Class 12

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 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 12th Maths Important Questions Chapter 11 Three Dimensional Geometry

Three Dimensional Geometry Important Questions

Three Dimensional Geometry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
The FZ – plane divides the line segment joining the points P (- 2, 4, 7) and Q (3, -5, 8) in the ratio:
(a) 2 : 3
(b) 1 : 2
(c) 2 : 5
(d) 3 : 4
Answer:
(a) 2 : 3

Question 2.
If a line makes an angle \(\frac { \pi }{ 4 } \) with both the positive direction of the x – axis, and y – axis, then the angle made by the line with positive direction of the Z – axis is:
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { \pi }{ 2 } \)
Answer:
(d) \(\frac { \pi }{ 2 } \)

Question 3.
Equation of the line passing through the points (2, 3, 4) and (1, -2, 3) are:
(a) \(\frac { x-2 }{ 1 } \) = \(\frac { y-2 }{ -5 } \) = \(\frac { z-4 }{ 1 } \)
(b) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ -1 } \)
(c) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ 5 } \) = \(\frac { z-4 }{ -1 } \)
(d) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ 1 } \)
Answer:
(b) \(\frac { x-2 }{ -1 } \) = \(\frac { y-3 }{ -5 } \) = \(\frac { z-4 }{ -1 } \)

Question 4.
Angle between the planes x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0 is:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { 3\pi }{ 2 } \)
(d) π
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 5.
Equation to the plane which cuts off intercepts 2, 3, -4 on the axis is:
(a) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 0
(b) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = -1
(c) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 1
(d) None of these.
Answer:
(c) \(\frac{x}{2}\) + \(\frac{y}{3}\) – \(\frac{z}{4}\) = 1

Question 2.
Fill in the blanks:

1. Direction cosine of unit vector \(\frac { 1 }{ \sqrt { 14 } } \) ( \(\hat { i } \) + 2\(\hat { j } \) + 3\(\hat { k } \) ) are ………………………….
2. Direction cosine of X – axis are ………………………….
3. Angle between the diagonals of cube is ………………………….
4. Angle between the straight lines \(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{-1}\) and \(\frac{x}{3}\) = \(\frac{y}{4}\)
   = \(\frac{z}{5}\) is ………………………….
5. If the lines \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ k } \) and \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ k } \) are coplaner, then k …………………………..
6. If the line makes an angle α, β, γ with the coordinate axis respectively, then cos²α + cos²β + cos² γ = ……………………………..
7. Intercept cut by the plane 2x + y – z = 5 on axis is ……………………………..

Answer:

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

Question 3.
Write True/False:

1. The points A (1, 2, 3), B (4, 0,4) and C (-2, 4, 2) are collinear.
2. The angle between lines whose direction ratios are (3,4, 5) and (4, -3, 5) is 30°.
3. Angle between the lines 2x = 3y = —z and 6x = -y = -4z is 90°.
4. Shortest distance between two intersecting lines always zero.
5. Angle between the lines \(\frac { x+1 }{ 3 } \) = \(\frac { y+1 }{ 2 } \) = \(\frac { z+2 }{ 4 } \) and plane 2x + y – 3z + 5 = 0 is cos^-1 (\(\frac { 4 }{ \sqrt { 406 } } \) )
6. Straight line \(\frac { x-2 }{ 1 } \) = \(\frac { y+1 }{ -2 } \) = \(\frac { z-4 }{ 1 } \) is parllel to the plane x + 3y + 5z = 4
7. Equation of plane parallel to the X – axis is ax + by + d = 0.

Answer:

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

Question 4.
Match the following:

Answer:

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

Answer:

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

Question 5.
Write the answer in one word/sentence:

1. Find the direction ratios of the normal of the plane x + 2y + 3z + 4 = 0.
2. Find the equation of the plane which intercepts unit length from the coordinate axis.
3. Find the equation of the plane which is perpendicular on FOZ-plane.
4. Find the angle between the planes x + 2y + z + 7 = 0 and 2x + y – z + 13 = 0.
5. Find the distance between the parallel planes 2x – 2y + z + 3= 0 and 4x – 4y + 2z.
6. Find the angle between the lines \(\frac{x}{2}\) = \(\frac{y}{-1}\) = \(\frac{z}{1}\) and \(\frac{x}{1}\) = \(\frac{y}{1}\) = \(\frac{z}{2}\)
7. If a line makes α, β, γ with the positive direction of the axis, then find the value of sin²α +sin² β + sin²γ.

Answer:

1. (1, 2, 3)
2. x + y + z = 1
3. by + cz + d = 0
4. \(\frac { \pi }{ 3 } \)
5. \(\frac{1}{6}\)
6. \(\frac { \pi }{ 3 } \)
7. 2

Three Dimensional Geometry Short Answer Type Questions

Question 1.
Find the direction cosine of the line joining two points (-2, 4, -5) and (1, 2, 3)? (NCERT)
Answer:
Given points A (- 2, 4, – 5) and B (1, 2, 3).
Direction ratio of AB = 1 + 2, 2 – 4, 3 + 5
= 3, -2, 8

Direction cosine of AB

Question 2.
Find direction cosines of the line which makes an angle 90°, 135° and 45° with X, Y and Z axis respectively? (NCERT)
Solution:
Given α = 90°, β = 135°, λ = 45°
cos α = cos 90° = 0
cos β = cos 135° = cos (180° – 45°)
= – cos 45° = – \(\frac { 1 }{ \sqrt { 2 } } \)
cos λ = cos 45° = \(\frac { 1 }{ \sqrt { 2 } } \)
Direction cosines are cos α, cos β, cos λ
i.e; 0, – \(\frac { 1 }{ \sqrt { 2 } } \), \(\frac { 1 }{ \sqrt { 2 } } \)

Question 3.
A line OP, makes the angle 120° and 60° with X – axis and Y – axis respectively find the angle made by line with Z – axis?
Solution:
Given α = 120°, β = 60°.
Let the angle made by the line with Z – axis be γ then
cos²α + cos²β + cos² γ = 1
⇒ cos² 120° + cos² 60° + cos² γ = 1

Hence, the required angle is 45° or 135°.

Question 4.
Show that the points (2, 3, 4), (-1, -2, 1) and (5, 8, 7) are collinear? (NCERT)
Solution:
Given Points are A (2, 3, 4), B (-1,-2, 1), C (5, 8, 7)
Direction ratio of AB x₂ – x₁, y₂ – y₁, z₂ – z₁
= – 1 – 2, – 2 – 3, 1 – 4
= – 3, – 5, – 3 = – (3, 5, 3)
Direction ratio of BC x₂ – x₁, y₂ – y₁, z₂ – z₁
= 5 + 1, 8 + 2, 7 – 1
= 6, 10, 6 = 2 (3, 5, 3)
It is clear that direction ratio of AB and BC are proportional, hence AB is parallel to BC but point B is common to both AB and BC therefore A, B, C are collinear. Proved.

Question 5.
If direction cosine of a line are cos α, cos β and cos γ then prove that:
cos 2α + cos 2β + cos 2γ = – 1.
Solution:
L.H.S. = cos2α + cos2β + cos2γ
= 2cos²α – 1 + 2cos²β – 1 + 2cos²γ – 1
= 2 (cos²α + cos²β + cos²γ) – 3
= 2 × 1 – 3, [∵ cos²α + ²β + cos²γ – 1]
= -1
= R.H.S. Proved.

Question 6.
Prove that the line passing through the points (1, -1, 2) and (3, 4, -2) are perpendicular to the lines passing through the points (0, 3, 2) and (3, 5, 6). (NCERT)
Solution:
Direction ratio’s of the line joining points (1, -1, 2) and (3, 4, -2)
= 3 – 1, 4 + 1, – 2 – 2
= 2, 5, – 4 = a₁, b₁, c₁ (let)
Direction ratios’s of the line joining the points (0, 3, 2) and (3, 5, 6)
= 3 – 0, 5 – 3, 6 – 2
= 3,2,4
= a₂, b₂, c₂ (let)
Lines will be perpendicular if
a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ 2.3 + 5.2 – 4.4 = 0
⇒ 16 – 16 = 0
⇒ 0 = 0. Proved.

Question 7.
Prove that the line joining the points A (1, 2, 3) and B (2, 3, 5) are parallel to then line joining the points C (-1, 2, -3) and D ( 1, 4, 1)?
Solution:
Direction ratio of line AB
= 2 – 1, 3 – 2, 5 – 3
= 1, 1, 2
= a₁, b₁, C₁
Direction ratio of line CD
= 1 + 1, 4 – 2, 1 + 3
= 2, 2, 4
= a₂, b₂, c₂
Where \(\frac { a_{ 1 } }{ a_{ 2 } } \) = \(\frac { b_{ 1 } }{ b_{ 2 } } \) = \(\frac { c_{ 1 } }{ c_{ 2 } } \)
= \(\frac{1}{2}\) = \(\frac{1}{2}\) = \(\frac{2}{4}\) or \(\frac{1}{2}\)
Hence line AB and CD are parallel. Proved.

Question 8.
Find the angle between the lines \(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\) and \(\frac{x – 5}{4}\) = \(\frac{y – 2}{1}\) = \(\frac{z – 3}{8}\)? (NCERT)
Solution:
Equation of the lines are
\(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\)
and \(\frac{x – 5}{4}\) = \(\frac{y – 2}{1}\) = \(\frac{z – 3}{8}\)
a₁ = 2, b₁ = 2, c₁ = 1
a₂ = 4, b₂ = 1, c₂ = 8
Let the angle between them = θ

Question 9.
Prove that lines
\(\frac{x – 5}{7}\) = \(\frac{y + 2}{- 5}\) = \(\frac{z}{1}\) and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) are perpendicular? (NCERT)
Solution:
\(\frac{x – 5}{7}\) = \(\frac{y + 2}{- 5}\) = \(\frac{z}{1}\)
and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\)
Where a₁ = 7, b₁ = -5, c₁ = 1
a₂ = 1, b₂ = 2, c₂ = 3
a₁a₂ + b₁b₂ + c₁c₂ = 7(1) + (- 5)(2) + 1 × 3
= 10 -10 = 0
Hence, the lines are perpendicular. Proved.

Question 10.
Find the cartesion equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by \(\frac { x+3 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+8 }{ 8 } \). (NCERT)
Solution:
Equation of given line
\(\frac { x+3 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+8 }{ 8 } \) ………. (1)
Direction ratio of (1) is 3, 5, 8
Equation of line passing through point (- 2, 4, -5)
\(\frac { x+2 }{ a } \) = \(\frac { y-4 }{ b } \) = \(\frac { z+5 }{ c } \) ……………….. (2)
Direction ratio of eqn. (2) is a, b, c
Eqns. (1) and (2) are parallel
∴\(\frac{a}{3}\) = \(\frac{b}{5}\) = \(\frac{c}{8}\) = k
a = 3k, b = 5k, c = 8k
Putting the value of a, b, c in eqn. (2) we get
\(\frac { x+2 }{ 3k } \) = \(\frac { y-4 }{ 5k } \) = \(\frac { z+5 }{ 8k } \)
\(\frac { x+2 }{ 3 } \) = \(\frac { y-4 }{ 5 } \) = \(\frac { z+5 }{ 8 } \).

Question 11.
Find the eqution of straight line passing through point (1, 2, 3) are parallel to \(\frac{x-6}{12}\) = \(\frac{y-2}{4}\) = \(\frac{z+7}{5}\)?
Solution:
Equation of line passing through point (x₁, y₁, z₁) and direction ratio a, b, c is

Question 12.
Find the angle between the lines \(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{3}\) and \(\frac{x}{4}\) = \(\frac{y}{5}\) = \(\frac{z}{0}\)?
Solution:
Equation of given lines
\(\frac{x}{1}\) = \(\frac{y}{0}\) = \(\frac{z}{3}\) …………………. (1)
and \(\frac{x}{4}\) = \(\frac{y}{5}\) = \(\frac{z}{0}\) ……………………… (2)
Where a₁ = 1, b₁ = 0, c₁ = 3 and a₂ = 4, b₂ = 5, c₂ = 0
Let θ be the angle between eqns. (1) and (2)

Question 13.
Find the condition that the lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ are perpendicular?
Solution:
x = ay + b and z = cy + 4
⇒ \(\frac { x-b }{ a } \) = \(\frac { y }{ 1 } \) and \(\frac { z-d }{ c } \) = \(\frac { y }{ 1 } \)
Hence the corresponding equation
\(\frac { x-b }{ a } \) = \(\frac { y }{ 1 } \) = \(\frac { z-d }{ c } \) ………………….. (1)
Again x = a’y + b’ and z = c’y + d’ equation is
\(\frac { x-b^{ ‘ } }{ a^{ ‘ } } \) = \(\frac { y }{ 1 } \) = \(\frac { z-d^{ ‘ } }{ c^{ ‘ } } \) ………………… (2)
Eqns. (1) and (2) are perpendicular.
a × a’ + 1 × 1 + c × c’ = 0
⇒ aa’ + cc’ + 1 = 0 is the required condition.

Question 14.
(A) Find the angle between the lines \(\vec { r } \) = ( \(\hat { i } \) + 3\(\hat { j } \) + 5\(\hat { k } \) ) + t ( \(\hat { i } \) – 2\(\hat { j } \) – 3\(\hat { k } \) ) and \(\vec { r } \) = ( \(\hat { i } \) – 2\(\hat { j } \) + 5\(\hat { k } \) ) + s(2\(\hat { i } \) – 2\(\hat { j } \) + \(\hat { k } \) )?
Solution:

(B) Find the angle between lines \(\vec { r } \) = (3\(\hat { i } \) + 4\(\hat { j } \) – 2\(\hat { k } \) ) + t(-\(\hat { i } \) + 2\(\hat { j } \) + \(\hat { k } \) ) and \(\vec { r } \) = ( \(\hat { i } \) – 7\(\hat { j } \) – 2\(\hat { k } \) ) + s( \(\hat { i } \) + 3\(\hat { j } \) + 2\(\hat { k } \) )
Solution:
Solve like Q. 14. (A)

Question 15.
Find angle between two planes 2x – y + z = 6 and x + y + 2z = 7?
Solution:
Given planes are:
2x – y + z = 6 ……………. (1)
and x + y + 2z = 7 ………………… (2)
Direction ratio of eqn. (1) = 2, -1, 1 1 ⇒ A₁, B₁, C₁
Direction ratio of eqn. (2) = 1, 1, 2 ⇒ A₂, B₂, C₂
Let θ be the angle between them

Question 16.
(A) If plane 3x – 6y – 2z = 7 and 2x + y – kz = 5 are perpendicular to each other. Find the value of k?
Solution:
Equation of given planes are
3x – 6y – 2z = 7 ………………………… (1)
and 2x + y – kz = 5 …………………… (2)
Where a₁ = 3, b₁ = -6, c₁ = -2, and a₂ = 2, b₂ = 1, c₂ = -k
Planes (1) and (2) are perpendicular
∴ a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ (3) (2) + (-6) (1) + (-2) (-k) = 0
⇒ 6 – 6 + 2k = 0
⇒ 2k = 0
⇒ k = 0.

(B) For which value of k the planes 2x + ky + z + 9 = 0 and 5x + 3y – 4z – 6 = 0 are perpendicular?
Solution:
Solve like Q.16 (A)
[Answer: k = -2]

(C) Prove that the planes x + 2y + 3z = 6 and 5x – 3y + z = 1 are perpendicular?
Solution:
Given equation of planes are:
x + 2y + 3z = 6 ……………………. (1)
and 3x – 3y + z = 1 ……………………… (2)
Direction ratio of eqn. (1) = 1, 2, 3
Direction ratio of eqn. (2) = 3,- 3, 1
∴ a₁a₂ + b₁b₂ + c₁c₂ = (1)(3) + 2(-3) + (3)(1)
= 3 – 6 + 3 = 0
Hence planes (1) and (2) are perpendicular. Proved.

Question 17.
A plane meets the co – ordinate axes at points A, B and C. If the centroid of ∆ABC is (2, – 1, 3) then find the equation of the plane?
Solution:
Let A(a, 0, 0), B(0, b, 0), C(0, 0, c)
∵ OA = a, OB = b, OC = c
Centroid (given) is (2,- 1, 3)

Question 18.
A plane meets the coordinate axes in A, B, C such that the centroid of ∆ABC is (a, b, c). Find the equation of the plane is \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 3?
Solution:

Question 19.
Find the equation of plane which is parallel to the plane 2x + 3y – z = 8 and passes through the point (1,  2, 3)?
Solution:
Any plane parallel to the plane
2x + 3y – z = 8
⇒ 2x + 3y – z + λ = 0 ………………….. (1)
∵ It passes through the point (1,2, 3), then
2(1) + 3(2) – 3 + λ = 0
⇒ 2 + 6 – 3 + λ = 0
⇒ 8 – 3 + λ = 0
⇒ 5 + λ = 0
⇒ λ = -5 ……………………… (2)
∴ From eqns. (1) and (2), we have
2x + 3y – 5 = 0.

Question 20.
Find the direction cosines of normal to the plane 2x + 4y + 4z = 9?
Solution:
Equation of given plane is:
2x + 4y + 4z = 9 …………………… (1)
The direction ratios of normal to the given plane are 2, 4, 4
Hence the direction cosines of the normal are

Question 21.
Find the equation of plane on which the length of perpendicular drawn from origin is 5 unit and the direction cosines of normal to it are \(\frac { 1 }{ \sqrt { 3 } } \), \(\frac { 1 }{ \sqrt { 3 } } \), – \(\frac { 1 }{ \sqrt { 3 } } \)?
Solution:
Equation of plane in normal form
lx + my + nz = p
Given: l = \(\frac { 1 }{ \sqrt { 3 } } \), m = \(\frac { 1 }{ \sqrt { 3 } } \), n = – \(\frac { 1 }{ \sqrt { 3 } } \), p = 5
Putting in eqn. (1),
\(\frac { 1 }{ \sqrt { 3 } } \) x + \(\frac { 1 }{ \sqrt { 3 } } \) y – \(\frac { 1 }{ \sqrt { 3 } } \) z = 5
⇒ x + y – z = 5\(\sqrt{3}\)

Question 22.
Find the equation of plane on which length of perpendicular drawn from origin is 4 units and whose direction cosines are in proportional to 2, -3, 6?
Solution:
Let the equation of plane
lx + my + nz = p
Given:
p = 4, a = 2, b = -3, c = 6

Putting in eqn. (1),
\(\frac{2}{7}\) x – \(\frac{3}{7}\) y + \(\frac{6}{7}\) z = 4
⇒ 2x – 3y + 6z = 28.

Question 23.
Find the equation of plane on which length of perpendicular drawn from origin is 5 unit and the direction ratios of normal to the plane are 2, 3, 6.
Solution:
Solve as Q. No. 22.
[Answer: 2x + 3y + 6z = 35]

Question 24.
Find the equation of the plane which passes through the point (1, – 2, 3) and perpendicular to line whose direction ratios are 2, 1, -1?
Solution:
We know that equation of the plane passing through a point (x₁, y₁, z₁) is
a (x – x₁) + b( y – y₁) + c (z – z₁) = 0
Where, a, b and c are the direction ratios of normal to the plane
Here the point (x₁, y₁, z₁) is (1, -2, 3)
and the direction ratios a, b, c are 2, 1, -1. Therefore by eqn. (1) we get
2. (x – 1) + 1. ( y + 2) -1. (z – 3) = 0
⇒ 2x + y – z – 2 + 2 + 3 = 0
⇒ 2x + y – z + 3 = 0
Which is the required equation of the plane.

Question 25.
(A) Find the perpendicular distance of the plane 6x – 3y + 2z – 14 = 0 from origin?
Solution:
Equation of plane,
6x – 3y + 2z – 14 = 0
The length of perpendicular drawn from (0, 0, 0) to the plane

(B) Find the length of perpendicular drawn from (1, 2, 0) on the plane 4x + 3y + 12z + 16 = 0?
Solution:
Equation of plane,
4x + 3y + 12z + 16 = 0
The length of perpendicular drawn from (1, 2, 0) to the plane

(C) Find the length of perpendicular drawn from (7, 14, 5) to the plane 2x + 4y – z?
Solution:
Solve as Q.No. 25 (B)
Answer:
3\(\sqrt{21}\) unit.

