Tamil Nadu 12th Economics Model Question Paper 1 English Medium

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TN State Board 12th Economics Model Question Paper 1 English Medium

General Instructions:

  1. The question paper comprises of four parts.
  2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  3. All questions of Part I, II, III and IV are to be attempted separately.
  4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each.
    These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about one or two sentences.
  6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in above three to five short sentences.
  7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in detail Draw diagrams wherever necessary.

Time: 3.00 Hours
Maximum Marks: 90

PART – I

Choose the correct answer. Answer all the questions: [20 × 1 = 20]

Question 1.
Indicate the contribution of J M Keynes to economics……..
(a) Wealth of Nations
(b) General Theory
(c) Capital
(d) Public Finance
Answer:
(b) General Theory

Question 2.
Per capita income is obtained by dividing the National income by………..
(a) Production
(b) Population of a country
(c) Expenditure
(d) GNP
Answer:
(b) Population of a country

Question 3.
Assertion (A): The Expenditure method is called outlay method.
Reason (R): This method is used only private sector.
(a) Both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation to ‘A’
(b) Both ‘A’ and ‘R’ are true but ‘R’ is not the correct explanation to ‘A’
(c) ‘A’ is true but ‘R’ is false
(d) ‘A’ is true but ‘R’ is false
Answer:
(c) ‘A’ is true but ‘R’ is false

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 4.
The component of aggregate demand is………..
(a) Personal demand
(b) Government expenditure
(c) Only export
(d) Only import
Answer:
(b) Government expenditure

Question 5.
If the Keynesian consumption function is C= 10 + 0.8 Y then, if disposable income is Rs 1000, what is amount of total consumption?
(a) Rs 0.8
(b) Rs 800
(c) Rs 810
(d) Rs 0.81
Answer:
(c) Rs 810

Question 6.
Which of the following is correctly matched:
(a) J.M. Clark – Ceteris Paribus
(b) J.M. Keynes – Psychological law of consumption
(c) R.F. Khan – Accelerator model
(d) Duesenberry – Laissez-faire
Answer:
(b) J.M. Keynes – Psychological law of consumption

Question 7.
………. inflation is in no way dangerous to the economy.
(a) Walking
(b) Running
(c) Creeping
(d) Galloping
Answer:
(a) Walking

Question 8.
Match the following and choose the correct answer by using codes given below:
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 1
Codes
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 2
Answer:
(a) A-3, B-4, C-1, D-2

Question 9.
Expansions of ATM.
(a) Automated Teller Machine
(b) Adjustment Teller Machine
(c) Automatic Teller mechanism
(d) Any Time Money
Answer:
(a) Automated Teller Machine

Question 10.
Commercial Banks create credit in favour of the……..
(a) consumers
(b) business men
(c) customers
(d) agriculturists
Answer:
(c) customers

Question 11.
Net export equals ………
(a) Export x Import
(b) Export + Import
(c) Export – Import
(d) Exports of services only
Answer:
(c) Export – Import

Question 12.
Assertion (A): Price of a commodity is measured by the amount of labour required to produce it.
Reason (R): Trade is one of the Demerit.
(a) Both ‘A’ and ‘R’ are true and ‘R’ is the correct explanation to ‘A’
(b) Both ‘A’ and ‘R’ are true but ‘R’ is not the correct explanation to ‘A’
(c) ‘A’ is true but ‘R’ is false
(d) ‘A’ is false but ‘R’ is true
Answer:
(c) ‘A’ is true but ‘R’ is false

Question 13.
Which of the following is not the member of SAARC?
(a) Pakistan
(b) Sri Lanka
(c) Bhutan
(d) China
Answer:
(d) China

Question 14.
IBRD otherwise called the………..
(a) IMF
(b) SDR
(c) SAF
(d) World Bank
Answer:
(d) World Bank

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 15.
Finance Commission determines …………
(a) The finances of Government of India
(b) The resources transfer to the states
(c) The resources transfer to the various departments
(d) None of the above
Answer:
(b) The resources transfer to the states

Question 16.
Match the following and choose the correct answer by using codes given below:
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 3
codes
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 4
Answer:
A-2, B-4. C-1, D-3

Question 17.
………. is the current increase in temperature of the Earth’s surface as well as its atmosphere.
(a) Globe warming
(b) Global warming
(c) Globe spoiled
(d) Temperature warming
Answer:
(b) Global warming

Question 18.
The supply side vicious circle of poverty suggests that poor nations remain poor because
(a) Saving remains low
(b) Investment remains low
(c) There is a lack of effective government
(d) a and b above
Answer:
(d) a and b above

Question 19.
……….planning pertains to the allocation of resources in terms of men, materials and machinery.
(a) Financial
(b) Physical
(c) Functional
(d) Structural
Answer:
(b) Physical

Question 20.
The word ‘statistics’ is used as……….
(a) Singular
(b) Plural
(c) Singular and Plural
(d) None of above
Answer:
(c) Singular and Plural

PART – II

Answer any seven question in which Question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
What do you mean by Capitalism?
Answer:
Capitalism, is total freedom and private ownership of means of production. Capitalistic economy is also termed as a free economy (Laissez faire, in Latin) or market economy where the role of the government is minimum and market determines the economic activities.

Question 22.
Write the formula for calculating GNP.
Answer:
GNP at market prices means the gross value of final goods and services produced annually in a country plus net factor income from abroad
(C + I + G + (X – M) + (R – P)

Question 23.
Write the headlines of difficulties in Measuring National Income.
Answer:
Difficulties in Measuring National Income:

  1. Transfer payments
  2. Difficulties in assessing depreciation allowance
  3. Unpaid services
  4. Income from illegal activities
  5. Production for self-consumption and changing price
  6. Capital Gains
  7. Statistical problems.

Question 24.
What is effective demand?
Answer:
The starting point of Keynes theory of employment and income is the principle of effective demand.  Effective demand denotes money actually spent by the people on products of industry. The money which entrepreneurs receive is paid in the form of rent, wages, interest and profit. Therefore effective demand equals national income.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 25.
Define “Unemployment”.
Answer:
Unemployment: When there are people, who are willing to work and able to work but cannot find suitable jobs.

Question 26.
Define average propensity to save (APS).
Answer:
Average Propensity to Save (APS):
The average propensity to save is the ratio of saving to income. APS is the quotient obtained by dividing the total saving by the total income. In other words, it is the ratio of total savings to total income. It can be expressed algebraically in the form of equation as under
APS = \(\frac{S}{Y}\)
Where, S = Saving; Y = Income

Question 27.
Define “Ceteris paribus”.
Answer:
Ceteris paribus (constant extraneous variables): The other variables such as income distribution, tastes, habits, social customs, price movements, population growth, etc. do not change and consumption depends on income alone.

Question 28.
State briefly the functions of SAARC.
Answer:
Functions of SAARC:
The main functions of SAARC are as follows.

  1. Maintenance of the co operation in the region
  2. Prevention of common problems associated with the member nations.
  3. Ensuring strong relationship among the member nations.
  4. Removal of the poverty through various packages of programmes.
  5. Prevention of terrorism in the region.

Question 29.
What are the components of GST?
Answer:
Components of GST:
The component of GST are of 3 types. They are: CGST, SGST & IGST.

  1. CGST: Collected by the Central Government on an intra-state sale (e.g. Within state/ union territory)
  2. SGST: Collected by the State Government on an intra-state sale (e.g. Within state/ union territory)
  3. IGST: Collected by the Central Government for inter-state sale (e.g. Maharashtra to Tamil Nadu)

Question 30.
What are environmental goods? Give examples.
Answer:
Environmental goods are typically non-market goods, including clear air, clean water, landscape, green transport infrastructure (footpaths, cycle ways, greenways, etc.), public parks, urban parks, rivers, mountains, forests, and beaches.

Concerns with environmental goods focus on the effects that the exploitation of ecological systems have on the economy, the well-being of humans and other species, and on the environment.

PART – III

Answer any seven question in which Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Describe the socialistic econonomy.
Answer:
Socialistic Economy (Socialism):

  1. The Father of Socialism is Karl Marx. Socialism refers to a system of total planning, public ownership and state control on economic activities.
  2. Socialism is defined as a way of organizing a society in which major industries are owned and controlled by the government.
  3. A Socialistic economy is also known as ‘Planned Economy’ or ‘Command Economy’.
  4. In a socialistic economy, all the resources are owned and operated by the government.
  5. Public welfare is the main motive behind all economic activities. It aims at equality in the distribution of income and wealth and equal opportunity for all.
  6. Russia, China, Vietnam, Poland and Cuba are the examples of socialist economies. But, now there are no absolutely socialist economies.

Question 32.
Discuss the limitations of Macro Economics.
Answer:
Macro economics suffers from certain limitations. They are:

  1. There is a danger of excessive generalisation of the economy as a whole.
  2. It assumes homogeneity among the individual units.
  3. There is a fallacy of composition. What is good of an individual need not be good for nation and vice versa. And, what is good for a country is not good for another country and at another time.
  4. Many non-economic factors determine economic activities; but they do not find place in the usual macroeconomic books.

Question 33.
Give short note on Expenditure method.
Answer:
The Expenditure Method (Outlay method):

  1. The total expenditure incurred by the society in a particular year is added together.
  2. To calculate the expenditure of a society, it includes personal consumption expenditure, net domestic investment, government expenditure on consumption as well as capital goods and net exports.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 34.
Describe the types of unemployment.
Answer:
The following are the types of unemployment.
Types of unemployment:

  1. Cyclical Unemployment
  2. Frictional Unemployment
  3. Technical Unemployment
  4. Disguised Unemployment
  5. Seasonal Unemployment
  6. Educated Unemployment
  7. Structural Unemployment

1. Cyclical Unemployment:

  • This unemployment exists during the downturn phase of trade cycle in the economy.
  • In a business cycle during the period of recession and depression, income and output fall leading to widespread unemployment.
  • It is caused by deficiency of effective demand.
  • Cyclical unemployment can be cured by public investment or expansionary monetary policy.

2. Seasonal Unemployment:

  • This type of unemployment occurs during certain seasons of the year.
  • In agriculture and agro based industries like sugar, production activities are carried out only in some seasons.
  • These industries offer employment only during that season in a year. Therefore people may remain unemployed during the off season.
  • Seasonal unemployment happens from demand side also; for example ice cream industry, holiday resorts etc.

3. Frictional Unemployment (Temporary Unemployment):

  • Frictional unemployment arises due to imbalance between supply of labour and demand for labour.
  • This is because of immobility of labour, lack of necessary skills, break down of machinery.
  • shortage of raw materials etc.
  • The persons who lose jobs and in search of jobs are also included under frictional unemployment.

4. Educated Unemployment:

  • Sometimes educated people are underemployed or unemployed when qualification does
    not match the job.
  • Faulty education system, lack of employable skills, mass student turnout and preference for white collar jobs are highly responsible for educated unemployment in India.

5. Technical Unemployment:

  • Modem technology being capital intensive requires less labourers and contributes to technological unemployment.
  • Now a days, invention and innovations lead to the adoption of new techniques there by the existing workers are retrenched.
  • Labour saving devices are responsible for technological unemployment.

6. Structural Unemployment:

  • Structural unemployment is due to drastic change in the structure of the society.
  • Lack of demand for the product or shift in demand to other products cause this type of unemployment.
  • For example rise in demand for mobile phones has adversely affected the demand for cameras, tape recorders etc.
  • So this kind of unemployment results from massive and deep rooted changes in economic structure.

7. Disguised Unemployment:

  • Disguised unemployment occurs when more people are than what is actually required.
  • Even if some workers are withdrawn, production does not suffer.
  • This type of unemployment is found in agriculture.
  • A person is said to be disguisedly by unemployed if his contribution to output is less than what he can produce by working for normal hours per day.
  • In this situation, marginal productivity of labour is zero or less or negative.

Question 35.
Differentiate autonomous and induced investment.
Answer:

Sl.No Autonomous Investment Induced Investment
1 Independent Planned
2 Income inelastic Income elastic
3 Welfare motive Profit Motive

Question 36.
What is money supply?
Answer:

  1. Money supply means the total amount of money in an economy.
  2. It refers to the amount of money which is in circulation in an economy at any given time.
  3. Money supply plays a crucial role in the determination of price level and interest rates.
  4. Money supply viewed at a given point of time is a stock and over a period of time it is a flow.

Question 37.
Distinguish between money market and capital market.
Answer:

S.No. Money Market Capital Market
1. Money market is the mechanism through which sthort term funds are loaned and borrowed. It designates financial instittutions which handle the purchase, sale and transfer of short term credit instruments. Capital Market is a part of financial system which is concerned with raising capital by dealing in shares, bonds and other long term investments.
2. Commercial banks, acceptance houses, Non Banking Financial Institutions and the Central Bank are the institutions catering to the requirements of short term funds in the money Market. The market where investment instruments like bonds, equities and mortgages are traded is known as the capital market.

Question 38.
List out the achievements of ASEAN.
Answer:
The ASEAN Declaration states the aims and purposes of the Association as:

  1. To accelerate the economic growth, social progress and cultural development in the region;
  2. To promote regional peace and stability and adherence to the principles of the United Nations Charter;
  3. To promote cooperation among the members of ASEAN through the exchange of knowledge and experience in the field of public sector auditing.
  4. To provide a conducive environment and facilities for research, training, and education among the members
  5. To serve as a centre of information and as an ASEAN link with other international organizations.

Question 39.
Give two examples for direct tax.
Answer:

  1. Equity: Direct taxes are progressive i.e. rate of tax varies according to tax base. For example, income tax satisfies the canon of equity.
  2. Certainity: Canon of certainty can be ensured by direct taxes. For example, an income tax payer knows when and at what rate he has to pay income tax.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 40.
Define economic planning.
Answer:
Economic Planning is “collective control or suppression of private activities of production and exchange”. – Robbins
“Economic Planning in the widest sense is the deliberate direction by persons in-charge of large resources of economic activity towards chosen ends”. – Dalton

PART – IV

Answer all the questions. [7 x 5 = 35]

Question 41 (a).
Compare the feature among Capitalism, Secularism and Mixedism.
Answer:
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 5

[OR]

(b) What are the difficulties involved in the measurement of national income?
Answer:
Difficulties in Measuring National Income:
In India, a special conceptual problem is posed by the existence of a large, unorganised and non-monetised subsistence sector where the barter system still prevails for transacting goods and services. Here, a proper valuation of output is very difficult.

Transfer payments:

  1. Government makes payments in the form of pensions, unemployment allowance, subsidies, etc. These are government expenditure.
  2. But they are not included in the national income.
  3. Because they are paid without adding anything to the production processes.
  4. During a year, Interest on national debt is also considered transfer payments because it is paid by the government to individuals and firms on their past savings without any productive work.

Difficulties in assessing depreciation allowance:

  1. The deduction of depreciation allowances, accidental damages, repair and replacement charges from the national income is not an easy task.
  2. It requires high degree of judgment to assess the depreciation allowance and other charges.

Unpaid services:

  1. A housewife renders a number of useful services like preparation of meals, serving, tailoring, mending, washing, cleaning, bringing up children, etc.
  2. She is not paid for them and her services are not directly included in national income.

Income from illegal activities:

  1. Income earned through illegal activities like gambling, smuggling, illicit extraction of . liquor, etc., is not included in national income.
  2. Such activities have value and satisfy the wants of the people but they are not considered as productive from the point of view of society.

