Prasanna

Students can download 6th Social Science Term 3 Civics Chapter 2 Local Bodies: Rural and Urban Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Civics Solutions Term 3 Chapter 2 Local Bodies: Rural and Urban

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
……………… is set up with several village panchayats
(a) Panchayat Union
(b) District Panchayat
(c) Taluk
(d) Revenue village
Answer:
(a) Panchayat Union

Question 2.
_______ is National panchayat Raj day
a) January 24
(b) July 24
(c) November 24
(d) April 24
Answer:
(d) April 24

Question 3.
The oldest urban local body in India is ………………
(a) Delhi
(b) Chennai
(c) Kolkata
(d) Mumbai
Answer:
(b) Chennai

Question 4.
_________ District has the highest number of Panchayat Unions.
(a) Vellore
(b) Thiruvallore
(c) Villupuram
(d) Kanchipuram
Answer:
(c) Villupuram

Question 5.
The head of a corporation is called a ………………
(a) Mayor
(b) Commissioner
(c) Chair Person
(d) President
Answer:
(a) Mayor

II. Fill in the blanks

1. ……………… is the first state in India to introduce town Panchayat.
2. The Panchayat Raj Act was enacted in the year ………………
3. The tenure of the local body representative is ……………… years.
4. ……………… is the first municipality in Tamil Nadu.

Answer:

1. Tamil Nadu
2. 1992
3. 5
4. Walajahphet Municipality

III. Match the following Answer

Answer:
1. – d
2. – c
3. – a
4. – b

IV. Answer the following

Question 1.
Is there any corporation in your district? Name it.
Answer:
Yes, Tirunelveli Corporation.

Question 2.
What is the need for local bodies?
Answer:
In order to fulfill the requirements of the people and to involve them directly in governance, there is a need for an effective system of local bodies.

Question 3.
What are the divisions of a rural local body?
Answer:
The rural bodies are categorized into village panchayats, Panchayat Unions, and District panchayats.

Question 4.
What are the divisions of an Urban local body?
Answer:
The Urban local bodies are Categorized into City Municipal Corporations, Municipalities and Town Panchayats.

Question 5.
Who are the representatives elected in a Village Panchayat?
Answer:
The Elected Representatives in a Village Panchayat

1. Panchayat president
2. Ward members
3. Councillor
4. District Panchayat Ward Councillor

Question 6.
List out a few functions of corporations.
Answer:

1. Drinking-Water Supply
2. Street Light
3. Maintenance of Clean Environment
4. Primary Health facilities
5. Corporation Schools
6. Birth and Death registration etc.

Question 7.
List out a few means of revenue of village Panchayats.
Answer:

1. House tax
2. Professional tax
3. Tax on shops
4. Water charges
5. Specific fees for property tax Funds from Central and State Governments etc.

Question 8.
When are Grama Sabha meetings convened? What are the special on those days?
Answer:

1. The Grama Sabha meetings are Convened on January 26, May 1, August 15, and October 2.
2. These days are celebrated as National festival days every year.

Question 9.
What are the special features of the Panchayat Raj system?
Answer:

1. Special features of Panchayat Raj
2. Grama Sabha
3. Three-tier local body governance
4. Reservations
5. Panchayat Elections
6. Tenure
7. Finance Commission
8. Account and Audit etc.

Question 10.
What is the importance of Grama Sabha?
Answer:

1. Grama Sabha is essential for the effective functioning of Village Panchayat.
2. It enhances public participation in the planning and implementation of schemes for social benefit.

V. HOTS

Question 1.
Local bodies play an important role in the development of villages and cities. How?
Answer:

1. India is a vast nation. It is very difficult for a single government to run the entire Country.
2. Our Constitution has provided for three separate levels of government.
3. Union government
4. State government
5. Local government
6. The Local government takes care of the local administration of cities and villages.
7. The main jobs of these bodies are
8. Keeping an area clean
9. Construction of roads and schools
10. arrangements for water and electricity etc.

VI. Activities

1. Prepare a questionnaire to interview a local body representative.
2. Discuss; If there is a contribution to the improvement of your school by local body representatives.
3. If I were a local body representative, I would
4. Find out the number of local bodies in your district and list them.
   

Question 1.
Prepare …………… representative
1. What is your plan to bring in a good drainage system?
2. When will the bridgework Connecting Maharaja Nagar and Thyagaraja Nagar be completed?
3. How many months it will take to bring street lights in our area?

Question 2.
Discuss: If there representatives:
1. Local body members of our area have Contributed much to the upliftment of our school.
2. They met the VIPS, the Common public, and the old Students to meet out the needs of the school.
3. From the Sponsorship they supplied things like Computer Laboratory equipment and books for the library.

Question 3.
If I were …………… them.
1. I would take steps to eradicate Dengu and all sorts of infectious diseases.
2. I would try to maintain a proper drainage system.
3. I would bring in proper roads and lightings etc.

Samacheer Kalvi 6th Social Science Local Bodies: Rural and Urban Additional Important Questions and Answers

I. Fill in the blanks Answer

1. There are ………………. Corporations in Tamil Nadu.
2. The Chennai Corporation was founded in ……………….
3. ………………. District has the most number of municipalities
4. A ………………. Panchayat is between a Village and a city
5. The ………………. Panchayats are the local bodies of Villages.
6. The ………………. and ………………. Districts have the lowest number of Panchayat Unions.
7. The Constituencies are also called……………….

Answer:

1. Twelve
2. 1688
3. Kanchipuram
4. Town
5. Village
6. Nilgris and Perambalur
7. Wards

II. Choose the Correct answer

Question 1.
advocated Panchayat Raj as the foundation of India’s Political System ……………….
(a) Jawaharlal Nehru
(b) Mahatma Gandhi
(c) Rajendra Prasad
Answer:
(b) Mahatma Gandhi

Question 2.
A Bio element Officer (BDQ) is the administrative head of a __________
(a) Village Panchayat
(b) District Panchayat
(c) Panchayat Union
(d) Town Panchayat
Answer:
(c) Panchayat Union

Question 3.
In the 2011 Local Bodies election, percent Seats were won by women ……………….
(a) 38
(b) 28
(c) 48
Answer:
(a) 38

Question 4.
Discretionary function of a Village Panchayat __________
(a) Cleaning roads
(b) Libraries
(c) Water supply
(d) Street lighting
Answer:
(b) Libraries

Question 5.
The Tamil Nadu State Election Commission is situated in ……………….
(a) Chennai
(b) Coimbatore
(c) Trichy
Answer:
(a) Chennai

III. Answer the following briefly

Question 1.
Write about the officials of Municipal Corporation.
Answer:

1. A City Municipal Corporation has a Commissioner, who is an IAS Officer.
2. Government Officials are deputed as Commissioners for municipalities.
3. The administrative officer of a Municipality is an Executive officer (EO)

Question 2.
Explain briefly about Panchayat Union.
Answer:

1. Many Village Panchayats join to form a Panchayat Union.
2. A Councillor is elected from each Panchayat.
3. The Councillors will elect a Panchayat union Chairperson among themselves.
4. A Vice-Chairperson is also elected.
5. A Block Development Officer (BDO) is the administrative head, of a Panchayat Union.
6. The Services are provided on the Panchayat Union level.

Question 3.
List out the Strength of local bodies in Tamil Nadu.
Answer:
Tamil Nadu:

1. Village Panchayats – 12, 524
2. Panchayat Unions – 385
3. District Panchayats – 31
4. Town Panchayats – 561
5. Municipalities – 125
6. City Municipal Corporation – 12

IV. Answer the following in detail

Question 1.
Discuss the District Panchayat
Answer:

1. A District Panchayat is formed in every district.
2. A district is divided into wards on the basis of a 50,000 population.
3. The ward members are elected by the village panchayats.
4. The members of the District Panchayat elect the District Panchayat Committee Chairperson.
5. They provide essential services and facilities to the rural population.
6. They also provide planning and execution of development programmes for the district.

Question 2.
List out the Works carried out by local bodies during natural disasters and outbreaks of diseases.
Answer:

1. Rescuing the public and settle them in a safer places.
2. Arranging food packets and pure drinking water.
3. Assisting them with medical Aids.
4. Creating awareness about the clean environment to the public.
5. Keeping the medicines in the upto date conditions.
6. Preventing them from getting panic against the diseases and make the situation, calm.

V. Mind map

Students can download 6th Social Science Term 3 Civics Chapter 1 Democracy Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Civics Solutions Term 3 Chapter 1 Democracy

Samacheer Kalvi 6th Social Science Democracy Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The early man settled near ……………….. and practiced agriculture.
(a) plains
(b) bank of rivers
(c) mountains
(d) hills
Answer:
(b) bank of rivers

Question 2.
The birth place of democracy is ________
(a) China
(b) America
(c) Greece
(d) Rome
Answer:
(c) Greece

Question 3.
……………….. is celebrated as the International Democracy Day.
(a) September 15
(b) October 15
(c) November 15
(d) December 15
Answer:
(a) September 15

Question 4.
Who has the right to work in a direct Democracy?
(a) Men
(b) Women
(c) Representatives
(d) All eligible voters
Answer:
(d) All eligible voters

II. Fill in the blanks 

1. Direct Democracy is practiced in ………………
2. The definition of democracy is defined by ………………
3. People choose their representatives by giving their ………………
4. In our country ……………… democracy is in practice.

Answer:

1. Switzerland
2. Abraham Lincoln
3. Votes
4. Parliamentary

III. Answer the following

Question 1.
What is Democracy?
Answer:

1. The citizens of a country select their representatives through elections.
2. Thus they take part in the direct governance of a country. This is termed democracy.

Question 2.
What are the types of democracy?
Answer:

1. There are various types of democracy in practice around the world.
2. Among those, direct democracy and representative democracy are the most popular forms of government.

Question 3.
Define: Direct Democracy
Answer:

1. In a direct democracy, only the citizens can make laws.
2. All changes have to be approved by the citizens.
3. The politicians only rule over parliamentary procedure. Eg. Switzerland.

Question 4.
Define Representative Democracy.
Answer:

1. In a representative democracy, all the members should be represented by a group of representatives.
2. To select their representative’s elections are held.
3. On behalf of people, these representatives obtain the power to take decisions in a democratic manner.
4. This is termed Representative Democracy.

Question 5.
What are the salient features of our constitution that you have understood?
Answer:

1. The Constitution of India guides the Indians in all aspects and maintains law and order.
2. It ensures freedom, equality and justice to everyone.
3. It defines the political principles, the structure of government, the powers and responsibilities.
4. It fixes the Rights and Duties and Directive Principles of the Citizens.
5. It is the longest written constitution in the world.

IV. HOTS

Question 1.
Compare and contrast direct democracy and representative democracy.
Answer:

V. Activity

Question 1.
Find out your area’s representative’s names and write down

1. MP
2. MLA
3. Local body member

Answer:

1. MP – KRPPrabakara
2. MLA – TPM Mohideenkhan
3. Local body member – A. Radhakrishnan

Question 2.
Discuss about the merits and demerits of democracy.
Answer:
The merits of democracy are :

1. A democratic government is better form of government because it is more accountable form of government.
2. Democracy improves the quality of decision making,
3. Democracy enhances the dignity of citizens.
4. Poor and least educated has the same status as the rich and educated.
5. Democracy allows us to correct own mistake.

Demerits:

1. Leaders keep on changing leading to instability.
2. Democracy in all about political competition and power play and there is no scope for mortality.
3. So many people have to be consulted in a democracy that it lead to delays.
4. Democracy leads to corruption for it is based on electoral competition.
5. Ordinary people do not know what is good for them, they should not decide anything

Samacheer Kalvi 6th Social Science Democracy Additional Important Questions and Answers

I. Fill in the blanks Answer

1. The UNO General Assembly resolved to observe 15th September as the International Day of Democracy in ………………
2. ……………… constitution is the longest written constitution in the world.
3. The Drafting committee of the Constituent Assembly was headed by ………………
4. In India, all the people above ……………… years of age enjoy universal Adult Franchise.
5. The oldest and longest functioning parliament in the world is ………………

Answer:

1. 2007
2. Indian
3. Dr. B.R. Ambedkar
4. 18
5. The Iceland Democracy

II. Choose the Correct answer

Question 1.
The Chief Architect of our constitution is ………………
(a) Dr. Rajendra Prasad
(b) Dr. B.R. Ambedkar
(c) Dr. S. Radhakrishnan
Answer:
(b) Dr. B.R. Ambedkar

Question 2.
USA follows ______
(a) Direct democracy
(b) Representative democracy
(c) Monarchy
(d) Dictatorship
Answer:
(b) Representative democracy

Question 3.
Presidential Democracy is practised in ………………
(a) USA
(b) Canada
(c) (a) and (b)
Answer:
(c) (a) and (b)

Question 4.
Presidential democracy is followed in
(a) USA
(b) India
(c) England
(d) Switzerland
Answer:
(a) USA

Question 5.
The Constitution of India guarantees ……………… fundamental rights to its citizens.
(a) 6
(b) 9
(c) 8
Answer:
(a) 6

III. Match the following

Answer:
1. – d
2. – c
3. – b
4. – e
5. – a

IV. Answer the following questions

Question 1.
What is a Government
Answer:
A group of people with the authority to govern a country is called government.

