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MP Board Class 12th Economics Important Questions Unit 10 Balance of Payments

Balance of Payments Important Questions

Balance of Payments Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Structure of balance of payment includes which account:
(a) Current account
(b) Capital account
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 2.
Balance of trade means :
(a) Capital transactions
(b)Import and export of goods,
(c) Total credit and debit
(d) All of the above.
Answer:
(b)Import and export of goods,

Question 3.
Measures to improve adverse balance of payment includes :
(a) Currency devaluation
(b) Import substitution
(c) Exchange control
(d) All of the above.
Answer:
(d) All of the above.

Question 4.
Foreign Exchange Rate is determined by :
(a) Demand of foreign currency
(b) Supply of foreign currency
(c) Demand and supply in foreign exchange market
(d) None of these.
Answer:
(c) Demand and supply in foreign exchange market

Question 5.
Types of Foreign Exchange Market are :
(a) Spot market
(b) Forward market
(c) Both (a) and (b)
(d) None of these.
Answer:
(c) Both (a) and (b)

Question 2.
Fill in the blanks:

1. Bretton woods system is also known as ………………… border system.
2. There is ………………… relation between foreign exchange rate and the supply of foreign exchange.
3. By devaluation, the value of currency …………………
4.  ………………… items are included in the balance of trade.
5. Balance of payment always remains …………………
6. The value of currency of one country with that of the currency of another country is called …………………

Answer:

1. Adaptable
2. Direct
3. Reduces
4. Visible
5. Balanced
6. Exchange rate.

Question 3.
State true or false :

1. Balance of trade includes both visible and invisible items.
2. Balance of trade is a part of Balance of payments.
3. Devaluation is declared by the government.
4. Balance of payment is always balanced.
5. For export promotion, help of devaluation is taken.
6. The increasing population in developing countries has direct impact on economic growth.
7. Export promotion is one of the ways of correcting Balance of payments.

Answer:

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

Question 4.
Match the following :

Answer:

1. (b)
2. (c)
3. (a)
4. (e)
5. (d)

Question 5.
Answer the following one word/ sentence:

1. New trade policy was declared in which year?
2. What will be the effect of devaluation of Indian currency on Indian imports?
3. In the long run, for what do the importers pay?
4. What does capital account imply?
5. What is the exchange of currency of one country in currency of another country called?

Answer:

1. 1991
2. Costly
3. Exports
4. International exchange and Indebtness
5. Ex – change Rate.

Balance of Payments Very Short Answer Type Questions

Question 1.
What is Balance of Payment?
Answer:
The balance of payment of a nation consists of the payments made, within a particular period of time between the residents of the country and the residents of foreign countries.

Question 2.
What do you mean by foreign exchange rate?
Answer:
Meaning:
The rate at which one currency buys or exchanges another currency is known as the rate of exchange. It simply expresses its external value or purchasing power. Foreign exchange rate between the currency units of two countries means the number of units of one national currency that are needed to buy one unit of other national currency.

Question 3.
Differentiate between Balance of Trade and Balance of Payment.
Answer:
Difference between Balance of Trade and Balance of Payment:
Balance of Trade:
The difference between exports and imports is called balance of trade.

Balance of Payment:
The difference between the total receipts of foreign exchange and total payment of foreign exchange is called balance of payment.

Question 4.
What do you mean by unfavourable balance of payment? Explain.
Answer:
Unfavourable balance of payment:
It is also called the deficit balance of payment. It refers to the situation when the total liability for payments of a nation exceeds the total receipts from foreign countries. Hence certain additional transactions are necessary to balance it such as export of gold, withdrawal of deposits in foreign banks etc.

Question 5.
What do you mean by capital account?
Answer:
Capital account is that account of balance of payment which records all such transactions between the residents of a country and rest of the world which cause a change in asset or liability of the country.

Question 6.
What do you mean by current account of balance of payment?
Answer:
Current account is that account of balance of payment which records imports and exports of goods and services and unilateral transfers. It includes visible, invisible and unilateral transfers.

Question 7.
What do you mean by managed floating system?
Answer:
Managed floating system is a mixture of flexible exchange rate system and fixed rate system. (The float part + managed part). Under this system the central bank tend to intervene to buy and sell foreign currencies in an attempt to reduce fluctuation in exchange rate.

Question 8.
What do you mean by protection?
Answer:
When by ending the freedom of foreign trade the ban is put on import and export of goods it is called protection.

Question 9.
Write one advantage of open economy?
Answer:
Investors get the choice of selection among domestic products and foreign goods.

Question 10.
What do you mean by dumping?
Answer:
When the goods are excess than demand then the seller sells this goods in foreign countries on lower rate it is called dumping.

Question 11.
Write points in favour of fixed exchange rate.
Answer:

1. To encourage international trade.
2. Formation of capital.
3. Encouragement to foreign capital
4. Encouragement to export countries.

Question 12.
What do you mean by foreign trade multiplier?
Answer:
Foreign exchange rate multiplier tells us how many times increase takes place in national income by increasing in export.

Question 13.
What does a balance of payment record?
Answer:
The balance of payment records the transaction in goods and services and assets between residents of a country with the rest of the world.

Question 14.
What is dirty floating?
Answer:
When managed floating in the absence of rules and guidelines are implemented, it is called dirty managed floating system.

Question 15.
What do you mean by import and export?
Answer:
Import:
When goods and services are brought from the foreign countries to the domestic countries it is called import.

Export:
Goods and services are send to foreign countries from domestic countries.

Question 16.
What does foreign exchange market include?
Answer:
Foreign exchange market includes banks specialised foreign exchange dealers, brokers, government agencies through which the currency of one country can be exchanged for that of other country.

Question 17.
What do you mean by flexible exchange rate?
Answer:
Flexible rate of exchange:
It is freely determined by the prices of supply and demand in the internal market.

Question 18.
What do you understand by managed floating rate system?
Answer:
It is a hybrid of a fixed exchange rate and flexible exchange rate system. In this system, central bank intervenes in the foreign exchange market to restrict the fluctuations in the exchange rate within certain limits. The aim is to keep exchange rate close to desired target values.

Question 19.
Give arguments in favour of fixed exchange rate.
Answer:
Following points can be studied in favour of fixed exchange rate:

1. Encouragement to international trade.
2. Encouragement to foreign capital.
3. Capital formation.

Question 20.
Write points against fixed exchange rate.
Answer:
Following are the points against fixed exchange rate :

1. Controlled economy.
2. Encouragement to corruption.
3. Sudden change in exchange rate.
4. Unfavourable effect on economic development.

Question 21.
Explain two merits and two demerits of fixed foreign exchange rate.
Answer:
(a) Two merits of fixed foreign exchange rate :

1. Fixed foreign exchange rate ensures stability in exchange rate. The exporters and importers do not have to operate under uncertainty about the exchange rate. Thus, it promotes foreign trade.
2. It also promotes capital movements.

(b) Two demerits of fixed foreign exchange rate :

1. Under this system, countries with deficit in balance of payment run down the stock of gold and foreign currencies. This can create serious problem for them.
2. There may be undervaluation of currency.

Question 22.
Write down the advantages of fixed exchange rate system.
Answer:
Advantages of fixed exchange rate system :

1. This system ensures stability in the international money market/ exchange market.
2. It encourages international trade.
3. It promotes bilateral trade agreements.
4. It avoids speculation.
5. It keeps the government under pressure to combat inflation.

Balance of Payments Short Answer Type Questions

Question 1.
What do you mean by fixed exchange rate? Write three points against it.
Answer:
A fixed exchange rate is a regime applied by a country whereby the government or central bank ties the official exchange rate to another country’s currency or the price of gold. The purpose of a fixed exchange rate system is to keep a currency’s value within a narrow band.

Following are the different points against the fixed exchange rate:
1. Controlled economic system:
For fixed rate of exchange it is compulsory to have strict control over economic system. If it is not possible then we will have to make changes in exchange rate.

2. Unfavourable effect on economic progress :
The main aim of fixed rate of exchange is to maintain stability in exchange rate. In this situation sometimes national income, employment policy, price level etc. are considered as secondary.

3. Corruption:
To maintain fixed exchange rate many restrictions are imposed in the country. Due to strict restrictions there is always possibility of corruption in the society.

4. Sudden change in the exchange rate:
Some times it becomes evitable to make changes in exchange rate. To keep the exchange rate stable some times currency of the nation becomes weak. In such stuation sometimes the there is devaluation. It has adverse effect on foreign trade and balance of payment.

Question 2.
What is meant by flexible exchange rate? Give arguments in favour and against flexible exchange rate.
Answer:
A system in which rate of exchange is determined by the sources of demand and supply of different currencies in foreign exchange market.

Following points which can be studied in favour of flexible exchange rate :

1. Independent economic policy : If the exchange rates are elastic any country can make their domestic economic policies internal policies) independently.

2. Implementation of monetary policy independently : If the rate of exchange is flexible monetary policy in the nation can be independently and effectively by changing the monetary policy.

Following points which can be studied against the flexible exchange rate :

1. Adverse economic effect:
If the flexible exchange rate is there then the feeling of insecurity comes in the minds of traders. It has an adverse effect on the foreign trade of the country. The tendency of gambling increases. If the exchange rate is reduced the inflation increases. The level of employment opportunities also go down.

2. Misuse of the resouroes:
If exchange rate goes on changing very often then the resources have to distributed again and again sometimes the resources are used in export sometimes for domestic industries. Due to these changes there is always wastage of resources.

Question 3.
What do you mean by balance of payment? What are the causes of adverse balance payment in India?
Or
Explain four causes of adverse balance of payment.
Answer:
Meaning of balance of payment:
The balance of payment of a nation consists of the payments made, within a particular period of time between the residents of the country and the residents of foreign countries.

Following are the causes of adverse balance of payment in India:
1. Increase in imports of petroleum products:
Oil producing countries are increasing the price of petroleum products every year. Along with it the consumption of petroleum products is also increasing day by day. Due to this import is done on lkrge scale.

2. Increase in import of machines:
Due to economic planning there is rapid growth in industrialization and progress in agricultural development. For this the need of machines was felt. Due to this reason more import has to be done.

3. Less increase in export:
The export did not increase according to expectations which is one of the causes of unfavourable balance of payment.

4. International loans and investment:
India has taken loan for developmental purpose. To repay the principal amount and interest, foreign exchange has to be paid. It gave rise to the situation of adverse balance of payment.

Question 4.
Suggest measures to improve the condition of adverse balance of payment
Answer:
Measures or method to correct adverse balance of payment : Following steps should be taken to improve adverse balance of payment in India :

1. Encouragement to export:
The government should encourage the export.

• The trade policy should be export oriented
• For this export tax should be reduced or some concession should be given for export of some goods
• Economic assistance should be provided to the industries of the country
• Advertisement should be done in foreign countries for the products.

2. Reduction in import:
India should reduce imports. For this import duty should be raised so, that imported goods will become expensive and people will be discouraged to purchase them. Domestic products should be discouraged reduce the imports. It will give rise to favourable balance of payment.

3. Foreign debt:
To remove adverse balance payment government can take foreign loan. But taking loan is a temporary solution of it.

4. Exchange control:
For keeping the balance payment exchange control is one method. By controlling exchange we can reduce import and increase export.

Question 5.
Give arguments in favour and against fixed exchange rate.
Or
Give arguments in favour of fixed exchange rate.
Answer:
Follo’wing are the points which can be studied in favour of fixed exchange rate :

1. Encouragement to international trade:
Under fixed exchange rate both importer and exporter know about the amount he has to pay and how much he will be getting. Thus in fixed exchange rate international trade develops in a balance way. Here there is less risk.

2. Encouragement to foreign capital:
If the exchange rate is stable foreign exchange can easily flow into the country because investor is not scared of getting less amount than what is fixed. There is no fear of bearing of loss if the rate of exchange is reduced.

3. Capital formation:
If the foreign exchange rate is fixed there is a favourable effect on internal condition of the country. There is no fear of inflation. In industry demand of capital is increased, savings is also increased. Thus rate of capital formation increases. It gives rise to the development of the country.

4. Exchange system:
If the exchange rate is fixed it does not give encouragement to the tendency of gambling. Thus government can control the exchange system in proper way.

5. Essential for export countries:
Some of the country depend on the income coming from export. Half the national income of such countries comes from exports. For countries like England, Denmark, Japan etc. fixed rate of exchange is very essential otherwise there will be adverse effect on its development.

Following points are there against the fixed rate of exchange :

1. Controlled economic system:
For fixed rate of exchange it is compulsory to have strict controlled over economic system. If it is not possible then we will have to make changes in exchange rate.

2. Unfavourable effect on economic progress:
The main aim of fixed rate of exchange is to maintain stability in exchange rate. In this situation sometimes national income, employment policy, price level etc. are considered as secondary.

3. Corruption:
To maintain fixed exchange rate many restriction are imposed in the country. Due to strict restriction there is always possibility of corruption in the society.

Question 6.
Differentiate between Balance of Trade and Balance of Payment.
Answer:
Differences between Balance of Trade and Balance of Payment:

Balance of Trade:

• The difference between exports and imports is called balance of trade.
• It refers to detailed description of imports and exports only.
• It includes visible items only.
• It may be favourable and unfavourable.
• If balance of trade is not favourable it is not a cause of great concern.
• Balance of trade is a part of balance payment.

Balance of Payment:

• The difference between the total receipts of foreign exchange and total payment of foreign exchange is called balance of payment.
• It comprises not only exports and imports but also services, capital, gold etc.
• It includes visible as well as invisible items both.
• It is always balanced.
• If the balance of payment is not favourable it is a cause of great concern for the nation.
• The concept of balance of payment is broader.

Question 7.
Differentiate between devaluation and depreciation.
Answer:
Differences between devaluation and depreciation:
Devaluation means lowering the value of one’s currency in terms of foreign currency. In this case, the domestic value of currency remains constant but its value in terms of foreign currencies fall. On the other hand, the fall in the price of foreign exchange under flexible exchange rate is known . as depreciation. For instance, if the equilibrium rupee – dollar exchange rate was Rs. 45 and now it has become Rs.50 due to rise in demand for dollars, then the rupee has depreciated against dollar.

Question 8.
What is the marginal propensity to import when M = 60 + 0.06Y? What is the relationship between the marginal propensity to import and the aggregate demand function?
Answer:
M = 60 + 0.6Y (Given)
M = \(\overline { m } \) + mY
Hence, m = 0.6
Where, m = Marginal propensity to import

Relationship:
There is positive relationship between marginal propensity to import and aggregate demand function. Marginal propensity of income spent. Thus,
m = \(\frac {∆m}{∆n}\)

Question 9.
Explain, why:
G -T = (S^g -1) – (X – M)?
Answer:
G -T = (Sf -1) – (X – M)
Here, G = Government expenditure
T = Taxes
S^g = Saving of government
I – Investment
S^g-I = Net Saving
X = Exporters
M = Importers
X – M = Balance of trade.
The given equation states that net government expenditure equals net government savings and balance of trade. It implies that net government expenditure is financed by government savings and trade deficit. Hence, the given equation is correct.

Question 10.
Should a current account deficit be a cause for alarm? Explain.
Answer:
When a country runs a current account deficit, then he must see whether there has been a decrease in saving, increase in investment or an increase in the budget deficit. There is reason to worry about a country’s long prospects of the trade deficit reflects smaller savings or a larger budget deficit. The deficit could reflect higher private or government consumption. In such cases, the country’s capital stock will not rise rapidly enough to yield enough growth it needs to repay its debt. There is less cause to worry, if the trade deficit reflects a rise in investment, which would build the capital stock more quickly and increase future output.

Question 11.
Distinguish between Balance of Trade and Balance on Current Account.
Answer:
Differences between Balance of Trade and Balance on Current Account:

Balance of Trade Account:

• Balance of trade account records the difference between value of imports and exports of material goods (visible items). lateral transfer (visible and invisible)
• Balance of trade is a part of balance on current account. So, it is a narrow concept.

Balance on Current Account:

• Balance on current account records the difference between receipts and payments of foreign exchange on account of goods, services and uni – items).
• Balance on current account is a wide concept.

Question 12.
If inflation is higher in country A than in country B and the exchange rate between the two countries is fixed, what is likely to happen to the trade balance between two countries ?
Answer:
Effect of inflation on the trade balance:
As the inflation is higher in country A than country B, so the prices of country A will be higher as compared to those of country B. In this case exports of A country will fail. The aggregate demand will fall and output and income will fall. Comparatively less price in country B will make its products less expensive and hence again increases w.e.f. net export and domestic output and income. The trade balance of country A will become deficit.

Balance of Payments Long Answer Type Questions

Question 1.
Write the components of Balance of payment.
Answer:
The Balance of payment of a nation consist of the payments made, within a particular period of time between the residents of that country and the residents of foreign countries. (MPBoardSolutions.com) In other words, it is an account of transactions involving receipts from foreigners on one side and payments to foreigners on the other side. The farmer relates to the international income of a country, they are called “credits” and since the later relates to the international out go, they are call “debits”.

Balance of payment includes all other payments apart from export and import. For example fee of banks, interest, profit, transfer of capital etc. are also included in balance of payment.

Question 2.
Write the main components of capital accounts.
Answer:
The main components of capital accounts are :

1. Borrowings and lending to and from abroad: It includes :
All transactions relating to borrowings from abroad by private sectors, government, etc.
All transactions of lending to abroad by private sectors and government.

2. Investment to and from abroad:
Investments by rest of the world in shares of
Indian companies, real estate in India etc.
Investments by Indian residents in shares of foreign companies, real estate abroad, etc.

3. Change in foreign exchange reserves:
The foreign exchange reserves are the financial assets of the government held in the central bank.

Question 3.
Explain the factors affecting fluctuations in foreign exchange.
Answer:
The factors affecting fluctuations in foreign exchange :

1. Banking related effects:
Banks through their functions affect the exchange rate. If the commercial bank float bank draft and other credit letters in large quantity then the demand for foreign exchange increases and the exchange rate of the country currency decreases. On the other hand, when foreign exchange bank floats credit letters against the country then the demand for home currency increases and the exchange rate becomes favourable for the country.

2. Change in prices:
In comparative view the change in prices results in the change of exchange rate of the country.

3. Impact of imports and exports:
Changes in the import and export quantity of country has a direct impact on the countries exchange rate if export increases in comparison to import the demand for foreign exchange. (MPBoardSolutions.com) If exports increase in comparison to imports the demand for foreign exchange increases and the countries exchange rate becomes favourable. But on the other hand if imports increase then the demand for country currency increases in the foreign country and this becomes unfavourable for the country.

4. Impact of speculation:
The changing trend in speculation trends also have an impact on exchange rate. In short period the.high rate of exchange leads to speculation tendencies. The uncertainty of exchange rate in international money market also encourages speculative motive.

5. Flow of capital:
The flow of capital from a country also affects the exchange rate. Flow of capital from one country to another to earn high profits is possible in short period or flow of capital to foreign countries for investment in the long period in the foreign country is also possible.

Question 4.
Explain visible and invisible export and import.
Answer:
Visible Import and Exports:
Such goods are included in visible imports and exports whose account is maintained in the register of ports. By seeing them we can find out the values of import and export done throughout the year. (MPBoardSolutions.com) In it only the export and imports of goods are kept.

Invisible Import and Exports:
In invisible export and import we include services which exchange are included. They are banking services, insurance, shipping services, education in foreign, medical facilities, tourism, interest, profit, military assistance, foreign donation, penalty etc. whose account is not maintained on ports are included in it.

Question 5.
Suppose C =100 + 0.75Y, 0T = 500, G = 750, taxes are 20% of income. X = 150, M = 100 + 0.2Y. Calculate equilibrium income, the budget deficit or surplus and the trade deficit or surplus.
Answer:
C = 100 + 0.75 YD
Here, C = 100, C = 0.75, I = 500, G = 750, X = 150, M = 100 + 0.2Y
Tax income (r) = 20%
Income (Y) = C + C (1 – t) Y +1 + G + (X – M)
or Y = 100 + 0.75 (1 – 0.2)Y + 500 + 750 + (150 – 100 – 0.2Y)
or Y = 100 + 0.75(0.8) Y + 500 + 750 + 150 – 100 – 0.2Y
or Y= 100 + 0.6Y+ 1300 – 0.2Y
or = 1400 + 0.4Y
or Y – 0.4 Y = 1400
or 0.6Y = 1400
or Y = \(\frac {1400}{0.6}\) = 2333
Deficit budget = Govt. expenditure Tax – (G) – Tax
= 750 – 2333 of 20 %
= 750 – 467 = 283
M = 100 + 0.2Y
= 100 + 0.2 (2333)
= 100 + 467 = 567
So, Trade deficit = M – X = 567 – 150 = 417.

MP Board Class 12th Economics Important Questions

MP Board Class 12th Physics Important Questions Chapter 4 Moving Charges and Magnetism

Moving Charges and Magnetism Important Questions

Moving Charges and Magnetism Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The magnetic effect of electric current was discovered by :
(a) Flemming
(b) Faraday
(c) Ampere
(d) Oersted.
Answer:
(d) Oersted.

Question 2.
A moving charge produces :
(a) Only electric field
(b) Only magnetic field
(c) Both electic and magnetic field
(d) Neither electric nor magnetic field.
Answer:
(c) Both electic and magnetic field

Question 3.
The SI unit of magnetic field intensity is :
(a) N/m
(b) Gauss or oersted
(c) N/A – m
(d) weber x metre .
Answer:
(c) N/A – m

Question 4.
Amperes circuital rule is :

Answer:

Question 5.
The magnetic field produced at the centre of a circular coil carrying current is :
(a) In the plane of plane
(b) Perpendicular to the plane of coil
(c) At 45° from the plane of coil
(d) At 60° from the plane of coil.
Answer:
(b) Perpendicular to the plane of coil

Question 6.
The force on a charge moving in a uniform magnetic field is zero if the direction of motion of charge is :
(a) Perpendicular to the magnetic field
(b) At 45° from magnetic field
(c) At 60° from the magnetic field
(d) Parallel to the magnetic field.
Answer:
(d) Parallel to the magnetic field.

Question 7.
The torque on a current carrying loop in a uniform magnetic field is maximum when the plane of loop is :
(a) Parallel to the magnetic field
(b) Perpendicular to the magnetic field
(c) At 45° from the magnetic field
(d) At 60° from the magnetic field.
Answer:
(a) Parallel to the magnetic field

Question 8.
To measure the current in a circuit we use :
(a) Voltmeter
(b) Galvanometer
(c) Ammeter
(d) Voltameter.
Answer:
(c) Ammeter

Question 2.
Fill in the blanks :

1. The SI unit of permeability is ……………..
2. SI unit of magnetic field is ……………..
3. Dimensional formula of magnetic field is ……………..
4. A current carrying solenoid behaves like a ……………..
5. The lorentz force on a charged particle in a uniform magnetic field is given as ……………..
6. The force between two parallel conductors carrying current in same direction is …………….. in nature.
7. The resistance of an ideal ammeter is ……………..

Answer:

1. ampere²
2. newton/ampere
3. [MT^-2 A^-1 ]
4. Bar magnet
5. q\(\vec { (v } ×\vec { B) } \)
6. Attractive
7. Infinite

Question 3.
Match the Column :
I.

Answer:

1. (c)
2. (e)
3. (d)
4. (a)
5. (b)

II.

Answer:

1. (d)
2. (e)
3. (b)
4. (a)
5. (c)

Question 4.
Write the answer in one word/sentence

1. State Ampere’s circuital law.
2. Which material is used for the suspension wire of a moving coil galvanometer and why?
3. What should be the resistance of an ideal voltmeter and ammeter?
4. What happens if a voltmeter is connected in series to the circuit?
5. What is a cyclotron?
6. What do you mean by magnetic effect of current?
7. State Maxwell’s right – hand screw rule.
8. What happens if a voltmeter is connected in series to the circuit?

Answer:
1. Ampere’s circuital law states that the line integral of magnetic field \(\vec { B }\) around any closed path is equal to Mo times the total current I enclosed by the path. Mathematically \(\oint { \vec { B. } } \vec { dl }\) = µ₀I

2. Phosphor bronze alloy is used as suspension wire because it has small restoring torque per unit twist and has a high tensile strength

3. An ideal voltmeter should have infinite resistance and the resistance of an ideal ammeter should be zero

4. The resistance of voltmeter is very high, therefore current will be decreased to almost zero

5. Cyclotron is a device used to accelerate positively charged particles (like protons, a particles). So that they can acquire sufficient energy to carry out nuclear disintegrations

6. When current is passed through any conductor, a magnetic field is produced around it. This phenomenon is called magnetic effect of current.

7. If a cork screw is turned so that it advances in the direction of current along the wire, then the direction in which the thumb rotates gives the direction of magnetic lines of force.

8. The resistance of voltmeter is very high, therefore current will be decreased to almost zero.

Moving Charges and Magnetism Very Short Answer Type Questions

Question 1.
Write practical unit of current and define it.
Answer:
The practical unit of current is ampere. One ampere of current is that current which produces a field of 10^-7 Wb/m², at the centre of the conductor of length 1m, placed in the form of an arc of a circle of radius 1 m.

Question 2.
Write S.I. unit of magnetic field intensity and define it.
Answer:
The S.I. unit of magnetic field intensity is newton / ampere x metre.
The intensity of the magnetic field is 1 newton ampere^-1 metre^-1. If 1 newton force acts on a conductor of length lm carrying a current of 1 ampere and held perpendicular to the magnetic field.

Question 3.
Write an expression for force acting on current – carrying conductor placed in magnetic field. Give the meaning of symbols used.
Answer:
The force on current – carrying conductor kept in magnetic field is given by :
F = IBl sinθ
Where, I = Current, l = Length of conductor, B = Field intensity and θ = Angle between the direction of magnetic field and the conductor.

Question 4.
What is the practical unit of current? Give its definition.
Answer:
If two parallel conductors situated at a perpendicular distance of 1 metre carry equal current in the same direction and exert an attractive force of 2 x 10^-7 newton on each other in air or vacuum, then the current on each conductor is equal to one ampere.

Question 5.
Why a soft iron core is kept in moving coil galvanometer?
Answer:
The magnetic lines of force crossed through the soft iron core. This increases the magnetic field and hence sensitivity of galvanometer. The soft iron core helps to make the magnetic field radial.

Question 6.
The pole pieces of magnet are cut concave in a galvanometer. Why?
Answer:
So that the magnetic field becomes radial, hence the plane of the coil becomes parallel to the magnetic field. Under this condition, the deflecting torque on the coil is maximum.

Question 7.
What happens if an ammeter is connected in parallel to the circuit? What is the resistance of an ideal ammeter?
Answer:
The resistance of an ammeter is very less, hence almost all the current will flow through the ammeter which may damage the ammeter. The resistance of an ideal ammeter is zero.

Question 8.
Why an ammeter is connected in series in an electric field?
Answer:
An ammeter measures the electric current of the circuit, hence all the current should pass through the ammeter. Therefore, it is connected in series.

Question 9.
The resistance of ammeter should be very small. Why?
Or
The resistance of an ideal ammeter is zero. Why?
Answer:
An ammeter measures the current of an electric circuit, therefore it is connected in series. If the resistance of ammeter is large, then it will decrease the current in the circuit. Thus, the resistance of an ammeter should be less or zero.

Question 10.
What will be the resultant magnetic field intensity of point O as shown in the figure?

Answer:
Due to the straight portion of the wire the resultant field will be zero and due to semi circle. The resultant field intensity will be
B = \(\frac { { µ }_{ 0 }I }{ 4R }\)

Moving Charges and Magnetism Short Answer Type Questions

Question 1.
What is second right – hand palm rule? Write its uses.
Answer:
Stretch out the palm of your right – hand such that the fingers are perpendicular to the direction of thumb.

If the thumb points the direction of current and the fingers point the direction of magnetic field, then the force acting on conductor will be in upward direction perpendicular to Direction i the palm.

Uses:
By it’s we can find out intensity of magnetic filed due topufrent carrying conductor.

Question 2.
Write Biot – Savart law for the magnetic field produced due to an element of a current – carrying conductor and explain the term used in it. Define the unit of current with the help of it
Answer:
Let AB be a conductor carrying current I. Consider a small line element dl of the conductor, due to which the magnetic field dB is produced at point P, then the strength of the magnetic field \(\vec { (dB) }\) depends on the following factors :

1. The field is directly proportional to current I.
i.e., dB ∝I

2. The field is directly proportional to the length of element,
i.e., dB ∝ dl

3. The field is directly proportional to the sine of angle between the line joining the point and dl.
i.e., dB ∝ sinθ

4. The field is inversely proportional to the square of the distance between the observation point and line element.
i.e.,
dB ∝ \(\frac { 1 }{ { r }^{ 2 } }\)
Combining all the four points,we get
dB ∝ \(\frac { Idl sinθ }{ { r }^{ 2 } }\)
or dB ∝ K\(\frac { Idl sinθ }{ { r }^{ 2 } }\) … (1)
Where, K is constant of proportionality. Its value depends upon the system of units
In C.G.S system, k = 1
∴ dB ∝ K\(\frac { Idl sinθ }{ { r }^{ 2 } }\) gauss … (2)
In M.K.S. system, K = \(\frac { { μ }_{ 0 } }{ 4π }\)
Where, μ₀ = Permeability of free space.
∴ dB = \(\frac { { μ }_{ 0 } }{ 4π }\) \(\frac { Idl sinθ }{ { r }^{ 2 } }\) Wb/m₂ … (3)
or dB =10^-7 [/latex] \(\frac { Idl sinθ }{ { r }^{ 2 } }\) Wb/m₂ … (4)
The relations given by eqns. (2), (3) and (4) are called Biot – Savart law.
The direction of the magnetic field \(\vec { dB }\) is always perpendicular to the plane containing \(\vec { dl }\) and \(\vec { r }\) and is given by the right – hand screw rule for the cross product of vectors.