Production for self-consumption and changing price:

  1. Farmers keep a large portion of food and other goods produced on the farm for self consumption.
  2. The problem is whether that part of the produce which is not sold in the market can be included in national income or not.

Capital Gains:

  1. The problem also arises with regard to capital gains.
  2. Capital gains arise when a capital asset such as a house, other property, stocks or shares, etc. is sold at higher price than was paid for it at the time of purchase.
  3. Capital gains are excluded from national income.

Statistical problems:

  1. There are statistical problems, too. Great care is required to avoid double counting. Statistical data may not be perfectly reliable, when they are compiled from numerous sources.
  2. Skill and efficiency of the statistical staff and cooperation of people at large are also equally important in estimating national income.

Question 42 (a).
Write short note on the implications of Say’s law.
Answer:
Implications of Say’s Law:

  1. There is no possibility for over production or unemployment.
  2. If there exist unutilized resources in the economy, it is profitable to employ them up to the point of full employment. This is true under the condition that factors are willing to accept rewards on a par with their productivity.
  3. As automatic price mechanism operates in the economy, there is no need for government intervention. (However, J.M. Keynes emphasized the role of the State)
  4. Interest flexibility brings about equality between saving and investment.
  5. Money performs only the medium of exchange function in the economy, as people will not hold idle money.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

[OR]

(b) State the propositions of Keynes’s Psychological Law of Consumption.
Answer:
Propositions of the Law:
This law has three propositions:
(i) When income increases, consumption expenditure also increases but by a smaller amount. The reason is that as income increases, we wants are satisfied side by side, so that the need to spend more on consumer goods diminishes. So, the consumption expenditure increases with increase in income but less than proportionately.

(ii) The increased income will be divided in some proportion between consumption expenditure and saving. This follows from the first proposition because when the whole of increased income is not spent on consumption, the remaining is saved. In this way, consumption and saving move together.

(iii) Increase in income always leads to an increase in both consumption and saving. This means that increased income is unlikely to lead to fall in either consumption or saving. Thus with increased income both consumption and saving increase.

Question 43 (a).
Bring out the methods of credit control.
Answer:
Methods of Credit Control:
1. Bank Rate Policy:
The bank rate is the rate at which the Central Bank of a country’ is prepared to re-discount the first class securities.

2. Open Market Operations:

  • In narrow sense, the Central Bank starts the purchase and sale of Government securities in the money market.
  • In Broad Sense, the Central Bank purchases and sells not only Government securities but also other proper eligible securities like bills and securities of private concerns.

3. Variable Reserve Ratio:
(i) Cash Reserves Ratio:

  • Under this system the Central Bank controls credit by changing the Cash Reserves Ratio.
  • For example, if the Commercial Banks have excessive cash reserves on the basis of which they are creating too much of credit,this will be harmful for the larger interest of the economy.
  • So it will raise the cash reserve ratio which the Commercial Banks are required to maintain with the Central Bank.

(ii) Statutory Liquidity Ratio:

  • Statutory Liquidity Ratio (SLR) is the amount which a bank has to maintain securities.
  • The quantum is specified as some percentage of the total demand and time liabilities (i.e., the liabilities of the bank which are payable on demand anytime, and those liabilities which are accruing in one month’s time due to maturity) of a bank.

[OR]

(b) Briefly explain the gains from International Trade Categories.
Answer:
Gains from International Trade:

  • International trade helps a country to export its surplus goods to other countries and secure a better market for it.
  • Similarly, international trade helps a country to import the goods which cannot be produced at all or can be produced at a higher cost.
  • The gains from international trade may be categorized under four heads.

I. Efficient Production:
International trade enables each participatory country to specialize in the production of goods in which it has absolute or comparative advantages. International specialization offers the following gains.

  1. Better utilization of resources.
  2. Concentration in the production of goods in which it has a comparative advantage.
  3. Saving in time.
  4. Perfection of skills in production.
  5. Improvement in the techniques of production.
  6. Increased production.
  7. Higher standard of living in the trading countries.

II. Equalization of Prices between Countries:
International trade may help to equalize prices in all the trading countries.

  1. Prices of goods are equalized between the countries (However, in reality it has not happened).
  2. The difference is only with regard to the cost of transportation.
  3. Prices of factors of production are also equalized (However, in reality it has not happened).

III. Equitable Distribution of Scarce Materials:
International trade may help the trading countries to have equitable distribution of scarce resources.

IV. General Advantages of International Trade:

  1. Availability of variety of goods for consumption.
  2. Generation of more employment opportunities.
  3. Industrialization of backward nations.
  4. Improvement in relationship among countries (However, in reality it has not happened).
  5. Division of labour and specialisation.
  6. Expansion in transport facilities.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 44 (a).
Write a note on SAARC.
Answer:
South Asian Association For Regional Co-Operation (SAARC):

  1. The South Asian Association for Regional Co-operation (SAARC) is an organisation of South Asian nations, which was established on 8 December 1985 for the promotion of economic and social progress, cultural development within the South Asia region and also for friendship and co-operation with other developing countries.
  2. The SAARC Group (SAARC) comprises of Bangaladesh, Bhutan, India, The Maldives, Nepal, Pakistan and Sri Lanka.
  3. In April 2007, Afghanistan became its eighth member.
  4. The basic aim of the organisation is to accelerate the process of economic and social development of member states through joint action in the agreed areas of cooperation.
  5. The SAARC Secretariat was established in Kathmandu (Nepal) on 16th January 1987.
  6. The first SAARC summit was held at Dhaka in the year 1985.
  7. SAARC meets once in two years. Recently, the 20th SAARC summit was hosted by Sri Lanka in 2018.

[OR]

(b) Briefly explain facilities offered by IMF.
Answer:
Facilities offered by IMF:
The Fund has created several new credit facilities for its members. Chief among them are:
(i) Basic Credit Facility:

  • The IMF provides financial assistance to its member nations to overcome their temporary difficulties relating to balance of payments.
  • A member nation can purchase from the Fund other currencies or SDRs, in exchange for its own currency, to finance payment deficits.
  • The loan is repaid when the member repurchases its own currency with other currencies or SDRs.
  • A member can unconditionally borrow from the Fund in a year equal to 25% of its quota.
  • This unconditional borrowing right is called the reserve tranche.

(ii) Extended Fund Facility:

  • Under this arrangement, the IMF provides additional borrowing facility up to 140% of the member’s quota, over and above the basic credit facility.
  • The extended facility is limited for a period up to 3 years and the rate of interest is low.

(iii) Compensatory Financing Facility:

  • In 1963, IMF established compensatory financing facility to provide additional financial assistance to the member countries, particularly primary producing countries facing shortfall in export earnings.
  • In 1981, the coverage of the compensatory financing facility was extended to payment problem caused by the fluctuations in the cost of cereal inputs.

(iv) Buffer Stock Facility:

  • The buffer stock financing facility was started in 1969.
  • The purpose of this scheme was to help the primary goods (food grains) producing countries to finance contributions to buffer stock arrangements for the stabilisation of primary product prices.

(v) Supplementary Financing Facility:
Under the supplementary financing facility, the IMF makes temporary arrangements to provide supplementary financial assistance to member countries facing payments problems relating to their present quota sizes.

(vi) Structural Adjustment Facility:

  • The IMF established Structural Adjustment Facility (SAF) in March 1986 to provide additional balance of payments assistance on concessional terms to the poorer member countries.
  • In December 1987, the Enhanced Structural Adjustment Facility (ESAF) was set up to augment the availability of concessional resources to low income countries.
  • The purpose of SAF and ESAF is to force the poor countries to undertake strong macroeconomic and structural programmes to improve their balance of payments positions and promote economic growth.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 45 (a).
What are the functions of a modern state?
Answer:
Functions of Modern State:
The modem state is a welfare state and not just police state. The state assumes greater roles by creating economic and social overheads, ensuring stability both internally and externally, conserving resources for sustainable development and so on.
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 6
(i) Defence:

  1. The primary function of the Government is to protect the people from external aggression and internal disorder.
  2. The government has to maintain adequate police and military forces and render protective
    services.

(ii) Judiciary:

  1. Rendering justice and settlement of disputes are the concern of the government.
  2. It should provide adequate judicial structure to render justice to all classes of citizens,

(iii) Enterprises:

  1. The regulation and control of private enterprise fall under the purview of the modem State.
  2. Ownership of certain enterprises and operating them successfully are the responsibilities of the government.

(iv) Social Welfare:
It is the duty of the state to make provisions for education, social security, social insurance, health and sanitation for the betterment of the people in the country.

(v) Infrastructure:
Modem States have to build the base for the economic development of the country by creating social and economic infrastructure.

(vi) Macro-economic policy:
The Government has to administer fiscal policy and monetary policy to achieve macro¬economic goals.

(vii) Social Justice:

  1. During the process of growth of an economy, certain sections of the society gain at the cost of others.
  2. The Government needs to intervene with fiscal measures to redistribute income.

(viii) Control of Monopoly:
Concentration of economic power is another evil to be corrected by the Government. So, the state intervenes through control of monopolies and restrictive trade practices to curb concentration of economic power.

[OR]

(b) What are the causes of water pollution?
Answer:
Water pollution is caused due to several reasons. Here are the few major causes of water pollution:
(i) Discharge of sewage and waste water:

  1. Sewage, garbage and liquid waste of households, agricultural runoff and effluents from factories are discharged into lakes and rivers.
  2. These wastes contain harmful chemicals and toxins which make the water poisonous for aquatic animals and plants.

(ii) Dumping of solid wastes:
The dumping of solid wastes and litters in water bodies cause huge problems.

(iii) Discharge of industrial wastes:
Industrial waste contains pollutants like asbestos, lead, mercury, grease oil and petrochemicals, which are extremely harmful to both people and environment.

(iv) Oil Spill:
Sea water gets polluted due to oil spilled from ships and tankers while travelling. The spilled oil does not dissolve in water and forms a thick sludge polluting the water.

(v) Acid rain:

  1. Acid rain is pollution of water caused by air pollution.
  2. When the acidic particles caused by air pollution in the atmosphere mix with water vapor, it results in acid rain.

(vi) Global warming:
Due to global warming, there is an increase in water temperature as a result aquatic plants and animals are affected.

(vii) Eutrophication:

  1. Eutrophication is an increased level of nutrients in water bodies.
  2. This results in bloom of algae in water.
  3. It also depletes the oxygen in water which negatively affects fish and other aquatic animal population.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

Question 46 (a).
Explain different sources of e-waste.
Answer:
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 7

[OR]

(b) What are the non-economic factors determining development?
Answer:
Non-Economic. Factors:

  1. Human Resource
  2. Technical Know-how
  3. Political Freedom
  4. Social Organization
  5. Corruption free administration
  6. Desire for Development
  7. Moral, ethical and social values
  8. Casino Capitalism
  9. Patrimonial Capitalism

Non-Economic Factors: ‘Economic Development has much to do with human endowments, social attitudes, political conditions and historical accidents. Capital is a necessary but not a sufficient condition of progress.
(i) Human Resources:

  • Human resource is named as human capital because of its power to increase productivity and thereby national income.
  • There is a circular relationship between human development and economic growth.
  • A healthy, educated and skilled labour force is the most important productive asset.
  • Human capital formation is the process of increasing knowledge, skills and the productive capacity of people.

(ii) Technical Know-how:
As the scientific and technological knowledge advances, more and more sophisticated techniques steadily raise the productivity levels in all sectors.

(iii) Political Freedom:
The process of development is linked with the political freedom.

(iv) Social Organization:
People show interest in the development activity only when they feel that the fruits of development will be fairly distributed.

(v) Corruption free administration:

  • Corruption is a negative factor in the growth process.
  • Unless the countries root-out corruption in their administrative system, the crony capitalists and traders will continue to exploit national resources.

(vi) Desire for development:
The pace of economic growth in any country depends to a great extent on people’s desire for development.

(vii) Moral, ethical and social values:

  • These determine the efficiency of the market, according to Douglas C. North.
  • If people are not honest, market cannot function.

(viii) Casino Capitalism:
If People spend larger proportion of their income and time on entertainment liquor and other illegal activities, productive activities may suffer, according to Thomas Piketty.

(ix) Patrimonial Capitalism:
If the assets are simply passed on to children from their parents, the children would not work hard, because the children do not know the value of the assets.

Question 47 (a).
Describe different types of Planning.
Answer:
Tamil Nadu 12th Economics Model Question Paper 1 English Medium 8
(i) Democratic Vs Totalitarian:
A form of rule in which the government attempts to maintain ‘total’ control over society, including all aspects of the public and private lives of its citizens.

(ii) Centralized Vs Decentralized:

  • Under centralized planning, the entire planning process in a country is under a central planning authority.
  • This authority formulates a central plan, fixes objectives, targets and priorities for every sector of the economy.
  • In other words, it is called ‘planning from above’.

(iii) Planning by Direction Vs Inducement:
Under planning by direction, there is a central authority which plans, directs and orders the execution of the plan in accordance with pre-determined targets and priorities.

(iv) Indicative Vs Imperative Planning:

  • Indicative planning is peculiar to the mixed economies. It has been in practice in France since the Monnet Plan of 1947-50.
  • In a mixed economy, the private sector and the public sector work together.
  • Under this plan, the outline of plan is prepared by the Government.
  • Then it is discussed with the representatives of private management, trade unions, consumer groups, finance institutions and other experts.

(v) Short, Medium and Long term Planning:

  • Short-term plans are also known as ‘controlling plans’.
  • They encompass the period of one year, therefore, they are also known as ‘annual plans’.

(vi) Financial Vs Physical Planning:
Financial planning refers to the technique of planning in which resources are allocated in terms of money while physical planning pertains to the allocation of resources in terms of men, materials and machinery.

(vii) Functional Vs Structural Planning:
Functional planning refers to that planning which seeks to remove economic difficulties by directing all the planning activities within the existing economic and social structure.

(viii) Comprehensive Vs Partial Planning:
General planning which concerns itself with the major issues for the whole economy is known as comprehensive planning whereas partial planning is to consider only the few important sectors of the economy.