Question 2.
How did Abraham Lincoln define democracy?
Answer:
Abraham Lincoln defined democracy as “Government of the people, by the people, and for the people”

Question 3.
What is meant by democratic decision making?
Answer:

1. In the system of democracy, the power to take decisions does not lie with the head.
2. All the members of the group hold open discussions and take final decisions only when everyone is convinced.
3. This is called democratic way of decision making.

IV. Answer the following in detail

Question 1.
What are the Aims of Democracy?
Answer:

1. To preserve and promote the dignity and fundamental rights of the individual
2. To achieve Social justice and Social development of the Community.
3. To establish the rule of law.
4. To enable the People to choose their government.
5. To work towards the development of the country with the help of People’s Participation.

V. Mind map

 

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Tamilnadu Samacheer Kalvi 6th Social Science Geography Solutions Term 3 Chapter 3 Understanding Disaster

Samacheer Kalvi 6th Social Science Understanding Disaster Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
……………… was the founder of Gupta dynasty.
(a) Chandragupta I
(b) Sri Gupta
(c) Vishnu Gopa
(d) Vishnugupta
Answer:
(b) Sri Gupta

Question 2.
Prayog prashasti was composed by ………………
(a) Kalidasa
(b) Amarasimha
(c) Harisena
(d) Dhanvantri
Answer:
(c) Harisena

Question 3.
The monolithic iron pillar of Chandragupta is at ………………
(a) Mehrauli
(b) Bhitari
(c) Gadhva
(d) Mathura
Answer:
(a) Mehrauli

Question 4.
……………… was the first Indian to explain the process of surgery.
(a) Charaka
(b) Sushruta
(c) Dhanvantri
(d) Agnivasa
Answer:
(b) Sushruta

Question 5.
……………… was the Gauda ruler of Bengal.
(a) Sasanka
(b) Maitraka
(c) Rajavardhana
(d) Pulikesin II
Answer:
(a) Sasanka

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : Chandragupta I crowned himself as a monarch of a large kingdom after eliminating various small states in Northern India.
Reason (R) : Chandragupta I married Kumaradevi of Lichchavi family.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct.
Answer:
(a) Both A and R are true and R is the correct explanation of A

Question 2.
Statement I : Chandragupta II did not have cordial relationship with the rules of South India.
Statement II : The divine theory of kingship was practised by the Gupta rulers.
(a) Statement I is wrong but statement II is correct.
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(a) Statement I is wrong but statement II is correct.

Question 3.
Which of the following is arranged in chronological order?
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya
(b) Chandragupta I – Vikramaditya – Srigupta – Samudragupta
(c) Srigupta – Samudragupta – Vikramaditya – Chandragupta I
(d) Vikramaditya – Srigupta – Samudragupta – Chandragupta I
Answer:
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya

Question 4.
Consider the following statements and find out which of the following statements (s) is/are correct.
(1) Lending money at high rate of interest was practised.
(2) Pottery and mining were the most flourishing industries,
(a) 1. is correct
(b) 2. is correct
(c) Both 1 and 2 are correct
(d) Both 1 and 2 are wrong
Answer:
(a) 1. is correct

Question 5.
Circle the odd one
(1)
Answer:
Samudragupta

(2)
Answer:
Harshacharita

III. Fill in the blanks Answer

1. …………….., the king of Ceylon, was a contemporary of Samudragupta
2. Buddhist monk from China …………….., visited India during the reign of Chandragupta II.
3. …………….. invasion led to the downfall of Gupta Empire.
4. …………….. was the main revenue to the Government.
5. The official language of the Guptas was ……………..
6. …………….., the Pallava king was defeated by Samudragupta.
7. …………….. was the popular king of Vardhana dynasty.
8. Harsha shifted his capital from …………….. to Kanauj.

Answer:

1. reign of
2. Fahien
3. Huns
4. Land tax
5. Sanskrit
6. Vishnugopa
7. Harsha Vardhana
8. Thaneswar

IV. State whether True of False

1. Dhanvantri was a famous scholar in the field of medicine.
2. The structural temples built during the Gupta period resemble the Indo – Aryan style.
3. Sati was not in practice in the Gupta Empire.
4. Harsha belonged to Hinayana school of thought
5. Harsha was noted for his religious intolerance.

Answer:

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

V. Match the following

A.

Answer:
b) 2,4,1,3,5

B.

Answer:
c) 3, 5,1, 2,4

VI. Answer in one or two sentences

Question 1.
Who was given the title Kaviraja? Why?
Answer:

1. Samudragupta was given the title Kaviraja.
2. Because he was a lover of poetry and music.

Question 2.
What are the two types of disasters? Give examples.
Answer:

1. Disaster can be classified as natural and man-made disaster.
2. Natural disaster: Earthquakes, Volcanoes, Tsunami, Cyclones, Floods, Landslides, Avalanches, Thunder and lightning.
3. Man made disaster: Fire, Destruction of building, Accidents in industries, Accident in transport, Terrorism, Stampede.

Question 3.
Explain the Divine Theory of Kingship.
Answer:

1. The Divine Theory of Kingship was practised by the Gupta rulers.
2. The king is the representative of God on earth. He is answerable only to God and not to anyone else.

Question 4.
Chennai, Cuddalore and Cauvery delta are frequently affected by floods. Give reason.
Answer:

1. In our State, Northeast Monsoon season starts from October. It will continue till December.
2. Every year, during this Northeast Monsoon season, low pressure depressions are formed in the Bay of Bengal.
3. The low pressure depressions are then transformed into cyclones and hit the coastal districts.
4. Heavy rain follows the depressions and cyclones.
5. Hence, Chennai, Cuddalore and Cauvery delta are often affected by floods

Question 5.
Who were the Huns?
Answer:

1. Huns were the nomadic tribes.
2. They were terrorising Rome and Constantinople.
3. The white Huns came to India through Central Asia.
4. They were giving trouble to all Indian frontier states.

Question 6.
Differentiate Landslide – Avalanches.
Answer:

Landslide

1. The movement of a mass of rocks, debris, soil etc., down slope is called landslide.

Avalanches

1. A large amount of ice, snow and rock falling quickly down the side of a mountain is called an Avalanches.

Question 7.
Name the books authored by Harsha.
Answer:

1. Ratnavali
2. Nagananda
3. Priyadharshika

VII. Answer the following briefly

Question 1.
Write a note on Prashasti.
Answer:

1. Prashasti is a Sanskrit word, meaning communication or in praise of.
2. Court poets flattered their kings listing out their achievements.
3. These accounts were later engraved on pillars so that the people could read them.

Question 2.
Give an account of Samudragupta’s military conquests.
Answer:

1. Samudragupta was a great general and he carried on a vigorous campaign all over the country.
2. He defeated the Pallava king Vishnugopa.
3. He conquered nine kingdoms in northern India.
4. He reduced 12 rulers of southern India to the status of feudatories and to pay tribute.
5. He received homage from the rulers of East Bengal, Assam, Nepal, the eastern part of Punjab and various tribes of Rajasthan.

Question 3.
Describe the land classification during the Gupta period.
Answer:
Classification of land during Gupta period.

1. Kshetra – Cultivable land
2. Khila – Wasteland
3. Aprahata – Jungle (or) Forest land
4. Vasti – Habitable land
5. Gapata saraha – Pastoral land

Question 4.
Write about Sresti and Sarthavaha traders.
Answer:
Sresti:
Sresti traders were usually settled at a standard place.

Sarthavaha:
Sarthavaha traders caravan traders who carried their goods to different places.

Question 5.
Highlight the contribution of Guptas to architecture.
Answer:

1. From the earlier tradition of rock-out shrines, the Guptas were the first to contruct temples.
2. These temples, adorned with towers and elaborate carvings, were dedicated ‘ to all Hindu deities.
3. The most notable rock – cut caves are found at Ajanta and Ellora, Bagh and Udaygiri.
4. The structural temples built during this period resemble the Dravidian style.

Question 6.
Name the works of Kalidasa.
Answer:

1. Kalidasa’s famous dramas were Sakunthala, Malavikagnimitra and Vikramaoorvashiyam.
2. Other sigrificant works were Meghaduta, Raghuvamsa, Kumarasambava and Ritusamhara

Question 7.
Estimate Harshvardhana as a poet and a dramatist.
Answer:

1. Harsha himself was a poet and dramatist.
2. Around him gathered a best of poets and artists.
3. His popular works are Ratnavali, Nagananda and Priyadharshika
4. His royal court was adorned by Banabhatta, Mayura, Hardatta and Jayasena.

VIII. HOTS

Question 1.
The gold coins issued by Gupta kings indicate.
Answer:
(a) the availability of gold mines in the kingdom
(b) the ability of the people to work with gold
(c) the prosperity of the kingdom
(d) the extravagant nature of kings.
Answer:
(c) the prosperity of the kingdom

Question 2.
The famous ancient paintings at Ajanta were painted on.
(a) walls of caves
(b) ceilings of temples
(c) rocks
(d) papyrus
Answer:
(a) walls of caves

Question 3.
Gupta period is remembered for.
(a) renaissance in literature and art
(b) expeditions to southern India
(c) invasion of Huns
(d) religious tolerance
Answer:
(a) renaissance in literature and art

Question 4.
What did Indian scientists achieve in astronomy and mathematics during the Gupta period?
Answer:

1. Invention of Zero and the cosequent evolution of the decimal system to the modern world.
2. Aryabhatta explained the true causes of solar and lunar eclipses. He was the, first Indian astronomer to declare that the earth revolves around its own axis.

IX. Student activity (For Students)

1. Stage any one of the dramas of Kalidasa in the classroom.
2. Compare and contrast the society of Guptas with that of Mauryas.

X. Life Skills (For Students)

1. Collect information about the contribution of Aryabhatta, Varahamihira and Brahmagupta to astronomy.
2. Visit a nearby ISRO centre to know more about satellite launching.

XI. Answer Grid

Question 1.
Who was Toromana?
Answer:
Answer:
Toromana was the chief of white Huns.

Question 2.
Name the high ranking officials of Gupta Empire.
Answer:
Dandanayakas and Maha dandanayakas

Question 3.
Name the Gupta kings who performed AsVamedha yagna.
Answer:
Samudragupta and Kumaragupta I

Question 4.
Name the book which explained the causes for the lunar and solar eclipses.
Answer:
Surya Siddhanta

Question 5.
Name the first Gupta king to find a place on coins.
Answer:
Samudragupta

Question 6.
Which was the main source of information to know about the Samudragupta’s reign?
Answer:
Allahabad Pillar

Question 7.
Harsha was the worshipper of in the beginning.
Answer:
Shiva

Question 8.
Universitv reached its fame during Harsha period.
Answer:
The Nalanda

Samacheer Kalvi 6th Social Science Understanding Disaster Additional Important Questions and Answers

I. Choose the Correct Answer

Question 1.
The successor of Sri Gupta …………….
(a) Kumaragupta I
(b) Skandagupta
(c) Vishnugupta
(d) Ghatotkacha
Answer:
(d) Ghatotkacha

Question 2.
‘Nitisara’ emphasises the importance of …………….
(a) Trade
(b) Military
(c) Agriculture
(d) Treasury
Answer:
(d) Treasury

Question 3.
The Huhs chief who crowned himself as king.
(a) Yasodharman
(b) Attila
(c) Mihirakula
(d) Toromana
Answer:
(d) Toromana

Question 4.
The Gupta coins were known as Dinara …………….
(a) Copper
(b) Silver
(c) Bronze
(d) Gold
Answer:

Question 5.
The place Harsha went to participate in the great Kumbhamela held.
(a) Allahabad
(b) Kasi
(c) Ayodhya
(d) Prayag
Answer:
(d) Prayag

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : The last of the great Guptas Narasimha Gupta I was paying tribute to Mihirakula.
Reason (R) : He stopped paying tribute as Mihirakula’s hostility towards Buddhism.
(a) Both A and R are true and R is the correct explantion of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct
(d) A is not correct but R is correct
Answer:
(b) Both A and R are correct but R is no correct explanation of A

Question 2.
Statement I : Criminal law was not more severe than that of the Gupta age.
Statement II : Death punishment was the punishment for violation of the laws and for plotting against the king.
(a) Statement I is wrong but statement II is correct
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(d) Both the statements are wrong

III. Fill in the blanks

1. In the assembly at ……………. Harsha distributed his weath.
2. The capital of China ……………. was a great centre of art and learning.
3. ……………. was wife of chandragupta I.
4. The military campaigns of kings were financed through revenue.
5. The peasants were required to pay various taxes and were reduced to the position of ……………..