Question 26.
Find the equation of plane which passes through the point (- 1, 2, 3) and parallel to the plane 3x + 4y – 5z = 52?
Solution:
The given plane is
3x + 4y – 5z = 52 …………………… (1)
A plane parallel to the given plane (1),
3x + 4y – 5z = λ ………………(2)
∵ Plane (2) passes through the point (-1, 2, 3)
∴ 3(-1) + 4(2) – 5(3) = λ
⇒ -3 + 8 – 15 = λ
⇒ -18 – 8 = λ
⇒ -10 = λ
∴ From eqn. (2), putting the value of λ
3x + 4y – 5z = -10
∴ The required plane is
3x + 4y – 5z + 10 = 0

Question 27.
(A) Find the intercepts made by the plane 3x + 4y – 7z = 84 from the co – ordinate axes?
Solution:

(B) Find the intercepts made by the plane 3x + 4y – 6z = 72 from the co – ordinate axes?
Solution:
Solve as Q. No. 27 (A).
[Answer: (24, 18, -12).]

Question 28.
Find the equation of plane parallel to the X – axis and cuts intercepts 5 and 7 from Y and Z – axis respectively?
Solution:
Let the equation of plane
\(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
The plane (1) is parallel to X – axis
a = ∞
Given b = 5, c = 7
∴ \(\frac{x}{∞}\) + \(\frac{y}{5}\) + \(\frac{z}{7}\) = 1
⇒ \(\frac{y}{5}\) + \(\frac{z}{7}\) = 1 [∵\(\frac{x}{∞}\) = 0]
⇒ 7y + 5z = 35.

Question 29.
Find the equation of the plane passing through the points (0, 0, 0) and perpendicular to the planes x + 2y – z = 1 and 3x – 4y + z = – 5?
Solution:
Equation of the plane passing through point (0,0,0)
a(x – 0) + b(y – 0) + c(z – 0) = 0
⇒ ax + by + cz = 0 ………………………… (1)
Plane (1) is perpendicular to the given planes.
x + 2y – z = 1 and 3x – 4y + z = -5
a + 2b – c = 0 ………………….. (2)
3a – 4b + c = 0 ………………………….. (3)
Solving eqns. (2) and (3),
\(\frac { a }{ 2-4 } \) = \(\frac { -b }{ 1+3 } \) = \(\frac { c }{ -4-6 } \)
⇒ \(\frac { a }{ -2 } \) = \(\frac { -b }{ 4 } \) = \(\frac { c }{ -10 } \)
⇒ \(\frac{a}{1}\) = \(\frac{b}{2}\) = \(\frac{c}{5}\) = k (say)
∴ a = k, b = 2k and c = 5k, where k ≠ 0.
∴ From eqn. (1),
k [x + 2y + 5z] = 0
∴ x + 2y + 5z = 0.

Question 30.
Find the equation of the plane parallel to X – axis and passes through the points (2, 3, -4) and (1, -1, 3)?
Solution:
Let the equation of the plane parallel to X – axis is
by + cz + d= 0 ………………………. (1)
It passes through the points (2, 3, – 4) and (1, -1, 3), then
3b – 4c + d = 0 ……………………….. (2)
and – b + 3c + d = o …………………….. (3)
Solving eqns. (2) and (3), we get
\(\frac { b }{ -4-3 } \) = \(\frac { c }{ -1-3 } \) = \(\frac { d }{ 9-4 } \)
⇒ \(\frac{b}{7}\) = \(\frac{c}{4}\) = \(\frac{d}{-5}\) = k (say)
∴ b = 7k, c = 4k, d = -5k
Put the values of b, c and d in eqn. (1), we get
7ky + 4kz – 5k = 0
⇒ 7y + 4z – 5 = 0
This is the required equation of the plane.

Question 31.
Prove that the distance between two parallel planes 2x – 2y + z + 3 = 0 and 4x – 4y + 2z + 5 = 0 is \(\frac{1}{6}\)?
Solution:
The equations of the plane are 4x – 4y + 2z + 5 = 0
2x – 2y + z + 3 = 0 ……………… (1)
and 4x – 4y + 2z + 5 = 0 ………….. (2)
The length of the perpendicular drawn from O (0, 0, 0) to the plane (1) is

and the length of the perpendicular drawn from O (0, 0, 0) to the plane (2) is

∴ The required distance MN = P₁ – P₂
= 1 – \(\frac{5}{6}\) = \(\frac{6-5}{6}\) = \(\frac{1}{6}\). Proved.

Question 32.
(A) Find the equation of the plane passes through the intersecting line of the planes x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and also through the point (1, 1, 1)?
Solution:
The given planes are
x + y + z – 6 = 0 …………………. (1)
and 2x + 3y + 4z + 5 = 0 ………………….. (2)
The equation of the plane passes through the intersecting line of the planes (1) and (2), is
(x – y + z – 6) + λ (2x + 3y + 4z + 5) = 0 ………………….. (3)
Because plane (3) passes through the point (1, 1, 1), then
(1 + 1 + 1 – 6) + (2 + 3 + 4 + 5) = 0
⇒ – 3 + λ (14) = 0
⇒ λ = \(\frac{3}{14}\)
Putting the value of λ in eqn.(3),
(x + y + z – 6) + \(\frac{3}{14}\) (2x + 3y + 4z + 5) = 0
⇒ 20x + 23y + 26z – 69 = 0.

(B) Find the equation of the plane passing through the intersecting line of the planes x + 2y + 3z = 5 and 2x – 4y + z = 3 and also through the point (0, 1, 0)?
Solution:
Solve as Q.No. 32. (A).
[Answer: 3x – 2y + 4z + 2 = 0.]

Question 33.
(A) Find the vector equation of a plane passing through the point 2\(\hat { i } \) – \(\hat { j } \) + \(\hat { k } \) and perpendicular to the vector 6\(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \)?
Solution:
Let the equation of plane be

(B) Find the vector equation of the plane which is at a distance 7 units from origin and perpendicular to the vector 4\(\hat { i } \) + 2\(\hat { j } \) – 3\(\hat { k } \)?
Solution:
Let the equation of plane be

(C) Find the distance of the point (2, – 1, 3) from the plane \(\vec { r } \). (3\(\hat { i } \) + 2\(\hat { j } \) – 6\(\hat { k } \) ) + 15 = 0?
Solution:

(D) Find the distance from the point (2\(\hat { i } \) – \(\hat { j } \) – 4\(\hat { k } \) ) to the plane \(\vec { r } \). (3\(\hat { i } \) – 4\(\hat { j } \) + 12\(\hat { k } \) ) = 19?
Solution:

Question 34.
Find the angle between the planes \(\vec { r } \). (2\(\hat { i } \) – 3\(\hat { j } \) + 4\(\hat { k } \) ) = 1 and \(\vec { r } \) (- \(\hat { i } \) + \(\hat { j } \) ) = 4?
Solution:

Three Dimensional Geometry Long Answer Type Questions – I

Question 1.
Find the perpendicular distance of line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) from the point (1, 6, 3)?
Solution:

Question 2.
Find the distance between two parallel lines \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) and \(\frac { x-2 }{ 4 } \) = \(\frac { y-3 }{ 6 } \) = \(\frac { z-4 }{ 8 } \)?
Solution:
Given lines:
\(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) …………………… (1)
and \(\frac { x-2 }{ 4 } \) = \(\frac { y-3 }{ 6 } \) = \(\frac { z-4 }{ 8 } \) ………………………… (2)
Any point P (1, 2, 3) lies on line (1). Now we find the length of perpendicular PM drawn from point P (1, 2, 3) to eqn. (2).
∵Eqn. (2) passes through point A (2, 3,4)

Question 3.
Find the equation plane passing through the line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 3 } \) = \(\frac { z-4 }{ -1 } \) and point (-6, 3, 2)?
Solution:
Let the equation of plane be
A(x – α) + B(y – β) + C(z – γ) = 0
Passes through line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 3 } \) = \(\frac { z-4 }{ -1 } \)
∴ A(x – 3) + B(y + 2) + C(z – 4) = 0 ………………………… (1)
Plane passes through point (-6, 3, 2)
∴ A(-6 – 3) + B(3 + 2) + C(2 – 4) = 0
⇒ -9A + 5B – 2C = 0
⇒ 9A – 5B + 2C = 0 …………………………… (2)
Direction ratio of plane →A, B, C
2A + 3B – C = 0
Solving eqns. (2) and (3)
9A – 5B + 2C = 0
2A + 3B – C = 0
\(\frac { A }{ 5-6 } \) = \(\frac { B }{ 4+9 } \) = \(\frac { C }{ 27+10 } \)
⇒ \(\frac { A }{ -1 } \) = \(\frac { B }{ 13 } \) = \(\frac { C }{ 37 } \)
Putting in eqn. (1)
-1(x – 3) + 13(y + 2) + 37(z – 4) = 0
⇒ -x + 3 + 13y + 26 + 37z – 148 = 0
⇒ -x + 13y + 37z – 119 = 0
⇒ x – 13y – 37z + 119 = 0.

Question 4.
Find the shortest distance between the straight lines \(\frac { x-3 }{ 3 } \) = \(\frac { y-8 }{ -1 } \) = \(\frac { z-3 }{ 1 } \) and \(\frac { x
+3 }{ -3 } \) = \(\frac { y+7 }{ 2 } \) = \(\frac { z-6 }{ 4 } \)?
Solution:
Here, x₁ = 3, y₁ = 8, z₁ = 3, x₂ = -3, y₂ = -7, z₂ = 6
a₁ = 3, b₁ = -1, c₁ = 1, a₂ = -3, b₂ = 2, c₂ = 4
Given lines are
\(\frac { x-3 }{ 3 } \) = \(\frac { y-8 }{ -1 } \) = \(\frac { z-3 }{ 1 } \) and \(\frac { x+3 }{ -3 } \) = \(\frac { y+7 }{ 2 } \) = \(\frac { z-6 }{ 4 } \)
We know that the shortest distance between 2 lines \(\frac { x-x_{ 1 } }{ a_{ 1 } } \) = \(\frac { y-y_{ 1 } }{ b_{ 1 } } \) = \(\frac { z-z_{ 1 } }{ c_{ 1 } } \) and \(\frac { x-x_{ 2 } }{ a_{ 2 } } \) = \(\frac { y-y_{ 2 } }{ b_{ 2 } } \) = \(\frac { z-z_{ 2 } }{ c_{ 2 } } \) is

Question 5.
Find the coordinates of point of intersection of the lines \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) and \(\frac { x-2 }{ 3 } \) = \(\frac { y-3 }{ 4 } \) = \(\frac { z-4 }{ 5 } \)?
Solution:
Let \(\frac { x-1 }{ 2 } \) = \(\frac { y-2 }{ 3 } \) = \(\frac { z-3 }{ 4 } \) = r
∴ x = 2r + 1, y = 3r + 2, z = 4r + 3
Suppose the above point is intersection point then it will satisfy line

Put in eqn.(1), we get
x = 2(-1) + 1, y = 3(-1) + 2, z = 4(-1) + 3
x = -1, y = -1, z = -1
∴ Intersection point (-1, -1, -1).

Question 6.
Find the coordinates of point of intersection of the lines x – 3 = \(\frac { y+4 }{ -3 } \) = \(\frac { z-5 }{ 3 } \) and x – 4 = \(\frac { y-5 }{ 3 } \) = \(\frac { z+6 }{ -4 } \)?
Solution:
Given lines are \(\frac { x-3 }{ 1 } \) = \(\frac { y+4 }{ -3 } \) = \(\frac { z-5 }{ 3 } \) ……………….. (1)
and \(\frac { x-4 }{ 1 } \) = \(\frac { y-5 }{ 3 } \) = \(\frac { z+6 }{ -4 } \) ……………………… (2)
x₁ = 3, y₁ = -4, z₁ = 5, l₁ = 1, m₁ = -3, n₁ = 3
x₂ = 4, y₂ = 5, z₂ = -6, l₂ = 1, m₂ = 3, n₂ = -4
If lines (1) and (2) are intersect, then

⇒ 1(12 – 9) -1(-36 + 33) + 1(27 – 33) = 0
⇒ 3 + 3 – 6 = 0
⇒ 0 = 0
Hence the lines (1) and (2) are intersect
Again let \(\frac { x-3 }{ 1 } \) = \(\frac { y+4 }{ 3 } \) = \(\frac { z-5 }{ 3 } \) = r
∴ x = r + 3, y = -3r – 4, z = 3r + 5 ……………………………… (3)
Suppose above point is point of intersection hence it will satisfy eqn. (2)

Putting r = -1 in eqn. (3),
We get x = -1 + 3, y = -3(-1) – 4, z = 3(-1) + 5
x = 2, y = -1, z = 2
∴Intersection point (2, -1, 2).

Question 7.
Find the shortest distance between the lines:
\(\vec { r } \) = \(\hat { i } \) + 2\(\hat { j} \) + 3\(\hat { k } \) + t(2\(\hat { i } \) + 3\(\hat { j } \) + 4\(\hat { k } \) ) and \(\vec { r } \) = 2\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) + s(3\(\hat { i } \) + 4\(\hat { j } \) + 5\(\hat { k } \) )?
Solution:

Question 8.
Prove that the lines \(\vec { r } \) = ( \(\hat { i } \) + \(\hat { j } \) – \(\hat { k } \) ) + λ(3\(\hat { i } \) r =(3\(\hat { i } \) – \(\hat { j } \) ) and \(\vec { r } \)  = (4\(\hat { i } \) – \(\hat { k } \) ) + μ (2\(\hat { i } \) + 3\(\hat { k } \) are intersect to each other. Also And the point of intersection?
Solution:
Given lines are

Equating the above lines (1) and (2)

and two lines intersect each other, if


1 + 3λ = 4 + 2μ …………….. (3)
1 – λ = 0 ………………….. (4)
and -1 = 3μ – 1 ………………… (5)
From eqn. (4) we get
1 – λ = 0 ⇒ λ = 1
From eqn.(5) we get λ and μ in eqns. (1) and (2) respectively, we get
μ = 0
The above values of

Question 9.
(A) Find the shortest distance between the lines:
\(\vec { r } \) = (3 – t) \(\hat { i } \) + (4 + 2t) \(\hat { j } \) + (t – 2) \(\hat { k } \)
and \(\vec { r } \) = (1 + s) \(\hat { i } \) + (3s – 7) \(\hat { j } \) + (2s – 2) \(\hat { k } \)?
Solution:

(B) Find the shortest distance between the lines:
\(\vec { r } \) = (λ – 1) \(\hat { i } \) + (λ + 1) \(\hat { j } \) + (1 + λ) \(\hat { k } \)
\(\vec { r } \) = (1 – µ) \(\hat { i } \) + (2µ – 1) \(\hat { j } \) + (µ + 2) \(\hat { k } \)?
Solution:
Solve as Q.No. 9(A)
[Answer: S.D. = \(\frac { 5 }{ \sqrt { 2 } } \) ]

(C) Find the shortest distance between the lines whose vector equations are
\(\vec { r } \) = (1 + 2λ) \(\hat { i } \) + (2 + 3λ) \(\hat { j } \) + (3 + 4λ) \(\hat { k } \)
\(\vec { r } \) = (2 + 3µ) \(\hat { i } \) + (3 + 4µ) \(\hat { j } \) + (4 + 5µ) \(\hat { k } \)?
Solution:

Question 10.
Find the equation of plane passing through the line of intersection of the planes x + 3y + 4z – 5 = 0 and 3x – 4y + 9z – 10 = 0 and perpendicular to the plane x + 2y = 0?
Solution:
Given equation of planes are
x + 3y + 4z – 5 = 0 ………………….. (1)
and 3x – 4y + 9z – 10 = 0 ………………………. (2)
The plane passing through the line of intersection of the planes (1) and (2)
(x + 3y + 4z – 5) + λ(3x – 4y + 9z – 10) = 0
⇒ (1 + 3λ) x + (3 – 4λ) y + (4 + 9λ) z – 5 – 10λ = 0 ………………… (3)
Again, given plane is
x + 2y = 0 ……………………… (4)
Plane (3) is perpendicular to the plane (4)
(1 + 3λ).1 + (3 – 4λ).2 + (4 + 9λ). 0 = 0
⇒ 1 + 3λ + 6 – 8λ = 0
⇒ λ = \(\frac{7}{5}\)
Putting the value of X in eqn. (3),
(x + 3y + 4z – 5) + \(\frac{7}{5}\) (3x – 4y + 9z – 10) = 0
⇒ 26x – 13y + 83z = 95.

Question 11.
Find the equation of planes which are parallel to the plane x – 2y + 2z = 3 and whose perpendicular distance from the point (1, 2, 3) is 1?
Solution:
The planes parallel to the given plane x – 2y + 2z = 3 is
x – 2y + 2z + λ = 0 ………………….. (1)
Perpendicular distance from the point (1, 2, 3) to the above plane is

Then substitute the value of 2 in eqn. (1), we have
x – 2y + 2z = 0, x – 2y + 2z = 6

Question 12.
(A) Find the equation of the plane passing through the points (1,-2, 4) and (3, -4, 5) and perpendicular to the plane x + y – 2z = 6?
Solution:
Equation of the plane passing through the point (1, -2, 4) is
A(x – 1) + B(y + 2) + C(z – 4) = 0 ………………………. (1)
It also passes through the point (3, -4, 5)
A(3 – 1) + 5(-4 + 2) + C(5 – 4) = 0
⇒ 2A – 2B + C = 0 …………………………. (2)
The given equation of plane is
x + y – 2z = 6 ……………………. (3)
The planes (1) and (3) are perpendicular
A + B – 2C = 0 ……………………. (4)
On solving by cross multiplication method, we get
\(\frac { A }{ 4-1 } \) = \(\frac { B }{ 1+4 } \) = \(\frac { C }{ 2+2 } \)
⇒ \(\frac{A}{3}\) = \(\frac{B}{5}\) = \(\frac{C}{4}\) = k
∴ A = 3k, B = 5k, C = 4k
Putting these values of A, B and C in eqn. (1), we get
⇒ 3k(x – 1) + 5k(y + 2) + 4k(z – 4) = 0
⇒ 3x – 3 + 5y + 10 + 4z – 16 = 0
⇒ 3x + 5y + 4z – 9 = 0.

(B) Find the equation of plane passing through the points (1, 1,-1) and (1, 1,-1) and perpendicular to the planes x + 2y + 2z = 9?
Solution:
Solve as Q.No. 12.(A)
[Answer: 2x + 2y – 3z + 3 = 0]

Question 13.
(A) Find the equation of plane passing through (1, 1, -1) and perpendicular to the planes x + 2y + 3z = 0 and 2x – 3y + 4z = 0?
Solution:
Let the equation of planes
A(x – x₁) + B(y – y₁) + C(z – z₁) = 0
It passes through (1, 1, -1)
A(x – 1) + B(y – 1) + C(z + 1) = 0 …………………. (1)
Given planes are
x + 2y + 3z = 0 …………………… (2)
2x – 3y + 4z = 0 ………………………. (3)
The plane (1) is perpendicular to eqns. (2) and (3),
∴ 1A + 2B + 3C = 0
and 2A – 3B + 4C = 0
Solving \(\frac { A }{ 8+9 } \) = \(\frac { B }{ 6-4 } \) = \(\frac { C }{ -3-4 } \)
⇒ \(\frac{A}{17}\) = \(\frac{B}{2}\) = \(\frac{C}{-7}\) = k
∴ A = 17k, B = 2k, C = -7k
By eqn. (1),
17k (x – 1) + 2k (y – 1) – 7k (z + 1) = 0
⇒ 17x + 2y – 7z – 26 = 0.

(B) Find the equation of plane passing through (2, 1, 4) and perpendicular to the planes 9x – 7y + 6z + 48 = 0 and x + y – z = 0?
Solution:
Solve as Q.No. 13 (A)
[Answer: x + 15y + 16z = 81]

Question 14.
Find the equation of the plane passing through the points (2, 2, -1), (3, 4, 2) and (7, 0, 6)?
Solution:
Equation of Plane passing through the points (2, 2, -1) is
A(x – 2) + B(y – 2) + C(z + 1) = 0 ………………….. (1)
Since plane given by eqn. (1) passes through the points (3,4, 2) and (7, 0, 6)
Hence it will satisfy eqn. (2),
We get A(3 – 2) + B(4 – 2) + C(2 + 1) = 0
A + 2B + 3C = 0 ………………………. (2)
and A(7 – 2) + B(0 – 2) + C(6 + 1) = 0
5A – 2B + 7C = 0 …………………….. (3)
Solving eqns. (2) and (3),
img 39
\(\frac { A }{ 14+6 } \) = \(\frac { B }{ 15-7 } \) = \(\frac { C }{ -2-10 } \)
\(\frac { A }{ 20 } \) = \(\frac { B }{ 15-7 } \) = \(\frac { C }{ -2-10 } \)
\(\frac { A }{ 20 } \) = \(\frac { B }{ 8 } \) = \(\frac { C }{ -12 } \)
Let \(\frac { A }{ 5 } \) = \(\frac { B }{ 2 } \) = \(\frac { C }{ -3 } \) = k
A = 5k, B = 2k, C = -3k
Let \(\frac{A}{5}\) = \(\frac{B}{2}\) = \(\frac{C}{-3}\) = k
A = 5k, B = 2k, C = -3k
Putting in eqn.(1),
We get 5k(x – 2) + 2k(y – 2) + (-3k) (z + 1) = 0
k (5x + 2y – 3z – 17) = 0
5x + 2y – 3z – 17 = 0.

