Tili Kannada Text Book Class 6 Solutions Nataka Karnataka Chapter 1 Kodi Nanna Balyava

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Tili Kannada Text Book Class 6 Solutions Nataka Karnataka Chapter 1 Kodi Nanna Balyava

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Siri Kannada Text Book Class 7 Solutions Gadya Chapter 7 Billa Habba

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2nd PUC Economics Question Bank Chapter 8 National Income Accounting

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Karnataka 2nd PUC Economics Question Bank Chapter 8 National Income Accounting

2nd PUC Economics National Income accounting One Mark Questions and Answers

Question 1.
Write the meaning of national income.
Answer:
National Income refers to the money value of all the final goods and services produced in an economy within an accounting year. It also includes the income from abroad factor.

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Question 2.
What are final goods?
Answer:
The final goods are those goods which are purchased for final utility or usage.

Question 3.
What do you mean by intermediate goods?
Answer:
The intermediate goods are those goods which are produced by one producer and used by another producer as a semi finished goods to produce some other final goods.

Question 4.
How do you get NDP?
Answer:
NDP is obtained by deducting depreciation cost from Gross Domestic Product (GDP) .

Question 5.
What is Per capita income?
Answer:
The per capita income is the average income of the population which is calculated by dividing the national income by the total population.

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Question 6.
What is economic welfare?
Answer:
Economic welfare is the satisfaction derived by an individual from the use of economic goods and services.

Question 7.
What are externalities?
Answer:
Externalities are those which are unintentional consequences of an economic action of a firm that accrue to another firm.

Question 8.
What are Transfer Payments?
Answer:
The payments made by the Government like old age pension, widow pension, scholarships etc. are called transfer payments.

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2nd PUC Economics National Income accounting Two Marks Questions and Answers

Question 1.
What are the differences between final goods and intermediate goods?
Answer:

Final Goods Intermediate Goods
(i) Final goods are those goods which are purchased for final utility or usage. (i) Intermediate goods are those goods which are produced by one producer and used by another producer to produce some other final goods
(ii) Example for final goods are TV, mobile, sugar, car, shoes etc. (ii) Examples are raw cotton, wood etc.

Question 2.
Differentiate between consumer goods and capital goods with examples.
Answer:

Consumer Goods

Capital Goods

(i) These are the goods which are purchased for consumption. (i) These are the goods which are used to produce other products.
(ii) Example food, clothes, TV, mobile, sugar, car, shoes etc. (ii) Examples are machinery, tools, roads.

Question 3.
Between net investment and capital, which is stock and which is a flow.
Answer:
Between net investment and capital, net investment is stock and capital is a flow.

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Question 4.
Write the meaning of circular flow of income.
Answer:
The circular flow of income illustrates the process where by the national income of an economy flows in a circular manner continuously between different sectors.

Question 5.
Classify the following into stocks and flows: Bank deposits, salary, wealth, food grain stock, exports, imports, foreign exchange reserves, national income.
Answer:

Stocks

Flows

Bank deposits, wealth, food grain stock, foreign exchange reserves. Salary, exports, imports, national income.

Question 6.
Name the factors of production.
Answer:
There are four factors of production viz.. Land, Labour, Capital and Organisation.

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Question 7.
Name the factor payments.
Answer:
The factor payments are Rent, Wages, Interest and Profit.

Question 8.
How does GNP differ from GDP?
Answer:
Gross Domestic Product Gross National Product

Gross Domestic Product

Gross National Product

  • It is the aggregate value of final goods and services produced within the country during a year.
  • It is a narrow concept.
  • It is the aggregate money value of all final goods and services produced by a country including income from abroad.
  • It is wider concept

Question 9.
Differentiate between nominal national income and real national income.
Answer:

Nominal National Income

Real National Income

  • When it is expressed in current prices, it is called nominal income.
  • It does not give a clear picture of the condition of the economy.
  • It is expressed in terms of base year prices/constant prices.
  • It provides a clear picture of the condition of the economy.

Question 10.
Mention the methods of measuring National Income.
Answer:
There are three methods of measuring national income viz.,

  1. Income Method
  2. Expenditure Method and
  3. Product method.

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Question 11.
What is the opinion of the International Monetary Fund (IMF) with regard to income from Foreign firms?
Answer:
The IMF is of the opinion that the income of a foreign firm should be included in the income of the country where it is located. But, profits earned by foreign firms should be credited to the parent country.

Question 12.
Nishanth is a lecturer in a College. He teaches his children at home. Are both teachings included in national income measurement? If not, why?
Answer:

  • Nishanth is a lecturer in a College. He gets his salary for his work. The salary is included in the national income.
  • Nishanth teaches his children at home. Here, he does not earn any income. So, it is not included in National Income.

Question 13.
What are externalities? Explain with an example.
Answer:
Externalities are those which are unintentional consequences of an economic action of a firm that accrue to another firm. For example, if there is establishment of an atomic power plant, it solves power crisis. It is a positive externality. But, the pollution caused by the power plant is harmful externality.

Question 14.
GDP is not a true indicator of welfare. Give reasons.
Answer:
Hie National income is not a reliable index of economic welfare because of the following reasons.

a) Inequality in the distribution of income.
b) Existence of non-monetary exchanges.
c) GDP does not consider whether the goods produced are useful or harmful.
d) GDP does not consider the manner of production.

Question 15.
What is simple economy?
Answer:
A simple economy is a clased economy in which there is no government or external trade or savings.

Question 16.
Mention any four difficulties in measuring National Income.
Answer:

  1. Neglect of transfer payments.
  2. Lack of reliable data.
  3. Income from illegal activities.
  4. Problem of double counting.

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2nd PUC Economics National Income accounting Five Marks Questions and Answers

Question 1.
Describe circular flow of income in a simple economy.
Answer:
Circular flow of income illustrates the process whereby the national income of an economy flows in a circular manner continuously between different sectors.

Now let us discuss about a circular flow of income in a simple economy.
A simple economy is a closed economy in which there is no Government or external trade or savings. A simple two-sector model economy is based on the following assumptions:

a) Existence of two sectors viz., household sector and producers.
b) Households are the owners of the factors of production.
c) Households receive income by selling the factor services.
d) There are no savings.
e) The firms produce entire produce to the households.
f) The economy is a closed economic system.

2nd PUC Economics Question Bank Chapter 8 National Income Accounting 1

In the above diagram, the household sector provides factors of production to the producers. The firms supply goods and services to the households. It is showtn in outer circle flow and is called real flow. The firms make factor payments like rent, wages, interest and profit to households as reward for factors of production. Households spend this income on buying goods and services. This is money is flow, which is shown in inner circle. In this way, production generates factor income, which is converted into expenditure.

The national income is calculated at three points as per the above chart viz., A, B, C. At point A, national income is calculated by adding all the factor payments and it is called income method. At point B, we take total expenditure incurred by the households on goods and services to calculate national income and it is called expenditure method. At point C, we measure the aggregate value of all the final goods and services produced by all the firms to calculate national income and it is called product method.

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Question 2.
Explain macroeconomic-identities GDP, NDP, GNP and NNP.
Answer:
a) Gross Domestic Product(GDP): GDP is the aggregate of the final goods and services produced in the domestic territory of a country during an accounting year.

b) Net Domestic Product ( NP): NDP refers to the market value of all final goods and services turned out in an economy during a given period of time after making allowance for depreciation charges. lt is obtained by subtracting depreciation from GDP.
NDP = GDP – Depreciation.

In simple words we can say that NDP is the net market value of final goods and services produced by its residents and non-residents within the domestic territory of a country in a year.

c) Gross National Product (GNP): It is the most important concept in N.I accounting. It is a National concept. GNP is defined as the total market value of all final goods and services produced in a country in a year’s time.

No allowance for wear and tear cost i.e., depreciation is made. While calculating the GNP, the money value of only the goods and services which are finally consumed by the people are to be taken into account. Hence, the value of all intermediary goods and inputs are to be excluded in order to avoid double or multiple counting.

The income received from foreign investment and from other factor services rendered abroad should be added to the gross domestic product of a country. Similarly, the income generated by the foreigner in a given country should be deducted from the GDP for the purpose of computing the GNP.

GNP = GDP + X – M

X= income earned by nationals abroad .
M= Income earned by foreigners in the given country.

GNP includes:

  • The value of all consumption goods which are currently produced.
  • The value of all capital goods currently produced.
  • Total government expenditure on buying various goods and services.
  • Net export value.
  • Net amount earned abroad.

GNP = (C+I+G) + (X-M) + (R-P)

d) Net National Product (NNP):
Net national product is the market value of the net output of final goods and services produced by the country during the relevant income period.
NNP = GNP – Depreciation.

Question 3.
Write a note on nominal national income and real national income.
Answer:
Nominal National Income: When the national income is expressed in the prices prevailing , in the year in which it is calculated it is called nominal national income. For example, if the national income of the year 2014 – 15 is calculated as per the prices of 2014-15, it becomes nominal national income.

Defects of Nominal National Income:

  • Nominal national Income does not give a clear picture of an economy. If there is rise in prices of goods and services due to scarcity, the nominal national income shows growth in GDP. But it is not true.
  • Difficulty in monitoring the changes in the price of all the goods and services.
  • Existence of non-monetary transaction hinders in the correct calculation of national income.