Tamil Nadu 12th Economics Model Question Paper 1 English Medium

[OR]

(b) Discuss the Economic planning of Democratic planning and Totalitarian planning.
Answer:
Democratic Vs Totalitarian:

  1. Democratic planning implies planning within democracy.
  2. People are associated at every step in the formulation and implementation of the plan.
  3. A democratic plan is characterized by the widest possible consultations with the various state governments and private enterprises at the stage of preparation.
  4. The plan prepared by the Planning Commission is not accepted as such.
  5. It can be accepted, rejected or modified by the Parliament of the country.
  6. Under totalitarian planning, there is central control and direction of all economic activities in accordance with a single plan.
  7. Consumption, production, exchange, and distribution are all controlled by the state. In authoritarian planning, the planning authority is the supreme body.
  8. It decides about the targets, schemes, allocations, methods and procedures of implementation of the plan.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5

Question 1.
Find the modulus of the following complex numbers.
(i) \(\frac{2i}{3+4i}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 1

(ii) \(\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 2
Modulus of z = |z| = \(\sqrt{4+4}\)
= √8
= 2√2

(iii) |(1 – i)10| = (|1 – i|)10
= \((\sqrt{1+1})^{10}=(\sqrt{2})^{10}=2^{5}=32\)
(iv) |2i(3 – 4i) (4 – 3i)|
= |2i| |3 – 4i| |4 – 3i|
= \(2 \sqrt{9+16} \sqrt{16+9}\)
= 2 × 5 × 5
= 50

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 2.
For any two complex numbers z1 and z2, such that |z1| = |z2| = 1 and z1 z2 ≠ -1, then show that \(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\) is real number.
Solution:
Given |z1| = |z2| = 1 and z1 z2 ≠ 1
|z1|² = 1 |z2|² = 1
z1 \(\bar{z}_{1}\) = 1 similarly z2 \(\bar{z}_{2}\) = 1
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 3
Since z = \(\bar{z}\), it is a real number.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 3.
Which one of the points 10 – 8i, 11 + 6i is closest to 1 + i.
Solution:
A (1 + i), B (10 – 8i), C (11 + 6i)
|AB| = |(10 – 8i) – (1 + i)|
= |10 – 8i – 1 – i|
= |9 – 9i|
= \(\sqrt{81+81}\)
= \(\sqrt{162}\)
= 9(1.414)
= 12.726
CA = |(11 + 6i) – (1 + i)|
= |11 + 6i – 1 – i|
= |10 + 5i|
= \(\sqrt{100+25}\)
= \(\sqrt{125}\)
C (11 + 6i) is closest to the point A (1 + i)

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 4.
If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.
Solution:
given |z| = 3
|z + 6 – 8i| ≤ |z| + |6 – 8i|
= 3 + \(\sqrt{6^2+8^2}\)
= 3 + \(\sqrt{100}\)
= 3 + 10 = 13
∴ |z + 6 – 8i| ≤ 13 ……….. (1)
|z + 6 – 8i| ≥ ||z| – |-6 + 8i||
= |3 – 10|
= |-7| = 7
∴ |z + 6 – 8i| ≥ 7 ………… (2)
from 1 and 2
we get 7 ≤ |z + 6 – 8i| ≤ 13
hence proved.

Question 5.
If |z| = 1, show that 2 ≤ |z² – 3| ≤ 4.
Solution:
|z| = 1 ⇒ |z|2 = 1
||z1| – |z2|| ≤ |z1 + z2| ≤ |z1| + |z2|
||z|2 – |-3|| ≤ |z2 – 3| ≤ |z|2 + |-3|
|1 – 3| ≤ |z2 – 3| ≤ 1 + 3
2 ≤ |z2 – 3| ≤ 4

Question 6.
If |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12
Solution:
Given |z| = 2
|z + 6 + 8i| = |z| + |6 + 8i|
= 2 + \(\sqrt{6^2+8^2}\)
= 2 + \(\sqrt{100}\)
= 2 + 10
= 12
∴ |z + 6 + 8i| ≤ 12 ……….. (1)
|z + 6 + 8i| ≥ ||z| – |-6 – 8i||
= |2 – 10|
= |-8|
= 8
|z + 6 + 8i| ≥ 8 ………… (2)
From 1 and 2 we get
8 ≤ |z + 6 + 8i| ≤ 12
Hence proved

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 7.
If z1 z2 and z3 are three complex numbers such that |z1| = 1, |z1| = 2, |z3| = 3 and |z1 + z2 + z3| = 1 show that |9z1 z2 + 4z1 z3 + z2 z3| = 6.
Solution:
|z1| = 1, |z1| = 2, |z3| = 3
|z1 + z2 + z3| = 1
Now |9z1 z2 + 4z1 z3 + z2 z3|
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 4
Hence proved.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 8.
If the area of the triangle formed by the vertices z, iz, and z + iz is 50 square units, find the value of |z|.
Solution:
The given vertices are z, iz, z + iz ⇒ z, iz are ⊥r to each other.
Area of triangle = \(\frac { 1 }{ 2 }\) bh = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |iz| = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |z| = 50
⇒ |z|2 = 100
⇒ |z| = 10
Aliter:
Given the area of triangle = 50 sq. unit
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 5
x² + y² = 100
|z|² = 100
|z| = 10

Question 9.
Show that the equation z³ + 2 \(\bar {z}\) = 0 has five solutions.
Solution:
Given z³ + 2 \(\bar {z}\) = 0
z³ = -2 \(\bar {z}\)
|z³| = |-2| |\(\bar {z}\)|
|z|³ = 2|z| [∵ |z| = |\(\bar {z}\)|
|z|³ – 2 |z| = 0
|z| [|z|² – 2] = 0
|z| = 0 |z|² = 2
z\(\bar {z}\) = 2
z = \(\frac{2}{\bar {z}}\) = ± √2 [∵ \(\bar {z}\) = \(\frac{-z^3}{2}\) ]
z = \(\frac{2}{(\frac{z^3}{-2})}\)
z4 = 4
It has 4 non zero solutions.
∴ Including z = 0 we have 5 solutions.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

Question 10.
Find the square roots of
(i) 4 + 3i
Solution:
|4 + 3i| = \(\sqrt {4^2+3^2}\) = \(\sqrt {16+9}\)
\(\sqrt {25}\) = 5
Let \(\sqrt {4+3i}\) = a + ib
squaring on both sides
4 + 3i = (a + ib)²
4 + 3i = (a² – b²) + 2 jab
Equating real and imaginary parts
a² – b² = 4, 2ab = 3
(a² + b²)² = (a² – b²)² + 4a² b²
= (4)² + (3)²
= 16 + 9 = 25
∴ a² + b² = 5
Solving a² – b² = 4 and a² + b² = 5.
we get a² = \(\frac {9}{2}\) , b² = \(\frac {1}{2}\)
a = ±\(\frac {3}{√2}\) and b = ±\(\frac {1}{√2}\)
∴ \(\sqrt {4 + 3i}\) = a + ib
= ±(\(\frac {3}{√2}\) + ±\(\frac {i}{√2}\))
Aliter:
Square root of 4 + 3i
formula method
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 6

(ii) -6 + 8i
Solution:
Let \(\sqrt {-6 + 8i}\) = a + ib
Squaring on both sides
-6 + 8i = (a + ib)²
-6 + 8i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -6 and 2ab = 8
Now (a² + b²)² = (a² – b²)² + 4a²b²
= (-6)² + (8)²
= 36 + 64 = 100
∴ a + b² = 10
Solving a² – b² = -6 and a² + b² = 10
we get 2a² = 4, b² = 8
a² = 2, b² = ±2√2
a = ±√2
∴ \(\sqrt {-6 + 8i}\) = ±√2 ± i 2√2
= ±(√2 + i 2√2)
Aliter:
square root of -6 + 8i
let a + ib = -6 + 8i
a = -6, b = 8
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 7

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

(iii) -5 – 12i
Solution:
Let \(\sqrt{-5-12 i}\) = a + ib
Squaring on both sides
-5 – 12i = (a+ib)²
-5 – 12i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -5, 2ab = -12
(a² + b²)² = (a² – b²)² + 4a²b²
= (-5)² + (-12)² = 169
∴ a² + b² = 13
Solving a²- b² = -5 and a² + b² = 13
we get a² = 4, b² = 9
a = ±2, b = ±3
Since 2ab = -12 < 0, a, b are of opposite signs.
∴ When a = ±2, b = ±3
Now \(\sqrt{-5-12 i}\) = ± (2 – 3i)
Aliter
Square root of -5 – 12i
Let a + ib = -5 – 12i
a = -5, b = -12
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5 8

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7

Question 1.
Write in polar form of the following complex numbers.
(i) 2 + i2 √3
Solution:
Let z + i2√3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 2 (+ve)
r sin θ = 2√3 (+ve)
r² cos² θ + r² sin² θ = (2)² + (2√3)²
r² = 4 + 12 = 16
|z| = r = 4
since cos cos θ and sin θ are positive ‘θ’ lies in 1st quadrant.
cos θ = \(\frac{1}{2}\), sin θ = \(\frac{√3}{2}\)
∴ θ = sin θ = \(\frac{π}{3}\) (or) θ = tan-1 |\(\frac{y}{x}\)|
= tan-1 |\(\frac{2√3}{2}\)|
= tan-1 √3 = \(\frac{π}{3}\)
∴ argument = 2kπ + \(\frac{π}{3}\)
∴ Polar form is z = r (cos θ + i sin θ)
2 + 2i√3 = 4 (cos (2kπ + \(\frac{π}{3}\)) + i sin(2kπ +\(\frac{π}{3}\))) k ∈ z

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

(ii) 3 – i √3
Solution:
Let z = 3 – i √3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 3 (+ve)
r sin θ = -√3 (-ve)
r² cos² θ + r² sin² θ = (3)² + (-√3)²
r² = 9 + 3 = 12
|z| = r = 2√3
since cos cos θ positive and sin θ in -ve so lies in IV quadrant.
cos θ = \(\frac{√3}{2}\), sin θ = \(\frac{-1}{2}\), θ = \(\frac{-π}{6}\)
argument = 2kπ – \(\frac{π}{6}\), k ∈ Z
polar from z = r(cos θ + i sin θ)
3 – i√3 = 2√3 (cos (2kπ – \(\frac{π}{6}\)) + i sin(2kπ – \(\frac{π}{6}\))) k ∈ Z

(iii) -2 – i 2 = r (cos θ + i sin θ)
Solution:
Let z = -2 – i2 = r(cos θ + i sin θ)
equating real and imaginary parts
r cos θ = -2
r sin θ = -2
r² cos² θ + r² sin² θ = (-2)² + (-2)²
r² = 4 + 4 = 8
r² = 8
|z| = r = 2√2
cos θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\), sin θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\)
since cos θ and sin θ both are in -ve so lies in III quadrant.
argument = 2kπ – 3\(\frac{π}{4}\)
as θ = \(\frac{π}{4}\) – π = –\(\frac{3π}{4}\)
polar from z = r(cos θ + i sin θ)
-2 – i2 = 2√2 (cos (2kπ – \(\frac{3π}{4}\)) + i sin(2kπ – \(\frac{3}{4}\))) k ∈ Z

(iv) \(\frac{i-1}{cos{\frac{π}{3}}+isin{\frac{π}{3}}}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 1
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 2

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 2.
Find the rectangular form of the complex numbers
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 3
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 4

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 5
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 6

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 3.
\(\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \cdots\left(x_{n}+i y_{n}\right)=a+i b\), show that
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 7
Solution:
Let (x1 + iy1) (x2 + iy2) (x3 + iy3) …… (xn + iyn) = a + ib
Taking modulus
|(x1 + iy1) (x2 + iy2) (x3 + iy3) …… (xn + iyn)| = |a + ib|
|x1 + iy1| |x2 + iy2| |x3 + iy3| …… |xn + iyn| = |a + ib|
\(\sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \ldots \sqrt{x_{n}^{2}+y_{n}^{2}}\) = \(\sqrt{a^2+b^2}\)
Squaring on both sides
\(\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)\) = a² + b²
Hence proved

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

(ii) \(\sum_{r=1}^{n}\) tan-1 (\(\frac{y_r}{x_r}\)) = tan-1 (\(\frac{b}{a}\)) + 2kπ, k ∈ Z
Solution:
Let (x1 + iy1) (x2 + iy2) (x3 + iy3) …… (xn + iyn) = a + ib
Taking arguments
arg [(x1 + iy1) (x2 + iy2) (x3 + iy3) …… (xn + iyn)] = arg (a + ib)
arg (x1 + iy1) + arg(x2 + iy2) + arg (x3 + iy3) …… + arg(xn + iyn) = arg(a + ib)
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 8

Question 4.
Given \(\frac{1+z}{1-z}\) = cos 2θ + i sin 2θ, show that To prove that z = i tan θ.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 9
Squaring on both sides
(1 + x)² + y² = (1 – x)² + y²
1 + 2x + x² + y² = 1 – 2x + x² +y²
x = 0
∴ z = 0 + iy = iy
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 10
∴ y = tan θ
hence z = iy
z = i tan θ

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 5.
If cos α + cos β + cos γ = sin α + sin β + sin γ = 0. then show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ) and
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ).
Solution:
Let a = cos α + i sin α = e
b = cos β + i sin β = e
c = cos γ + i sin γ = e
a + b + c = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)
⇒ a + b + c = 0 + i 0
⇒ a + b + c = 0
If a + b + c = 0 then a3 + b3 + c3 = 3abc
Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7 Q5
(cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ) = 3 [cos (α + β + γ) + i sin (α + β + γ)]
(cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ) = 3 cos (α + β + γ) + i 3sin(α + β + γ)
Equating real and Imaginary parts
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 6.
If z = x + iy and arg \(\left(\frac{z-i}{z+2}\right)\) = \(\)\frac{π}{4}, then show that x² + y³ + 3x – 3y + 2 = 0.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 11
2y – x – 2 = x² + 2x + y² – y
x² + y² + 2x + x – y – 2y + 2 = 0
⇒ x² + y² + 3x – 3y + 2 = 0
Hence proved

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1.
Write the following in the rectangular form:
(i) \(\overline { (5+9i)+(2-4i) } \)
Solution:
\(\overline { (5+9i)+(2-4i) } \)
= \(\overline {(5+9i)} \) + \(\overline {(2-4i)} \)
= 5 – 9i + 2 + 4i
= 7 – 5i

(ii) \(\frac {10-5i}{6+2i} \)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 1

(iii) \(\overline {3i} + \frac{2}{2-i}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 2

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re(\(\frac {1}{z} \))
Answer:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 3

(ii) Re(i\(\bar{z}\)) = Re[i(\(\overline{x+i y}\))]
= Re(ix + y)
= y
(iii) Im(3z + 4\(\bar{z}\) – 4i)
= Im (3(x + iy) + 4(x – iy) – 4i)
= Im (3x + 3iy + 4x – 4iy – 4i)
= Im (3x + 4 + i (3y – 4y – 4)
= Im (3x + 4x + i(-y – 4))
= Im [7x + i(-y – 4)]
= -y – 4
= -(y + 4)

Question 3.
If z1 = 2 – i and z2 = -4 + 3i, find the inverse of z1, z2 and \(\frac {z_1}{z_2} \)
Solution:
z1 = 2 – i, z2 = -4 + 3i
z1 z2 = (2 – i) (-4 + 3i)
= -8 + 3 + 4i + 6i
= -5 + 10i
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 4
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 5

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 4.
The complex numbers u, v, and w are related by \(\frac {1}{u}\) = \(\frac {1}{v}\) + \(\frac {1}{w}\). If v = 3 – 4i and w = 4 + 3i, find u in rectangular form.
Solution:
v = 3 – 4i, w = 4 + 3i
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 6

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 5.
Prove the following properties:
(i) z is real if and only if z = \(\overline {z}\)
Solution:
z is real iff z = \(\bar{z}\)
Let z = x + iy
z = \(\bar{z}\)
⇒ x + iy = x – iy
⇒ 2iy = 0
⇒ y = 0
⇒ z is real.
z is real iff z = \(\bar{z}\)

(ii) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2i}\)
Solution:
let z = x + iy
\(\overline {z}\) = x – iy
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 7
Hence proved

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 6.
Find the least value of the positive integer n for which (√3 + i)n (i) real, (ii) purely imaginary.
Solution:
Given (√3 + i)n
= (√3)² + 2i √3 + (i)²
= 3 + 2i √3 – 1
= 2 + 2i √3
= 2(1 + i√3)
put n = 3 or 4 or 5
then real part is not possible
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 8

which is purely real ∴ n = 6

(ii) (√3 + i)n
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 9
which is purely imaginary
∴ n = 3