Question 15.
prove that the points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar?
Solution:
Equation of plane passing through the points (0, -1, -1)
A(x – 0) + B(y + l) + C(z + 1) = 0 ………………………… (1)
It passes through the point (4, 5, 1)
A(4 – 0) + B(5 +1) + C(1 +1) = 0
⇒ 4A + 6B + 2C = 0
⇒ 2A + 3B + C = 0 ………………………………. (2)
Eqn. (1) also passes through (3, 9, 4),
A(3 – 0) + B(9 + 1) + C(4 + 1) = 0
⇒ 3A + 10B + 5C = 0 …………………………. (3)
Solving eqns. (2) and (3),
\(\frac { A }{ 15-10 } \) = \(\frac { B }{ 3-10 } \) = \(\frac { C }{ 20-9 } \) = k (let)
⇒\(\frac { A }{ 5 } \) = \(\frac { B }{ -7 } \) = \(\frac { C }{ 11 } \) = k
⇒ A = 5k, B = -7k, C = 11k
Putting in eqn. (1),
5k – 7k (y + 1) + 11k (z + 1) = 0
⇒ k(5x – 7y + 11z + 4) = 0
⇒ 5x – 7y + 11z + 4 = 0 ………………………… (4)
If the plane also passes through the point (-4, 4, 4) will satisfies eqn. (4),
5(- 4) – 7(4) + 11 (4) + 4 = 0
⇒ -20 – 28 + 44 + 4 = 0
⇒ 0 = 0
Hence given points are coplanar. Proved.

Question 16.
A variable plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1 is at a distance 1 unit from origin. It cuts co – ordinate axes at A, B, C. The centroid (x, y, z) satisfies the equation \(\frac { 1 }{ x^{ 2 } } \) + \(\frac { 1 }{ y^{ 2 } } \) + \(\frac { 1 }{ z^{ 2 } } \) = k? Find the value of k?
Solution:
Given equation of plane \(\frac{x}{a}\) + \(\frac{y}{b}\) + \(\frac{z}{c}\) = 1
OA = a, OB = b, OC = c
The coordinates of the point A, B, C are (a, 0, 0), (0, b, 0) and (0, 0, c).
The length of perpendicular drawn from origin to the plane (1) is 1.

Question 17.
Find the equation of plane passing through the line of intersection of the planes x + 2y + 3z = 4 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0?
Solution:
Given planes
x + 2y + 3z = 4 ………………….. (1)
2x + y – z + 5 = 0 ………………… (2)
Equation of plane passing through the line of intersection of the planes (1) and (2)
x + 2y + 3z – 4 + λ(2x + y – z + 5) = 0 …………………. (3)
⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z = 0
⇒ 10λ + 3λ – 6λ + 5 + 6 + 18 = 0
⇒ 7λ + 29 = 0
⇒ λ = \(\frac{-29}{7}\)
Putting the value of A in eqn. (3),
x + 2y + 3z – 4 – \(\frac{-29}{7}\) (2x + y – z + 5) = 0
⇒ 7x + 14y + 21z – 28 – 58x – 29y + 29z – 145 = 0
⇒ -51x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0.

Question 18.
Find the equation of plane passes through the line \(\frac { x-3 }{ 2 } \) = \(\frac { y+2 }{ 9 } \) = \(\frac { z-4 }{ -1 } \) and point (-6, 3, 2)?
Solution:
Point on given line D (3, -2,4)
∴ Eqn. of plane passing through point (3, -2, 4) is
A(x – 3) + B(y + 2) + C(z – 4) = 0 ……………………….. (1)
Plane (1) passes through point (-6, 3, 2),
– 9A + 5B – 2C = 0 …………………………. (2)
Direction ratio’s of given line is 2, 9, -1 and is parallel to plane (1),
2A + 9B – C = 0 ………………………. (3)
Now,
– 9A + 5B – 2C = 0
2A + 9B – C = 0
\(\frac { A }{ -5+18 } \) = \(\frac { B }{ -4-9 } \) = \(\frac { C }{ -81-10 } \)
⇒ \(\frac { A }{ 13 } \) = \(\frac { B }{ -13 } \) = \(\frac { C }{ -91 } \) = k
⇒ \(\frac { A }{ 1 } \) = \(\frac { B }{ -1 } \) = \(\frac { C }{ -7 } \) = k
A = k, B = -k, C = – 7k
Putting in eqn.(1), we get,
k(x – 3) – k (y + 2) – 7k(z – 4) = 0
⇒ x – y – 7z – 3 – 2 + 28 = 0
⇒ x – y – 7z + 23 = 0.

Question 19.
Find the vector equation of the line passing through (1,2,3) and parallel to the planes \(\vec { r } \). ( \(\hat { i } \) – \(\hat { j } \) + 2\(\hat { k } \) = 5 and \(\vec { r } \).(3\(\hat { i } \) + \(\hat { j } \) + \(\hat { k } \) ) = 6? (NCERT)
Solution:
Equation of line is:

Question 20.
Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines
\(\frac { x-8 }{ 3 } \) = \(\frac { y+19 }{ -16 } \) = \(\frac { z-10 }{ 7 } \) and \(\frac { x-8 }{ 3 } \) = \(\frac { y-20 }{ 8 } \) = \(\frac { z-5 }{ -5 } \)? (NCERT)
Solution:

Question 21.
A line makes angles α, β, γ and and with four diagonals of a cube then prove that:
cos²α + cos²β + cos² = \(\frac{4}{3}\)
Solution:
Taking three adjacent edges OA, OB, OC of the cube of side a as coordinate the coordinates of vertices of the cube are :
0 (0,0,0), A (a,0,0), B (0,a,0), R (0,0,a),
D (a,a,0), K (a,0,a), L (0,a,a), P (a,a,a)
Direction ratio of diagonal are a – 0, a – 0, a – 0, a, a, a direction cosines of OP are:

MP Board Class 12 Maths Important Questions

MP Board Class 12th Biology Important Questions Chapter 14 Ecosystem

Ecosystem Important Questions

Ecosystem Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
The flow of energy in ecosystem: (MP2011)
(a) Unidirection
(b) Bidirectional
(c) Tridirectional
(d) Four directional
Answer:
(a) Unidirection

Question 2.
Chief source of energy in ecosystem :
(a) Solar energy
(b) Green plants
(c) Food substances
(d) All of these.
Answer:
(a) Solar energy

Question 3.
The term ‘ecosystem’ was used for the first time by : (MP 2015)
(a) Tansley
(b) Odum
(c) Reiter
(d) Mishra and Puri.
Answer:
(a) Tansley

Question 4.
The pyramid of number of tree ecosystem is :
(a) Inverted
(b) Upright
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Inverted

Question 5.
Correct food chain is :
(a) Grass → Grasshopper → Frog → Snake → Hawk
(b) Grass → Frog → Snake → Peacock
(c) Grass → Peacock → Grasshopper → Hawk
(d) Grass → Snake → Rabbit.
Answer:
(a) Grass → Grasshopper → Frog → Snake → Hawk

Question 6.
Pyramid of biomass of lake ecosystem is :
(a) Upright
(b) Inverted
(c) Upright or inverted
(d) None of these.
Answer:
(b) Inverted

Question 7.
Vegetation present between two communities :
(a) Ecad
(b) Ecotype
(c) Ecotone
(d) None of these.
Answer:
(c) Ecotone

Question 8.
Xerosere is started from :
(a) Water
(b) Naked rock
(c) Swamp
(d) All of these.
(b) Naked rock

Question 9.
Serial development of plant community is called :
(a) Attack
(b) Succession
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Succession

Question 10.
Plants starting succession in any area are called :
(a) Pioneer
(b) Sere
(c) Displacer
(d) All of these.
Answer:
(a) Pioneer

Question 11.
In forest ecosystem, the pyramid of energy is : (MP 2012,17)
(a) Always inverted
(b) Always upright
(c) First upright then inverted
(d) None of these.
Answer:
(b) Always upright

Question 12.
The study of ecology of a species is called :
(a) Ecology
(b) Autecology
(c) Synecology
(d) None of these.
Answer:
(b) Autecology

Question 13.
Food chain starts from :
(a) Respiration
(b) Photosynthesis
(c) Decomposers
(d) Nitrogen fixation.
Answer:
(b) Photosynthesis

Question 14.
Man is:
(a) Autotrophic
(b) Carnivorous
(c) Herbivorous
(d) Omnivorous.
Answer:
(d) Omnivorous.

Question 15.
In any ecosystem, solar energy is conserved by :
(a) By producers
(b) By consumers
(c) By decomposers
(d) All of the above.
Answer:
(a) By producers

Question 16.
The two components of ecosystem are: (MP 2013)
(a) Organism and Plant
(b) Weeds and trees
(c) Frog and man
(d) Biotic and Abiotic.
Answer:
(d) Biotic and Abiotic.

Question 2.
Fill in the blanks:

1. The transitional zone present between to adjacent communities is called ………………….
2. All the plants of a particular area constitute …………………. of that place.
3. Only …………………. % energy is transferred from one trophic level to another.
4. Pyramid of …………………. is always upright. (MP 2013)
5.  …………………. and …………………. are the examples of omnivorous animals.
6. The term ‘ecosystem’ was proposed by ………………….
7. The chief source of energy ecosystems is (MP 2016)
8. N₂ fixing bacteria is called …………………. (MP 2009 Set A)
9. All ecosystems are depended for energy on …………………. (MP 2009 Set D)
10. In forest trees basically function as ………………….
11. Establishment of organisms in new habitat is called ………………….
12. A sereal development of plant community is called ………………….
13. Succession which takes place in baren rock is called ………………….
14. Succession in soil is called ………………….

Answer:

1. Ecotone
2. Flora
3. 10
4. Energy
5. Man, Pig
6. Tansley
7. Sunlight
8. Nitrifying bacteria
9. Sun (solar energy)
10. Producers
11. Ecesis
12. Succession
13. Lithosere
14. Psamosere.

Question 3.
Match the followings:
I.

Answer:

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

II.

Answer:

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

Question 4.
Write the answer in one word/sentances:

1. What is called the study of ecology of a species?
2. What is called the process of soil formation?
3. What type of energy pyramid found in nature?
4. Give one example of an omnivorous animal.
5. Who proposed the term ecosystem?
6. Who gave the 10% rule of ecosystem?
7. In an ecosystem, producer is known as.
8. What is the direction of the flow of energy in an ecosystem?
9. When many food chains operate simultaneously and interlock such patterns is termed as?
10. Give an example of autotrophic componant.
11. Which gas is used by green plants in photosynthesis?
12. Name the special nutrient which is present in metheonine amino acid.
13. Name the cycle in which nitrogen is converted into multiple chemicals and circulated among atmosphere.
14. Due to an ecosystem fungus and bacteria are known as.
15. What is T₂?

Answer:

1. Autecology
2. Pedogenesis
3. Upright
4. Man
5. A.G. Tansley
6. Lindman
7. Biotic componant
8. Leanear
9. Food web
10. Plants
11. CO₂
12. Sulfer
13. Nitrogen cycle
14. Decomposer
15. Herbivorous animal.

Ecosystem Very Short Answer Type Questions

Question 1.
Write the name and ratio of different components of biosphere.
Answer:
Name and ratio of different components of biosphere is:

Some gases are also found in biosphere example Helium, Neyon and Crypton are found in less quantity.

Question 2.
Differentiate between detritivore and decomposer.
Answer:
Detritivore are organisms which feed on detritus and break them into smaller particles, example earth worm. And decomposer are organisms which by secreting enzymes break down complex organic matter into in organic substance example some bacteria and fungi.

Question 3.
Explain consumers of ecosystem
Answer:

1. Producers – All the green plants.

2. Consumers – Depends on others for food.

• Primary consumer: Depends on plants called herbivores.
• Secondary consumers : Depends on herbivores for food.
• Tertiary consumers : Depends on secondary consumers.

3. Decomposers – They decomposed dead organic matter.

Question 4.
Differentiate between biome and ecosystem.
Answer:
An ecosystem is the interaction of living and non – living things in an environment. A biome is a specific geographic area notable for the species living there.

Question 5.
What are transducers according to some ecologists?
Answer:
Some ecologists call green plants as transducers of the ecosystem as they convert solar energy into chemical energy.

Question 6.
Name four submerged plants.
Answer:

1. Hydrilla
2. Vallisneria
3. Elodea
4. Potamogeton.

Question 7.
Name the stages of xerosere.
Answer:
Stages of xerosere:

1. Crustose lichen stage
2. Foliose lichen stage
3. Moss stage
4. Herb stage
5. Shrub stage
6. Climax forest stage.

Question 8.
When many food chain operate simultaneously and interlock such pattern is formed.
Answer:
Food web.

Question 9.
Name the ecosystem which shows most productivity.
Answer:
Tropical ecology.

Question 10.
What are fungi and bacteria called in an ecosystem?
Answer:
Micro – consumer or Decomposer.

Question 11.
Which part of the energy is transferred from one trophic level to other in ecosystem?
Answer:
10%.

Question 12.
Name the type of chemosynthetic bacteria.
Answer:
Autotrophic.

Question 13.
Name the word which is similar to ecosystem given by Prof. R. Mishra.
Answer:
Ecocosm.

Question 14.
Name the trophic level in which green plants are found.
Answer:
Primary trophic level.

Question 15.
Who gave the word transformer for producer?
Answer:
E.J. Kormondy.

Question 16.
Name any two sedimentary cycle.
Answer:

Phosphorus cycle
Sulphur cycle.

Question 17.
The energy pyramids are always.
Answer:
Upright.

Question 18.
Give examples of decomposers.
Answer:
Bacteria and Fungi.

Question 19.
Who gave 10% rule of energy?
Answer:
Lindeman.

Question 20.
Which form of nitrogen is absorbed by plants?
Answer:
In the form of nitrate ion (NO₃^–)

Ecosystem Short Answer Type Questions

Question 1.
Draw a pyramid of energy of grassland ecosystem.

Answer:

Question 2.
Explain nitrogen cycle in nature.

Answer:

Question 3.
Explain sulphur cycle by diagram.
Answer:

Question 4.
Explain the effect of light on plants.
Answer:
Effect of light on plants:
Light is the source of energy. It is essential for life. It is an important factor of an ecosystem. The existence of life on earth is because of light obtained from sun. Sunlight is essential for photosynthesis, help in preparation of food for the whole living world. Light effects biological activities of plants by its intensity, period and duration. Plants are classified into following two categories on the basis of requirement of light intensity:

1. Heliophytes – Plants, which can grow better in bright light are called heliophytes.
2. Sciophytes – Plants, which require relatively less of light and they can grow better in shades are called sciophytes.

Question 5.
Explain the meaning of food web and draw its diagram.
Answer:
Food Web:
In nature, foodchains are not isolated sequences, but are interrelated and interconnected with one another. When many foodchains operate simultaneously and interlock such pattern is termed as food web. Thus, the food web is a description of feeding connections between the organisms which make up a community. Energy passes through one trophic level to next via these food web links, example a rat feeds on various kinds of grains, fruits, stems, roots, etc.

A rat in its turn is consumed by a snake which is eaten by a falcon. The snakes feed on both frogs and rats. Thus, a network of food – chains exists and this is called food web. The food web gets more complicated because of variability in taste and preference availability and compulsion and several other factors at each level. For example, tigers normally do not feed on fishes or crabs but in Sunderbans they are forced to eat them.

Question 6.
Explain calcium cycle with well labelled diagram.
Answer:

Question 7.
Draw ecological pyramid of number of a tree ecosystem and grassland ecosystem.
Answer:

Question 8.
What do you mean by ecosystem? Describe the important components of a pond ecosystem.
Or
Write about role of decomposers in an ecosystem with example.
Answer:
Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Components of a pond ecosystem:
A pond ecosystem should contain all components of ecosystems like:

1. Producers:
Organism, which can synthesize their own food are included under producers, example Volvox, Pandorina, Oedogonium, Saggitaria, Utricularia, Azolla, Trapa, Lemna, Typha, Nymphaea etc. form the producer class of the pond ecosystem.

2. Consumers:

• Primary consumer – Animals, which feed on producers are included into this category example Daphnia, Cyclops, Paramoecium, Amoeba and small fishes.
• Secondary consumers – Primary consumers also serve as food for water snakes, a few tortoise, few types of fish etc. hence, these are carnivores.
• Tertiary consumers – Secondary consumers also serve as food for aquatic birds like kingfisher, cranes, big fish and these together form a top class carnivorous group and called as tertiary consumers.

3. Decomposers:
All producers and consumers die and accumulate on the floor of the pond. Even the waste material and faeces of these animals get accumulated on the floor of the pond. Similarly, the floor of pond is also occupied by decomposers, which include bacteria and fungi. These decomposers decompose complex organic compounds of their bodies into simpler forms which are finally mixed with soil of floor of ponds. These are again absorbed by the roots of producer plants and thus matter is recycled.

Question 9.
Explain pyramid of biomass of pond ecosystem.
Answer:
The biomass, i.e., the living weight of the organisms in the foodchain present at different trophic levels in an ecosystem forms the pyramid of biomass. When biomass of consumers is greater than biomass of producer then pyramid is called as inverted pyramid of biomass. example pyramid of biomass of pond ecosysyem is always inverted.

Ecosystem:
The system resulting from the interaction between organisms and their environment is called as ecosystem.

Question 10.
Distinguish between:

1. Grazing food chain and Detritus food chain
2. Production and Decomposition
3. Upright and Inverted pyramid
4. Food chain and Food web
5. Litter and Detritus,
6. Primary and Secondary productivity

Answer:
1. Differences between Grazing food chain and Detritus food chain :

Grazing food chain:

• Energy for the food chain comes from the sun.
• First trophic level organisms are producers.

Detritus food chain:

• Energy comes from detritus (organic matter).
• First trophic level organisms are detritivores and decomposers.

2. Differences between Production and Decomposition :

Production:

• It refers to the process of synthesis of organic compounds from inorganic substances utilising sunlight.
• Example: Plants perform the function of production of food.

Decomposition:

• It is the phenomenon of degradation of waste biomass.
• Example : Bacteria and fungi decompose dead organic matter.

3. Differences between Upright pyramid and Inverted pyramid:
Upright pyramid:
When the number of producers or theirbiomass is maximum in an ecosystem and it decreases progressively at each trophic level in a food chain, an uprightpyramid is formed.

Inverted pyramid:
When the number of individuals or their biomass at the producer level is minimum and it Increases progressively at each trophic level in a food chain, an inverted pyramid is formed.

4. Differences between Food chain and Food web:

Food chain:

• A food chain is a single pathway where energy is transferred from producers to successive orders of consumers.
• All food chains start with green plants which are the original source of all food.
• Energy flow is unidirectional.

Food web:

• A food web is a network of various food chains which are interconnected with each other like an interlocking pattern.
• It has many linkages and intercrosscs among producers and consumers.
• Energy flow in multidirectional.

5. Differences between Litter and Detritus:
Litter:
The dead remains of plants (leaves, flowers etc.) and animals excreta which falls on the surface of the earth in terrestrial ecosystems is called litter.

Detritus:
The dead remains of plants and animals constitute detritus. It is differentiated into litter fall (above ground detritus) and below ground detritus.

6. Differences between Primary and Secondary productivity:

Primary productivity:

• It is the rate at which organic matter is built up by producers.
• It is due to photosynthesis.
• Primary productivity is two types :
  Gross Primary Productivity (GPP) and Net Primary Productivity (NPP) GPP – R = NPP (R = loss in Respiration)

Secondary productivity:

• It is the rate of synthesis of organic matter by consumers.
• It is due to herbivory and predation.
• Secondary productivity is two types :
  Gross Secondary Productivity (GSP) and Net Secondary Productivity (NSP) NSP = GSP – R (Loss in Respiration)

Question 11.
What is primary productivity ? Give brief description of factors that affect primary productivity.
Answer:
The rate of biomass production is called productivity.
It is expressed in terms of g^-2yr^-1 or (kcal – m^-2) yr^-1 to compare the productivity of ecosystems. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).

Gross primary productivity of an ecosystem is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is utilized by plants in respiration. Gross primary productivity minus respiration losses (R), is the Net Primary Productivity (NPP). GPP – R = NPP

Primary productivity depends on:

• The plant species inhabiting a particular area.
• The environmental factors.
• Availability of nutrients.
• Photosynthetic capacity of plants.

Question 12.
Define decomposition and describe the processes and products of decomposition.
Answer:
Decomposition:
Decomposition is the process that involves the breakdown of complex organic matter or biomass from the body of dead plants and animals with the help of decomposers into inorganic raw materials such as carbon dioxide, water, and other nutrients.