Unit of electric current:
1. In C.GS. system : If dl= 1cm, r= 1cm, sinθ = 1 i.e., 9 = 90° and dB = 1 gauss, then from eqn. (2) I = 1 electromagnetic unit (e.m.u.).
For 9= 90°, the conductor should be taken as a part of circle, as the radius is always perpendicular to the circumference. Thus, 1 e.m.u. of current is that current which produces a field of 1 oersted at the centre of the conductor of length 1 cm, kept in the form of an arc of a circle of radius 1 cm.

2. In M.K.S. system: If dl =1m, r = 1m, sinθ = 1 and dB = 10^-7 Wb/m², then from eqn. (4)
I = 1 ampere.
Thus, 1 ampere of current is that current which produces a field of 10^-7 Wb/m², at the centre of tWconductor of length 1m.

Question 3.
Find the expression for magnetic field intensity for a Toroid.
Answer:
A solenoid bent into the form of closed ring is called toroidal solenoid to fig. (a).
In a toroidal solenoid, the magnetic field \(\vec { B }\) has a constant magnitude everywhere inside the toroid while it is zero in the open space interior (point P) and exterior (point Q) to the toroid.

The direction of magnetic field in – side is clockwise as per the right hand thumb rule for circular loops. Three circular Amperian loops are shown by dashed lines.

By symmetry, the magnetic field should be tangential to them and constant in magnitude for each of the loops.

1. For points in the open space interior to the toroid:
Let B₁ be the magnitude of the magnetic field along the Amperian loop 1 of radius r₁
Length of the loop L₁ = 2πr₁
As the loop encloses no current, I = 0
Applying Ampere’s circuital law.
B₁L₁ = μ₀I
or B₁2πr₁ = μ₀ × 0
Thus, the magnetic field at any point P in the open space interior to the toroid is zero.

2. For points inside the toroid:
Let B be the magnitude of the magnetic field along the Amperian loop 2 of radius r.
Length of loop 2, L₂ = 2π r
If N is the total number of turns in the toroid and I the current in the toroid, then total I the current enclosed by the loop 2 = NI.
Applying Ampere’s circuital law.
B × 2πr = μ₀ NI
or B = \(\frac { { μ }_{ 0 }NI }{ 2πr }\)
If r be the average radius of the toroid and n the number of turns per unit length, then
N = 2πrn
∴ B = \(\frac { { μ }_{ 0 }I }{ 2πr }\).2πrn
or B = μ₀I n.

3. For points in the open space exterior to the toroid:
Each turn of the torpid passes twice through the area enclosed by the Amperian loop 3. For each turn, the current coming out of the plane of paper is cancelled by the current going into the plane of paper.
Thus, I = 0 and hence B₃ = 0.

Question 4.
Write four similarities between Biot – savart law and coloumb’s inverse square law.
Answer:
The four similarities between both are :

1. Both laws obey inverse square law.
2. Wide range of field is given by both the law’s.
3. Principle of superpositions held for both the law.
4. Both the law is effected by the medium of surrounding of the conductor.

Question 5.
State and prove Ampere’s circuital law.
Answer:
Ampere’s circuital law:
Ampere’s circuital law states that the line integral of magnetic field B around any closed path is equal to μ₀ times the total current I enclosed by the path.

Mathematically \(\oint { \vec { B. } } \vec { dl }\) = µ₀I
Proof:
Consider an infinitely long straight conductor carrying current I. The magnetic lines of force are produced around the conductor as concentric circles.
The magnetic field due to this current – carrying infinite conductor at a distance a is given by
B = \(\frac { { μ }_{ 0 } }{ 4π }\) \(\frac {2I}{a}\), … (1)
(from Biot – Savart law)

Consider a circle of radius a. Let XY be a small element of length dl. \(\vec { dl }\) and \(\vec { B }\) are in
the same direction because direction of [/latex] and \(\vec { B }\) is along the tangent to the circle.
The line integral for the closed path will be

This proves Ampere’s circuital law.

Question 6.
Obtain an expression for the magnetic field due to a long straight current carrying conductor using Ampere’s circuital law.
Answer:
Consider an infinite long to conductor XY carrying current I as shown in the figure.
Magnetic field at P has to be found out. Distance between P and the wire is ‘a’. Draw an Amperian loop of radius a. Consider a line element RS = \(\vec { dl }\).

Let \(\vec { B}\) be the magnetic field at P; then the line integral of magnetic field along the circular path = \(\oint { Bdl }\) By Ampere’s circuital law,
\(\oint { Bdl }\) = µ₀
Where, I is the total current flowing in the Amperian loop.
Angle between \(\vec { B }\) \(\vec { dl }\) = 0

This is the strength of the magnetic field due to the conductor at a distance a.

Question 7.
Derive on expression for force acting on a current carrying conductor in a magnetic field.
Answer:
We know that a moving charge experiences a magnetic force in a uniform magnetic field. It can be extended for a current conductor placed in a uniform magnetic field, because electric current is the flow of free electrons. Let l be the length of a conductor, A its area of cross – section and n be the number density of free electrons.

Then no. of free electrons in the conductor N=nAl
Let conductor be lying along Y – axis and magnetic field B lying in YY – plane makes angle 6 with Y – axis. Current flows through the coil along the direction of X – axis. Electric current flows through a conductor due to unidirectional flow of free electrons. Therefore, magnetic force acting on each free electron
\(\oint { f }\) = -e\((\vec { { v }_{ d } } \times \vec { B } )\)
Where \(\vec { { v }_{ d } }\) is the drift velocity of electron and e is the charge on an electron. Therefore, total magnetic force acting on the conductor

I \(\vec { l }\) is a current element directed along the direction of current. I \(\vec { l }\) and vd are directed in opposite directions. Therefore, eqn. (2) may be written as

Case I : If θ = 0° , then sinθ = sin 0° = 0
Then from eqn. (4), F = 0
Thus, no magnetic force acts on a current carrying conductor lying parallel to magnetic field.

Case II : If θ = 90° ⇒ sinθ = sin90° = 1
Then p = JlB sin 90° or F = IlB (maximum)
Thus, a current carrying conductor placed perpendicular to a magnetic field experiences maximum magnetic force.

Question 8.
State Fleming’s left – hand rule.
Answer:
Stretch the forefinger, the middle finger and the thumb of your left – hand so that they are mutually perpendicular to each other. If the forefinger points the direction of magnetic field, the middle finger points the direction of current then the thumb indicates the direction of force acting on the conductor.

Question 9.
For circular motion of a charged particle in a uniform magnetic field obtain the expression for radius of path and periodic time of revolution.
Or
Discuss the motion of the charged particle in a uniform magnetic field with the initial velocity perpendicular to magnetic field.
Answer:
Consider a charged particle with charge q which is moving with a velocity v in a magnetic field of intensity B. Then, the maximum Lorentz force acting on the charged particle will be
F = qvB
and the direction of the force is perpendicular to v and B which is according to Fleming’s left – hand rule. So, no work will be done by the force on the charge because d W = Fd cosθ, here 0= 90°, hence d W = 0. It means that kinetic energy or the speed of the charged particle will be constant. So, the charge will move on a circular path.

Now, for circular motion Lorentz force provides the necessary centripetal force.
Let the mass of the charged particle be m and the radius of circular path be r.
Then, Lorentz force = Centripetal force
or qvB = \(\frac { { mv }^{ 2 } }{ r }\)
or r = \(\frac { mv}{ qB }\) … (1)
or r = \(\frac { p}{ qB }\) … (2)
Where, mv = p = Momentum of the particle.
Hence, from eqn. (2), the radius of circular path is directly proportional to the momentum.
Again, angular velocity, w = \(\frac { v}{ r }\) = \(\frac { qB}{ m }\)
Frequency, v = \(\frac { w}{ 2π }\) = \(\frac { qB}{ 2πm }\)
and T = \(\frac { 1}{ v }\)
= \(\frac {2πm}{qB}\)

Question 10.
When the current flows in opposite direction in two parallel wires, both repel each other, why?
Answer:
If direction of current are opposite in the conductor then according to Fleming left hand rule the direction of force acting on the conductor CD will be in the plane of the paper and opposite to conductor AB. The direction of force acting on AB due to CD will be perpendicular to AB and opposite to CD on the plane of paper. Obviously the conductor will be repel each other.

Question 11.
When the current flows in the same direction in two “ parallel wires. Both attract each other, why?
Answer:
In the direction of the current in both the conductors are same, then according to Fleming left hand rule, the direction of force acting on the conductor CD carrying current will be in the plane of paper perpendicular to conductor CD toward the conductor AB on the other hand the direction of force acting in conductor AB will be in the plane of paper perpendicular to conductor AB toward the conductor CD. Obviously the conductor AB and CD will attract each other.

Question 12.
Given two parallel wires carrying currents I₁ and I₂ are kept at a distance d apart. Obtain the expression for force exerted by one on (nt length of another. When will this force be attractive and when will it be repulsive?
Answer:
Let AB and CD be two parallel conductors kept at a distance d apart, current flowing through them is I₁ and I₂ respectively.
A magnetic field due to current I, is produced around AB.
∴ Intensity of magnetic field due to AB at a distance d is
B₁ = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { { 2I }_{ 1 } }{ d }\) Wb/m²
According to right – hand palm rule, the direction of this field is downwards, normal to the plane of the paper.

If another conductor CD carrying current I₂ in the same direction of I₁ is situated in the magnetic field of AB, then force on length l of CD is given by
F = I₂ /B₁ sin 90° = I₂/B₁
or F = I₂l.\(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { { 2I }_{ 1 } }{ d }\)
or F = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { { 2I }_{ 1 }{ I }_{ 2 }l }{ d } \).
Force acting per unit length on CD will be
\(\frac { F}{ l }\) = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { { 2I }_{ 1 }{ I }_{ 2 }}{ d } \).
This is the required expression.
The direction of magnetic field B, on wire CD is acting inwards. Hence, by Fleming’s left – hand rule, the force F₂ acting on CD will be directed towards AB, hence CD will come near to AB. So, the conductors attract each other, [see Fig. (a)] And when the current flows in opposite direction, they will repel each other, [see Fig. (b)].

Question 13.
Obtain an expression for the torque or the couple acting on a current loop, when it is placed in a magnetic field.
Or
Prove that the torque \(\oint { τ }\) acting bn a rectangular loop is given by \(\oint { τ }\) = \(\oint { m }\) \(\oint { B }\), where \(\oint { m}\) is the magnetic moment of the loop.
Answer:
Consider a rectangular coil ABCD of length ‘l’ and breadth ‘ b’ kept in a magnetic field of strength \(\oint { B }\). Let I be the amount of current flowing through the coil.
Force acting on AB is F₁ = BI/sin90° = BIL. By Fleming’s left hand rule, it comes out of the plane of paper.
Force acting on CD is F₂ = Bl/ sin 90° = Bll. By Fleming’s rule, it goes into the plane of paper.

The forces, F₁ and F₂ are equal in magnitude but act along different line of action, hence they constitute a torque or couple.
At any instant of time, the normal to the coil makes an angle θ w.r.t. magnetic field B. The perpendicular distance between the forces F₁ and F₂ is bsinθ.
Torque acting on the coil is
τ = Magnitude of force × Perpendicular distance between the forces
= BIl x b sinθ
= BI(lb)sinθ
τ = BIAsinθ (∵ lb = A = area of coil)
If the coil consists of N circles, then
τ = NBIAsinθ
This is the expression for the torque acting on a current loop when placed in a magnetic field.
But NIA = m, the magnetic moment of loop.
So τ = mBsinθ
In vector notation, torque r is given by
\(\oint { τ }\) = \(\oint { m }\) \(\oint { B }\), where \(\oint { m}\)
The direction of the torque r is such that it rotates the loop clockwise along the axis of suspension.

Question 14.
Explain the construction of moving coil galvanometer by drawing its diagram; Why is a soft – iron core kept in the moving-coil galvanometer ? Why the pole pieces are made concave?
Answer:
1. A permanent strong horseshoe magnet is taken. Its pole pieces NS are made of cylindrical soft – iron, cut in concave shape. A coil A, is suspended between the pole pieces, by a phosphor bronze wire F. The other end of the coil is connected to a spring Sp. The coil consists of insulated copper wire wound on an aluminium frame and a soft iron cylinder C is fixed within the coil, so that the coil can freely turn around it.

The wire F and spring S_p is connected to the terminals T₁ and T₂. Aconcave mirror M is attached to the wire F, so that the deflection of the coil can be measured with the help of lamp and scale arrangement. All the above arrangement is kept inside a non – magnetic box to protect it by dust, air, etc. The front portion of the box is made of glass and the base is provided by levelling screws.

2. Soft – iron core is used in moving coil galvanometer because :

• The permeability of soft – iron is very high, hence the field intensity increases.
• It makes the field radial.

3. By making pole pieces of magnet concave, the field is made radial, so that the plane of the coil becomes parallel to the magnetic field in all the positions.

Question 15.
Explain the principle of moving – coil galvanometer and find the expression for the current.
Or
What is the principle of moving – coil galvanometer? Prove that the current is proportional to the deflection of the coil.
Answer:
Principle:
Whenever a current is made to pass through a coil placed in a uniform magnetic field, then a torque acts on it which rotates the coil and tries to make it perpendicular to the direction of the field. Let ABCD be a rectangular coil, which is kept in a magnetic field B. such that the sides AB and CD are perpendicular, to field. Let the length of the coil AB be l and breadth BC be b. If I is the current flowing in the coil, then Lorentz force acting on AB and CD will be F = BIl.

The force acting on AB is F₁ which comes out of the plane of paper and on CD is F₂ which goes inside the plane of paper. As the forces are equal in magnitude and opposite in direction along different line of action, hence they produce a couple.

At any instant of time, the axis of coil (i.e., the normal to the coil) makes an angle θ with respect to the magnetic field, then the perpendicular distance between two forces F₁ and F₂ is bsinθ.

Now, torque = Magnitude of force x Perpendicular distance between two forces
= BIl × bsinθ = BIAsinθ
Where, A =lb = Area of the coil.
If N is the number of turns on the coil, then
τ = NBI Asin θ
Initially, when the plane of the coil lies parallel to the magnetic field, then the angle between the magnetic field and normal to the coil is 90°, i.e., θ = 90° therefore sinθ= 1.
∴τ_max =NBIA
This torque is the deflecting torque. Which rotates the coil. As a result, a restoring torque gets developed in the suspension wire which tries to restore the coil back to the initial position.
Let ϕ be the angle of twist and C is the couple for unit twist.
Restoring torque produced = Cϕ
Under equilibrium, deflecting torque = Restoring torque.
∴ NBIA = Cϕ
or I = \(\frac {C}{NBA}\)ϕ
or I = kϕ (where, k = \(\frac {C}{NBA}\)
or I ∝ϕ
This is the principle of moving – coil galvanometer.

Question 16.
What do you understand by the sensitivity of moving – coil galvanometer? Write its expression. On what factors does it depend and how?
Answer:
The current sensitivity of a moving coil galvanometer is defined as the deflection produced by unit current through the coil.
Let ϕ be the deflection produced due to current I, then
I = \(\frac {C}{NBA}\)ϕ
If the current through the galvanometer is I, which produces a deflection ϕ, then
\(\frac {ϕ}{I}\) = \(\frac {NBA}{C}\)
Sensitivity of galvanometer s = \(\frac {ϕ}{I}\)
= \(\frac {NBA}{C}\)

Sensitivity depends on the following factors :

1. N (No. of turns) should be greater. As the nufnber of turns increases, sensitivity increases.
2. For greater sensitivity, magnetic field should be greater. To increase the magnetic field B, a permanent horseshoe magnet must be used. By increasing B, sensitivity increases.
3. Area of the coil : If area increases, sensitivity increases.
4. C (Couple per unit twist) : The value of C should be less for more sensitivity.

Question 17.
What is meant by shunt? If the resistance of a galvanometer is R_g, then calculate the value of the shunt carrying current nth part of total current to pass through the galvanometer.
Answer:
Shunt:
A shunt is a thick copper wire which is joined in parallel with the coil of the galvanometer. It has very low resistance.
Let R_g be the resistance of galvanometer and S be the value of the shunt resistance.

Let I be the total current flow in the circuit. It gets divided as I_g and I_s: where I_g is the current flowing in galvanometer and I, in the shunt resistance (S).
∴ I = I_g + I_s … (1)
Now, potential difference across galvanometer = I_g.R_g
and Potential difference across shunt = I_s.S.
As galvanometer and shunt are in parallel, hence potential difference will be equal
i.e.,

This is the value of shunt resistance.
Adding 1 to both sides of eqn. (2), we get

The above equation gives the value of current flow through the galvanometer in turns of total current.
If the current passing through the galvanometer is nth part of the total current, then
\(\frac { { I }_{ g } }{ I }\) = \(\frac {1}{n}\)
Eqn. (4) becomes
\(\frac {1}{n}\) = \(\frac { s }{ { R }_{ g }+S }\)
or nS = R_g + S
or nS – S = R_g
or S(n -1) = R_g
or S = \(\frac { { R }_{ g } }{ (n-1) }\) … (6)
Hence, for the nth part of total current to pass through the galvanometer, the resistance of the shunt should be (n – l)th of the resistance of galvanometer.

Question 18.
How a galvanometer can be converted to ammeter and voltmeter?
Answer:
Conversion of galvanometer into ammeter:
Since, the coil of the galvanometer has low resistance, so to convert it into ammeter, a low resistance (called shunt) is joined in parallel, so that most of the current passes through the shunt and very less current passes through the coil of galvanometer.

Derive up to eqn. (3) from Short Ans. Type Q. No. 17.
Hence, by joining a shunt resistance of value
S = \(\frac { { I }_{ g }{ R }_{ g } }{ I-{ I }_{ g } }\)
So, the galvanometer gets converted to an ammeter.

Conversion of galvanometer into voltmeter:
The resistance of voltmeter is high, So to convert a galvanometer into a voltmeter, a high resistance wire is connected wire is connected in series to the coil of galvanometer, [see fig.(b)].

Suppose R_g be the resistance of galvanometer and R is the resistance of wire connected A in series. Let I be the current flowing in the galvanometer.
In order to convert a galvanometer into a voltmeter of range (0 – V) volts, we have from adjacent figure
The total potential difference between A and B will be
V = I_g(R_g + R)
or R_g + R = \(\frac { V }{ { I }_{ g } }\)
or R = \(\frac { V }{ { I }_{ g } }\) – R_g.
This is the expression and the value of the resistance required to convert the galvanometer to voltmeter of range 0 to V.

Question 19.
What is a shunt? Write its uses. What are advantages and disadvantages of shunt?
Answer:
Shunt:
It is a wire of low resistance, connected in parallel to the coil of a galvanometer.

Uses:
A galvanometer is converted into an ammeter by using a shunt.

Advantages:

1. It protects the coil of the galvanometer from burning as well as the breaking of the pointer.
2. As the shunt is connected in parallel, the resultant resistance becomes less. So, when the shunted galvanometer (ammeter) is joined in series, then the value of the current does not change.

Disadvantages:
Due to shunt, the sensitivity of galvanometer is reduced. So, it should be removed from the galvanometer when we have to obtain null point.

Moving Charges and Magnetism Long Answer Type Questions

Question 1.
Derive an expression for the intensity of magnetic field at a point on the axis of a circular current loop.
Or
Obtain an expression for the intensity of the magnetic field at a point on the axis of a circular coil.
Answer:
Magnetic field at a point on the axis of a circular current loop:
Let a be the radius of a circular loop and current 1 is flowing through it in the direction shown in the figure. A point P is considered on the axis of the loop, at a distance x from the centre O, at which the intensity of the magnetic field is to be determined.

The plane of the loop is normal to the plane of the paper and its axis OP lies on the plane of the paper. Let the loop be divided into so many small elements, each of length dl, let one of such small part is AB.
∴ The intensity of the magnetic field at P, due to dl, is given by

The direction of dB is along PR, normal to CP. Let ∠CPO = ϕ
Resolving dB in two parts, we get

1. dB sinϕ, along OP and
2. dB cosϕ, along PN, normal to OP.

If another small element dl is considered, diametrically opposite to AB, i.e., A’B’.
∴ Intensity of magnetic field at P, due to A’B’ will be
dB = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { I.dl }{ { r }^{ 2 } }\)
Again, resolving dB, we get

1. dB sin0, along OP and
2. dB cos0, along PN’, normal to OP.

As the directions of PN and PN’, are opposite, hence they will cancel the effect of each other. Similarly, all the resolved parts, perpendicular to OP will be cancelled out. But the components along the direction AP, will be summed up.
∴ Intensity of magnetic field due to the circular loop at point P will be

Question 2.
State Biot – Savart law and with the help of it derive an expression for magnetic field intensity at a point situated at a distance from a current carrying straight wire of infinite length.
Answer:
Biot – Savart law:
Let AB be a conductor carrying current I. Consider a small line element dl of the conductor, due to which the magnetic field dB is produced at point P, then the strength of the magnetic field \(\vec { (dB) }\) depends on the following factors :
(i) The field is directly proportional to current I.
i.e., dB ∝I
(ii) The field is directly proportional to the length of element,
i.e., dB ∝ dl

(iii) The field is directly proportional to the sine of angle between the line joining the point and dl.
i.e., dB ∝ sinθ
(iv) The field is inversely proportional to the square of the distance between the observation point and line element.
i.e.,
dB ∝ \(\frac { 1 }{ { r }^{ 2 } }\)
Combining all the four points,we get
dB ∝ \(\frac { Idl sinθ }{ { r }^{ 2 } }\)
or dB ∝ K\(\frac { Idl sinθ }{ { r }^{ 2 } }\) … (1)
Where, K is constant of proportionality. Its value depends upon the system of units
In C.G.S system, k = 1
∴ dB ∝ K\(\frac { Idl sinθ }{ { r }^{ 2 } }\) gauss … (2)
In M.K.S. system, K = \(\frac { { μ }_{ 0 } }{ 4π }\)
Where, μ₀ = Permeability of free space.
∴ dB = \(\frac { { μ }_{ 0 } }{ 4π }\) \(\frac { Idl sinθ }{ { r }^{ 2 } }\) Wb/m₂ … (3)
or dB =10^-7 [/latex] \(\frac { Idl sinθ }{ { r }^{ 2 } }\) Wb/m₂ … (4)
The relations given by eqns. (2), (3) and (4) are called Biot – Savart law.
The direction of the magnetic field \(\vec { dB }\) is always perpendicular to the plane containing \(\vec { dl }\) and \(\vec { r }\) and is given by the right – hand screw rule for the cross product of vectors.

Unit of electric current:
1. In C.GS. system : If dl= 1cm, r= 1cm, sinθ = 1 i.e., 9 = 90° and dB = 1 gauss, then from eqn. (2) I = 1 electromagnetic unit (e.m.u.).
For 9= 90°, the conductor should be taken as a part of circle, as the radius is always perpendicular to the circumference. Thus, 1 e.m.u. of current is that current which produces a field of 1 oersted at the centre of the conductor of length 1 cm, kept in the form of an arc of a circle of radius 1 cm.

2. In M.K.S. system: If dl =1m, r = 1m, sinθ = 1 and dB = 10^-7 Wb/m², then from eqn. (4)
I = 1 ampere.
Thus, 1 ampere of current is that current which produces a field of 10^-7 Wb/m², at the centre of tWconductor of length 1m.

Consider a long straight conductor XY which is on the plane Y of paper and current flow through it is I, which flows from X to Y. c Then magnetic field has to be found at position P, which is situated t at a distance a (distance measured perpendicularly from the wire) from the wire.
∴PC = a
To find out the total magnetic field at P, we have to first find out the magnetic field due to small line element \(\vec { dl }\), which is situated at a distance l from C. On integrating the magnetic field due to line element \(\vec { dl }\) , we can get the total magnetic field.
Let \(\vec { r }\) be the position vector of P with respect to the line element \(\vec { dl }\) and θ is the angle
between the line element \(\vec { dl }\) and \(\vec { r }\).
By Biot – Savart law,
dB = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { I.dl }{ { r }^{ 2 } }\)
We have to find out the magnetic field due to line element \(\vec { dl }\) at P, which is situated at r from \(\vec { dl }\) . The position of \(\vec { dl }\) can vary, hence θ can change, so we have to find out the value of sinθ and dl. For that, consider right angled triangle POC.

Putting the values of sinθ, r and dl from eqns. (2), (3) and (4) respectively in eqn. (1), we get

Eqn. (5) is the magnetic field due to line element dl. To find out the total magnetic field, integrating both sides under limits from ϕ₁ to ϕ₂ (-ϕ₁ is taken because it is clock wise or below the line joining CP and ϕ₂ is taken because it is anticlockwise or above the line joining CP).

For an infinitely long conductor and if the observation point is very near, then

Question 3.
Describe the cyclotron under the following points :

1. Construction
2. Principle and working process.

Or
What is a cyclotron? Write its principle. Obtain the expression for the cyclotron frequency and the maximum energy of the charged particle when it hits the target.
Answer:
Principle:
It is based on the principle that when a positively charged particle is made to move again and again in a high frequency electric field and using strong magnetic field, then it gets accelerated and acquires sufficiently large amount of energy.

Construction:
It consist of two hollow D – shaped metallic chambers D₁ and D₂ called dees. These dees are separated by a small gap where a source of positively charged particle is placed. Dees are connected to a high frequency oscillator, which provide high frequency electric field across the gap of the dees. This arrangement is placed between two poles of a strong electromagent. The magnetic field due to this electromagnet is perpendicular to the plane of the dees.

Working:
If a positively charged particle (proton) is emitted from O, when D₂ is negatively charged and the dee D₁, is positively charged, it will accelerate towards D₂. As soon as it enters D₂, it is shielded from the electric field by metallic chamber (enclosed space). Inside D₂, it moves at right angles to the magnetic field and hence describes a semi – circle inside it. After completing the semicircle, it enters the gap between the dees at the time when the polarities of the dees have been reversed.

Now, the proton is further accelerated towards D₁. Then it enters D₁ and again describes the semicircle due to the magnetic field which is perpendicular to the motion of the proton. This motion continues till the proton reaches the periphery of the dee system. At this stage, the proton is deflected by the deflecting plate which then comes out through the window and hits the target.

Theory:
When a proton (or any other positively charged particle) moves at right angle to the magnetic field B inside the dees, Lorentz force acts on it.
i. e., F = qvB sin 90° = qvB
Where, q = Charge of particle and v = Velocity of particle.
The force provides the necessary centripetal force \(\frac { m{ v }^{ 2 } }{ r }\) to the charged particle to move in a circular path of radius r.
∴ qvB = { m{ v }^{ 2 } }{ r }[/latex]
or r = \(\frac {mv}{qB}\) … (1)
Time taken to complete one semicircle inside a dee,
t = \(\frac {Distance}{Speed}\) = \(\frac { πr}{v}\)
or t = \(\frac { π}{v}\) × \(\frac { mv}{qB}\) [from eqn.(1)]
t = \(\frac { πm}{qB}\) … (2)
Thus, time taken to complete one semicircle does not depend upon radius of path. If T is the time – period of the alternate electric field, then the polarities of the dees changes in time 772.
i.e., \(\frac {T}{2}\) = t = \(\frac {πm}{qB}\)
or T = \(\frac {2πm}{qB}\) … (3)
So, cyclotron frequency or magnetic resonance frequency,
v = \(\frac {1}{T}\) = \(\frac {qB}{2πm}\) … (4)
Energy gained by a positively charged particle is given by
E = \(\frac {1}{2}\)mv²
From eqn (1), v = \(\frac {qBr}{m}\)
so, E = \(\frac {1}{2}\)m\(\frac { { q }^{ 2 }{ B }^{ 2 }{ r }^{ 2 } }{ { m }^{ 2 } } \)
= \(\frac { { q }^{ 2 }{ B }^{ 2 }{ r }^{ 2 } }{ { 2m } }\)
Maximum energy is gained by the positively charged particle when it is at the periphery of the dees (r is maximum), i.e.,
E_max = \(\frac { { q }^{ 2 }{ B }^{ 2 } }{ { 2m } }\) r²_max

Question 4.
Obtain an expression for magnetic field due to a solenoid using Ampere’s circuital law.
Answer:
Consider a very long solenoid having n turns per unit length carrying current I. The magnetic field inside the solenoid is uniform and directed along the axis of solenoid.

Consider a rectangular Amperian loop abed in a solenoid. Magnetic field B is uniform within the solenoid. Let the length of the Amperian loop be h.
∴Total number of turns in Amperian loop = nh.
The integral \(\oint { \vec { B } .\vec { dl } }\) is basically equal to the sum of four integrals

Comparing eqns.(2) and (3)
Bh = µ₀nhI
⇒ B = µ₀nI.

Question 5.
Explain the experiment to find reduction factor of a tangent galvanometer under following points :

1. Formula
2. Circuit diagram
3. Observation table
4. Any two precautions.

Answer:
1. Formula : i = ktanθ or k = \(\frac {i}{tanθ}\)
Where i = Current flowing through coil, and
θ = Deflection in magnetic needle.

2. Circuit diagram :
TG = Tangent Galvanometer,
K = Reversing key,
B = Cell, Rh = Rheostat,
A = Ammeter.

3. Observation table:

4. Precautions :

There should be no magnet near tangent galvanometer.
After aligning in magnetic meridian, the tangent galvanometer should not be moved.