Real National Income: As the nominal national income does not provide correct picture of an economy, the concept of real national income is developed. The real national income is expressed in terms of base year prices. While calculating National income, a particular year is taken as base year. The price level is assumed to be as 100 for the base year. The formula to calculate real national income is as follows:

2nd PUC Economics Question Bank Chapter 8 National Income Accounting 2

4. Describe any five problems in the measurement of national income.
Answer:
a) Problem of double counting: The greatest difficulty in calculating the national income is that of double counting, which arises from the failure to distinguish properly between a final and an intermediate product. There always exists the fear of a good or a service being included more than once. If it so happens, the national income would work out to be many times the actual.

b) Illegal activities: Income earned through illegal activities such as gambling, illegal extraction of wine, hoarding and black marketing is not included in national income. While calculating national income, such earnings are left out, so the national income works out to less than the actual.

c) Change in price: Another difficulty in calculating national income is that of price changes which fail to keep stable the measuring rod of money for national income. When the price level in the country rises, the national income also shows an increase even though the production might have fallen and vice versa. Thus due to price-changes the national income cannot be adequately measured.

d) Illiteracy: In developing countries, we find crores of people as illiterates. They do not keep proper accounts about the production and sales of their products. Under such circumstances, the estimates of production and earned incomes are simply a guess work.

e) Non-availability of Data: Adequate and correct production and cost data are not available. The data relating to crops, forestry, fisheries, animal husbandry and the activities of petty shop keepers, small enterprises, etc., are not counted. For estimating national income by the income method, data on unearned incomes and on persons employed in the service sector are not available.

Question 5.
Explain the relationship between national income and welfare.
Answer:
Alfred Marshall, Prof.A.C Pigou and J.R.Hicks say that there is a close relationship between economic welfare and national income, because both of them are measures in terms of money. When National income increases, total welfare also increases and vice versa.

The effect of national income on economic welfare can be studied in two ways:

  • By change in the size of national income.
  • By change in the distribution of National income.

i) The change in the size of National Income: The positive change in the national income increases its volume. As a result people consume more of goods and services, which leads to increase in the economic welfare. Whereas the negative change in national income, results in reduction of its volume. People get lesser goods and services for consumption which leads to decrease in economic welfare.

ii) The change in the distribution of National Income: The distribution of national income takes place in two ways, firstly, by transferring wealth from poor to rich and secondly from the rich to the poor. But, it is advisable to follow the second way in distribution of national income.

The actual relation between the distribution of national income and economic welfare concerns the transfer of wealth flow from the rich to the poor. The redistribution of wealth in favour of the poor is brought about by reducing the wealth of the rich and increasing the income of the poor.

Thus, the increase in national income leads to increase in economic welfare provided that the income of the poor increases instead of decreasing and they improve their standard of living and that the income of the rich decreases in such a way that their productive capacity, investment and capital accumulation do not decline.

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Question 6.
“GDP is not a barometer of economic welfare but only a rough indicator”. Analyse this statement.
Answer:
The GDP is not a satisfactory measure of economic welfare because the estimates of national income do not include certain services and production activities which affect welfare. Following are the factors which affect human welfare but not included in GDP estimates:
i) Non-market transactions: Some of the non-market transactions increase welfare but they are not included in national income estimates. The services of housewives within the home and community activities like welfare activities of NGOs influence the welfare of the people but they are not included in GDP.

ii) Consumption of harmful goods: The consumption of harmful goods like cigarettes, liquor, narcotic drugs etc may not bring welfare to the community but they are included in the GDP estimates.

iii) Unequal distribution of National Income: The increase in National Income may not always coincide with the increase in economic welfare of a country. This is mainly because of unequal distribution of income. The rich people may be having more share than the poor.

iv) Environmental concerns: The rapid industrialization and urbanization are causing a severe threat to environment. Many hazardous pollutants are being added to the atmosphere in the name of development which is not in favour of economic welfare. But these are included in GDP estimates.

v) Externalities: An externality is a cost or benefit conferred upon second or third parties as a result of acts of individual production and consumption. But the cost or benefit of an externality cannot be measured in money, terms because it is not included in market activities. For example, the pleasure one gets from his neighbour’s garden is an external benefit and external cost is environmental pollution caused by industries. Both are excluded from national income estimates.

vi) Leisure and work: One of the important things that affect the welfare of a society is leisure. But is not included in GDP. For example, longer working hours may make people unhappy because their leisure is reduced. On the contrary, shorter working hours per week may increase leisure and make people happy.

vii) Manner of production: The economic welfare also depends on the manner of production of goods and services. If goods are produced by child labour or by exploitation of workers, then the economic welfare cannot increase.

Keeping the above limitations in view, GDP cannot be used as a barometer of economic welfare.

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2nd PUC Economics National Income accounting Ten Marks Questions and Answers

Question 1.
Explain the difficulties in the measurement of National income.
Answer:
i) Problem of double counting: The greatest difficulty in calculating the national income is of double counting, which arises from the failure to distinguish properly between a final and an intermediate product. There always exists the fear of a good or a service being included more than once. If it so happens, the national income would work out to be many times the actual.

ii) Illegal activities: Income earned through illegal activities such as gambling, illegal extraction of wine, hoarding and black marketing is not included in national income. While calculating national income, such earnings are left out, so the national income works out to less than the actual.

iii) Change in price: Another difficulty in calculating national income is that of price changes which fail to keep stable the measuring rod of money for national income. When the price level in the country rises, the national income also shows an increase even though the production might have fallen and vice versa. Thus due to price-changes the national income cannot be adequately measured.

iv) Illiteracy: In developing countries we find crores of people as illiterates. They do not keep proper accounts about the production and sales of their products. Under such circumstances, the estimates of production and earned incomes are simply a guess work.

v) Non-availability of Data: Adequate and correct production and cost data are not available. The data relating to crops, forestry, fisheries, animal husbandry and the activities of petty shop keepers, small enterprises, etc., are not counted. For estimating national income by the income method, data on unearned inijomes and on persons employed in the service sector are not available.

vi) Goods kept for self consumption: In India and other developing countries, producers keep a large portion of products for self consumption. For example, farmers keep certain portion of foodgrains for themselves. Such goods do not enter the market. It is not included in national income estimates.

vii) Absence of occupational specialization: The absence of occupational specialization makes calculation of national income difficult. Many people work as part-time workers and as such they do not give complete information about all sources of their income.

viii) Transfer Payments: The payments made by the Government to senior citizens, widows, scholarships, etc, are neglected while calculating the national income. But these are a part of individual income and a part of Government expenditure.

ix) Income from foreign companies: According to IMF, the income of foreign company should be included in the income of the host country and the profits earned by foreign companies should be included in the parent country. But it may not give correct national income estimates.

x) Existence of non-market transactions: Many times people stitch their own clothes, grow vegetables in their own garden and prepare many items in their own houses. The value of all such productive activities does not enter the market transactions and hence are not included in the national income estimates.

xi) Absence of trained personnel: There is a lack of well trained, skilled and efficient persons and staff to collect, classify and analyse the various information in relation to national income accounting.

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Question 2.
Discuss the Macro-economic identities of national income Explain the important concepts of National Income.
Answer:
The Macro economic identities of national income are as follows:
a) Gross Domestic Product(GDP): GDP is the aggregate of the final goods and services produced in the domestic territory of a country during an accounting year. It is the total money value of all final goods and services produced within the country by the nationals of the country and by the foreign nationals staying in the country during a year. It does not include goods and services produced by non-resident Indians, It can be expressed as follows:
GDP = C + I + G + net ‘X’

where, C – Consumption expenditure of public,
I – Investment expenditure of private sector
G – Governments consumption and investment expenditure.
net X – Difference between value of exports and imports.

b) Net Domestic Product (NDP): NDP refers to the market value of all final goods and services turned out in an economy during a given period of time after making allowance for depreciation charges. It is obtained by subtracting depreciation from GDP.
NDP = GDP – Depreciation.

In simple words we can say that NDP is the net market value of final goods and services produced by its residents and non-residents with in the domestic territory of a country in a year.

c) Gross National Product (GNP):
It is the most important concept in N.I accounting. It-is a National concept. GNP is defined as the total market value of all final goods and services produced in a country in a year time.

No allowance for wear and tear cost i.e., depreciation is made. While calculating the GNP, the money value of only the goods and services which are finally consumed by the people are to be taken into account. Hence, the value of all intermediary goods and inputs are to be excluded in order to avoid double or multiple counting.

The income received from foreign investments and from other factor services rendered abroad should be added to the gross domestic product of a country. Similarly, the income generated by the foreigner in a given country should be deducted from the GDP for the purpose of computing the GNP.
GNP = GDP + X – M .
X= income earned by national abroad .
M= Income earned by foreigners in the given country.

GNP includes:

  • The value of all consumption goods which are currently produced.
  • The value of all capital goods currently produced.
  • Total Government expenditure on buying various goods and services.
  • Net export value
  • Net amount earned abroad.
    GNP = (C+I+G) + ( X – M ) + (R – P)

d) Net National Product (NNP):
Net national product is the market value of the net output of final goods and services produced by the country during the relevant income period.
NNP = GNP- Depreciation.

e) Personal Income (PI):
The concept of personal income refers to the sum of all the incomes actually received by the individual and households in a country during one year. It is the amount available to them for spending, paying taxes and saving purposes PI is less than NI because several deduction are made out of it.