Answer:

1. Prayag
2. Xi’an
3. Kumaradevi
4. surpluses revenue
5. serfs

IV. Match the following

Answer:
b) 4, 5, 2,1, 3

V. Answer in one or two sentences

Question 1.
Write a note on ‘Lichchhavi’.
Answer:

1. Lichchhavi was an old gana – Sanga and its territory lay between the Ganges and the Nepal Terai.
2. Chandragupta I married Kumaradevi of the famous and powerful lichchhavi family.

Question 2.
What do you know about ‘Kaviraja’?
Answer:

1. In one of the gold coins issued by Samudragupta he is portrayed playing harp (Veena).
2. He was a lover of poetry and music and for this, he earned the title ‘Kaviraja’.

Question 3.
What did the travel accounts of Fahien provide information about the condi¬tions of the people of Magadha?
Answer:

1. According to Fahien the people of Magadha were happy and prosperous.
2. Gaya was desolated. Kapilvasthu had become a jungle, but at Pataliputra people were rich and prosperous.

VII. Answer the following briefly

Question 1.
Name the officials employed by the Gupta rulers.
Answer:

1. High – ranking officials were called dandanayakas and mahadandnayakas.
2. The provinces known as deshas or bhuktis were administered by the governors designated as Uparikas. The districts such as vishyas, were controlled by vishyapatis. At the village level gramika and gramadhyaksha were the functionaries.
3. The military designations.
   Baladhikrita (Commander of infantry)
   Mahabaladhikrita (Commander of cavalry)
   Dutakas (spies)

Question 2.
Mention the importance of Forecasting and Early warning.
Answer:
(i) Weather forecasting, Tsunami early warning system, cyclonic forecasting and warning provide necessary information. This information help in reducing risks during disasters.

(ii) School Disaster Management Committee, Village Disaster Management Committee, State and Central government institutions take mitigation measures, together during disaster.

(iii) Newspaper, Radio, Television and social media bring updated information and give alerts on the vulnerable area, risk preparatory measures and relief measures including medicine.

VIII. Mind map

Students can Download Tamil Nadu 11th Physics Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 2 English Medium

General Instructions:

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

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be ………………..
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Hint:
Volume of the sphere, V = \(\frac{4}{3}\) πr³
\(\frac{∆V}{V}\) × 100 = 3 × (\(\frac{∆r}{r}\) × 100) = 3 × 2%
\(\frac{∆V}{V}\) × 100 = 6%
Answer:
(d) 6%

Question 2.
A ball is dropped from a building. It takes 4s to reach the ground. The height of the building is (use g= 10 m/s²)
(a) 20 m
(b) 40 m
(c) 80 m
(d) 75 m
Hint:
s = ut + \(\frac{1}{2}\) at² s = h, g = a, u = 0
h = \(\frac{1}{2}\) gt²
h = \(\frac{1}{2}\) × 10 × (4)²; h = 80m
Answer:
(c) 80 m

Question 3.
For inelastic collision between two spherical rigid bodies ……………………
(a) the total kinetic energy is conserved
(b) the total mechanical energy is not conserved
(c) the linear momentum is not conserved
(d) the linear momentum is conserved
Answer:
(d) the linear momentum is conserved

Question 4.
Two rods OA and OB of equal length and mass are lying on xy plane as shown in figure. Let I_x, I_y and I_z be the moments of inertia of the the rods about x, y and z axis respectively, then …………………….
(a) I_x = I_y > I_z
(b) I_x > I_y > I_z
(c) I_x = I_y < I_z
(d) I_z > I_y > I_x

Hint:
I_x = I_y = 2\(\left[\frac{\mathrm{M} l^{2}}{3} \sin ^{2} 45^{\circ}\right]\) = \(\frac { ml^{ 2 } }{ 3 }\)
I_z = \(\left[\frac{m l^{2}}{3}\right]\) = \(\frac { 2ml^{ 2 } }{ 3 }\)
Answer:
(c) I_x = I_y < I_z

Question 5.
The motion of a rocket is based on the principle of conservation of ………………….
(a) Linear momentum
(b) Mass
(c) Angular momentum
(d) Kinetic energy
Answer:
(a) Linear momentum

Question 6.
The work done by the Sun’s gravitational force on the Earth is ………………….
(a) Always zero
(b) Always positive
(c) Can be positive or negative
(d) Always negative
Answer:
(c) Can be positive or negative

Question 7.
A given glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown. If the tube is rotated with a constant angular velocity ω, then …………………
(a) Water levels in both sections A and B go up
(b) Water level in section A goes up and that in B comes down
(c) Water level in section B goes up and that in A comes down
(d) Water level remain same in both
Answer:
(a) Water levels in both sections A and B go up

Question 8.
The efficiency of a heat engine working between the freezing point and boiling point of water is …………………..
(a) 6.25%
(b) 20%
(c) 26.8%
(d) 12.5%
Hint:
Freezing point of water T_L = 0°C = 273K
Boiling point of water T_H – 100°C = 373K
∴ Efficiency, η = 1 – \(\frac { T_{ L } }{ T_{ H } } \) = 1- \(\frac{273}{373}\) = 0.2861; η = 26.8%.
Answer:
(c) 26.8%

Question 9.
Two waves represented by the following equation are travelling in the same medium
y₁ = 5 sin 2π (75 t – 0.25 x), y₂ = 10 sin 2π (150 – 0.25 x)
The intensity ratio of the two waves is ………………….
(a) 1 : 2
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
Hint:
I ∝ A² ⇒ \(\frac { I_{ 1 } }{ I_{ 2 } }\) = (\(\frac { A_{ 1 } }{ A_{ 2 } }\))² = (\(\frac{5}{2}\))² = \(\frac{1}{4}\)
Answer:
(b) 1 : 4

Question 10.
A man pushes a wall and fails to displace it. He does …………………..
(a) Negative work
(b) Positive but not maximum work
(c) No work at all
(d) Maximum work
Answer:
(c) No work at all

Question 11.
A car moving on a horizontal road may be thrown out of the road in taking a turn …………………
(а) By the gravitational force
(b) Due to lack of sufficient centripetal force
(c) Due to rolling frictional force between tyre and road
(d) Due to the reaction of the ground
Answer:
(b) Due to lack of sufficient centripetal force

Question 12.
The volume of a gas expands by 0.25 m³ at a constant pressure of 10³ N/m, the workdone is equal to ……………………..
(a) 250 W
(b) 2.5 W
(c) 250 N
(d) 250 J
Hint:
Workdone = P. ∆V = 10³ × 0.25 = 250 J
Answer:
(d) 250 J

Question 13.
When three springs of spring constants k₁, k₂, k₃ connected in parallel, then the resultant spring constant is ……………………
(a) K = k₁ + k₂ + k₃
(b) \(\frac{1}{K}\) = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(c) K = \(\frac { 1 }{ k_{ 1 } } +\frac { 1 }{ k_{ 2 } } +\frac { 1 }{ k_{ 3 } } \)
(d) K = k₁ – k₂ – k₃
Answer:
(a) K = k₁ + k₂ + k₃

Question 14.
The distance of the two planets from Sun are 10¹³ and 10¹² m respectively. The ratio of time period of the planets is …………………..
(a) 100
(b) \(\frac { 1 }{ \sqrt { 10 } } \)
(c) \(\sqrt{10}\)
(d) 10\(\sqrt{10}\)
Hint:
According to Kepler’s third law of planetary motion.
\(\frac { T_{ 1 } }{ T_{ 2 } } \) = \(\sqrt{\frac{R_{1}^{3}}{R_{2}^{3}}}\) = \(\sqrt{\frac{\left(10^{13}\right)^{3}}{\left(10^{12}\right)^{3}}}\) = \(\sqrt{\frac{10^{39}}{10^{36}}}\) = \(\sqrt{10^{3}}\) = 10\(\sqrt{10}\)
Answer:
(d) 10\(\sqrt{10}\)

Question 15.
The dimensional formula of planck’s constant is ………………..
(a) [M L² T^-1]
(b) [M L² T^-3]
(c) [M L T^-1]
(d) [M L³ T^-3]
Answer:
(a) [M L² T^-1]

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A particle is moving along a circular track of radius lm with uniform speed. What is the ratio of the distance covered and the displacement in half revolution?
Answer:
Distance covered by a particle = π × 1 = πm
Displacement covered by a particle = 2 × 1 = 2m
Ratio between distance and displacement

Question 17.
Give one argument in favour of the fact that frictional force is a non-conservative force?
Answer:
The direction of the frictional force is opposite to the direction of motion. When a body is moved, say from A to B and then back to A, work is required to be done both during forward and backward motion. So, the net workidone in a round trip is not zero. Hence, the frictional force is a non-conservative force.

Question Question 18.
Why does a gas not have a unique value of specific heat?
Answer:
This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating.

Question 19.
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest path in 15 min. Calculate the velocity of river water in km/h?
Answer:
Resultant velocity = \(\frac{1km}{(15/60)h}\) = 4 km/h
If v is velocity of river, then v² + 4² = 5² ⇒ v = \(\sqrt{2-16}\) =3 km/h

Question 20.
In a dark room would you be able to tell whether a given note had been produced by a Piano or a Violin?
Answer:
Yes, in a dark room we can easily identify a sound produced by a Piano or a Violin by using the knowledge of timber or quality of sound. The two sources even though having the same intensity and fundamental frequency will be associated with different number of overtones of different relative intensities. These overtones combine and produce different sounds which enables us to identify them.

Question 21.
What is mean by P – V diagram?
Answer:
PV diagram is a graph between pressure P and volume V of the system. The P-V diagram is used to calculate the amount of work done by the gas during expansion or on the gas during compression.

Question 22.
Why does a parachute descend slowly?
Answer:
The surface area of a parachute is much larger as compared to the surface area of stone. So, the air resistance in the case of a parachute is much larger than in the case of a stone. This explains as to why parachute descends slowly.

Question 23.
What is Brownian motion?
Answer:
The motion of the particles in a random and zig-zag mannar in a fluid is called Brownian motion.

Question 24.
Write a note on reverberation?
Answer:
The persistence of audible sound after the source has ceased to emit sound is called reverberation.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Write the rules for determining significant figure?
Answer:
Rules for counting significant figures:

Question 26.
Explain Joule’s experiment of the mechanical equivalent of heat?
Answer:
Joule’s mechanical equivalent of heat:
The temperature of an object can be increased by heating it or by doing some work on it. In the eighteenth century, James Prescott Joule showed that mechanical energy can be converted into internal energy and vice versa. In his experiment, two masses were attached with a rope and a paddle wheel as shown in Figure. When these masses fall through a distance h due to gravity, both the masses lose potential energy equal to 2 mgh.

When the masses fall, the paddle wheel turns. Due to the turning of wheel inside water, frictional force comes in between the water and the paddle wheel. This causes a rise in temperature of the water. This implies that gravitational potential energy is converted to internal energy of water.

The temperature of water increases due to the work done by the masses. In fact, Joule was able to show that the mechanical work has the same effect as giving heat. He found that to raise 1 g of an object by 1°C , 4.186 J of energy is required. In earlier days the heat was measured in calorie. 1 cal = 4.186 J This is called Joule’s mechanical equivalent of heat.

Question 27.
How do you classify the physical quantities on the basis of dimension?
Answer:
(I) Dimensional variables:
Physical quantities, which possess dimensions and have variable values are called dimensional variables. Examples are length, velocity, and acceleration etc.

(II) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.

(III) Dimensional Constant:
Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck’s constant etc.

(IV) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are n, e, numbers etc.

Question 28.
State laws of simple pendulum?
Answer:
Law of length:
For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
T ∝ \(\sqrt{l}\)

Law of acceleration:
For a fixed length, the time period of a simple pendulum is inversely proportional to square root of acceleration due to gravity.
T ∝ \(\frac { 1 }{ \sqrt { g } } \)

Question 29.
Explain super position principle for gravitational field?
Answer:
The total gravitational field at a point due to all the masses is given by the vector sum of the gravitational field due to the individual masses. This principle is known as superposition of
\(\overrightarrow{\mathrm{E}}_{\text {total }}=\overrightarrow{\mathrm{E}}_{1}+\overrightarrow{\mathrm{E}}_{2}+\ldots \overrightarrow{\mathrm{E}}_{n}=-\frac{\mathrm{G} m_{1}}{r_{1}^{2}} \hat{r}_{1}-\frac{\mathrm{G} m_{2}}{r_{2}^{2}} \hat{r}_{2}-\ldots \cdot \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)
\(\overrightarrow{\mathrm{E}}_{\mathrm{total}}=-\sum_{i=1}^{n} \frac{\mathrm{G} m_{n}}{r_{n}^{2}} \hat{r}_{n}\)

Question 30.
Write a note on static friction?
Answer:
Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
0 ≤ f_s ≤ µ_sN
where, µ_s – coefficient of static friction
N – Normal force then, equation shows that f_s can take any value from 0 & µ_sN. If object is at rest, when no external force acts on it then f_s = 0. If object is at rest, also external force acts on it, then f_s = F_ext
But still the static friction f_s is less than µ_sN when object begins to slide, the static friction (f_s) acting on the object attains maximum.

