cH_2(g) – ΔH_cC₂H_6(c)
-37.1k cal = ΔH_cC₂H₄-68.32-(-368.4)
ΔH_cC₂H₆ =-337.18 kcal.

Question 6.
The following thermochemical equations represent combustion of ammonia and hydrogen:

4NH₃(g) + 3O₂(g) → 6H₂O(l) + 2N₂(g); ΔH = -1516 kJ
2H₂(g) + O₂(g) → 2H₂O(l); ΔH = -572kJ
Calculate enthalpy of formation of ammonia.
Answer:
Required equation is  \(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)H_2(g) → NH_3(g) ΔH = ?
4NH_3(g) + 3O_2(g) → 2N_2(g) + 6H₂O_2(l) ΔH = -1516 kJ …(l)
2H_2(g) + O_2(g) → 2H₂O_(l) ΔH = -572 kJ …(2)
Reserving equation (1) and dividing by 4 we get:
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)2H₂O_2(l) → NH_3(g)+ \(\frac { 3 }{ 4 }\)O_2(g) ΔH = +379 kJ …(3)
Multiplying equation (2), by 3/4 we get,
\(\frac { 3 }{ 2 }\)H_2(g) + \(\frac { 3 }{ 4 }\)O₂ → \(\frac { 3 }{ 2 }\)H₂O_(l) ΔH = -572 × \(\frac { 3 }{ 4 }\) = -429kJ
Adding (3) and (4) we get ,
\(\frac { 1 }{ 2 }\)N_2(g) + \(\frac { 3 }{ 2 }\)H_2(g) → NH_3(g) ;ΔH = +379 – 429 = -50 kJ/mol

Question 7.
The equilibrium constant at 25° C for the process.
CO³⁺(aq) + 6NH₃(aq) ⇌ [CO(NH₃)₆]³(aq) is. 2.5 × 10⁶. Calculate the value of ΔG° at 25° C. (R = 8.314 JK1 mol^-1). In which direction is the reaction spontaneous under standard conditions ?
Answer:
ΔG° =-2.303 RT log K .
= -2.303 × 3.134 × 298 log (2.5 x 10⁶)
= -5705.8 [0.3980 + 6.0000] = -5705.8 × 6.3980 = -36.505 k/mol.
The reaction is spontaneous in forward direction under standard conditions.

Question 8.
What is the value of equilibrium constant for the following reaction at 400 E?
2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
ΔH⁰ = 77.5kJmol^-1, R = 8.3124 J Mol^-1K^-1, AS = 135 J K^-1mol^-1
Answer:
ΔG = ΔH – TΔS
ΔG = 77.5 × 1000 J – 400 K × 135 J K^-1 = 77500 J – 54000 J = 23500 J
ΔG = -2.303 RT log K
23500 J = -2.303 × 8.314 × 400 K log K

Log k =- 3.068 k = antilog(-3.008)
= Antilog (0.932-4) = Antilog (0.93) × 10^-4 = 8055 × 10^-4

Question 9.
The standard free energy change for a reaction is -212.3 kJ mol^-1. If the enthalpy of the reaction is -216.7 kJ mol^-1, Calculate the entropy change for the reaction.
Answer:
ΔG° = -212.3 kJ mol^-1, ΔH⁰ = -216.7 kJ mol^-1, ΔS° = ?
T = 298 K (Because standard free energy is measured at 298 K)
ΔG° = -212.3 kJ mol^-1, ΔH° = 216.7 kJ mol^-1, ΔS° = ?
-212.3 kJ = 216.7 kJ – 298 K × ΔS°

Question 10.
Calculate the standard free energy change AG° for the reaction.
2HgO(s) → 2Hg(l) + O₂(g)
ΔH° = OlkJmol^-1 at 298 K, S°_(Hg0) = 72.0JK^-1 mol^-1
Answer:
ΔS = 2S°_Hg + S°_(O2) – 25°_Hgo
= (2 × 77.4 + 205 – 2 × 72.0) JK ^-1mol^-1 = 215.8 jK^-1 mol^-1
ΔG° = Δ_rH° – TΔ_rS°

Question 11.
From the data given below at 298 K for the reaction :
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Calculate the enthalpy of formation of CH₄(g) at 298 K.
Enthalpy of reaction is = -89.5 kJ
Enthalpy of formation of CO₂(g) – -393kJ mol^-1
Enthalpy of formation of H₂O(l ) = -286.0 kJ mol^-1
Answer:
ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g)
-890.5 kJ = -393.5 kJ +2 × -286 kJ – Δ H_fCH₄(g) – 0
ΔH_fCH₄ – 75.0kJ