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

Question 7.
Show that
(i) (2 + i√3)10 – (2 – i√3)10 is purely imaginary.
Solution:
Let z = (2 + i√3)10 – (2 – i√3)10
Let Z
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 10
= (2 – i√3)10 – (2 + i√3)10
= -[(z + i√3)10 – (2 – i√3)10]
= -z
(2 + i√3)10 – (2 – i√3)10 is purely imaginary

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4

(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}\) + \(\left(\frac{20-5 i}{7-6 i}\right)^{12}\) is real
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.4 11
∴ z is real.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6

Question 1.
If 2 = x + iy is a complex number such that \(\left|\frac{z-4 i}{z+4 i}\right|\) = 1 show that the locus of z is real axis.
Solution:
Let z = x + iy
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6 1
Squaring on both sides
x² + (y – 4)² =  x² + (y + 4)²
simplifying
We get y = 0
Which is real axis

Question 2.
If z = x + iy is a complex number such that Im \(\left(\frac{2 z+1}{i z+1}\right)\) = 0, show that the locus of z is 2x² + 2y² + x – 2y = 0.
Solution:
Let z = x + iy
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6 2
2y(1 – y) – x(2x + 1) = 0
⇒ 2y – 2y² – 2x² – x = 0
∴ The locus is 2x² + 2y² – 2y + x = 0
Hence proved

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

Question 3.
Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:
(i) [Re (iz)]² = 3
Solution:
z = x + iy
[Re(iz)]2 = 3
⇒ [Re[i(x + iy]]2 = 3
⇒ [Re(ix – y)]2 = 3
⇒ (-y)2 = 3
⇒ y2 = 3

(ii) Im[(1 – i)z + 1] = 0
⇒ Im [(1 – i)(z + iy) + 1] = 0
⇒ Im[x + iy – ix + y + 1] = 0
⇒ Im[(x + y + 1) + i(y – x)] = 0
Considering only the imaginary part
y – x = 0 ⇒ x = y

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

(iii) |z + i| = |z – 1|
⇒ |x + iy + i| = | x + iy – 1|
⇒ |x + i(y + 1)| = |(x – 1) + iy|
Squaring on both sides
|x + i(y + 1)|2 = |(x – 1) + iy|2
⇒ x2 + (y + 1)2 = (x – 1)2 + y2
⇒ x2 + y2 + 2y + 1 = x2 – 2x + 1 + y2
⇒ 2y + 2x = 0
⇒ x + y = 0

(iv) \(\bar {z}\) = z-1
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6 3
x² + y² = 1, x² + y² = -1 which cannot be true.
∴ x² + y² = 1

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
Solution:
Let z = x + iy
|z – 2 – i| = 3
⇒ |x + iy – 2 – i| = 3
⇒ |(x – 2) + i(y – 1)| = 3
⇒ \(\sqrt{(x-2)^{2}+(y-1)^{2}}=3\)
Squaring on both sides
(x – 2)2 + (y – 1)2 = 9
⇒ x2 – 4x + 4 + y2 – 2y + 1 – 9 = 0
⇒ x2 + y2 – 4x – 2y – 4 = 0 represents a circle
2g = -4 ⇒ g = -2
2f = -2 ⇒ f = -1
c = -4
(a) Centre (-g, -f) = (2, 1) = 2 + i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=3\)
Aliter: |z – (2 + i)| = 3
Centre = 2 + i
radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2
⇒ |2x + i2y + 2 – 4i| =2
⇒ |(2x + 2) + i(2y – 4)| = 2
⇒ |2(x + 1) + 2i(y – 2)| = 2
⇒ |(x + 1) + i(y – 2)| = 1
⇒ \(\sqrt{(x+1)^{2}(y-2)^{2}}=1\)
Squaring on both sides,
x2 + 2x + 1 + y2 + 4 – 4y – 1 = 0
⇒ x2 + y2 + 2x – 4y + 4 = 0 represents a circle
2g = 2 ⇒ g = 1
2f = -4 ⇒ f = -2
c = 4
(a) Centre (-g, -f) = (-1, 2) = -1 + 2i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-4}=1\)
Aliter: 2|(z + 1 – 2i)| = 2
|z – (-1 + 2i)| = 1
Centre = -1 + 2i
radius = 1

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

(iii) |3(x + iy) – 6 + 12i| = 8
⇒ |3x + i3y – 6 + 12i| = 8
⇒ |3(x – 2) + i3 (y + 4)| = 8
⇒ 3|(x – 2) + i (y + 4)| = 8
⇒ \(3 \sqrt{(x-2)^{2}+(y+4)^{2}}=8\)
Squaring on both sides,
9[(x – 2)2 + (y + 4)2] = 64
⇒ x2 – 4x + 4 + y2 + 8y + 16 = \(\frac{64}{9}\)
⇒ x2 + y2 – 4x + 8y + 20 – \(\frac{64}{9}\) = 0
x2 + y2 – 4x + 8y + \(\frac{116}{9}\) = 0 represents a circle.
2g = -4 ⇒ g = -2
2f = 8 ⇒ f = 4
c = \(\frac{116}{9}\)
(a) Centre (-g, -f) = (2, -4) = 2 – 4i
(b) Radius = \(=\sqrt{g^{2}+f^{2}-c}=\sqrt{4+16-\frac{116}{9}}=\sqrt{\frac{180-116}{9}}=\frac{8}{3}\)
Aliter:
|z – 2 + 4i| = \(\frac{8}{3}\)
⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)
Centre = 2 – 4i, Radius = \(\frac{8}{3}\)

Question 5.
Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases:
(i) |z – 4| = 16
Solution:
Let z = x + iy
|x + iy – 4| – 16
|(x – 4) + iy| = 16
\(\sqrt{(x – 4)² + y²}\) = 16
∴ Squaring on both sides
(x – 4)² + y² = 256
x² – 8x + 16 + y² – 256 = 0
x² + y² – 8x – 240 = 0
The locus of the point is a circle.

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.6

(ii) |z – 4|² – |z – 1|² = 16.
Solution:
|x + iy – 4|2 – |x + iy – 1|2 = 16
⇒ |(x – 4) + iy|2 – |(x – 1) + iy|2 = 16
⇒ [(x – 4)2 + y2] – [(x – 1)2 + y2] = 16
⇒ (x2 – 8x + 16 + y2) – (x2 – 2x + 1 + y2) = 16
⇒ x2 + y2 – 8x + 16 – x2 + 2x – 1 – y2 = 16
⇒ -6x + 15 = 16
⇒ 6x + 1 = 0
The locus of the point is a straight line.

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Students can Download Tamil Nadu 12th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 5 English Medium

Instructions:

  1.  The question paper comprises of four parts.
  2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  3. questions of Part I, II. III and IV are to be attempted separately
  4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
  6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
  7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If | adj(adj A) | = |A|9, then the order of the square matrix A is _______.
(a) 3
(b) 4
(c) 2
(d) 5
Answer:
(b) 4

Question 2.
If |z1| = 1, |z2| = 2, |z3| = 3 and |9z1z2 + 4z1z2 + z2z3| = 12, then the value of |z1 + z2+ z3| is ________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 3.
The value of \(\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}\) is ________.
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 1
Answer:
(a) cis \(\frac{2 \pi}{3}\)

Question 4.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to _____.
(a) tan2 α
(b) 0
(c) -1
(d) tan 2α
Answer:
(c) -1

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 5.
If \(\cot ^{-1} x=\frac{2 \pi}{5}\) for some x∈R, the value of tan-1 x is _______.
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 2
Answer:
(c) \(\frac{\pi}{10}\)

Question 6.
The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is ____.
(a) 10
(b) \(2 \sqrt{5}\)
(c) 6
(d) 4
Answer:
(b) \(2 \sqrt{5}\)

Question 7.
The length of the L.R. of x2 = -4y is _______.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 8.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ______.
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 9.
The distance from the origin to the plane \(\vec{r} \cdot(2 \vec{i}-\vec{j}+5 \vec{k})=7\) is _____.
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 3
Answer:
(a) \(\frac{7}{\sqrt{30}}\)

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 10.
The number given by the Mean value theorem for the function \(\frac{1}{x}\), x ∈ [1, 9] is ______.
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Answer:
(c) 3

Question 11.
f is a differentiable function defined on an interval I with positive derivative. Then f is ______.
(a) increasing on I
(b) decreasing on I
(c) strictly increasing on I
(d) strictly decreasing on I
Answer:
(c) strictly increasing on I

Question 12.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is ________.
(a) 0.4 cu.cm
(b) 0.45 cu.cm
(c) 2 cu.cm
(d) 4.8 cu.cm
Answer:
(d) 4.8 cu.cm

Question 13.
If u(x, y) = \(e^{x^{2}+y^{2}}\), then \(\frac{\partial u}{\partial x}\) is equal to _______.
(a) \(e^{x^{2}+y^{2}}\)
(b) 2xu
(c) x2u
(d) y2u
Answer:
(b) 2xu

Question 14.
The value of \(\int_{0}^{\infty} e^{-3 x} x^{2} d x\) is _______.
(a) \(\frac{7}{27}\)
(b) \(\frac{5}{27}\)
(c) \(\frac{4}{27}\)
(d) \(\frac{2}{27}\)
Answer:
(d) \(\frac{2}{27}\)

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 15.
\(\int_{0}^{a} f(x) d x\) is _____.
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 4
Answer:
(b) \(\int_{0}^{a} f(a-x) d x\)

Question 16.
The integrating factor of the differential equation \(\frac{d y}{d x}\) + P(x) y = Q (x) is x, then P(x) ______.
(a) x
(b) \(\frac{x^{2}}{2}\)
(c) \(\frac{1}{x}\)
(d) \(\frac{1}{x^{2}}\frac{1}{x^{2}}\)
Answer:
(c) \(\frac{1}{x}\)

Question 17.
The order and degree of the differential equation p\(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 3}+x^{1 / 4}=0\) are respectively _______.
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Answer:
(a) 2, 3

Question 18 .
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only
(c) III only
(d) II and III
Answer:
(a) I and II

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 19.
If p is true and q is false then which of the following is not true?
(a) p → q is false
(b)p ∨ q is true
(c)p ∧ q is false
(d) p ↔ q is true
Answer:
(d) p ↔ q is true

Question 20.
The operation * defined by a*b = \(\frac{a b}{7}\) is not a binary operation on ______.
(a) Q+
(b) Z
(c) R
(c) C
Answer:
(b) Z

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Using elementary transformations find the inverse of the following matrix \(\left[\begin{array}{ll}
4 & 7 \\
3 & 0
\end{array}\right]\)
Answer:
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 5

Question 22.
If Z1 = 1 – 3i, z2 = -4i, and z3 = 5, show that (z1 + z2) + z3 = Z1 + (z2 + z3)
Answer:
z1 = 1 – 3i, z2 = 4i, z3 = 5
(z1 + z2) + z3 = (1 – 3i – 4i) + 5(1 – 7i) + 5
= 6 – 7i …..(1)
z1 + (z2 + z3) = (1 – 3i) + (-4i + 5)
= 6 – 7i …..(2)
from(1)& (2)we get
∴ (z1 + z2) + z3 = z1 + (z2 + z3)

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 23.
Find a polynomial equation of minimum degree with rational coefficients, having 2+ \(\sqrt{3}\) i as
a root.
Answer:
Given roots is (2 + \(\sqrt{3}\) i)
∴ The other root is (2 – \(\sqrt{3}\) i), since the imaginary roots with real co-efficient occur as conjugate
pairs.
x2 – x(S.O.R) + P.O.R = 0 ⇒ x2 – x(4) + (4 + 3) = 0
x2 – 4x + 7 = 0.

Question 24.
Evaluate: \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+17 x+29}{x^{4}}\right)\)
Answer:
This is an indeterminate of the form (\(\frac{\infty}{\infty}\)). To evaluate this limit, we apply l’Hôpital Rule.
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 6

Question 25.
Let g(x, y) = 2y + x2, x = 2r – s, y = r2 + 2s, r, s ∈R. Find \(\frac{\partial g}{\partial r}, \frac{\partial g}{\partial s}\).
Answer:
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 7

Question 26.
Evaluate: \(\int_{0}^{\frac{\pi}{2}}\left(\sin ^{2} x+\cos ^{4} x\right) d x\)
Answer:
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 8

Question 27.
Find the differential equation corresponding to the family of curves represented by the equation y = Ae8x + Be-8x, where A and B are arbitrary constants.
Answer:
y = Ae8x + Be-8x Where A and B are arbitrary constants.
Differentiate with respect to ‘x’
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 9

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 28.
If F(x) = \(\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\) – ∞ < x < ∞ is a distribution function of a continuous variable X, find P (0 ≤ x ≤ 1).
Answer:
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 10

Question 29.
Show that p → q and q → p are not equivalent.
Answer:
Truth table for p → q
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 11
Truth table for q → p
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 12
The entries in the column corresponding to p → q and q → p are not identical, hence they are not equivalent.

Question 30.
Show that the lines \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) and \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) are parallel.
Answer:
We observe that the straight line \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) is parallel to the vector \(4 \hat{i}-6 \hat{j}+12 \hat{k}\) and the straight line \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) is parallel to the vector \(-2 \hat{i}+3 \hat{j}-6 \hat{k}\).
Since \(4 \hat{i}-6 \hat{j}+12 \hat{k}=-2(-2 \hat{i}+3 \hat{j}-6 \hat{k})\) the two vectors are parallel, and hence the two straight lines are parallel.

Part – III

II. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\)

Question 32.
Find the square roots of – 15 – 8i

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 33.
Find the sum of the squares of the roots of ax4 + bx3 + cx2 + dx + e = 0, a ≠ 0

Question 34.
For what value of x, the inequality \(\frac{\pi}{2}\) < cos-1 (3x – 1) < π holds?

Question 35.
Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\). Also, find the equation of the perpendicular.

Question 36.
Evaluate \(\int_{0}^{\infty} \frac{x^{n}}{n^{x}} d x\), where n is a positive integer ≥ 2.

Question 37.
The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to die velocity at that time. Find the velocity after 2 seconds of switching off the engine.

Question 38.
The probability that Mr.Q hits a target at any trial is \(\frac{1}{4}\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) at least one time.

Question 39.
Consider the binary operation * defined on the set A = {a, b, c, d} by the following table:
Tamil Nadu 12th Maths Model Question Paper 5 English Medium 13
Is it commutative and associative?

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 40.
Evaluate the following limit, if necessary use l’Hopital Rule. \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\)

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) Find the inverse of A = \(\left[\begin{array}{lll}
2 & 1 & 1 \\
3 & 2 & 1 \\
2 & 1 & 2
\end{array}\right]\) by Gauss-Jordan method
[OR]
(b) Find the point of intersection of the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).

Question 42.
(a) Suppose z1, z2 and z3 are the vertices of an equilateral triangle inscribed in the circle.
|z| = 2. If z1 = 1 + i\(\sqrt{3}\), then find z2 and z3.
[OR]
(b) If A = \(\left[\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right]\) show that A2 – 3A – 7I2 = O2. Hence Find A-1.