Personal income = National income – undistributed profit – social security contribution + transfer payment
The concept of PI helps us to know the potential purchasing power of people.

f) Disposable personal Income:
The entire PI accounting to individual or house hold in not available for consumption purpose.
A part of PI have to be paid to the Government by way of personal direct tax. Hence, that part of the personal direct taxes is called as disposable personal income.
DI = PI – Personal Direct Tax.

Disposable income can either be spent entirely of a part of the income can be saved. Therefore, the Personal Disposable Income may be written as:
PDI = Consumption Expenditure + Savings.

g) Nominal and Real National Income:
When the national income is expressed in the prices prevailing in the year in which it is calculated it is called nominal national income. For example, if the national income of the year 2014 – 15 is calculated as per the prices of 2014 – 15, it becomes nominal national income.

The real national income is expressed in terms of base year prices. While calculating National income, a particular year is taken as base year. The price level is assumed to be as 100 for the base year. The formula to calculate real national income is as follows:

2nd PUC Economics Question Bank Chapter 8 National Income Accounting 3

h) Per Capita Income:
Per capita income refers to the income of an individual person. It is the average income of the people of a country. The per capita income is calculated by dividing the national income by population. Thus,
2nd PUC Economics Question Bank Chapter 8 National Income Accounting 4

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Question 3.
Explain the methods of measuring National Income.
Answer;
There are three methods of measuring national income, they are as follows:

a) Product methods or output method:
Under this method, a census of all goods and services are conducted to get the correct picture of total national production.
While,calculating total volume of goods and service, the following four items are to be included.

  1. All kinds of consumption goods and services.
  2. Gross domestic investment, which includes inventories, capital formation, construction of houses etc.
  3. Production in the public sector.
  4. Export minus imports.

b) Income method:
National income is the result of the combined and co-operative efforts put in by all factors of production. Alter employing them, we have to remunerate them in the form of rent, wages, interest and profits.

This method may be represented in the following equation.
Y = (r + w + i + p) + (X – M) + (R – P), where r – rent, w – wages, i – interest, p – profit, X – exports, M imports,  R – receipts, P – payments.

c) Expenditure method:
Incomes earned by factor inputs are spent on buying different goods and services. If we add the total expenditure incurred by all people in a years’ time, then we get total income of the people. Income determines the expenditures. All kinds of expenditures are to be taken into account while calculating the national income of a country. They are

i. Personal consumption expenditures of all people on all kinds of goods and services.
ii. Gross domestic investment or investment expenditures made by all businessmen in a year.
iii. Gross Governments’expenditure on all kinds of goods and services.
iv. Net foreign investment, exports – imports.

This method may be represented with the help of the following equation.
Y = (C +I + G) + (X – M) + (R – P), where, C – Consumption, I – Investment, G – Government’s Investment, X – exports, M – Imports, R – Receipts and P – Payments.

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2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm

You can Download Chapter 5 Dissolution of Partnership Firm Questions and Answers, Notes, 2nd PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm

2nd PUC Accountancy Dissolution of Partnership Firm NCERT Textbook Questions and Answers

2nd PUC Dissolution of Partnership Firm Short Answer Questions With Answers

Question 1.
State the difference between dissolution of partnership and dissolution of partnership firm.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 1

Question 2.
State the accounting treatment for:
1. Unrecorded assets
2. Unrecorded liabilities
Answer:
For realisation of any unrecorded assets including goodwill, if any
Bank A/c Dr.
To Realisation A/c
10. For settlement of any unrecorded liability Realisation A/c Dr.
To Bank A/c

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Question 3.
On dissolution, how will you deal with partner’s loan if it appears on the
(a) assets side of the balance sheet,
(b) liabilities side of balance sheet.
Answer:
Treatment partners loan: While writing entry related to loan, we have to b serve that loan paid by the partner, then firm liable to repay. At the time of dissolution the loan amount should be transferred to partners capital a/c credti side.

Incase of firm has paid has paid loan to partnership firm at appears in assets side of balance sheet. Such amount should be transferred t respective/concmed partners capital a/c debit side.

Question 4.
Distinguish between firm’s debts and partner’s private debts.
Answer:
Private Debts and Firm’s Debts: Where both the debts of the firm and private debts of a partner co-exist, the following rules, as stated in Section 49 of the Act, shall apply.
(a) The property of the firm shall be applied first in the payment of debts of the firm and then the surplus, if any, shall be divided among the partners as per their claims, which can be utilised for payment of their private liabilities.

(b) The private property of any partner shall be applied first in payment of his private debts and the surplus, if any, may be utilised for payment of the firm’s debts, in case the firm’s liabilities exceed the firm’s assets.

It may be noted that the private property of the partner does not include the personal properties of his wife and children. Thus, if the assets of the firm are not adequate enough to pay off firm’s liabilities, the partners have to contribute out of their net private assets

Question 5.
State the order of settlement of accounts on dissolution.
Answer:
Settlement of Accounts: In case of dissolution of a firm, the firm ceases to conduct business and has to settle its accounts. For this purpose, it disposes off all its assets for satisfying all the claims against it. In this context it should be noted that, subject to agreement among the partners, the following rules as provided in Section 48 of the
Partnership Act 1932 shall apply.
(a) Treatment of Losses
Losses, including deficiencies of capital, shall be paid:

  • first out of profits,
  • next out of capital of partners, and
  • lastly, if necessary, by the partners individually in their profits sharing ratio.

Question 6.
On what account realisation account differs from revaluation account
Answer:

Revalution a/c Realisation
It is prepared at the time of admission retirement or death of a partner It is prepared at the time of closure of business i.e dissolution
Revised only these assets and liability are revalued All  assets and external liabilities are record which are payable
Unrecorded assets liabilities will not be given much importance All unrecorded assets and liabilities are recorded on respective side.

2nd PUC Dissolution of Partnership Firm Long Answer Questions With Answers

Question .
What is meant by dissolution of partnership firm?
Answer:
Dissolution of a partnership firm may take place without the intervention of court or by the order of a court, in any of the ways specified later in this section.
It may be noted that dissolution of the firm necessarily brings in dissolution of the partnership. Dissolution of a firm takes place in any of the following ways:
1. Dissolution by Agreement: A firm is dissolved :
(a) with the consent of all the partners or
(b) in accordance with a contract between the partners.

2. Compulsory Dissolution: A firm is dissolved compulsorily in the following cases:
(a) when all the partners or all but one partner, become insolvent, rendering them incompetent to sign a contract;
(b) when the business of the firm becomes illegal; or
(c) when some event has taken place which makes it unlawful for the partners to carry on the business of the firm in partnership, e.g., when a partner who is a citizen of a countiy becomes an alien enemy because of the declaration of war with his countiy and India.

KSEEB Solutions

3. On the happening of certain contingencies: Subject to contract between the partners, a firm is dissolved :
(a) if constituted for a fixed term, by the expiry of that term;
(b) if constituted to carry out one or more ventures, by the completion thereof;
(c) by the death of a partner;
(d) by the adjudication of a partner as an insolvent.

4. Dissolution by Notice: In case of partnership at will, the firm may be dissolved if any one of the partners gives a notice in writing to the other partners, signifying his intention of seeking dissolution of the firm.

5. Dissolution by Court: At the suit of a partner, the court may order a partnership firm to be dissolved on any of the following grounds:
(a) when a partner becomes insane;
(b) when a partner becomes permanently incapable of performing his duties as a partner;
(c) when a partner is guilty of misconduct which is likely to adversely affect the business of the firm;
(d) when a partner persistently commits breach of partnership agreement;
(e) when a partner has transferred the whole of his interest in the firm to a third party;
(f) when the business of the firm cannot be carried on except at a loss; or
(g) when, on any ground, the court regards dissolution to be just and equitable.

Question 2.
What is a Realisation Account?
Answer:
When the firm is dissolved, its books of account are to be closed and the profit or loss arising on realisation of its assets and discharge of liabilities is to be computed. For this purpose, a Realisation Account is prepared to ascertain the net effect (profit or loss) of realisation of ‘ assets and payment of liabilities which may be is transferred to partner’s capital accounts in their profit sharing ratio.