The various processes involved in decomposition are as follows :

1. Fragmentation:
It is the first step in the process of decomposition. It involves the breakdown of detritus into smaller pieces by the action of detritivores such as earthworms.

2.Leaching:
It is a process where the water soluble nutrients go down into the soil layers and get locked as unavailable salts.

3. Catabolism:
It is a process in which bacteria and fungi degrade detritus through various enzymes into smaller pieces.

4. Humification:
The next step is humification which leads to the formation of a dark coloured colloidal substance called humus, which acts as reservoir of nutrients for plants.

5. Mineralization:
The humus is further degraded by the action of microbes, which finally leads to the release of inorganic nutrients into the soil. This process of releasing inorganic nutrients from the humus is known as mineralization. Decomposition produces a dark coloured, nutrient rich substance called humus. Humus finally degrades and releases inorganic raw materials such as CO₂, water, and other nutrient in the soil.

Question 13.
Given account of explain energy flow in ecosystem.
Answer:
Energy flow:
In the ecosystem, energy is transferred in an orderly sequence. The flow of solar energy from producers to consumers and to decomposers subsequently in an ecosystem is known as energy flow. Energy flow is an ecosystem is always unidirectional. Sun is the sole source of solar energy in an ecosystem. Green plants utilize this energy in photosynthesis and convert it in the form of chemical energy and store it.

Plants utilize maximum part of this energy to do its biological functions. Some of it is converted into heat and released in the environment. Remaining part of the energy is stored in various components of the body. When a consumer eats these producer plants, the energy is then transferred into its body.

In any food chain energy flows from primary producers to primary consumers, from primary consumers to secondary consumers and secondary consumers to tertiary consumers and so on. Because every organism of a trophic level continuously converts chemical energy into heat, there is always a loss of energy with each step in a food – chain. According to an estimate only 10% of the total energy obtained is transferred from one trophic level to another.

Question 14.
Write important features of a sedimentary cycle in an ecosystem.
Answer:
Sedimentary cycles have their reservoirs in the earth’s crust or rocks. Nutrient elements are found in the sediments of the earth. Elements such as sulphur, phosphorus, potassium, and calcium have sedimentary cycles. Sedimentary cycles are very slow. They take a long – time to complete their circulation and are considered as less perfect cycles. This is because during recycling, nutrient elements may get locked in the reservoir pool, thereby taking a very long – time to come out and continue circulation. Thus, it usually goes out of circulation for a long – time.

Ecosystem Long Answer Type Questions

Question 1.
Describe various components of ecosystem.
Answer:
Ecosystem has following components:

1. Abiotic components:
The physical conditions of the ecosystem depends upon latitude, overall climatic and edaphic factors. The physical deficiencies are overcome by artificial irrigation and use of fertilizers. Thus, like natural ecosystem all organic, inorganic substances and climatic factors together forms abiotic component of the ecosystem.

• Organic substances : Such as carbohydrates, proteins, lipids, etc.
• Inorganic substances : Such as C, H, N, P, K, Ca, I, etc.
• Climatic factors : Such as temperature, light, water, humidity, wind, pH, minerals, soil structure, etc.

2. Biotic factors:
Three types are there

(a) Producers :
This type of crop (dominant species) depends upon the climate, season and the choice of the farmer.

(b) Consumers:

• Primary consumers : Insects, beetles, fishes, etc.
• Secondary consumers : Frog and fishes.
• Tertiary consumers : Snakes and cranes.

(c) Decomposers:
Bacteria, fungi etc.

Question 2.
Explain consumer components ecosystem of a pond in brief.
Answer:
Consumers:
They feed on producers directly or indirectly. It is of following categories:

1. Primary consumers:
Varied forms of zooplanktons are found in the water surface. Most of them are unicellular protists, such as Amoeba, Paramoecium whereas some are multicellular crustaceans, such as Daphnia, Cyclops, etc. Animals which are found under the surface of water are called as benthos, such as many types of fishes, crustaceans, molluscs, insects, beetles, etc. They too feed on producers.

2. Secondary consumers:
They feed on primary consumers, example big fishes, water snakes, etc.

3. Tertiary consumers:
They feed on secondary consumers, example kingfisher, cranes, omnivorous man, etc.

Question 3.
Describe carbon cycle in an ecosystem.
Answer:
Carbon Cycle:
Importance of carbon:
Carbon is considered as the basis of life. Carbon is the most important constituent of proteins, fats and nucleic acids which form the essential constituents of protoplasm.

Sources of carbon – The three sources of carbon in non – living world are:

• The carbon dioxide of the air and that which is dissolved in water.
• The rocks of the earth crust containing carbonates.
• The fossil fuel like coal and petroleum.

Recycling of carbon:
Carbon dioxide (CO₂) is the main source of carbon for the living beings. The carbon of coal, graphite, petroleum are insoluble and carbonates are not available to the organism until they are burnt or chemically changed. Most of the carbon dioxide enters the living world through photosynthesis. In this process green plants trap CO₂ from atmosphere and convert it into carbohydrates by using water and solar energy.

The amount of the carbon fixed by photosynthesis is nearly 7 x 10¹³ kg/year. The organic compounds synthesized in photosynthesis are passed from plants to the herbivores and carnivores. It is estimated that one hectare of a healthy forest produces about 10 tonnes of oxygen and absorbs 30 tonnes of carbon dioxide annually.

Question 4.
What do you mean by trophic levels? What are ecological pyramids ? Explain various types of ecological pyramids.
Answer:
Trophic level:
In an ecosystem, the producer consumer arrangement is a kind of structure known as trophic structure and each food level in the food chain is called as trophic level or energy level. In other words each level of food in food chain is called its trophic level. The first trophic level (T₁) in an ecosystem is occupied by producers. Herbivores (primary consumers) form second trophic level (T₂), secondary consumers form third trophic level (T₃), tertiary consumers form fourth trophic level (T₄) and decomposers form fifth trophic level (T₅) in an ecosystem.

Food or Ecological pyramids:
If we express the organisms of various trophic levels according to their number, biomass and ratio of energy stores within it, then we obtain a cone or pyramid like structure which is known as food or ecological pyramid. Ecological pyramids represent the trophic structure and function of an ecosystem. In base and successive trophic levels the tiers which make up the apex. Ecological pyramids are of the following three types:

1. Pyramid of biomass
2. Pyramid of number
3. Pyramid of energy.

1. Pyramid of Biomass:
Biomass is the dry weight of living organisms per unit of space. The ecological pyramid, which shows the quantitative relationship of the standing crop at each trophic level/The pyramid of biomass shows gradual reduction in biomass at each trophic level from base to apex.

The pyramid of biomass may be :

• Upright – example all terrestrial ecosystems.
• Inverted – example all aquatic ecosystems.

2. Pyramid of Number:
The ecological pyramid which shows the number of individual organisms at each trophic level. It represents numerical relationship between different trophic level of a food chain. In this pyramid more abundant species from the first trophic level and from the base of pyramid and the less abundant species remain near the top. The pyramid of number may be:

• Upright : example grassland, pond, forest ecosystem.
• Inverted : example ecosystem of tree.

3. Pyramid of Energy:
It indicates the total amount of energy at each trophic level of the food chain. At each producer level, the total energy available is relatively more than at the higher trophic levels because of the loss of the energy at each trophic level. Thus, there is a gradual loss of energy at each trophic level. The pyramid of energy of each types of ecosystem is always upright.

Question 5.
What is meant by terrestrial biomes? What are its types? Explain any one biomes in detail.
Answer:
Terrestrial biome:
Large area occupying ecosystems in nature are called biomes. If biomes are on land than they are called terrestrial biomes.
Terrestrial biomes may be :

1. Forest biomes – They may be as below :

• Topical rain forest
• Cold tropical forest
• Taiga forest.

2. Grassland biomes – They may be as below:

• Tropical rain forest
• Cold tropical forest

3. Desert biomes

4. Tundra biomes.

Grassland biome – Grassland biome or ecosystem has long grasses, Its, land is fertile. It receives approximately 25 to 75 cm average rainfall. Its component are:

1. Abiotic component:
All organic, inorganic substances and climatic factors together form abiotic component.

2. Biotic component:

• Producers – Grasses, herb, shrubs.
• Primary consumers – Herbivore like cow, buffalo, goats, sheep, deer, rabbit, rat insect.
• Secondary consumers – Carnivore animals which eat primary consumers, like snake, birds, foxes, jackal etc.
• Tertiary consumers – These organism which eat secondary consumers because no other one eats them, like Hawk, Peacock etc.
• Decomposers – Micro fungus, Bacteria, Actinomycetes are decomposer of grassland biomes and recycle the material back to soil and used by producers.

Question 6.
What are biogeochemical cycles? Write in short sulphur and calcium cycle.
Answer:
Biogeochemical cycles:
All living organisms get matter from the biosphere components i.e., lithosphere, hydrosphere and atmosphere. Essential elements or inorganic substances are provided by earth and are required by organisms for their body building and metabolism, they are known as biogeochemicals or biogenetic nutrients.

Sulphur cycle:
Producers (green plants) need sulphur in the form of sulphates from soil or from water (aquatic plants). The animals get sulphur through food. Some animals get sulphur from water also. Sulphur is found in three amino acids hence, sulphur is component of most proteins, some vitamins and enzymes. Plants pick up sulphur in the form of sulphates. They are converted to organic form mostly as component of some amino acids. It is found in nature as element and also as sulphates in soil, water and rocks. After the death of plants and animals, they are decomposed by microbes like Asperigillus, Neurospora and Escherichia releasing hydrogen sulphide.

Calcium cycle:
Calcium is slowly released from the rocks by water and wind action. These are either blown into the air or absorbed by plants through their roots. Animals obtain it directly as compounds and also from plants. Calcium is released from plant and animal bodies by decomposition after death. Molluscs and Corals deposit a large quantity of calcium in their shells and skeletons making it unavailable for quick cycling.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 1 The Solid State

The Solid State Important Questions

The Solid State Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Due to Frankel defect, density of ionic solids :
(a) Decreases
(b) Increases
(c) Does not change
(d) It changes.
Answer:
(c) Does not change

Question 2.
In CsCl each Cl is surrounded by how many Cs :
(a) 8
(b) 6
(c) 4
(d) 2.
Answer:
(a) 8

Question 3.
Frenkel defect is not shown by :
(a) AgBr
(b) AgCl
(c) KBr
(d) ZnS.
Answer:
(c) KBr

Question 4.
In NaCl crystal number of oppositely charged ions situated at equal distance are:
(a) 8
(b) 6
(c) 4
(d) 2
Answer:
(b) 6

Question 5.
Best conductor of electricity is :
(a) Diamond
(b) Graphite
(c) Silicon
(d) Carbon (non – crystalline).
Answer:
(b) Graphite

Question 6.
Which type of point defect is found in NaCI crystal of KCl crystal: (MP 2009 Set D)
(a) Frenkel defect
(b) Schottky defect
(c) Lattice defect
(d) Impurity defect.
Answer:
(b) Schottky defect

Question 7.
How many space lattices (Bravais lattice) can be obtained from various crystal systems :
(a) 7
(b) 14
(c) 32
(d) 230.
Answer:
(b) 14

Question 8.
Diamond is a :
(a) H – bond solid
(b) Ionic solid
(c) Covalent solid
(d) Glass
Answer:
(c) Covalent solid

Question 9.
The Co – ordination number of Ca²⁺ ions in fluoride structure is :
(a) 4
(b) 6
(c) 8
(d) 3.
Answer:
(c) 8

Question 10.
8 : 8 Co – ordination number is found in which compound :
(a) MgO
(b) A1203
(C) CsCl
(d) All of these
Answer:
(C) CsCl

Question 11.
Co – ordination number of body centred cubic cell is :
(a) 8
(b) 12
(c) 6
(d) 4
Answer:
(a) 8

Question 12.
Density of unit cell is :
(a) \(\frac { ZM }{ { a }^{ 3 }{ N }_{ 0 } } \)
(b) \(\frac { Z{ N }_{ 0 } }{ { a }^{ 3 }M } \)
(c) \(\frac { { N }_{ 0\quad }{ a }^{ 3 } }{ Z } \)
(d) \(\frac { Z }{ M{ N }_{ 0 } } \)
Answer:
(a) \(\frac { ZM }{ { a }^{ 3 }{ N }_{ 0 } } \)

Question 13.
The number of tetrahedral voids in unit cell of cubic close packing :
(a) 4
(b) 8
(c) 6
(d) 2
Answer:
(b) 8

Question 14.
Intra – ionic distance of CsCl will be :
(a) a
(b) \(\frac {a}{2}\)
(c) \(\frac { \sqrt { 3 } a }{ 2 } \)
(d) \(\frac { 2a }{ \sqrt { 3 } } \)
Answer:
(c) \(\frac { \sqrt { 3 } a }{ 2 } \)

Question 15.
Number of atoms in a body centred cubic unit cell is : (MP 2011)
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 16
Which of the following is Bragg equation :
(a) nλ = 2ϕ sinθ
(b) nλ = 2d sinθ
(c) nλ = sinθ
(d) n\(\frac {θ}{2}\) = \(\frac {d}{2}\) sinθ.
Answer:
(b) nλ = 2d sinθ

Question 17.
Constituents of covalent crystal is :
(a) Atom
(b) Molecule
(c) Ion
(d) All of these.
Answer:
(a) Atom

Question 18.
Number of Na atom present in the unit cell of NaCI crystal is : (MP 2012)
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 19.
What type of magnetic substance are Fe, Co, Ni: (MP 2012,18)
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.
Answer:
(b) Ferromagnetic

Question 20.
The correct example of Frenkel defect is : (MP 2012)
(a) NaCI
(b) CsCl
(c) KCl
(d) AgCl.
Answer:
(d) AgCl.

Question 21.
Dry ice (solid CO₂) is a/an : (MP 2012)
(a) Ionic crystal
(b) Covalent crystal
(c) Molecular crystal
(d) Metallic crystal.
Answer:
(c) Molecular crystal

Question 22.
Co – ordination number of Cs in CsCl: (MP2015)
(a) Like Cl i.e., 8
(b) Unlike Cl i.e., 6
(c) Unlike Cl i.e., 8
(d) Like Cl i.e., 6.
Answer:
(a) Like Cl i.e., 8

Question 23.
Structure of NaCI crystal: (MP 2015)
(a) Tetragonal
(b) Cubic
(c) Orthorhombic
(d) Monoclinic.
Answer:
(b) Cubic

Question 24.
Each Na⁺ion in NaCI crystal is surrounded by :
(a) Three Cl^– ions
(b) Eight Cl^– ions
(c) Four Cl^– ions
(d) Six Cl^– ions.
Answer:
(d) Six Cl^– ions.

Question 25.
For increasing of electro conductivity in a solid crystal, mixing of impurities is known as : (MP2016)
(a) Schottky defect
(b) Frenkel defect
(c) Doping
(d) Electronic defect.
Answer:
(c) Doping

Question 26.
Which type of lattice is found in KCl crystal:
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Simple tetragonal.
Answer:
(a) Face centred cubic

Question 27.
Number of atoms in a body centred cubic unit cell of a monoatomic substance is :
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 28.
Radius ratio limit for tetrahedral symmetry is :
(a) 0155
(b) 0.414
(c) 0.732
(d) 0.225
Answer:
(d) 0.225

Question 29.
The defect produced due to a cation and an anion vacancy in a crystal lattice is known as :
(a) Schottky defect
(b) Frenkel defect
(c) Crystal defect
(d) Ionic defect
Answer:
(a) Schottky defect

Question 30.
If co – ordination number of Cs⁺ is 8 in CsCl then co – ordination number of Cl^– ion is :
(a) 8
(b) 4
(c) 6
(d) 12
Answer:
(a) 8

Question 2.
Answer in one word / sentence :

1. Give two examples of metallic crystal.
2. Give two examples of covalent crystal.
3. Give two examples of ionic crystal.
4. What is the co – ordination number of F⁺ ion in CaF₁?
5. What is the value of co – ordination number of hexagonal close packing structure?
6. What is the type of structure of NaCl crystal?
7. Give an example of body centred cubic cell.
8. Give an example of a compound which has both Schottky and Frenkel type of defect.
9. What types of crystal is SiC? (MP 2011)
10. Write Bragg equation. (MP2017)
11. What is effect on the density of a substance or crystal due to Schottky defect?
12. Write the formula of radius ratio.
13. Give two examples of amorphous or non – crystalline solid.
14. F – centres give colour of crystal due to whose presence? (MP 2018)

Answer:

1. Copper, Nickel
2. Diamond, Graphite
3. NaCl, NaNO₃
4. 4
5. 12
6. Cubic
7. CsCl
8. AgBr
9. Covalent solid
10. nλ = 2d sinθ
11. Due to Schottky defect density of substance decreases
12. Radius ratio = \(\frac { radiusofcation\quad { r }^{ + } }{ radiusofanion\quad { r- } } \)
13. Glass, plastic
14. Due to presence of free electron.

Question 3.
Fill in the blanks :

1. The defect produced due to removal of a cation and an anion from a crystal lattice is called …………………. (MP 2018)
2. If in a crystal lattice a cation leaves its lattice site and occupies a space in the interstitial site then the defect is called ………………….
3. The cause of electric conduction of NaCl in its molten state are its ………………….
4. Due to …………………. defect the density of crystal decreases
5. Total …………………. types of crystal system are there.
6.  …………………. proposed the concept of atom for the first time.
7. The ratio of the cation and anion present in a crystal is known as ………………….
8. The process of adding small amount of impurities in an element or compound is called …………………. (MP 2018)
9. Total 14 types of unit cells are there which are known as ………………….
10. In NaCl crystal structure, co – ordination number of both Na⁺and Cl^– ion is ………………….
11.  …………………. defect is found in ZnS and AgCl crystal.
12. Due to Schottky defect, density of crystal ………………….
13. In metallic solids, conductivity is due to the presence of ………………….
14. Point defects are found in …………………. crystals.
15. Substances which are attracted in magnetic field are called ………………….
16. For a unit cell, if r = \(\frac { a }{ \sqrt { 8 } } \), then it will be …………………. type of unit cell.
17. Conductivity of semiconductor …………………. on increasing temperature.

Answer:

1. Schottky defect
2. Frenkel defect
3. Free ions
4. Schottky
5. Seven
6. Kannad
7. Radius ratio
8. Doping
9. Bravais lattice
10. Six
11. Frenkel
12. Decreases
13. Free electron
14. Ionic
15. Paramagnetic substance
16. Fcc
17. Increases.

Question 4.
Match the following:
I. (MP2014)

Answer:

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

II.

Answer:

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

III.

Answer:

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

IV. (MP2017)

Answer:

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

MP Board Class 12th Chemistry Important Questions

In this article, we will share MP Board Class 12th Hindi Swati Solutions पद्य Chapter 2 वात्सल्य और स्नेह Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 12th Hindi Swati Solutions पद्य Chapter 2 वात्सल्य और स्नेह

वात्सल्य और स्नेह अभ्यास

वात्सल्य और स्नेह अति लघु उत्तरीय प्रश्न

प्रश्न 1.
कृष्ण ने दही का दोना कहाँ छिपा लिया?
उत्तर:
बालकृष्ण ने दही का दोना अपनी पीठ के पीछे छिपा लिया है।

प्रश्न 2.
दूध पीने से चोटी बढ़ती है, यह सुझाव कृष्ण को किसने दिया था?
उत्तर:
गाय का दूध पीने से चोटी बढ़ती है, यह सुझाव यशोदा माता ने बालकृष्ण को दिया था।

प्रश्न 3.
सूरदास किस भाषा के कवि हैं?
उत्तर:
सूरदास ब्रजभाषा के कवि हैं।

प्रश्न 4.
‘मोसों कहत मोल को लीनो’ यह कथन किसने कहा है? (2016)
उत्तर:
यह कथन बलदाऊ की शिकायत करते हुए कृष्ण यशोदा माता से कहते हैं कि बलदाऊ उन्हें मोल का खरीदा हुआ बताते हैं।

प्रश्न 5.
“घर का पहरे वाला” से कवि का क्या आशय है? (2015)
उत्तर:
“घर का पहरे वाला” से कवि का तात्पर्य घर के रखवाले से है अर्थात् देश की रक्षा करने वाला प्रहरी।

प्रश्न 6.
‘ममता की गोद’ किसे कहा गया है?
उत्तर:
बहन को ‘ममता की गोद’ कहा गया है।

वात्सल्य और स्नेह लघु उत्तरीय प्रश्न

प्रश्न 1.
अपने को निर्दोष सिद्ध करने के लिए कृष्ण ने यशोदा को क्या-क्या तर्क दिए? (2013)
उत्तर:
अपने को निर्दोष सिद्ध करने के लिए कृष्ण ने निम्न तर्क दिए-

1. मैंने दही नहीं खाया है,बल्कि ग्वालवालों ने मेरे मुँह पर लपेट दिया है।
2. तेरा दधि का बर्तन इतने ऊँचे छींके पर लटका हुआ है।
3. मेरे छोटे-छोटे हाथ हैं जो उस छींके तक नहीं पहुँच सकते।

प्रश्न 2.
“गोरे नन्द जसोदा गोरी, तुम कत स्याम सरीर” यह पंक्ति किसने किससे और क्यों कही?
उत्तर:
उक्त पंक्ति बलदाऊ ने कृष्ण से कहीं। उन्होंने यह पंक्ति अपने इस पक्ष को दृढ़ करने के लिए कहीं कि तुझे तो मोल लिया है। यदि तू नन्द और यशोदा माँ का पुत्र होता तो काला क्यों होता ? वे दोनों तो गोरे हैं। यहाँ इस तथ्य को उजागर किया गया है कि गोरे माता-पिता की सन्तान भी गोरी होती है।

प्रश्न 3.
मुँह से मिट्टी निकालने के लिए यशोदा ने कौन-सा उपाय किया?
उत्तर:
कृष्ण के मुँह से मिट्टी निकालने के लिए यशोदा ने कसकर उनकी बाँह पकड़ ली और मारने के लिए एक हाथ में डंडी उठा ली। उनका भाव यह था कि डर के मारे कृष्ण अपने मुख से मिट्टी बाहर निकाल देगा।

प्रश्न 4.
‘मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी’ से कवि का क्या आशय है?
उत्तर:
कवि कहता है कि भाई का जीवन तो क्रीड़ा-कौतूहल आदि से भरा रहा, लेकिन बहन गम्भीर रहते हुए भी आनन्द देने वाली बनी रही। बहन के होते हुए कभी भी भाई के प्रेम और आनन्द में कमी नहीं आई।.