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 43.
(a) Solve the equation 3x3 – 16x2 + 23x – 6 = 0 if the product of two roots is 1.
[OR]
(b) The mean and variance of a binomial variate X are respectively 2 and 1.5.
Find (i) P(X = 0) (ii) P(X = 1) (iii) P(X ≥ 1)

Question 44.
(a) Find the value of \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right)\)
[OR]
(b) Find, by integration, the volume of the solid generated by revolving about y-axis the region bounded between the curve y = \(\frac{3}{4} \sqrt{x^{2}-16}\), x ≥ 4, the y-axis, and the lines y = 1 and y = 6.

Question 45.
(a) A semielliptical archway over a one-way road has a height of 3m and a width of 12m. The truck has a width of 3m and a height of 2.7m. Will the truck clear the opening of the archway?
[OR]
(b) For the function f(x, y) = \(\frac{3 x}{y+\sin x}\) find the fx, fy, and show that fxy = fyx.

Question 46.
(a) Derive the equation of the plane in the intercept form.
[OR]
(b) Let A be Q\{1}. Define * on A by x * y = x +y – xy. Is * binary on A ? If so, examine the existence of identity, existence of inverse properties for the operation * on A .

Tamil Nadu 12th Maths Model Question Paper 5 English Medium

Question 47.
(a) Find the angle between the rectangular hyperbola xy = 2 and the parabola x2 + 4y = 0.
[OR]
(b) A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C, and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Students can Download Tamil Nadu 12th Physics Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 5 English Medium

General Instructions:

  • The question paper comprises of four parts.
  • You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
  • All questions of Part I, II, III, and IV are to be attempted separately.
  • Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each.
    These are to be answered by choosing the most suitable answer from the given four
    alternatives and writing the option code and the corresponding answer
  • Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
    in about one or two sentences.
  • Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
    in about three to five short sentences.
  • Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
    in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
If voltage applied on a capacitor is increased from V to 2V, Choose the correct conclusion,
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 2.
The electric field in the region between two concentric charged spherical shells
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 3.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be …………
(a) R
(b) 2R
(c) \(\frac{\mathbf{R}}{4}\)
(d) \(\frac{\mathbf{R}}{2}\)
Answer:
(c) \(\frac{\mathbf{R}}{4}\)

Question 4.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 5.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 6.
Faraday’s law of electromagnetic induction is related to the …………….
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 7.
The dimension of \(\frac{1}{\mu_{0} \varepsilon_{0}}\) is ……….
(a) [L T-1]
(b) [L2 T-2]
(c) [L-1 T]
(d) [L-2 T2]
Answer:
(b) [L2 T-2]

Question 8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,…………..
(a) 2D
(b) \(\frac{\mathrm{D}}{2}\)
(c) \(\sqrt{2} \mathrm{D}\)
(d) \(\frac{\mathrm{D}}{\sqrt{2}}\)
Answer:
(a) 2D
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 1

Question 9.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 10.
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 x 1026 W. The number of photons received by the human eye per second on the average from sunlight is of the order of ……………..
(a) 1045
(b) 1042
(c) 1054
(d) 1051
Answer:
(a) 1045
Hint:
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 2

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 11.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint: As the intensity of incident light increases, photoelectric current increases.

Question 12.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint: In the condition of no deflection \(\frac{e}{m}=\frac{\mathrm{E}^{2}}{2 v \mathrm{B}^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt{208}\) = 14.4 times

Question 13.
A forward biased diode is treated as
(a) An open switch with infinite resistance
(b) A closed switch with a voltage drop of 0V
(c) A closed switch in series with a battery voltage of 0.7V
(d) A closed switch in series with a small resistance and a battery.
Answer:
(d) A closed switch in series with a small resistance and a battery.

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating
signal is called …………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956?
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell
Answer:
(c) Engelberger

Part – II

Answer any six questions in which Q. No 17 is compulsory.

Question 16.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from  – q to +q. The SI unit of dipole moment is coulomb meter (cm).
\(\vec{p}=q a \hat{i}-q a(-\hat{i})=2 q a \hat{i}\)

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 17.
Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC-1. The angle 0 is 60°. Suppose 0 becomes zero, what is the electric flux?
Answer:
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 3

Question 18.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Question 19.
What is meant by wattles current?
Answer:
The component of current (IpMS sin φ), which has a phase angle of \(\frac{\pi}{2}\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell’s window.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 21.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It’s unit is electron volt (eV).

Question 22.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 23.
What is the angular momentum of an electron in the third orbit of an atom?
Answer:
Here n = 3; h = 6.6 x 10-34 Js
Angular momentum.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 4

Question 24.
Explain the current flow in a NPN transistor
Answer:

  • The conventional flow of current is based on the direction of the motion of holes
  • In NPN transistor, current enters from the base into the emitter.

Part – III

Answer any six questions in which Q.No. 27 is compulsory. [6×3 = 18]

Question 25.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
\(C=\frac{Q}{V} \text { or } Q \propto V\)
The SI unit of capacitance is coulomb per volt or farad (F).

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 26.
Distinguish between drift velocity and mobility
Answer:

SNo. Drift Velocity Mobility
1. The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. Mobility of an electron is defined as the magnitude of the drift velocity per unit electric field.
2. \(\overrightarrow{\mathrm{V}}_{d}=\vec{a} \tau\) \(\mu=\frac{e \tau}{m} \text { or } \mu=\frac{\left|\overrightarrow{\mathrm{V}}_
{d}\right|}{\overrightarrow{\mathrm{E}}}\)
3. It’s unit is ms-1. It’s unit is m2 v_1 s_1

Question 27.
A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth’s magnetic field is 25 x 10-6 T then, calculate the current which gives a deflection of 60°.
Answer:
The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10-6 T
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 5

Question 28.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

  • By changing the magnetic field B
  • By changing the area A of the coil and
  • By changing the relative orientation 0 of the coil with magnetic field

Question 29.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
Answer:
L = 400 x 10-3 H; Ieff = 6 x 10-3A; f= 1000 Hz
Inductive reactance, XL = Lω = L x 2πf = 2 x 3.14 x 1000 x 0.4 = 2512Ω
Voltage across L, V – IXL = 6 x 10-3 x 2512 =15.072 V (RMS)

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Question 30.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).

(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 31.
Write down the draw backs of Bohr atom’model.
Answer:
Limitations of Bohr atom model
The following are the drawbacks of Bohr atom model

  • Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for
    complex atoms.
  • When the spectral lines are closely examined, individual lines of hydrogen spectrum is
    accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
  • Bohr atom model fails to explain the intensity variations in the spectral lines.
  • The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 32.
What do you mean by leakage current in a diode?
Answer:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in μA, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.

Question 33.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication Wireless communication
It is a point-to-point communication. It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres. It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected. These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV. Ex. mobile, radio or TV broadcasting and satellite communication.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field: Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec{E}\)  in the direction of the field and charge -q will experience a force -q \(\vec{E}\) in a direction opposite to the field. Since the external field \(\vec{E}\) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 6
Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
The magnitude of the total torque
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 7
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 8
where θ is the angle made by \(\vec{p} \text { with } \overrightarrow{\mathrm{E}}\) .
Since p = 2aq, the torque is written in terms of the vector product as \(\vec{\tau}=\vec{p} \times \vec{E}\)
The magnitude of this torque is τ = pE sin θ and is maximum
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field [/latex]\overrightarrow{\mathrm{E}}[/latex] . Once \(\overrightarrow{\mathrm{p}}\) is aligned with \(\overrightarrow{\mathrm{E}}\), the total torque on the dipole becomes zero.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

[OR]

(b) Explain the equivalent resistance of a parallel resistor network.
Answer:
Resistors in parallel: Resistors are in parallel when they are connected across the same potential difference as shown in fig. (a).
In this case, the total current I that leaves the battery in split into three separate paths. Let I1, I2 and I3 be the current through the resistors R1 ,R2 and R3 respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I1 + I2 + I3 ………………… (1)
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 9
Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 10
Substituting these values in equation (1) we get.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 11

Here Rp is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 35.
(a) Obtain a relation for the magnetic induction at a point along the axis of a circular coil ‘ carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d \bar{l} \) at C and D Let r be the vector joining the current element (I \(d \bar{l}\)) at C to the point P.
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar{l} \) is
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 12
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 13
The magnitude of magnetic field due to current element I \(d \bar{l} \) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\) ) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar{l} \) around the loop, \(d \bar{B} \) sweeps out a cone, then the net magnetic field \(\overrightarrow{\mathrm{B}}\) at point P is
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 13
Using Pythagorous theorem r2 = R2 + Z2 and integrating line element from 0 to 2πR, we get
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 14
Note that the magnetic field \(\bar{B} \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

[OR]

(b) Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed hire. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 15
Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 16
Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule.

Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Flence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.

Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum _ can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 17

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 18

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

[Or]

(b) Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus: The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.

The light passing through one cut in the wheel will, get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working: The angular speed c rotation of the toothed wheel was increased from zero to a value co until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light: The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
\(v=\frac{2 d}{t}\) ………………… (1)

The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed ω of the toothed wheel. The angular speed ω of the toothed wheel when the light disappeared for the first time is,
\(\omega=\frac{\theta}{t}\) ………………….. (2)

Here, θ is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 19
Rewriting the above equation for t
\(t=\frac{\pi}{\mathrm{N} \omega}\) ……………….. (3)
Substituting t from equation (3) in equation (1)
\(v=\frac{2 d}{\pi / \mathrm{N} \omega}\)
\(v=\frac{2 d \mathrm{N} \omega}{\pi}\) …………….(4)
Fizeau had some difficulty to visually estimate the minimum intensity blocked by the adjacent tooth, and his value for speed of light was very value.

Question 37.
(a) List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:

  • For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
  • Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.
  • Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
  • For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
  • There is no time lag between incidence of light and ejection of photoelectrons.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

[OR]

(b) Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:

  • Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.
  • The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:

  • The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only \(0.7 \% \text { of }_{92}^{235} \mathrm{U} \text { and } 99.3 \% \text { are only }^{238} \mathrm{g}_{2} \mathrm{U}\) . So the \(_{ 92 }^{ 235 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\) .
  • In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:

  • The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced
  • Most of the reactors use water, heavy water (D20) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium 20

 

Control rods:

  • The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.
  • Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.
  • If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:

  • For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Cooling system:

  • The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.
  • This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 38.
(a) Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. A solar cell is of two types: p-type and n-type.

Both types use a combination of p-type and n-type Silicon which together forms the p-n junction of the solar cell. The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards n-type Silicon and holes move towards p-type Silicon layer.

The electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact. Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load.

Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For  high power applications, solar panels and solar arrays are used.
Tamil Nadu 12th Physics Model Question Paper 5 English Medium 21

Applications:

  • Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
  • Solar cells are used in satellites and space applications
  • Solar panels are used to generate electricity.

(b) Elaborate on the basic elements of communication system with the necessary block diagram.
Answer:
Elements of an electronic communication system
Information (Baseband or input signal): Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.

Input transducer
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

Amplifier:
The transducer output is very weak and is amplified by the amplifier. Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 x 108 ms-1).

Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

Tamil Nadu 12th Physics Model Question Paper 5 English Medium

Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation.

Range:
It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

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TN State Board 12th Biology Model Question Paper 5 English Medium

General Instructions:

    1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
    2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
    3. All questions of Part I, II, III and IV are to be attempted separately.
    4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
    5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
    6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
    7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Choose the correct statement(s) about tenuinucellate ovule
(a) Sporogenous cell is hypodermal
(b) Ovules have fairly large nucellus
(c) Sporogenous cell is epidermal
(d) Ovules have single layer of nucellus tissue
Answer:
(a) Sporogenous cell is hypodermal

Question 2.
The genotype of a plant showing the dominant phenotype can be determined by _______.
(a) Back cross
(b) Test cross
(c) Dihybrid cross
(d) Pedigree analysis
Answer:
(b) Test cross

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 3.
An allo-hexaploidy contains ________.
(a) Six different genomes
(b) Six copies of three different genomes
(c) Two copies of three different genomes
(d) Six copies of one genome
Answer:
(b) Six copies of three different genomes

Question 4.
Match the following column I with column II
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 1
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 2
Answer:
(D) 1 – c, 2 – d, 3 – a, 4 – b

Question 5.
Assertion (A): Incubation is followed by Inoculation.
Reason (R): Explant is inoculated to media.
(a) Both A and R are correct but R is not a correct explanation to A
(b) R explains A
(c) A is correct R is incorrect
(d) Both A and R are incorrect
Answer:
(b) R explains A

Question 6.
Environment of any community is called
(a) Paratope
(b) Biotope
(c) Ecotope
(d) Epitope
Answer:
(b) Biotope

Question 7.
Which of the following is not a sedimentary cycle?
(a) Nitrogen cycle
(b) Phosphorus cycle
(c) Sulphur cycle
(d) Calcium cycle
Answer:
(a) Nitrogen cycle

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 8.
The plants which are grown in silvopasture system are
(a) Sesbania and Acacia
(b) Solanum and Crotolaria
(c) Clitoria and Begonia
(d) Teak and sandal
Answer:
(a) Sesbania and Acacia

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What are multiple alleles? Give an example.
Answer:
Three or more alternative forms of a gene that occupy the same locus and control the expression of a single trait.
E.g: ABO blood group

Question 10.
List out the benefits of herbicide tolerant crops.
Answer:
Advantages of Herbicide Tolerant Crops:

  • Weed control improves higher crop yields.
  • Reduces spray of herbicide.
  • Reduces competition between crop plant and weed.
  • Use of low toxicity compounds which do not remain active in the soil.
  • The ability to conserve soil structure and microbes.

Question 11.
Soil formation can be initiated by biological organisms. How?
Answer:
Soil formation is initiated by the biological weathering process. Biological weathering takes place when organisms like bacteria, fungi, lichens and plants help in the breakdown of rocks through the production of acids and certain chemical substances.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 12.
Where did Montreal Protocol was held? State its objectives.
Answer:
The International treaty called the Montreal Protocol (1987) was held in Canada on substances that deplete ozone layer and the main goal of it is gradually eliminating the production and consumption of ozone depleting substances and to limit their damage on the Earth’s ozone layer.

Question 13.
What is Heterosis?
Answer:
The superiority of the F1 hybrid in performance over its parents is called heterosis or hybrid vigour. Vigour refers to increase in growth, yield, greater adaptability of resistance to diseases, pest and drought.

Question 14.
Write the cosmetic uses of Aloe.
Answer:
Aloe gel are used as skin tonic. It has a cooling effect and moisturising characteristics and hence used in preparation of creams, lotions, shampoos, shaving creams, after shave lotions and allied products. It is used in gerontological applications for rejuvenation of aging skin. Products prepared from aloe leaves have multiple properties such as emollient, antibacterial, antioxidant, antifungal and antiseptic. Aloe vera gel is used in skin care cosmetics.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
List out the functions of tapetum.
Answer:

  • It supplies nutrition to the developing microspores.
  • It contributes sporopollenin through ubisch bodies thus plays an important role in pollen wall formation.
  • The pollenkitt material is contributed by tapetal cells and is later transferred to the pollen surface.
  • Exine proteins responsible for ‘rejection reaction’ of the stigma are present in the cavities of the exine. These proteins are derived from tapetal cells.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 16.
What do you mean by Embryoids? Write its application.
Answer:
Somatic embryogenesis is the formation of embryos from the callus tissue directly and these embryos are called Embryoids or from the in vitro cells directly form pre-embryonic cells which differentiate into embryoids.