Question 31.
A small metal ball falls in liquid with a terminal velocity of V. If a ball of radius twice of first ball but same mass falls through a same medium, calculate the terminal velocity with which it falls?
Answer:
Given v = \(\frac{2 r^{2} \rho g}{9 \eta}\)
mass = \(\frac{4}{3}\) πr³ ρ = \(\frac{4}{3}\)π(2r)³ ρ¹ or ρ¹ = \(\frac{ρ}{8}\)
Terminal velocity of second ball is
v¹ = \(\frac{2(2 r)^{2}\left(\frac{\rho}{8}\right) g}{9 \eta}\) = \(\frac{v}{2}\)
v¹ = \(\frac{v}{2}\)

Question 32.
Derive an expression for co-efficient of performance of refrigerator?
Answer:
Coefficient of performance (COP) (β):
COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.
COP = β = \(\frac { Q_{ L } }{ W } \) …………………… (1)
From the equation Q_L + W = Q_H
β = \(\frac{Q_{L}}{Q_{H}-Q_{L}}\)
β = \(\frac{1}{\frac{\mathrm{Q}_{\mathrm{H}}}{\mathrm{Q}_{\mathrm{L}}}-1}\) ………………. (2)
But we know that \(\frac { Q_{ H } }{ Q_{ L } } \) = \(\frac { T_{ H } }{ T_{ L } } \)
Substituting this equation (1) we get β = \(\frac{1}{\frac{T_{H}}{T_{L}}-1}\) = \(\frac{\mathrm{T}_{\mathrm{L}}}{\mathrm{T}_{\mathrm{H}}-\mathrm{T}_{\mathrm{L}}}\)

Question 33.
Derive an expression for Laplace’s correction?
Answer:
Laplace’s correction:
In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast. Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat.

Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………… (4)

where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume.

Differentiating equation (4) on both the sides, we get
\(\mathrm{V}^{\gamma} d \mathrm{P}+\mathrm{P}\left(\gamma \mathrm{V}^{\gamma-1} d \mathrm{V}\right)=0\)
or
\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………… (5)

where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (5) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_a = \(\sqrt{\frac{B_{A}}{\rho}}=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\gamma v_{T}}\)

Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_a = (\(\sqrt{1.4}\)) (280m s^-1) = 331.30 m s^-1, which is very much closer to experimental data.

PART – IV

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

Question 34 (a).
Explain different types of error?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows

(i) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(ii) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(iii) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(iv) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(v) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer. Who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs.

For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.

If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ………………. a_n. The arthematic mean is
\(a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots a_{n}}{n}\) (or) \(a_{m}=\frac{1}{n} \sum_{i=1}^{i=n} a_{i}\)

[OR]

(b) By using equations of motion, derive an expression for range and maximum height reached by the object thrown at an oblique angle θ with respect to the horizontal direction?
Answer:

This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.

(Oblique projectile):
Examples:

1. Water ejected out of a hose pipe held obliquely.
2. Cannot fired in a battle ground.

Consider an object thrown with initial velocity at an angle 0 with the horizontal.
Then,
\(\vec { u } \) = u_x\(\hat { i } \) + u_y\(\hat { j } \)
where u_x = u cos θ is the horizontal component and u_y = u sin θ the vertical component of velocity.

Since the acceleration due to gravity is in the direction opposite to the direction of vertical component u_y, this component will gradually reduce to zero at the maximum height of the projectile.

At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (u_x = u cos θ) remains the same till the object reaches the ground.

Hence after the time f, the velocity along horizontal motion v_x = u_x + a_xt = u_x = u cos θ.
The horizontal distance travelled by projectile in time t is s_x = u_xt + \(\frac{1}{2}\)a_xt²
Here, s_x = x, u_x = u cos θ, a_x = 0.

Thus, x = u cos θ.t or t = \(\frac{x}{u cosθ}\) …………….. (1)
Next, for the vertical motion v_y= u_y + a_yt
Here u_y = u sin θ, ay = – g (acceleration due to gravity acts opposite to the motion).
Thus, v_y = u sin θ – gt
The vertical distance travelled by the projectile in the same time t is
Here, s_y = y, u_y = u sin θ, a_x = – g Then,
y = u sin θ t – \(\frac{1}{2}\)gt² ……………….. (2)
Subsitute the value of t from equation (i) and equation (ii), we have
y = \(u \sin \theta \frac{x}{u \cos \theta}-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\)
y = \(x \tan \theta-\frac{1}{2} g \frac{x^{2}}{u^{2} \cos ^{2} \theta}\) ………………… (3)
Thus the path followed by the projectile is an inverted parabola.

Maximum height (h_max):
The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
\(v_{y}^{2}=u_{y}^{2}+2 a_{y} s\)
Here, u_y = u sin θ, a = -g, s = h_max, and at the maximum height v_y = 0
Hence, (0)² = u² sin² θ = 2gh_max
or
h_max = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) …………….. (4)

Time of flight (T_f):
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight.
This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that s_y = u_yt + \(\frac{1}{2}\) a_yt²
Here, s_y = y = 0 (net displacement in y-direction is zero), u_y = u sin θ, a_y = -g, t = T_f.
Then 0 = u sinθ T_f – \(\frac{1}{2}\)gT^2_f
T_f = 2u \(\frac{sin θ}{g}\) …………………. (5)

Horizontal range (R):
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write

Range R = Horizontal component of velocity x time of flight = u cos θ × T_f = \(\vec{r}_{1} \times \vec{r}_{2}\)

The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum,
sin 2θ = 1. This implies 2θ = π/2
or θ = \(\frac{π}{4}\)
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range is given by.
R_max = \(\frac { u^{ 2 } }{ g }\) ……………….. (6)

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows: Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let 0 be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between R and A. Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

(I) Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.
From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴ AN = B cos θ and sin θ = \(\frac{BN}{B}\) ∴BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B² sin² θ
⇒ R² = A² + B ² (cos² θ + sin² θ) + 2AB cos θ
⇒ R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

(II) Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
IF \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA + AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1(\(\frac{B \sin \theta}{A+B \cos \theta}\))

[OR]

(b) Arrive at an expression for velocity of objects in one dimensional elastic collision?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁u₁ + m₂v₁ ………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) ………………. (2)
Further,

Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ……………….. (3)
After simplifying and rearranging terms,
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
m₁(u₁ + v₁) = m₂(v₂ + u₂) (v₂ – u₂) …………………. (4)
Dividing equation (4) by (2) gives,
\(\frac{m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)}{m_{1}\left(u_{1}-v_{1}\right)}=\frac{m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)}{m_{2}\left(v_{2}-u_{2}\right)}\)
u₁ + v₁ = v₂ + u₁
Rearranging, u₁ – u₂ = v₂ – v₁ ………………… (5)
Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………. (6)
or v₂ = u₁ + v₁ – u₁ ………………. (7)

To find the final velocities v₁ and v₂:
Substituting equation (7) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – 2a₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ = m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂) u₁ + 2m₂u₂ = (m₁ + m₂)v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂ ……………. (8)
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final velocity of m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ ………………. (9)

Question 36 (a).
Discuss the variation of g with change in altitude and depth?
Answer:
When an object is on the surface fo the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the objecf would have been mg. Elowever, the object experiences an additional centrifugal force due to spinning of the Earth.

This centrifugal force is given by mω²R’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g’ = g – ω²2 R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Variation of g with depth:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points. The part of the Earth which is above the radius (R_e – d) do not contribute to the acceleration. The result is proved earlier and is given as

g’ = \(\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}_{\mathrm{e}}-d\right)^{2}}\) ………………. (1)
Here M’ is the mass of the Earth of radius (R_e – d) Assuming the density of the earth ρ be constant,
ρ = \(\frac{M}{V}\) …………………. (2)

where M is the mass of the Earth and V its volume, Thus,

Here also g’ < g. As depth increases, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

[OR]

(b) Explain in detail the maxwell Boltzmann distribution function?
Answer:
Maxwell-Boltzmann: In speed distribution function:-
Consider an atmosphere, the air molecules-are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed.

In the previously we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of 5 ms^-1 to 10 ms^-1 or 10 ms^-1
to 15 ms^-1 etc. In general our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell’s speed distribution function.

The above expression is graphically shown as follows:

From the figure (1), it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure (1). It can be seen that the rms speed is greatest among the three.

1. The area under the graph will give the total number of gas molecules in the system.
2. Figure 2 shows the speed distribution graph for two different temperatures.

As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

Question 37 (a).
What is meant by simple harmonic motion?
Answer:
A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.

(b) The acceleration due to gravity on the surface of the moon is 1.7 ms^-2. What is the time period of simple pendulum on the moon if its time period on the Earth is 3.5 s? Given g on Earth = 9.8 ms^-2?
Answer:

(c) A man with wrist watch on his hand falls from the top of the tower. Does the watch , give correct time during the free fall?
Answer:
Yes. Because the working of a wrist watch does pot depend on gravity at that place but depends on spring action.

[OR]

(d) State Wien’s law?
Answer:
When the animals feed cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

(e) Normal human body of the temperature is 98.6°F. During high fever if the temperature increases to 104°F. What is the change in peak wavelength that emitted by our body (Assume human body is a black body)?
Answer:
Normal human body temperature (T) = 98.6°F.
Convert Fahrenheit into Kelvin, \(\frac{F-32}{180}\) = \(\frac{K-273}{100}\)
So, T = 98.6°F = 310K
From Wien’s displacement law
Maximum wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9348 × 10^-9 m
λ_max = 9348 nm (at 98.6°F)
During high fever, human body temperature
T = 104°F = 313K
Peak wavelength λ_max = \(\frac{b}{T}\) = \(\frac { 2.898\times 10^{ -3 } }{ 313 } \)
λ_max = 9259 × 10^-9 m
λ_max = 9259 (at 104°F)

(f) Animals curl into a ball, when they feel very cold. Why?
Answer:
When animals feel cold, they curl their bodies into the ball so as to decrease the surface area of their bodies. As total energy radiated by a body varies directly as the surface area of the body, the loss of heat due to radiation would be reduced.

Question 38 (a).
Explain the horizontal oscillations of spring?
Answer:
Horizontal oscillations of a spring-mass system: Consider a system containing a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant k placed on a smooth horizontal surface (frictionless surface) as shown in figure.

Let x₀ be the equilibrium position or mean position of mass m when it is left undisturbed. Suppose the mass is displaced through a small displacement x towards right from its equilibrium position and then released, it will oscillate back and forth about its mean position x₀. Let F be the restoring force (due to stretching of the spring) which is proportional to the amount of displacement of block. For one dimensional motion, . mathematically, we have.
F ∝ x
F = -kx ……………… (1)

where negative sign implies that the restoring force will always act opposite to the direction of the displacement. This equation is called Hooke’s law. Notice that, the restoring force is linear with the displacement (i.e., the exponent of force and displacement are unity).

This is not always true; in case if we apply a very large stretching force, then the amplitude of oscillations becomes very large (which means, force is proportional to displacement containing higher powers of x) and therefore, the oscillation of the system is not linear and hence, it is called non-linear oscillation.

We restrict ourselves only to linear oscillations throughout our discussions, which means Hooke’s law is valid (force and displacement have a linear relationship). From Newton’s second law, we can write the equation for the particle executing simple harmonic motion.

\(m \frac{d^{2} x}{d t^{2}}=-k x\) ……………….. (1)
\(\frac{d^{2} x}{d t^{2}}=-\frac{k}{m} x\) ……………….. (2)
Comparing the equation with simple harmonic motion equation, we get
ω² = \(\frac{k}{m}\) ………………….. (3)

which means the angular frequency or natural frequency of the oscillator is
ω = \(\sqrt{\frac{k}{m}} \mathrm{rad} s^{-1}\) ……………….. (4)

The frequency of the oscillation is
f = \(\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \mathrm{Hertz}\) …………………… (5)
and the time period of the oscillation is
T = \(\frac{1}{f}\) = 2π \(\sqrt{m/k}\) seconds …………………. (6)

[OR]

(b) What is capillarity? Obtain an expression for the surface tension of a liquid by capillary rise method?
Answer:
In a liquid whose angle of contact with solid is less than 90° suffers capillar rise. On the other hand, in a liquid whose angle of contact is greater than 90°, suffers capillary fall. The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Practical application of capillarity:

1. Due to capillary action, oil rises in the cotton within an earthen lamp. Likewise, sap raises from the roots of a plant to its leaves and branches.
2. Absorption of ink by a blotting paper.
3. Capillary action is also essential for the tear fluid from the eye to drain constantly.
4. Cotton dresses are preferred in summer because cotton dresses have fine pores which act as capillaries for sweat.