Hence, all assets (other than cash in hand bank balance and fictitious assets, if any), and all external liabilities are transferred to this account. It also records the sale of assets, and payment of liabilities and realisation expenses; The balance in this account is termed as profit or loss on realisation which is transferred to partners’ capital accounts in thier profit sharing ratio

Question 3.
Reproduce the format of Realisation Account.
Answer:

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 2

Question 4.
How deficiency of Crditors is paid off?
Answer:
For settlement with the creditor through transfer of assets when a creditor accepts an asset in full and final settlement of his account, journal entry needs to be recorded. But, if the creditor accepts an asset only as part payment of his/her dues, the entry will be made for cash payment only. For example, a creditor to whom ₹ 10,000 was due accepts office equipment worth ₹ 8,000 and is paid ₹ 2,000 in cash, the following entry shall be made for the payment of ₹ 2,000 only.
Realisation A/c Dr.
To Bank A/c
However, when a creditor accepts an asset whose value is more than the amount due to him, he/she will pay cash to the frim for the difference for which the entry will be:
BankA/c Dr.
To Realisation A/c
For payment of realisation expenses
(a) When some expenses are incurred and paid by the firm in the process of realisation of assets and payment of liabilities:
Realisation A/c Dr.
To Bank A/c

2nd PUC Dissolution of Partnership Firm Numerical Questions

Question 1.
Journalise the following transactions regarding Realisation expenses:
(a) Realisation expenses amounted to ₹ 2,500.
(b) Reahisation expenses amounting to ₹ 3,000 were paid by Ashok, one of the partners.
(c) Realisation expenses ₹ 2,300 borne by Tarun, personally.
(d) Amit, a partner was appointed to realize the assets, at a cost of ? 4,000. Theactual amount of Realisation amounted to ₹ 3,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 3

KSEEB Solutions

Question 2.
Record necessary journal entries in the following cases:
(a) Creditors worth ₹ 85,000 accepted ₹ 40,000 as cash and Investment worth ₹ 43,000, in full settlement of their claim.
(b) Creditors were ₹ 16,000. They accepted Machinery valued at ₹ 18,000 in settlement of their claim.
(c) Creditors were ₹ 90,000. They accepted Buildings valued ₹ 1,20,000 and paid cash to the firm ₹ 30,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 4

Question 3.
There was an old computer which was written-off in (the books of accounts in the pervious year. The same has been taken over by a partner Nitin for ₹ 3,000. Journalise the transaction, supposing. That the firm has been dissolved.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 5

Question 4.
What journal entries will be recorded for the following transactions on the dissolution of a firm:
(a) Payment of unrecorded liabilities of ₹ 3,200.
(b) Stock worth ₹ 7,500 is taken by a partner Rohit.
(c) Profit on Realisation amounting to ₹ 18,000 is to be distributed between the partners Ashish and Tarun in the ratio of 5 : 7.
(d) An unrecorded asset realised ₹ 5,500.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 6

Question 5.
Give journal entries for the following transactions
1. To record the Realisation of various assets and liabilities.
2. A firm has a Stock of ₹ 1,60,000. Aziz, a partner took over 50% of the Stock at a of 29%
3. was sold at a profit of 30% on cost.
4. (book value ₹ 1,60,000) sold for ₹ 3,00,000 through a bro charged 2%, commission on the deal.
5. Plant and Machinery (book value ₹ 60,000) was handed over to a creditor at an agreed valuation of 10% less than the book value
6. Investment whose face value was ₹ 4,000 was realized at 50%
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 7
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 8

KSEEB Solutions

Question 6.
How will you deal with the realisation expenses of the firm of Rashim and Bindiya in the following cases:
1. Realisation expenses amounts to ₹ 1,00,000,
2. Realisation expenses amounts to ₹ 30,000 are paid by Rashini, a partner,
3. Realisation expenses are to be borne by Rashim for which he will be paid ₹ 70,000
as remuneration for completing the dissolution process. The actual expenses Incurred by Rashim were ₹ 1,20,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 9
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 10

Question 7.
The book value of assets (other than cash and bank) transferred to Realisation Account is ₹ 1,00,000. 50% of the assets are taken over by a partner Atul, at a discount of 20%; 40% of the remaining assets are sold at a profit of 30% on cost; 5% of the balance being obsolete, realised nothing and remaining assets are handed over to a Creditor, in full settlement of his claim. You are required to record the journal entries for realisation of – assets.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 11

Question 8.
Record necessary journal entries to record the following Unrecorded assets ann liabilities in the books of Paras and Priya:
1. There was an old furniture in the firm which had been written-off completely in the books. This was sold for ₹ 3,000.
2. Ashish, an old customer whose account for ₹ 1,000 was written-off as bad in the previous year, paid 60%, of the amount.
3. Paras agreed to takeover the firm’s goodwill (not recorded in the books of the ‘ firm), at a valuation of ₹ 30,000,
4. There was aa old typewriter which had been written-off completely from the books. It was estimated to realize ₹ 400. It was taken away by Priya at an estimated price less 25%.
5. There were 100 shares of ₹ 10 each in Star Limited acquired at a cost of ₹ 2,000 which had been written-off completely from the books. These shares are valued ₹ 6 each and divided among the partners in their profit sharing ratio.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 12

Question 9.
All partners wishes to dissolve the firm. Yastin, a partner wants that her loan of ₹ 2,00,000 must be paid off before the payment of capitals to the partners. But, Amart, another partner wants that the capitals must be paid before the payment of Yastin’s loan. You are required to settle the conflict giving reasons.
Answer:
As per section 48 of Partnership Act 1932, at the time of dissolution, loans and advances from the partners must be paid off before the settlement of their capital accounts. Hence, Yastin’s argument is correct that her loan of ₹ 2,00,000 must be paid off before the payment of partners capital.

KSEEB Solutions

Question 10.
What journal entries would be recorded for the following transactions on the dissolution of a firm after various assets (other than cash) on the third party liabilities have been transferred to Realisation account
1. Arti took over the Stock worth ₹ 80,000 at ₹ 68,000.
2. There was unrecorded Bike of ₹ 40,000 which was taken over By Mr. Karim.
3. The firm paid ₹ 40,000 as compensation to employees. ,
4. Sundry creditors amounting to ₹ 36,000 were settled at a discount of 15%.
5. Loss on realisation ₹ 42,000 was to be distributed between Arti and Karim in the ratio of 3:4.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 14

Question 11.
Rose and Lily shared profits in the ratio of 2 : 3. Their Balance Sheet 2014 was as follows:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 15
Rose and. Lily decided to dissolve the firm on the above date. Assets except bills receivables) realised ₹ 4,84,000. Bills Receivable were taken over by Rose at ₹ 30,000. Creditors agreed to take ₹ 38,000. Cost of Realisation was ₹ 2,400. There was a Motor Cycle in the firm which was bought out of the firm’s money, was not shown in the books of the firm. It was now sold for ₹ 10,000. There was a contingent liability in respect of outstanding electric bill of ₹ 5,000, Bill Receivable taken over by Rose at ₹ 33,000.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 16
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 17

KSEEB Solutions

Question 12.
Shilpa, Meena and Nanda decided to dissolve their partnership oh March 31, 2014. Their profit, sharing ratio was 3:2:1 and their Balance Sheet was as under:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 18
The stock of value of ₹ 41,660 are taken over by Shilpa for ₹ 35,000 and she agreed to discharge bank loan. The remaining stock was sold at ₹14,000 and debtors amounting toRs, 10,000 realised ₹ 8,000, land is sold for ₹ 1,10,000. The remaining debtors realized 50% at their book value. Cost of realization amounted to ₹ 1,200. There was a typewriter not recorded in the books worth ₹ 6,000 which were taken over by one of the Creditor at this value. Prepare Realisation A/c.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 19
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 20

Question 13.
Surjit and Rahi were sharing profits (losses) in the ratio of 3:2, their Balance Sheet as on March 31, 2014 is as follows:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 21

The firm was dissolved on March 31, 2014 on the following terms:
1. Surjit agreed to take the investments at ₹ 8,000 and to pay Mrs. Surojit’s loan.
2. Other assets were realised as follows: Stock ₹ 5,000, Debtors ₹ 18,500, Furniture ₹ 4,500, Plant ₹ 25,000
3. Expenses on realisation amounted to ₹ 1,600.
4. Creditors agreed to accept ₹ 37,000 as a final settlement.
You are required to prepare Realisation account, Partner’s Capital account and Bank account.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 22
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 23

KSEEB Solutions

Question 14.
Rita, Geeta and Ashish were partners in a firm sharing profits/losses in the ratio of 3:2:1. On March 31,2014 their balance sheet was as follows:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 24
On the date of above mentioned date the firm was dissolved:
1. Rita was appointed to realise the assets. Rita was to receive 5% commission on the rate of assets (except cash) and was to bear all expenses of realisation,
2. Assets were realised as follows:
Debtors 30,000
Stock 26,000
Plant 42,750
3. Investments were realised at 85% of the book value,
4. Expenses of realisation amounted to ₹ 4,100,
5. Firm had to pay ₹ 7,200 for outstanding salary not provided for earlier,
6. Contingent liability in respect of bills discounted with the bank was also materialized and paid off ₹ 9,800,’Prepare Realisation account, Capital Accounts of Partner’s and Cash Account.
Answer:

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 25
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 26

Question 15.
Anup and Suntit are equal partners in a firm. They decided to dissolve the parutership on December 31, 2014. When the balance sheet is as under
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 27
The Assets were realised as follows:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 28
The Creditors were paid ₹ 25,500 in full settlement. Expenses of Realisation amount to ₹ 2,500.
Prepare Realisation Account, Bank Account, Partners Capital Accounts to close the books of the firm.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 29
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 30

KSEEB Solutions

Question 16.
Ashu and Harish are partners sharing profit and losses as 3:2. They decided to dissolve the firm on December 31. 2014. Their balance sheet oil the above date was:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 31
Asha is to take over the building at ₹ 95,000 and Machinery and Furniture is take over by Harish at value of ₹ 80,000. Ashu agreed to pay Creditor and Harish agreed to meet Bank overdraft. Stock and Investments are taken by both partner in profit sharing ratio. Debtors realised for ₹ 46,000, expenses of realisation amounted to ₹ 3,000: Prepare necessary ledger account.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 32
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 33

Question 17.
Sanjay, Tarun and Vineet shared profit in the ratio of 3:2:1. On December 31,2012 their balance sheet was as follows

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 34
On this date the firm was dissolved. Sanjay was appointed to realise the assets. Sanjay was to receive 6% commission on sale of assets (except cash) and was to bear all expenses of realisation. Sanjay realised the assets as follows: Plant ₹ 72,000, Debtors ₹ 4,000, Furniture ₹ 18,000, Stock 90% of the book value, Investments ₹ 76,000 and Bills receivable ₹ 31,000, Expenses of realisation amounted to ₹ 4,500. Prepare Realisation Account, Capital Accounts and Cash Account
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 35
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 36

Question 18.
The following is the Balance Sheet of Gupta and Sharma as on December.