प्रश्न 5.
बहन को भाई का ध्रुवतारा’ क्यों कहा गया है? (2009, 11, 14, 17)
उत्तर:
भाई तो एक अल्हड़ जीवन या यों कहें कि लापरवाह जीवन बिताता रहा, लेकिन बहन अपने लक्ष्य के लिए ध्रुवतारे की तरह अटल खड़ी हुई दिखाई दी। बहन ने भाई को मुसीबतों के समय उत्साह दिया और उसकी देश-भक्ति को जाग्रत कर देश के प्रति अपने कर्तव्य को पूर्ण करने में सहयोग दिया।

वात्सल्य और स्नेह दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
कृष्ण माता यशोदा से बलदाऊ की क्या-क्या शिकायतें करते हैं?
उत्तर:
कृष्ण माता यशोदा से शिकायत करते हैं कि बलदाऊ मुझे बहुत चिढ़ाते हैं। मझसे यह कहते हैं कि तुझे यशोदा माता ने जन्म नहीं दिया है,बल्कि तुझे किसी से मोल लिया है। में इस गुस्से में खेलने भी नहीं जाता हूँ। बार-बार मुझसे पूछते हैं कि तेरे माता-पिता कौन हैं और कहते हैं कि नन्द बाबा और यशोदा तो दोनों गोरे हैं,यदि तू उनका पुत्र है तो साँवला क्यों है? फिर सभी ग्वालों को सिखा देते हैं और मेरी तरफ देखकर ताली मारकर हँसते हैं और मुझे खिझाते हैं और तू मुझे ही मारना जानती है,बलदाऊ को कभी नहीं मारती।

प्रश्न 2.
‘सूर वात्सल्य के चितेरे हैं’ इस कथन पर अपने विचार व्यक्त कीजिए।
उत्तर:
‘सूर वात्सल्य के चितेरे हैं’ यह कथन अक्षरशः सत्य है। इनका वात्सल्य वर्णन अद्वितीय है। कृष्ण के बाल रूप और उनकी बाल-सुलभ क्रीड़ाओं का जैसे विशद वर्णन सूरदास ने किया है, वैसा सम्पूर्ण विश्व साहित्य में कहीं नहीं मिलता। कृष्ण का घुटनों चलना,मणियों के खम्भ में अपने प्रतिबिम्ब को माखन खिलाना, माँ से जिद करना,खेलते समय खिसिया जाना आदि विविध बाल क्रीड़ाओं का वर्णन सूरसागर में चित्रित है। उदाहरण देखिए-
“मैया मैं तो चंद खिलौना लैहों।
जैहों लोटि धरनि पै अब ही तेरी गोद न ऐहौं।”

बच्चों की पारस्परिक होड़ का चित्रण भी सूरदास ने अनूठे रूप में किया है। माता यशोदा उनसे कहती है कि दूध पीने से चोटी बढ़ती है तो वह इस लालच में रोज दूध पी लेते हैं। जब उन्हें चोटी बढ़ती हुई दिखाई नहीं देती तो वह कहते हैं-
“मैया कबहि बढ़ेगी चोटी।
काचो दूध पियाबति पचि-पचि देत न माखन रोटी ॥
कितनी बार मोहि दूध पिबत भई यह अजहूँ है छोटी॥”

प्रश्न 3.
बहन को ‘चिनगारी’ तथा भाई को ‘ज्वाला’ बताने के पीछे कवि का क्या आशय (2009)
उत्तर:
गोपाल सिंह नेपाली राष्ट्रीय चेतना जाग्रत करने वाले कवि हैं। उन्होंने भाई-बहन के प्रेम को राष्ट्रीय प्रेम के रूप में अभिव्यक्त किया है। अपनी मातृभूमि की रक्षार्थ बहन चिनगारी के रूप में कार्य करेगी तो भाई ज्वाला बन कर देश की रक्षार्थ तत्पर रहेगा। बहन की क्रोध रूपी चिनगारी जलकर भयानक ज्वाला का रूप ले लेगी और मातृभूमि के दुश्मन को जलाकर राख कर देगी। दूसरी ओर भाई का क्रोध तो स्वयं ज्वाला बनकर देश की रक्षा करेगा। भाई बहन को कहता है कि देश की रक्षा के लिए हम दोनों को सजग और सक्रिय रहना है। कवि आश्वस्त है कि भाई और बहन दोनों मिलकर अपने देश की रक्षा के लिए सतर्क रहेंगे और देश के दुश्मन के लिए ज्वाला बनकर उसका सर्वनाश करने में सक्षम होंगे।

प्रश्न 4.
कवि ने भाई-बहन के स्नेह को किन-किन प्रतीकों के माध्यम से अभिव्यक्त किया है?
उत्तर:
कवि ने भाई-बहन के स्नेह को देश-प्रेम में परिवर्तित कर नये-नये प्रतीकों का सहारा लिया है। बहन के प्रेम के प्रतीक हैं-चिनगारी,हहराती गंगा,बसन्ती चोला,कराल क्रान्ति, राधारानी, आँगन की ज्योति,ममता की गोद, बहन की बुद्धि, नदी की धारा, ध्रुवतारा के रूप में बताया है। दूसरी तरफ भाई के प्रेम को ज्वाला, बेहाल झेलम, सजा हुआ लाल, विकराल, वंशीवाला, घर का पहरेदार,प्रेम का पुतला, जीवन का क्रीड़ा कौतुक, क्रियाशीलता, एक लहर बताया है। इन प्रतीकों के माध्यम से कवि ने राष्ट्रीय चेतना के साथ-साथ मानवीय जीवन की ऊष्मापूर्ण अनुभूतियों का प्रभावशाली वर्णन किया है। इस कविता में कवि ने देश-प्रेम का उच्च आदर्श उद्घाटित किया है।

प्रश्न 5.
भाई-बहन’ कविता के माध्यम से कवि क्या संदेश देना चाहता है?
उत्तर:
भाई-बहन’ कविता के माध्यम से कवि ने नये-नये प्रतीकों के द्वारा भाई बहन के प्रेम को राष्ट्रीय प्रेम के परिवेश में व्यक्त किया है। भाई अपनी बहन को सम्बोधित कर कहता है कि हम दोनों को देश की रक्षा के लिए सजग और सक्रिय होना है। बहन की बुद्धि और भाई की क्रियाशीलता मिलकर जीवन और राष्ट्र को आनन्दमय बना देगी। भाई-बहन दोनों मिलकर आजादी के गीत गाकर लोगों में देश की रक्षा के प्रति जागति पैदा कर देश के प्रति अपने कर्तव्य को पूरा कर सकते हैं। भाई-बहन के प्रेम को राष्ट्रीय प्रेम से जोड़कर उदात्त बना दिया गया है। कवि संदेश दे रहा है कि हम अपने आनन्द में ही मस्त होकर देश की रक्षा के अपने कर्त्तव्य को कहीं भूल न जाएँ। यहाँ सभी को सजग रहकर एक सच्चे देशभक्त प्रहरी की तरह अपने देश की रक्षा करनी है।

देश के लिए बसन्ती चोला धारण करना पड़े तो करें और विकराल स्वरूप धारण करना पड़े तो करें तथा अपनी बुद्धि और बल के सहारे देश के लिए अपना बलिदान कर दें। जब देश की आजादी का प्रश्न हो, तो अपनी क्रोध रूपी ज्वाला को जलाकर सामने आये अरि (दुश्मन) को नष्ट कर दें। किसी भी कीमत पर देश की आजादी को बचाना है। यदि मातृभूमि तुम्हें अपनी रक्षा के लिए आवाज दे, तो सभी सुख-सम्पत्ति को भूल कर एक सच्चे देश-भक्त की तरह आगे आयें और अपना सर्वस्व अर्पण करके भी अपनी मातृभूमि की रक्षा करें। कवि देश में एकता बनाये रखने का भी संदेश देता है।

प्रश्न 6.
सन्दर्भ सहित व्याख्या कीजिये
(अ) मैया मैं नाही……. शिव विरंचि बौरायो।
(ब) मैया कबहिं बढ़ेगी……. हरि हलधर की जोटी।
(स) भाई एक लहर ……. भाई का ध्रुवतारा है।
उत्तर:
(अ) सन्दर्भ :
प्रस्तुत पद ‘वात्सल्य और स्नेह’ से सूरदास द्वारा रचित ‘सूर के बालकृष्ण’ नामक शीर्षक से उद्धृत किया गया है।

सन्दर्भ :
इसमें बालकृष्ण अपनी माता यशोदा से अपने प्रति की गई दही खाने की शिकायत को नकारते हुए बड़े ही बुद्धिकौशल से बाल सुलभ उत्तर देते हैं।

व्याख्या :
श्रीकृष्ण माता यशोदा से कहते हैं कि, हे माता! मैंने दही नहीं खाया है। मुझे याद आ रहा है कि इन सभी सखाओं ने मिलकर मेरे मुख पर दही लपेट दिया था। तू जानती है कि इतने ऊँचे टँगे हुए छींके पर दही का बर्तन रखा हुआ है। तू देख सकती है कि मेरे छोटे-छोटे हाथ हैं। इन छोटे हाथों से मैं कैसे उस दही के बर्तन को प्राप्त कर सकता हूँ। फिर नन्दकुमार बालकृष्ण ने दोने को पीठ के पीछे छिपाते हुए और मुख पर लगे हुए दही को पोंछते हुए उपर्युक्त बातें कहीं। यहाँ उनकी बाल सुलभ चतुरता का प्रदर्शन किया गया है। उन्हें भान है कि मुख पर लगे हुए दही से और हाथ में लगे दोने से उनकी चोरी पकड़ी जाएगी, अतः मुख को साफ कर लिया और दोने को पीठ के पीछे छुपा लिया।

यशोदा जी सब समझ गईं, लेकिन पीटने वाली लकड़ी को फेंक कर मुस्कराने लगी और मनमोहन बालकृष्ण को अपने गले से लगा लिया। श्रीकृष्ण के बाल विनोद के आनन्द ने माता यशोदा के मन को मोहित कर लिया। यहाँ श्रीकृष्ण के प्रति उनकी भक्ति का प्रताप दर्शाया गया है। सूरदास कहते हैं कि यशोदा जी और बालकृष्ण के सख को देखकर शिव और ब्रह्मा भी विवेक रहित होकर मोहित हो गए। अर्थात् श्रीकृष्ण यशोदा जी को अपनी बाल लीलाओं का जो सुख दे रहे हैं उससे किसी को भी ईर्ष्या हो सकती है और वह भ्रमित हो सकता है। निरंजन निराकार परब्रह्म साकार रूप में आकर यशोदा जी के आँगन में उनके प्रमोद के लिए जो क्रीड़ाएँ कर रहे हैं, ऐसा सुख शिव-विरंचि को भी दुर्लभ है।

(ब) सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बालकृष्ण अपनी माता यशोदा से यह शिकायत कर रहे हैं कि उनको कितने ही दिन दूध पीते हो गए, लेकिन उनकी चोटी बड़ी नहीं हुई है।

व्याख्या :
बालकृष्ण यह जानने को उत्सुक हैं कि उनकी चोटी कब बढ़ेगी। उनकी माता उन्हें यह भरोसा दिलाकर दूध पिलाती थीं कि दूध पीने से उनकी चोटी बढ़ जाएगी। वह माता से पूछते हैं कि हे माता मुझे कितनी ही बार (बहुत समय) दूध पीते हुए हो गईं, लेकिन यह चोटी अभी तक छोटी है। तेरे कथनानुसार मेरी चोटी बलदाऊ की चोटी के समान लम्बी और मोटी हो जाएगी, काढ़ते में, गुहते में,नहाते समय और सुखाते समय नागिन के समान लोट जाया करेगी। तू मुझे कच्चा दूध अधिक मात्रा में पिलाती है तथा माखन और रोटी नहीं देती है। माखन और रोटी बालकृष्ण को प्रिय हैं, लेकिन वे नहीं मिलते और चोटी बढ़ने की लालसा से उन्हें गाय का कच्चा दूध पीना पड़ता है। सूरदास कहते हैं कि माता बलाएँ लेने लगी और कहने लगी कि हरि हलधर दोनों भाइयों की जोड़ी चिरंजीव हो।

(स) सन्दर्भ :
पूर्ववत्।

प्रसंग :
भाई और बहन एक-दूसरे के सहायक हैं। हर उन्माद में बहन को भाई का ही सहारा है। दोनों मिलकर आजादी के गीत गाकर मातृभूमि के प्रति अपने कर्तव्य को पूर्ण कर सकते हैं।

व्याख्या :
कवि गोपाल सिंह नेपाली कहते हैं कि यहाँ पर भाई एक लहर के रूप में है, तो बहन नदी की एक धारा है। दोनों साथ-साथ रहकर कल्याण के मार्ग पर चलकर सभी का सहारा बनेंगे। गंगा-यमुना के संगम में पानी की अधिकता होने से कभी बाढ़ जैसी स्थिति हो जाती है और किनारे डूबने लगते हैं। ऐसे पागलपन भरे अवसर पर बहन को भाई का ही एकमात्र सहारा है। बेफिक्री के समय में भी भाई अपनी बहन के लिए ध्रुवतारे की तरह अडिग है। कुछ समय ऐसे आते हैं जब हमें अपने संयम को दृढ़ रखना है। भाई और बहन दोनों मिलकर आजादी के गीत गाकर ही मातृभूमि के प्रति अपने कर्तव्य को पूरा कर सकते हैं। अपनी मुसीबतों को झेलकर और बलिदानों के माध्यम से पत्थर-हृदय देश के दुश्मनों को चेतावनी देते हैं कि हम हर हाल में देश की रक्षा के लिए तत्पर हैं।

वात्सल्य और स्नेह काव्य सौन्दर्य

प्रश्न 1.
निम्नलिखित शब्दों के मानक रूप लिखिए
उत्तर:

प्रश्न 2.
निम्नलिखित पंक्तियों में अलंकार पहचान कर लिखिए
(अ) काढ़त गुहत न्हवावत ओछत, नागिन सी भुई लोटी।
(ब) मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी।
(स) काचो दूध पिआवत पचि-पचि, देत न माखन रोटी।
उत्तर:
(अ) उपमा अलंकार
(ब) अनुप्रास अलंकार
(स) पुनरुक्तिप्रकाश अलंकार।

प्रश्न 3.
निम्नलिखित शब्दों में से तत्सम और तद्भव शब्द छाँटकर लिखिए
ज्योति, उन्माद, बहन, कलंक, जननी,माटी,मैया, पूत,तात, पत्थर।
उत्तर:

प्रश्न 4.
गोद राखि चचुकारि दुलारति पुन पालति हलरावति।
आँचर ढाँकि बदन विधु सुंदर थन पय पान करावति॥
उपर्युक्त पंक्तियों में प्रयुक्त रस बताइये।
उत्तर:
वात्सल्य रस।

प्रश्न 5.
इस पाठ में से पुनरुक्ति प्रकाश और अनुप्रास अलंकार के उदाहरण छाँट कर लिखिए।
उत्तर:
(i) पुनरुक्तिप्रकाश-तू चिनगारी बनकर उड़ री,जाग-जाग मैं ज्वाल बनूँ।
(ii) अनुप्रास अलंकार-तू भगिनी बन क्रान्ति कराली,मैं भाई विकराल बनूँ।

सूर के बालकृष्ण भाव सारांश

‘सूर के बालकृष्ण’ नामक पदों के रचयिता भक्त कवि ‘सूरदास’ हैं। सूरदास ने शिशु सुलभ मुद्राओं, क्रीड़ाओं और शिशु के स्वभावगत सौन्दर्य को भी अपनी उदात्त और आकर्षक छवियों में प्रकट किया है।

वात्सल्य प्रेम में प्रेम का निश्छल,उदार और शिशु सुलभ स्वभाव प्राप्त होता है। शिशु के सुन्दर स्वरूप, आकर्षक मुद्राओं और उसके अबोध व्यवहारों के प्रदर्शन से यह वात्सल्य भाव दर्शक के हृदय में संचरित होता है। हिन्दी साहित्य में वात्सल्य भाव की रचनाओं की कमी है, किन्तु जितनी भी रचनाएँ प्राप्त होती हैं, वे वत्सलता के प्रभावी स्वरूप को प्रकट करती हैं। सूरदास ने शिशु सुलभ भुद्राओं,क्रीड़ाओं को तो अपने काव्य में स्थान दिया ही है साथ ही शिश के स्वभावगत सौन्दर्य को भी अपनी आकर्षक छवियों में प्रस्तुत किया है। शिशु कृष्ण अपने वाक्यचातुर्य से अपने दही खाने वाली बात को काट देते हैं। शिशु सहज विश्वासी होता है। शिशु कृष्ण से यदि माँ ने कह दिया कि गाय का दूध पीने से चोटी बढ़ती है तो कृष्ण खूब दूध पीते हैं और रोज-रोज माँ से पूछते हैं कि चोटी क्यों नहीं बढ़ रही? यहाँ शिशु की अधीरता दिखाई देती है। बच्चों में पारस्परिक चिढ़ाने का भाव है। कृष्ण अपनी माँ से बलराम के चिढ़ाने की शिकायत करते हैं। बाल सुलभ चेष्टाओं में मिट्टी खाने की प्रवृत्ति भी समाहित है। सूरदास ने इस तथ्य का आकर्षक वर्णन अपने पदों में किया है। सूरदास की भाषा में चमत्कृत कर देने वाला प्रवाह इन पदों में उपलब्ध है।

सूर के बालकृष्ण संदर्भ-प्रसंग सहित व्याख्या

(1) मैया मैं नाहीं दधि खायो।
ख्याल परे ये सखा सबै मिलि, मेरे मुख लपटायो।
देखि तही सीके पर भाजन, ऊँचे धर लटकायो।
तुही निरखि नान्हे कर अपने, मैं कैसे करि पायो।
मुख दधि पोंछि कहत नंदनंदन, दोना पीठि दुरायो।
डारि सॉट मुसुकाई तबहि, गहि सुत को कंठ लगायो।
बाल विनोद मोद मन मोह्यो, भक्ति प्रताप दिखायो।
सूरदास प्रभु जसुमति के सुख, शिव विरंचि बौरायो।

शब्दार्थ :
ख्याल परे = याद आया; लपटायो = लपेट दिया है; भाजन = बर्तन; निरखि = देख; नान्हे = छोटे,कर = हाथ; दोना = पत्तों का बना पात्र; साँटि = पीटने की डंडी; सुत = पुत्र; कंठ = गला; मोद = प्रसन्नता; मोह्यो = मोहित हो गया; विरंचि = ब्रह्माजी।

सन्दर्भ :
प्रस्तुत पद ‘वात्सल्य और स्नेह’ से सूरदास द्वारा रचित ‘सूर के बालकृष्ण’ नामक शीर्षक से उद्धृत किया गया है।