Surface Tension by capillary rise method:
The pressure difference across a curved liquid air interface is the basic factor behind the rising up of water in a narrow tube (influence of gravity is ignored). The capillary rise is more dominant in the case of very fine tubes.

But this phenomenon is the outcome of the force of surface tension. In order to arrive a relation between the capillary rise (h) and surface tension (T), consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.

The surface tension force F_T acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, it resolved into two components

1. Horizontal component T sin θ and
2. Vertical component T cos θ acting upwards, all along the whole circumference of the meniscus.

Total upward force = (T cos θ) (2nr) = 2nrT cos θ)

where θ is the angle of contact, r is the radius of the tube. Let ρ be the density of water and h be the height to which the liquid rises inside the tube. Then,

The upward force supports the weight of the liquid column above the free surface, therefore,

If the capillary is a very fine tube of radius (i.e., radius is very small) then \(\frac{r}{3}\) can be neglected when it is compared to the height h. Therefore,
T = \(\frac{r \rho g h}{2 \cos \theta}\)
Liquid rises through a height h
h = \(\frac{2 \mathrm{T} \cos \theta}{r \rho g} \Rightarrow h \alpha \frac{1}{r}\)
This implies that the capillary rise (h) is inversely proportional to the radius (r) of the tube, i.e., the smaller the radius of thd tube greater will be the capillarity.

Students can download 6th Social Science Term 3 Geography Chapter 2 Globe Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Geography Solutions Term 3 Chapter 2 Globe

Samacheer Kalvi 6th Social Science Globe Text Book Back Questions and Answers

I. Fill in the blanks

1. The line of latitude which is known as the Great Circle is ……………..
2. The imaginary lines drawn horizontally on Earth from the West to East are called ……………..
3. The 90° lines of latitude on the Earth are called ……………..
4. The Prime Meridian is also called ……………..
5. The world is divided into …………….. time zones.

Answer:

1. Equator
2. lines or parallels of latitude
3. Poles
4. Greenwich meridian
5. 24

II. Choose the best answer

Question 1.
The shape of the Earth is ……………..
(a) Square
(b) Rectangle
(c) Geoid
(d) Circle
Answer:
(c) Geoid

Question 2.
The North Pole is _______.
(a) 90° N Latitude
(b) 90° S latitude
(c) 90° W Longitude
(d) 90° E longitude
Answer:
(a) 90° N Latitude

Question 3.
The area found between 0° and 180° E lines of longitude is called ……………..
(a) Southern Hemisphere
(b) Western Hemisphere
(c) Northern Hemisphere
(d) Eastern Hemisphere
Answer:
(d) Eastern Hemisphere

Question 4.
The 23 ° N line of latitude is called _______.
(a) Tropic of Capricorn
(b) Tropic of Cancer
(c) Arctic Circle
(d) Antarctic Circle
Answer:
(b) Tropic of Cancer

Question 5.
180° line of longitude is ……………..
(a) Equator
(b) International Date Line
(c) Prime Meridian
(d) North Pole
Answer:
(b) International Date Line

Question 6.
The Sun is found overhead the Greenwich Meridian at ________
(a) 12 midnight
(b) 12 noon
(c) 1 p.m.
(d) 11 a.m.
Answer:
(b) 12 noon

Question 7.
A day has ……………
(a) 1240 minutes
(b) 1340 minutes
(c) 1440 minutes
(d) 1140 minutes
Answer:
(c) 1440 minutes

Question 8.
Which of the following lines of longitude is considered for the Indian Standard Time?
(a) 82 1/2° E
(b) 82 1/2° W
(c) 811/2° E
(d) 81 1/2° W
Answer:
(a) 82 1/2° E

Question 9.
The total number of lines of latitude are ……………
(a) 171
(b) 161
(c) 181
(d) 191
Answer:
(c) 181

Question 10.
The total number of lines of longitude are ________
(a) 370
(b) 380
(c) 360
(d) 390
Answer:
(c) 360

III. pic the odd one Answer

Question 1.
North Pole, South Pole, Equator, International Date Line.
Answer:
International Dateline

Question 2.
Tropic of Capricorn, Tropic of Cancer, Equator, Prime Meridian.
Answer:
Prime Meridian

Question 3.
Torrid Zone, Time Zone, Temperate Zone, Frigid Zone
Answer:
Time Zone

Question 4.
Royal Astronomical Observatory, Prime Meridian, Royal Astronomical Greenwich Meridian, International Date Line.
Answer:
Royal Astronomical observatory

Question 5.
10° North, 20° South, 30°, North, 40° West
Answer:
40° west.

IV. Match the following

Answer:
1. – d
2. – c
3. – b
4. – a

V. Examine the following statements

1. The Earth is spherical in shape.
2. The shape of the Earth is called a geoid.
3. The Earth is flat.
Look at the options given below and choose the correct answer
(a) 1 and 3 are correct
(b) 2 and 3 are correct
(c) 1 and 2 are correct
(d) 1, 2, and 3 are correct
Answer:
(c) 1 and 2 are correct

VI. Examine the following statements

Question 1.
Statement I: The lines of latitude on Earth are used to find the location of a place and define the heat zones on Earth.
Statement II: The lines of longitudes on Earth are used to find the location of a place and to calculate time.
Choose the correct option:
(a) Statement I is correct; II is wrong
(b) Statement I is wrong; II correct
(c) Both the statements are correct
(d) Both the statements are wrong
Answer:
(a) Statement I is correct; II is wrong

VII. Name the following

Question 1.
The imaginary lines are drawn horizontally on Earth
Answer:
Lines of latitude/parallels

Question 2.
The imaginary lines are drawn vertically on Earth.
Answer:
Lines of longitude /Meridians

Question 3.
The three dimensional model of the Earth.
Answer:
Globe

Question 4.
India is located in this hemisphere based on lines of longitude.
Answer:
Southern Hemisphere

Question 5.
The network of lines of latitude and longitude.
Answer:
Earth grid / Geographic grid

VIII. Answer briefly

Question 1.
What is a geoid?
Answer:

1. The Earth cannot be compared with any other geometrical shape as it has a very unique shape.
2. Hence, its shape is called a geoid (earth shaped).

Question 2.
What is the local time?
Answer:

1. When the sun is overhead on a particular line of longitude, it is 12 moon at all the places located on that line of longitude.
2. This is called local time.

Question 3.
How many times would the sun pass overhead a line of longitude?
Answer:
The sun is overhead on a line of longitude only once a day.

Question 4.
What are lines of latitude and longitude?
Answer:

1. There are imaginary lines which are drawn on the globe horizontally and vertically to find a location and calculate distance and time.
2. These imaginary lines are called lines of latitudes and longitudes.

Question 5.
Name the four hemispheres of the Earth.
Answer:

1. Northern Hemisphere,
2. Southern Hemisphere,
3. Eastern Hemisphere and
4. Western Hemisphere.

IX. Give reason

Question 1.
The 0° line or longitude is called the Greenwich Meridian.
Answer:
The 0° line of longitude is called the Greenwich Meridian because it passes through Greenwich.

Question 2.
The regions on Earth between North & South lines of latitude (66 Vi°) and poles (90°) is called Frigid Zone.
Answer:

1. From the Arctic circle (66 ‘A° N) to the North Pole (90° N) and from the Antarctic circle (66 Vi0 S) to the South Pole (90° S) the sun’s rays full further inclined, through out the year.
2. The temperature is very low.
3. Hence this region is known as Frigid Zone.

Question 3.
The International Date Line runs zigzag.
Answer:
If the International Date line is drawn straight, two places in the same country would have different dates. So the International date line is found Zig Zag to avoid confusion.

X. Answer in detail

Question 1.
What are the uses of globe?
Answer:

1. Since the Earth is huge and we live on a very area, we are not able to see the Earth as a whole.
2. But when we travel to space, we can see the Earth as a whole.
3. So, in order to see the shape of the Earth as a whole and to know its unique features, a three-dimensional model of the Earth was created with a specific scale in the name of the globe.

Question 2.
How are the hemispheres divided on the basis of lines of latitude and longitude? Explain with diagrams.
Answer:
Northern Hemisphere:
Northern Hemisphere. The area of the Earth found between the Equator (Oo) and the North pole (90oN) is called the Northern Hemisphere.

Southern Hemisphere:
Southern Hemisphere. The area of the Earth from the Equator (0°) and the South pole (90°S) is called the Southern Hemisphere.

Eastern Hemisphere:
Eastern Hemisphere. The part of the Earth between the 0° line of longitude and 180° East line of longitude is known as the Eastern Hemisphere.

Western Hemisphere
Western Hemisphere. The part of the Earth from Oo line of longitude and 180° West line of longitude is called as Western Hemisphere.

Question 3.
What are the significant lines of latitude? Explain the zones found between them.
Answer:

1. The significant lines of latitude are
2. Equator 0°
3. Tropic of cancer 23 1/2° N
4. Tropic of Capricorn 23 1/2° S
5. Arctic Circle 66 1/22°N
6. Antarctic Circle 66 1/2° S
7. North Pole 90° N
8. South Pole 90° S
9. Based on the amount of heat received from the sun, our world is broadly divided into three heat zones.

1. The Torrid Zone : On both sides of the equator, the region lying between the Tropic of cancer and Tropic of Capricorn is called the Torrid zone. This zone gets the direct rays of the sun all the year round. Hence the climate is hot and humid.

2. The Temperature Zone : This zone is neither hot nor very cold. It lies between the Torrid zone and Frigid zone in both the hemispheres.

3. The Frigid Zone : The region lying between the Arctic circle and North pole and between Antarctic circle and south pole is called the frigid zone. It receives very slanting rays of the sun and is therefore very cold.

Question 4.
Explain : Indian Stanard Time.
Answer:

1. The longitudinal extent of India is from 68° 7’E to 97° 25’E
2. Twenty nine lines of longitude pass through India.
3. Having 29 standard time is not logical
4. Hence 821/2° E line of longitude is observed as the prime meridian to calculate the Indian standard Time (IST)

XI. Activity

There are five Posistion marked on the grid given below. Look at them carefully and fill the blanks with reference to the lines of latitude and longitude. The first one is done for you.

1. The latitudinal and longitudinal reference of point A 40° N 30°W
2. The latitudinal and longitudinal reference of point B 20° N 10°W
3. The latitudinal and longitudinal reference of point C 10° N 20°W
4. The latitudinal and longitudinal reference of point D 40° N 50°W
5. The latitudinal and longitudinal reference of point E 20° N 20°W

Samacheer Kalvi 6th Social Science Globe Additional Important Questions and Answers

I. Fill in the blanks Answer

1. The surface area of the Earth is ………….. million square kilometres.
2. ………….. was the first person to draw the lines of latitude and longitude on a map
3. The Royal Astronomical observatory is located at …………..
4. The directions on the ground are always shown with respect to the …………..
5. There are four ………….. directions.

Answer:

1. 510.1
2. Ptolemy
3. Greenwich
4. North
5. Cardinal

II. Answer the following questions

Question 1.
Who said this? The stars in the sky seem to move towards the west because of the Earth’s rotation on its axis
Answer:
Aryabhata – The Indian astronomer

Question 2.
Who wrote the book ‘Geographia’?
Answer:
Ptolemy, (a Greco – Roman mathematician (astronomer and geographer)

Question 3.
Name the country that has 7 time zones.
Answer:
Russia

III. Answer in briefly

Question 1.
Define Meridian
Answer:
The word meridian is derived from the latin word ‘Meridianus’. It means mid day. So, meridian means the postion of the sun overhead at the place at noon.

Question 2.
What is an axis?
Answer:

1. An imaginary line around which a large round object such as the Earth turns.
2. The Earth rotates on its axis between the north and South Poles.

Question 3.
What is hemisphere?
Answer:
A half of the earth usually as divided into Northern and Southern halves by the equator or into Western and Eastern halves by an imaginary line passing through the poles.