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 37
The firm was dissolved on December 31, 2014 and asset realised and settlements of liabilities as follows:
(a) The realisation of the assets were:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 38
(b) Investment was taken over by Gupta at agreed value of ₹ 36,000 and agreed to pay of Mrs. Gupta’s loan.
(c) The Sundry Creditors were paid off less 3% discount.
(d) The realisation expenses incurred amounted to ₹ 1,200.
Journalise the entries to be made on the dissolution and prepare Realisation Account, Bank Account and Partners Capital Accounts.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 39
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 40
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 41
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 42

KSEEB Solutions

Question 19.
Ashok, Babu and Chetan are in partnership sharing profit in the proportion of 1/2. 1/3,1/6 respectively. They dissolve the partnership of the December 31, 2014 when the balance sheet of the firm as under:

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 43
The Machinery was taken over by Babu for ₹ 45,000, Ashok took over the investment for ₹ 40,000 and Freehold property took over by Chetan at ₹ 55,000. The remaining Assets realised as follows: Sundry Debtors ₹ 56,500 and Stock ₹ 36,500. Sundry Creditors were settled at discount of 7%. A Office computer, not shown in the books of accounts realised ₹ 9,000. Realisation expenses amounted to ₹ 3,000.
Prepare Realisation Account, Partners Capital Account, Bank Account.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 44
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 45
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 46

Question 20.
The following is the Balance sheet of Tanu and Manu, who shares profit and losses in the ratio of 5:3, On December 31,2014:

2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 47
On the above date the firm is dissolved and the following agreement was made:
Tanu agree to pay the bank loan and took away the sundry debtors. Sundry creditors accepts stock and paid ₹10,000 to the firm. Machinery, is taken over by Manu for ₹ 40,000 and agreed to pay of bills payable at a discount of 5%. Motor car was taken over by Tanu for ₹ 60,000. Investment realised ₹ 76,000 and fixtures ₹ 4,000. The expenses of dissolution amounted to ₹ 2,200.
Prepare Realisation Account, Bank Account and Partners Capital Accounts.
Answer:
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 48
2nd PUC Accountancy Question Bank Chapter 5 Dissolution of Partnership Firm - 49

KSEEB Solutions

Siri Kannada Text Book Class 7 Solutions Gadya Chapter 6 Chagali Iruve

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Tili Kannada Text Book Class 6 Solutions Padya Chapter 2 Nanna Desha Nanna Jana

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2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds

You can Download Chapter 9 Coordination Compounds Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds

2nd PUC Chemistry Coordination Compounds NCERT Textbook Questions and Answers

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
The valencies of the elements in cobalt (III) chloride and in ammonia seem to be completely satisfied. However, the two substances react to form a stable compound with the formula COCl3·6NH3. The compound is known as a coordination compound.

  1. In co-ordination compounds, metal atoms exhibit two types of valencies. These are primary and secondary valencies.
  2. The ligands attached to the primary valencies are ionizable and represent the oxidation state of the metal. The ligands attached to the secondary valencies are non-ionizable and these correspond to its co-ordination number.
  3. Every metal has a fixed number of secondary valencies because its co-ordination number is fixed.
  4. Metal tends to satisfy both these valencies. The primary valencies are satisfied by negative ions (anions) whereas secondary valencies are satisfied by either negative ions or neutral molecules.

Question 2.
FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why?
Answer:
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 1
Both compounds i.e. FeSO4 (NH4)2SO4.6H2O and [Cu(NH3)4]SO4.5H2O fall under the category of addition compounds with only one major difference i.e. the former is an example of a double salt, while the latter is a coordination compound. A double salt is an addition compound that is stable in the solid-state bnK that which breaks up into its constituent ions in the dissolved state. These compounds exhibit individual properties of their constituents for e.g. FeSO4 (NH4)2SO4.6H2O breaks into Fe2+ NH4+, and SO42- ions.

Hence, it gives a positive test for Fe2+ ions. A coordination compound is an addition compound that retains its identity In the solid as well as in the dissolved state. However, the individual properties of the constituents cost. This happens because [Cu(NH3)]4]SO4.5H2O does not show the test for Cu2+ The ions present in the solution of [Cu(NH3)4]SO4.5H2O are [Cu(NH3)4]2+ and SO42-

KSEEB Solutions

Question 3
Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic complexes;
Answer:
(i) Coordination entity:
A coordination entity is an electrically charged radical or species carrying a positive or negative charge. In a coordination entity, the central atom or ion is surrounded by a suitable number of neutral molecules or negative ions (called, ligands).
For examples:
[Ni(NH3)6]2+, [Fe(CN)6]4+ = cationic complex
[PtCl4]2-, [Ag(CN)2] – anionic complex
[Ni (CO)4], [CO (NH3)4Cl2] = neutral complex

(ii) ligand:
The neutral molecules or negatively charged ions that surround the metal atom in a coordination entity or a coordinate complex are known as ligands.
For example
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 2
Cl, OH. Ligands are usually polar in nature and possess at least one unshared pair of valence electrons.

(iii) Coordination number:
The total number of ligands (either neutral molecules or negative ions) that get attached to the central metal atom in the coordination sphere is called the coordination number of the central metal atom. It is also referred to as its legacy.
For example:
(a) In the complex K2[PtCl6], there are six chloride ions attached to Pt in the coordinate sphere. Therefore, the coordination number of Pt is 6.
(b) similarly, in the complex [Ni (NH3)4 Cl2], the coordination number of the central atom (Ni) is 6.

(iv) Coordination polyhedron:
Coordination polyhedrons, about the central atom, can be defined as the spatial arrangement of the ligands that are directly attached to the central metal ion in the coordination sphere.
For example:
(a) Square planar
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 3
(b) Square planar
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 4

(v) Homoleptic complexes:
These are those complexes in which the metal ion is bound to only one kind of a donor group For e.g.:
[CO (NH3)6]3+
[pt Cl4]2

(vi) Heteroleptic complexes:
Heteroleptic complexes are those complexes where the central metal is bound to more than one type of donor group.
For e.g.:
[CO(NH3)4C2]+
[CO(NH3)5 Cl]2+

Question 4.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
A ligand may contain one or more unshared pairs of electrons which are called the donor sites of ligands. Now depending on the number of these donor sites, ligands can be classified as follows.
(a) Unidentate ligands: Ligands with only one donor sites are called unidentate ligands.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 5
(b) Didentate ligands: Ligands that have two donor sites are called didentate ligands.
For e.g.:
(a) Ethane – 1, 2 – diamine.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 6
(b) enolic acid
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 7
c) Ambidentate ligand:
Ligands that can attach themselves to the central metal atom through two different atoms are called ambidentate ligands.
For Example :
(a)
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 8
(b)

2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 9

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [CO(H2O)(CN)(en)2]2+
(ii) [COBr2(en)2]+
(iii) [PtCl4]2-
(iv) K3[Fe(CN)6]
(v) [Cr(NH3)3Cl3]
Answer:
(i) [CO(H2O) (CN) (en)2]2+
Let the oxidation number of CO be x.
The charge on the complex is + 2

2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 10

(ii) [PtCl4]2-
x + 4 (-1) = -2 x = +2

(iii) [COBr2 (en)2]+
x + 2 (-1) + 2.(0) = + 1
x = + 3

(iv) K3[Fe(CN)6]
x + 6 (-1) = -3
x = + 3

(v) [Cr (NH3)3 Cl3]
x + 3 (0) +3 (-1) = 0
x – 3 = 0
x = 3

KSEEB Solutions

Question 6.
Using IUPAC norms write the formula for the following:

  1. Tetrahydroxozincate (II)
  2. Potassium tetrachloridopalladate(II)
  3. piamminedichloridoplatinum(II)
  4. Potassium tetracyanonickelate(II)
  5. Pentaamminenitrito-O-cobalt(III)
  6. Hexaamminecobalt(III) sulphate
  7. Potassium tri(oxalato)chromate(III)
  8. Hexaammineplatinum(IV)
  9. Tetrabromidocuprate(II)
  10. Pentaamminenitrito-N-cobalt(III)

Answer:

  1. [Zn (O4)]2-
  2. K2 [Pd Cl4]
  3. [Pt (NH3)2 Cl2]
  4. K2[Ni(CN)4]
  5. [CO (NH3)5 (ONO)]2+
  6. [CO (NH3)6]2(SO4)3
  7. K3 [Cr (C2O4)3]
  8. [Pt (NH3)6]4+
  9. [Cu (Br4)2-
  10. [CO (NO2) (NH3)5]2+

Question 7.
Using IUPAC norms write the systematic names of the following:

  1. [CO(NH3)6]Cl3
  2. [Pt(NH3)2Cl(NH2CH3)]Cl
  3. [Ti(H2O)6]3+
  4. [CO(NH3)4Cl(NO2)]Cl
  5. [Mn(H2O)6]2+
  6. [NiCl4]2-
  7. [Ni(NH3)6]Cl2
  8. [CO(en)3]3+
  9. [Ni(CO)4]

Answer:

  1. Hexaamminecobalt (III) chloride.
  2. Diamine chlorido (methylamine) platinum (II) chloride.
  3. Hexaaquatitanium (III) ion.
  4. Tetraamminichloridonitrite – N – cobalt (HI) chloride.
  5. Flexaquamanganese (II) ion.
  6. Tetrachloridononickeiate (II) ion.
  7. Hexaammmenickei (11) chloride,.
  8. Tris (ethane -1, 2 – diammine) cobalt (III) ion.
  9. Tetracarbonylnickel (O)

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 11
(a) Geometric isomerism:
This type of isomerism is common in heteroleptic complexes It arises dire to the different possible geometric arrangements of the li sands.
For Example
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 12

(b) Optical isomerism:
This type of isomerism arises in chiral molecules Isomers are mirror images of each other and are non-superimposable.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 13

(c) Linkage isomerism:
This type of isomerism is found in complexes that contain ambidentate ligands.
For example:
[CO (NH3)5 (NO2)]Cl2 and [CO (NH3)5 (ONO)] Cl2

(d) Coordination isomerism:
This type of isomerism arises when the ligands are inter-changed between cationic and anionic entities of different metal ions present in the complex.
[CO (NH3)6] [Cr (CN)6] and [CO (CN)6][Cr (CN)6]

(e) Ionisation isomerism:
This type of isomerism arises when a counter ion replaces a ligand within the coordination sphere. Thus, complexes that have the same composition, but furnish different ions when dissolved in water are called ionisation isomers.
For example
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 14

KSEEB Solutions

Question 9.
How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3-
(ii) [CO(NH3)3Cl3]
Answer:
(i) For [Cr(C2O4)3]3- No geometric isomer is possible as it is a bidentate ligand.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 15
(ii) [CO(NH3)3Cl3]
Two geometrical isomers are possible.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 16

Question 10.
Draw the structures of optical isomers of:
(1) [Cr(C2O4)3]3-
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Answer:
(i) [Cr(C2O4)3]3-
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 17
(ii) [PtCl2(en)2]2+
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 18

(iii) [Cr(NH3(en)2]2+
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 19
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 20

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [COCl2(en)2]+
(ii)[CO(NH3)Cl(en)2]2+
(iii) [CO(NH3)2Cl2(en)]+
Answer:
(i) [COCl2(en)2]+
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 21
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 22
(ii)[CO(NH3)Cl(en)2]2+
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 23
(iii) [CO(NH3)2Cl2(en)]+
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 24

Question 12.
Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
[Pt (NH3) (Br) (eC) (Pg)]
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 25

From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely show optical isomerisation. They do so only in presence of unsymmetric chelating agents.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous CuSO4 exists as [Cu(H2O)4]SO4. It is blue in colour due to the presence of [CU(H2O)4]2+ ions.
(i) When KF is added
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 26
(ii) When KCl is added.

[Cu (H2O)4]2+ 4Cl → [CuCl4]2- + 4H2O
In both cases, the weak field ligand water is replaced by the F and Cl ions.

KSEEB Solutions

Question 14.
What is the coordination entity formed when an excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?
Answer:
On mixing the aqueous solutions of KCN and CuSO4, the complex formed is potassium tetracyanocuprate(II). Since CN” ions are strong ligands, the complex is quite stable. It is evident from the stability constant value (K = 2·0 x 1027). It is not cleaved by H2S gas when passed through the aqueous solution and no precipitate of CuS is formed.
[CU(H2O)4]2+ + 4CN → [CU(CN)4]2- + 4H2O.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6]4-
(ii) [FeF6]3-
(Hi) [CO(C2O4)3]3-
(iv) [COF6]3-
Answer:
(i) [Fe(CN)6]4-
In the above coordination complex. Iron exists in +2 oxidation state.
Fe2+: electronic configuration is 3d6
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 27
As CN is a strong field ligand, it causes pairing of the unpaired 3d electrons.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 28
Since there are six ligands aound the central metal ion, the most feasible hybridisation is d2sp3 6 electrons pairs from CN ions occupy the six hybrid d2sp3 orbitais .
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 29
Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF6]3-
Oxidation state of Fe is + 3
Orbitais of Fe+3 ion.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 30

There are 6F ions. Thus, it will undergo d2Sp3 or sp3d2 hybridisation. As F is a weak ligand, it does not cause pairing of electrons in the 3d orbitals. Hence the most feasible hybridisation is Sp3d2.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 31
Hence the geometry of the complex is found to be octahedral.

(iii) [CO(C2O4)3]3-
Cobalt exists + 3 oxidation state in the complex.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 32
Oxalate is a weak field ligand. Therefore, it cannot cause pairing of the 3d orbital electrons.
As there are 6 ligands, hybridisation has to be either Sp3d2 or d2Sp3 hybridization.
As there are no pairing of electrons Sp3d2 is the most stable hybridisation.

2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 33
Oxalate being a bidentate ligand donates 2 pairs of electrons each.
Hence, the geometry of the complex is found to be octahedral.

(iv) [COF6]3-
cobalt exists in the +3 oxidation state.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 34
Again, fluoride ion is a weak ligand. It cannot cause the pairing of the 3d electrons. Sp3d2 hybridised orbitals of CO3+ ion are
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 35
Hence, the geometry of the complex is octahedral and paramagnetic.

KSEEB Solutions

Question 16.
Draw a figure to show the splitting of d orbitals in an octahedral crystal field.
Answer:
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 36

The splitting of d orbitals in an octahedral field takes place in such a way that dx2-y2, dz2 experience a rise in energy and form the e.g. level, dry, dyz, dzx experience a fall in energy and form t2g level.

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called the spectrochemical series.
The ligands with a small value of CFSE (Δ0) are called weak field ligands whereas those with a large value of CFSE (Δ0) are called strong field ligands.

KSEEB Solutions

Question 18.
What is crystal field splitting energy? How does the magnitude of ∆o decide the actual configuration of d orbitals in a coordination entity?
Answer:
When ligands approach a particular metal ion, the d-orbitals split into two sets, one with lower energy while the other with higher energy. The difference of energy between the two sets of orbitals is known as crystal field stabilising energy (CFSE).

  • If ∆0 < P, the configuration will be t32ge1g and it is there in case of weak field ligands and high spin complex will be formed.
  • If ∆0 > P, the configuration will be t42ge0g and it is there in case of strong-field ligands and low spin complex will be formed.

Question 19.
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
Answer:
Cr is in the +3 oxidation state i.e. d3 configuration. Also, NH3 is a weak field ligand that does not cause the pairing of the electrons in the 3d orbitals.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 37
Therefore, it undergoes d2sp3 hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, ibis paramagnetic in nature.
In [Ni CCN)4]2-, Ni exists in the +2 oxidation state.
i.e, d8 Configuration.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 38

CN is a strong field ligand. It causes the pairing of 3d orbital electrons. Then Ni2+ undergoes dsp2 hybridisation.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 39
As there are no unpaired electrons, it is diamagnetic.

Question 20.
A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain.
Answer:
In [Ni(H2O)6]2+ is a weak field ligand. Therefore there are unpaired electrons in Ni2+. In this complex, the cl electrons from the lower energy level can be excited to the higher energy level if the possibility of d- d transition is present. Hence [Ni(H2O)6]2+ is colored.

In [Ni(CN)4]2- CN is a strong field ligand. Therefore, d- d transition is not possible due to the absence of unpaired electrons (CN initiates pairing of d electrons). Hence it is colourless.

Question 21.
[Fe(CN)6]4- and [Fe(H2O)6]2- are of different colours in dilute solutions. Why?
Answer:
In both the complexes, Fe is in +2 oxidation state with cf configuration. This means that it has four unpaired electrons: Both CN ion and H20 molecules act as ligands occupy different relative positions in the spectrochemical series. They differ in crystal field splitting energy (Δ0). Quite obviously, they absorb radiations corresponding to different wavelengths and frequencies from the -visible region of light, (VIBGYOR) and the transmitted colours are also different. This means that the complexes have different colours in solutions.