Applications:

  • Somatic embryogenesis provides potential plantlets which after hardening period can establish into plants.
  • Somatic embryoids can be used for the production of synthetic seeds.
  • Somatic embryogenesis is now reported in many plants such as Allium sativum, Hordeum vulgare, Oryza sativa, Zea mays and this is possible in any plant.

Question 17.
Give a comparative account on Seaweed liquid fertilizer.
Answer:
Seaweed liquid fertilizer (SLF) contains cytokinin, gibberellins and auxin apart from macro and micro nutrients. Most seaweed based fertilizers are made from kelp (brown algae) which grows to length of 150 metres. Liquid seaweed fertilizer is not only organic but also ecofriendly. The alginates in the seaweed that reacts with metals in the soil and form long, crosslinked polymers in the soil.

These polymers improve the crumbing in the soil, swell up when they get wet and retain moisture for a long time. They are especially useful in organic gardening which provides carbohydrates for plants. Seaweed has more than 70 minerals, vitamins and enzymes. It promotes vigorous growth. Improves resistance of plants to frost and disease. Seeds soaked in seaweed extract germinate much rapidly and develop a better root system.

Question 18.
Write a brief note on Detritus food chain.
Answer:
Detritus food chain is a type of food chain which begins with dead organic matter which is an important source of energy. A large amount of organic matter is derived from the dead plants, animals and their excreta. This type of food chain is present in all ecosystems. The transfer of energy from the dead organic matter, is transferred through a series of organisms called detritus consumers (detritivores)- small carnivores – large (top) carnivores with repeated eating and being eaten respectively. This is called the detritus food chain.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 3

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 19.
Mention the name of man-made cereal. How it is developed?
Answer:
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 4
Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

  1. Tetraploidy: Crosses between diploid wheat and rye.
  2. Hexaploidy: Crosses between tetraploid wheat Triticum durum (macaroni wheat) and rye.
  3. Octoploidy: Crosses between hexaploid wheat T. aestivum (bread wheat) and rye.

Hexaploidy Triticale hybrid plants demonstrate characteristics of both macaroni wheat and rye. For example, they combine the high-protein content of wheat with rye’s high content of the amino acid lysine, which is low in wheat.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Describe the structure of a Cicer seed (dicot seed) with labelled diagram.
Answer:
Structure of a Cicer seed as an example for Dicot seed The mature seeds are attached to the fruit wall by a stalk called funiculus. The funiculus disappears leaving a scar called hilum. Below the hilum a small pore called micropyle is present. It facilitates entry of oxygen and water into the seeds during germination.

Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 5
In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root.

The other end of the axis called embryonic shoot is the plumule. Embryonal axis above the level of cotyledon is called epicotyl whereas the cylindrical region between the level of cotyledon is called hypocotyl. The epicotyl terminates in plumule whereas the hypocotyl ends in radicle.

[OR]

(b) (i) Explain the three types of hyperploidy.
(ii) List out the significance of ploidy.
Answer:
(i) (a) Trisomy: Addition of single chromosome to diploid set is called Simple trisomy (2n+l). Trisomics were first reported by Blackeslee (1910) in Datura stramonium (Jimson weed). But later it was reported in Nicotiana, Pisum and Oenothera. Sometimes addition of two individual chromosome from different chromosomal pairs to normal diploid sets are called Double trisomy (2n + 1 + 1).
(b) Tetrasomy: Addition of a pair or two individual pairs of chromosomes to diploid set is called tetrasomy (2n+2) and Double tetrasomy (2n+2+2) respectively. All possible tetrasomics are available in Wheat.
(c) Pentasomy: Addition of three individual chromosome from different chromosomal pairs to normal diploid set are called pentasomy (2n+3).

(ii)

  • Many polyploids are more vigorous and more adaptable than diploids.
  • Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
  • Autopolyploids usually have increase in fresh weight due to more water content.
  • Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.
  • Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 21.
a. (i) Define tissue culture.
(ii) Explain the basic concepts involved in plant tissue culture.
Answer:
(i) Growing plant protoplasts, cells, tissues or organs away from their natural or normal environment, under artificial condition, is known as Tissue Culture.

(ii) Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.
Totipotency: The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

[OR]

(b) What is soil profile? Explain the characters of different soil horizons.
Answer:
Soil is commonly stratified into horizons at different depth. These layers differ in their physical, chemical and biological properties. This succession of super-imposed horizons is called soil profile.

Horizon Description
O-Horizon (Organic horizon) Humus It consists of fresh or partially decomposed organic matter.
O1 – Freshly fallen leaves, twigs, flowers and fruits.
O2 – Dead plants, animals and their excreta decomposed by micro-organisms.
Usually absent in agricultural and deserts.
A-Horizon (Leached horizon)
Topsoil – Often rich in humus and minerals.
It consists of top soil with humus, living creatures and in-organic minerals.
A1 – Dark and rich in organic matter because of mixture of organic and mineral matters.
A2 – Light coloured layer with large sized mineral particles.
B-Horizon (Accumulation horizon) (Subsoil – Poor in humus, rich in minerals) It consists of iron, aluminium and silica rich clay organic compounds.
C – Horizon (Partially weathered horizon) Weathered rock Fragments – Little or no plant or animal life. It consists of parent materials of soil, composed of little amount of organic matters without life forms.
R – Horizon
(Parent material) Bedrock
It is a parent bed rock upon which underground water is found.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Assertion (A): Asexual reproduction is called blastogenic reproduction.
Reason (R): It is accomplished by mitotic and meiotic divisions.
(a) A and R are correct
(b) A is correct but R is incorrect
(c) Both A and R are incorrect
(d) R is the correct explanation for A
Answer:
(b) A is correct but R is incorrect

Question 2.
The mature sperms are stored in the ________.
(a) Seminiferous tubules
(b) Vas deferens
(c) Epididymis
(d) Seminal vesicle
Answer:
(c) Epididymis

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 3.
A contraceptive pill prevents ovulation by _______.
(a) blocking fallopian tube
(b) inhibiting the release of FSH and LH
(c) stimulating the release of FSH and LH
(d) causing immediate degeneration of released ovum.
Answer:
(b) inhibiting the release of FSH and LH

Question 4.
Patau’s syndrome is also referred to as ________.
(a) 13-Trisomy
(b) 18-Trisormy
(c) 21-Trisormy
(d) None of these
Answer:
(a) 13-Trisomy

Question 5.
Cyclosporin – A is an immunosuppressive drug produced from ________.
(a) Aspergillus niger
(b) Manascus purpureus
(c) Penicillium notatum
(d) Trichoderma polysporum
Answer:
(d) Trichoderma polysporum

Question 6.
PCR proceeds in three distinct steps governed by temperature, they are in order of ________.
(a) Denaturation, Annealing, Synthesis
(b) Synthesis, Annealing, Denaturation
(c) Annealing, Synthesis, Denaturation
(d) Denaturation, Synthesis, Annealing
Answer:
(a) Denaturation, Annealing, Synthesis

Question 7.
Match column I with column II
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 6
(a) A – 4, B – 5, C – 2, D – 3, E – 1
(b) A – 3, B – 1, C – 4, D – 2, E – 5
(c) A – 2, B – 3,C – 1, D – 5, E – 4
(d) A – 5, B – 4, C – 2, D – 3, E – 1
Answer:
(a) A – 4, B – 5, C – 2, D – 3, E – 1

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 8.
Total number of mega biodiversity countries in the world is _______.
(a) 12
(b) 15
(c) 17
(d) 19
Answer:
(c) 17

Part – II

Answer any four of the following questions.  [4 × 2 = 8]

Question 9.
Why are the offsprings of oviparous animals are at a greater risk as compared to offsprings of viviparous organisms?
Answer:
Oviparous animals are egg-layers. The eggs containing embryo are laid out of their body and are highly susceptible to environmental factors (temperature, moisture etc.) and predators. Whereas, in viviparous animals, the embryo develops inside the body of female and comes out as young ones. Hence offsprings of oviparous animals are at risk compared to viviparous animal.

Question 10.
What is “let-down reflex”?
Answer:
Oxytocin causes the “Let-Down” reflex the actual ejection of milk from the alveoli of the mammary glands. During lactation, oxytocin also stimulates the recently emptied uterus to contract, helping it to return to pre – pregnancy size.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 11.
In E.coli, three enzymes β- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer:
In the absence of lactose, the repressor protein binds to the operator and prevents the transcription of structural gene by RNA polymerase, hence the enzymes are not produced. However, there will always be a minimal level of lac operon expression even in absence of lactose.

Question 12.
Compare relative dating with absolute dating.
Answer:
Relative dating is used to determine a fossil by comparing it to similar rocks and fossils of known age. Absolute dating is used to determine.the precise age of a fossil by using radiometric dating to measure the decay of isotopes.

Question 13.
Write the name of causative agent, infection site, mode of transmission and any two symptoms of Chikungunya.
Answer:
Causative agent – Alpha virus
Infection site – Nervous system
Mode of transmission – Aedes aegypti (Mosquito)
Symptoms – Fever, headache, joint pain and swelling.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 14.
Define the following terms.
(a) Eutrophication (b) Algal Bloom
Answer:
Eutrophication refers to the nutrient enrichment in water bodies leading to lack of oxygen . and will end up in the death of aquatic organisms. Algal Bloom is an excess growth of algae due to abundant excess nutrients imparting distinct color to water.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Write a short note on foeto scope.
Answer:
Foetoscope is used to monitor the foetal heart rate and other functions during late pregnancy and labour. The average foetal heart rate is between 120 and 160 beats per minute. An abnormal foetal heart rate or pattern may mean that the foetus is not getting enough oxygen and it indicates other problems. A hand-held doppler device is often used during prenatal visits to count the foetal heart rate. During labour, continuous electronic foetal monitoring is often used.

Question 16.
Under which condition does a microbe gains resistance against antibiotic?
Answer:
Antibiotic resistance occurs when bacteria develop the ability to defeat the drug designed to kill or inhibit their growth. It is one of the most acute threat to public health. Antibiotic resistance is accelerated by the misuse and over use of antibiotics, as well as poor infection prevention control. Antibiotics should be used only when prescribed by a certified health professional.

When the bacteria become resistant, antibiotics cannot fight against them and the bacteria multiply. Narrow spectrum antibiotics are preferred over broad spectrum antibiotics. They effectively and accurately target specific pathogenic organisms and are less likely to cause resistance.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 17.
State Fisher and Race hypothesis.
Answer:
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 7
In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh+positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

Question 18.
Extinction of Dodo bird led to the danger of Calvaria tree – Support your answer.
Answer:
An example for co-extinction is the connection between Calvaria tree and the extinct bird of Mauritius Island, the Dodo. The Calvaria tree is dependent on the Dodo bird for completion of its life cycle. The mutualistic association is that the tough homy endocarp of the seeds of Calvaria tree are made permeable by the actions of the large stones in birds gizzard and digestive juices thereby facilitating easier germination. The extinction of the Dodo bird led to the imminent danger of the Calvaria tree coextinction.

Question 19.
Whether PCR can be done for RNA molecules? Yes or No? Explain.
Answer:
The PCR technique can also be used for amplifications of RNA in which case it is referred to as reverse transcription PCR (RT-PCR). In this process the RNA molecules (mRNA) must be converted to complementary DNA by the enzyme reverse transcriptase. The cDNA then serves as the template for PCR.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Define the following terms with an example
(i) Hologamy
(ii) Isogamy
(iii) Anisogamy
(iv) Merogamy
(v) Paedogamy
Answer:
(i) Hologamy: In Hologamy, the adult individuals do not produce gametes, but they themselves act as gametes and fuse to form new individuals.
E.g.: Trichonympha

(ii) Isogamy : Fusion of morphologically and physiologically similar gametes.
E.g.: Monocystis

(iii) Anisogamy : Fusion of morphologically and physiologically dissimilar gametes.
E.g.: Vertebrates.

(iv) Merogamy : Fusion of small sized morphologically different gametes (merogametes)

(v) Paedogamy : Fusion of young individuals produced immediately after the mitotic division of adult parent cell.

[OR]

(b) Write in detail about cervical cancer.
Answer:
Cervical cancer is caused by a sexually transmitted virus called Human Papilloma virus (HPV). HPV may cause abnormal growth of cervical cells or cervical dysplasia. The most common symptoms and signs of cervical cancer are pelvic pain, increased vaginal discharge and abnormal vaginal bleeding. The risk factors for cervical cancer include

  1. Having multiple sexual partners
  2. Prolonged use of contraceptive pills

Cervical cancer can be diagnosed by a Papanicolaou smear (PAP smear) combined with an HPV test. X-Ray, CT scan, MRI and a PET scan may also be used to determine the stage of cancer. The treatment options for cervical cancer include radiation therapy, surgery and chemotherapy.

Modem screening techniques can detect precancerous changes in the cervix. Therefore screening is recommended for women above 30 years once in a year. Cervical cancer can be prevented with vaccination. Primary prevention begins with HPV vaccination of girls aged 9-13 years, before they become sexually active. Modification in lifestyle can also help in preventing cervical cancer. Healthy diet, avoiding tobacco usage, preventing early marriages, practicing monogamy and regular exercise minimize the risk of cervical cancer.

Tamil Nadu 12th Biology Model Question Paper 5 English Medium

Question 21.
(a) Point out the differences between active and passive immunity.
Answer:
Active Immunity:

  1. Active immunity is produced actively by host’s immune system.
  2. It is produced due to contact with pathogen or by its antigen.
  3. It is durable and effective in protection.
  4. Immunological memory is present.
  5. Booster effect on subsequent dose is possible.
  6. Immunity is effective only after a short period.

Passive Immunity:

  1. Passive immunity is received passively and there is no active host participation.
  2. It is produced due to antibodies obtained from outside.
  3. It is transient and less effective.
  4. No memory.
  5. Subsequent dose is less effective.
  6. Immunity develops immediately.

[OR]

(b) How is the amplification of a gene sample of interest carried out using PCR?
Answer:
Denaturation, renaturation or primer annealing and synthesis or primer extension, are the three steps involved in PCR. The double stranded DNA of interest is denatured to separate into two individual strands by high temperature . This is called denaturation. Each strand is allowed to hybridize with a primer (renaturation or primer annealing). The primer template is used to synthesize DNA by using Taq – DNA polymerase.

During denaturation the reaction mixture is heated to 95°C for a short time to denature the target DNA into single strands that will act as a template for DNA synthesis. Annealing is done by rapid cooling of the mixture, allowing the primers to bind to the sequences on each of the two strands flanking the target DNA. During primer extension or synthesis the temperature of the mixture is increased to 75°C for a sufficient period of time to allow Taq DNA polymerase to extend each primer by copying the single stranded template.