Moving Charges and Magnetism Numerical Questions

Question 1.
The frequency of a cyclotron oscillator is 10 MHz.What should be the magnetic field required to accelerate a proton (e = 1.6 x 10^-19C, m =1.67 x 10^-27kg)
Solution:
f = \(\frac {qB}{2πm}\)
or B = \(\frac {2πmf}{q}\)
Given : m = 1.67 x 10^-27kg, f = 10 x 10⁶Hz, q = e = 1.6 x 10^-19

Question 2.
A particle having 100 times charge of an electron is revolving in a circle of radius 0-8 metre in each second. Calculate the intensity of magnetic field.
Solution:
Formula: B = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2\pi I }{ R }\)
Given : q = 100 x 1.6 x 10^-19
= 1.6 x 10^-17 coulomb, R = 0.8 metre
Putting the value in formula, we get I = \(\frac {q}{t}\)
\(\frac { 1.6\times { 10 }^{ -17 } }{ 1 }\) = 1.6 x 10^-17
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2\pi \times 1.6\times { 10 }^{ -17 } }{ 0.8 }\) or B = 10^-2µ₀

Question 3.
Calculate the torque on a 20 turns square coil of side 10 cm carrying a current of 12 A, when placed, making an angle of 30° with the magnetic field of 0.8 T.
Solution:
Given : A = 100cm² 100 x 10^-4m² = 10^-2m² n = 20, I = 12A, B = 0.8T, ϕ = 30°
Formula: τ = nIABsinϕ
= 20 x 12 x 10² x 0. 8sin30°
= 20 x 12 x 0.8 x 10² x \(\frac {1}{2}\)
= 96 x 10² = 0.96 N – m.

Question 4.
Resistance of a Galvanometer is 50 ohms, when 0.01 Acurrent flows through it, full scale defections is obtained. How it can be converted into

1. 5 A range ammeter and
2. 5 volts range voltmeter.

Solution:
1. Given : G = 50 ohm, i_g = 0.01 A, I = 5A
Formula : S = \(\frac { { I }_{ g }G }{ I-{ I }_{ g } }\)
Putting the value in the formula are get = \(\frac {0.01 x 50}{5-0.01}\) = \(\frac {0.5}{4.99}\)

2. Given : V = 5 vol
Formula : R = \(\frac { V }{ { I }_{ g } }\) – G
Putting the value in the formula are get
R = \(\frac {5}{0.01}\) – 50 = 500 – 50 = 450 ohms.

Question 5.
Find the magnitude of magnetic field at the centre of a circular coil of radius 10 cm having 100 turns. Current flowing through the coil is 1A. (NCERT Solved Example)
Solution:
Given, R = 10 cm = 10 x 10^-2m; N = 100; I = 1A;
B = \(\frac { { \mu }_{ 0 }NI }{ 2R }\) = \(\frac { 4\pi \times { 10 }^{ -7 }\times 100\times 1 }{ 2\times 10\times { 10 }^{ -2 } }\)
B = 2π x 10^-4
= 6.28 x 10^-4 tesla.

Question 6.
10 A current is flowing through a straight conductor. Determine the intensity of magnetic field at a distance 10 m from it.
Solution:
Formula :B = \(\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2I }{ d }\)
Given: I = 10 Aandd = 10m.
Substituting the values in the formula, we get
B = 10^-7 x \(\frac {2×10}{10}\)
∴B = 2 x 10^-7Wb/m².

Question 7.
An ammeter of resistance 99 Ω, gives full – scale deflection with 10^-4A current. What arrangement is required to measure a current of 1A by it? Calculate the resistance of ammeter.
Solution:

Question 8.
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5A. What is the magnitude of the magnetic field inside the solenoid? (NCERT)
Solution:
Given, l = 0.5 m, r = 1 cm = 1 x 10^-2 m, N = 500; I = 5A Number of turns per unit length
Number of turns per unit length
n = \(\frac {N}{l}\) = \(\frac {500}{0.5}\)
Here l >> r
∴Magnetic field inside the solenoid
B = µ₀nI
= 4π x 10^-7 x \(\frac {500}{0.5}\) x 5
B = 6.28 x 10^-3tesla.

Question 9.
A wire through which a current of 8A is flowing makes an angle of 30° with the direction of magnetic field of 0.15 tesla. Calculate the force acting per unit length of wire.
Solution:
Given : I = 8A, B = 0.15T, θ = 30°, F =?
formula: F = BIlsinθ
⇒ \(\frac {F}{l}\) = BIlsinθ
= 0.15 x 8 x sin30°
= 0.15 x 8 x \(\frac {1}{2}\) = 0.6N/m.

Question 10.
Two parallel wires A and B are carrying currents 10 A and 2 A respectively, in opposite directions. If the length of the wire A is infinite and length of B is 1 metre, calculate the force on B situated at a normal distance of 10 cm from A.
Solution:
Formula : \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ d } \times l\)
= 10^-7\(\frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ d } \times l\)
Given : I₁ = 10 A, I₂ = 2A , l = 1 m, d = 0.1 m
∴ F = 10^-7 × \(\frac {2×10×2×1}{0.1}\)
= 400 x 10^-7= 4.0 x 10^-5N.

MP Board Class 12th Physics Important Questions

MP Board Class 11th Maths Important Questions Chapter 11 Conic Sections

Conic Sections Important Questions

Conic Sections Objective Type Questions

(A) Choose the correct option :

Question 1.
Coordinates of the focus of the parabola y = 2x² + x are:
(a) (0, 0)
(b) (\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 4 }\))
(c) (- \(\frac { 1 }{ 4 }\), 0)
(d) ( – \(\frac { 1 }{ 4 }\), \(\frac { 1 }{ 8 }\))
Answer:
(c) (- \(\frac { 1 }{ 4 }\), 0)

Question 2.
In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 a > b, the relation between a, b an eccentricity e is:
(a) b² = a²(1 – e²)
(b) b² = a²(e² – 1)
(c) a² = b²(1 – e²)
(d) a² = b²(e² – 1)
Answer:
(a) b² = a²(1 – e²)

Question 3.
The length of latus rectum of ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:
(a) \(\frac { { 2a }^{ 2 } }{ b }\)
(b) \(\frac { { 2b }^{ 2 } }{ a }\)
(c) \(\frac { { a }^{ 2 } }{ b }\)
(d) \(\frac { { b }^{ 2 } }{ a }\)
Answer:
(b) \(\frac { { 2b }^{ 2 } }{ a }\)

Question 4.
The eccentricity of the parabola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(d) 1

Question 5.
The eccentricity of the ellipse is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(a) Less than 1

Question 6.
The eccentricity of the hyperbola is:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(b) Greater than 1

Question 7.
In a ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, represent a circle then its eccentricity will be:
(a) Less than 1
(b) Greater than 1
(c) 0
(d) 1
Answer:
(c) 0

Question 8.
The sum of focal distances from any point on the ellipse is:
(a) Equal to major axis
(b) Equal to minor axis
(c) The distance between two foci
(d) Equal to latus rectum.
Answer:
(a) Equal to major axis

Question 9.
The differecne of the focal distances from any point on the hyperbola is:
(a) Equal to its conjugate axis
(b) Equal to its transverse axis
(c) The distance between two foci
(d) Equal to its latus rectum.
Answer:
(b) Equal to its transverse axis

Question 10.
The value of the eccentricity of ellipse 25x² + 16y² = 400 is:
(a) \(\frac { 3 }{ 5 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 2 }{ 5 }\)
(d) \(\frac { 1 }{ 5}\)
Answer:
(a) \(\frac { 3 }{ 5 }\)

Question 11.
Equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent a circle if:
(a) a = b, c = 0
(b) f = g, h = 0
(c) a = b, h = 0
(d) f = g, c = 0
Answer:
(a) a = b, c = 0

Question 12.
Area of triangle whose centre (1,2) and which is passes through the point (4,6) will be:
(a) 5π
(b) 10π
(c) 25π
(d) 25π²
Answer:
(c) 25π

Question 13.
The circle passing through (1, – 2) and touching the X – axis at (3,0), also passes through the point:
(a) (2, – 5)
(b) (5, – 2)
(c) (- 2, 5)
(d) (- 5, 2)
Answer:
(a) (2, – 5)

Question 14.
The length of the diameter of the circle which touches the X – axis at the point (1,0) and passes through the point (2,3) is:
(a) \(\frac { 10 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) \(\frac { 6 }{ 5 }\)
(d) \(\frac { 5 }{ 3 }\)
Answer:
(a) \(\frac { 10 }{ 3 }\)

Question 15.
Eccentricity of the hyperbola 3x² – y² = 4 :
(a) 1
(b) 2
(c) – 2
(d) \(\sqrt {2}\)
Answer:
(b) 2

(B) Match the following :

Answer:

1. (c)
2. (e)
3. (b)
4. (a)
5. (d)
6. (i)
7. (h)
8. (f)
9. (j)
10. (g)

(C) Fill in the blanks :

1. The length of the latus rectum of the parabola y² = 4ax is ……………
2. The centre of the ellipse \(\frac { { (x-1) }^{ 2 } }{ 9 } +\frac { { (y-2) }^{ 2 } }{ 4 }\) = 1 will be ……………
3. The vertex of the parabola (y – 2)² = 4a(x -1) is ……………
4. The lines \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = m and \(\frac { x }{ a }\) + \(\frac { y }{ b }\) = \(\frac { 1 }{ m }\) meets always at ……………
5. If an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, a > b and its eccentricity is e, then the foci will be
6. Standard form of equation of parabola is ……………
7. Parametric equation of a circle x² + y² = 4 is ……………
8. A line y = x + a\(\sqrt {2}\) touches the circle x² + y² = a² point ……………
9. A line y = mx + c touches the circle x² + y² = a² if c = ……………
10. Vertex of the parabola 3y² + 6y – 4x + 11 = 0 is ……………
11. Equation 2x² + 2y² – 12x – 16y + 4 = 0 represent a point circle if k = ……………
12. Radius of circle 3x² + 3y² – 5x – 6y + 4 = 0 is ……………

Answer:

1. 4a
2. (1, 2)
3. (1, 2)
4. Hyperbola
5. (± ae, 0)
6. y² = 4ax
7. x = 2cosθ
8. (- \(\frac { a }{ \sqrt { 2 } }\), \(\frac { a }{ \sqrt { 2 } }\) )
9. ±a\(\sqrt { 1+{ m }^{ 2 } }\)
10. (5, 1)
11. 50, 12
12. \(\sqrt {61}\)

(D) Write true / false :

1. Conic section is a locus of the point whose the ratio between the distance from the fixed point and distance from the fixed line, this ratio is called eccentricity of the conic section.
2. The ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 has two directrics, the equation of directrics are x = ± \(\frac { a }{ e }\); Where a > b and y = ± \(\frac { b }{ e }\) ; where b > a.
3. The foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are (0, ± be) where a < b.
4. A circle drawn by taking major axis of the ellipse as diameter is called auxiliary circle of the ellipse.
5. The locus of intersection point of the lines bx + ay = abt and bx – ay = \(\frac { ab}{ t }\) will be a ellipse.
6. The focus of a parabola x² = – 16y will be (0, – 4).
7. Equation x² + y² – 6x + 8y + 50 = 0 represent a circle.
8. Circle x² + y² = 9 and x² + y² + 8y + c = 0 touches externally if c = 15 .
9. Eccentricity of hyperbola is 1.
10. Minimum distance between line y – x = 1 and curve x = y² is \(\frac { 3\sqrt { 2 } }{ 8 }\)

Answer:

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. True
9. False
10. True

(E) Write answer in one word / sentence :

1. If the circle x² + y² + 2ax + 8y +16 = 0, touches X – axis, then the value of α will be.
2. Coordinate of focus of parabola x² = – 10y will be.
3. Write the equation of a circle whose centre is (2,2) and passes through the point (4, 5).
4. The centre of a circle is (5, 7) and touches Y – axis, then its radius will be.
5. If the radius of a circle x² + y² – 6x + ky – 25 = 0 is \(\sqrt {38}\) the value of k will be.
6. Vertex of the parabola y = x² – 2x + 3 will be.
7. Equation of a parabola whose vertex (0, 0) and focus (0, 3) will be.
8. Length of major axis of ellipse 9x² + 16y² = 144 will be.
9. Eccentricity of an ellipse whose latus rectum in half of its minor axis will be.
10. Equation of hyperbola whose one focus in (4, 0 ) and corresponding equation of directrix x = 1 will be.

Answer:

1. ± 4
2. (0, \(\frac { – 5 }{ 2 }\))
3. x² + y² – 4x – 4y – 5 = 0
4. 7, 5
5. ± 4
6. (1, 2)
7. x² = 12y
8. 6
9. \(\frac { \sqrt { 3 } }{ 2 }\)
10. \(\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } \) = 1

Conic Sections Long Answer Type Questions

Question 1.
Find the equation of circle which touches the X – axis at a distance of 4 units in the negative direction and makes intercept of 6 units on positive direction of Y – axis.
Solution:
Here OA = CM = 4, BD = 6.
Length of perpendicular drawn from centre C on BD.
Then, BM = MD = 3
In right angled ∆ CMB,
CB² = CM² + BM²
= 4² + 3²
= 16 + 9 =25
⇒ CB = 5

∴ CA = Radius of circle = CB = 5
∴ Centre of circle (- 4, 5) and radius = 5
Hence, equation of circle :
(x + 4)² + (y – 5)² = 5²
⇒ x² + 8x + 16 + y² – 10y + 25 = 25
⇒ x² + y² + 8x – 10y + 16 = 0

Question 2.
Find the equation of circle which touches Y – axis at a distance of 4 units and makes intercept of 6 units on Y – axis?
Solution:
Given : OP = 4, AB = 6, PC = AC = radius.
CM ⊥ AB ∴ AM = BM = \(\frac { 6 }{ 2 }\) = 3
OP = CM = 4
In right angled ∆ AMC,

AC² = AM² + CM²
= (3)² + (4)² = 9 + 16 = 25
∴ AC = 5
From figure PC – OM= 5 = radius
Centre of circle is (5, 4) and radius = 5.
Hence, required equation of circle :
(x – 5)² + (y – 4)² = (5)²
x² – 10x + 25 + y² – 8y + 16 = 25
x² + y² – 10x – 8y + 16 = 0.

Question 3.
ABCD is a square. Supposing AB and AD as the coordinate axes. Find the equation of the circle circumscribing the square if each side of square is of length l.
Solution:
Taking AB and AD as X – axis and Y – axis respectively
Given : AB = BC = CD = DA = 1
M is mid point of AB.
N is mid point of AD.
AM = \(\frac { l }{ 2 }\), AN = \(\frac { l }{ 2 }\) = OM
In ∆OAM,
OA² = AM² + OM²
= \(\frac { l }{ 2 }\)² + \(\frac { l }{ 2 }\)²
= \(\frac{l^{2}}{4}+\frac{l^{2}}{4} = \frac{l^{2}}{2}\)
∴ Radius = OA = \(\frac{l}{\sqrt{2}}\)
Centre of circle (AM, OM) = ( \(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\) )

Required equation of circle is :
(x – \(\frac { l }{ 2 }\) )² + (y – \(\frac { l }{ 2 }\) )² = \(\frac{l^{2}}{2}\)
Centre of circle (AM, OM) = (\(\frac { l }{ 2 }\), \(\frac { l }{ 2 }\))
Required equation of circle is :
(x – \(\frac { l }{ 2 }\))² + (y – \(\frac { l }{ 2 }\))² = \(\frac{l^{2}}{2}\)
⇒ x² – lx + \(\frac{l^{2}}{4}\) + y² – ly + \(\frac{l^{2}}{4}\) = \(\frac{l^{2}}{2}\)
⇒ x² + y² – l(x + y) = 0
⇒ x² + y² = l(x + y)

Question 4.
Find the equation of the circle passing through the points (4, 1) and (6, 5). Whose centre lies on line 4x + y = 16. (NCERT)
Solution:
Let the equation of circle be
x² + y² + 2gx + 2 fy + c = 0 …. (1)
It passes through points (4, 1) and (6, 5).
∴ 8g + 2f + c + 17 = 0 …. (2)
and 12g + 10f + c + 61 = 0 …. (3)
Centre of circle (1) is (- g, – f) which lies on line 4x + y = 16.
∴ – 4g – f – 16 = 0
⇒ 4g + f + 16 – 0 …. (4)
Subtracting equation (2) from equation (3), we get
4g + 8f + 44 = 0
⇒ g + 2f + 11 = 0 …. (5)
On solving equation (4) and (5), g = – 3, f = – 4
Put g = – 3 and f = – 4 in equation (2),
– 24 – 8 + C + 17 = 0
⇒ c = 15
Put values of g, f and c in equation (1), then required equation of circle is :
x² + y² – 6x – 8y + 15 = 0.

Question 5.
Find the equation of the circle which passes through the points (2, 3) and (- 1, 1) whose centre lies on line x – 3y – 11 = 0. (NCERT)
Solution:
Let the equation of circle is :
x² + y² + 2gx + 2fy + c = 0 …. (1)
∵Points (2, 3) and (- 1, 1) lies on equation (1),
∴ (2)² + (3)² + 2g(2) + 2f(3) + c = 0
⇒ 4 + 9 + 4g + 6f + c + 13 = 0
4g + 6f + c + 13 = 0 …. (2)
and (-1)² + (l)² – 2g + 2f + c = 0
⇒ 1 + 1 – 2g + 2f + c = 0
⇒ – 2g + 2f + c + 2 = 0 …. (3)

Putting the value of g, f and c in equation (1), then required equation of circle will be :
x² + y² + 2(\(\frac { -7 }{ 2 }\))x + 2(\(\frac { 5 }{ 2 }\))y + c = 0 …. (1)
x² + y² – 7x + 5y -14 = 0.

Question 6.
Find the equation of circle whose radius is 5, centre is on Y – axis and which passes through point (2, 3).
Solution:
Centre of circle is on X – axis, so k = 0.
Let the equation of circle be :
(x – h)² + (y – k)² = a²
Here a = 5
(x – h)² + (y – 0)² = (5)²
(x – h)² + y² = 25
Circle (1) passes through point (2, 3),
∴ (2 – h)² + (3)² = 25
⇒ (2 – h)² = 25 – 9 = 16 = (4)²
⇒ 2 – h = ± 4

Question 7.
y = mx is a chord of the circle whose radius is ‘a’ and its diameter is X – axis. Origin is one of the limiting points of the chord. Show that the equation to a circle whose diameter is the given chord is given by the equation (1 + m²) (x² + y² ) – 2a (x + my) = 0. Solution:
Equation of circle whose radius is a and centre is (a, 0) will be
(x – a)² + y² = a²
⇒ x² – 2ax + y² + a² = a²
⇒ x² – 2ax + y² = 0 …. (1)
Equation of given line is :
y = mx …. (2)
Now, equation of circle passing through the intersection of eqns. (1) and (2) will be :
x² + y² – 2ax + λ(y – mx) = 0 … (3)
Centre of co – ordinate of circle (3) are (\(\frac { λm + 2a }{ 2 }\), \(\frac { λ }{ 2 }\))
∵ Centre lies on line y = mx.
∴ – \(\frac { λ }{ 2 }\) = m\(\frac { λm + 2a }{ 2 }\)
⇒ λ = \(\frac{-2 a m}{1+m^{2}}\)

Put the value of λ in Equation (3),
x² + y² – 2ax + \(\frac{-2 a m}{1+m^{2}}\) (y – mx) = 0
⇒ (l + m²)(x² + y²) = 2ax + 2am²x + 2amy – 2am²x
⇒ (l + m²)(x² + y²) = 2a(x + my)
⇒ (l + m²)(x² + y²) – 2a(x + my) = 0
Which is required equation of circle.

Question 8.
If the straight line x cos α + y sin α = p cuts a circle x² + y² = a² in two points M and N, then show that the equation of the circle whose diameter is MN will be x² + y² – a² = 2p(x cos α + y sin α – p).
Solution:
Given : Equation of line is :
x cos α + y sin α = p …. (1)
and Equation of circle is
x² + y² = a² …. (2)
Now, equation of circle passing through the intersection of line (1) and circle (2) at points M and N is :
x² + y² – a² + λ(x cos α + y sin α – p) = 0 …. (3)
If MN is diameter of above circle then centre is :
(- \(\frac { λ }{ 2 }\)cos α, – \(\frac { λ }{ 2 }\)sin α)
Which is lies on line x cos α + y sin α = p.
– ( \(\frac { λ }{ 2 }\)cos α )cos α + (- \(\frac { λ }{ 2 }\)sin α)sin α = p
⇒ – \(\frac { λ }{ 2 }\)[cos² α + sin² α] = p
⇒ λ = – 2p
Put the value of λ in equation (3), then required equation of circle is
x² + y² – a² – 2p(x cos α + y sin α – p) = 0
⇒ x² + y² – a² = 2p(x cos α + y sin α – p)

Question 9.
Find the following equation of parabola : (i) co – ordinates of focus, (ii) axis, (iii) equation of directrix, (iv) length of Iatus rectum. (NCERT)
(A) y² = 12x
Solution:
Equation of parabola : y² = 12x
Comparing with y² = 4 ax,
4a = 12 ⇒ a = 3
∴Co – ordinates of focus (a, 0) = (3, 0).
Axis of parabola = X – axis.
Equation of directrix is x = – a ⇒ x = – 3.
Length of latus rectum = 4a = 4 x 3 = 12.

(B) x² = 6y.
Solution:
Equation of parabola: x² = 6y
Comparing with x² = 4ay
4a = 6 a ⇒ 3/2
∴ Co – ordinate of focus (0, a) = (0, 3/2).
Axis of parabola = Y – axis.
Equation of directrix is y = – a ⇒ y = – 3/2.

(C) y² = – 8x
Solution:
Equation of parabola : y² = – 8x
Comparing with y² = – 4ax
– 4a – = – 8 ⇒ a = 2
∴ Co – ordinates of focus (- a, 0) = (- 2, 0).
Axis of parabola = X – axis.
Equation of directrix is x = a ⇒ x = 2.
Length of latus rectum 4a = 4 x 2 = 8.

(D) x² = – 16y
Solution:
Equation of parabola : x² = – 16y
Comparing with x² = – 4ay
– 4a = – 16 ⇒ a = 4
∴ Co – ordinate of focus (0, – a) = (0, – 4)
Axis of parabola = Y – axis
Equation of directrix is y = a ⇒ y = 4
Length of latus rectum = 4a = 16.

Question 10.
An equilateral triangle inscribed in the parabola y² = 4ax, where one vertex is at the vertex of parabola. Find the length of the side of triangle. (NCERT)
Solution:
Let the equation of parabola is y² = 4ax.
Let APQ be the equilateral triangle whose vertex A(0, 0), P(h, k) and Q(h, – k).
AP² = (h – 0)² + (k – 0)²
= h² + k²
⇒ AP = \(\sqrt{h^{2}+k^{2}}\)
Similarly, AQ = \(\sqrt{h^{2}+k^{2}}\)

Again, PQ = \(\sqrt{(h-k)^{2}+(k+h)^{2}}\)
= \(\sqrt{(2k)^{2}}\) = 2k
∴ AP = PQ
⇒ \(\sqrt{h^{2}+k^{2}}\) = 2k
⇒ h² + k² = 4k²
⇒ h² = 3k²
⇒ h = \(\sqrt {3}\).k
∵ Point P(h, k) lies on parabola y² = 4ax.
k² = 4ah = 4a.\(\sqrt {3}\)k
⇒ k = 4a\(\sqrt {3}\), [∵ k ≠ 0]
Hence, length of side PQ = 2k = 2.(4a\(\sqrt {3}\)) = 8a\(\sqrt {3}\).

Question 11.
If a parabola reflector is 20 cm in diameter and 5 cm deep. Find the focus. (NCERT)
Solution:
Taking vertex of parabola reflector at origin and X – axis along the axis of parabola.

Equation of parabola y² = 4ax …. (1)
Given : OS = 5 cm, AB = 20 cm, AS = 10 cm
∴ Co – ordinate of A will be (5, 10).
∴ (10)² = 4a x 5
⇒ 100 = 20a
⇒ a = 5
∴ OS = 5 cm
Co – ordinates of focus S will be (5, 0).

Question 12.
An arch is in the form of a parabola with its vertical axis. The arch is 10 m high and 5 m wide at the base. How wide it is 2 m vertex of the parabola. (NCERT)
Solution:
Let the equation of parabola is :
x² = 4 ay …. (1)
Given : AB = 5 metre
AF = BF = \(\frac { 5 }{ 2 }\) metre
OE = 2 metre
OF = 10 metre
Co – ordinate of A will be (\(\frac { 5 }{ 2 }\), 10)
This point lies on parabola, hence it will be satisfy equation (1),
∴ \(\frac { 5 }{ 2 }\)² = 4a x 10
⇒ \(\frac { 25 }{ 4 }\) = 4 x a x 10
⇒ a = \(\frac { 25 }{ 4 × 4 × 10 }\)
⇒ a = \(\frac { 5 }{ 32 }\)

Put the value of a in Equation (1),
∴ x² = 4 x \(\frac { 5 }{ 32 }\)y
⇒ x² = \(\frac { 5 }{ 8 }\)y
Let EC = k
OE = 2
Co – ordinate of C will be (k, 2) and it will satisfy equation of parabola.
We get k² =\(\frac { 5 }{ 8 }\) x 2
⇒ k² = \(\frac { 5 }{ 4 }\)
⇒ k = \(\frac{\sqrt{5}}{2}\)
DE = 2EC
2 x \(\frac{\sqrt{5}}{2}\) = \(\sqrt {5}\)
= 2.23 metre (approx.)

Question 13.
In each of the following ellipse. Find the co – ordinates of the foci and vertices, the length of major axis and minor axis, the eccentricity and the length of latus rectum of the ellipse. (NCERT)
(A) = \(\frac{x^{2}}{36}+\frac{y^{2}}{16}\) = 1.
Solution:
Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
We get, a² = 36 ⇒ a = 6, b² = 16 ⇒ b = 4
Here a > b.
∴ b² = a²(1 – e²)
⇒ 16 = 36(1 – e²)
⇒ 1 – e² = \(\frac { 16 }{ 36 }\) = \(\frac { 4 }{ 9 }\)
⇒ e² = 1 – \(\frac { 4}{9 }\) = \(\frac { 5 }{ 9 }\)
∴ Eccentricity e = \(\frac{\sqrt{5}}{3}\)
Foci (± ae, o) = (± 6 × \(\frac{\sqrt{5}}{3}\), o)
= (± 2\(\sqrt {5}\), 0)
Vertices (± a, 0) = (± 6, 0)
Length of major axis = 2a = 2 x 6 = 12.
Length of minor axis = 2b = 2 x 4 = 8.
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 × 16 }{ 6 }\) = \(\frac { 16 }{ 3 }\).

(B) \(\frac{x^{2}}{4}+\frac{y^{2}}{25}\) = 1.
Solution:
Comparing with standard form of ellipse, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1
We get, a² = 36 ⇒ a = 6, b² = 16 ⇒ b = 4
Here a < b.
∴ a² = b²(1 – e²)
⇒ 4 = 25(1 – e²)
⇒ 1 – e² = \(\frac { 4 }{ 25 }\)
⇒ e² = 1 – \(\frac { 4}{25 }\) = \(\frac { 21 }{ 25 }\)
∴ Eccentricity e = \(\frac{\sqrt{21}}{5}\)
Foci (0, ± b) = (0, ± 5 × \(\frac{\sqrt{21}}{5}\))
= (0, ± \(\sqrt {21}\))
Vertices (0, ± b) = (0, ± 5)
Length of major axis = 2b = 2 x 5 = 12.
Length of minor axis = 2a = 2 x 2 = 8.
Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac{2 \times 2^{2}}{5}\) = \(\frac { 2 × 4 }{ 5 }\) = \(\frac { 8 }{ 5 }\).

Question 14.
Find the equation of hyperbola whose foci is (± 4, 0) and length of latus rectum is 12. (NCERT)
Solution:
Foci of hyperbola (± 4, 0).
Hence equation of hyperbola will be :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Foci (±ae, 0) = (± 4, 0)
∴ ae = 4
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = 12
⇒ b² = 6a
We know, b² = a²(e² – 1)
⇒ 6a = a²e² – a²
⇒ 6a = 4² – a²
⇒ 6a = 16 – a²
⇒ 6a = 16 – a²
⇒ a² + 6a – 16 = 0
⇒ a² + 8a – 2a – 16 = 0
⇒ a(a – 2)(a + 8) = 0
⇒ a = 2, a = – 8, (∵ a cannot be negative)
∴ a = 2
b² = 6a = 6 x 2 = 12
⇒ b = \(\sqrt {12}\)
Putting values of a and b in equation (1), then required equation of hyperbola will be :
\(\frac{x^{2}}{2^{2}}-\frac{y^{2}}{(\sqrt{12})^{2}}\) = 1
⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{12}\) = 1
⇒ 3x² – y² = 12.

Question 15.
Find the axis, foci, directrix, eccentricity and the latus rectum of the ellipse 9x² + 4y² = 36.
Solution:
Given equation of ellipse
9x² + 4y² = 36
⇒ \(\frac{x^{2}}{4}+\frac{y^{2}}{9}\) = 1
Here, b² > a² or b > a
∴ Major axis = 2.3 = 6
Minor axis = 2.2 = 4
Now,
a² = b²(1 – e²)
(2)² = (3)²(1 – e²)
⇒ 4 = 9(1 – e²)
⇒ \(\frac { 4 }{ 9 }\) = 1 – e²
⇒ e² = 1 – \(\frac { 4 }{ 9 }\) = \(\frac { 9 – 4 }{ 9 }\) = \(\frac { 5 }{ 9 }\)
∴ e = \(\frac{\sqrt{5}}{3}\)
Co – ordinate of foci = (0, ± be )
= (0, ± 3. \(\frac{\sqrt{5}}{3}\))
= (0, ± \(\sqrt {5}\)).
Co – ordinate of vertex = (0, ± b )
= (0 ± 3 ).
Length of latus rectum = \(\frac{2 a^{2}}{b}\)
= \(\frac { 2.4 }{ 3 }\) = \(\frac { 8 }{ 3 }\).