Question 22.
Discuss the nature of bonding in, metal carbonyls.
Answer:
The metal-carbon bonds in’ metal carbonyls have both σ and π characteristics. A σ bond is formed when the carbonyl carbon donates a lone pair of electrons to the vacant orbital of metal. A π bond is formed by the donation of a pair of vacant anti-bonding π orbital (also known as back bonding of the carbonyl group). The σ bond strengthens the π bond and vice-versa. This synergic effect is created due to this metal-ligand bonding. This synergic effect strengthens the bond between CO and metal.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 40

KSEEB Solutions

Question 23.
Give the oxidation state, d orbital occupation, and coordination number of the central metal ion in the following complexes:
(i) K3[CO(C2O4)3]
(ii) cis-[Cr(en)2Cl2]Cl
(iii) (NH4)2[COF4]
(iv) [Mn(H2O)6]SO4
Answer:
(i) K3[CO(C2O4)3]
The central metal ion is CO
coordination number: 6
oxidation state: +3
d orbital occupation : t62g eg0

(ii) cis-[Cr(en)2Cl2]Cl
central metal: Cr
coordination number: 6
oxidation state:+3 .
d orbital occupation : t32g eg0

(iii) (NH4)2[COF4]
central metal: Co oxidation state: +2 coordination number: 4
d orbital occupation : eg4 t32g

(iv) [Mn(H2O)6]SO4
Central metal: Mn
coordination number: 6
oxidation state : + 2
d orbital occupancy : t32g eg2

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration, and coordination number. Also, give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2].3H2O
(ii) [CO(NH3)5Cl.]Cl2
(iii) CrCl3(py)3
(iv) Cs[FeCl4]
(v) K4[Mn(CN)6]
Answer:
(i) Potassium diaquadioxatatochromate (III) trihydrate oxidation state of chromium = +3
Electronic configuration: 3d3: t32g
coordination number: 6
Shape: octahedral
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 41
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 42

(ii) [Co (NH3)5Cl]Cl2
Pentaamminechloridocobalt (III) chloride
oxidation state =+3
coordination number = 6
shape: octahedral
electronic configuration : d6: t62g
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 43
Magnetic moment = 0 (∵ n = 0)

(iii) CrCl3(py)3
Trichloridopyridinechromium (III)
oxidation state = +3
electronic configuration : d3: t32g
coordihation number = 6
shape = octahedral
Stereochemistry:
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 44

Both isomers are optically active. Therefore, a total of 4 isomers exist.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 45

(iv) Cs [Fe Cl4]
IVPAC name: Caesium tetrachloroferrate(III)
oxidation state = +3
coordination number = 4
shape: tetrahedral
stereochemistry: optically inactive
Magnetic moment
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 46

(v) K4[Mn(CN)6]
Potassium hexacyanomanganate (II)
oxidation state =+2
electronic configuration = d5 : t52g
coordination number = 6
shape: octahedral
stereochemistry: optically inactive
Magnetic moment
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 47

KSEEB Solutions

Question 25.
What is meant by the stability of a coordination compound in solution? State the factor which governs the stability of complexes.
Answer:
The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium stability can be expressed quantitatively in terms of stability constant pr formation constant.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 48
For this reaction, the greater the value of the stability constant, the greater is the proportion of ML3 in the solution.
stability can be two types

(a) Thermodynamic stability:
The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.

(b) Kinetic stability:
This helps in determining the speed with which the transformation will occur to attain the state of equilibrium.
Factors that affect the stability of a complex are:

  • charge on the central metal ion: The greater the charge on the central metal ion, the greater the stability of the complex.
  • Basic nature of ligand: A more basic ligand will form a more stable complex.

(c) Presence of chelate rings: Chelation increases the stability of complexes

Question 26.
What is meant by the chelate effect? Give an example.
Answer:
When a di-or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands.
For example:
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 49
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 50

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry and
(iii) analytical chemistry
(iv) extraction/metallurgy of metals.
Answer:
Application of coordination compounds.
(a) Used in qualitative and quantitative analysis.
(b) Determination of hardness of water
(c) Extraction of Au and Ag through complex formation
(d) Refining of Ni through complex formation
(e) Use of Rhodium complex in hydrogenation ofalkene.
(f) Use of EDTA in lead poisoning etc.

KSEEB Solutions

Question 28.
How many ions are produced from the complex CO(NH3)6Cl2 in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Answer:
(iii) 3
The given complex can be written as [CO(NH3)6]Cl2 Thus, [CO (NH3)6]+ along with two Cl ions are produced.

Question 29.
Amongst the following ions which one has the highest magnetic moment value?
(i) [Cr(H2O)6]3+
(ii) [Fe(H2O)6]2+
(iii) [Zn(H2O)6]2+
Answer:
(i) n = 3
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 51

(ii) n = 4
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 52

(iii) n = 0
μ = 0
Hence [FeCH2O)6]2+ has the highest magnetic moment value.

Question 30.
The oxidation number of cobalt in K[CO(CO)4] is
(i) +1 (ii) +3 (iii) -1 (iv) -3
Answer:
We know that CG is a neutral ligand and K carries a charge of + 1. Therefore, the complex is written as K+ [Co (CO)4]. Therefore, the oxidation number of Co in the given complex is -1.
Hence (III) is correct.

Question 31.
Amongst the following, the most stable complex is
(i) [Fe(H2O)6]3+
(ii) [Fe(NH3)6]3+
(iii) [FeiC2O4)3]3-
(iv) [FeCl6]3-
Answer:
We know that the stability of a complex substance increases by chelation. Therefore, the most stable complex is [Fe (C2O4)3]3-
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 53

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO2)6]4-, [Ni(NH3)6]2+, [Ni(H2O)6]2+?
Answer:
The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increase in the spectrochemical series as follows.
H2 O < NH3 < NO2
Thus, the amount of crystal field splitting observed will be in the following order.
2nd PUC Chemistry Question Bank Chapter 9 Coordination Compounds - 54
Hence, the wavelengths of absorption in the visible region will be in the order
[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni (NO2)6]4-

KSEEB Solutions

Tili Kannada Text Book Class 6 Solutions Gadya Chapter 2 Kugutide Pakshi

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2nd PUC Physics Question Bank Chapter 7 Alternating Current

You can Download Chapter 7 Alternating Current Questions and Answers, Notes, 2nd PUC Physics Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Physics Question Bank Chapter 7 Alternating Current

2nd PUC Physics Alternating Current NCERT Text Book Questions and Answers

Question 1.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 1

Question 2.
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 2

Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 3

Question 4.
A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 4

Question 5.
In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Answer:
For Ex.7.3- The current lags behind the emf by a phase angle \(\frac { \pi }{ 2 } \), when a.c. flows through an inductor.Therefore, power absorbed by inductor over a complete cycle is,
2nd PUC Physics Question Bank Chapter 7 Alternating Current 5

KSEEB Solutions

Question 6.
Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0H, C = 32 µ F and R = 10Ω . What is the Q-value of this circuit?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 6

Question 7.
A charged 30 pF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 7

Question 8.
Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
q=6mC = 6×10-3C
Initially, energy is stored in the capacitor and is given by
2nd PUC Physics Question Bank Chapter 7 Alternating Current 8
The initial energy stored in the capacitor is shared by the inductor during oscillations of the circuit, but the total energy remains the same even at a later time.

Question 9.
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µ F is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer:
When XL = Xc, Z = R
p = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
= \(\frac { 200×200 }{ 20 }\)
= 2KW

Question 10.
A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]
Answer:
L = 200 p H = 200×10-6H
Frequency range of MW broadcast
= 800 KHz to 1200 KHz
= 8×105Hz to 12 ×105Hz .
For tuning the radio, the natural frequency of the LC circuit should be equal to the frequency of the signal.
Now,
2nd PUC Physics Question Bank Chapter 7 Alternating Current 9
Therefore, variable capacitor should be of range 88 pF to 198 pF.
2nd PUC Physics Question Bank Chapter 7 Alternating Current 10

Question 11.
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 µF, R – 40 Ω
2nd PUC Physics Question Bank Chapter 7 Alternating Current 11

(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 12

(b) At resonance, the impedance of the LCR circuit is simply equal to R
∴ at resonance, Z = R = 40 Ω
The rms value of current at resonance frequency is,
2nd PUC Physics Question Bank Chapter 7 Alternating Current 13

2nd PUC Physics Alternating Current Additional Exercises

Question 12.
An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored
(d) completely electrical (i.e., stored in the capacitor)?
(e) completely magnetic (i.e., stored in the inductor)?
(f) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Answer:
L = 20mH = 20×103 H
C = 50 µF = 50 x 10-6 F
charge on the capacitor initially,
2nd PUC Physics Question Bank Chapter 7 Alternating Current 14
(d) The total energy is shared equally between the inductor and the capacitor, when the energy stored in capacitor becomes equal to one half of the total energy. If q is charge on the capacitor at that instant, then
2nd PUC Physics Question Bank Chapter 7 Alternating Current 15

(e) When resistance is introduced, the whole of the energy will be dissipated in the form of heat. The introduction of resistance produces damped oscillations.

KSEEB Solutions

Question 13.
A coil of inductance 0.50 H and resistance 100 Q is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 16

Question 14.
Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady-state?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 17
2nd PUC Physics Question Bank Chapter 7 Alternating Current 18
In Ex-7.13, H was obtained that, when F = 50H, I = 1.289 A However, when frequency is very high (≈10kHz), the current drops to 1.08 x 10-2A (nearly zero). Hence at very high frequency in an a.c. circuit, an inductor nearly amounts to an open circuit.
For d.c. supply, F = 0
∴ Y = 2πFL = 2π x 0 x 0.50 = 0
Thus in a d.c. circuit, inductor offers no resistance i.e. it behaves like a conductor.