Question 12.
Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given : Lattice energy of NaCl = – 777.8 kJ mol^-1), Hydration energy
= -774.1 kJ mol^-1 and ΔS = 0.043 kJK^-1 mol^-1 at 298 K.)
Answer:
ΔH = Hydration energy – Lattice energy
ΔH = -774.1 kJ mol^-1 (-777.8 kJ mol^-1) = 3.7 kJ mol^-1
ΔG = ΔH – TΔS = +3.7 kJ- 298 × 0.043 kJ = +3.7 kJ- 12.81 kJ
ΔG = -9.11 kJ moH

Question 13.
For the equilibrium, PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g) at 298 K, K = 1.8 × 10 ^-7.
Calculate ΔG° for the reaction (R = 8.314JK 1mol 1)
Answer:
PCl₅(g) ⇌ PCl₂(g) + Cl₂(g), K = 1.8 × 10^-7
ΔG° = -2.303RTlogK = -2.303 × 8.314JK^-1mol^-1 × 298K × log(1.8 × 10^-7)
= -2.303 × 8.314 × 298[log 1.8 + logl0^-7 ] = -19.147 × 298 [0.2553 – 7.000]

Question 14.
Calculate the entropy change involved in conversation of 1 mole of water at 373 K to vapours at the same temperature. Latent heat of vaporation of water = 2.257 kJ g1.
Answer:
k = 1.8 × 10^-7
ΔH = 2.257 × 18kJ= 40.626kJ mol^-1

Question 15.
Calculate ΔH_f of HCl if bond energy of H – H bond is 437 kJ Cl – Cl bond is 244, and H – C is 433 kJ mol1.
Answer:
\(\frac { 1 }{ 2 }\)H2(g) + \(\frac { 1 }{ 2 }\) cl₂(g) > HCl(g)
ΔH = \(\frac { 1 }{ 2 }\)B_{H – H} + \(\frac { 1 }{ 2 }\)B_{O – O} = \(\frac { 1 }{ 2 }\) × 437 + \(\frac { 1 }{ 2 }\) × 244 -433
= 218.5 kJ + 122 kJ – 433 kJ = -92.5 kJ mol^-1

Question 16.
Calculate bond energy of C – H bond if ΔHc of CH₄ is -891 kJ, ΔHc of C (s) is – 394 kJ, ΔHc of H₂(g) is -286 kJ, heat of sublimation of C(s) is 717 kJ, heat of dissociation of H₂ is 436 kJ.
Answer:
CH₄(g) + 2O₂(g) → C0₂(g) + 2H₂O(l) ; ΔH = -891 kJ
C(s) + O₂(g) → CO₂(g) ;ΔH = -394 kJ
H₂(g) + \(\frac { 1 }{ 2 }\) O₂(g) → H₂O(l) ; ΔH = -286 kJ
C(s) → C(g) ;ΔH = +717 kJ
H₂(g) → 2H(g);Δ H = +436 kJ
Target equation is CH₄ +4H(g)
Reversing (ii),Multiply (iii) by 2 and reverse, Multiply eqn. (v) 2 and then adding all together we get:
CH₄ → C(g) + 4H(g) ΔH = -891kJ + 394kJ + 575kJ + 717kJ + 872kJ
ΔH = 1664 kJ
Energy required to break 4(C – H) bond = 1664 kJ
Energy required to break one (C – H) bond =\frac{1664}{4} = 416 kJ mol^-1

Question 17.
What would be heat released when :
(i) 0.25 mole of HCl in solution is neutralized 0.25 mole of NaOH solution ?
(ii) 0.5 mole of HNO₃ in solution is mixed with 0.2 mole of KOH solution ?
(iii) 200 cm³ of 0.2 M HCl is mixed with 300 cm3 of 0.1 M NaOH solution ?
(iv) 400 cm³ of 0.2 M H₂SO₄ is mixed with 300 cm3 of 0.1 M NaOH solution ?
Answer:
(i) H⁺(0.25mole) + OH^–(0.25mole) → H₂O(0.25mole)
Heat released = 0.25 × 57.1 = 14.3 kJ
(ii) 0.5 mole HNO₃ (aq) + 0.2 mole KOH(aq)
The net reaction is (0.2 mole HNO₃ reacts with only 0.2 mole KOH)
0.2 mole is limiting reagent.
0.2 mole of H⁺ + 0.2 mole of OH^– → 0.2 mole of H₂O
0.3 mole of H⁺ will remain unreacted.
Heat evolved = 0.2 × 57.1 = 11.4 kJ

(iii)

0.03 mole of OH^– is limiting reactant
Heat evolved = 0.03 × 57.1 = 1.71 kJ

(iv)

0.03 is limiting reagent Heat evolved = 0.03 × 57.1 = 1.7 kJ

(v) Mass of solution = 200 + 300 = 500 g in (iii) part (Assuming d = 1 g cm 3)
q = m × c × T 1.71 × 1000 J = 500 g × 4.18 × T
T = 0.82 K For (iv), T = \(\frac{1.7 \times 1000}{700 \times 4.18}\) = 0.58K