Question 16.
(A) Find the equation of ellipse whose vertices are (± 5, 0) and foci (± 4, 0).
Solution:
Given : Vertices are (± 5, 0) and foci are (± 4, 0)
∴ a = 5
and ae = 4
⇒ 5e = 4
⇒ e = \(\frac { 4 }{ 5 }\)
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a > b …. (1)
∴ From b² = a²(1 – e²),
b² = 5²(1 – (\(\frac { 4 }{ 5 }\))²)
b² = 25(1 – \(\frac { 16 }{ 25 }\))
b² = 25 x \(\frac { 9 }{ 25 }\) = 9
Putting values of a and b in equation (1), the required equation of ellipse will be :
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1
⇒ 9x² + 25y² = 225

(B) Find the equation of ellipse whose vertices are (0, ± 13) and foci is (0, ± 5).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)
Vertices of ellipse = (0, ± b) = (0, ± be)
∴ b = 13
Foci = (0, ± 5) = (0, ± be)
∴ be = 15
⇒ 13 x 3 = 5
⇒ e = \(\frac { 5 }{ 13 }\)
Now, a² = b²(1 – e²)
⇒ a² = 13²[1 – ( \(\frac { 5 }{ 13 }\))² ]
⇒ a² = 169[1 – \(\frac { 25 }{ 169 }\) ]
⇒ a² = 169\(\frac { 169 – 25 }{ 169 }\)
⇒ a² = 144
⇒ a = 12.
Putting values of a and b in equation (1), then required equation of ellipse will be :
\(\frac{x^{2}}{(12)^{2}}-\frac{y^{2}}{(13)^{2}}\) = 1
⇒ \(\frac{x^{2}}{144}+\frac{y^{2}}{169}\) = 1

Question 17.
Find the equation of ellipse whose centre is at (0, 0), major axis on the Y – axis passing through the points (3, 2) and (1, 6).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where a < b …. (1)
∵ Equation (1) passes through points(3, 2) and (1, 6)
∴ \(\frac{9}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)
and \(\frac{1}{a^{2}}+\frac{36}{b^{2}}\) = 1 …. (3)
Multiply equation (2) by 9, we get

Putting value of a² in equation (2), we get
\(\frac { 9 }{ 10 }\) + \(\frac{4}{b^{2}}\) = 1
⇒ \(\frac{4}{b^{2}}\) 1 – \(\frac { 9 }{ 10 }\)
⇒ \(\frac{4}{b^{2}}\) = \(\frac { 1}{ 10 }\)
⇒ b² = 40
Putting values of a² and b² in equation (1), then required equation of ellipse will be :
\(\frac{x^{2}}{10}+\frac{y^{2}}{40}\) = 1

Question 18.
Find the equation of ellipse whose major axis on the X – axis which passes through the points (4, 3) and (6, 2).
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
∵ It passes through points(4, 3) and (6, 2)
∴\(\frac{36}{a^{2}}+\frac{4}{b^{2}}\) = 1 …. (2)
and \(\frac{16}{a^{2}}+\frac{9}{b^{2}}\) = 1 …. (3)
Subtracting equation (3) from equation (2), we get
\(\frac{20}{a^{2}}-\frac{5}{b^{2}}\) = 1
⇒ \(\frac{4}{a^{2}}-\frac{1}{b^{2}}\) = 1
⇒ \(\frac{4}{a^{2}} = \frac{1}{b^{2}}\)
⇒ a² = 4b²
Putting value of a² in equation (2), we get
\(\frac{36}{4b^{2}}+\frac{4}{b^{2}}\) = 1
⇒ \(\frac{9}{b^{2}}+\frac{4}{b^{2}}\) = 1
⇒ 9 + 4 = b²
⇒ b² = 13
Putting values of a² and b² in equation (1), hence
Required equation of ellipse \(\frac{x^{2}}{52}+\frac{y^{2}}{13}\) = 1

Question 19.
An arch is the form of a semi ellipse. It is 8 m wide and 2 m high of the centre. Find the height of the arch at a point 1-5 m from one end. (NCERT)
Solution:
Let the equation of ellipse is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
Here 2a = 8 ⇒ a = 4, b = 2
Putting the values of a and b, we get

\(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{2^{2}}\) = 1 …. (1)
\(\frac{x^{2}}{16}+\frac{y^{2}}{4}\) = 1
Given :
AP = 1.5m, OA = \(\frac { 8 }{ 2 }\) = 4m,
OP = OA – AP = 4 – 1.5 = 2.5m.
Let PQ = k
∴ Co – ordinate of Q will be (2.5, k), which will satisfy ellipse’s equation.
Hence,
\(\frac{(2.5)^{2}}{16}+\frac{k^{2}}{4}\) = 1
⇒ \(\frac { 6.25}{ 16 }\) + \(\frac{k^{2}}{4}\) = 1
⇒ \(\frac{k^{2}}{4}\) = 1 – \(\frac { 6.25 }{ 16 }\)
⇒ \(\frac{k^{2}}{4}\) = \(\frac { 16 – 6.25 }{ 16 }\)
⇒ k² = \(\frac { 9.75 }{ 4 }\)
⇒ k² = 2.437
⇒ k = 1.56metre (approx).

Question 20.
A rod of length 12 cm moves with its ends always touching the co – ordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the X – axis. (NCERT)
Solution:
Let AB be the rod of length 12 cm which make an angle θ with X – axis.
∴ ∠BAO = θ
AB = 12 cm
AP = 3 cm, then PB = 9 cm

In ∆PNA,
sin θ = \(\frac {PN }{ PA }\) = \(\frac { y }{ 3 }\)
In ∆PMB,
cos θ = \(\frac { PM }{ PB }\) = \(\frac { x }{ 9 }\)
sin²θ + cos²θ = \(\frac { y }{ 3 }\)² + \(\frac { x }{ 9 }\)²
⇒ 1 = \(\frac{y^{2}}{9}+\frac{x^{2}}{81}\)
Hence required equation is :
\(\frac{x^{2}}{81}+\frac{y^{2}}{9}\) = 1

Question 21.
Find the eccentricity, co – ordinate of foci, equation of directrix and length of Iatus rectum of ellipse 4x² + y² – 8x + 2y + 1 = 0.
Solution:
4x² + y² – 8x + 2y + 1 = 0
⇒ 4x² – 8x + y² + 2y +1 = 0
⇒ 4x² – 8x + (y + 1)² = 0
⇒ 4(x² – 2x) + (y + 1)² = 0
⇒ 4(x² – 2x + 1) + (y + 1)² = 4
⇒ 4(x² – 2x + 1) + (y + 1)² = 4
⇒ \(\frac{(x-1)^{2}}{1}+\frac{(y+1)^{2}}{4}\) or \(\frac{X^{2}}{1}+\frac{Y^{2}}{4}\) = 1
Here, b > a
a² = b²(1 – e²)
⇒ 1 = (1 – e²)
⇒ \(\frac { 1 }{ 4}\) = 1 – e²
⇒ e² = 1 – \(\frac { 1 }{ 4}\) = \(\frac { 3 }{ 4}\)
⇒ Eccentricity e = \(\frac{\sqrt{3}}{2}\)
Co – ordinate of foci (0, ± be)
= (0, ±2. \(\frac{\sqrt{3}}{2}\)
= (0, ± \(\sqrt {3}\) )
Here X = 0, Y = ± \(\sqrt {3}\)
∴ x – 1 = 0, y + 1 = ± \(\sqrt {3}\)
⇒ x = 1, y = – 1 ± \(\sqrt {3}\)
foci = (1 ± \(\sqrt {3}\) – 1)
Equation of directrix Y = ± \(\frac { b }{ e}\)
⇒ Y = ± \(\frac{2}{\sqrt{3}}\).2
⇒ Y = ± \(\frac{4}{\sqrt{3}}\)
Y + 1 = \(\frac{4}{\sqrt{3}}\), (∵ Y = y+1)
⇒ y = ± \(\frac{4}{\sqrt{3}}\) – 1
Length of latus rectum = \(\frac{2 a^{2}}{b}\)
= 2. \(\frac { 1 }{ 2 }\) = 1.

Question 22.
Find the vertices, co-ordinate of foci, eccentricity and length of latus rectum of hyperbola :
(A) 9y² – 4x² = 36.
Solution:
Given : 9y² – 4x² = 36
⇒ \(\frac{9 y^{2}}{36}-\frac{4 x^{2}}{36}\) = 1
⇒ \(\frac{y^{2}}{4}-\frac{x^{2}}{9}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
b² = 4 ⇒ b = 2, a² = 9 ⇒ a = 3
Let e is the eccentricity of hyperbola.
Then, a² = b²(e² – 1)
⇒ 9 = (e² – 1)
⇒ \(\frac { 9 }{ 4 }\) = e² – 1
⇒ e² = \(\frac { 9 }{ 4 }\) + 1 = \(\frac { 13 }{ 4 }\)
∴ Eccentricity e = \(\frac{\sqrt{3}}{2}\)
Vertices = (0, ± b) = (0, ± 2)
Foci = (0, ± be) = (0, ± 2 x \(\frac{\sqrt{3}}{2}\)) = (0, ±\(\sqrt {3}\))
Length of latus rectum = \(\frac{2 a^{2}}{b}\) = \(\frac { 2 x 9 }{ 2 }\) = 9

(B) 16x² – 9y² = 576.
Solution:
Given : 16x² – 9y² = 576
⇒ \(\frac{16 x^{2}}{576}-\frac{9 y^{2}}{576}\) = 1
⇒ \(\frac{x^{2}}{36}-\frac{y^{2}}{64}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 …. (1)
Here, a² = 36 ⇒ a = 6, b² = 64 ⇒ b = 3
We Know that, b² = a²(e² – 1)
64 = 36(e² – 1)
⇒ \(\frac { 64 }{ 36 }\) = e² – 1
⇒ \(\frac { 16 }{ 9 }\) = e² – 1
⇒ e² = \(\frac { 16 }{ 9 }\) + 1 = \(\frac { 25 }{ 9 }\)
∴ Eccentricity e = \(\frac{5}{3}\)
Vertices = (± a, 0) = (± 6, 0)
Foci = (± ae, 0) = (± 6 x \(\frac { 5 }{ 3 }\)) = (± 10, 0)
Length of latus rectum = \(\frac{2 b^{2}}{a}\) = \(\frac { 2 x 64 }{ 6 }\) = \(\frac { 64 }{ 3 }\).

(c) 5y² – 9x² = 36.
Solution:
Given : 5y² – 9x² = 36
⇒ \(\frac{5 y^{2}}{36}-\frac{9 x^{2}}{36}\) = 1
\(\frac{y^{2}}{\frac{36}{5}}-\frac{x^{2}}{4}\) = 1
Comparing the above equation with the standard form of hyperbola
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
Here b² = \(\frac { 36 }{ 5 }\) ⇒ b = \(\frac{\sqrt{6}}{5}\), a² = 4 ⇒ a = 2
We know that, a² = b²(e² – 1)
⇒ 4 = \(\frac{\sqrt{36}}{5}\)(e² – 1)
⇒ e² – 1 = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)
⇒ e² = 1 + \(\frac { 5 }{ 9 }\) = \(\frac { 14 }{ 9 }\)

Question 23.
Find the equation of hyperbola whose foci are (0, ± \(\sqrt {10}\)) and which passes through point (2,3).
Solution:
Foci of hyperbola are (0, ±\(\sqrt {10}\)).
∴Form of hyperbola is :
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1 …. (1)
Foci (0, ± be) = (0, ± \(\sqrt {10}\))
be = \(\sqrt {10}\)
Equation (1) passes through point (2, 3).
∴ \(\frac{9}{b^{2}}-\frac{4}{a^{2}}\)
⇒ 9a² – 4b² = a²b²
We know that, a² = b² (e² – 1)
⇒ a² = b²e² – b²
a² = ( \(\sqrt {10}\))² – b²
⇒ a² = 10 – b²
⇒ b² = 10 – a²
Putting value of b² in equation (2),
9a² – 4(10 – a²) = a² (10 – a²)
⇒ 9a² – 40 + 4a² = 10a² – a⁴
⇒ 13a² – 40 = 10a² – a⁴
⇒ a⁴ + 13a² – 10a² – 40 = 0
⇒ a⁴ + 3a² – 40 = 0
⇒ a⁴ + 8a² – 5a² – 40 = 0
⇒ a²(a² + 8) – 5(a² + 8) = 0
⇒ (a² – 5)(a² + 8) = 0
a² = 5, a² = – 8
∵ The value of a cannot be negative.
∴ a² = 5
b² = 10 – a²
⇒ b²  = 10 – 5
⇒  b² = 5
Putting values of a² and b² in equation (1), then required equation of hyperbola will be :
\(\frac{y^{2}}{5}-\frac{x^{2}}{5}\) = 1
⇒ y² – x² = 5.

Question 24.
Find the equation of hyperbola in which the distance between foci is 8 and distance between directrix is 6.
Solution:
Let the equation of hyperbola is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) … (1)
and Eccentricity of hyperbola is e and foci are (ae, 0) and (- ae, 0) and latus rectum
are x = \(\frac { a }{ e }\) and x = – \(\frac { a }{ e }\)
Distance between foci = 2ae
Distance between latus rectum = \(\frac { 2a }{ e }\)
According to question, 2ae = 8
and \(\frac { 2a }{ e }\) = 6
Multiplying equation (2) and (3),
4a² = 48 ⇒ a² = 12
⇒ a = 2\(\sqrt {3}\)
Putting value of a in equation (2),
2.2\(\sqrt {3}\)e = 8 ⇒ e = \(\frac{2}{\sqrt{3}}\)
b² = a²(e² – 1)
= 12( \(\frac { 4 }{ 3 }\) – 1) = 4
Putting values of a² and b² in equation (1), the required equation of hyperbola is :
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\)
⇒ x² – 3y² = 12.

Question 25.
Find the centre, eccentricity, foci and length of latus rectum of hyperbola 9x² – 16y² + 18x + 32y – 151 = 0.
Solution:
Equation of hyperbola is :
9x² – 16y² + 18x + 32y – 151 = 0
⇒ 9x² + 18x – 16y² + 32y = 151
⇒ 9(x² + 2x) – 16(y² – 2y) = 151
⇒ 9(x + 1)² – 16(y – 1)² = 151 – 16 + 9
⇒ 9(x + 1)² – 16(y – 1)² = 144
⇒ \(\frac{9(x+1)^{2}}{144}-\frac{16(y-1)^{2}}{144}\)
⇒ \(\frac{(x+1)^{2}}{16}-\frac{(y-1)^{2}}{9}\) = 1
Let x + 1 = X and y – 1 = Y, then equation of hyperbola is
\(\frac{X^{2}}{16} – \frac{Y^{2}}{9}\) = 1
∴ a² = 16 ⇒ a = 4 and b² = 9 ⇒ b = 3.
∴ Centre is (- 1, 1).
For eccentricity = e
b² = a²(e² – 1)
⇒ 9 = 16(e² – 1)
⇒ \(\frac { 9 }{ 16 }\) = e² – 1
⇒ e² = 1 + \(\frac { 9 }{ 16 }\) = \(\frac { 25 }{ 16 }\)
⇒ e = \(\frac { 5 }{ 4 }\)
For foci, X = ± ae, Y = 0
⇒ x +1 = ± 4 x \(\frac { 5 }{ 4 }\), y – 1 = 0
⇒ x + 1 = ± 5, y = 1
⇒ x = 4, – 6, y = 1
∴ Foci are (4, 1) and (6, 1).
Equation of directrix is X = ± \(\frac { a }{ e }\)
⇒ x +1 = ± \(\frac { 4 }{ 5/4 }\)
⇒ x = ± \(\frac { 16 }{ 5 }\) – 1
⇒ x = ± \(\frac { 11 }{ 5 }\) and x = – \(\frac { 21 }{ 5 }\)
⇒ 5x = 11 and 5x + 21 = 0.

Question 26.
If e and ex are the eccentricity of hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) and \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1, then prove that: \(\frac{1}{e^{2}}+\frac{1}{e_{1}^{2}}\) = 1.
Solution:
Equation of hyperbola is
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
and
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = 1
For eccentricity e of equation (1),
b² = a²(e² – 1)
⇒ \(\frac{b^{2}}{a^{2}}\) = (e² – 1)
⇒ e ² = 1 + \(\frac{b^{2}}{a^{2}}\) = \(\frac{a^{2}+b^{2}}{b^{2}}\)
⇒ \(\frac{1}{e_{1}^{2}}\) = \(\frac{a^{2}}{a^{2}+b^{2}}\)
Again for eccentricity e of equation (2),

Question 27.
On a level plain the crack of the rifle and the thud of the ball striking the target are heard at the same instant, prove that the locus of the hearer is a hyperbola.
Solution:
Let P be the situation of hearer and T be the situation of the rifle and S is target. Let the velocity of the ball be v₁ and the velocity of sound be v₂.

Then,
Time to reach the ball from T to S = \(\frac{T S}{v_{1}}\)
Time to reach the sound from S to P = \(\frac{S P}{v_{2}}\)
and Time to reach the sound from T to P = \(\frac{T P}{v_{2}}\)
∴ The crack of the rifle and the thud of the ball are heard at the same instant.
∴ \(\frac{T S}{v_{1}}\) + \(\frac{S P}{v_{2}}\) = \(\frac{T P}{v_{2}}\)
⇒ \(\frac{T P}{v_{2}}\) – \(\frac{S P}{v_{2}}\) = \(\frac{T S}{v_{1}}\)
⇒ TP – SP = \(\frac{v_{2}}{v_{1}}\)
⇒ PT – PS = A constant (∵ v₂, v₂, TS are constant)
Hence locus of point P is hyperbola whose foci is T and S.

MP Board Class 11th Maths Important Questions

 

MP Board Class 12th Maths Important Questions Chapter 7a समाकलन

समाकलन Important Questions

समाकलन वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –
1. ∫ \(\frac { sec^{ 2 }x }{ 1+tanx } \) का मान है –
(a) log_e (1 + tan x) + c
(b) tan x + c
(c) – cot x + c
(d) log_e x + c.
उत्तर:
(a) log_e (1 + tan x) + c

प्रश्न 2.
∫ \(\frac { x }{ 4+x^{ 4 } } \) dx का मान है –
(a) \(\frac{1}{4}\) tan^-1 x² + c
(b) \(\frac{1}{4}\) tan^-1 ( \(\frac { x^{ 2 } }{ 2 } \) )
(b) \(\frac{1}{2}\) tan^-1 ( \(\frac { x^{ 2 } }{ 2 } \) )
(d) इनमें से कोई नहीं।
उत्तर:
(a) \(\frac{1}{4}\) tan^-1 x² + c

प्रश्न 3.
यदि ∫ \(\frac { 2^{ 1/x } }{ x^{ 2 } } \) dx = k(2)^1/x + c हो, तो k का मान है –
(a) \(\frac { -1 }{ log_{ e }2 } \)
(b) – log_e2
(c) – 1
(d) \(\frac{1}{2}\)
उत्तर:
(a) \(\frac { -1 }{ log_{ e }2 } \)

प्रश्न 4.
∫ \(\frac { e^{ x }(1+x) }{ cos^{ 2 }(xe^{ x }) } \) dx का मान है –
(a) 2log_e cos(xe^x) + c
(b) sec(xe^x) + c
(c) tan(xe^x) + c
(d) tan(x + e^x) + c
उत्तर:
(c) tan(xe^x) + c

प्रश्न 5.
यदि ∫ x sin x dx = – x cos x + α हो, तो α का मान होगा –
(a) sin x + c
(b) cos x + c
(c) c
(d) इनमें से कोई नहीं।
उत्तर:
(a) sin x + c

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए –


उत्तर:

1. \(\frac{1}{a}\) tan^-1 ( \(\frac{x}{a}\) ) + c
2. log [x + \(\sqrt { x^{ 2 }-a^{ 2 } } \) ] + c
3. log [x + \(\sqrt { a^{ 2 }-x^{ 2 } } \) ] + c
4. \(\frac{x}{2}\) \(\sqrt { a^{ 2 }-x^{ 2 } } \) + \(\frac { a^{ 2 } }{ 2 } \) sin^-1( \(\frac{x}{a}\) ) + c
5. log(sec x + tan x) + c
6. sin^-1 x + \(\sqrt { 1-x } \) + c
7. tan x + sec x

प्रश्न 3.
निम्न कथनों में सत्य/असत्य बताइए –

उत्तर:

1. सत्य
2. सत्य
3. असत्य
4. सत्य
5. असत्य
6. असत्य।

प्रश्न 4.
सही जोड़ी बनाइये –

उत्तर:

1. (a)
2. (d)
3. (e)
4. (b)
5. (c)
6. (f).

प्रश्न 5.
एक शब्द/वाक्य में उत्तर दीजिए –

1. ∫ e^x(sin x + cos x) dx का मान क्या है?
2. ∫ \(\frac { cotx }{ log(sinx) } \) dx का मान क्या है?
3. ∫ \(\frac { dx }{ sin^{ 2 }x+4cos^{ 2 }x } \) का मान क्या है?
4. ∫ cosec x dx का मान क्या है?
5. ∫ log x dx का मान क्या है?
6. ∫ \(\frac { dx }{ \sqrt { 4-x^{ 2 } } } \)

उत्तर:

1. e^x sin x
2. log log sin x
3. \(\frac{1}{2}\) tan^-1 ( \(\frac { tanx }{ 2 } \) )
4. log tan \(\frac{x}{2}\)
5. x(log x + 1)
6. sin^-1 \(\frac{x}{2}\)

समाकलन अति लघु उत्तरीय प्रश्न

प्रश्न 1.
∫ \(\frac { e^{ tan-1x } }{ 1+x^{ 2 } } \) dx का मान ज्ञात कीजिये।
उत्तर:
e^tan-1x + c

प्रश्न 2.
∫e^log(sinx) dx का मान ज्ञात कीजिये।
उत्तर:
– cos x + c

प्रश्न 3.
∫2 sin x cos x dx का मान ज्ञात कीजिये।
उत्तर:
sin² x + c

प्रश्न 4.
∫\(\frac { dx }{ a^{ 2 }-x^{ 2 } } \) का मान लिखिये।
उत्तर:
\(\frac{1}{2a}\) log \(\frac{a+x}{a-x}\)

प्रश्न 5.
∫\(\frac { dx }{ x^{ 2 }-a^{ 2 } } \) का मान लिखिये।
उत्तर:
\(\frac{1}{2a}\) log \(\frac{x-a}{x+a}\)

प्रश्न 6.
∫e^x ( \(\frac{1}{x}\) – \(\frac { 1 }{ x^{ 2 } } \) ) का मान ज्ञात कीजिये।
उत्तर:
\(\frac { e^{ x } }{ x } \) + c

प्रश्न 7.
∫\(\frac { 1 }{ log_{ x }e } \) dx का मान ज्ञात कीजिये।
उत्तर:
x log \(\frac{x}{e}\) + c

प्रश्न 8.
यदि ∫\(\frac { dx }{ (1+x)\sqrt { x } } \) = f (x) + c, जहाँ c एक स्वेच्छ अचर है, तब फलन f (x) क्या है?
उत्तर:
2 tan^-1 x

प्रश्न 9.
∫\(\frac { 1+logx }{ x } \) dx का मान ज्ञात करने में कौन-सा प्रतिस्थापन उचित रहेगा?
उत्तर:
1 + log x = t

प्रश्न 10.
यदि ∫\(\frac { 1 }{ 1+xsinx } \) dx = tan ( \(\frac{x}{2}\) + a) + b है, तब a और b क्या होंगे?
उत्तर:
a = – \(\frac { \pi }{ 4 } \), b = 3

प्रश्न 11.
∫\(\frac { sin\sqrt { x } }{ \sqrt { x } } \) dx का मान ज्ञात करने के लिये उचित प्रतिस्थापन क्या होगा?
उत्तर:
x = t²

प्रश्न 12.
∫sin(ax + b) dx का मान ज्ञात कीजिये।
उत्तर:
– \(\frac{1}{a}\) cos(ax + b)

प्रश्न 13.
∫\(\frac{1}{ax+b}\) का मान ज्ञात कीजिये।
उत्तर:
\(\frac{1}{a}\) log(ax + b)

प्रश्न 14.
∫tan² x dx का मान ज्ञात कीजिये।
उत्तर:
tan x – x + c

प्रश्न 15.
\(\frac { 1 }{ \sqrt { a^{ 2 }-x^{ 2 } } } \) का मान लिखिये।
उत्तर:
sin^-1 ( \(\frac{x}{a}\) ) + c

प्रश्न 16.
∫e^log_ex2 dx का मान ज्ञात कीजिये।
उत्तर:
\(\frac { x^{ 3 } }{ 3 } \) + c

प्रश्न 17.
∫e^logx dx का मान ज्ञात कीजिये।
उत्तर:
\(\frac { x^{ 2 } }{ 2 } \) + c

प्रश्न 18.
∫e^x [f (x) + f'(x)] dx का मान ज्ञात कीजिये।
उत्तर:
e^{x f (x) + c}

प्रश्न 19.
∫(1 + x + \(\frac { x^{ 2 } }{ 2! } \) + ………………..) dx का मान कितना है?
उत्तर:
e^x

प्रश्न 20.
∫e^x (log x + \(\frac{1}{x}\) ) dx का मान ज्ञात कीजिये।
उत्तर:
e^{x log x + c}

समाकलन लघु उत्तरीय प्रश्न

पश्न 1.
मूल्यांकन कीजिये –
∫\(\frac { cos2x+2sin^{ 2 }x }{ cos^{ 2 }x } \) dx (CBSE 2018)
हल:
∫\(\frac { cos2x+2sin^{ 2 }x }{ cos^{ 2 }x } \) dx

प्रश्न 2.
∫\(\frac { 1-sinx }{ cos^{ 2 }x } \) dx का मान ज्ञात कीजिये। (NCERT)
हल:
माना I = ∫\(\frac { 1-sinx }{ cos^{ 2 }x } \) dx

प्रश्न 3.
∫\(\frac { 2-3sinx }{ cos^{ 2 }x } \) dx का मान ज्ञात कीजिये। (NCERT)
हल:
माना I = ∫\(\frac { 2-3sinx }{ cos^{ 2 }x } \) dx

प्रश्न 4.
∫sin^-1(cos x) dx का मान ज्ञात कीजिये। (NCERT)
हल:
माना I = ∫sin^-1(cos x) dx

प्रश्न 5.
∫\(\frac { dx }{ 1+cos2x } \) का मान ज्ञात कीजिए।
हल:
माना I = ∫\(\frac { dx }{ 1+cos2x } \)

प्रश्न 6.
∫tan^-1 xdx का मान ज्ञात कीजिए।
हल:
माना I = ∫tan^-1 x. 1 dx

प्रश्न 7.
∫sin²x dx का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 8.
∫\(\frac { cosx }{ cos(x-a) } \) dx का मान ज्ञात कीजिए।
हल:
माना I = ∫\(\frac { cosx }{ cos(x-a) } \) dx
x – α = t ⇒ x = t + α
dx = dt

= cos α ∫dt – sin α ∫ tan t dt = cos α.t – sin α. log sec t + c
= cos α. (x – a) – sin α log sec (x – a) + c

प्रश्न 9.
(A) ∫\(\frac { 1 }{ \sqrt { 1+cosx } } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { 1 }{ \sqrt { 1+cosx } } \) dx

(B)∫\(\sqrt { 1+sin2x } \) dx का मान  ज्ञात कीजिए।
हल:
माना
I = ∫\(\sqrt { 1+sin2x } \) dx

(C) ∫\(\sqrt { 1+cos2x } \) dx का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 9 (B) की भाँति हल करें।

(D) ∫\(\sqrt { 1-sin2x } \) dx का मान ज्ञात कीजिए।
उत्तर:
प्रश्न क्र. 9 (B) की भाँति हल करें।

प्रश्न 10.
(A) ∫\(\frac { e^{ x }(1+x) }{ cos^{ 2 }(xe^{ x }) } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { e^{ x }(1+x) }{ cos^{ 2 }(xe^{ x }) } \) dx
माना xe^x = t ⇒ e^x (1 + x) dx = dt
∴ I = ∫\(\frac { dt }{ cos^{ 2 }t } \) = ∫sec² t dt = tan t = tan(xe^x)

(B)
∫\(\frac { e^{ tan-1x } }{ 1+x^{ 2 } } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { e^{ tan-1x } }{ 1+x^{ 2 } } \) dx
= ∫e^x dt, [माना tan^-1 x = t, \(\frac { 1 }{ 1+x^{ 2 } } \) dx = dt]
= e^t = e^tan-1x.