Question 15.
A 100 pF capacitor in series with a 40 fit resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 19
2nd PUC Physics Question Bank Chapter 7 Alternating Current 20

Question 16.
Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain. the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady-state.
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 21
2nd PUC Physics Question Bank Chapter 7 Alternating Current 22
Or φ = 0.2° ≈ 0° ;
In absence of capacitor, current through the circuit
\(=\frac { { E }_{ v } }{ R } =\frac { 110 }{ 40 } =2.75A\)
It follows that when frequency is very high (≈ 12kHz) current in the circuit is same, whether capacitor is connected in the circuit or not. In other words, at very high frequency, the capacitor behaves like a conductor in an a.c. circuit.
For d.c. supply,f=0
2nd PUC Physics Question Bank Chapter 7 Alternating Current 23
Thus in d.c, circuit, a capacitor amounts to an open circuit (offer infinite resistance).

Question 17.
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the’ current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
Answer:
From the phasor diagram, it can be obtained that when the three elements L, C and R are arranged in parallel, the impedance of the circuit is given by,
2nd PUC Physics Question Bank Chapter 7 Alternating Current 24
will be maximum and equal to R. Likewise, the current amplitude will be minimum.
L = 5H;C = 80 pF = 80 x 10-6F; R = 40q
2nd PUC Physics Question Bank Chapter 7 Alternating Current 25

Question 18.
A circuit containing an 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance . of the circuit is negligible.
(a) Obtain the current amplitude and ran values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit?
[‘Average’ implies ‘averaged over one cycle’.]
Answer:
L = 80mH = 80×10-3H
C = 60pF = 60 x 10-6F
E = 230 V
F = 50 Hz
2nd PUC Physics Question Bank Chapter 7 Alternating Current 26
2nd PUC Physics Question Bank Chapter 7 Alternating Current 27
2nd PUC Physics Question Bank Chapter 7 Alternating Current 28

Question 19.
Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Ans:
R = 15Ω
L = 80 mH = 80 x 10-3 H
2nd PUC Physics Question Bank Chapter 7 Alternating Current 29
Average power transferred to both L and C is zero. It can be checked that the average power of the circuit is equal to the average power transferred to R.

KSEEB Solutions

Question 20.
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 O is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source are the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Answer:
L = 0.12H
C = 480nF = 480 x 10-9F
R = 23 Ω
Ev= 230 V
(a) Current amplitude is maximum at resonant frequency.
The resonant frequency is given by 1
2nd PUC Physics Question Bank Chapter 7 Alternating Current 30
(b) Average power will also be maximum at resonant frequency i.e. at 4166.7 rad s-1 P =E I cos Φ
2nd PUC Physics Question Bank Chapter 7 Alternating Current 31
2nd PUC Physics Question Bank Chapter 7 Alternating Current 32

Question 21.
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 (an F, and R = 7.4 Ω It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full, width at half maximum’ by a factor of 2. Suggest a suitable way.
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 33
2nd PUC Physics Question Bank Chapter 7 Alternating Current 34

The simplest way to improve the sharpness of the resonance of the circuit is by reducing its ‘ full width at half maximum by a factor of 2 is to reduce the value of R to \(\frac { R }{ 2 } \) i.e. by making
\(R=\frac { 7.4 }{ 2 } =3.7\Omega \)

Question 22.
Answer the following questions:
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?

(b) A capacitor is used in the primary circuit of an induction coil.                               ‘

(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across

(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.

(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Ans:
(a) Yes, the first part is true but it is incorrect for r.m.s. voltage because the potential difference across various components may not be in phase.

(b) When the primary circuit of the induction coil breaks, high voltage is induced which is used to charge the capacitor. This avoids sparking in the circuit.

2nd PUC Physics Question Bank Chapter 7 Alternating Current 35

XL has zero value for d.c and a high value for a.c while Xc has an infinite value for d.c and a low value for a.c. When the voltage signal consisting of a superposition of .d.c voltage and an a.c voltage is applied to the series combination of a capacitor and an inductor, the inductor will oppose a.c, whereas it will allow easy path to d.c. Therefore, a.c voltage will appear across L. For a similar reason, d.c. the voltage will appear across C.

(d) For d.c, XL = 0. It means, insertion of an iron core in choke or inductor does not affect XL. However, in case of a.c. XL = ωL = 2ΠvL. When iron core in a choke is inserted, value of L increases. Hence, current decreases and the brightness of the lamp decreases.

For d.c, XL = 0. It means, insertion of an iron core in choke or inductor does not affect XL. However, in case of a.c. XL = coL = 2n When iron core in a choke is inserted, value of L increases. Hence, current decreases and the brightness of the lamp decreases.

(e) Choke is an inductor which reduces the supply voltage to an appropriate value to operate the tube without any power loss. However, when the resistor is used, power is dissipated in the form of heat.

Question 23.
A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 36

Question 24.
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 mV. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms-2).
Answer:
2nd PUC Physics Question Bank Chapter 7 Alternating Current 37

Question 25.
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant.
Answer:
Length of the wire line = 15 x 2 = 30 km
Resistance of the wire line, R = 30 x 0.5 = 15Ω
The voltage at which power is transmitted – 4000 V
Power transmitted, P 800 kW = 8 x 105 W
2nd PUC Physics Question Bank Chapter 7 Alternating Current 38

The plant generates power at 440 V and it has to be stepped up so that after a voltage drop of 3000 V across the wireline, the power at 4000 V is received at the substation in the town. Therefore, the step-up transformer at the plant ranges from 440 V to (4000 + 3000) V i.e.440 V – 7000 V.

KSEEB Solutions

Question 26
Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?
Answer:
Length of the wire line = 15 ×2 = 30Km.
Resistance of the wire line, R = 30 x 0.5=15q
Voltage at which power is transmitted = 40000 V
Power transmitted, P 800KW = 8 x 105 W
2nd PUC Physics Question Bank Chapter 7 Alternating Current 39

(b) Power that must be supplied by plant = 800 + 6 = 806 kW

(c) Voltage drop across the wireline = 15 x 20 = 3000 V
Therefore, percentage power loss
\(=\frac { 6 }{ 806 } \times 100=42.86%\)
Since the power loss at high voltage transmission is negligible as compared to that at low voltage, the high voltage transmission of power is preferred.

2nd PUC Physics Alternating Current Additional Questions and Answers

Question 1.
Define root mean square value of an alternating current.  (CBSE 2010)
Answer:
It is that steady current, which when passed through resistance for a given time will produce the same amount of heat as the alternating current does in the same resistance and at the same time.

Question 2.
Why d.c. voltmeter and d.c. the ammeter cannot read a.c?
Answer:
It is because the average value of a.c. is zero over a complete cycle.

Question 3.
The frequency of a.c. the source is doubled. How do R, XL, and Xc get affected?
Answer:
There is no effect of doubling the frequency of the a.c. source on R. Since XL = 2ΠfL, the value of XL becomes double on doubling the frequency of the a.c. source.
Further \({ X }_{ c }=\frac { 1 }{ 2\pi fc } \)
∴The value of Xc becomes half on doubling the frequency of the a.c. source.

Question 4.
In an oscillating LC- circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic field is
(A) \(\frac { Q }{ 2 } \)
(B) \(\frac { Q }{ \sqrt { 3 } } \)
(C) \(\frac { Q }{ \sqrt { 2 } } \)
(D) Q
Answer:
(C) \(\frac { Q }{ \sqrt { 2 } } \)
Let C be the capacitance of the capacitor. When the charge on the capacitor is Q, the energy stored inside it is given by
\({ U }_{ C }\frac { { Q }^{ 2 } }{ 2 } \)
Let Q1 be a charge on the capacitor when the energy is stored equally between the electric and magnetic fields. Then,
\({ U }_{ c }\frac { { U }_{ c } }{ 2 } \)
2nd PUC Physics Question Bank Chapter 7 Alternating Current 40

Question 5.
In a circuit, L, C, and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is
2nd PUC Physics Question Bank Chapter 7 Alternating Current 41
Answer:
φ = 45°
In LCR circuit, the phase difference between current and emf is given by
2nd PUC Physics Question Bank Chapter 7 Alternating Current 42

Question 6.
The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be
(A) X
(B) 2X
(C) 4X
(D) \(\frac { X }{ 4 } \)
Answer:
Xc = X
\( \frac { 1 }{ 2\pi fC } =X\)
When both capacitance and frequency are doubled, the reactance will become,

2nd PUC Physics Question Bank Chapter 7 Alternating Current 43

Question 7.
The coil of a choke in a circuit
(A) Increases the current
(B) Decreases the current
(C) Does not change the current
(D) Has high resistance to d.c circuit
Answer:
(B) Decreases the current

Question 8.
A normal domestic electric supply is an alternating current, whose average value is
(A) zero
(B) half the peak value
(C) the peak value multiplied by \(\left( \frac { \pi }{ 2 } \right) \)
(D) The peak value divided by \(\left( \frac { \pi }{ 2 } \right) \)
Answer:
(D) The peak value divided by \(\left( \frac { \pi }{ 2 } \right) \)

Question 9.
The equation of an a.c. voltage is V = 200 sin 50 mt . Then, the rms value of voltage is
(A) \(100\sqrt { 2 } V\)
(B) \(200\sqrt { 2 } V\)
(C) 100 V
(D) 400 V
Answer:
V0 = 200
2nd PUC Physics Question Bank Chapter 7 Alternating Current 44

Question 8.
In a series LCR – circuit, what is the potential drop across resistance at resonance, when it is operating at 220 V?
Answer:
At resonance, the impedance of the circuit is equal to the resistance in the circuit. Therefore, the potential drop across the resistance at resonance is equal to the operating i.e. 220 V.

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