सन्दर्भ :
इसमें बालकृष्ण अपनी माता यशोदा से अपने प्रति की गई दही खाने की शिकायत को नकारते हुए बड़े ही बुद्धिकौशल से बाल सुलभ उत्तर देते हैं।

व्याख्या :
श्रीकृष्ण माता यशोदा से कहते हैं कि, हे माता! मैंने दही नहीं खाया है। मुझे याद आ रहा है कि इन सभी सखाओं ने मिलकर मेरे मुख पर दही लपेट दिया था। तू जानती है कि इतने ऊँचे टँगे हुए छींके पर दही का बर्तन रखा हुआ है। तू देख सकती है कि मेरे छोटे-छोटे हाथ हैं। इन छोटे हाथों से मैं कैसे उस दही के बर्तन को प्राप्त कर सकता हूँ। फिर नन्दकुमार बालकृष्ण ने दोने को पीठ के पीछे छिपाते हुए और मुख पर लगे हुए दही को पोंछते हुए उपर्युक्त बातें कहीं। यहाँ उनकी बाल सुलभ चतुरता का प्रदर्शन किया गया है। उन्हें भान है कि मुख पर लगे हुए दही से और हाथ में लगे दोने से उनकी चोरी पकड़ी जाएगी, अतः मुख को साफ कर लिया और दोने को पीठ के पीछे छुपा लिया।

यशोदा जी सब समझ गईं, लेकिन पीटने वाली लकड़ी को फेंक कर मुस्कराने लगी और मनमोहन बालकृष्ण को अपने गले से लगा लिया। श्रीकृष्ण के बाल विनोद के आनन्द ने माता यशोदा के मन को मोहित कर लिया। यहाँ श्रीकृष्ण के प्रति उनकी भक्ति का प्रताप दर्शाया गया है। सूरदास कहते हैं कि यशोदा जी और बालकृष्ण के सख को देखकर शिव और ब्रह्मा भी विवेक रहित होकर मोहित हो गए। अर्थात् श्रीकृष्ण यशोदा जी को अपनी बाल लीलाओं का जो सुख दे रहे हैं उससे किसी को भी ईर्ष्या हो सकती है और वह भ्रमित हो सकता है। निरंजन निराकार परब्रह्म साकार रूप में आकर यशोदा जी के आँगन में उनके प्रमोद के लिए जो क्रीड़ाएँ कर रहे हैं, ऐसा सुख शिव-विरंचि को भी दुर्लभ है।

काव्य सौन्दर्य :

1. बालकृष्ण की बाल सुलभ चतुरता दर्शनीय है।
2. ब्रजभाषा का प्रयोग।
3. अनुप्रास अलंकार का प्रयोग।
4. वात्सल्य रस है।

(2) मैया कबहिं बढ़ेगी चोटी।
किती बार मोहि दूध पिअत भई, यह अजहूँ है छोटी।
तू जो कहति बल की बेनी, ज्यौं, है है लॉबी मोटी।
काढ़त गुहत न्हवावत ओछत, नागिन सी भुइँ लोटी।
काचो दूध पिआवत पचि-पचि देत न माखन रोटी।
सूर श्याम चिरजीवौ दोऊ भैया, हरि हलधर की जोटी।। (2010, 15)

शब्दार्थ :
किती बार = कितनी बार; बल = बलदाऊ; बेनी = चोटी; काढ़त गुहत = काढ़ते में और गुहते में; न्हवावत = नहाने में; ओछत = सुखाने में; पचि-पचि = खूब; हरि हलधर = कृष्ण बलराम; जोटी = जोड़ी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बालकृष्ण अपनी माता यशोदा से यह शिकायत कर रहे हैं कि उनको कितने ही दिन दूध पीते हो गए, लेकिन उनकी चोटी बड़ी नहीं हुई है।

व्याख्या :
बालकृष्ण यह जानने को उत्सुक हैं कि उनकी चोटी कब बढ़ेगी। उनकी माता उन्हें यह भरोसा दिलाकर दूध पिलाती थीं कि दूध पीने से उनकी चोटी बढ़ जाएगी। वह माता से पूछते हैं कि हे माता मुझे कितनी ही बार (बहुत समय) दूध पीते हुए हो गईं, लेकिन यह चोटी अभी तक छोटी है। तेरे कथनानुसार मेरी चोटी बलदाऊ की चोटी के समान लम्बी और मोटी हो जाएगी, काढ़ते में, गुहते में,नहाते समय और सुखाते समय नागिन के समान लोट जाया करेगी। तू मुझे कच्चा दूध अधिक मात्रा में पिलाती है तथा माखन और रोटी नहीं देती है। माखन और रोटी बालकृष्ण को प्रिय हैं, लेकिन वे नहीं मिलते और चोटी बढ़ने की लालसा से उन्हें गाय का कच्चा दूध पीना पड़ता है। सूरदास कहते हैं कि माता बलाएँ लेने लगी और कहने लगी कि हरि हलधर दोनों भाइयों की जोड़ी चिरंजीव हो।

काव्य सौन्दर्य :

1. बच्चों को बहाने से दूध पिलाने के तथ्य को उजागर किया गया है।
2. ब्रजभाषा का सुन्दर प्रयोग हुआ है।
3. पुनरुक्तिप्रकाश,उपमा अलंकार का प्रयोग।

(3) मैया, मोहिं दाऊ बहुत खिझायो।
मोसों कहत मोल को लीनो, तोहि जसुमति कब जायो।
कहा कहौं यहि रिस के मारे, खेलन हौं नहिं जात।
पुनि-पुनि कहत कौन है माता, को है तुमरो तात।।
गोरे नन्द जसोदा गोरी, तुम कत स्याम सरीर।
चुटकी दै दै हँसत ग्वाल सब, सिखे देत बलवीर।।
तू मोही को मारन सीखी, दाऊ कबहुँ न खीझै।
मोहन को मुख रिस समेत लखि, जसुमति सुनि-सुनि रीझै।।
सुनहु कान्ह बलभद्र चबाई, जनमत ही को धूत।
सुर-श्याम मो गोधन की सौं, हों माता तू पूत।।

शब्दार्थ :
खिझायो = चिढ़ाना; जायो = पैदा किया; लखि = देखकर, रिसके = गुस्से में; तात = पिता; चबाई = चुगलखोर; धूत = धूर्त; सौं = सौगन्ध।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
बलदाऊ और कृष्ण अन्य ग्वाल-बालों के साथ बाहर खेलने जाते हैं तो बलदाऊ भिन्न-भिन्न प्रकार से उन्हें चिढ़ाते हैं। यहाँ यही शिकायत कृष्ण माता यशोदा जी से कर रहे हैं।

व्याख्या:
श्रीकृष्ण अपने बड़े भाई बलभद्र की शिकायत करते हुए अपनी माता से कहते हैं कि हे माता बलभद्र भाई मुझे बहुत चिढ़ाते हैं। मुझसे कहते हैं कि तुझे यशोदा जी ने जन्म नहीं दिया है, तुझे तो किसी से मोल लिया है। मैं क्या बताऊँ इस गुस्से के कारण मैं खेलने भी नहीं जाता। मुझसे बार-बार पूछते हैं कि तेरे माता-पिता कौन हैं। तू नन्द-यशोदा का पुत्र तो हो नहीं सकता क्योंकि तू साँवले रंग का है जबकि नन्द और यशोदा दोनों गोरे हैं। ऐसी मान्यता है कि गोरे माता-पिता की सन्तान भी गोरी होती है। यह तर्क बलदाऊ ने इसलिए दिया ताकि कृष्ण इसको काट न सके। इस बात पर सभी ग्वाल-बाल ताली दे-देकर हँसते हैं।

सभी ग्वाल-बालों को बलदेव सिखा देते हैं और सभी हँसते हैं, तब मैं खीझ कर रह जाता हूँ। तू सिर्फ मुझे ही मारना सीखी है दाऊ से कभी कुछ भी नहीं कहती कृष्ण के गुस्से से भरे मुख को बार-बार देखकर उनकी रिस भरी बातें सुन-सुनकर यशोदा जी अत्यन्त प्रसन्न होती हैं। मोद भरे मुख से यशोदा जी बोलीं, हे कृष्ण! बलदाऊ तो चुगलखोर है और जनम से ही धूर्त है। सूरदास जी कहते हैं, यशोदाजी कहने लगीं-मुझे गोधन (अपनी गायों) की सौगन्ध है,मैं माता हूँ और तू मेरा पुत्र है। इस कथन ने कृष्ण के गुस्से को दूर कर दिया।

काव्य सौन्दर्य :

1. ब्रजभाषा में अनूठा माधुर्य भर दिया है।
2. माता और पुत्र के प्रश्नोत्तर तार्किक दृष्टि से उत्तम हैं।
3. अनुप्रास व पुनरुक्तिप्रकाश अलंकार का प्रयोग है।

4. मो देखत, जसुमति तेरे ढोटा, अबहिं माटी खाई।
यह सुनिकै रिसि करि उठि धाई, बांह पकरि लै आई।।
इन कर सों भज गहि गाढ़े, करि इक कर लीने सांटी।
मारति हौ तोहिं अबहिं कन्हैया, वेग न उगिलौ माटी॥
ब्रज लरिका सब तेरे आगे, झूठी कहत बनाई।
मेरे कहे नहीं तू मानति, दिखरावौं मुख बाई॥ (2016)

शब्दार्थ :
ढोटा = पुत्र; भुइँ = भूमि; मुख बाई = मुख फैलाकर; गहि गाढ़े = जोर से पकड़कर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
ग्वाल-बालों ने यशोदा माँ से यह शिकायत की है कि कान्हा ने मिट्टी खाई है। उनका विश्वास कर माँ उसे पकड़ लेती है।

व्याख्या :
कृष्ण के साथ खेलने वाले बालकों ने माता यशोदा से शिकायत की कि कान्हा ने मिट्टी खाई है। यह सुनकर माता को बड़ा क्रोध आया और कृष्ण को बाँह से पकड़ लिया। उनके हाथ से कहीं छूट कर भाग न जाय इसलिए कसकर बाँह पकड़ ली और एक हाथ में मारने के लिए एक डंडी ले ली। माता यशोदा ने कृष्ण से कहा हे कृष्ण ! मैं तुझे अभी मारूँगी नहीं तो जल्दी से मुँह से मिट्टी उगल दो। यह सुनकर बालकृष्ण ने डरते हुए अपनी माँ से कहा-ये सब ब्रज के ग्वाल-वाल तेरे पास आकर मेरी झूठी शिकायत लगाते हैं। मैं जानता हूँ कि तू मेरा कहना तो मानेगी नहीं इसलिए में अपना मुंह खोलकर दिखाता हूँ कि मुँह में माटी कहाँ है।

काव्य सौन्दर्य :

1. बाल सुलभ चेष्टाओं का सुन्दर वर्णन है।
2. ब्रजभाषा का सुन्दर प्रयोग किया है।
3. बच्चों के माटी खाने के तथ्य को उजागर किया गया है।
4. अनुप्रास अलंकार प्रयुक्त हुआ है।

भाई-बहन भाव सारांश

प्रस्तुत कविता ‘भाई-बहन’ सुप्रसिद्ध कवि ‘गोपाल सिंह नेपाली’ द्वारा रचित है। इसमें कवि ने भाई बहन के प्रेम को राष्ट्रीय प्रेम के परिवेश में व्यक्त किया है।

गोपाल सिंह नेपाली ने राष्ट्रीय चेतना के साथ-साथ जीवन की अनुभूतियों का वर्णन अपनी कविताओं में किया है। प्रस्तुत कविता में भाई-बहिन के प्रेम को राष्ट्र-प्रेम के रूप में व्यक्त किया है। यहाँ भाई अपनी बहन को सम्बोधित कर राष्ट्र की रक्षा हेतु समर्पित होने का आग्रह कर रहा है। इस कविता में अनेक प्रतीकों के माध्यम से राष्ट्र-प्रेम को उजागर किया गया है। बहन की बुद्धि और भाई की क्रियाशीलता मिलकर ही जीवन और राष्ट्र को आनन्दमय बना सकेगी। दोनों मिलकर आजादी के गीत गाकर अपने राष्ट्र के प्रति अपने प्रेम को प्रकट कर सकते हैं। कविता में दिए गए प्रतीक और बिम्ब मौलिक हैं।

भाई-बहन संदर्भ-प्रसंग सहित व्याख्या

1. तू चिनगारी बनकर उड़ री, जाग-जाग मैं ज्वाल बनें,
तू बन जा हहराती गंगा, मैं झेलम बेहाल बून,
आज बसन्ती चोला तेरा, मैं भी सज लूं लाल बनूँ
तू भगिनी बन क्रान्ति कराली, मैं भाई विकराल बनूँ
यहाँ न कोई राधारानी, वृन्दावन, वंशीवाला;
तू आँगन की ज्योति बहन री, मैं घर का पहरेवाला।

शब्दार्थ :
हहराती गंगा = कलकल ध्वनि करती गंगा; चोला = वेश; भगिनी = बहन; विकराल = भयानक।

सन्दर्भ :
प्रस्तुत पंक्तियाँ ‘वात्सल्य और स्नेह’ पाठ के ‘भाई-बहन’ शीर्षक कविता से ली गई हैं। इसके रचयिता गोपाल सिंह नेपाली हैं।

प्रसंग :
यहाँ पर भाई अपनी बहन से अपने देश की रक्षा के लिए बसन्ती चोला पहनने और अपने हृदय के अन्दर की ज्वाला को जलाए रखने को प्रेरित कर रहा है।

व्याख्या :
कविवर गोपाल सिंह नेपाली कहते हैं कि एक भाई अपनी बहन से देश की रक्षा के लिए जाग्रत रहने का आह्वान कर रहा है। भाई कहता है कि हे बहन! तू चिनगारी बनकर आकाश में उड़ और मैं ज्वाला बनकर देश-रक्षा के लिए जाग्रत रहूँ। तू कल-कल ध्वनि करती हुई गंगा बनकर बह और मैं बेहाल की तरह चलने वाली झेलम नदी बन कर बहूँ तू देश की रक्षा के लिए बसन्ती रंग के वस्त्र पहन ले और मैं भी देश के वीरों के रूप में सज जाऊँ। तू दुश्मन के लिए भयंकर क्रान्ति बन जा और मैं भयानक वीर बन जाऊँ ताकि उसकी दृष्टि हमारे देश पर न पड़े। यहाँ सिर्फ देश की रक्षा का सवाल है, सभी की एकता का प्रश्न है। यहाँ न तो वृन्दावन अलग है,न राधारानी अलग है और न नन्दलाल अलग है। सभी देश की रक्षा के लिए उद्यत हैं। भाई अपनी बहन से कहता है कि वह आँगन की ज्योति बनकर प्रकाश करे और वह घर का पहरेदार बनेगा। हम दोनों अपने प्रेम को देश-प्रेम पर न्यौछावर करते हैं।

काव्य सौन्दर्य :

1. भाषा साहित्यिक होने के साथ-साथ व्यावहारिक है।
2. देश-भक्ति का अनूठा समर्पण है।
3. अनुप्रास अलंकार है।
4. कविता में मधुरता है।

(2) बहन प्रेम का पुतला हूँ मैं, तू ममता की गोद बनी;
मेरा जीवन क्रीड़ा-कौतुक तू प्रत्यक्ष प्रमोद भरी;
मैं भाई फूलों में भूला, मेरी बहन विनोद बनी;
भाई की गति, मति भगिनी की दोनों मंगल-मोद बनी
यह अपराध कलंक सुशीले, सारे फूल जला देना।
जननी की जंजीर बज रही, चल तबियत बहला देना।

शब्दार्थ :
पुतला = मूर्ति; विनोद = मनोरंजन; प्रमोद = आनन्द; भगिनी = बहन।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
यहाँ पर भाई अपनी बहन को इस बात के लिए प्रेरित करता है कि यदि अपने देश की परतन्त्रता पुकार रही हो तो अपने सारे सुख भुलाकर उसकी रक्षा के लिए तैयार हो जाना।

व्याख्या :
भाई कहता है कि मैं तो प्रेम की मूर्ति बनकर रहा हूँ और तू ममता की गोद बनकर रही है। अर्थात् तूने हर किसी को अपनी ममता की छाँव में बैठाया है। मेरा जीवन क्रीड़ा और कौतूहल बनकर रहा है और तू साक्षात् आनन्दमयी बन कर रही है। मैं अपने जीवन के सुखों में खोया रहा और मेरी बहन मनोरंजन में डूबी रही। भाई की क्रियाशीलता और बहन की बुद्धि दोनों मिलकर आनन्द का साधन बनीं। भाई अपनी बहन को चेतावनी देते हुए कहता है कि हे बहन! हमारे सुख और आनन्द कहीं अपराध और कलंक न बन जायँ, इसलिए सभी सुखों को तिलांजलि देकर भारतमाता की परतन्त्रता की बेड़ियों की आवाज को सुन और चलकर उसे धैर्य दिला कि हम सब मिलकर उसे स्वतन्त्र करेंगे। हम सभी सुखों को अपने देश की रक्षा के लिए अर्पण कर देंगे।

काव्य सौन्दर्य :

1. अनेक प्रतीकों के माध्यम से कवि ने इस कथ्य को प्रकट किया है।
2. अनुप्रास अलंकार की छटा दृष्टव्य है।
3. वीर रस का पुट दिया गया है।

(3) भाई एक लहर बन आया, बहन नदी की धारा है।
संगम है, गंगा उमड़ी है, डूबा कूल-किनारा है;
यह उन्माद बहन को अपना, भाई एक सहारा है;
यह अलमस्ती, एक बहन ही भाई का ध्रुवतारा हैं;
पागल घड़ी, बहन-भाई है, वह आजाद तराना है।
मुसीबतों से बलिदानों से पत्थर को समझाना है। (2009)

शब्दार्थ :
तराना = ताल स्वर; अलमस्त = मतवाला, बेफ्रिक; उन्माद = पागलपन।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
भाई और बहन एक-दूसरे के सहायक हैं। हर उन्माद में बहन को भाई का ही सहारा है। दोनों मिलकर आजादी के गीत गाकर मातृभूमि के प्रति अपने कर्तव्य को पूर्ण कर सकते हैं।

व्याख्या :
कवि गोपाल सिंह नेपाली कहते हैं कि यहाँ पर भाई एक लहर के रूप में है, तो बहन नदी की एक धारा है। दोनों साथ-साथ रहकर कल्याण के मार्ग पर चलकर सभी का सहारा बनेंगे। गंगा-यमुना के संगम में पानी की अधिकता होने से कभी बाढ़ जैसी स्थिति हो जाती है और किनारे डूबने लगते हैं। ऐसे पागलपन भरे अवसर पर बहन को भाई का ही एकमात्र सहारा है। बेफिक्री के समय में भी भाई अपनी बहन के लिए ध्रुवतारे की तरह अडिग है। कुछ समय ऐसे आते हैं जब हमें अपने संयम को दृढ़ रखना है। भाई और बहन दोनों मिलकर आजादी के गीत गाकर ही मातृभूमि के प्रति अपने कर्तव्य को पूरा कर सकते हैं। अपनी मुसीबतों को झेलकर और बलिदानों के माध्यम से पत्थर-हृदय देश के दुश्मनों को चेतावनी देते हैं कि हम हर हाल में देश की रक्षा के लिए तत्पर हैं।

काव्य सौन्दर्य :

1. भाषा सरल और मधुर है।
2. भाई-बहन का स्नेह देश-प्रेम से मिलकर उदात्त बन गया है।
3. बिम्ब और प्रतीक नवीन रूप में दर्शाए गए हैं।

MP Board Class 12th Hindi Solutions

MP Board Class 12th Maths Important Questions Chapter 5A Continuity and Differentiability

Continuity and Differentiability Important Questions

Continuity And Differentiability Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
If x = at², y = 2at, then \(\frac{dy}{dx}\) will be:
(a) t
(b) t²
(c) \(\frac{1}{t}\)
(d) \(\frac { 1 }{ t^{ 2 } } \)
Answer:
(c) \(\frac{1}{t}\)

Question 2.
If y = 2\(\sqrt { cot(x^{ 2 }) } \) ,then \(\frac{dy}{dx}\) will be:

Answer:
(a)

Question 3.
The value of \(\frac{d}{dx}\) (x³ + sin x²) is to be:
(a) 3x² + cos x²
(b) 3x² + x sin x²
(c) 3x² + 2x cos x²
(d) 3x² + x cos x²
Answer:
(c) 3x² + 2x cos x²

Question 4.
The value of \(\frac{d}{dx}\) a^x is to be:
(a) a^x
(b) a^x log_a e
(c) a^x log_e a
(d) \(\frac { a^{ x } }{ log_{ e }a } \)
Answer:
(c) a^x log_e a

Question 5.
If y = 500e^7x + 600e^-7x, then the value of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) will be:
(a) 45 y
(b) 47 y
(c) 49 y
(d) 50 y
Answer:
(c) 49 y

Question 2.
Fill in the blanks:

1. Differential coefficient of cos x⁰ w.r.t x is ……………………………..
2. Differential coefficient of e^log_ea w.r.t x is ……………………………….
3. Differential coefficient of log_e a w.r.t x is ………………………………
4. Differential coefficient of a^x w.r.t x is ……………………………….
5. Differential coefficient of sin 3x w.r.t x is ……………………………….
6. If y = sin^-1 (2x \(\sqrt { (1-x^{ 2 }) } \)), then \(\frac{dy}{dx}\) = ………………………………
7. Differential coefficient of sin x w.r.t. cos x is …………………………………..
8. The value of \(\frac{d}{dx}\) (log tan x) is ………………………………………
9. Differential coefficient of log (log sin x) ………………………………
10. If x = y\(\sqrt { (1-y^{ 2 }) } \), then \(\frac{dy}{dx}\) will be …………………………………
11. n^th differentiation of sin x will be ………………………………………
12. If y = \(\sqrt { x+\sqrt { x+……..\infty } } \), then \(\frac{dy}{dx}\) will be ……………………………….
13. If x = r cos θ, y = r sin θ, then \(\frac{dy}{dx}\) will be ………………………..
14. Differential coefficient of e^x w.r.t \(\sqrt { x } \) will be ………………………….