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA, This process is also called DNA amplification.
Tamil Nadu 12th Biology Model Question Paper 5 English Medium 8

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 5 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

  1. The question paper comprises of four parts
  2. You are to attempt all the parts An internal choice of questions is provided wherever applicable
  3. All questions of Part I, II, III, and IV are to be attempted separately
  4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences
  6. Question numbers 25 to 33 in Part III are three-marks questions. These are to be answered in about three to five short sentences
  7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to …………..
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Question 2.
Compound used for propellant is
(a) BN
(b) H2B4O7
(c) B2H6
(d) Borax
Answer:
(c) B2H6

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 3.
P4O6 reacts with cold water to give …………..
(a) H3PO3
(b) H4P2O7
(c) HPO3
(d) H3PO4
Answer:
(a) H3PO3

Question 4.
Which one of the following elements show high positive electrode potential?
(a) Ti2+
(b) Mn2+
(c) CO2+
(d) Cr2+
Answer:
(c) CO2+

Question 5.
As per IUPAC guidelines, the name of the complex [Co(en)2(ONO)Cl]Cl is …………..
(a) chlorobisethylenediaminenitritocobalt(III) chloride
(b) chloridobis(ethane-l,2-diamine)nitro k-Ocobaltate(III) chloride
(c) chloridobis(ethane-l,2-diammine)nitrito k-Ocobalt(II) chloride
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride
Answer:
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride

Question 6.
Solid NH3 solid CO2 are examples of …………………..
(a) Covalent solids
(b) polar molecular solids
(c) molecular solids
(d) ionic solids
Answer:
(b) polar molecular solids

Question 7.
After 2 hours, a radioactive substance becomes\(\left(\frac{1}{16}\right)^{4}\) of original amount. Then the half life ( in min ) is ………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 1

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 8.
What is the decreasing order of strength of bases?
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 2
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 3

Question 9.
Which electrolyte is used in Leclanche cell?
(a) ZnSO4 + CuSO4
(6) NH4Cl + ZnCl2
(e) NaCl + CuSO4
(d) MnSO4 + MnO2
Answer:
(6) NH4Cl + ZnCl2

Question 10.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………………
(a) Cataphoresis
(b) Eleætrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 11.
In the following sequence of reactions,
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 4
(a) Butanal
(b) n-butyl alcohol
(c) propan-1-ol
(d) Propanal
Answer:
(c) propan-1-ol

Question 12.
Of the following, which is the product formed when cyclohexanone undergoe’s aldol condensation followed by heating?
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 5
Answer:
(a)
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 6

Question 13.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 7
Answer:
(a) A – 2, B – 1, C – 4, D – 3

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 14.
Insulin, a hormone chemically is ………………..
(a) Fat
(b) Steroid
(c) Protein
(d) Carbohydrates
Answer:
(c) Protein

Question 15.
The role of phosphate in detergent powder is
(a) control pH level of the detergent water mixture
(b) remove Ca2+ and Mg2+ ions from water that causes hardness of water
(c) provide whiteness to the fabric
(d) more soluble in soft water
Answer:
(b) remove Ca2+ and Mg2+ ions from water that causes hardness of water

Part – II

Answer any six questions. Question No. 23 is compulsory. [6 x 2 = 12]

Question 16.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

  • Concentration of the ore
  • Extraction of crude metal
  • Refining of crude metal

Question 17.
Give the uses of Borax.
Answer:

  • Borax is used for the identification of coloured metal ions.
  • In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
  • It is also used as a flux in metallurgy and also acts as a good preservative.

Question 18.
Differentiate primary valency and secondary valency.
Answer:
Primary Valency :

  1. The primary valence of a metal ion positive in most of the cases and zero in certain cases.
  2. The primary valence is always satisfied by negative ions.
  3. The primary valences are non directional
  4. Example: In COCl3.6NH3, the primary valence of cobalt is +3 Example: In CoCl3.6NH3, the secondary valence of cobalt +3

Secondary Valency :

  1. The secondary valence as the coordination number.
  2. The secondary valence is satisfied by negative ions, neutral molecular or positive ions.
  3. The secondary valences are directional
  4. Example: In COCl3.6NH3, the secondary valence of cobalt is 6

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 19.
Write a note about molecular solids.
Answer:

  • In molecular solids, the constituents are neutral molecules. They are held together by weak vander waals forces.
  • Molecular solids are soft and they do not conduct electricity. Eg., Solid CO2

Question 20.
What are the limitations of Arrhenius concept?
Answer:

  • Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.
  • This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 21.
Write a note on catalytic poison.
Answer:
Catalytic poison: Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.
For example, in the reaction, 2SO2 + O2 → 2SO3 with a Pt catalyst, the poison is AS2O3. i. e., AS2O3 destroys the activity of pt. AS2O3 blocks the activity of the catalyst. So, the activity is lost.

Question 22.
Convert phenyl magnesium bromide to phenyl methanol (or) How would you prepare phenyl methanol from Grignard reagent?
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 8

Question 23.
Identify compounds A,B and C in the following sequence of reactions.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 9
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 10

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 24.
Why cannot Vitamin C be stored in our body?
Answer:
Vitamin C is water soluble, therefore it is readily excreted in urine and hence cannot be stored in the body.

Part – III

Answer any six questions. Question No. 29 is compulsory. [6 x 3 = 18]

Question 25.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that – contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 26.
Discuss the Commercial method to prepare Nitric acid.
[OR]
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.

4NH3 + 5O2 → 4NO + 6H2O + 120 kJ
2NO + O2 → 2NO2

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.

6NO2 + 3H2O → 4HNO3 + 2NO + H2O

Question 27.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:

  • Actinoid contraction is greater from element to element than lanthanoid cintraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
  • Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
  • In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only. ,
  • Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 28.
What are the examples of first order reaction?
Answer:
(i) Decompostion of dinitrogen pentoxide
2N2O5(g) → 2NO2(g) + 1/2 O2(g)

(ii) Decomposition of thionylchloride
SO2Cl2(g) → SO2(g) + Cl2(g)

(iii) Decompostion of H2O2 in aqueous solution.
H2O2(aq) → H2O(l) + 1/2 O2(g)

(iv) Isomerisation of cyclopropane to propene

Question 29.
Complete the following reaction.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 11
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 12

Question 30.
How will you calculate degree of dissociation of weak electrolytes and dissociation constant using Kohlrausch’s law?
Answer:
(i) The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity in infinite dilution using the formula \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)

(ii) According to Ostwald’s dilution law \(\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{C}}{1-\alpha}\)
Substituting α value in the above equation
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 13

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 31.
What are the characteristics of adsorption?
Answer:

  • Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid. .
  • Adsorption is always accompanied by decrease in free energy. When AG reaches zero, the equilibrium is attained.
  • Adsorption is a spontaneous process.
  • When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
    ∆G = ∆H – T ∆S where ∆G = change in free energy
    ∆H = change in enthalpy
    ∆S = change in entropy
    .’. ∆H = ∆G + T∆S
  • Adsorption is exothermic and it is a quick process.
  • If simultaneous adsorption and absorption take place, it is termed as ‘sorption’ and sorption of gases on metal surface is called occlusion.

Question 32.
What are the uses of cellulose?
Answer:

  • Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
  • Gun cotton – nitrated ester of cellulose an explosive is prepared from cellulose.
  • Cellulose act as food for animals.

Question 33.
How will you prepare PHBV? Give its use?
Answer:
(i) The biodegrable polymer PHBV (Poly hydroxy butyrate-co hydroxyl valerate) is prepared by the polymerisation of monomers 3 – hydroxy butanoic acid and 3 – hydroxy pentanoic acid.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 14
(ii) It is used in orthopaedic devices and in controlled release of drugs.

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) Which type of ores can be concentrated by froth floatation method?
Give two examples for such ores. (2)
(ii) Explain the variation In E°M3+/M2+ series. (3)
[OR]
(b) (i) Mention the uses of silicon tetrachtoride. (2)
(ii) What are all the conditions that are necessary for catenation? (3)
Answer:
(a) (i) Suiphide ores can be concentrated by froth floatation method.
e.g., (i) Copper pyrites (CuFeS2) (ii) Zinc blende (ZnS) (iii) Galena (PbS)

(ii)

  • In transition series, as we move down from Ti to Zn, the standard reduction potential E° M3+/M2+value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu2+
  • M3+/M2+ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d5 configuration in Mn2+ and completely filled d10 configuration in Zn2+.
  • The standard electrode potential for the M3+ /M2+ half cell gives’the relative stability between M3+ and M2+.
  • The high reduction potential of Mn3+ / Mn2+ indicates Mn2+ is more stable than Mn3+.
  • For Fe3+/Fe2+ the reduction potential is 0.77 V, and this low value indicates that both Fe3+ and Fe2+ can exist under normal condition.
  • Mn3+ has a 3d4 configuration while that of Mn2+ is 3d5. The extra stability associated with a half filled d sub-shell makes the reduction of Mn3+ very feasible [E° = +1.5 IV]

[OR]

(b) (i) Silicon tetrachloride is used in the production of semiconducting silicon.
It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

(ii) Essential condition for catenation:

  • The valency of elements is greater than or equal to two.
  • Element should have an ability to bond with itself.
  • The self bond must be as strong as its bond with other elements.
  • Kinetic inertness of catenated compound towards other molecules.

Question 35.
(a) (i) Discuss the manufacture of chlorine. (3)
(ii) What is inert pair effect? (2)
[OR]
(b) (i) Calculate the magnetic moment of Ti3+ and V4+. (2)
(ii) Draw all possible geometrical isomers of the complex [CO(en)2Cl2]+ and identify the optically active isomer. (3)
Answer:
(a) (i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na+ and Cl ions are formed. Na+ ion reacts with OH ions of water and forms sodium hydroxide.

Hydrogen and chlorine are liberated as gases.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 15

Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 16
The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 17

(ii) In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

[OR]

(b) (i) Ti (Z = 22) Ti3+3d1
V (Z = 23) V4+ 3d1
.’. \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mu_{\mathrm{B}}\) So they are paramagnetic.

(ii) 1. Cis – [Co(en)2Cl2]+
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 18
The coordination complex [Co(en)2Cl2]+ has three isõmers two optically active cis forms and the optically inactive trans form.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 19

Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium

Question 36.
(a) (i) Calculate the number of atoms in a fee unit cell.
(ii) How do nature of the reactant influence rate of reaction?
[OR]
(b) (i) Account for the acidic nature of HClO4
In terms of Bronsted – Lowry theory, identify its conjugate base.
(ii) IS it possible to store copper sulphate in an iron vessel for a long time?
Given \(\mathbf{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\) and \(\mathbf{E}_{\mathbf{F e}^{2+} \mathbf{F e}}^{s}=+\mathbf{0 . 4 4 V}\)
Answer:
(a) (i) Number of atoms in a fcc unit cell,
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 20
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 21

(ii) Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Let us compare the following two reactions that we carried out in volumetric analysis.

  1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO4
  2. Redox reaction between oxalic acid and KMnO4

The oxidation of oxalate ion by KMnO4 is relatively slow compared to the reaction between KMnO4 and Fe2+ . In fact heating is required for the reaction between KMnO4 and Oxalate ion and is carried out at around 60°C.

The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts. If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantaneously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 22

[OR]

(b) (i) HClO4 ⇌ H+ + ClO4

According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

Let us consider the stabilities of the conjugate bases ClO4 , CIO3 , ClO2 and CIO formed from these acid HClO4, HClO3 , HClO2, HOCl respectively. These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

The charge stabilization increases in the order, ClO < ClO2 < ClO3 < CIO4 . This means ClO4 will have maximum stability and therefore will have minimum attraction for H+. Thus ClO4 will be weakest base and its conjugate acid HClO4 is the strongest acid.

ClO4 is the conjugate base of the acid HClO4.

(ii) E0cell = E0ox + E0red = 0.44 V +0.34 V = 0.78 V
These +ve E0cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 37.
(a) (i) Explain the formation of water with copper catalyst by intermediate compound formation theory. (3)
(ii) O-nitro phenol is slightly soluble in water where as P-nitro phenol is more soluble. Give reason. (2)
[OR]
(b) (i) What happens when the following alkenes are subjected to reductive ozonolysis.
1. propene
2.1-Butene
3. Isobutylene (3)
(ii) What are reducing and non – reducing sugars? (2)
Answer:
(a) (i) Formation of water due to the reaction of H2 and O2 in the presence of Cu can be given as
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 23

(ii) O-nitro phenol is slightly soluble in water and more volatile due to intra molecular hydrogen bonding, whereas P-nitro phenol is more soluble in water and less volatile due to intermolecular hydrogen bonding.

[OR]

(b)
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 24

(ii) Reducing sugars: Those carbonhydrates which contain free aldehyde or ketonic group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosacchaides whether aldose or ketone are reducing sugars.

Non-reducing sugars: Cabohydrates which do not reduce Tollen’s reagent and Fehling’s solution are called non-reducing sugars. Example Sucrose. They do not have free aldehyde group.

Question 38.
(a) A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2 (B) on heating with liquid ammonia followed by treating with Br2 /KOH gives (C) which on treating with NaNO2 and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D. (5)
[OR]

(b) Explain the mechanism of cleansing action of soaps and detergents.(5)
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 25

[OR]

(b) Mechanism of cleansing action of soaps and detergents:

  • The cleansing action of both soaps and detergents from their ability to lower the surface tension of water, to emulsify oil or grease and to hold them in a suspension in water.
  • This ability is due to the structure of soaps and detergents.
  • In water a sodium soap dissolves to form soap anions and sodium cations. For example, the following chemical equation shows the ionisation of sodium palmitate.
    Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 26
  • A soap anion consists of a long hydrocarbon chain with a carboxylate group on one end. The hydrocarbon chain, which is hydrophobic, is soluble in oils or grease. The ionic part is the carboxylate group which is hydrophilic, is soluble in water.
    Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 27
  • In water, detergent dissolves to form detergent anions and sodium cations. For example the following chemical equations show the ionisation of sodium alkyl sulphate and sodium alkyl benzene sulphate.
    Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium - 28

6. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy stain:

  • A soap or detergent anion consists of a hydrophobic part and a hydrophilic part.
  • Soap or detergent reduces the surface tension of water. Therefore the surface of the cloth is wetted thoroughly.
  • The hydrophobic parts of the soap or detergents anions are soluble in grease.
  • The hydrophilic parts of the anions are soluble in water.
  • Scrubbing or mechanical agitation helps to pull the grease away from the cloth and the
  • grease is broken into smaller droplets.
  • Repulsion between the droplets causes the droplets to be suspended in water, forming an emulsion.
  • Thus the droplets do not coagulate or redeposit on the cloth. Rinsing washes away the droplets.