IV. Choose the correct answer

Question 1.
23 1/220° N and S 66 1/220° N and S lines of latitudes are called ……………..
(a) Low latitudes
(b) Middle latitudes
(c) High latitudes
Answer:
(b) Middle latitudes

Question 2.
The Sun’s rays fall vertically in this region.
(a) Frigid Zone
(b) Temperate Zone
(c) Torrid Zone
Answer:
(c) Torrid Zone

Question 3.
High latitudes are there between
(a) 66 1/2°N and S – 90°N and S
(b) 231/2°N and S and 66f°N and S
(c) 0° – 23 1/2°N and S
(d) North poles
Answer:
(a) 66 1/2° N and S – 90°N and S

Question 4.
Latitudes are also known as ……………..
(a) Nilavangu
(b) Ahalangu
(c) Nettangu
Answer:
(b) Ahalangu

Question 5.
Lines of latitudes merge
(a) at poles
(b) at the equator
(c) at International Date line
(d) do not merge
Answer:
(d) do not merge

V. Match the following

Answer:
1. – b
2. – d
3. – e
4. – a
5. – c

VI. Answer in detail

Question 1.
Draw the Heat Zones of the Earth
Answer:

VII. Mind map

Students can Download Tamil Nadu 11th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 1 English Medium

General Instructions:

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

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec average velocity is ………………..
(a) 0
(b) 2 m/s
(c) 4 m/s
(d) 2π m/s
Hint:
Displacement of the cyclist in half revolution is
d = diameter of the circular track
i.e., d= 80 m
Time taken, t = 40 s
Average velocity, V =  = \(\frac{80}{40}\)
V = 2 m/s
Answer:
(b) 2 m/s

Question 2.
A wheel has angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 seconds it has rotated through an angle of (in radian) ………………..
(a) 10
(b) 12
(c) 4
(d) 6
Answer:
(a) 10

Question 3.
If the origin of co-ordinate system lies at the centre of mass. The sum of the moments of the masses of the system about the centre of mass is …………………
(a) May be greater than zero
(b) May be less than zero
(c) May be equal to zero
(d) Always zero
Answer:
(d) Always zero

Question 4.
Dimensional formula for co-efficient of viscousity
(a) ML^-2 T^-2
(b) ML^-2 T^-1
(c) ML^-1 T^-1
(d) M^-1 L^-1 T^-1
Answer:
(c) ML^-1 T^-1

Question 5.
Action and reaction …………………
(a) Acts on same object
(b) Acts on two different objects
(c) Have resultant not zero
(d) Acts on the same direction
Answer:
(b) Acts on two different objects

Question 6.
A spring is stretched by applying load to its free end. The strain produced in the spring is …………………
(a) Volumetric
(b) Shear
(c) Longitudinal
(d) Longitudinal and shear
Answer:
(d) Longitudinal and shear

Question 7.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 ms^-2
(d) 25 ms^-2

Hint:
τ = F × r
Iα = F × r
MR² × α = 30 × \(\frac{40}{100}\); \(\frac { 3\times 40\times 40\times \alpha }{ 100\times 100 } \) = 12
\(\frac { 3\times 16\times \alpha }{ 100 } \) = 12; α = 25 rad/s²
Answer:
(b) 25 rad s^-2

Question 8.
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (E is total energy) ………………
(a) \(\frac{2}{3}\) E
(b) \(\frac{1}{3}\) E
(c) \(\frac{1}{4}\) E
(d) \(\frac{1}{2}\) E
Hint:
PE = \(\frac{1}{2}\) kx²
⇒ \(P E_{V_{2}}=\frac{1}{2} K\left(\frac{A}{2}\right)^{2}=\frac{1}{4}\left(\frac{1}{2} K A^{2}\right)\)
Answer:
(c) \(\frac{1}{4}\) E

Question 9.
A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad/s and 7.5 m/s² respectively. Amplitude of the oscillation is ………………
(a) 0.36
(b) 0.28
(c) 0.61
(d) 0.53
Hint:
x = A sin ωt
∴ a = \(\frac { d^{ 2 }x }{ dt^{ 2 } } \) = -Aω² sinωt
∴Maximum acceleration |a_max| = Aω²
Now Aω² = 7.5
A = \(\frac { 7.5 }{ \omega ^{ 2 } } \) = \(\frac { 7.5 }{ (3.5)^{ 2 } } \) = 0.61
Answer:
(c) 0.61

Question 10.
If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved, then its fundamental frequency will become ………………….
(a) \(\frac{n}{4}\)
(b) \(\sqrt{2n}\)
(c) n
(d) \(\frac { n }{ \sqrt { 2 } } \)
Hint:

Answer:
(c) n

Question 11.
The theory of refrigerator is based on …………….
(a) Joule-Thomson effect
(b) Newton’s particle theory
(c) Joule’s effect
(d) None of the above
Answer:
(d) None of the above

Question 12.
Work done by 0.1 mole of a gas at 27°C to double its volume at constant pressure is ………………..
(a) 54 cal
(b) 60 cal
(c) 546 cal
(d) 600 cal
Hint:
Workdone (W) = – p.dv = nRT
= 0.1 × (0.2 cal) × (273 + 27) = 0.1 × 2 × 300
W = 60 cal
Answer:
(b) 60 cal

Question 13.
When a lift is moving upwards with acceleration a, then time period of simple pendulum in it will ………………..
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)
(b) 2π\(\sqrt { \frac { g+a }{ l } } \)
(c) \(\frac{1}{2π}\)\(\sqrt { \frac { 1 }{ g+a } } \)
(d) \(\frac{1}{2π}\)\(\sqrt { \frac { g+a }{ l } } \)
Answer:
(a) 2π\(\sqrt { \frac { 1 }{ g+a } } \)

Question 14.
A disc is rotating with angular speed ω. If a child sits on it, what is conserved?
(a) Linear momentum
(b) Angular momentum
(c) Kinetic energy
(d) Potential energy
Answer:
(b) Angular momentum

Question 15.
The vectors \(\vec { A } \) and \(\vec { B } \) are such that |\(\vec { A } \) + \(\vec { B } \)| = |\(\vec { A } \) – \(\vec { B } \)|. The angle between the two vector is ………………..
(a) 45°
(b) 60°
(c) 75°
(d) 90°
Hint:
The angle between two vector is always 90°.
Answer:
(d) 90°

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
Velocity – time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement – time graph for the motion of the object?
Answer:
The given graph shows that the velocity of the object is constant. That is, the velocity of the object is not changing, so the acceleration of the object is zero. Since the acceleration of an object is given by


Displacement – time graph for the motion of the object is shown in the figure above.

Question 17.
Can a body subjected to a uniform acceleration always move in a straight line?
Answer:
It will be a straight line in one dimensional motion but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform acceleration.

Question 18.
Calculate the viscous force on a ball of radius 1mm moving through a liquid of viscosity 0.2 Nsm^-2 at a speed of 0.07 ms^–
Answer:
Radius of the ball (a) = 1mm = 1 × 10^-3m
Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2
Speed of the ball (v) = 0.07 ms^-1
According to Stoke’s law
Viscous force F = 6 π η av
= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07
= 0.26376 × 10^-3 = 2.64 × 10^-4N

Question 19.
Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10ms^-2)
Answer:
Given data: F = 30 N, load (m) 2 kg; height = 10m, g = 10 ms^-2
Gravitational forcc F = mg = 30 N
The distance moved h = 10 m
Work done on the object W = Fh = 30 × 10 = 300 J

Question 20.
Why are shockers used in automobiles like car?
Answer:
In the event of jump or jerk, the lime of action of force increases. Since the product of force aid time is constant in a given situation. therefore the force decreases.

Question 21.
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighted 250 N on the surface?
Answer:
As g_d = g (1 – \(\frac{d}{R}\)) ⇒ mg_d = mg(1 – \(\frac{d}{R}\))
Here d = \(\frac{R}{2}\)
∴mg_d = (250) × (1 – \(\frac { R/2 }{ R } \)) = 250 × \(\frac{1}{2}\) = 125N

Question 22.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors \(\vec { A } \) and \(\vec { B } \) are perpendicular to each other than their scalar product \(\vec { A } \).\(\vec { B } \) = O because cos 90° = 0. Then the vectors \(\vec { A } \) and \(\vec { B } \) are said to be mutually orthogonal.

Question 23.
An air bubble of radius r in water is at a depth of h below the water surface at some instant. If P is atmospheric pressure and d and T are the density and surface tension of ater respectively. Calculate the pressure P inside the bubble?
Answer:
Excess ot pressure inside the air bubble in water = \(\frac{2T}{r}\)
∴ Total pressure inside the air bubble
= atmospheric pressure + pressure due to liquid column + Excess pressure due to surface tension
= P + hρg + \(\frac{2T}{r}\)

Question 24.
Define beats?
Answer:
Formation of beats: When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second.
n = |f₁ – f₂| per second.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Define centripetal acceleration and give any two examples?
Answer:
The acceleration that is directed towards the centre of the circle along the radius and perpendicular to the velocity of the particle is known as centripetal or radial or normal acceleration.
Example:-

1. In the case of planets revolving round the Sun or the moon revolving round the earth, the centripetal force is provided by the gravitational force of attraction between them.
2. For an electron revolving round the nucleus in a circular path, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Question 26.
Write any six properties of vector product of two vectors?
Answer:
(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec { A } \) and \(\vec { B } \), even though the vectors \(\vec { A } \) and \(\vec { B } \) may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec { A } \) × \(\vec { B } \) ≠ \(\vec { B } \) × \(\vec { A } \). But, \(\vec { A } \) × \(\vec { B } \) = – \(\vec { B } \) × \(\vec { A } \) .
Here it is worthwhile to note that |\(\vec { A } \) × \(\vec { B } \)| = |\(\vec { B } \) × \(\vec { A } \)| = AB sin θ i.e., in the case of the product vectors \(\vec { A } \) × \(\vec { B } \) and \(\vec { B } \) × \(\vec { A } \), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin θ = 1, i.e., θ = 90° i.e., when the vectors \(\vec { A } \) and \(\vec { B } \) are orthogonal to each other.
(\(\vec { A } \) × \(\vec { B } \))_max = AB\(\hat { n } \)

(IV) The vector product of two non-zero vectors will be minimum when sin θ = 0, i.e., θ = 0° or θ = 180°
(\(\vec { A } \) × \(\vec { B } \))_min = 0
i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector
\(\vec { A } \) × \(\vec { A } \) = AA sin 0° \(\hat { n } \) = \(\vec { 0 } \)
In physics the null vector \(\vec { 0 } \) is simply denoted as zero.

(VI) The self-vector products of unit vectors are thus zero.
\(\hat { i } \) × \(\hat { i } \) = \(\hat { j } \) × \(\hat { j } \) = \(\hat { k } \) × \(\hat { k } \) = 0

Question 27.
Show that the pressure of the gas is equal to two third of mean kinetic energy per unit volume?
Answer:
The internal energy of the gas is given by
U = \(\frac{3}{2}\) NkT
The above equation can also be written as
U = \(\frac{3}{2}\) PV
Since PV = NkT
P = \(\frac{2}{3}\) \(\frac{U}{V}\) = \(\frac{2}{3}\) u
From the equation (1), we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density (u = \(\frac{U}{V}\))
Writing pressure in terms of mean kinetic energy density using equation.
P = \(\frac{1}{3}\) nm\(\overline { V^{ 2 } } \) = \(\frac{1}{3}\) ρ\(\overline { V^{ 2 } } \)
where ρ = nm = mass density (Note n is number density)
Multiply and divide R.H.S of equation (2) by 2, we get
P = \(\frac{2}{3}\)(\(\frac{ρ}{2}\) \(\overline { V^{ 2 } } \))
P = \(\frac{2}{3}\) \(\overline { KE } \)
From the equation (3), pressure is equal to \(\frac{2}{3}\) of mean kinetic energy per unit volume.

Question 28.
Derive an expression for gravitational potential energy?
Answer:
The gravitational tbrce is a conservative force and hence we can define a gravitational potential energy associated with this conservative force field.
Two masses m₁ and m₂, are initially separated by a distance r’. Assuming m₁ to be fixed in its position. work must be done on m₂ to move the distance from r’ to r.


To move the mass m₂, through an infinitesimal displacement d\(\vec { r } \) from \(\vec { r } \) to \(\vec { r } \) + d\(\vec { r } \) , work has to be done externally. This infinitesimal work is given by
dW = \(\vec { F } \)_ext . d\(\vec { r } \) ……………….. (1)
The work is done against the gravitational force, therefore,
\(\vec { F } \)_ext = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \) ………………. (2)
Substituting equation (2) in (1), we get
dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).d\(\vec { r } \) ………………… (3)
d\(\vec { r } \) = dr \(\hat { r } \) ⇒ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \hat { r } \).(dr \(\hat { r } \))
\(\hat { r } \).\(\hat { r } \) = 1 (Since both are unit vectors)
∴ dW = \(\frac { Gm_{ 1 }m_{ 2 } }{ r^{ 2 } } \) dr …………….. (4)
Thus the total work done for displacing the particle from r’ to r is

This work done W gives the gravitational potential energy difference of the system of masses and m₁ and m₂ when the separation between them are r and r’ respectively.