प्रश्न 11.
∫\(\frac { dx }{ 1-sinx } \) का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 12.
∫\(\frac{logx}{x}\) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac{logx}{x}\) dx
माना log x = t ⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫t dt = \(\frac { t^{ 2 } }{ 2 } \) + c = \(\frac { (logx)^{ 2 } }{ 2 } \) + c

प्रश्न 13.
(A) ∫\(\frac{dx}{1-cosx}\) का मान ज्ञात कीजिए।
हल:
माना

(B) ∫\(\frac{dx}{1+cosx}\) का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 14.
∫\(\frac { dx }{ 1+sinx } \) का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 15.
∫\(\frac { cos\sqrt { x } }{ \sqrt { x } } \) dx का मान ज्ञात कीजिए।
हल:
माना
∫\(\frac { cos\sqrt { x } }{ \sqrt { x } } \) dx
माना \(\sqrt{x}\) = t ⇒ \(\frac { dx }{ 2\sqrt { x } } \) = dt ⇒ \(\frac { dx }{ \sqrt { x } } \) = 2dt
∴ I = ∫cos t(2dt)
= 2∫cos t dt = 2 sin t + c = 2 sin \(\sqrt{x}\) + c

प्रश्न 16.
∫\(\frac { 1-cos2x }{ 1+cos2x } \) dx का मान ज्ञात कीजिए।
हल:
माना

= tan x – x + c

प्रश्न 17.
\(\frac { x^{ 4 } }{ x^{ 2 }+1 } \) का समाकलन x के सापेक्ष कीजिए।
हल:
माना

प्रश्न 18.
∫\(\frac { sec^{ 2 }(logx) }{ x } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { sec^{ 2 }(logx) }{ x } \) dx
log x = t रखने पर,
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫sec² tdt
⇒ I = tan t + c
⇒ I = tan(log x) + c

प्रश्न 19.
∫\(\frac { sin(logx) }{ x } \) dx का मान ज्ञात कीजिए।
हल:
माना
log x = t रखने पर,
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫sin t dt = – cos t + c
⇒ I = – cos(log x) + c

प्रश्न 20.
∫\(\frac { cos(logx) }{ x } \) dx का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 19 की भाँति हल करें।

प्रश्न 21.
(A) मान ज्ञात कीजिए – ∫tan² x dx?
हल:
I = ∫tan² x dx = ∫(sec² x – 1) dx
= ∫sec² x dx – ∫1dx = tan x – x

(B)
मान ज्ञात कीजिए – ∫cot² x dx?
हलः
I = ∫cot² x dx = ∫(cosec² x – 1 dx
= ∫cosec² x dx – ∫1. dx = – cot x – x

प्रश्न 22.
∫\(\frac { sinx }{ 1+cosx } \) dx का मान ज्ञात कीजिये।
हल:
I = ∫\(\frac { sinx }{ 1+cosx } \) dx
= ∫\(\frac{1}{t}\) dt, (1 + cos x = t रखने पर sin x dx = dt)
= log t
= log (1 + cos x)

प्रश्न 23.
∫\(\frac { sin^{ =1 }x }{ \sqrt { 1-x^{ 2 } } } \) dx का मान ज्ञात कीजिये।
हल:
I = \(\frac { sin^{ =1 }x }{ \sqrt { 1-x^{ 2 } } } \) dx
= ∫t dt (sin^-1 x = t रखने पर ⇒ \(\frac { 1 }{ \sqrt { 1-x^{ 2 } } } \) dx = dt)
= \(\frac { t^{ 2 } }{ 2 } \)
= \(\frac{1}{2}\) (sin^-1 x)²

समाकलन दीर्घ उत्तरीय प्रश्न – I

प्रश्न 1.
∫\(\sqrt { \frac { a+x }{ a-x } } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\sqrt { \frac { a+x }{ a-x } } \)
पुनः माना x = a cos θ ⇒ dx = – a sinθ dθ

प्रश्न 2.
मान ज्ञात कीजिए –
∫[ \(\frac { 1 }{ (logx)^{ 2 } } \) – \(\frac { 2 }{ (logx)^{ 3 } } \) ] dx?
हल:
माना
I = ∫[ \(\frac { 1 }{ (logx)^{ 2 } } \) – \(\frac { 2 }{ (logx)^{ 3 } } \) ] dx
पुनः माना log x = t ⇒ x = e^t ⇒ dx = e^t dt

प्रश्न 3.
∫sin^-1 x dx का मान ज्ञात कीजिए।
हल:
माना

पुनः माना 1 – x² = t ⇒ -2x dx = dt

प्रश्न 4.
∫cos^-1 x dx का मान ज्ञात कीजिए।
हल:
I = ∫cos^-1 x dx

प्रश्न 5.
(A)
∫\(\frac { x^{ 2 } }{ 1+x } \) dx का मान ज्ञात कीजिए।
हल:
माना

(B) ∫\(\frac { x }{ 1+x^{ 4 } } \) dx का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 6.
∫\(\frac { 1 }{ sinx-cosx } \) dx का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 7.
∫\(\frac { dx }{ e^{ x }+1 } \) का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 8.
∫sec³ xdx का मान ज्ञात कीजिए।
हल:
माना I = ∫sec³ xdx = ∫secx. sec² xdx
= sec x ∫sec² xdx – ∫ [ \(\frac{d}{dx}\) sec x ∫sec² xdx] dx, [खण्डशः समाकलन द्वारा]
= sec x tan x – ∫sec x tan x tan x dx
= sec x tan x – ∫sec x tan² x dx
= sec x tan x – ∫sec x(sec² x – 1)dx
= sec x tan x – ∫sec³ x dx + ∫sec x dx
⇒ I = sec x tan x – I + log(sec x + tan x)
⇒ 2I = sec x tan x + log(sec x + tan x)
⇒ I = \(\frac{1}{2}\) [sec x tan x + log(sec x + tan x)]

प्रश्न 9.
∫\(\frac { dx }{ x^{ 2 }-a^{ 2 } } \) का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 10.
∫\(\frac { 3x }{ (x-2)(x+1) } \) dx का मान ज्ञात कीजिए।
हल:
माना ∫\(\frac { 3x }{ (x-2)(x+1) } \) = \(\frac { A }{ x-2 } \) + \(\frac { B }{ x+1 } \) ……………….. (1)
⇒ \(\frac { 3x }{ (x-2)(x+1) } \) = \(\frac { A(x+1)+B(x-2) }{ (x-2)(x+1) } \)
⇒ 3x = A(x + 1) + B(x – 2)
⇒ 3x = (A + B)x + (A – 2B) …………………. (2)
समी. (2) के दोनों पक्षों में x के गुणांकों तथा अचर पदों की तुलना करने पर,
3 = A + B
0 = A – 2B
3 = 3B
⇒ B = 1
तथा A = 2B = 2
∴ ∫\(\frac { 3x }{ (x-2)(x+1) } \) dx = ∫\(\frac { 2 }{ x-2 } \) dx + ∫\(\frac { dx }{ x+1 } \)
= 2 log(x – 2) + log(x + 1) + c

प्रश्न 11.
∫\(\frac { x^{ 2 }+1 }{ x^{ 4 }-x^{ 2 }+1 } \) dx का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 12.
∫\(\frac { x^{ 2 }+1 }{ x^{ 4 }+x^{ 2 }+1 } \) dx का मान ज्ञात कीजिए।
हल:
प्रश्न क्र. 11 की भाँति हल करें।
उत्तर: \(\frac { 1 }{ \sqrt { 3 } } \) tan^-1 ( \(\frac { x^{ 2 }-1 }{ \sqrt { 3.x } } \) ) + c

प्रश्न 13.
मान ज्ञात कीजिए –
∫\(\frac { dx }{ \sqrt { x^{ 2 }+2x+3 } } \)?
हल:
माना

प्रश्न 14.
फलन \(\frac { 1 }{ 1+sin^{ 2 } } \) का x सापेक्ष समाकलन कीजिए।
हल:
माना
I = ∫\(\frac { 1 }{ 1+sin^{ 2 } } \) dx

प्रश्न 15.
∫\(\frac { cos2x }{ (cosx+sinx)^{ 2 } } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { cos2x }{ (cosx+sinx)^{ 2 } } \) dx


cos x + sin x = रकने पर
\(\frac{d}{dx}\) (cos x + sin x) = \(\frac{dt}{dx}\)
⇒ (-sin x + cos x) dx = dt
∴ t = ∫\(\frac{dt}{t}\)
⇒ I = log t + c
अतः I = log(cos x + sin x) + c

प्रश्न 16.
∫\(\frac { 1+tanx }{ x+logsecx } \) dx का मान ज्ञात कीजिए।
हल:
माना I = ∫\(\frac { 1+tanx }{ x+logsecx } \) dx
x + log sec x = t रकने पर
\(\frac{d}{dx}\) [x + log(sec x)] = \(\frac{dt}{dx}\)
⇒ \(\frac{d}{dx}\) x + \(\frac{d}{dx}\) log(sec x) = \(\frac{dt}{dx}\)
sec x = u रकने पर
1 + \(\frac{d}{dx}\) log u = \(\frac{dt}{dx}\)
⇒ 1 + \(\frac{d}{du}\) log u \(\frac{du}{dx}\) = \(\frac{dt}{dx}\)
⇒ 1 + \(\frac{1}{u}\)\(\frac{d}{dx}\) sec x = \(\frac{dt}{dx}\)
⇒ 1 + \(\frac{1}{secx}\) × sec x tan x = \(\frac{dt}{dx}\)
⇒ (1 + tan x)dx = dt
∴ I = ∫\(\frac{dt}{t}\)
⇒ I = log t + c
अतः I = log(x + log sec x) + c

प्रश्न 17.
∫\(\frac { cotx }{ log(sinx) } \) dx का मान ज्ञात कीजिए।
log(sin x) = t रकने पर
\(\frac{d}{dx}\) log (sin x) = \(\frac{dt}{dx}\)
sin x = u रकने पर

प्रश्न 18.
∫\(\frac { 2cosx-3sinx }{ 6cosx+4sinx } \) dx का मान ज्ञात कीजिए। (NCERT)
हल:
माना
I = ∫\(\frac { 2cosx-3sinx }{ 6cosx+4sinx } \) dx
6 cos x + 4 sin x = t रखने पर,
\(\frac{d}{dx}\) (6 cos x + 4 sin x) = \(\frac{dt}{dx}\)
⇒ – 6 sin x + cos x = \(\frac{dt}{dx}\)
⇒ 2(2 cos x – 3 sin x) = \(\frac{dt}{dx}\)
⇒ (2 cos x – 3 sin x)dx = \(\frac{1}{2}\) dt
⇒ I = \(\frac{1}{2}\) ∫dt
⇒ I = \(\frac{1}{2}\) log t + c
अतः I = \(\frac{1}{2}\) log(6 cos x + 4 sin x) + c

प्रश्न 19.
∫e^3logx (x⁴ + 1)^-1 dx का मान ज्ञात कीजिए।
हल:
माना

x⁴ + 1 = t रखने पर,

प्रश्न 20.
∫\(\frac { dx }{ x-\sqrt { x } } \) का मान ज्ञात कीजिए। (NCERT)
हल:
माना

\(\sqrt{x}\) – 1 = t रखने पर,

प्रश्न 21.
∫\(\frac { dx }{ 1+3sin^{ 2 }x } \) का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { dx }{ 1+3sin^{ 2 }x } \)

समाकलन दीर्घ उत्तरीय प्रश्न – II

प्रश्न 1.
∫sin^-1 ( \(\frac { 2x }{ 1+x^{ 2 } } \) ) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫sin^-1 ( \(\frac { 2x }{ 1+x^{ 2 } } \) ) dx
पुनः माना x = tan θ ⇒ dx = sec²θ dθ

प्रश्न 2.
फलन \(\frac { e^{ mtan-1x } }{ (1+x^{ 2 })^{ 3/2 } } \) का x के सापेक्ष समाकलन कीजिए।
हल:
माना tan^-1 x = t ⇒ x = tan t
∴ dx = sec² t dt

प्रश्न 3.
∫\(\frac { x^{ 2 }tan^{ -1 }x }{ 1+x^{ 2 } } \) dx का मान ज्ञात कीजिए।
हल:
माना I = ∫\(\frac { x^{ 2 }tan^{ -1 }x }{ 1+x^{ 2 } } \) dx
माना x = tan θ ⇒ θ = tan^-1 x
⇒ dx = sec² θ dθ

प्रश्न 4.
मान ज्ञात कीजिए – ∫tan^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) dx?
हल:
माना
I = ∫tan^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) dx
माना x = tan θ ⇒ dx = sec² θ dθ

प्रश्न 5.
∫\(\frac { xtan^{ -1 }x }{ (1+x^{ 2 })^{ 3/2 } } \) dx मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { xtan^{ -1 }x }{ (1+x^{ 2 })^{ 3/2 } } \) dx
पुनः माना x = tan θ ⇒ dx = sec² θ dθ

प्रश्न 6.
∫\(\frac{dx}{3+2cosx}\) का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac{dx}{3+2cosx}\)

प्रश्न 7.
मान ज्ञात कीजिए –
∫\(\frac{dx}{4+5cosx}\)?
हल:
प्रश्न का. 6 की बीत हल करे।

प्रश्न 8.
मान ज्ञात कीजिए –
∫\(\frac{dx}{5-3cosx}\)?
हल:
प्रश्न क्र. 6 की भाँति हल करें।

प्रश्न 9.
∫\(\frac{1}{4+5sinx}\) dx का मान ज्ञात कीजिए।
हल:
माना

माना tan\(\frac{x}{2}\) = t ⇒ sec² \(\frac{x}{2}\).\(\frac{1}{2}\) dx = dt

प्रश्न 10.
∫\(\frac { e^{ x }(1+sinx) }{ (1+cosx) } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { e^{ x }(1+sinx) }{ (1+cosx) } \) dx

प्रश्न 11.
∫\(\frac { xe^{ x } }{ (1+x)^{ 2 } } \) dx का मान ज्ञात कीजिए।
हल:
माना

प्रश्न 12.
∫\(\frac { 2cosx }{ (1-sinx)(1+sin^{ 2 }) } \) dx का मान ज्ञात कीजिए।
हल:
माना
I = ∫\(\frac { 2cosx }{ (1-sinx)(1+sin^{ 2 }) } \) dx
sin x = t रखने पर, cos x dx = dt

⇒ 2 = A(1 + t²) + (1 – t)(Bt + 1)
⇒ 2 = A + A² + Bt – Bt² + 1 – t
⇒ 2 = (A – B)t² + (B – 1)t + (A + 1)
गुणांकों की तुलना करने पर,
∴ A – B = 0
B – 1 = 0
तथा A + 1 = 2
B = 1, A = 1

प्रश्न 13.
∫\(\frac { dx }{ e^{ x }-1 } \) का मान ज्ञात कीजिए।
हल:
I = ∫\(\frac { dx }{ e^{ x }-1 } \) = \(\frac { e^{ x }dx }{ e^{ x }(e^{ x }-1) } \)
माना e^x = t, तब e^xdx = dt

प्रश्न 14.
∫\(\frac { dx }{ x(x^{ n }+1) } \) का मान ज्ञात कीजिए।
हल:
माना

xⁿ = t रखने पर,

A और B के मान समी. (1) में रखने पर,

प्रश्न 15.
∫( \(\sqrt{tanx}\) + \(\sqrt{cotx}\) ) dx का मान ज्ञात कीजिए। (NCERT)
हल:
माना

पुनः माना sin x – cosx = t
(cos x + sin x)dx = dt
चूंकि (sin x – cos x)² = t²
sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
I = \(\sqrt{2}\) ∫\(\frac { dt }{ \sqrt { 1-t^{ 2 } } } \)
= \(\sqrt{2}\) .sin^-1t
= \(\sqrt{2}\).sin^-1(sin x – cos x)

MP Board Class 12 Maths Important Questions

MP Board Class 12th Maths Important Questions Chapter 10 Vector Algebra

Vector Algebra Important Questions

Vector Algebra Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Unit vector parallel to the resultant vector of vectors 2\(\hat { i } \) + 4\(\hat { j } \) – 5\(\hat { k } \) and \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) is:
(a) – \(\hat { i } \) – \(\hat { j } \) + 8\(\hat { k } \)
(b) \(\frac { 3\hat { i } +6\hat { j } -2\hat { k } \quad }{ 7 } \)
(c) \(\frac { -\hat { i } -+8\hat { k } \quad }{ \sqrt { 69 } } \)
(d) \(\frac { -\hat { i } +2\hat { j } -8\hat { k } \quad }{ \sqrt { 69 } } \)

Question 2.
If \(\vec { O } \)A = a, \(\vec { O } \)B = b and C is a point on AB such that \(\vec { A } \)C = 3AB, then \(\vec { O } \)C is equal to:
(a) 3\(\vec { a } \) – 2\(\vec { b } \)
(b) 3\(\vec { b } \) – 2\(\vec { a } \)
(c) 3\(\vec { a } \) – \(\vec { b } \)
(d) 3\(\vec { b } \) – \(\vec { a } \)

Question 3.
If \(\vec { a } \) and \(\vec { b } \) are two vectors such that |\(\vec { a } \)| = 2, |\(\vec { b } \)| = 1 and \(\vec { a } \).\(\vec { b } \) = \(\sqrt { 3 } \), then the angle between them is:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 4 } \)
(c) \(\frac { \pi }{ 6 } \)
(d) \(\frac { \pi }{ 7 } \)

Question 4.
Area of parallelogram whose adjacent sides are \(\hat { i } \) – 2\(\hat { j } \) + 3\(\hat { k } \) and 2\(\hat { i } \) + \(\hat { j } \) – 4\(\hat { k } \) is:
(a) 3\(\sqrt{6}\)
(b) 4\(\sqrt{6}\)
(c) 5\(\sqrt{6}\)
(d) 6\(\sqrt{6}\)

Question 5.
If \(\vec { a } \) = \(\vec { b } \) + \(\vec { c } \), then \(\vec { a } \).( \(\vec { b } \) × \(\vec { c } \) ) is equal to:
(a) 2\(\vec { a } \). ( \(\vec { b } \) + \(\vec { c } \) )
(b) 0
(c) \(\vec { b } \) = ( \(\vec { a } \) + \(\vec { c } \) )
(d) None of these

Question 2.
Fill in the blanks:

1. Sum or difference of two vectors is always a ………………………….
2. Addition of vectors obeys ………………………….
3. ( \(\vec { a } \) + \(\vec { b } \) ) + \(\vec { c } \) = \(\vec { a } \) + …………………………..
4. Addition of two vectors can be obtained from ……………………………..
5. Position vector of point (1,2, 3) w.r.t. the origin will be ……………………………..
6. If \(\vec { a } \) and \(\vec { b } \) are parllel then \(\vec { a } \) × \(\vec { b } \) = …………………………..
7. If \(\vec { a } \) and \(\vec { b } \) are parallel then \(\vec { a } \) × \(\vec { a } \) = ………………………..
8. The unit vector in the direction of vector \(\vec { a } \) will be ……………………………
9. The projection of \(\vec { b } \) along the direction of \(\vec { a } \) will be ……………………………..
10. If the vectors 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) 3\(\hat { i } \) + p\(\hat { j } \) + 5\(\hat { k } \) are coplanar then value of p will be …………………………….
11. A force 2\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), acts at a point A whose position vector 2\(\hat { i } \) – \(\hat { j } \) The moment of the force with respect to the origin will be ………………………………..
12. The area of the parallelogram will be …………………………. whose diagonals are 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) and \(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \).

Answer:

1. New vector
2. Commutative and associative law
3. ( \(\vec { b } \) + \(\vec { c } \) )
4. Traingle law of vector addition
5. \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
6. collinear
7. \(\vec { O } \)
8. \(\frac { \vec { a } }{ |\vec { a } | } \)
9. \(\frac { \vec { a } .\vec { b } }{ |\vec { a } | } \)
10. -4
11. \(\hat { i } \) + 2\(\hat { j } \) + 4\(\hat { k } \)
12. 5\(\sqrt { 3 } \) sq. unit.

Question 3.
Write True/False:

1. The sum of the vectors determined by the sides of a triangle taken in order is zero.
2. If \(\vec { a } \) and \(\vec { b } \) are two non collinear vectors, then |\(\vec { a } \) + \(\vec { b } \)| ≥ |\(\vec { a } \) + \(\vec { b } \)|
3. A vector whose initial and terminal points are coincident is called unit vector.
4. If the position vector of the points P and Q are \(\hat { i } \) + 3\(\hat { j } \) – 7\(\hat { k } \) and 5\(\hat { i } \) – 2\(\hat { j } \) + 4\(\hat { k } \) respectively, then the value of |\(\vec { P } \)Q| is 9\(\sqrt { 2 } \).
5. If |\(\vec { a } \) + \(\vec { a } \)|=|\(\vec { a } \) – \(\vec { b } \)|, then \(\vec { a } \) × \(\vec { b } \) = \(\vec { 0 } \).
6. The value of \(\vec { a } \).( \(\vec { a } \) × \(\vec { b } \) ) is zero.
7. If the vectors \(\hat { i } \) – λ\(\hat { j } \) + \(\hat { k } \) and \(\hat { i } \) – \(\hat { j } \) + 5\(\hat { k } \) are mutually perpendicular, then the value of λ is 6.

Answer:

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

Question 4.
Write the answer is one word/sentence:

1. If \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are the position vectors of the vectors of the ∆ABC, then write the formula for area of ∆ABC.
2. If \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + 3\(\hat { k } \),\(\vec { b } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) and \(\vec { c } \) = \(\hat { j } \) + \(\hat { k } \), then find the value of [ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ]
3. Find the angle between two vector 3\(\hat { i } \) – 2\(\hat { j } \) + 4\(\hat { k } \) and \(\hat { i } \) – \(\hat { j } \) + 5\(\hat { k } \).
4. Find the value of \(\hat { i } \) × ( \(\hat { j } \) + 3\(\hat { k } \) ) + \(\hat { j } \) × ( \(\hat { k } \) + \(\hat { i } \) ) + \(\hat { k } \) × ( \(\hat { i } \) + \(\hat { j } \) )
5. Find the projection of \(\vec { a } \) in the direction of \(\vec { b } \).
6. If \(\vec { a } \) and \(\vec { b } \) are mutually perpendicular vector then find, the value ( \(\vec { a } \) + \(\vec { b } \) ) ²

Answer:

1. \(\frac{1}{2}\) |\(\vec { a } \) × \(\vec { b } \) + \(\vec { b } \) × \(\vec { c } \) + \(\vec { c } \) × \(\vec { a } \)|
2. 12
3. cos^-1 \(\frac { 25 }{ \sqrt { 783 } } \)
4. 0
5. \(\frac { \vec { a } .\vec { b } }{ |\vec { b } | } \)
6. |\(\vec { a } \)|² + |\(\vec { b } \)|²

Question 4.
Match the Column:

Answer:

1. (d)
2. (e)
3. (a)
4. (b)
5. (c)
6. (g)
7. (f).

Vector Algebra Very Short Answer Type Questions

Question 1.
Given vectors \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = – 2\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) and \(\vec { c } \) = \(\hat { i } \) – 6\(\hat { j } \) – 7\(\hat { k } \). Then find the value of |\(\vec { a } \) + \(\vec { b } \) + \(\vec { c } \)|? (NCERT, CBSE 2012)
Solution:

Question 2.
Find the unit vector in the direction of sum of the vectors \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { b } \) = – \(\hat { i } \) + \(\hat { j } \) + 3\(\hat { k } \)?
Solution:

Question 3.
Find the vector in the direction of vector \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) which has magnitude 7 units? (NCERT)
Solution:
\(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \)
Unit vector in the direction of given vector a is:

The vector having magnitude be equal to 7:

Question 4.
Prove that the vectors 2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) and -4\(\hat { i } \) + 6\(\hat { j } \) – 8\(\hat { k } \) are collinear? (NCERT)
Solution:

Hence vector \(\vec { a } \), \(\vec { b } \) are collinear. Proved.

Question 5.
Find direction cosine of the vector \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)? (NCERT)
Solution:

Question 6.
If \(\vec { a } \) = 2 \(\hat { i } \) – 3 \(\hat { j } \) + \(\hat { k } \) and \(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \), then find \(\vec { a } \) – \(\vec { b } \)?
Solution:

Question 7.
If \(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { j } \) – \(\hat { k } \), then find |2\(\vec { a } \) – \(\vec { b } \)|?
Solution:

Question 8.
If the vector \(\vec { a } \) = 2\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) = \(\hat { i } \) – 4\(\hat { j } \) + λ\(\hat { k } \) are perpendicular then find the value of λ?
Solution:
The given vectors are perpendicular
Hence \(\vec { a } \) . \(\vec { b } \) = 0
(2\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) ). ( \(\hat { i } \) – 4\(\hat { j } \) + λ\(\hat { k } \) ) = 0
⇒ 2 – 4 + λ = 0
⇒ λ = 2.

Question 9.
(A) Prove that the vectors 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and –\(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) are perpendicular to each other?
Solution:

L.H.S = – 2 – 3 + 5
= 0 = R.H.S. Proved

(B) Prove that vector 3\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) and 2\(\hat { i } \) + \(\hat { j } \) – 4\(\hat { k } \) are perpendicular?
Solution:
Solve as Q.No. 9(A)

Question 10.
If \(\vec { a } \) = 4\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) p\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) are perpendicular. Find the value of p?
Solution:
\(\vec { a } \) = 4\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) p\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
\(\vec { a } \) and \(\vec { b } \) are perpendicular
\(\vec { a } \).\(\vec { b } \) = 0
∴ 4p – 2 + 3 = 0
⇒ 4p = -1
⇑ p = – \(\frac{1}{4}\)

Question 11.
(A) Find the angle between the vectors (2\(\hat { i } \) + 3\(\hat { j } \) – 4\(\hat { k } \) ) and (3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) )?
Solution:
Let \(\vec { a } \) = 2\(\hat { i } \) + 3\(\hat { j } \) – 4\(\hat { k } \), \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \)
Let θ be the angle between them

(B) Find the angle between vectors \(\vec { a } \) = 2\(\hat { i } \) – 2\(\hat { j } \) – \(\hat { k } \) and \(\vec { b } \) = 6\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \)?
Solution:
Solve as Q.No. 11(A)

(C) If \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) = 3\(\hat { i } \) – 4\(\hat { j } \) – 4\(\hat { k } \), then find their dot product and angle between them?
Solution:

If θ be the angle between them

Question 12.
If |\(\bar { |a| } \) = 10, \(\bar { |b| } \) = 2 and \(\bar { a } \). \(\bar { b } \) = 2 and \(\bar { a } \). \(\bar { b } \) = 12, then find the value of |\(\bar { a } \) × \(\bar { b } \)?
Solution:

Question 13.
If \(\vec { a } \) = \(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) – \(\hat { k } \), then find \(\vec { a } \) × \(\vec { b } \)?
Solution:

Question 14.
A force \(\vec { F } \) = 4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) is acting along the direction \(\vec { d } \) = – \(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \)? Find the work done by the force?
Solution:
\(\vec { d } \) = – \(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \), \(\vec { F } \) = 4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) (given)
∴ Work done by force
W = \(\vec { F } \). \(\vec { d } \)
= (4\(\hat { i } \) – 3\(\hat { j } \) + 2\(\hat { k } \) ). (-\(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \) )
= -4 + 9 + 10 = 15 unit.

Question 15.
If |\(\vec { a } \) + \(\vec { b } \)| = |\(\vec { a } \) – \(\vec { b } \)|, then prove that \(\vec { a } \) and \(\vec { b } \) are perpendicular?
Solution:

Since dot product is zero. So vectors \(\vec { a } \) and \(\vec { b } \) are perpendiculars. Proved.

Question 16.
If \(\vec { a } \) and \(\vec { b } \) are two vectors such that |\(\vec { a } \)| = 2, |\(\vec { b } \)| = 3 and \(\vec { a } \). \(\vec { a } \) = 3, then find angle between \(\vec { a } \) and \(\vec { b } \)?
Solution:
Solve as Q.No. 17

Question 17.
If |\(\vec { a } \)| = 4, |\(\vec { b } \)| = 4 and \(\vec { a } \). \(\vec { b } \) = 6, then find the angle between \(\vec { a } \) and \(\vec { b } \)?
Solution:

Question 18.
If \(\vec { a } \) and \(\vec { b } \) are two two vectors such that |\(\vec { a } \)| = 2, |\(\vec { b } \)| = 7 and \(\vec { a } \) ×
\(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 6\(\hat { k } \), then find the angle between \(\vec { a } \) and \(\vec { b } \)?
Solution:
Let θ be the angle between \(\vec { a } \) and \(\vec { b } \)

Question 19.
Find cosine angle between vectors 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \) and \(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \)?
Solution:

Question 20.
Find the area of the parallelogram whose two adjacent sides are represented by the vectors \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)?
Solution:


= 2(1 – 2) + 3(-1-4) + 1(1 + 2)
= -2 – 15 + 3
= -14 cubic unit

Question 21.
Prove that:
\(\hat { i } \).( \(\hat { j } \) × \(\hat { k } \) + ( \(\hat { i } \) × \(\hat { k } \)). \(\hat { j } \) = 0?
Solution:

Question 22.
If vectors \(\vec { a } \) and \(\vec { b } \) are perpendicular then prove that |\(\vec { a } \) + \(\vec { b } \)|² = |\(\vec { a } \)|² + |\(\vec { b } \)|²?
Solution:
We know that
|\(\vec { a } \) + \(\vec { b } \)|² = |\(\vec { a } \)|² + |\(\vec { b } \)|²
Vector \(\vec { a } \) and \(\vec { b } \) are perpendicular, then
\(\vec { a } \). \(\vec { b } \) = 0
⇒ |\(\vec { a } \) + \(\vec { a } \)|² + |\(\vec { a } \)|² + |\(\vec { b } \)|². Proved.

Question 23.
Prove that:
\(\vec { a } \) × ( \(\vec { b } \) + \(\vec { c } \) ) + \(\vec { b } \) × ( \(\vec { c } \) + \(\vec { a } \) ) + \(\vec { c } \) × ( \(\vec { a } \) + \(\vec { b } \) ) = \(\vec { 0 } \)?
Solution:

Question 24.
Find the work done by the force \(\vec { F } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) in the direction \(\vec { d } \) = 3\(\hat { i } \) + 2\(\hat { j } \) + 5\(\hat { k } \)?
Solution:
W = \(\vec { F } \). \(\vec { d } \)
= (2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) ). (3\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) )
= 6 – 2 + 3 = 7 unit.

Question 25.
If modulus of two vectors \(\vec { a } \) and \(\vec { a } \) are equal and angle between them is 60° and their dot product is \(\frac{9}{2}\) find their modulus? (CBSE 2018)
Solution:

Question 26.
Find the area of the parallelogram whose adjacent sides are given by vectors \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) = 3\(\hat { i } \) + 4\(\hat { j } \) – \(\hat { k } \)?
Solution:

Question 27.
If \(\vec { a } \) = 4\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – 2\(\hat { k } \) then find the value of |2\(\vec { b } \) × \(\vec { a } \)|?
Solution:

Question 28.
If \(\vec { a } \) = 4\(\hat { i } \) + 3\(\hat { j } \) + 3\(\hat { k } \) and \(\vec { b } \) = 3\(\hat { i } \) + 2\(\hat { k } \) then, find the value of |\(\vec { b } \) × 2\(\vec { a } \)|?
Solution:

Question 29.
If \(\vec { a } \) = 2\(\hat { i } \) + \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { b } \) = 5\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), then find the magnitude of vector \(\vec { b } \) in the direction of \(\vec { a } \)?
Solution:

Question 30.
If \(\vec { a } \) = \(\hat { i } \) + 3\(\hat { j } \) – 2\(\hat { k } \), \(\vec { b } \) = – \(\hat { j } \) + 3\(\hat { k } \) then find the value |\(\vec { a } \) × \(\vec { b } \)|?
Solution:

Vector Algebra Short Answer Type Questions

Question 1.
Prove that: A(-2\(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) ), B( \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) ) and C(7\(\hat { i } \) + 0\(\hat { j } \) – \(\hat { k } \) ) are coplanar? (NCERT)
Solution:
Let O be the origin then position vector of A, B and C is

Hence vector \(\vec { A } \)B and \(\vec { B } \)C are parallel but \(\vec { A } \)B and \(\vec { B } \)C has common point B. Hence points A, B and C are coplanar.