Answer:

1. – \(\frac { \pi }{ 180 } \) sin x⁰
2. 0
3. 0, 4
4. log_e a.a^x
5. cos 3x
6. \(\frac { 2 }{ \sqrt { 1-x^{ 2 } } } \)
7. – cot x
8. 2 cosec 2x
9. \(\frac { cotx }{ logsinx } \)
10. \(\frac { \sqrt { 1-y^{ 2 } } }{ 1-2y^{ 2 } } \)
11. sin (\(\frac { n\pi }{ 2 } \) + x)
12. \(\frac { 1 }{ 2y-1 } \)
13. – cot θ
14. 2\(\sqrt { x } \)e^x.

Question 3.
Write True/False:

1. Differential coefficient of e^log_ex is \(\frac{1}{x}\)?
2. If f(x) = \(\sqrt { x } \); x>0, then value of f'(2) is \(\frac { 1 }{ 2\sqrt { 2 } } \)?
3. Any function f(x) is said to be differentiatiable at any point x = a when Lf'(a) ≠ Rf'(a)?
4. Differential coefficient of sec^-1a w.r.t x is 0?
5. If y = Ae^mx + Be^-mx, then \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = – m²y?
6. If y = sin^-1( \(\frac { x-1 }{ x+1 } \) ) + cos^-1 ( \(\frac { x-1 }{ x+1 } \) ), then \(\frac{dy}{dx}\) = 0?
7. Every differentiatiable function is continous?
8. Differential coefficient of a^2x is a^2x log_ea?

Answer:

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

Question 4.
Match the Column:

Answer:

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

Question 5.
Write the answer in one word/sentence:

1. Find differential coefficient of \(\frac { 6^{ x } }{ x^{ 6 } } \) w.r.t. x?
2. Find differential coefficient of y = ^log_etanxw.r.t x?
3. Find n^th derivative of a^x?
4. If y = sin(ax + b), then find the value of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)?
5. If x² + y² = sin xy, then find the value of \(\frac{dy}{dx}\)?
6. Find differential coefficient of log tan \(\frac{x}{2}\) w.r.t. x?
7. Find differential coefficient of sin^-1 \(\frac { 2x }{ 1+x^{ 2 } } \) w.r.t. x?
8. Find differential coefficient of e^-log_ex w.r.t x?

Answer:

1. \(\frac { 6^{ x } }{ x^{ 6 } } \) [log 6 – \(\frac{6}{x}\) ]
2. sec² x
3. a^x(log_ea)ⁿ
4. -a²y
5. \(\frac { ycosxy-2x }{ 2y-xcosxy } \)
6. cosec x,
7. \(\frac { 2 }{ 1+x^{ 2 } } \)
8. – \(\frac { 1 }{ x^{ 2 } } \)

Continuity And Differentiability Short Answer Type Questions

Question 1.
Find all the points of discontinuity of f, when f is defined as:
f(x) = \(\left\{\begin{array}{lll}
{2 x+3,} & {\text { if }} & {x \leq 2} \\
{2 x-3,} & {\text { if }} & {x>2}
\end{array}\right.\) (NCERT)
Solution:
For x< 2, f(x) = 2x + 3 is polynomial function.
Hence, for x < 2, f(x) is continuous function. For x > 2, f(x) = 2x – 3 is polynomial function.
Hence, x > 2, f(x) is continous.
Now, we shall examine the continuty of f(x) at x = 2 only.
Put x = 2 + h,
When x → 2, then h → 0

= 2(2 + 0) – 3 = 4 – 3 = 1.
Put x = 2 – h,
When x → 2, then h → 0

= 2(2 – 0) + 3 = 7
f(2) = 2(2) + 3 = 7

Hence, f(x) is discontinous at x = 2 only.

Question 2.
Find all the points of discontunity of f, when f is defined as follow:
f(x) = \(\left\{\begin{array}{ccc}
{\frac{|x|}{x},} & {\text { if }} & {x \neq 0} \\
{0} & {\text { if }} & {x=0}
\end{array}\right.\). (NCERT)
Solution:
Hence, we shall examine the continuty of f(x) at x = 0 only,
Put x = 0 + h,
When x → 0, then h → 0

Put x = 0 – h,
When x → 0, then h → 0

= -1
Given: f(0) = 0

Hence, the given function f(x) is discontinous at x = 0.

Question 3.
Examine the continuty of function f(x) at point x = 0?
f(x) = \(\left\{\begin{array}{cc}
{\frac{1-\cos x}{x^{2}},} & {x \neq 0} \\
{\frac{1}{2},} & {x=0}
\end{array}\right.\)
Solution:
f(x) = \(\frac { 1-cosx }{ x^{ 2 } } \), when x ≠ 0.
Put x = 0 + h, when x → 0, then h → 0

Again, put x = 0 – h, when x → 0, then h → 0

Hence, f(x) is continous at x = 0.

Question 4.
Function f is defined as:
f(x) = \(\left\{\begin{aligned}
\frac{|x-4|}{x-4} ; & x \neq 4 \\
0 ; & x = 4
\end{aligned}\right.\)
Then prove that function f is continous function for all points except x = 4?
Solution:

Hence, for x = 4 the function f(x) is dicontinous

When x < 4 then f(x) = -1 which is constant function Hence, it is continous function when x > 4 then f(x) = 1 which is constant function.
Hence, it is continous function
The given functions f is continous at all points except x = 4. Proved.

Question 5.
Find the value of k for which the function
f(x) = \(\left\{\begin{array}{c}
{\frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2}} \\
{3, \text { if } x=\frac{\pi}{2}}
\end{array}\right.\)
is contionuous at x = \(\frac { \pi }{ 2 } \). (NCERT)
Solution:
Put x = \(\frac { \pi }{ 2 } \) + h,
When x → \(\frac { \pi }{ 2 } \), then h → 0


\(\frac{k}{2}\) × 1 = \(\frac{k}{2}\)
Put x = \(\frac { \pi }{ 2 } \) – h
When x → \(\frac { \pi }{ 2 } \), then h → 0

\(\frac{k}{2}\) × 1 = \(\frac{k}{2}\)
Given that f ( \(\frac { \pi }{ 2 } \) ) = 3
The given function is continous,

\(\frac{k}{2}\) = \(\frac{k}{2}\) = 3
k = 6.

Question 6.
Find the value of k, if function
f(x) = \(\left\{\begin{array}{lll}
{k x+1,} & {\text { if }} & {x \leq \pi} \\
{\cos x,} & {\text { if }} & {x>\pi}
\end{array}\right.\) is continous at x = π? (NCERT)
Solution:
Put x = π + h,
When x → π, then h → 0

= cos (π + 0)
= -1.
Put x = π – h,
When x → π, then h → 0

= k(π – 0) + 1
= πk + 1
f(π) = kπ + 1
∴ The given function is continous at x = π

-1 = kπ + 1 = kπ + 1
kπ = -2
k =- \(\frac { 2 }{ \pi } \)

Question 7.
Function f is continous at x = 0:
f (x) = \(\left\{\begin{array}{c}
{\frac{1-\cos k x}{x \sin x} ; x \neq 0} \\
{\frac{1}{2} \quad ; x=0}
\end{array}\right.\) Find the value of k?
Solution:
Given:
f(x) = \(\frac { 1-coskx }{ xsinx } \), x ≠ 0


Given: f(0) = \(\frac{1}{2}\)

Question 8.
Find the relationship between a and b so the following function f defined by:
f(x) = \(\left\{\begin{array}{lll}
{a x+1,} & {\text { if }} & {x \leq 3} \\
{b x+3,} & {\text { if }} & {x>3}
\end{array}\right.\) is continous at x = 3. (NCERT; CBSE 2011)
Solution:
Put x = 3 + h,
When x → 3, then h → 0

= b(3 + 0) + 3
= 3b + 3
Put x = 3 – h,
When x → 3, then h → 0

= a(3 – 0) + 1
= 3a + 1
f(3) = 3a + 1
The given function is continous at x = 3.

3b + 3 = 3a + 1 = 3a + 1
3a + 1 = 3b + 3
3a = 3b + 2
a = b + \(\frac{2}{3}\).

Question 9.
Prove that the funcion f(x) = |x – 1|, x ∈ R is not differentiable at x = 1? (NCERT)
Solution:
Given:

f(1) = 1 – 1 = 0
Put x = 1 – h, when x → 1, then h → 0

Put x = 1 + h, when x → 1, then h → 0

Lf'(1) ≠ Rf'(1)

Question 10.
Show that the function:
f(x) = \(\left\{\begin{aligned}
x-1, & \text { if } x<2 \\
2 x-3, & \text { if } x \geq 2
\end{aligned}\right.\), is not differentaible at point x = 2?
Solution:
We know that:
RHD =

From the above, it is clear that,
LHD at x = 2 ≠ RHD at x = 2
∴ f(x) is not differentiable at x = 2. Proved.

Question 11.
Determine the function of defined by
f(x) = \(\left\{\begin{array}{cc}
{x^{2} \sin \frac{1}{x},} & {\text { when } x \neq 0} \\
{0,} & {\text { when } x=0}
\end{array}\right.\) in continous function? (NCERT)
Solution:
Here, f(0) = 0


= 0 × a finite quantiy, [∵ sin \(\frac{1}{h}\) is between -1 and 1]
= 0

Hence, the given function is continous at x = 0.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Physics Important Questions Chapter 5 Magnetism and Matter

Magnetism and Matter Important Questions

Magnetism and Matter Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
If a bar magnet is placed with its north pole pointing towards geographical north and south pole pointing towards geographical south, then the neutral point will be:
(a) Situated on the axial line
(b) Situated on the equatorial line
(c) Situated at a place where is neither axial line nor equatorial line
(d) Not formed at all.
Answer:
(b) Situated on the equatorial line

Question 2.
Magnetic moment is which type of physical quantity :
(a) Scalar
(b) Vector
(c) Neutral
(d) None of these.
Answer:
(b) Vector

Question 3.
If a bar magnet of magnetic moment M is divided in two equal parts, the magnetic moment of each part will be :
(a) 2M
(b) \(\frac {M}{2}\)
(c) M
(d) Zero
Answer:
(b) \(\frac {M}{2}\)

Question 5.
‘weber’ is the unit of:
(a) Magnetic moment
(b) Magnetic induction
(c) Magnetic field
(d) Magnetic flux
Answer:
(d) Magnetic flux

Question 6.
The two magnetic lines of force :
(a) Meet each other at the poles
(b) Meet at the neutral point
(c) Never meet
(d) All the above statements are correct.
Answer:
(c) Never meet

Question 7.
The intensity of magnetic field is defined as :
(a) Magnetic moment per unit volume
(b) Magnetic induction force acting on unit magnetic pole
(c) Number of magnetic lines of force passing per unit area
(d) Number of lines of force passing through unit volume.
Answer:
(c) Number of magnetic lines of force passing per unit area

Question 8.
A magnetic needle is placed in a non – uniform magnetic field. The needle will experience :
(a) A force without any torque
(b) A torque without any force
(c) A force and a torque
(d) Neither a torque nor a force.
Answer:
(c) A force and a torque

Question 9.
The ratio of magnetic field intensities at equal distance in end on position and broadside on position of a small bar magnet is :
(a) 1 : 4
(b) 1 : 2
(c) 1 : 1
(d) 2 : 1.
Answer:
(d) 2 : 1.

Question 10.
A magnetic dipole of magnetic moment M is placed in a magnetic field of intensity B with its axis along the magnetic field. The work done in rotating it by 180° is :
(a) -MB
(b) MB
(c) Zero
(d) +2 MB
Answer:
(d) +2 MB

Question 11.
If the net magnetic moment of individual atom of a substance is zero, the substance is :
(a) Diamagnetic
(b) Paramagnetic
(c) Ferromagnetic
(d) Non – magnetic.
Answer:
(a) Diamagnetic

Question 12.
Electromagnets are made up of:
(a) Paramagnetic substances
(b) Soft iron
(c) Steel
(d) Diamagnetic substances.
Answer:
(b) Soft iron

Question 13.
At equator the total intensity of earth’s magnetic field is equal to :
(a) V
(b) H
(c) Both
(d) None of these.
Answer:
(b) H

Question 14.
The resultant intensity of earth’s magnetic field at a place is given by :
(a) \(\frac {H}{V}\)
(b) \(\frac {V}{H}\)
(c) \(\sqrt { { H }^{ 2 }+V^{ 2 } }\)
(d) \(\sqrt { { H }^{ 2 }-V^{ 2 } }\)
Answer:
(c) \(\sqrt { { H }^{ 2 }+V^{ 2 } }\)

Question 15.
The value of angle of dip near the magnetic poles is :
(a) 90°
(b) 45°
(c) 30°
(d) Zero.
Answer:
(a) 90°

Question 16.
The south pole of earth’s magnet is :
(a) Near the geographic north pole
(b) Near the geographic south pole
(c) In geographic east
(d) In geographic west.
Answer:
(a) Near the geographic north pole

Question 17.
The angle of dip at equator is :
(a) 90°
(b) 30°
(c) 0°
(d) 45°
Answer:
(c) 0°

Question 18.
In a plane perpendicular to the magnetic meridian, a dip needle :
(a) Will be horizontal
(b) Will be vertical
(c) Will be inclined at angle of dip at that place
(d) Will be inclined at any angle.
Answer:
(b) Will be vertical

Question 2.
Fill in the blanks :

1. The SI unit of pole strength is ……………………….
2. The direction of magnetic moment of a magnet is always from ………………………. to ………………………. pole.
3. The SI unit of magnetic moment is ……………………….
4. The magnetic lines of force are ………………………. curve.
5. The tangent drawn at any point of a magnetic line of force gives ……………………….
6. The magnetic field produced due to a solenoid is same as that produced by a ……………………….
7. A magnet is also called a ……………………….
8. At same distance, magnetic field intensity in broadside on position is ………………………. the intensity in end on position.
9. Nowadays magnetic lines of force are called ……………………….
10. At ………………………. point the resultant intensity of magnetic field is zero.
11. The temperature at which ferromagnetic substance is converted into paramagnetic substance is known as ……………………….
12. The strength of ………………………. magnet can be changed.
13. The vertical component of earth’s magnetic field at a place becomes zero where angle of dip is ……………………….
14. The angle of dip from equator to poles lies between ………………………..
15. ………………………. substances can easily be magnetized.

Answer:

1. ampere x metre
2. south; north
3. ampere x metre²
4. Closed
5. Direction of magnetic field
6. Bar magnet
7. Magnetic dipole
8. Half
9. Magnetic field line
10. Neutral
11. Curie temperature
12. Electro
13. Zero
14. Zero to 90°
15. Ferro – magnetic.

Question 3.
Match the Columns :
I.

Answer:

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

II.

Answer:

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

III.

Answer:

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

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

1. Name the elements or parameters of earth’s magnetic field.
2. What is the value of angle of dip at poles and equator?
3. How is the relative permeability (μ_r) ofthe material related to susceptibility (x_m)?
4. Give two examples of diamagnetic substance.
5. Name any two paramagnetic substances.

Answer:

1.

• Declination
• Angle of dip
• Horizontal component of earth’s magnetic field

2. Angle of dip at poles is 90° and at equator it is 0°

3. μ_r = 1 + x_m

4. Zinc and Bismuth

5. Aluminium and Manganese.

Magnetism and Matter Very Short Answer Type Questions

Question 1.
Define effective length of a magnet?
Answer:
The distance between two poles of magnet is called its effective length.

Question 2.
What do you mean by intensity of magnetic field? Write its SI unit. Is it scalar or vector?
Answer:
Intensity of magnetic field:
The intensity of field at a point is defined by the force experienced by a unit north pole, placed-at that point.
Its SI unit is tesla or weber metre^-2. Magnetic field is a vector.

Question 3.
Define magnetic lines of force.
Answer:
1st definition:
The magnetic lines of force are the curves in the magnetic field, on which if a unit north pole is placed, then it will follow the imaginary curve drawn.

2nd definition:
“A magnetic line of force is a smooth curve in a magnetic field such that the tangent at any point on it gives the direction of the magnetic field at that point.”

Question 4.
Can two magnetic lines of force intersect?
Or
Magnetic lines of force do not intersect each other, why?
Answer:
No. If the two magnetic lines of force intersect, then there will be two tangents and hence two directions of magnetic field at the point of intersection. This is impossible.

Question 5.
Define magnetic moment. Write its SI unit. Is it a scalar or a vector?
Answer:
The product of pole strength (m) and effective length (2l) of the magnet is called magnetic moment (M).
If m be the pole strength and 2l be the effective length, then
M = m x 2l
SI unit of magnetic moment is weber x metre. It is a vector quantity having a direction from south pole to north pole.

Question 6.
What is a diamagnetic substance?
Answer:
A substance which when placed in a magnetizing field develops very weak magnetization in the opposite direction of the applied field is called diamagnetic substance.

Question 7.
What is paramagnetic substance?
Answer:
A substance which when placed in a magnetizing field develops weak magnetism in the direction of the applied field is called paramagnetic substance.

Question 8.
What is paramagnetism?
Answer:
The atoms or molecules of some materials (e.g., Al, CuCl₂) have non – zero magnetic moment. When such a substance is placed in a magnetic field \(\vec { B }\), the individual magnetic dipoles align in the direction of \(\vec { B }\). There is net magnetization in the direction of \(\vec { B }\) and proportional to \(\vec { B }\) This is called paramagnetism.

Question 9.
What are ferromagnetics or ferromagnetic substances?
Answer:
Ferromagnetics are the substances, which when placed in a magnetic field are strongly magnetized in the direction of the magnetizing field. Example Fe. Ni, Co etc.

Question 10.
Write about the number of electrons in diamagnetic and paramagnetic substances.
Answer:
The number of electrons in diamagnetic substances are in even number and in paramagnetic substances electrons are in odd numbers.

Question 11.
Does the magnetism of paramagnetic salts depend upon temperature? Give reason.
Answer:
Yes, with the increase of temperature its magnetism decreases. When a paramagnetic salt is placed in a magnetic field then on each elementary magnet, a torque acts which tends to bring them in the direction of magnetic field. When the temperature is increased the thermal agitation opposes this tendency, hence the paramagnetism is decreased.

Question 12.
Define magnetic intensity. Give its SI unit.
Answer:
Magnetic intensity is the ability of a magnetizing field to magnetize a material and is defined as the number of ampere turns flowing around unit length of solenoid required to produce magnetic induction B0 inside it.
H = \(\frac { { B }_{ 0 } }{ { \mu }_{ 0 } }\)
SI unit of H is Am^-1

Question 13.
Define magnetic permeability. State its SI unit.
Answer:
Magnetic permeability is defined as the ratio of magnetic induction B to the magnetizing field intensity H ie., μ = \(\frac {B}{H}\)
SI unit is TmA^-1.

Question 14.
Why the magnetic property increases in paramagnetic substances with cooling?
Answer:
When a paramagnetic substances is kept in an external magnetic field, then on each elementary magnet a torque acts which tries to bring them parallel to the direction of magnetic field. The thermal vibrations opposes it. If the temperature is decreased, then thermal vibrations decreases, hence the magnetic property increases.

Question 15.
Why is diamagnetism independent of temperature?
Answer:
The induced magnetic moment in diamagnetic sample is always opposite to the magnetizing field, no matter what the internal motion of atom is.

Question 16.
What is Curie point?
Answer:
Curie point is the temperature above which a ferromagnetic substance becomes paramagnetic.