Question 29.
A satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by satellite due to Earth’s gravitational force?
Answer:
g’ = g(1 – \(\frac { 2h }{ R_{ e } } \))
= g(\(\frac { 1-2\times 1600\times 10^{ 3 } }{ 6400\times 10^{ 3 } } \)) = g(1 – \(\frac{2}{4}\))
g’ = g (1 – \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = g (1- \(\frac{1}{2}\)) = \(\frac{g}{2}\)
g’ = \(\frac{8}{2}\) (or) g’ = \(\frac{9.8}{2}\) = 4.9ms^-2

Question 30.
Explain the v ariation of a g with latitude?
Answer:
When an object is on the surface fo the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by mωR’.
\(\mathrm{OP}_{z}, \cos \lambda=\frac{\mathrm{PZ}}{\mathrm{OP}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}\)
R’ = R cos λ
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
\(a_{\mathrm{PQ}}=\omega^{2} \mathrm{R} \cos \lambda=\omega^{2} \mathrm{R} \cos ^{2} \lambda\)
Since R’ = R cos λ

Therefore, g = g – ω²R cos² λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g’ = g, it is maximum. At the equator, g’ is minimum.

Question 31.
A bullet of mass 50g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms^-2
Answer:
m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁ The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\)
v = \(\frac{0.05 u_{1}+(0.45 \times 0)}{(0.05+0.45)}\) = \(\frac{0.05}{0.50}\)u₁
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
v = \(\sqrt{2gh}\)
v = \(\sqrt{2 \times 10 \times 1.8}\) = \(\sqrt{36}\)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
6 = \(\frac{0.05}{0.50}\)u₁ or u₁ = \(\frac{0.05}{0.50}\) × 6 = 10 × 6
u₁ = 60ms^-1

Question 32.
If the piston of a container is pushed fast inward. Will the ideal gas equation be valid in the intermediate stage? If not, why?
Answer:
When the piston is compressed so quickly that there is no time to exchange heat to the surrounding, the temperature of the gas increases rapidly. In this intermediate stage the ideal gas equation be not valid. Because this equation can be relates the pressure, volume and temperature of thermodynamic system at equilibrium.

Question 33.
Calculate how many times more intense is 90 dB sound compared to 40 dB sound?
Answer:
Given Data:
L = log \(\frac { I }{ I_{ 0 } } \) = log I – log I₀
We get 90 dB = 9 B = log I₁ – log I₀ ……………….. (1)
40 dB = 4 B = logI₂ – logI₀ ………………….. (2)
Subtract (2) from (1)
50 dB = 5B = log I₁ – logI₂
5 = log₁₀ (\(\frac{I_{1}}{I_{2}}\))
\(\frac{I_{1}}{I_{2}}\) = 10⁵

PART – IV

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

Question 34 (a)
Obtain an expression for the time period T of a simple pendulum. The time period T depends on

1. Mass of the bob(m)
2. Length of the pendulum (l)
3. Acceleration due to gravity (g) at the place where the pendulum is suspended, (constant k = 2π)

Answer:
Example:
An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon

1. Mass m of the bob
2. Length l of the pendulum and
3. Acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is k = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c ⇒ a = 0, b = 1/2, and c = -1/2
From the above equation
T = \(\mathrm{k} \mathrm{m}^{0} l^{1}=g^{-1}-2\)
T = \(k\left(\frac{1}{g}\right)^{1}\) = \(k \sqrt{1 / g}\)
Experimentally k = 2π, hence
T = \(2 \pi \sqrt{1 / g}\)

[OR]

(b) Obtain an expression for the escape speed in detail?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v_i the initial total energy of the object is
\(\mathrm{E}_{i}=\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) ………………. (1)
where, M_E is the mass of the Earth and R_E the radius of the Earth.
The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_F = 0
E_i = E_f ……………….. (2)
Substituting (1) in (2) we get,

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e i.e..

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 35 (a).
Derive an expression for loss of kinetic energy in perfect inelastic collision?
Answer:
In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let KE_i be the total kinetic energy before collision and KE_f be the total kinetic energy after collision.
Total kinetic energy before collision,
\(\mathrm{KE}_{i}=\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}\) …………….. (1)
Total kinetic energy after collision
KE_f = \(\frac{1}{2}\) (m₁ + m₂)v² …………….. (2)
Then the loss of kinetic energy is
Loss of KE, ∆Q = KE_f – KE_i = \(\frac{1}{2}\) (m₁ + m₂)v² – \(\frac{1}{2}\) m₁ u₁² – \(\frac{1}{2}\) m₂ u₂² ………………. (3)
Substituting equation v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}\) in equation (3), and on simplying (expand v by using the algebra) (a + b)² = a² + b² + 2ab, we get
Loss of KE, ∆Q = \(\frac{1}{2}\) \(\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)\) (u₁ – u₂)²

(b) A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. Calculate the initial velocity of the shell?
Answer:
Given Data :
m = 200 gm = 0.2 kg; M = 4 kg.
Energy generated = 1.05 KJ = 1.05 × 10³ J
According to law of Conservation of linear momentum
mv = Mv’
∴v’ = (\(\frac{m}{M}\)) v
Total K.E of the gun and bullet

[OR]

(c) State parallel axis theorem?
Answer:
Parallel axis theorem: Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²

(d) Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about

1. The diameter of the disc
2. The axis, tangent to the disc and parallel to its diameter
3. The axis through the centre of the disc and perpendicular to its plane

Answer:
1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m
Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR²
Id = \(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m²

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc
= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = \(\frac{5}{4}\) × 0.5 × 1
= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc
= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²

Question 36 (a).
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it?
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum.

When two particles interact with each other, they exert equal and opposite forces on each other. The particle 1 exerts force \(\vec { F } \)₁₂ on particle 2 and particle 2 exerts an exactly equal and opposite force \(\vec { F } \)₁₂ on particle 1 according to Newton’s third law.
\(\vec { F } \)₂₁ = –\(\vec { F } \)₁₂ …………… (1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as
\(\vec { F } \)₁₂ = \(\frac{d \bar{p}_{1}}{d t}\) and \(\vec { F } \)₂₁ = \(\frac{d \vec{p}_{2}}{d t}\) ……………… (2)

Here \(\vec { P } \)₁ is the momentum of particle 1 which changes due to the force \(\vec { F } \)₁₂ exerted by particle 2. Further \(\vec { P } \)₂ is the momentum of particle 2. This changes due to \(\vec { F } \)₂₁ exerted by particle 1.
Substitute equation (2) in equation (1)

It implies that \(\vec { P } \)₁ + \(\vec { P } \)₂ = (constant vector always)

\(\vec { P } \)₁ + \(\vec { P } \)₂ is the total linear momentum of the two particles ( \(\vec { P } \)_tot = \(\vec { P } \)₁ + \(\vec { P } \)₂). It is also called as total linear momentum of the system. Flere, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system ( \(\vec { P } \)_tot) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec { P } \)₁ and \(\vec { P } \)₂ can vary,
in such a way that \(\vec { P } \)₁ + \(\vec { P } \)₂ is a constant vector.

The forces \(\vec { F } \)₁₂ and \(\vec { F } \)₁₂ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

To find the recoil velocity of a gun when a bullet is fired from it:
Consider the firing of a gun. Here the system is Gun + bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec { P } \)₁ be the momentum of the bullet and \(\vec { P } \)₂ the momentum of the gun before firing. Since initially both are at rest.

\(\vec { P } \)₁ = 0, \(\vec { P } \)₂ = 0.

Total momentum before firing the gun is zero, \(\vec { P } \)₁ + \(\vec { P } \)₂ = 0

According to the law of conservation of linear momentum, total linear momemtum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec { P } \)₁ + \(\vec { P } \)₂. To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec { P } \)₂ to \(\vec { P } \)₂. Due to the conservation of linear momentum, \(\vec { P } \)₁+ \(\vec { P } \)₂‘= 0. It implies that \(\vec { P } \)₁‘ = –\(\vec { P } \)₂ the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum \(\vec { P } \)₂. It is called ‘recoil momentum’. This is an example of conservation of total linear momentum.

[OR]

(b) Derive an expression for escape speed?
Answer:
Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v;, the initial total energy of the object is
\(E_{i}=\frac{1}{2} M v_{i}^{2}-\frac{G M M_{E}}{R_{E}}\) ………………. (1)
where, M_E is the mass of the Earth and RE the radius of the Earth. The term \(-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.

When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be nonzero.
E_f = 0

According to the law of energy conservation,
E_i – E_f = 0 …………….. (2)

Substituting (1) in (2) we get,
\(\frac{1}{2} \mathrm{M} v_{i}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{F}}}=0\)
\(\frac{1}{2} \mathrm{M} v_{i}^{2}=\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) …………….. (3)

Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v_i with v_e, i.e.,

From equation (4) the escape speed depends on two factors acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g (9.8 ms^-2) and R_e = 6400 km, the escape speed of the Earth is v_e = 11.2 kms^-1. The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity.

Question 37 (a).
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s) …………….. (1)

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount of T in time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt
\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) = -a(T – T_s) ……………….. (4)

Where a is some positive constant.
From equation (3) and (4)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides, we get
\(\mathrm{T}=\mathrm{T}_{\mathrm{s}}+b_{2} e^{\frac{-a}{m s}}\) ……………. (6)
Here b₂ = _e^b1 = Constant

[OR]

(b) Derive an expression for pressure exerted by the gas on the wall of the container?
Answer:
Expression for pressure exerted by a gas:
Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side l.
The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of energy, but a change in momentum occurs.

The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision. the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

A molecule of mass in moving with a velocity \(\vec { v } \) having components (v_x v_y, v_z) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with sanie speed and its x-component is reversed. This is shown in the figure. The components of velocity of the molecule after collision are (-v_x, v_y, v_z)
The x-component of momentum of the molecule bêfore collision = mv_x
The x-component of momentum of the molecule after collision = -mv_x
The change in momentum of the molecule in x direction
= Final momentum – initial momentum = -mv_x – mv_x = -2mv_x
According to law of conservation of linear momentum, the change in momentum of the wall = 2mv_x

The number of molecules hitting the right side wall in a small interval of time ∆t.

The molecules within the distance of v_x ∆t from the right side wall and moving towards the right will hit the wall in the time interval &. The number of molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume (Av_x∆t) and number density of the molecules n). Here A is area of the wall and ii is number of molecules per
unit volume \(\frac{N}{V}\) We have assumed that the number density is the same throughout the cube.

Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

Te no.of molecules that hit the right side wall in a time interval ∆t

= \(\frac{n}{2} \mathrm{A} v_{x} \Delta t\) ……………….. (1)
In the same interval of time ∆t, the total momentum transferred by the molecules
\(\Delta \mathrm{P}=\frac{n}{2} \mathrm{A} v_{x} \Delta t \times 2 m v_{x}=\mathrm{A} v_{x}^{2} m n \Delta t\) ……………….. (2)

From Newton’s second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

F = \(\frac{\Delta p}{\Delta t}=n m \mathrm{A} v_{x}^{2}\) ……………….. (3)
Pressure P = force divided by the area of the wall

P = \(\frac{F}{A}\) = nmv²_x …………………. (4)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v²_x by the average \(\bar{v}_{x}^{2}\) in equation (4)

P = nm\(\bar{v}_{x}^{2}\) ………………… (5)

Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \(\bar{v}_{x}^{2}\) = \(\bar{v}_{y}^{2}\) = \(\bar{v}_{z}^{2}\). The mean square speed is written as

\(\bar{v}^{2}\) = \(\bar{v}_{x}^{2}\) + \(\bar{v}_{y}^{2}\) + \(\bar{v}_{z}^{2}\) = 3\(\bar{v}_{x}^{2}\)
\(\bar{v}_{x}^{2}\) = \(\frac{1}{3}\) \(\bar{v}^{2}\)
Using this in equation (5), we get
P = \(\frac{1}{3}\)nm \(\bar{v}^{2}\) or P = \(\frac{1}{3}\) \(\frac{N}{V}\) m\(\bar{v}^{2}\) as [n = \(\frac{N}{V}\)] ……………… (6)

Question 38 (a).
Explain with graphs the difference between work done by a constant force and by a variable force. Arrive at an expression for power and velocity. Give some examples for the same?
Answer:
Work done by a constant force: When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW = (F cos θ) dr ………………. (1)
The total Work done in producing a displacement from initial position r_i to final position r_f is,

The graphical representation of the work done by a constant force is shown in figure given below. The area under the graph shows the work done by the constant force.

Work done by a variable force:
When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation.

dW = F cos θ dr [F cos θ is the component of the variable force F]
where, F and θ are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation,
W = \(\int_{r_{i}}^{r_{f}} d \mathrm{W}=\int_{r_{i}}^{r_{f}} \mathrm{F} \cos \theta d r\) ………………. (4)

A graphical representation of the work done by a variable force is shown in figure given below. The area under the graph is the work done by the variable force.