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Chemistry Model Question Paper 4 English Medium

Time: 3 Hours
Maximum Marks: 70

Instructions:

  1. The question paper comprises of four parts
  2. You are to attempt all the parts An internal choice of questions is provided wherever applicable
  3. All questions of Part I, II, III, and IV are to be attempted separately
  4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
  5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences
  6. Question numbers 25 to 33 in Part III are three-marks questions. These are to be answered in about three to five short sentences
  7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Depressing agents used to separate ZnS from PbS is …………..
(a) NaCN
(b) NaCl
(c) NaNO3
(d) NaNO2
Answer:
(a) NaCN

Question 2.
The basic structural unit of silicates is
(a)(SiO3)2-
(b) (SiO4)2-
(c) (SiO)
(d) (SiO4)4-
Answer:
(d) (SiO4)4-

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 3.
………………… is a pungent smelling gas.
(a) Ammonia
(b) Nitric acid
(c) Fluorite
(d) Sodium chloride
Answer:
(a) Ammonia

Question 4.
Consider the following statement.
(i) All the actinoids are non radioactive.
(ii) Neptunium and other heavier elements are produced by artificial transformation of naturally occurring elements by nuclear reactions.
(iii) Most.of the actinoids have long half lives.
Which of the above statements is/are not correct.
(a) i only
(b) i and ii
(c) ii and iii
(d) i and iii
Answer:
(d) i and iii

Question 5.
How many geometrical isomers are possible for [Pt (Py) (NH3) (Br) (Cl)]?
(a) 3
(b) 4
(c) 0 3
(d) 15
Answer:
(a) 3

Question 6.
Metal excess defect is possible in……………….
(a) AgCl
(b) AgBr
(c) KCl
(d) Fes
Answer:
(c) KCl

Question 7.
The correct difference between first and second order reactions is that ……………………
(a) A first order reaction can be catalysed; a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.
Answer:
(b) The half life of a first order reaction does not depend on [A0]; the half-life of a second order reaction does depend on [A0].
Solution:
For a first order reaction
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 1

Question 8.
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.24 × 10-4 mol L-1 solubility product of Ag2C2O4 is
(a) 2.42 × 10-8 mol3 L-3
(b) 2.66 × 10-12mol3 L-3
(c) 4.5 × 10-11mol3 L-3
(d) 5.619 × 10-12mol3 L-3
Answer:
(d) 5.619 × 10-12mol3 L-3
Solution:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 2

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 9.
Assertion(A): Copper Sulphate can be stored in a Zinc vessel.
Reason (R): Zinc is less reactive than Copper.
(a) Both A and R are correct
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are wrong

Question 10.
Fog is colloidal solution of……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
cTamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 3
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 4
(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 5
Answer:
(c) benzene diazonium chloride

Question 13.
Which one of the following is the IUPAC name of CH3 – CH2 – CH2 CN?
(a) Propiono nitrile
(b) Butane cyanide
(c) Isobutyro nitrile
(d) Butane nitrile
Answer:
(d) Butane nitrile

Question 14.
The correct corresponding order of names of four aldoses with configuration given below Respectively is ……………
(a) L-Erythrose, L-Threose, L-Eiythrose, D-Threose
(b) D-Threose, D-Erythrose, L-Threose, L-Erythrose
(c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer:
(d) D-Erythrose, D-Threose, L-Erythrose, L-Threose

Question 15.
Tranquilisers are substances used for the treatment of ………………
(a) cancer
(b) AIDS.
(c) mental diseases
(d) blood infection
Answer:
(c) mental diseases

Part – II

Answer any six questions. Question No. 22 is compulsory. [6 × 2 = 12]

Question 16.
What are leaching process?
Answer:
This method is based on the solubility of the ore in a suitable solvent and the reactions in aqueous solution. In this method, the crushed ore is allowed to dissolve in a suitable solvent, the metal present in the ore is converted to its soluble salt or complex while the gangue remains insoluble. This process is also called chemical method.

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 6

Question 18.
Write the biologically importance of coordination compounds.
Answer:
Biological important of coordination compounds:

(i) Ared blood corpuscles (RBC) is composed of heme group, which is Fe2+– Porphyrin complex, it plays an important role in carrying oxygen from lungs to tissues and carbon dioxide from tissues to lungs.

(ii) Chlorophyll, a green pigment present in green plants and algae, is a coordination complex containing Mg2+ as central metal ion surrounded by a modified Porphyrin ligand called corrin ring. It plays an important role in photosynthesis, by which plants converts CO2 and water into carbohydrates and oxygen.

(iii) Vitamin B12(cyanocobalamine) is the only vitamin consist of metal ion. it is a coordination complex in which the central metal ion is CO+ surrounded by Porphyrin like ligand.

(iv) Many enzymes are known to be metal complexes, they regulate biological processes. For example, Carboxypeptidase is a protease enzyme that hydrolytic enzyme important in digestion, contains a zinc ion coordinated to the protein.

Question 19.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

  • Nature and state of the reactant .
  • Concentration of the reactant
  • Surface area of the reactant
  • Temperature of the reaction
  • Presence of a catalyst

Question 20.
What is meant by standard reduction potential? What is its application?
Answer:

  • The standard reduction potential (E°) is a measure of the oxidising tendency of the species.
  • The greater the E° value means greater is the tendency shown by the species to accept electrons and undergo reduction.
  • So higher the E° values, lesser is the tendency to undergo corrosion.

Question 21.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

  • It is reversible
  • It has low heat of adsorption
  • It has weak van der Waals forces of attraction with adsorbent.
  • It increases with increase in pressure.
  • It forms multimolecular layer.

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 7
This reaction follows SN1 mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 8

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 9
(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 10
This is the reason why soaps do not work in hard water.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 × 3 = 18]

Question 25.
Write the application of Iron (Fe).
Answer:

  • Iron is one of the most useful metals and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
  • Cast iron is used to make pipes, valves and pumps stoves etc.
  • Magnets can be made of iron and its alloys and compounds.

An important alloy of iron is stainless steel, and it is very resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H2SO4.H2O) and di (H2SO4.H2O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 11

Question 27.
Give evidence that [CO(NH3)gCl]SO4 and [CO(NH3)5SO4]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO3 solution and BaCl2 solution. When Cl ions are the counter ions, a white precipitate will be obtained with AgNO3 solution. If SO42- ions are the counter ions, a white precipitate will be obtained with BaCl2 solution.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 12

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 13z = k [x]m [y]n …… (1)
8z = k [4x]m [y]n…… (2)
I 6z = k [4x]m [4y]n…… (3)
Dividing Eq (2) by Eq (1) we get,
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 14
1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 15
16 = 4m . 4n
16 = 42.4n
16/16 = 4n
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]m[y]n
k [x]1.5 [y]0
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 16Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 17

  • HF and F , HS and H2S are two conjugate acid – base pairs.
  • F is the conjugate base of the acid HF (or) HF is the conjugate acid of F .
  • H2S is the conjugate acid of HS (or) HS is the conjugate base of H2S.

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 18

  • HPO42- and PO43- , SO3-2 and HSO3 are two conjugate acid – base pairs.
  • PO43- is the conjugate base of the acid HPO42- (or)
    HPO42- is the conjugate acid of PO43-.
  • HSO3 is the conjugate acid of SO3-2 (or) SO3-2 is the conjugate base of HSO3

(iii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 19

  • NH4+ and NH3, CO32- and HCO3 are two conjugate acid – base pairs.
  • HCO3 is the conjugate of acid CO32- (or) CO32- is the conjugate bases of HCO3
  • NH3 is the conjugate base of NH4+ (or) NH4+ is the conjugate acid of NH3.

Question 30.
In the following fields, how adsorption is applied?
Answer:
(i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer:
(i) Medicine: – Drugs cure diseases by adsorption on body tissues.

(ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by pine oil.

(iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances used for fixing dyes onto the fabric.

(iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after getting absorbed on precipitate. It is used to indicate the end point of filtration.

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 31.
Identify A, B, C, and D
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 20
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 21
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 22

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 23

Question 33.
Differentiate soap and detergents?
Soap :

  1. Soaps are sodium or potassium salt of long chain fatty acid.
  2. Soaps are made from animal (or) plant fats and oils.
  3. Soaps have lesser cleansing action.
  4. Soaps are bio degradable.
  5. Soaps are less effective in hard water.
  6. They have a tendency to form a scum in hard water.
  7. Example: Sodium palmitate.

Detergent :

  1. Detergent is sodium salt of alkyl hydrogen sulphate or alkyl benzene sulphonic acid.
  2. Detergents are made from petrochemicals.
  3. Detergents have more cleansing action.
  4. Detergents are non-bio degradable.
  5. Detergents are more effective even in hard water.
  6. They do not form scum with hard water.
  7. Example: Sodium lauryl sulphate.

Part – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) (i) Explain the role of carbon monoxide in the purification of nickel? (2)
(ii) Describe the structure of diborane. (3)
[OR]
(b) (i) What type of hybridisation occur in 1. BrF5. 2. BrF3 (2)
(ii) Most of the transition metals act as catalyst. Justify this statement. (3)
Answer:
(a) (i) During the purification of Nickel by Mond’s process, carbon monoxide (CO) is used to convert impure nickel to nickel carbonyl.
Nickel carbonyl is an unstable compound. Heating to higher temperature decomposes it to give pure nickel.

(ii) In diborane two BH2 units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to
form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds.
1. e. two three centred B-H-B bonds utilise two electrons each. Hence, these bonds are three centre- two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure.

In dibome, the boron is sp3 hybridised. Three of the four sp3 hybridised orbitals contains single electron and the fourth orbital is empty. Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled Is orbital of hydrogen.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 24

[OR]

(b) (i) 1. BrF5. is a AX5 type. Therefore is has sp3 d2 hybridisation. Hence, BrF5 molecule has square pyramidal shape.
2. BrF3 is a AX3 type. Therefore it has sp3 d hybridisation. Hence, BrF3 molecule has T-shape.

(ii) Many industrial processes use transition metals or their compounds as catalysts. Transition metal has energetically available d orbitals that can accept electrons from reactant molecule or metal can form bond with reactant molecule using its ‘d’ electrons.

For example, in the catalytic hydrogenation of an alkene, the alkene bonds to an active site by using its n electrons with an empty d orbital of the catalyst.

Question 35.
(a) (i) Why tetrahedral complexes do not exhibit geometrical isomerism. (2)
(ii) Explain about the importance and application of coordination complexes. (3)
[OR]
(b) Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
(a) (i) In tetrahedral geometry

  • AH the four ligands are adjacent or equidistant to one another.
  • The relative positions of donor atoms of ligands attached to the central metal atom are same with respect to each other.
  • It has plane of symmetry

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 25

(ii) 1. Phthalo blue – a bright blue pigment is a complex of copper (II) ion and it is used in printing ink and packaging industry.

2. Purification of Nickel by Mond’s process involves formation of [Ni (CO)4 ] which yields 99.5% pure on decomposition.

3 . EDTA is used as a chelating ligand for the separation of lanthanides, in softening of hard water and also in removing lead poisoning.

4. Coordination complexes are used in the extraction of silver and gold from their ores
by forming soluble cyano complex. These cyano complexes are reduced by zinc to
yield metals. This process is called Mac – Arthur Forrest cyanide process.

5. Some metal ions are estimated more accurately by complex fonnation. For eg., Ni present in Nickel chloride solution is estimated accurately forming an insoluble, complex called [Ni (DMG)2].

6. Many of the complexes are used as catalyst in organic and inorganic reactions. For e.g.,
(i) Wilkinson’s Catalyst – [(PPh3)3, Rh Cl] is used for hydrogenation of alkenes.
(ii) Ziegler – Natta Catalyst [TiCl4 + Al (C2H5) ]3 is used in the polymerisation of ethene.

7. In photography, when the developed film is washed with sodium thio sulphate solution (hypo), the negative film gets fixed. Undecomposed AgBr forms a soluble complex called sodium dithio sulphate argentate (I) which can be removed easily by washing the film with water.

AgBr + 2 Na2 S2O3 → Na3 [Ag (S2O3)2] + 2NaBr

[OR]

(b) (i) AAAA type of three dimensional packing:
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 26

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

(ii) ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure. The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 27

(iii) ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer. Third layer gives different arrangement. Fourth Layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (Le.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated with the addition of more layers.

In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12.

Voids: The empty spaces between the three dimensional layers are known as voids. There are
two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void:
A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom r Layer a layer is a hole between four spheres. The spheres are arranged at the vertices Layerc of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 28

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10-3s-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of kspof two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10-15 and 6 × 10-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t1/2 = 0.693/ k
t1/2= O.693/1.54 ×10-3s -1 = 450s

(ii) 1. Lyophobic sols are precipitated readily even with small amount of electrolytes. But they are stabilised by the addition of small amount of lyophillic colloid.

2. A smaLl amount of gelatine sol is added to gold sol to protect gold sol.

3. Gold number is a measure of protecting power of a colloid. Gold number is defined as the number of milligrams of hydrophillic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of ImI of 10% NaCI solution. Smaller the gold number, greater the protective power.

[OR]

(b) (i) 1. Buffer is a solution which consists of a mixture of weak acid and its conjugate base (or) a weak base and its conjugate acid.

2. This buffer solution resists drastic changes in its pH upon addition of a small quantities of acids (or) bases and this ability is called buffer action.

3. Acidic buffer solution : Solution containing acetic acid and sodium acetate.
Basic buffer solution : Solution containing NH4O and NH4Cl.

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 29
Ni(OH)2 is more soluble than AgCN.

Question 37.
(a) Describe about lead storage battery construction and its uses. (5)
[OR]
(b) (i) What happens when m-cresol is treated with acidic solution of sodium dichromate? (3)
(ii) Formic acid is more stronger than acetic acid. Justify this statement. (2)
(a) Lead storage battery :
1. Anode : Spongy lead
Cathode : Lead plate bearing PbO2
Electrolyte : 38% by mass of H2SO4 with density 1.2 g/ml

2. Oxidation occurs at the anode

Pb(s) → Pb2+ (aq) + 2e …………(1)

The Pb2+ ions combine with SO42- to form PbSO4 precipitate

Pb2+(aq) + SO42- → PbS4(s) …………….(2)

3. Reduction occurs at the cathode

PbO2(s) + 4H+(aq) + 2e – → Pb2+(aq) + 2H2O ………….(3)

The Pb2+ ions also combines with SO42- ions to form sulphuric acid to form PbSO4 Precipitate

Pb2+(aq) + SO42-(aq) → PbSO4 …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb(s) + PbO2(s) + 4H+ + 2SO42-(aq) → 2PbSO4(s) + 2H2O(l)

5. The emf of a single cell is about 2V. Usually six such cells are combined in series to produce 12 volts.

6. The emf of the cell depends on the concentration of H2SO4. As the cell reaction uses SO42- ions, the concentration H2SO4 decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H2SO4 is regenerated.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 30
The above reaction is exactly the reverse of redox reaction which takes place while discharging.

8. Uses: Lead storage battery is used in automobiles, trains, inverters.

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 31

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH3, – C2H5, – C3H7
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 32

Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium

Question 38.
(i) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating
forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’
of molecular formula C6H7N. Write the structures and IUPAC names of compound
A,BandC. (3)
(ii) Lactose act as reducing sugar. Justify this statement. (2)
[OR]
(b) (i) Explain the mechanism of enzyme action? (3)
(ii) Which chemical is responsible for the antiseptic properties of dettol? (2)
Answer:
(a) (i) Step-1: To find out the structure of compounds‘B’and‘C’.
1. Since compound ‘C’ with molecular formula C6H7N is formed from compound ‘B’ on treatment with Br2 + KOH (i.e, ., Hoffmann bromamide reaction). Therefore, compound ‘B’ must be an amide and ‘C’ must be an amine. The only amine having the molecular formula C6H7N is C6H5NH2 (i.e., aniline or benzenamine).

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C6H5CONH2). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 33

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 34

(ii) Lactose is a disaccharide and contains one galactose unit and one glucose unit.
In lactose, the β-D galactose and β-D glucose are linked by β-1, 4 – glycosidic bond.
The aldehyde carbon is not involved in the glycosidic bond hence it retains its reducing property and is called a reducing sugar.

[OR]

(b) (i) Enzymes are bio catalysts that catalyse a specific bio chemical reaction. They generally activate the reaction by reducing the activation energy by stabilising the transition state.

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.
Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium - 35

(ii) (a) Chloroxylenol and (b) Terpineol are the chemicals responsible for the antiseptic properties of dettol. But among these two, chloroxylenol plays more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.