Question 2.
If position vectors of points A, B, C and D are 2\(\hat { i } \) + 4\(\hat { k } \), 5\(\hat { i } \) + 3\(\sqrt { 3 } \) \(\hat { j } \) + 4\(\hat { k } \), -2\(\sqrt { 3 } \) \(\hat { j } \) + \(\hat { k } \) then prove that:
CD||AB and CD = \(\frac{2}{3}\) \(\vec { A } \)B?
Solution:
Let O be the origin

Question 3.
If G is centroid of ∆ABC, then prove that:
\(\vec { G } \)A + \(\vec { G } \)B + \(\vec { G } \)C = \(\vec { 0 } \)?
Solution:
Let vectors of vertices A,B and C of ∆ABC are \(\vec { a } \), \(\vec { b } \) and \(\vec { c } \) respectively.
∴ Position vector of centroid G = \(\frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } \)

Question 4.
Using vectors prove that the medians of traiangle are concurrent?
Solution:
Let medium of ∆ABC are AD, BE and CF.
Let \(\vec { a } \), \(\vec { b } \) and \(\vec { c } \) be the positive vector of points A, B and C respectively.

Now position vector of a point dividing the median AD in the ratio 2 : 1 is

Position vector of a point which divides median BE in the ratio of 2 : 1 is

Position vector of a point which divides median BE in the ratio of 2 : 1 is

Hence, medians of triangle meets at point G it means concurrent whose position vector is \(\frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } \). Point G is centroid of traingle.

Question 5.
A vector \(\vec { O } \)P, makes angle 45° with OX and 60° with OY. Find the angle made by \(\vec { O } \)P with OZ?
Solution:
Let angle made by vector \(\vec { O } \)P with axes OX, OY and OZ are α, β, γ respectively. then
α = 45°,
β = 60°
∴ l = cos α = cos 45° = \(\frac { 1 }{ \sqrt { 2 } } \)
m = cos β = cos 60° = \(\frac{1}{2}\)
and n = cos γ
We know that
l² + m² + n² = 1

Question 6.
Find the vector \(\vec { a } \) which makes an angle with X – axis, F – axis and Z – axis respectively are \(\frac { \pi }{ 4 } \), \(\frac { \pi }{ 2 } \) and angle θ and its magnitude is 5\(\sqrt { 2 } \)?
Solution:
Given:
α = \(\frac { \pi }{ 4 } \),
β = \(\frac { \pi }{ 2 } \), γ = θ
∴l = cos \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \), m = cos \(\frac { \pi }{ 2 } \) = 0, n = cos θ.
We know that


Direction cosine of vector \(\frac { 1 }{ \sqrt { 2 } } \), 0 , \(\frac { 1 }{ \sqrt { 2 } } \)

Question 7.
Prove that:
( \(\vec { a } \) × \(\vec { b } \) )² = a²b² – ( \(\vec { a } \).\(\vec { b } \) )²?
Solution:
L.H.S = ( ( \(\vec { a } \) × \(\vec { b } \) )² = ( \(\vec { a } \) × \(\vec { b } \) ).( \(\vec { a } \) ×  \(\vec { b } \) )
= (ab sin θ\(\hat { n } \) ). (ab sin θ \(\hat { n } \) ) = a² b² sin²θ,
= a² b² (1 – cos²θ)
= a² b² – a² b²cos²θ
= a² b² – (ab cos θ)²
= a² b² – ( \(\vec { a } \). \(\vec { b } \) )² = R.H.S Proved.

Question 8.
If \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) then find the value of \(\vec { a } \) × ( \(\vec { b } \) × \(\vec { c } \) )?
Solution:

Question 9.
Find the volume of parallel cuboid whose vectors of three faces are denoted by: \(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \)?
Solution:

Question 10.
If \(\vec { a } \) = 2\(\hat { i } \) – 3\(\hat { j } \) + \(\hat { k } \), \(\vec { b } \) = \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) and \(\vec { c } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) then find the value of [ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ]
Solution:

Question 11.
If \(\vec { a } \) = 3\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) and \(\vec { c } \) = \(\hat { i } \) – 2\(\hat { j } \) + 2\(\hat { k } \) then, find the value of \(\vec { a } \), \(\vec { b } \), \(\vec { c } \)?
Solution:
Solve like Q.No.10.

Question 12.
If \(\vec { a } \) = \(\hat { i } \) – 2\(\hat { j } \) + 3\(\hat { k } \), \(\vec { b } \) = – \(\hat { i } \) + 3 \(\hat { j } \) – 4 \(\hat { k } \) and \(\vec { c } \) = \(\hat { i } \) – 3\(\hat { j } \) + 5\(\hat { k } \) then prove that \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are coplanar?
Solution:
If \(\vec { a } \), \(\vec { b } \), \(\vec { c } \) are coplanar then

Question 13.
Prove that 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) and 3\(\hat { i } \) – 4\(\hat { j } \) + 5k are coplanar?
Solution:
Let \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) and 3\(\hat { i } \) – 4\(\hat { j } \) + 5\(\hat { k } \)

Question 14.
(A) Find the value of λ for which the vectors λ\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \), 2\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) and 2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \) are coplanar?
Solution:
Let \(\vec { a } \) = λ\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) and 2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \) are coplanar?
Solution:
Let \(\vec { a } \) = λ\(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \), \(\vec { b } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \), \(\vec { c } \) = 2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \)
Given vector are coplanar if
[ \(\vec { a } \) \(\vec { b } \) \(\vec { c } \) ] = 0
\(\left|\begin{array}{lll}
{\lambda} & {3} & {2} \\
{2} & {2} & {3} \\
{2} & {3} & {4}
\end{array}\right|\) = 0
⇒ λ(8 – 9) -2(12 – 6+ 2 (9 – 4) = 0
⇒ -λ – 12 + 10 = 0
⇒ λ = -2.

(B) Find the value of λ for which given vectors are coplanar
\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), 2\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \), λ\(\hat { i } \) – \(\hat { j } \) + λ\(\hat { k } \)
Solution:
Solve like Q.No. 14 (A)
Answer:
λ = 1

(C) Find the value of λ for which the given vectors are coplanar 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \), \(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \) and 3\(\hat { i } \) + λ\(\hat { j } \) + 5\(\hat { k } \)?
Solution:
Solve like Q.No. 14 (A)
Answer:
λ = – \(\frac{18}{5}\)

Question 15.
If the angle between two unit vectors \(\vec { a } \) and \(\vec { b } \) is θ then prove that:
cos \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\bar { a } \) + \(\bar { b } \)| is θ then prove that:
sin \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\bar { a } \) – \(\bar { b } \)|
Solution:

Question 16.
The angle between two vectors \(\vec { a } \) and \(\vec { b } \) is θ then prove that:
sin \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\bar { a } \) – \(\bar { b } \)|
Solution:
sin \(\frac { \theta }{ 2 } \) = \(\frac{1}{2}\) |\(\bar { a } \) – \(\bar { b } \)|

Question 17.
In any traiangle prove that ABC?
(A) ac cos B – bc cos A = a² – b²?
(B) 2(bc cos A + ca cos B + ab cos C) = a² + b² + c²?
Solution:

Question 18.
In ∆ABC prove by vector method c = acosB + bcosA?
Solution:
In ∆ABC

⇒ c² = ac cos B + bc cos A
⇒ c² = c(a cos B + b cos A)
⇒ c = a cos B + b cos A. Proved.

Question 19.
In ∆ABC prove by vector method
b² = a² + c² – 2ac cos B?
Solution:
In ∆ABC we know that

Question 20.
In ∆ABC Prove the following:
(A) a² = b² + c² – 2bc cos A?
(B) c² = a² + b² – 2ab cos C?
Solution:
Solve like Q.No. 19

Question 21.
(A) Find the unit vector normal to the vector \(\vec { a } \) = 2\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \)
= 4\(\hat { i } \) + 4\(\hat { j } \) – 7\(\hat { k } \)?
Solution:

(B) Find the unit vector normal to the vectors \(\vec { a } \) = \(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) =
\(\hat { i } \) + 2\(\hat { j } \) – \(\hat { k } \)?
Solution:
Solve like Q.No. 21 (A)

(C) Find the unit vector normal to the vectors \(\vec { a } \) = 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) and \(\vec { b } \) = 2\(\hat { i } \) + 3\(\hat { j } \) – \(\hat { k } \)?
Solution:
Solve like Q.No. (A)
Answer:

Question 22.
Find the unit vector normal to the vectors \(\vec { a } \) = 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and \(\vec { b } \) = 3\(\hat { i } \) – 4\(\hat { j } \) – \(\hat { k } \)?
Solution:
Solve like Q.No. 21 (A)
Answer:

Question 23.
Find the area of parallelogram whose digonals are 3\(\hat { i } \) + \(\hat { j } \) – 2\(\hat { k } \) and \(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \)?
Solution:
ABCD is parallelogram whose diagonals are \(\vec { A } \)C = \(\vec { d } \)₁ and \(\vec { B } \)D = \(\vec { d } \)₂

Question 24.
By vector method prove that the square of the hypotenuse of a right angle triangle is equal to the sum of the square of the other two sides?
Solution:
Let OAB be a right angled triangle at O. Taking O as the origin. Let the position vector of \(\vec { a } \) and \(\vec { b } \) be a and b respectively then \(\vec { O } \)A = \(\vec { a } \) and \(\vec { O } \)B = \(\vec { b } \) and ∠BOA = 90°.
∴\(\vec { a } \). \(\vec { b } \) = 0

Question 25.
Find the moment of force 5\(\hat { i } \) + \(\hat { k } \) passing through the point 9\(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) about the point 3\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \)?
Solution:

Moment of the force \(\vec { F } \) about the point O = \(\vec { r } \) × \(\vec { F } \)

Question 26.
(A) Prove that:

Solution:

(B) Prove that:

Question 27.
Prove that:

Question 28.
Two forces are represented by vectors \(\vec { p } \) = 4\(\hat { i } \) + \(\hat { j } \) – 3\(\hat { k } \) and \(\vec { Q } \) = 3\(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) displace a particle from points (1,2,3) to point2? (5,4,1)? Find the work done by the forces?
Solution:

Question 29.
Two forces 4\(\hat { i } \) + 3\(\hat { j } \) and 3\(\hat { i } \) + 2\(\hat { j } \) are acting on a particle, Due to the forces the particle is displaced from the point \(\hat { i } \) + 2\(\hat { j } \) to the point 5\(\hat { i } \) + 4\(\hat { j } \)? Find the work done by the forces?
Solution:

Question 30.
Alorce of 6 units along the direction of vector 2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) acts on a partical? The partical is displaced from point \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) to 5\(\hat { i } \) + 3\(\hat { j } \) + 7\(\hat { k } \). Find the work done by the force?
Solution:
Unit vector parllel to vector 2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \)

Question 31.
Prove that |\(\vec { a } \) – \(\vec { b } \) \(\vec { b } \) – \(\vec { c } \) \(\vec { c } \) – \(\vec { a } \)| = 0?
Solution:
We know

MP Board Class 12 Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 14 Mathematical Reasoning

Mathematical Reasoning Important Questions

Mathematical Reasoning Objective Type Questions

(A) Choose the correct option :

Question 1.
If p and H are statements then determine of equivalance statement is :
(a) p ⇔ q
(b) p ∨ q
(c) p ∧ q
(d) None of these.
Answer:
(a) p ⇔ q

Question 2.
Examine of the following sentence is not statement:
(a) The sum of interior angles of a quadrilateral is 360°
(b) There are only three sides in a triangle
(c) You have long live
(d) The sum of three and three is 6.
Answer:
(c) You have long live

Question 3.
Examine that which of the following is correct part of a negative statement:
(a) 3 + 5 > 9
(b) Each square is a rectangle
(c) Equilateral triangle is an isosceles triangle
(d) None of these.
Answer:
(a) 3 + 5 > 9

Question 4.
Which of the following is a statement:
(a) n is a real numbers
(b) Let us go
(c) Switch of the fan
(d) 3 is a natural number.
Answer:
(d) 3 is a natural number.

Question 5.
The connective in the statement “3 + 5 > 9 or 3 + 5 < 9” is :
(a) >
(b) <
(c) or
(d) and.
Answer:
(c) or

Question 6.
Examine which of the following sentence statement:
(a) Each square is a rectangle
(b) Door closed
(c) God bless you
(d) Oh ! I passed.
Answer:
(a) Each square is a rectangle

Question 7.
The opposite statement of the statement p => q is :
(a) ~ q ⇔ p
(b) q ⇒ P
(c) ~ p ⇒ q
(d) None of these.
Answer:
(b) q ⇒ P

Question 8.
The negative of the statement “51 is not a multiple of 3” is :
(a) 51 is an odd number
(b) 51 is not an odd number
(c) 51 is a multiple of 2
(d) 51 is a multiple of 3.
Answer:
(d) 51 is a multiple of 3.

Question 9.
The contrapositive of the statement “ If p, then q ” is :
(a) If q, then ~ P
(b) If ~ p, then ~ q
(c) If p, then ~ q
(d) If ~ q, then ~ p
Answer:
(d) If ~ q, then ~ p

(B) Write answer in one word / sentence :

1. Write the component statements of the compound statement:
“13 is an odd number and a prime number.”
Answer:
p : Number 13 is prime; q: Number 13 is odd.

2. Write the negation of the statement:
“Everyone who lives in India is an Indian.”
Answer:
Every one who live in India is an Indian

3. Identify the connective in the following compound statement:
“It is raining or the Sun is shining.”
Answer:
or.

4. From the biconditional statement p ⇔ q, where :
p ≡ A triangle is an equilateral.
q ≡ All three sides of a triangle are equal.
Answer:
A triangle is an equilateral triangle if and only if all three sides of the triangle are equal.

5. If p ≡ Mathematics is hard, q ≡ 4 is even number, then write the formula p ∨ q in logical sentences.
Answer:
Mathematics is hard or 4 is even number.

6. If p ≡ the question paper is hard, q ≡ I will fail in the examination, then write the statement in symbolically, “The question paper is hard if and only if I will fail in the examination.”
Answer:
p ⇔ q.

7. State whether the statement is exclusive or inclusive :
“All integers are positive or negative.”
Answer:
Exclusive.

Mathematical Reasoning Very Short Answer Type Questions

Question 1.
Which of the following sentences are statements? Give reasons for your answer: (NCERT)

1. There are 35 days in a month.
2. Mathematics is difficult.
3. The sum of 5 and 7 is greater than 10.
4. The square of a number is an even number.
5. The sides of a quadrilateral have equal lengths.
6. Answer this question.
7. The product of (- 1) and 8 is 8.
8. The sum of all interior angles of a triangle is 180°.
9. Today is a windy day.
10. All real numbers are complex numbers.

Answer:

1. A month has 30 or 31 days. It is false to say that a month has 35 days, hence it is a statement.
2. Mathematics may be difficult for one but may be easy for the others. Hence it is not a statement.
3. It is true that sum of 5 and 7 is greater than 10. Hence it is a statement.
4. The square of a number may be even or it may be odd. Squaring of an odd number is always odd and square of even number is always even. Hence it is not a statement.
5. A quadrilateral may have equal lengths as it may be a rhombus or a square or the quadrilateral may have unequal sides
6. like parallelogram, hence it is not a statement.
7. It is an order, hence it is not a statement.
8. It is false because the product of (- 1) and 8 is – 8, hence it is a statement. It is true because sum of three angles of a triangles is 180°. Hence it is a statement.
9. It is a windy day. It is not clear that about which day it is said. Thus it can’t be concluded whether it is true or false. Hence it is not a statement.
10. It is true that all real numbers are complex numbers. All real numbers can be expressed as a + ib, hence it is a statement.

Question 2.
State whether the ‘or’ used in the following statements is exclusive or inclusive. Give reasons for your answer: (NCERT)

1. Sun rises or Moon sets.
2. To apply for driving licence, you should have a ration card or a passport.
3. All integers are positive or negative.

Answer:

1. When Sun rises, the Moon sets. One of happening will take place, hence ‘or’ is exclusive.
2. To apply for a driving licence either a ration card or a passport or both can be used, hence ‘or’ is inclusive.
3. All integers are positive or negative. An integers cannot be both positive or negative, hence ‘or’ is exclusive.

Question 3.
Rewrite the following statement ‘if then’ in five different ways conveying the same meaning:
If a natural number is odd than its square is odd. (NCERT)
Answer:

1. A natural number is odd implies that its square is odd.
2. A natural number is odd only if its square is odd.
3. If the square of a natural number is not odd, then the natural number is also not odd.
4. For a natural number to be odd, its necessary that its square is odd.
5. For a square of a natural number to be odd, if it is sufficient that the number is odd.

Question 4.
Give statement in (a) and (b) identify the statements given below as contra – positive or converse of each other: (NCERT)
(a) If you live in Delhi, then you have winter clothes.
(i) If you do not have winter clothes, then you do not live in Delhi.
(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is parallelogram, then its diagonal bisect each other.
(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a parallelogram.
(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram
Answer:
(a) (i) Contrapositive statement, (ii) Converse statement.
(b) (i) Contrapositive statement, (ii) Converse statement.

Question 5.
Which of the following sentence is a statement:

1. New Delhi is a capital of India.
2. How are you?

Answer:

1. Its a statement.
2. Its not a statement.

Question 6.
Find the component statement of the following statements :
1. The sky is blue and grass is green.
2. All the rational numbers are real number and all the real numbers are complex number.
Solution:
1. The components are:
p : The sky is blue.
q : The grass is green.

2. The component statement are as follows:
p : All the rational numbers are real number.
q : All the real numbers are complex number.

Question 7.
If p = It is 7 o’clock
q = The train is late,
then write the following symbols in the form of statement:

1. q ∨ ~ p
2. p ∧ q
3. ~ (p ∧ q)
4. p ∧ ~ q

Answer:

1. Either the train is late or it is not 7 o’clock.
2. It is 7 o’clock, but train is late.
3. It is not true that it is 7 o’clock and the train is late.
4. It is 7 o’clock, but train is not late.

Question 8.
If p ≡ He is intelligent and q ≡ He is strong, then write the following statement symbolically with the help of logical connectives:

1. He is intelligent and strong.
2. He is intelligent but not strong.
3. He is neither intelligent nor strong.

Answer:

1. p ∧ q
2. p ∧ ~ q
3. ~ q ∧ ~ q

Mathematical Reasoning Short Answer Type Questions

Question 1.
Show that the statement p : If x is a real number such that x3 + 4x = 0, then x is ‘0’ is true by :

1. Direct method
2. Method of contradiction
3. Method of contrapositive

Solution:
1. Direct method :
Given :
x³ + 4x = 0 ⇒ x(x² + 4) = 0
x = 0 or x³ + 4 = 0
x is real number.
∴ x² + 4 ≠ 0
∴ x = 0.

2. Method of contradiction :
Let x ≠ 0
Let x = p, (where p is a real number)
p is one the root of equation x³ + 4x = 0.
∴ p³ + 4p = 0
⇒ P(P² + 4) = 0
⇒ P = o
and p² + 4 ≠ 0, (It is a real number)
∴ p = 0.

3. Method of contrapositive :
Let x = 0 is not true and x = p ≠ 0
∴p being the root of x³ + 4x = 0
∴ p³ + 4 p = 0
⇒ P(p² + 4) = 0
p = 0 and p² + 4 = 0
⇒ P(p² + 4) ≠ 0, if p is not true
∴ x = 0 is one root of equation x³ + 4x = 0.

Question 2.
Show that the statement ‘for any real number a and b, a² = b² implies that a = b’ is not true by giving a counter example.
Solution:
Let a = 3 and b = -3 then a, b are real numbers.
Here a² = b² but a b
Hence a, b ∈ R and a² = b²
⇒ a = b, statement is not true.

Question 3.
Find the components of the following compound statement and check whether they are true or false.
(i) The number 3 is prime or odd.
(ii) All the integers are positive or negative.
(iii) Number 100 is divisible by the numbers 3, 11 and 5.
Answer:
(i) p : Number 3 is prime.
q : Number 3 is odd.
Here p and q both are true.

(ii) p : All the integers are positive. q : All the integers are negative.
Clearly p and q both are false.

(iii) p : The number 100 is divisible by 3.
q : The number 100 is divisible by 11.
r : The number 100 is divisible by 5.
p is false, q is false and r is true. p, q and r are false statement.

Question 4.
Show that the following statement is true by the method of contrapositive :
p : If x is an integer and x² is even, then x is also even.
Solution:
P – If x is an integer and x² is even, then x is also even.
Let q : x is an integer and x2 is even.
r : x is even.
To prove that P is true by contrapositive method, we assume that r is false and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that
x is not an integer or x² is not even.
x is not even implies that x² is also not even.
Therefore, statement q is false.
Thus, the given statement p is true.

Question 5.
For each of the following compound statements first identify the connecting words and then break it into compound statements: (NCERT)
(i) All rational numbers are real and all real numbers are not complex.
(ii) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.
Solution:
(i) The connecting word in the compound statement is ‘and’.
p : All the rational numbers are real number.
q: AH the real numbers are not complex number.

(ii) The connecting word in the compound statement is ‘and’.
p : x = 2 is the root of equation 3x² – x – 10 = 0.
q : x = 3 is the root of equation 3x² – x – 10 = 0.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 10 Straight Lines

Straight Lines Long Answer Type Questions

Question 1.
The slope of a line is double the slope of another line. If tangent of the angle between them is \(\frac {1}{3}\), then find the slope of the line.
Solution:
Let the angle between two lines be θ.

⇒ 1 + 2m² = 3m
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1 (m – 1) = 0
⇒ (m – 1)(2m – 1) = 0
⇒ m = 1,\(\frac {1}{2}\)

Question 2.
If three points (h, 0), (a, b) and (0, k) lies on same line, then prove that \(\frac {a}{h}\) + \(\frac {b}{k}\) = 1. (NCERT)
Solution:
Given points are A(h, 0), B(a, b) and C(0, k).

ab = (b – k)(a – h)
⇒ ab = ab – bh – ak + hk
⇒ bh + ak = hk
⇒ \(\frac {bh}{hk}\) \(\frac {ak}{hk}\) \(\frac {hk}{hk}\), (dividing both sides by hk)
⇒ \(\frac {a}{h}\) + \(\frac {b}{k}\) = 1.

Question 3.
Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6). (NCERT)
Solution:
Gradient of AB = m = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
Where x₁ = 2 , y₁ = 5, x₂ = – 3, y₂ = 6
∴ m = \(\frac {6 – 5}{- 3 – 2}\) = \(\frac {1}{- 5}\)

Let the gradient of CD = m₁,
AB ⊥ CD
m x m₁ = – 1
⇒ \(\frac {-1}{5}\) x m₁ = – 1
⇒ m₁ = 5
Equation of line CD will be :
y – y₁ = m(x – x₁)
Here x₁ = – 3, y₁ = 5, m = 5 .
⇒ y – 5 = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0.

Question 4.
Find the equation of straight line passing through the point (2, 2) and cutting off intercepts the axes whose sum is 9. (NCERT)
Solution:
Let the required equation of line is :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 …. (1)
Given: a + b = 9 …. (2)
Line (1) passes through point (2, 2).
∴ \(\frac {2}{a}\) + \(\frac {2}{b}\) = 1
\(\frac {2a + 2b}{ab}\) = 1
2a + 2b = ab
Put b = 9 – a from equation (2), we get
2a + 2(9 – a) = a(9 – a)
⇒ 2a + 18 – 2a = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a (a – 6) – 3(a – 6) = 0
⇒ (a – 3)(a – 6) = 0
∴ a = 3 or a = 6
When a = 3, then b = 9 – a = 9 – 3 = 6
Hence the required equation is :
\(\frac {x}{3}\) + \(\frac {y}{6}\) = 1
⇒ 2x + y = 6.
When a = 6, then b = 9 – 6 = 3
Hence the required equation is :
\(\frac {x}{6}\) + \(\frac {y}{3}\) = 1
⇒ x + 2 y = 6.

Question 5.
Find the equation to that straight line which passes through the point (7,9) and such that the portion between the axes is divided by the point in the ratio 3 : 1.
Solution:
Let the required equation of the line is :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 …. (1)
Let the required line cuts the axes at points A (a, 0) and B(0, b),
and Point P intercept the line AB internally in ratio 3 : 1.

Putting values of a and b in eqn. (1), we get
\(\frac {x}{28}\) + \(\frac {y}{12}\) = 1
\(\frac {3x + 7y}{84}\) = 1
Hence required equation is 3x + 7y = 84.

Question 6.
Find the equation of the line which passes through the point (1, 2) and makes a right angled triangle with axes of area \(\frac {9}{2}\) sq units.
Solution:
Area of (∆AOB) = \(\frac {1}{2}\)a.b
⇒ \(\frac {9}{2}\) = \(\frac {1}{2}\)a.b
⇒ ab = 9

Let the equation of line \(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 which passes through point (1, 2)
∴ \(\frac {1}{a}\) + \(\frac {2}{b}\) = 1
From equation (1), b = \(\frac {9}{a}\)
∴ \(\frac {1}{a}\) + \(\frac {2}{9}\) = 1
⇒ 9 + 2a² = 9a
⇒ 2a² – 9a + 9 = 0
⇒ 2a² – 3a – 6a + 9 = 0
⇒ a(2a – 3) – 3(2.a – 3) = 0
⇒ (2a – 3)(a – 3) = 0
∴ b = 6, 3
Equation of line \(\frac {x}{3/2}\) + \(\frac {y}{6}\) = 1
⇒ \(\frac {2x}{3}\) + \(\frac {y}{6}\) = 1
⇒ 12x + 3y = 18
⇒ 4x + y = 6
or \(\frac {x}{3}\) + \(\frac {y}{3}\) = 1
⇒ x + y = 3

Question 7.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then prove that \(\frac { 1 }{ { p }^{ 2 } }\) = \(\frac { 1 }{ { a }^{ 2 } }\) + \(\frac { 1 }{ { b }^{ 2 } }\) (NCERT)
Solution:
Equation of given line be :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1
⇒ bx + ay = ab
⇒ bx + ay – ab = 0 …. (1)
Let the form of required equation be :
xcosα + ysinβ – p = 0 …. (2)
Comparing equation (1) and (2),

Question 8.
Find the angle between the lines.
y = (2 – \(\sqrt {3}\))x + 6 and y = (2 + \(\sqrt {3}\))x – 8.
Solution:
y = (2 – \(\sqrt {3}\))x + 6 … (1)
m₁ = 2 – \(\sqrt {3}\)
Second line y = (2 + \(\sqrt {3}\))x – 8
m₂ = 2 + \(\sqrt {3}\)
Let θ be the angle between them,

Question 9.
If the perpendicular drawn from origin on line y = mx + c intersect the line at point (- 1, 2), then find the value of m and c. (NCERT)
Solution:
Let the equation of AB be :
y = mx + c … (1)
Gradient of eqn. (1) is m and OP is perpendicular to AB

Gradient of OP m₁ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
m₁ = \(\frac {2 – 0}{- 1 – 0}\) = – 2
∵ Op ⊥ AB
∴ m × m₁ = – 1
⇒ m(- 2) = – 1
⇒ m = \(\frac {1}{2}\)
put value of m in equation (1),
y = \(\frac {1}{2}\)x + c
Above equation passes through point (-1, 2), hence it will satisfy it.
2 = \(\frac {- 1}{2}\) + c
⇒ c = 2 + \(\frac {1}{2}\) = \(\frac {5}{2}\)
∴ m = \(\frac {1}{2}\), c = \(\frac {5}{2}\)

Question 10.
Find the coordinate of the foot of perpendicular from the point (- 1, 3) to the line 3x – 4y – 16 = 0.
Solution:
Equation of line CD is :
3x – 4y – 16 = 0  … (1)
Equation of line AB is perpendicular to equation (1),
4x + 3y + c = 0  … (2)

Equation (2) passes through point (- 1, 3).
∴ 4(- 1) + 3 x 3 + c = 0
– 4 + 9 + c = 0
c = – 5
Put c = – 5 in equation (2), we get 4x + 3y – 5 = 0
Now the intersection of lines (1) and (3) is the foot of perpendicular B
3x – 4y – 16 = 0
4x + 3y – 5 = 0
By cross multiplication method,

Question 11.
Find the equation of line parallel to y – axis and drawn through the point of intersection of line x – 7y + 5 = 0 and 3x + y = 0. (NCERT)
Solution:
Equation of given lines are :
x – 7y + 5 = 0 … (1)
3x + y = 0 … (2)
Equation of straight line passing through the intersection of equation (1) and (2) is :
x – 7y + 5 + λ(3x + y) = 0 … (3)

Put A = 7 in equation (3), we get
x – 7y + 5 + 7(3x + y) = 0
⇒ x – 7y + 5 + 21x + 7y = 0
⇒ 22x + 5 = 0.

Question 12.
Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point. (NCERT)
Solution:
Equation of given lines are :
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2x – y – 3 = 0 … (3)
From eqns. (1) and (3),
3x + y – 2 = 0
2x – y – 3 = 0
Solving by cross multiplication method,

Lines intersect at the one point, hence point (1, – 1) will satisfy equation (2). p(1) – 2 x 1 – 3 = 0
p = 5.