Question 17.
At any point on the surface of earth, horizontal component of earth magnetic field and vertical component of it are equal. What will be the angle of dip at that point?
Solution:
According to question H = V
Or B_H = B_v
But tanθ = \(\frac { { B }_{ v } }{ { B }_{ H } }\)
Or tanθ = \(\frac { { B }_{ v } }{ { B }_{ v } }\)= 1 =tan45°
θ = 45°
Angle of dip will be 45°.

Question 18.
When a bar magnet is cut into two equal pieces perpendicular to its axis, then what will be its charge in magnetic moments.
Answer:
In this position, magnetic moment of each pieces will be M’ = m’ x 2l
But m’= \(\frac {M}{2}\)
∴ M’ = \(\frac {m}{2}\) x 2l
= \(\frac {M}{2}\)
Therefore magnetic moment will become half of its initial value.

Magnetism and Matter Short Answer Type Questions

Question 1.
Write Coulomb’s law of magnetism and define the unit magnetic pole with its help.
Answer:
Coulomb’s law:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of pole strength and inversely proportional to the square of the distance between them and acts along the line joining the roles.
Let m₁ and m₂ be the pole strengths and d be the distance between them, then

Unit pole:
If F = 10^-7N, d = 1m and m₁=m₂ = m, then putting the values in eqn. (2),
we get
10^-7 = 10^-7\(\frac { { m }^{ 2 } }{ 1 } \) ⇒ m = ±1
Thus, if two similar poles are kept 1m apart in vacuum and repei each other by a force of 10^-7N, then the poles are called unit poles.

Question 2.
What is end – on – position or axial position? Derive an expression for the intensity of field at a point on the axis of a bar magnet. What is the direction of resultant field?
Or
Derive an expression for the intensity of field at a point on the axial position of a bar magnet.
Answer:
End – on – position:
The point where the intensry of the magnetic field is to be found, is on the magnetic axis, then this point is called end – on – position.

Let NS be a bar magnet of pole strength m and effective length 2l. Consider a point P on its axis at a distance d from the centre of the magnet O. Magnetic field at P has to be found out.
Now, the intensity of field at P due to N – pole :
B₁ = \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ N{ P }^{ 2 } }\),(along\(\vec { NP } \))
NP = OP – ON =d – l
∴ B₁ =\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ (d-1)^{ 2 } }\) … (1)
Similarly, the intensity of field at P due to S – pole :
B₂ = \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ S{ P }^{ 2 } } \),(along\(\vec { PS }\))
or ∴ B₁ =\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { m }{ (d+1)^{ 2 } }\) … (2)
Since, B₁ and B₂ are acting in opposite direction and B₁ > B₂
∴Resultant field B = B₁ – B₂ (along \(\vec { NP }\))
Putting the values from eqns. (1) and (2),

This is the required expression for the intensity of field on the axis.
Again, if the magnet is small i.e., I << d
By neglecting l.
\(\frac { { \mu }_{ o } }{ 4\pi } .\frac { 2m }{ ({ d }^{ 2 })^{ 2 } }\)
or \(\frac { { \mu }_{ o } }{ 4\pi } .\frac { 2M }{ ({ d })^{ 3 } }\)
In CGS units, B = \(\frac { 2M }{ ({ d })^{ 3 } }\)
The direction of resultant field is along the magnetic axis from south pole to north pole.

Question 3.
What is broad – side – on position or equatorial position? Derive an expression for the intensity at a point on broad – side – on position of a bar magnet. What will be the direction of resultant field?
Or
Determine the force on a unit north pole, kept on the broad-side-on position of a small bar magnet.
Answer:
Broad – side – on position:
When the point where the intensity of the magnetic field has to be found lies on the perpendicularbisector of magnetic axis, i.e., on the neutral axis, then it is called broad – side – on position.

Magnetic field is the force experienced by unit north pole placed at that point.
Hence, B = \(\frac {F}{m}\)
If m = 1, then B = F.
Let NS be a bar magnet of pole strength m and effective length 2l and magnetic moment M = m2l.
Consider a point P at a distance d from the centre O of a magnet on its neutral axis. Let unit north pole be placed at P.
Now, the intensity of field at P, due to N – pole will be :

Resolving B₁ and B₂ into its components, we have B₁, cosθ along NS and B₂sinθ⊥ to NS along OP. Also B₂ cosθ along NS and B₂ sinθ⊥ to NS along PO.
But B₁ = B₂
⇒ B₁ sinθ= B₂ sinθ
Since, their directions are opposite and their magnitudes are equal, hence they cancel each other.
The resultant field is therefore

Question 4.
What are magnetic lines of force ? Write down its properties.
Answer:
Magnetic lines of force :
1st definition:
The magnetic lines of force are the curves in the magnetic field, on which if a unit north pole is placed, then it will follow the imaginary curve drawn.

2nd definition:
“A magnetic line of force is a smooth curve in a magnetic field such that the tangent at any point on it gives the direction of the magnetic field at that point.”

Properties of lines of force :

1. They are closed and continuous curves.
2. Outside the magnet the direction is from north to south and inside the magnet the direction is from south to north.
3. The tangent drawn at any point on the curve gives the direction of the resultant field at that point.
4. They do not intersect each other. If two lines of force intersect at a point, then there would be two tangents at that point hence, the resultant force would have two directions; which is not possible, therefore the lines of force do not intersect.
5. They are dense near the poles where the magnetic field is strong and get separated where the magnetic field is weak.
6. They repel each other in the direction, perpendicular to it, therefore the like poles repel each other.
7. They experience tension along the lines of force therefore, unlike poles attract each other.
8. They behave just like a stretched elastic string.

Question 5.
Derive an expression for the torque acting on a bar magnet placed in a uniform magnetic field, making angle 0with the field and hence define magnetic moments with its help.
Answer:
Let NS be a bar magnet, placed in a uniform magnetic field of intensity B, making an angle θ with the field. Suppose m be the pole strength and 2l be the effective length, Force acting on each pole will be mB.

On the N – pole this force will be along the direction of field, whereas on S – pole this will be opposite to the direction of field. As two equal and opposite forces are acting on it along different line of action, hence a couple acts on it which tries to bring the magnet along the direction of magnetic field. This couple is called ‘restoring couple’ or ‘restoring torque’.

Restoring torque is defined as the product of magnitude of any one of the forces and the perpendicular distance between them.
∴ τ = Force x Perpendicular distance
or Torque, τ = mB x SP … (1)

Also, in ANPS, we get sinθ = \(\frac {SP}{NS}\)
or SP = NSsinθ = 2lsinθ
Putting the value of SP in eqn. (1),
τ = mB x 2l sinθ
But m x 2l = M (magnetic moment)
∴ τ = mB sinθ
In vector form :
\(\vec { τ }\) = \(\vec { M }\) x \(\vec { B }\)
and the direction of \(\vec { τ }\) will be perpendicular to the plane containing \(\vec { M }\) and \(\vec { B }\).

Definition of magnetic moment :
As τ = MBsinθ
If the magnet is held perpendicular to the field, then 0= 90° or sinθ = 1 then the torque acting on the magnet will be maximum, if the strength of the applied field is 1 i.e., B = 1,then
τ_max = M
Hence, magnetic moment is numerically equal to the maximum torque acting on the bar magnet when it is held perpendicular in a uniform magnetic field of unit intensity.

Question 6.
Compare to a bar magnet and a current – carrying solenoid.
Answer:
Comparison of a bar magnet and a solenoid :
Bar magnet:

• It attracts magnetic substances.
• When it is suspended freely it rests in the direction of N – S.
• It has two poles.
• Like poles of magnet repel and unlike poles attract.

Solenoid:

• It also attracts magnetic substances.
• It also rests in N – S direction if suspended freely.
• It has also two poles.
• Like poles of solenoid also repel and unlike poles attract.

Question 7.
Explain how does an atom behave as a magnetic dipole. Derive an expression for the magnetic dipole moment of the atom. Also define Bohr magneton.
Or
Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus.
Answer:
Magnetic dipole moment of a revolving electron:
In hydrogen – like atoms, an electron revolves around the nucleus. Its motion is equivalent to a current loop which possesses a magnetic dipole moment = IA. As shown in Fig., consider an electron revolving anticlockwise around a nucleus in an orbit of radius r with speed v and time – period T.

Equivalent current,

According to right hand thumb rule, the direction of the magnetic dipole moment of the revolving electron will be perpendicular to the plane of its orbit and it the downward direction, as shown in Fig.
Also, the angular momentum of the electron due to its orbital motion is
I = m_evr  … (2)
The direction of / is normal to the plane of the electron orbit and in the upward direction, as shown in Fig.
Dividing equation. (1) by equation. (2), we get

The above ratio is a constant called gyromagnetic ratio. Its value is 8.8 x 10¹⁰Ckg^-1.
So

The negative sign shows that the direction of \(\vec { l }\) is opposite to that of \(\vec { { \mu }_{ 1 } }\) According to Bohr’s quantization condition, the angular momentum of an electron in any permissible orbit is,
l = \(\frac { nh }{ 2π }\) , where n =1,2 ,3, ………….
∴ µ₁ = n(\(\frac { eh }{ 4\pi { m }_{ e } }\))
This equation gives orbital magnetic moment of an electron revolving in nth orbit.

Bohr magneton:
It is defined as the magnetic moment associated with an electron due to its orbital motion in the first orbit of hydrogen atom. It is the minimum value of µ₁, which can be obtained by putting n = 1 in the above equation. Thus Bohr magneton is given by
µ_B= (µ₁)_min = \(\frac { eh }{ 4\pi { m }_{ e } }\) = 9.27 x 10^-24Am².

Question 8.
Derive an expression for work done in rotating a bar magnet in uniform magnetic field through 8 angle.
Answer:
Let a bar magnet of effective length 2l and of magnetic moment M be kept in uniform magnetic field B. When a bar magnet is rotated through some angle in the magnetic field then some work has to be done against moment of restoring couple.
If magnet is rotated through dQ angle then work done, dW = τ dθ,
(where τ is moment of restoring couple)
or dW = MB sinθ dθ … (1)
When magnet is rotated from θ₁ to θ₂, then work done is given by :

This is the required expression.

Question 9.
Establish the expression for potential energy of a bar magnet placed in a uniform magnetic field.
Answer:
The potential energy of the bar magnet in any orientation is the work done by the external agent to turn the dipole from its zero position (θ = 90°) to that orientation (θ = θ°)
dW = MB sinθ dθ … (1)
Amount of work done to rotate the bar magnet from zero position (θ = π/2) to an arbitrary position (θ = θ) will be obtained by integrating equation  (1) under proper limit.

Question 10.
Define the magnetic elements of earth’s magnetic field at a place.
Or
Establish relation between element of earth’s magnetic field?
Answer:
Elements of earth’s magnetic field:
The earth’s magnetic field at a place can be completely described by three parameters which are called elements of earth’s magnetic field. They are declination, dip and horizontal component of earth’s magnetic field.

1. Magnetic declination:
The angle between the geographical meridian and the magnetic meridian at a place is called the magnetic declination (α) at that place, Or, it is the angle which a compass needle (free to swing in a horizontal planb) makes with the geographic north – south direction.

2. Angle of dip or magnetic inclination:
The angle made by the earth’s total magnetic field \(\vec { B }\) with the horizontal direction in the magnetic meridian is called angle of dip (δ) at any place. Or, it is the angle which a dip needle (free to swing in the plane of the magnetic meridian) makes with the horizontal.

At the magnetic equator, the dip needle rests horizontally so that the angle of dip is zero at the magnetic equator. The dip needle rests vertically at the magnetic poles so that the angle of dip is 90° at the magnetic poles. At all other places, the dip angle lies between 0° and 90°.

3. Horizontal component of earth’s magnetic field:
It is the component of the earth’s total magnetic field \(\vec { B }\) in the horizontal direction in the magnetic meridian. If δ is the angle of dip at any place, then the horizontal component of earth’s field \(\vec { B }\) at that place is given by
B_H = Bcosδ
At the magnetic equator,
δ = 0°,B_H = Bcos0°= B
At the magnetic poles,
δ = 90°,B_H =5cos90°= 0
Thus the value of BH is different at different places on the surface of the earth.

Question 11.
Prove tanδ = \(\frac { { B }_{ v } }{ { B }_{ H } }\) and B = \(\sqrt { { B }_{ H }^{ 2 }+{ B }_{ v }^{ 2 } }\) where symbol have there usual meaning.
Answer:
Relations between elements of earth’s magnetic field:
Fig. shows the three elements of earth’s magnetic field. If 8 is the angle of dip at any place, then the horizontal and vertical components of earth’s magnetic field B at that place will be

B_H = Bcosδ .. (1)
and B_v = Bsinδ
\(\frac { { B }_{ v } }{ { B }_{ H } }\) = \(\frac { Bsinδ }{ Bcosδ}\)
\(\frac { { B }_{ v } }{ { B }_{ H } }\) =tanδ .. (2)
Also
B²_H + B²_v = B²(cos²δ + sin²δ) = B²
or B = \(\sqrt { { B }_{ H }^{ 2 }+{ B }_{ v }^{ 2 } }\) .. (3)
Equations (1), (2) and (3) are the different relations between the elements of earth’s magnetic field.

Question 12.
Compare the magnetic properties of soft iron and steel.
Answer:
Comparison of magnetic properties of soft iron and steel:
Soft iron:

• In soft iron, greater magnetism can be produced, than steel. Its magnetic nature is greater than steel.
• Soft iron does not retain magnetism for longer time. Its retaintivity is less.
• The magnetization and demagnetization of soft iron are easy.
• Temporary magnets are made by soft iron.

Steel:

• In steel, less magnetism can be produced than soft iron, its magnetic nature is less than soft iron.
• Steel retains magnetism for longer time. Its retaintivity is greater than soft iron.
• The magnetization and demagnetization of steel are difficult.
• Permanent magnets are made by soft steels.

Magnetism and Matter Long Answer Type Questions

Question 1.
Answer the following regarding terrestrial magnetism quantities :

1. Three quantities are required to express a vector completely write the name of that there independent quantity.
2. At which place of south India the angle of dip in 18° will you expect more value of angle of dip at britain ?
3. If you draw lines of forces at Melbourne city of Austrilia. This lines of forces will go inside the earth or outside.
4. The magnetic needle which is free to revolve in vertical plane. If it is kept at geographical north or south pole then in which direction it will revolve?

Answer:
1.

• Angle of declination
• Angle of dip
• Horizontal component of Earth magnetic field.

2. Britain is near magnetic pole of earth, therefore angle of dip at Britain is more than the angle of dip at south India (approx. 70°).

3. The magnetic lines of force at Melbourne city of Austrilia will go outside.

4. Geographical North pole or South pole is just situated vertically to the direction of earth magnetic field. Therefore the magnetic needle will be independent to revolve in vertical plane

Question 2.
Compare the magnetic properties of paramagnetic substance and ferromagnetic substance on any three points.
Answer:
Comparison :

Question 3.
What do you mean by magnetic field intensity. Derive an expression for magnetic field due to a bar magnet in general position. How is this formula used to find magnetic field in

1. Axial position
2. Equatorial position.

Answer:
The force experienced by unit north pole at any point in the magnetic field is known as magnetic field intensity. NS is a bar magnet of magnetic moment \(\vec { M }\), we have to find out magnetic field intensity at P, which is situated at θangle from axis of magnet.

Now, we divide M into two components

• Mcosθ
• Msinθ

For Mcosθ point ‘P’ lies in axial position, therefore magnetic field at P due to M cos G component is :
B₁ = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2Msin\theta }{ { d }^{ 3 } }\) … (1)

For M sinθ point P lies on equatorial position :
B₂ = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { Msin\theta }{ { d }^{ 3 } }\) … (2)
∵ B₁is perpendicular to B₂ , the resultant of B₁ and B₂ is given by :

This is the required expression.
(i) For axial position θ = 0° => cos 0°= 1
∴From eqn. (3),
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2M }{ { d }^{ 3 } } \)
From equatorial position θ = 90° ⇒ cos90° = 0
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { M }{ { d }^{ 3 } } \)

Magnetism and Matter Numerical Questions

Question 1.
Magnetic wire of magnetic moment ‘M’ is bent in the shape of L, at one third of its length. What will be the new magnetic moment.
Solution:

Question 2.
The length of magnetic wire is L and its magnetic moment is M. If it is bent in the form of semi – circle then what will be its new magnetic moment?
Solution:
Initial magnetic moment of magnetic wire M = mL
If it is bent in the form of semi – circle then
L = πr ⇒ r = \(\frac {L}{π}\)
New magnetic moment M’ = m x 2r
M’ = m x \(\frac {2L}{π}\)
or M’ = \(\frac {2M}{π}\)

Question 3.
The distance between two magnetic poles of pole strengths ‘m’ and ‘4m’ is 3m. Find the distance of point in between them at which magnetic field intensity is zero.
Solution:

Question 4.
If the pole strength of each pole of two similar magnetic poles is made two times and distance between them becomes half of its initial value then how will magnetic force acting between them change?
Solution:

Force becomes 16 times its initial value.

Question 5.
Obtain the earth’s magnetization. Assume that the earth’s field can be approximated by a giant bar magnet of magnetic moment 8.0 x 10²²Am². The earth’s radius is 6400 km. [NCERT]
Solution:
Here magnetic moment m = 8.0 x 10²² Am²
Radius of the Earth R = 6400 km = 6.4 x 10⁶m

Question 6.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetizing current IM. [NCERT]
Solution:
Here n = 1000 tums/m, I = 2A, μ_r = 400
1. H = nI = 1000 x 2 = 2x 10³ Am^-1

2. M = x_mH = (μ_r -1)H
= (400 – 1) x 2 x 10³ ≈ 8 x 10⁵Am^-1

3. B = μH = μ_rμ₀H
= 400 x 4π x 10^-7 x 2 x 10³ T = 1.0T .

4. As M = nI_M
∴ I_M = \(\frac {M}{n}\)
= \(\frac { 8\times { 10 }^{ 5 } }{ 1000 }\)
= 8 x 10² A.

Question 7.
A short bar magnet placed with its axis at 30° experiences a torque of 0.016 Nm in an external field of 800G

1. What is the magnetic moment of the magnet?
2. What is the work done by an external force in moving it from its most stable to most unstable position?
3. What is the work done by the force due to the external magnetic field in the process mentioned in part (b)?
4. The bar magnet is replaced by a solenoid of cross – sectional area 2 x 10^-4 and 1000 turns, but the same magnetic moment Determine the current flowing through the solenoid. [NCERT]

Solution:
1. Here θ = 30°, B = 800G = 800 x 10^-4T, τ = 0.016Nm
Magnetic moment,
m = \(\frac {τ }{B sinθ}\) =\(\frac { 0.016 }{ 800\times { 10 }^{ -4 }\times sin30° }\)
= 0.40 Am².

2. For most stable position θ = 0°and for most unstable position θ = 180°. So the required work done by the external force,
W = mB (cos 180°- cos0°) = 2mB
= 2 x 0.40 x 800 x 10^-4
=0.064J

3. Here the displacement and the torque due to the magnetic field are in opposition. So the work done by the magnetic fied due to the external magnetic field is,
W_B = 0.064J

4. Here A =2 x 10^-4m², N = 1000
Magnetic moment of solenoid,
m_s = m = 0.40 Am²
But m_s = NIA.
∴ current, I = \(\frac { { m }_{ s } }{ NA }\) = \(\frac { 0.40 }{ 1000\times 2\times { 10 }^{ -4 } }\)

Question 8.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26G and the dip angle is 60°. What is the magnetic field of the Earth in this location?
Solution:
Here B_H= 0.26G, δ = 60°
As B_H = Bcosδ
∴ B = \(\frac { { B }_{ H } }{ cos\delta }\)
= \(\frac {0.26}{cos60°}\)
= \(\frac {0.26}{0.5}\)
= 0.52G

Question 9.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5 cm at a distance of 50 cm from its mid – point? The magnetic moment of the bar magnet is 0.40 Am². [NCERT]
Sol. Here m = 0.40 Am²
r = 50 cm = 0-50 m, 21 = 5-0 cm
Clearly, the magnet is a short magnet (l<<r)

Question 10.
A planar loop of irregular shape encloses an area of 7.5 x 10^-4 m² and carries a current of 12 A. The sense of flow of current appears to be clockwise to an observer. What is the magnitude and direction of the magnetic moment vector associated with the current loop? [NCERT]
Solution:
Here A = 7.5 x 10^-4m², l = 12A
Magnetic moment associated with the loop is
m = IA = 12 x 7.510^-4 = 9.0 x 10^-3 JT^-1
Applying right hand rule, the direction of magnetic moment is along the normal to the plane of the loop away from the observer.

MP Board Class 12th Physics Important Questions