Question 13.
Find the equation of a line drawn perpendicular to the line \(\frac {x}{4}\) + \(\frac {y}{6}\) = 1 through the point where it meets the y – axis. (NCERT)
Solution:
Given equation is :
\(\frac {x}{4}\) + \(\frac {y}{6}\) = 1
Line (1) meets the F – axis at point (0, 6).
From eqn. (1), \(\frac {6x + 4y}{24}\) = 1
⇒ 6x + 4y = 24
⇒ 6x + 4y – 24 = 0 … (2)
Equation of line perpendicular to eqn. (2),
4x – 6y + c = 0 … (3)
Above equation passes through point (0, 6).
∴ 4 x 0 – 6 x 6 + c = 0
c = 36
Put value of c = 36 in equation (3),
4x – 6y + 36 = 0
2x – 3y + 18 = 0.

Question 14.
For what value of k the equation (k – 3) x (4 – k²)y + k² – 7k + 6 = 0 is :

1. Parallel to X – axis
2. Parallel to y – axis
3. Passes through origin.

Solution:
Equation of given line is :
(k – 3)x – (4 – k²)y + k² – 7k+6 = 0
(4 – k²)y = (k – 3)x + k² – 7k + 6
⇒ y = \(\frac { (k – 3)x }{ 4 – { k }^{ 2 } }\) + \(\frac { { k }^{ 2 } – 7k + 6 }{ 4 – { k }^{ 2 } }\)

1. Line parallel to X – axis :
∴ m = 0
⇒ \(\frac { k – 3 }{ 4 – { k }^{ 2 } }\)
⇒ k – 3 = 0
⇒ k = 3.

2. Line parallel to Y – axis :
∴ m = \(\frac {1}{0}\)
⇒ \(\frac { k – 3 }{ 4 – { k }^{ 2 } }\)
⇒ 4 – k² = 0
⇒ k²= 4
⇒ k = ± 2

3. Line passing through origin, if c = 0.
∴ \(\frac { { k }^{ 2 } – 7k + 6 }{ 4 – { k }^{ 2 } }\)
⇒ k² – 7k+6 = 0
⇒ k² – 6k – k+6 = 0
⇒ k(k – 6) – 1(k – 6) = 0
⇒ (k – 6)(k – 6) = 0
∴ k = 1, 6

Question 15.
The line passing through the points (h, 3) and (4,1) meets the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Equation of given line is :
7x – 9y – 19 = 0
⇒ 9y = 7x – 19
⇒ y = \(\frac {7}{9}\)x – \(\frac {19}{9}\) … (1)
Gradient of eqn. (1) m₁ = \(\frac {7}{9}\)
Equation of line passing through points (h, 3) and (4, 1),
m₂ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
m₂ = \(\frac {1 – 3}{4 – h}\) = \(\frac {- 2}{4 – h}\)
∵ Lines are perpendicular.
∴m₁ x m₂= – 1
\(\frac {7(- 2)}{9(4 – h)}\) = – 1
⇒ – 14 = – 36 + 9h
⇒ 9h = 36 – 14
⇒ 9h = 22
∴ h =\(\frac {22}{9}\).

Question 16.
If the lengths of the perpendiculars drawn from origin on the straight lines xcosθ – y sinθ = kcos2θ and xsecθ + y cosec θ = k are p and q then prove that:
p² + 4q² = k²
Solution:
Equation of given lines :
xsecθ + ycosecθ = k … (1)
and x cosθ – y sinθ = k cos2θ … (2)
Length of perpendicular from origin on equation (1),

Adding equation (3) and (4),
4q² + p² = k² sin² 2θ +k² cos² 2θ
= k²(sin² 20 + cos²2θ)
⇒ p² + 4q² = k² .

Question 17.
Find the equation to the straight line passing through the intersection of lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 and makes an equal intercept on both the axes.
Solution:
Equation of given lines are :
4x + 7y – 3 = 0 … (1)
2x – 3y + 1 = 0 … (2)
Equation of straight line passing through intersection of lines (1) and (2),
4x + 7y – 3 + λ(2x – 3y + 1) = 0 … (3)
⇒ x(4 + 2 λ) + y(7 – 3 λ) – 3 + λ = 0
⇒ x(4 + 2 λ) + y(7 – 3 λ) = 3 – λ

Put λ = \(\frac {3}{5}\) in equation (3),
4x + 7y – 3 + \(\frac {3}{5}\)(2x – 3y + 1) = 0
⇒ 20x + 35y – 15 + 6x – 9y +3 = 0
⇒ 26x + 26y = 12
⇒ 13x + 13y = 6.
⇒ 13x + 13y – 6 = 0.

Question 18.
Find the distance of point (-1, 1) from the line 12 (x + 6) = 5 (y – 2). (NCERT)
Solution:
Equation of given line is :
12(x + 6) = 5(y – 2)
⇒ 12x + 72 = 5y – 10
⇒ 12x – 5y + 82 = 0
Length of perpendicular

Question 19.
Find the points on the X – axis, whose distance from the line \(\frac {x}{3}\) + \(\frac {y}{4}\) = 1 are 4 units. (NCERT)
Solution:
Equation of given line is :
\(\frac {x}{3}\) + \(\frac {y}{4}\) = 1
⇒ 4x + 3y = 12
⇒ 4x + 3y – 12 = 0
Let the point on X – axis is (h, 0).
Length of perpendicular drawn from point (h, 0) is 4.

Taking (+) sign,
4h = 20 + 12 = 32
⇒ h = 8
Taking (-) sign,
4h = – 20 + 12
⇒ 4h = – 8
⇒ h = – 2
Point on X – axis are ( – 2, 0) and (8, 0).

Question 20.
Find the distance between two parallel lines : 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Solution:
Equation of given lines are :
15x + 8y – 34 = 0
15x + 8y + 31 = 0
Put x = 0 in eqn. (1),
15(0) + 8y – 34 = 0
⇒ 8y = 34
y = \(\frac {34}{8}\) = \(\frac {17}{4}\)
Point (0, \(\frac {17}{4}\)) is on line (1),
Length of perpendicular from point (0, \(\frac {17}{4}\)) on equation (2),

= \(\frac {65}{17}\) units
Hence the distance between lines is \(\frac {65}{17}\) units

Question 21.
The length L (in centimeters) of a copper rod is a linear function of its Celsius temperature C. In an experiment. If L = 124.942 when C= 20 and L = 125.134 when C = 110. Express L in terms of C. (NCERT)
Solution:
Given : L = 124.942 when C = 20
L = 125.134 when C= 110
In coordinate form points (20, 124.942) and (110, 125.134) are two points.
Required equation of straight line :

Question 22.
The owner of a milk store finds that he can sell 980 L of milk each week at Rs. 14/L and 1220 L of milk each week at Rs. 16/L. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17 L? (NCERT)
Solution:
Let price and litre be denoted in ordered pair (x, y), two points are (14, 980) and (16,1220).
∴ Required equation of straight line will be :
y – y₁ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)(x – x₁)
⇒ y – 980 = \(\frac {1220 – 980}{16 – 14}\)(x – 14)
⇒ y – 980 = \(\frac {240}{2}\)(x – 14)
⇒ y – 980 = 120 (x – 14)
⇒ y – 980 = 120 x – 120 x 14
⇒ y = 120 x – 1680 + 980
⇒ y = 120 x – 700
When x = 17, then
y = 120 x 17 – 700
⇒ y = 2040 – 700
⇒ y = 1340
Hence, he will sell weekly 1340 L milk at the rate of Rs. 17 L.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 2 Relations and Functions

Relations and Functions Important Questions

Relations and Functions Objective Type Questions

(A) Choose the correct option :

Question 1.
If A = {2, 4, 5}, B = {7, 8, 9}, then n(A × B) =
(a) 6
(b) 9
(c) 3
(d) 0.
Answer:
(b) 9

Question 2.
If A = { 1, 2, 3, 4, 5} and 5 = {2, 3, 6, 7}, then the number of element in (A × B)∩(B × A) is:
(a) 4
(b) 5
(c) 10
(d) 20.
Answer:
(a) 4

Question 3.
If A and B are two non – empty sets, then:
(a) A × B = {(a, b) : a ∈ B, b ∈ A}
(b) A × B = {(a, b) : a ∈ A, b ∈ B}
(c) {(a, b) : (a, b) ∈ A, (a, b) ∈ B}
(d) None of these.
Answer:
(b) A × B = {(a, b) : a ∈ A, b ∈ B}

Question 4.
If f(x) = log \(\frac { 1+x }{ 1-x } \), then find f [ \(\frac { { 2x } }{ 1+{ x }^{ 2 } }\) ] =
(a) [f(x)]²
(b) [f(x)]³
(c) 2 f(x)
(d) 3 f(x).
Answer:
(c) 2 f(x)

Question 5.
Let A = {1,2} and B = {3,4}, then the number of relation from A to B will be:
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(d) 16

Question 6.
The range of the function f(x) = \(\sqrt { x-1 } \) is:
(a) [1, ∞)
(b) [0, ∞)
(c) (0, ∞)
(d) (1, ∞)
Answer:
(b) [0, ∞)

Question 7.
If f(x) = \(\frac { x^{ 2 }-1 }{ x^{ 2 }+1 } \), then f ( \(\frac{1}{x}\) ) is:
(a) f(x)
(b) – f(x)
(c) f(-x)
(d) \(\frac { 1 }{ f(x) } \)
Answer:
(b) – f(x)

Question 8.
Domain of the function f(x) = \(\frac { 1 }{ \sqrt { 2x-3 } } \) is:
(a) R – { \(\frac{3}{2}\) }
(b) ( \(\frac{3}{2}\), ∞)
(c) [ \(\frac{3}{2}\), ∞)
(d) None of these.
Answer:
(b) ( \(\frac{3}{2}\), ∞)

(B) Match the following :

Answer:

1. 1. (d)
   2. (a)
   3. (c)

1. (b)
2. (c)

(C) Fill in the blanks:

1. If A = {1, 2} and B = {3, 4, 5}, then the number of subsets of A × B is ………………………….
2. Range of the function f = {(2, 1), (3, 1), (4, 1), (5, 1)} is …………………………..
3. Range of the function f(x) = 11 – 7 sinx is …………………………..
4. If f(x) = x² and g(x) = x + 1, ∀ x ∈ R , then (f + g)x is …………………………
5. If f(x) = 1 – cosx, then the value of f ( \(\frac { \pi }{ 4 } \) ) …………………………….
6. Domain of the function f(x) = \(\frac { 1 }{ \sqrt { (1 – x)(x – 2) } } \) is ……………………………
7. If relation R = {(1, 3), (3, 3), (4, 5)} then the value of R^-1 is ……………………………

Answer:

1. 64
2. {1}
3. [4, 18]
4. x² + x + 1
5. 1 – \(\frac { 1 }{ \sqrt { 2 } } \)
6. (1, 2)
7. {(3, 1), (3, 3), (5, 4)}.

(D) Write true/false :

1. If A, B, C are three sets, then the value of A × (B∪C) is (A∪B) × (A∪C).
2. If A = {x : x² – 5x + 6 = o}, B = {2,4}, C = {4,5}, then A × (B∩C) = {(2, 4), (3, 4)}.
3. The relation R = {(2, 1), (3, 2), (4, 3), (5, 4)} is a function.
4. If a relation on Z is R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4}, then domain of R is {0, ± 1, ± 2}.
5. Domain of the function f(x) = \(\sqrt { a^{ 2 }-x^{ 2 } } \), a > 0 is [0, a].

Answer:

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

(E) Write answer in one word/sentence:

1. If f(x) = x² and g(x) = x + 3, x ∈ R, then the value of (fog)(2).
2. Range of function f(x) = sin x.
3. If mapping f : R → R is defmd by f(x) = x² + 1 the value of f^-1 (26) is:
4. If A = {1, 2, 3} and B = {5, 7}, then the value of A×B is:
5. Domain of the function f(x) = \(\sqrt { 3-2x } \) is:
6. If function f(x) = \(\frac { { x }^{ 2 } }{ 1-{ x }^{ 2 } }\), then the value of f(sin θ) is :

Answer:

1. 25
2. [-1, 1]
3. {-5, 5}
4. {(1, 5), (1, 7), (2, 5), (2, 7), (3, 5), (3, 7)}
5. (-∞, \(\frac { 3 }{ 2 }\))
6. tan² θ

Straight Lines Very Short Answer Type Questions

Question 1.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in ( A x B). (NCERT)
Solution:
Given : n(A) = 3, B = {3, 4, 5}, n (B) = 3
∴ n (A x B) = n (A) x n (B) = 3 x 3
⇒ n (A x B) = 9.

Question 2.
If G = {7, 8} and H = {5, 4, 2}, then find G x H. (NCERT)
Solution:
G x H = { 7, 8} x {5, 4, 2}
G x H= { (7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}.

Question 3.
If A x B = { (a, x), (a, y), (b, x), (b, y)} then find A and 5. (NCERT)
Solution:
Given:
A x B = {(a, x), (a, y), (6, x), (6, y)}
A = {a, b} and S = {x, y}.

Question 4.
If A = {1, 2} and B = {3, 4} then find A x B. (NCERT)
Solution:
A x B = { 1, 2} x {3, 4}
A x B = { ( 1, 3), (1, 4), (2, 3), (2, 4)}

Question 5.
The figure shows the relationship between the set P and Q. Write the relation : (NCERT)

1. In set builder form
2. In roster form
3. Find its domain
4. Find its range.

Solution:

1. Set builder form, R = {(x, y) : y = x – 2, x ∈ P and y ∈ Q}.
2. Roster form, R = {(5, 3), (6, 4), (7, 5)}
3. Domain = {5, 6, 7}
4. Range = {3, 4, 5}.

Question 6.
Let A = (1, 2, 3, 4, 6} and R be the relation on A defined by
{(a, b) : a, b ∈ A, b is exactly divisible by a } :

1. Write R in roster form,
2. Find the domain of R
3. Find the range of R. (NCERT)

Solution:
Given: A = {1, 2, 3, 4, 6}
1. R = {{a, b): a, be A, b is exactly divisible by a }
R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}.

2. Domain = {1, 2, 3, 4, 6}.

3. Range = {1, 2, 3, 4, 6}.

Question 7.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B. (NCERT)
Solution: A = {x, y, z}, B= {1, 2}
n(A) = 3, n(B) = 2
No. of relations from A to B = 2^mn = 23^3/2 = 2⁶
= 64.

Question 8.
A function f is defined by f (x) = 2x – 5, write down the following values : (NCERT)

1. f(0)
2. f(7)
3. f(-3).

Solution:
Given: f(x) = 2x – 5
1. Put x = 0,
f(0) = 2(0) – 5 = 0 – 5 = – 5.

2. Put x = 7,
f(7) = 2 x 7 – 5 = 14 – 5 = 9.

3. Put x = – 3,
f(- 3) = 2(- 3) – 5 = – 6 – 5 = – 11.

Question 9.
The function ‘t’ which maps temperature in degree celcius into temperature in degree fehrenheit, it is defined as t(c) = \(\frac { 9c }{ 5 }\) +32. Find the following : (NCERT)

1. t (0)
2. t (28)
3. t (- 10)
4. Find c, when t (c) = 212.

Solution:
Given :
t(c) = \(\frac { 9c }{ 5 }\) + 32
1. Put c = 0,
t(0) = \(\frac { 9 × 0 }{ 5 }\) + 32
⇒ t(0) = 32

2. Put c = 28,
t(28) = \(\frac { 9 × 28 }{ 5 }\) + 32 = \(\frac { 252}{ 5 }\) + \(\frac { 32}{ 1 }\)
⇒ t(28) = \(\frac { 252 + 160}{ 5 }\) = \(\frac { 412}{ 5 }\)

3. Put c = – 10,
t(- 10) = \(\frac { 9 x (- 10)}{ 5 }\) + 32
⇒ t(- 10) = – 18 + 32 = 14

4. Put t(c) = 212,
212 = \(\frac { 9c }{ 5 }\) + 32
⇒ \(\frac { 9c }{ 5 }\) = 212 – 32
⇒ \(\frac { 9c }{ 5 }\) = 180
⇒ c = \(\frac { 180 x 5 }{ 9 }\)

Question 10.
Find the domain and range of the function f(x) = |x|.
Solution:
f(x) = – |x|, f(x) < 0
Domain of f = R.
Range of f = {y : y ∈ R, y ≤ 0} = (- ∞, 0].

Straight Lines Short Answer Type Questions

Question 1.
Find the domain and range of the function f(x) = \(\sqrt { 9 – { x }^{ 2 } }\).
Solution:
Given : f(x) = \(\sqrt { 9 – { x }^{ 2 } }\)
Value of f(x) is real, if f(x) ≥ 0
9 – x² ≥ 0
⇒ – (x² – 9) ≥ 0
⇒ x² – 9 ≤ 0
⇒ (x + 3)(x – 3) ≤ 0
∴ Domain = [- 3, 3].

Question 2.
If f(x) = x², then find \(\frac { f(1.1) – f(1)}{ (1.1 – 1)}\)
Solution:
f(x) = x²
f(1.1) = (1.1)² = 1.21
f(1) = 1² = 1

Question 3.
Find domain of function f(x) = \(\frac { { x }^{ 2 } + 2x + 1 }{ { x }^{ 2 } – 8x + 12 }\)
Solution:
Given function is :

f(x) will be defined if,
x² – 8x + 12 ≠ 0
⇒ x² – 6x – 2x + 12 ≠ 0
⇒ x(x – 6) – 2(x – 6) ≠ 0
⇒ (x – 2)(x – 6) ≠ 0
x ≠ 2 and x ≠ 6
Domain of function = R – {2, 6}.

Question 4.
Let f : g → R → R be defined by f(x) = x + 1 and g(x) = 2x – 3 respectively, then find :

1. f + g
2. \(\frac { f }{ g }\)

Solution:
Given: f(x) = x + 1, g (x) = 2x – 3
1. (f + g)x = f(x) + g(x)
= x + 1 + 2x – 3
∴ (f + g)x = 3x – 2

2. \(\frac { f }{ g }\)(x) = \(\frac { f(x)}{ g(x) }\) = \(\frac { x +1 }{ 2x – 3}\)

Question 5.
If f(x) = x² – \(\frac { 1 }{ { x }^{ 2 } }\), then prove that:
f(x) + f\(\frac { 1 }{ x }\) = 0
Solution:

Question 6.
If f(x) = \(\frac { { x }^{ 2 } }{ 1-{ x }^{ 2 } }\), then prove that:
f(sinθ) = tan² θ.
Solution:

Question 7.
If f(x) = x³ + 3x + tanx, then prove that f(x) is an odd function.
Solution:
Given : f(x) = x³ + 3x + tanx
f(- x) = (- x)³ + 3(- x) + tan(- x)
= – x³ – 3x – tanx
= – (x³ + 3x + tanx)
= – f(x).
Hence, f(x) is an odd function.

Question 8.
If f(x) = x² + 2xsinx + 3, then prove that f(x) is an even function.
Solution:
Given: f(x) = x² + 2x sin x + 3
f(- x) = (- x)² + 2(- x) sin(- x) + 3
= x² + 2xsinx + 3, [∵ sin(- x) = – sinx]
= f(x).
Hence, f(x) is an even function

Question 9.
If f(x) = x², g(x) = x + 2, ∀ x ∈R, then find gof and fog. Is gof = fog.
Solution:
Given : f(x) = x², g(x) = x + 2
fog(x) = f[g(x)]
= f(x + 2) = (x+2)².
gof(x) = g[f(x)]
= g(x²) = x² + 2
Hence, fog(x) ≠ gof(x).

Question 10.
If f(x) = e^2x and g(x) = log \(\sqrt {x}\), x> 0, then find the value of fog(x) and gof(x).
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 11th Hindi Swati Solutions गद्य Chapter 11 जीने की कला

जीने की कला अभ्यास

जीने की कला अति लघु उत्तरीय प्रश्न

प्रश्न 1.
जीवमात्र किसका अवतार है?
उत्तर:
जीवमात्र ईश्वर का अवतार है, लेकिन लौकिक भाषा में सबको अवतार नहीं माना जाता है।

प्रश्न 2.
ईश्वर रूप हुए बिना मनुष्य को क्या नहीं मिलता? (2016)
उत्तर:
ईश्वर रूप हुए बिना मनुष्य को सुख नहीं मिलता एवं शान्ति का अनुभव नहीं होता।

प्रश्न 3.
गाँधीजी के अनुसार गीता में किस युद्ध का वर्णन किया गया है? (2017)
उत्तर:
गाँधीजी के अनुसार गीता में महाभारत के युद्ध का वर्णन किया गया है।

प्रश्न 4.
हृदय में भीतर के युद्ध को रसप्रद बनाने के लिए की गई कल्पना क्या है?
उत्तर:
हृदय में भीतर के युद्ध को रसप्रद बनाने के लिए हृदय-मन्थन, अर्थात् ज्ञान प्राप्त कर भक्त बनकर आत्मदर्शन की कल्पना की गई है।

प्रश्न 5.
महात्मा गाँधीजी की दृष्टि में निषिद्ध क्या है?
उत्तर:
महात्मा गाँधीजी की दृष्टि में फलासक्ति ही निषिद्ध है।

जीने की कला लघु उत्तरीय प्रश्नोत्तर

प्रश्न 1.
महाभारत के ऐतिहासिक पात्रों का उपयोग व्यास ने किस हेतु के किया है?
उत्तर:
महाभारत के रचयिता मुनि वेदव्यास ने अपने महाभारत महाकाव्य में ऐतिहासिक पात्रों का प्रयोग किया है क्योंकि महाभारत का युद्ध ऐतिहासिक है तथा युद्धभूमि कुरुक्षेत्र भी इतिहास प्रसिद्ध स्थान है। ऐसे में पात्रों का भी ऐतिहासिक होना अनिवार्य था। इस ऐतिहासिक घटना को प्रमाणित करने तथा कृष्ण के उपदेश को भी सत्यता की कसौटी पर खरा उतारने के लिए ऐतिहासिक पात्रों का होना अनिवार्य था। अतः व्यास का यही हेतु था।

प्रश्न 2.
पात्रों की अमानुषी और अतिमानुषी उत्पत्ति का वर्णन करके व्यास ने किस उद्देश्य की पूर्ति की है?
उत्तर:
गीता का निष्काम कर्म तथा फलासक्ति जैसा सिद्धान्त एक अमानुषी मस्तिष्क की उपज हो सकती है तो उस सिद्धान्त को समझने के लिए भी ऐसा ही अमानुषी पात्र चाहिए था। इसी कारण कृष्ण ने समस्त ब्रह्माण्ड की रचना करके उन्हें (कृष्ण को) अतिमानुषी बताया। जिसे देखकर अर्जुन निष्काम कर्म के तत्त्व को समझा। रचनाकार का जो उद्देश्य था कि संसार को निष्काम कर्म में संलग्न कर मोह से छुड़ाया जाए, वह पूरा हुआ।

प्रश्न 3.
गीता की शिक्षा का आचरण करने वाले मनुष्य का स्वभाव कैसा होता है? (2014)
उत्तर:
गाँधीजी ने बताया है कि गीता की शिक्षा का आचरण करने वाले मनुष्य को स्वभाव से ही सत्य और अहिंसा का पालन करना पड़ता है। गीता की शिक्षा है जो कर्म आसक्ति के बिना हो ही न सकें वे सब कर्म छोड़ने लायक हैं।’ यह सुवर्ण नियम मनुष्य को कई धर्म संकटों से बचाता है। अतः मनुष्य झूठ, हत्या, असत्य को त्यागने वाला बन जाता है, जिससे उसका जीवन सरल तथा शान्तिपूर्ण बन जाता है।

प्रश्न 4.
रीति को दृष्टि में रखकर गीता के मूलमन्त्र का जिज्ञासु क्या कर सकता है?
उत्तर:
गीता सूत्र ग्रन्थ न होकर एक महान ग्रन्थ है। इस ग्रन्थ के भावों की गहराई में हम जितना उतरेंगे उतने ही उसमें से नए और सुन्दर अर्थ निकलेंगे। गीता जनसमाज के लिए है, अतः गीता में आए हुए महान शब्दों के अर्थ सदैव बदलेंगे। अर्थ व्यापक भी बनेंगे। परन्तु गीता का मूलमन्त्र कभी नहीं बदलेगा। गीता का मूलमन्त्र है-कर्म के फल में आसक्ति न होना तथा आत्मदर्शन करना। यह मन्त्र जिस रीति से जीवन में अपनाया जा सके उस रीति को दृष्टि में रखकर जिज्ञासु गीता के महाशब्दों का मनचाहा अर्थ कर सकता है।

प्रश्न 5.
अवतार का क्या अर्थ है?
उत्तर:
अवतार का अर्थ है-शरीरधारी विशिष्ट पुरुष। सभी जीवमात्र ईश्वर के अवतार हैं, परन्तु सामान्य भाषा में समस्त जीवों को अवतार नहीं कहते। जो प्राणी (पुरुष) अपने युग में सर्वश्रेष्ठ धर्मवान पुरुष होता है, उसे आने वाली पीढ़ी अवतार के रूप में पूजती है क्योंकि उस अवतार रूपी प्राणी में कोई दोष नहीं होता है। इस प्रकार दोष रहित धर्मवान पुरुष ही अवतार होता है।

जीने की कला दीर्घ उत्तरीय प्रश्नोत्तर

प्रश्न 1.
गीता के अध्ययन से गाँधीजी को क्या अनुभूति हुई?
उत्तर:
गीता के अध्ययन से गाँधीजी को अनुभव हुआ कि गीता में मोक्ष और सांसारिक व्यवहार के बीच कोई भेद नहीं है, परन्तु धर्म को व्यवहार में उतारा गया है। अतः आसक्ति से किया कर्म छोड़ देने लायक है। गीता “आसक्ति त्यागने” को कहती है गीता की इस शिक्षा का आचरण करने वाला व्यक्ति स्वभाव से ही सत्य व अहिंसा का पालन करने लग जाता है, झूठ व लालच से दूर रहने लगता है तब उसके जीवन में सरलता व शान्ति आ जाती है। इसके विपरीत हिंसा या असत्य के पीछे परिणाम की इच्छा रहती है, अतः फलासक्ति में मनुष्य की रुचि बढ़ जाती है, जिससे वह हत्या, झूठ, व्यभिचार में रुचि लेने लगता है। अत: गाँधीजी ने अनुभव किया कि अनासक्ति व फलासक्ति के अभाव में कर्म करना ही मोक्ष है। इसी कारण गाँधीजी ने फलासक्ति को निषिध कहा है। गाँधीजी के अनुसार गीता की रचना परिवार के मामूली झगड़े निपटाने के लिए नहीं बल्कि प्राणी को स्थित प्रज्ञ व्यवहार हेतु हुई है।

प्रश्न 2.
आत्म दर्शन से सम्बन्धित गाँधीजी का दृष्टिकोण स्पष्ट कीजिए।
उत्तर:
आत्मदर्शन से सम्बन्धित गाँधीजी का दृष्टिकोण है कि मनुष्य को सुख व शान्ति पाने के लिए ईश्वर रूप बनना होगा। ईश्वर रूप बनने के लिए किए जाने वाले प्रयत्न का ही नाम सच्चा और एकमात्र पुरुषार्थ है और वही आत्मदर्शन है। आत्मदर्शन का अर्थ है-आत्मा के बारे में निरूपण करने वाला शास्त्र। आत्मा के बारे में निरूपण करना समस्त धर्म ग्रन्थों का विषय है। ठीक उसी प्रकार गीता भी आत्मा के बारे में निरूपण करती है अर्थात् आत्मदर्शन के विषय में बताती है।

गाँधीजी कहते हैं कि गीताकार ने आत्मदर्शन का प्रतिपादन करने के लिए गीता की रचना नहीं की। बल्कि गीता आत्मा को पाने के लिए उत्सुक प्राणी को आत्मा का स्वरूप बताती है। आत्मा को पहचानने का एक अनोखा उपाय कर्म के फल का त्याग है। प्रत्येक कर्म में दोष होता है, उस दोष को मन, वचन और काया से ईश्वर को अर्पित करके अर्थात् कर्म के फल का त्याग करके दूर किया जा सकता है। साथ ही, उस कर्म के फल-त्याग में हृदय मन्थन भी हो। अतः संक्षिप्त रूप में गाँधीजी ने कहा है कि ज्ञान प्राप्त करना, भक्त होना ही आत्म-दर्शन है।

प्रश्न 3.
गाँधीजी के अनुसार गीता का उद्देश्य क्या है?
उत्तर:
गाँधीजी के अनुसार गीता का उद्देश्य आत्मार्थी को आत्मदर्शन करने का एक अद्वितीय उपाय बताना तथा आत्मा को पाने के लिए उत्सुक प्राणी को आत्मा के स्वरूप का ज्ञान कराना है। इस आत्मदर्शन को प्राप्त करने का अद्वितीय उपाय है कर्म के फल का त्याग। इसी केन्द्र बिन्दु के आस-पास गीता का सारा विषय गाँथा गया है। देह कर्म से मुक्त नहीं हो सकती और कर्म दोष से मुक्त नहीं हो सकता, दोष से मुक्त हुए बिना मुक्ति (मोक्ष) नहीं मिल सकती।

तब मुक्ति पाने के लिए इस दोष से कैसे छूटा जा सकता है? इस प्रश्न का उत्तर गीता ने दिया है-“निष्काम कर्म करके, यज्ञार्थ कर्म करके, कर्म के फल का त्याग करके, सारे कर्म मन, वचन और काया से ईश्वर को अर्पित करके” कर्म को दोष मुक्त किया जा सकता है। इस क्रिया में केवल बुद्धि ही नहीं बल्कि हृदय-मन्थन भी आवश्यक है। इस प्रकार किए गए कर्म से मनुष्य को सिद्धि मिलती है। यही कर्म प्राणी का धर्म और मोक्ष है। इसी मोक्ष की प्राप्ति करना गीता का उद्देश्य है।

प्रश्न 4.
गाँधीजी ने फल त्यागी किसे कहा है?
उत्तर:
गीता में आत्मदर्शन का उपाय बताया गया है-कर्म के फल का त्याग। जो व्यक्ति कर्म के फल का त्याग करता है वही फल-त्यागी है। कर्म के फल का त्याग केवल कह देने से नहीं होता, बुद्धि या बुद्धि के प्रयोग से नहीं होता। वह हृदय-मन्थन से ही होता है। हृदय-मन्थन किए कर्म से ही सिद्धि मिलती है। दूसरी ओर फल त्याग का अर्थ कर्म के परिणाम के विषय में लापरवाह रहना भी नहीं है।

परिणाम का और साधना का विचार करना तथा दोनों का ज्ञान होना अत्यन्त आवश्यक है। इतना करने के बाद जो मनुष्य परिणाम की इच्छा किए बिना साधन में तन्मय रहता है, वह फलत्यागी कहा जाता है। गीताकार ने कर्मफल के त्याग का सिद्धान्त संसार के सामने रखा है। यह स्वर्णिम नियम मनुष्य को अनेक धर्म संकटों से बचाता है। झूठ, असत्य जैसी बुराइयों से बचाता है। अत: गाँधीजी के अनुसार कर्म के फल को त्यागने वाला ही फलत्यागी है।

प्रश्न 5.
गीताकार ने किस भ्रम को दूर कर दिया है?
उत्तर:
गीताकार ने धर्म और अर्थ के परस्पर विरोधी भ्रम को दूर कर दिया है। सामान्यतः यह माना जाता है कि धर्म और अर्थ परस्पर विरोधी हैं। “व्यापार आदि सांसारिक व्यवहारों में धर्म का पालन नहीं हो सकता, धर्म के लिए स्थान नहीं हो सकता। धर्म का उपयोग केवल मोक्ष के लिए ही किया जा सकता है। धर्म के स्थान पर धर्म शोभा देता है, अर्थ के स्थान पर अर्थ शोभा देता है।” गाँधीजी कहते हैं कि गीताकार ने इस भ्रम को दूर कर दिया है। उन्होने मोक्ष और सांसारिक व्यवहार के बीच ऐसा कोई भेद नहीं रखा है; परन्तु धर्म को व्यवहार में उतारा है। गीता कहती है “जो धर्म व्यवहार में नहीं उतारा जा सकता वह धर्म ही नहीं है।” इस प्रकार गीताकार ने मोक्ष और सांसारिक व्यवहार के बीच के भेद को आसक्ति रहित कर्म के द्वारा दूर कर दिया है।

प्रश्न 6.
निम्नलिखित पंक्तियों का भाव पल्लवन कीजिए
(क) “जहाँ देह है वहाँ कर्म तो है ही।”
उत्तर:
‘महात्मा गाँधी’ ने ‘जीने की कला’ निबन्ध में गीता के अनासक्ति कर्म की व्याख्या की है। गाँधीजी ने कहा है कि देहधारी प्राणी की आत्मा उसका कर्म है। इस संसार में जन्म लेने वाला प्राणी एक पल भी बिना कर्म के नहीं रह सकता। कर्म से कभी भी मनुष्य को मुक्ति नहीं मिल सकती। जब मनुष्य सोता है तब भी वह साँस लेने, स्वप्न देखने आदि का कर्म करता रहता है तो जागने की अवस्था में वह भला बिना कर्म के कैसे रह सकता ह? अतः कर्म ही मनुष्य की जीवित अवस्था का प्रमाण है।

(ख) “कर्म के बिना किसी को सिद्धि प्राप्त नहीं हुई।”
उत्तर:
‘जीने की इच्छा’ निबन्ध में महात्मा गाँधी’ ने बताया है कि निष्काम कर्म करने से मनुष्य नर से नरश्रेष्ठ बन जाता है। गीता ने ज्ञानियों, भक्तों तथा नर को कर्म करने का सन्देश दिया है। साथ ही गीताकार ने कहा है कि कर्म से ही सिद्धि प्राप्त होती है। सिद्धि का अर्थ है आत्मा की मुक्ति। कहा गया है कि बिना कर्म के शक्तिशाली शेर तक के मुँह में हिरन नहीं जाता तो आत्मा की सिद्धि के लिए आसक्ति रहित कर्म करना अनिवार्य है। कर्म सिद्धि का पर्याय है।

जीने की कला भाषा अध्ययन

प्रश्न 1.
निम्नलिखित शब्दों की शुद्ध वर्तनी लिखिएगिता, ऐतिहसिक, धरम, फलसक्ति, सुत्रग्रन्थ।
उत्तर:
गिता = गीता। ऐतिहसिक = ऐतिहासिक। धरम = धर्म। फलसक्ति = फलासक्ति। सुत्रग्रन्थ = सूत्रग्रन्थ।

प्रश्न 2.
निम्नलिखित शब्दों को वाक्यों में प्रयोग कीजिएविश्वास, अहिंसा, चिन्तन-मनन, अभिलाषा, स्वभाव।
उत्तर:

1. विश्वास – अवतार में विश्वास करना प्राणी की उदात्त आध्यात्मिक अभिलाषा का सूचक है।
2. अहिंसा – सत्य और अहिंसा बापू के शस्त्र थे।
3. चिन्तन – मनन-भगवद्गीता में निष्काम कर्म पर चिन्तन-मनन किया गया है।
4. अभिलाषा – मनुष्य की अन्तिम उदात्त आध्यात्मिक अभिलाषा मुक्ति दिलाती है।
5. स्वभाव – क्रोधी स्वभाव का व्यक्ति कभी विनम्र नहीं होता है।

प्रश्न 3.
दिए गए शब्दों के विलोम शब्द लिखिएउत्पत्ति, निरर्थकता, सच्चा, विधि-निषिद्ध।
उत्तर:

प्रश्न 4.
रिक्त स्थानों की पूर्ति कीजिए

1. महाभारत में ………….. सर्वोच्च स्थान पर विराजती है।
2. अद्वितीय उपाय है …………. के फल का त्याग।
3. गीता का ………… आत्मदर्शन करने का एक अद्वितीय उपाय बताना है।

उत्तर:

1. गीता
2. कर्म
3. उद्देश्य आत्मार्थी को।

प्रश्न 5.
निम्नलिखित वाक्यांश के लिए एक शब्द लिखिए
उत्तर:

1. संसार से सम्बन्धित = सांसारिक
2. दोष से मुक्त = निर्दोष
3. काम से रहित = निष्काम
4. जानने की इच्छा रखने वाला = जिज्ञासु
5. धर्म का पालन करने वाला = धर्मवान।

जीने की कला पाठ का सारांश

श्रीमद्भगवतगीता को ‘गीता मैया’ कहने वाले महात्मा गाँधी ने ‘जीने की कला’ निबन्ध में मानव बुद्धि को व्यावहारिक बनाने का सूत्र बताने के लिए गीता को आधार माना है। संसार में बुद्धि दो प्रकार की होती है-भौतिक तथा स्थित प्रज्ञ। गीता में स्थित प्रज्ञ बुद्धि को श्रेष्ठ माना है। गीता की रचना पारिवारिक झगड़े निपटाने या अवतारवाद की स्थापना के लिए नहीं हुई। समस्त धर्म ग्रन्थों तथा गीता का विषय आत्म दर्शन है परन्तु कृष्ण कृत गीता आत्म-दर्शन करने के लिए; एवं आत्म-दर्शन करने का उपाय बताती है।

वह अद्वितीय उपाय है, कर्म के फल का त्याग। देहधारी कर्म से मुक्त नहीं हो सकता और कर्म में दोष अवश्य होते हैं। कर्म में दोष न होने का अर्थ है ‘निष्काम कर्म करके, यंज्ञार्थ कर्म करके, कर्म के लिए फल का त्याग करके, सारे कर्म कृष्णार्पण करके अर्थात् मन, वचन और काया को ईश्वर में होम करके।’ यह निष्कामता स्थिर बुद्धि तथा हृदय-मन्थन से ही उत्पन्न होती है, गीता में इसी को आत्म-दर्शन कहा गया है। परिणाम की इच्छा किए बिना साधना (कर्म) में लीन रहना ही फल का त्याग है। फल के त्याग का अर्थ कर्म के परिणाम के प्रति लापरवाह होना नहीं है।

गीता में दूसरी बात बताई गई है कि धर्म और अर्थ परस्पर विरोधी नहीं हैं। व्यापार आदि सांसारिक व्यवहारों में धर्म का पालन नहीं हो सकता, धर्म का उपयोग केवल मोक्ष के लिए ही किया जा सकता है। परन्तु गीताकार ने इस भ्रम को दूर कर दिया है। उन्होंने मोक्ष और सांसारिक व्यवहार के बीच ऐसा कोई भेद नहीं रखा है, परन्तु धर्म को व्यवहार में उतारा है। अतः जो कर्म अनासक्ति यानि लगाव रहित न हो उसे त्याग देना चाहिए। यह स्वर्णिम नियम मनुष्य को अनेक धर्म संकटों, जैसे हत्या, झूठ, व्यभिचार आदि से बचाता है। इससे जीवन सरल बन जाता है और सरलता से शान्ति मिलती है और गीता की इसी शिक्षा को अपनाने वाले मनुष्य स्वभाव से सत्य और अहिंसा का पालन करते हैं।

गीता सूत्र ग्रन्थ न होकर समाज के लिए एक महान ग्रन्थ है जिसमें जितना डूबेंगे उतने ही अर्थ मिलेंगे। साथ ही ये अर्थ प्रत्येक युग में बदलेंगे तथा व्यापक बनेंगे। लेकिन मूल मन्त्र कभी नहीं बदलेगा। इसीलिए गीता करने योग्य और न करने योग्य कर्म बताने वाला संग्रह-ग्रन्थ भी नहीं है क्योंकि जो कर्म एक के लिए मान्य है वह दूसरे के लिए अमान्य हो सकता है। उसी प्रकार एक काल या एक देश में जो कर्म विहित है वह दूसरे देश तथा दूसरे काल में अमान्य हो सकता है। अतः फलासक्ति निषिद्ध है और अनासक्ति विहित है। इसीलिए गीता नर को नरश्रेष्ठ बनने का मार्ग दिखाती है, जीने की कला सिखाती है।

जीने की कला कठिन शब्दार्थ

सर्वोच्च = सबसे ऊँचा। भौतिक-सांसारिक। स्थित प्रज्ञ- स्थिर बुद्धि वाला। औचित्य = उचित। अनौचित्य = अनुचित। लौकिक = सांसारिक। उदात्त = उदार। आत्म-दर्शन = आत्मा के बारे में निरूपण करने वाला शास्त्र। प्रतिपादन = स्थापित। आत्मार्थी = आत्मा को पाने के लिए उत्सुक। अद्वितीय = अनोखा। देह = शरीर। मुक्त = स्वतन्त्र। काया = शरीर। निष्काम कामना से रहित। हृदय-मन्थन = मन में अच्छे-बुरे की पहचान । सिद्धि = सफलता। तन्मय = लीन। फलासक्ति = फल में लगाव। अनासक्ति = आसक्ति रहित, लगाव रहित। आसक्ति = लगाव। सुवर्ण = सुन्दर। त्याज्य = छोड़ने योग्य। सूत्र = नियम। साधा = अपनाया। जिज्ञासु = जानने की इच्छा रखने वाला। विहित = मान्य, अनिवार्य। निषिध = अमान्य। अमानुषी = जो मनुष्य से सम्बन्धित न हो। अतिमानुषी = मानव धर्म से परे सिद्धि देवी।

जीने की कला संदर्भ-प्रसंग सहित व्याख्या

1. अवतार का अर्थ है शरीरधारी विशिष्ट पुरुष। जीवमात्र ईश्वर के अवतार हैं, परन्तु लौकिक भाषा में सबको अवतार नहीं कहते। जो पुरुष अपने युग में सबसे श्रेष्ठ धर्मवान पुरुष होता है, उसे भविष्य की प्रजा अवतार के रूप में पूजती है। इसमें मुझे कोई दोष नहीं मालूम होता।

सन्दर्भ :
प्रस्तुत गद्यांश हमारी पाठ्य-पुस्तक के जीने की कला’ नामक पाठ से अवतरित है। यह पाठ महात्मा गाँधी की पुस्तक ‘गीता माता से संग्रहीत एवं सम्पादित है।

प्रसंग :
लेखक ने गीता की रचना का कारण बताते हुए उस विशिष्ट पुरुष की कल्पना की है जो निर्दोष होने के कारण अवतार कहा जाता है।

व्याख्या :
शरीरधारी पुरुषों में जो पुरुष विशेष होता है, उसे अवतार कहते हैं। इस संसार में जितने भी प्राणी हैं, वे सब ईश्वर के अवतार हैं अर्थात् भगवान की एक महान रचना है। परन्तु संसार में सारे प्राणियों को अवतार नहीं कहते। संसार उस प्राणी को अवतार मानता है जो सब प्राणियों से अलग होता है। कुछ अनोखे-अनुपम कार्य करता है, धर्म का पालनकर्ता होता है, चरित्रवान होता है, उसके प्रति सबके मन में श्रद्धा होती है, समाज के लिए भलाई करता है अर्थात् हर दृष्टि से सबका प्रिय होता है, वही व्यक्ति भविष्य में अवतार माना जाता है। अपने समय में उसे श्रेष्ठ पुरुष माना जाता है। उस श्रेष्ठ पुरुष में कोई भी बुराई नहीं होती। अवतार सदैव अवगुणों से दूर रहता है। अवतार सर्वगुण सम्पन्न होता है। उसकी बुराई करने वाला कोई नहीं होता।

विशेष :

1. भाषा सरल, बोधगम्य एवं प्रभावपूर्ण खड़ी बोली है।
2. विशिष्ट, श्रेष्ठ, लौकिक जैसे संस्कृत के शब्दों का प्रयोग है।
3. व्याख्यात्मक शैली है।
4. लघु वाक्य सूत्र जैसे प्रतीत होते हैं।

2. मुक्ति केवल निर्दोष मनुष्य को ही मिलती है। तब कर्म के बंधन से अर्थात् दोष के स्पर्श से कैसे छूटा जा सकता है? इस प्रश्न का उत्तर गीता जी ने निश्चयात्मक शब्दों में दिया है। ‘निष्काम कर्म करके, यज्ञार्थ कर्म करके, कर्म के फल का त्याग करके, सारे कर्म कृष्णार्पण करके अर्थात् मन, वचन और काया को ईश्वर में होम कर।’

सन्दर्भ :
पूर्ववत्।

प्रसंग :
देह के साथ कर्म और कर्म के साथ दोष जुड़ा है, अतः इस दोष को दूर करके मोक्ष मिलता है। गीता में दोष दूर करने का उपाय बताया है कि बिना फल की इच्छा से कर्म करना ही दोषमुक्त होता है।

व्याख्या-गाँधीजी कहते हैं कि इस संसार में जन्म लेकर प्रत्येक व्यक्ति को कर्म करना पड़ता है। कर्मों में दोष होता है। इन्हीं दोषों के कारण मनुष्य को मुक्ति नहीं मिलती। प्राणी इसी कारण कर्म के बन्धन में बँधता चला जाता है। इस कर्म के बन्धन से मुक्त होने का उपाय या निर्दोष कर्म अपनाने का तरीका गीता बताती है। गीता में बताया गया है कि निष्काम कर्म, अर्थात् बिना किसी फल की इच्छा से किया गया कर्म मुक्ति दिलाता है। कर्म के फल का त्याग करके अर्थात् अपने समस्त कार्य कृष्णार्पण अर्थात् मन, वचन और काया को ईश्वर को अर्पित करके, भले-बुरे फल की कामना न करके, निरन्तर कर्म करते रहने से सांसारिक बंधन टूट जाते हैं और आत्मा मुक्त हो जाती है। इसी आत्म-दर्शन को गाँधीजी गीता का मूल मानते हैं।

विशेष :

1. संस्कृत शब्दावली युक्त खड़ी बोली।
2. प्रश्नोत्तर व सूत्रात्मक शैली का प्रयोग है।
3. भावों की गहनता व बोधगम्यता है।
4. होम कर’ जैसे मुहावरे का प्रयोग है।

3. गीता के मत के अनुसार जो कर्म आसक्ति के बिना हो ही न सकें वे सब त्याज्य हैं-छोड़ देने लायक हैं। यह सुवर्ण नियम मनुष्य को अनेक धर्म संकटों से बचाता है। इस मत के अनुसार हत्या, झूठ, व्यभिचार आदि कर्म स्वभाव से ही त्याज्य हो जाते हैं। इससे मनुष्य जीवन सरल बन सकता है और सरलता से शांति का जन्म होता है।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
राष्ट्रपिता गाँधी के अनुसार गीता में बताया गया है कि आसक्ति से मुक्त कर्म ही धर्म है। व्यवहार में लाया जाने वाला धर्म ही सच्चा धर्म है। यही धर्म हमें अमानुषिक कर्मों से बचाता है।

व्याख्या :
गीता में एक शब्द आता है आसक्ति, जिसका अर्थ होता है लगाव। गीता में बताया गया है कि जिस कर्म में आसक्ति होती है अर्थात् कर्म में लगाव या मोह होता है, उस कर्म को त्यागना ही मनुष्य का धर्म है। आसक्ति के बिना किया गया कर्म, मनुष्य को धर्म संकटों से बचाता है। श्रेष्ठ फल देने वाले कर्म को मनुष्य करना चाहता है। आसक्ति से रहित होकर मनुष्य जब कर्म करता है तो वह अनेक सांसारिक बुराइयों; जैसे-झूठ, दुराचार, हत्या इत्यादि से बच जाता है और वह स्वभाव से सत्य और अहिंसा का पालन करने वाला बन जाता है। जब मनुष्य आसक्तिपूर्ण कर्म को त्याग देता है तब उसका मन परिणाम के लिए बेचैन नहीं होता जिससे जीवन में सन्तोष व शान्ति का आगमन होता है, सन्तोष मन को शान्ति देता है। जीवन में सरलता व शान्ति के आने पर मनुष्य का व्यवहार तथा आचरण बहुत ही कोमल व साफ-सुथरा हो जाता है जिससे मुक्ति का मार्ग खुल जाता है।

विशेष :

1. तत्सम शब्दों के साथ खड़ी बोली का प्रयोग है।
2. विचारात्मक तथा व्याख्यात्मक शैली।
3. मनुष्य को आसक्ति से दूर रहने का उपदेश।

MP Board Class 11th Hindi Solutions

MP Board Class 11th Hindi Swati Solutions गद्य महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

बहु-विकल्पीय प्रश्न

प्रश्न 1.
सच्चे उत्साही कहलाते हैं (2008)
(i) पलायनवादी
(ii) कायर
(iii) कर्म सौन्दर्य के उपासक
(iv) कर्म से विरक्त।
उत्तर:
(iii) कर्म सौन्दर्य के उपासक

प्रश्न 2.
साहसपूर्ण आनन्द की उमंग का नाम है- (2008)
(i) भय
(ii) क्रोध
(iii) उत्साह
(iv) घृणा।
उत्तर:
(iii) उत्साही

प्रश्न 3.
शिरीष फूलता है
(i) शरद ऋतु में
(ii) वर्षा ऋतु में
(iii) जेठ मास में
(iv) बसंत में।
उत्तर:
(ii) जेठ मास में

प्रश्न 4.
आजादी की लड़ाई में बड़े घर से आशय था (2009)
(i) महल
(ii) जेलखाना
(iii) बहुत बड़ी हवेली
(iv) विशाल मकान।
उत्तर:
(ii) जेलखाना

प्रश्न 5.
स्नेहबन्ध कहानी में सास का हृदय परिवर्तन होता है (2008)
(i) मीता की सम्पन्नता देखकर
(ii) मीता का सौन्दर्य देखकर
(iii) मीता की सामाजिक प्रतिष्ठा देखकर,
(iv) मीता की सेवा सुश्रूषा तथा कर्त्तव्य भावना देखकर।
उत्तर:
(iv) मीता की सेवा सुश्रूषा तथा कर्त्तव्य भावना देखकर।

प्रश्न 6.
स्नेहबन्ध कहानी में किया गया है (2008)
(i) सास-बहू की अनबन का चित्रण
(ii) सास-बहू के सम्बन्धों का मर्मस्पर्शी चित्रण
(iii) सास-बहू के अस्तित्व का चित्रण,
(iv) सास-बहू के बीच श्रेष्ठता का चित्रण।
उत्तर:
(ii) सास-बहू के सम्बन्धों का मर्मस्पर्शी चित्रण

प्रश्न 7.
‘मर्यादा’ एकांकी में निहित है (2009, 13)
(i) पारिवारिक आदर्श
(ii) सामाजिक आदर्श
(iii) धार्मिक आदर्श
(iv) ऐतिहासिक आदर्श।
उत्तर:
(i) पारिवारिक आदर्श

प्रश्न 8.
‘जननी जन्मभूमिश्च’ निबन्ध में मिश्रजी ने मातृभूमि का महत्व बताया है (2012)
(i) कल्पना के द्वारा
(ii) ऐतिहासिक प्रसंगों के द्वारा
(ii) परम्पराओं के द्वारा
(iv) नकल के द्वारा।
उत्तर:
(ii) ऐतिहासिक प्रसंगों के द्वारा

प्रश्न 9.
जननी जन्मभूमिश्च में लेखक ने जन्मभूमि को निरूपित किया है- (2008)
(i) स्वर्ग के समान
(ii) स्वर्ग से भी श्रेष्ठ
(iii) सबसे सुन्दर
(iv) सबसे महत्त्वपूर्ण।
उत्तर:
(ii) स्वर्ग से भी श्रेष्ठ

प्रश्न 10.
सतपुड़ा पर्वत की सबसे ऊँची चोटी है (2008)
(i) एवरेस्ट
(ii) माउण्ट आबू
(ii) कारगिल
(iv) धूपगढ़।
उत्तर:
(iv) धूपगढ़।

प्रश्न 11.
धूपगढ़ स्थित है (2010, 15)
(i) पचमढ़ी
(ii) भोजपुर
(iii) साँची
(iv) रायगढ़।
उत्तर:
(i) पचमढ़ी

प्रश्न 12.
नीरा के पिता को पढ़ने का शौक था (2008, 11,14)
(i) अखबार
(ii) पत्रिका
(iii) पत्र
(iv) उपन्यास।
उत्तर:
(i) अखबार

प्रश्न 13.
अवतार का अर्थ है
(i) जीवमात्र
(ii) विशिष्ट पुरुष
(iii) सामान्यजन
(iv) भगवान।
उत्तर:
(ii) विशिष्ट पुरुष

प्रश्न 14.
भोलाराम का जीव ढूँढ़ने धरती पर कौन गया? (2013)
(i) बच्चे
(ii) यमदूत
(iii) स्त्री
(iv) नारद।
उत्तर:
(iv) नारद।

प्रश्न 15.
संयुक्त परिवार की नींव है (2016)
(i) सद्भावना और त्याग
(ii) भाईचारा
(iii) सूझ-बूझ से
(iv) इन सबके सामंजस्य से।
उत्तर:
(iv) इन सबके सामंजस्य से।

प्रश्न 16.
कोणार्क मंदिर किस देवता से सम्बन्धित है? (2017)
(i) सूर्य
(1) चन्द्रमा
(ii) श्रीराम
(iv) श्रीकृष्ण।
उत्तर:
(i) सूर्य

प्रश्न 17.
शंकर की माताजी का नाम था
(i) उभय भारती
(ii) आर्यम्बा
(iii) अहिल्या
(iv) आत्री।
उत्तर:
(ii) आर्यम्बा

प्रश्न 18.
शंकराचार्य ने किस मत का प्रचार किया?
(i) अद्वैत
(ii) सनातन
(iii) द्वैत
(iv) हिन्दू।
उत्तर:
(i) अद्वैत

रिक्त स्थान पूर्ति

1. उत्साह के बीच …………. संचरण होता है। (2009)
2. शिरीष अपना पोषण …………. से प्राप्त करता है। (2009)
3. सतपुड़ा पर्वत की सबसे ऊँची चोटी………… है। (2014)
4. धूपगढ़ ………… में स्थित है। (2008, 09)
5. महादेवी वर्मा ने पहली पुस्तक ………… पढ़ी थी। (2009)
6. नीरा के पिता कुली बनकर ……….. गये थे। (2008, 09)
7. कोणार्क मन्दिर …………. देवता से सम्बन्धित है। (2008)
8. कोणार्क के मन्दिर में कलश की स्थापना ………. के सहयोग से सम्भव हो सकी। (2008)
9. सुमन जगदीश के घर …………. सुनने के लिए आई थी। (2010)
10. भोलाराम ने दरख्वास्तों पर …………. नहीं रखा था।
11. यमदूत को चकमा ………… के जीव ने दिया था। (2016)
12. शिरीष की तुलना ……….. से की गई है। (2011)
13. शिरीष …………. के आगमन से फूलना प्रारम्भ हो जाता है। (2013, 17)
14. ‘नीरा’ एक …………. है। (2013)
15. ‘गंगा और गंगा के कछार को मेरा सलाम कहें’ कथन ………. है। (2012)
16. शंकर के गुरु …………. थे।
17. भारत की आत्मा …………. में निवास करती है।

उत्तर:

1. साहस का
2. वायुमण्डल
3. धूपगढ़
4. पचमढ़ी
5. पंचतन्त्र
6. मॉरीशस,
7. सूर्य
8. धर्मपद
9. रामायण
10. वजन
11. भोलाराम
12. अवधूत
13. बसन्त ऋतु
14. कहानी
15. परदेशी का
16. गोविन्दपाद
17. वाराणसी।

सत्य / असत्य

1. उत्साह की गिनती दुर्गुण में होती है। (2011)
2. अवधूतों के मुख से संसार की सबसे सरस रचनाएँ निकली हैं। (2009)
3. फूल की अभिलाषा देवताओं पर चढ़ाए जाने की होती है।
4. नीरा के पिता को गीता पढ़ने का शौक था। (2015)
5. नीरा के पिता को अपनी बेटी के भविष्य की चिन्ता थी। (2009)
6. ‘स्नेहबंध’ कहानी में नारी हृदय के परिवर्तन का स्वाभाविक अंकन हुआ है। (2013)
7. ‘मेरे बचपन के दिन’ रेखाचित्र विधा से सम्बन्धित है। (2009)
8. भोलाराम का जीव पेंशन की दरख्वास्तों में अटका था। (2009)
9. कला के सम्बन्ध में आचार्य विशु और धर्मपद के विचारों में अन्तर है। (2008)
10. भोलाराम हृदयगति रुकने के कारण मरा। (2009)
11. भोलाराम का जीव ढूँढ़ने के लिए धरती पर यमराज आये। (2017)
12. श्रीमती मालती जोशी सुप्रसिद्ध आलोचक हैं। (2012)
13. संयुक्त परिवार की नींव धन पर आधारित होती है। (2012, 14)
14. ‘जननी जन्मभूमिश्च’ निबन्ध में लेखक ने स्वर्ग को श्रेष्ठ माना है। (2013)
15. सतपुड़ा की सबसे ऊँची चोटी एवरेस्ट है। (2013)
16. धूपगढ़ पचमढ़ी पर स्थित है। (2016)
17. शंकर के गुरु गौड़पादाचार्य थे।
18. शंकर ने शास्त्रार्थ में मंडन मिश्र और उनकी विदुषी पत्नी उभय भारती को पराजित किया था।

उत्तर:

1. असत्य
2. सत्य
3. असत्य
4. असत्य
5. सत्य
6. सत्य
7. असत्य
8. सत्य
9. सत्य
10. असत्य
11. असत्य
12. असत्य
13. असत्य
14. असत्य
15. असत्य
16. सत्य
17. असत्य
18. सत्य।

जोड़ी मिलाइए

I.

उत्तर:
1. → (ङ)
2. → (क)
3. → (घ)
4. → (ग)
5. → (ख)
6. → (च)।

II.

1. → (ख)
2. → (क)
3. → (घ)
4. → (ग)
5. → (च)
6. → (ङ)।

एक शब्द / वाक्य में उत्तर

प्रश्न 1.
उत्साह के बीच किसका संचरण होता है?
उत्तर:
धृति और साहस का

प्रश्न 2.
जननी और जन्मभूमि किससे अधिक श्रेष्ठ हैं? (2009, 15)
उत्तर:
स्वर्ग से

प्रश्न 3.
माँ का दूध कैसा होता है?
उत्तर:
अनमोल

प्रश्न 4.
यमदूत को किसने चकमा दिया था? (2014)
भोलाराम के जीव ने

प्रश्न 5.
महादेवी वर्मा अपने जेब खर्च के पैसे क्यों बचाती थीं?
उत्तर:
देश के लिए

प्रश्न 6.
धर्मपद कौन था? (2009)
उत्तर:
आशुशिल्पी युवक

प्रश्न 7.
साँझ की आँखों में करुणा किस रूप में प्रकट होती है?
उत्तर:
ओस कणों के रूप में

प्रश्न 8.
नीरा की माँ का क्या नाम था? (2017)
उत्तर:
कुलसम

प्रश्न 9.
विदेश न जाने के पीछे मीता का क्या भाव था? (2009)
उत्तर:
खर्च बचाना

प्रश्न 10.
‘मर्यादा’ एकांकी में परिवार की किस व्यवस्था का समर्थन हुआ है?
उत्तर:
संयुक्त परिवार

प्रश्न 11.
गीता का प्रमुख उद्देश्य क्या है?
उत्तर:
आत्मार्थी को आत्मदर्शन का एक अद्वितीय उपाय बताना

प्रश्न 12.
कोणार्क मन्दिर कहाँ स्थित है? (2012)
उत्तर:
उड़ीसा प्रान्त में पुरी के निकट समुद्र तट पर

प्रश्न 13.
‘मेरे बचपन के दिन’ पाठ किस विधा में लिखा गया है? (2012)
उत्तर:
संस्मरण

प्रश्न 14.
शिरीष अपना पोषण रस किससे खींचता है? (2013)
उत्तर:
वायुमण्डल से

प्रश्न 15.
शंकर ने किस नाम से पाँच श्लोकों की रचना की?
उत्तर:
मनीषपंचक

प्रश्न 16.
अपने लक्ष्य की पूर्ति के लिए अन्त में शंकराचार्य कहाँ पहुँचे?
उत्तर:
केदारनाथ

MP Board Class 11th Hindi Solutions