Expression for power and velocity

The work done by a force \(\vec { F } \) for a displacement \(\bar { dr } \) is
W = ∫\(\vec { F } \).\(\vec { dr } \) ……………. (1)
Left hand side of the equation (1) can be written as
W = ∫dW = ∫\(\frac{dW}{dt}\) (multiplied and divided by dt) ………………… (2)
Since, velocity is \(\vec { v } \) = \(\frac{d \vec{r}}{d t}\); \(\vec { dr } \) = \(\vec { v } \) dt. Right hand side of the equation (I) can be written as

Substituting equation (2) and equation (3) in equation (1), we get

This relation is true for any arbitrary value of di. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{dW}{dt}\) – \(\vec { F } \).\(\vec { v } \) = 0 Or \(\frac{dW}{dt}\) = \(\vec { F } \).\(\vec { v } \)
Hence power P = \(\vec { F } \).\(\vec { v } \)

[OR]

(b) Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dt}\) ∝ (T – T_s)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding
From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in
time dt, then the amount of heat lost is,
dQ = msdT ………………. (2)
Dividing both sides of equation (2) by dt

\(\frac{dQ}{dt}\) = \(\frac{msdT}{dt}\) ……………….. (3)
From Newton’s law of cooling
\(\frac{dQ}{dt}\) ∝ (T – T_s)
\(\frac{dQ}{dt}\) -a(T – T_s) …………………. (4)
Where a is a positive constant.
From equation (3) and (4)
– a(T – T_s) = ms \(\frac{dT}{dt}\)
\(\frac{d T}{T-T_{s}}\) = -a\(\frac{a}{ms}\) dt ………………. (5)
Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
\(\mathrm{T}=\mathrm{T}_{s}+b_{2} e^{\frac{-a}{m s}}\) ………………… (6)
Here b₂ = e^b1 = constant

Students can download 6th Social Science Term 3 Geography Chapter 1 Asia and Europe Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science Geography Solutions Term 3 Chapter 1 Asia and Europe

Samacheer Kalvi 6th Social Science Asia and Europe Text Book Back Questions and Answers

I. Choose the correct Answer

Question 1.
Which is not the Western margin of Asia?
(a) Black Sea
(b) Mediterranean Sea
(c) Red Sea
(d) Arabian Sea
Answer:
(d) Arabian Sea

Question 2.
The Intermontane …………… plateau is found between Elbruz and Zagros.
(a) Tibet
(b) Iran
(c) Deccan
(d) The Yunnan
Answer:
(b) Iran

Question 3.
Equatorial climate:
(i) Uniform throughout the year.
(ii) The average / mean rainfall is 200 mm.
(iii) The average temperature is 10°C.
(iv) Of the statements give above.
(a) i alone is correct
(b) ii and iii are correct
(c) i and iii are correct
(d) i and ii are correct
Answer:
(a) i alone is correct

Question 4.
Match list I correctly with list II and select your answer from the codes given below.

Codes:
(a) 2, 3, 4, 1
(b) 4, 3 , 2, 1
(c) 4, 3, 1, 2
(d) 2, 3, 1, 4
Answer:
(a) 2, 3, 4, 1

Question 5.
India is the leading producer of ______
(a) Zinc
(b) Mica
(c) Manganese
(d) Coal
Answer:
(b) Mica

Question 6.
The natural boundary between Spain and France is ……………
(a) The Alps
(b) The Pyrenees
(c) The Carpathian
(d) The Caucasus
Answer:
(b) The Pyrenees

Question 7.
The western and north-western Europe enjoys mild and humid climate.
Choose the correct option:
(a) These regions are found near the equator
(b) It is influenced by the North Atlantic Drift
(c) It is surrounded by mountains
(d) All of the above
Answer:
(b) It is influenced by the North Atlantic Drift

Question 8.
Which of the following statements is incorrect?
(a) Europe produces electricity from hydel power.
(b) All the rivers of Europe originate in the Alps.
(c) Most of the rivers in Europe are used for inland navigation.
(d) The fivers of Europe are perennial in nature.
Answer:
(b) All the rivers of Europe originate in the Alps.

Question 9.
Choose the incorrect pair.
(a) The Meseta – Spain
(b) The Jura – France
(c) The Pennine – Italy
(d) The Black Forest – Germany
Answer:
(c) The Pennines – Italy

Question 10.
Which country in Europe has a very low density of population?
(a) Iceland
(b) The Netherlands
(c) Poland
(d) Switzerland
Answer:
(a) Iceland

II. Fill in the blanks

1. The Taurus and the Pontine ranges radiate from the …………….
2. The wettest place in the world is …………….
3. Iran is the largest producer of ……………. in the world.
4. Europe connected with south and south east Asia by …………….
5. The national dance of Philippines is …………….
6. The second highest peak in Europe is …………….
7. The type of climate that prevails in the central and eastern parts of Europe is …………….
8. The important fishing ground in North Sea is ……………..
9. The density of population in Europe is ……………..
10. The river ……………. passes through nine countries of Europe.

Answer:

1. Armenian
2. Mawsynram
3. dates
4. the Suez canal
5. Tinikling
6. the Mont Blanc
7. Continental type
8. Dogger Bank
9. 34 persons/Km²
10. Danube

III. Match the following

Answer:
1. – d
2. – a
3. – e
4. – b
5. – c

IV. Let us learn

Question 1.
Assertion (A) : Italy has dry summers and rainy winters
Reason (R) : It is located in the Mediterranean region
(a) Both A and R are individually true and R is the correct explanation for A
(b) Both A and R are individually true but R is not the correct explanation for A
(c) A is true, but R is false
(d) A is false, but R is true
Answer:
(d) A is false, but R is true

Question 2.
Places marked as 1, 2, 3 and 4 in the given map are noted for the following plains.
(a) Indo – Gangetic plain
(b) Manchurian plain
(c) Mesopotamian
(d) Great plains of China
Match the plains with the notation on the map and select the correct answer using the codes given below.
Codes:

Answer:
b) 2 1 3 4

Question 3.
In the given outline map of Asia, the shaded areas indicate the cultivation of ……………..

(a) Sugarcane
(b) Dates
(c) Rubber
(d) Jute
Answer:
(b) Dates

V. Answer in Brief

Question 1.
Name the important intermontane plateaus found in Asia.
Answer:
The plateau of Anatolia, The plateau of Iran, and the plateau of Tibet are the important intermontane plateaus found in Asia.

Question 2.
Write a short note on the monsoon climate.
Answer:

1. The south, southeast and eastern parts of Asia are strongly influenced by monsoon winds.
2. Summer is hot and humid while winter is cool and dry.
3. The summer monsoon winds bring heavy rainfall to India, Bangladesh, Indo-China, Philippines and Southern China.

Question 3.
How does physiography play a vital role in determining the population of Asia?
Answer:

1. In Asia, the population is unevenly distributed because of various physical features.
2. China and India alone cover three-fifth of Asia’s population.
3. River plains and industrial regions have a high density of population, whereas low density is found in the interior parts of Asia.

Question 4.
Name the ports found is Asia.
Answer:
Tokyo, Shanghai, Singapore, Hong Kong, Chennai, Mumbai, Karachi and Dubai are the important seaports in Asia.

Question 5.
Asia is called the ‘Land of Contrasts – Justify.
Answer:
The biggest continent Asia is called “the land of contrasts”.
Because:

1. Asia is the biggest continent has different types of land features such as mountain, plateau, plain, valley, bay, island etc.
2. It has different climatic conditions from the equator to polar region.
3. Apart from this many races, languages, religions and cultures are followed by people who live in Asia. Therefore, Asia is called ‘the land of contrasts’.

Question 6.
Name the important mountains found in the Alpine system.
Answer:
The important mountain ranges in the Alpine system are the Sierra Nevada, the Pyrenees, the Alps, the Apennines, the Dinaric Alps, the Caucasus and the Carpathian.

Question 7.
What are the important rivers of Europe?
Answer:
The important rivers of Europe are Volga, Danube, Dnieper, Rhine, Rhone, Po, and Thames.

Question 8.
Name a few countries which enjoy the Mediterranean type of climate.
Answer:
The Mediterranean and sub-mediterranean climate regions in Europe are found in much of southern Europe, mainly in southern Portugal, most of Spain, the southern coast of France, Italy, the Croatian Coast, much of Bosnia, Montenegro, Kosovo, Serbia, Albania, Macedonia, Greece and the Mediterranean Islands.

Question 9.
Give a short note on the population of Europe.
Answer:

1. Europe is the third-most populous continent, after Asia and Africa. The population density in Europe is.34 persons / km2.
2. High population density is often associated with the coalfields of Europe.
3. Other populous areas are sustained by mining, manufacturing, commerce, offering large market, labour forces and productive agriculture.
4. Monaco, Malta, San Marino, and the Netherlands are the most densely populated countries; Iceland and Norway have a very low density of population.

Question 10.
Name the important festivals celebrated in Europe.
Answer:

1. The Europeans celebrate both religious and holiday festivals.
2. Christmas, Easter, Good Friday, the Saint Day, Redentore, Tomatina and Carnival are the important festivals of Europe.

VI. Distinguish

Question 1.
Intermontane plateaus and southern plateaus.
Answer:
Intermontane plateaus:

1. Intermontane plateaus are found in the mountain ranges.
2. Eg, The plateau of Anatolia, The plateau of Iran, and the plateau of Tibet

Southern plateaus:

1. The southern plateaus are relatively lower than the northern plateaus.
2. Eg. Arabian Plateau, Deccan Plateau, Shan Plateau, and the Yunnan Plateau.

Question 2.
Cold desert and hot desert.
Answer:
Cold desert:

1. A barren or desolate area especially sandy region of little rainfall, featuring cold dry winter.
2. Cold deserts are found in the antarctic, Green land, Western China, Turkartan the Gopi crodi desert in Mongolia.

Hot desert:

1. A barren or desolate area often the sandy region of little rainfall.
2. The largest hot desert is the Sahara desert.

Question 3.
Tundra and Taiga.
Answer:

Thunder:

1. The Arctic and northern Scandinavian highland have Tundra type of vegetation.
2. The winters are very long and severe summers are very short and warm.
3. It is the land with few animals like polar bear, reindeer and walrus
4. No trees. Lowest form of vegetation like Mosses and Lichen only available.

Taiga:

1. They are found in the south of the Tundra region in Norway, Sweden, Finland, Germany, Poland and Austria.
2. The winters are long and cold. Summers are short and warm.
3. It is the land of fur bearing animals. Eg. Mink, silver fox, squirrel etc.
4. Pine, fir, spruce and larch are the important tree varieties.

Question 4.
The North-Western highlands and Alpine mountain range
Answer:
The North-Western Highlands:

1. This region includes the mountains and plateaus of Norway, Sweden, Finland, Scotland, and Iceland.
2. This region has a fjord coast. It was created by glaciations.
3. A lot of lakes here serve as reservoirs for producing hydroelectricity.

The Alpine mountain range:

1. The Alpine mountain system consists of a chain of young fold mountains found in the southern part of Europe.
2. The Sierra Nevada, the Pyrenees, the Alps, the Apennines, the Dinaric Alps, the Caucasus and the
3. Carpathian is the important mountain ranges.
4. The Pyrenees is a natural boundary between Spain and France.

VII. Give Reasons

Question 1.
Asia is the leading producer of rice.
Answer:
Because:

1. In Asia, India has the largest area of arable lands.
2. Agriculture is intensively practiced in the riverine plains of Asia
3. China and India are the leading producers of rice in the world.

Question 2.
Asia is the largest and most populous continent in the world.
Answer:
Because:

1. Most of the land of Asia in the northern hemisphere has different physical and cultural features.
2. Lofty mountains, plateaus, plains, islands and peninsulas are the major physiographic features.
3. Many perennial rivers flow through different parts. These river valleys are the cradles of ancient civilizations.
4. River plains and industrial regions have high density of population. Population density 143 persons/Km².
5. Asia covers about 30 percent of the world’s population.

Question 3.
Although Western Europe is located in the high latitudes, it has a moderate climate.
Answer:
Because:

1. The western part has a mild, generally humid climate, influenced by the North Atlantic Drift.
2. North Atlantic Drift is a Warm ocean current which brings warmth to the Western Europe.

VIII. Answer in Paragraph

Question 1.
Give an account of the drainage system in Asia.
Answer:

1. The rivers of Asia originate mostly from the central highlands.
2. The Ob, Yenise and Lena are the major rivers flow towards the north and drain into the Arctic Ocean. They remain frozen during winter.
3. Many perennial rivers like Brahmaputra, Indus, Ganga and Irrawaddy originate from the high mountains.
4. They do not freeze during winter.
5. The Euphrates and Tigris flow in west Asia.
6. The Amur, Huang He, Yangtze and Mekong flow in the south and southeastern parts. The Yangtze is the longest river in Asia.

Question 2.
Describe the mineral sources found in Asia.
Answer:

1. Asia has a variety of mineral deposits like Iron, Coal, Manganese, Bauxite, Zinc, Tungsten, Petroleum, Tin etc. Oil and Natural Gas found in the west Asian countries.
2. Iron Ore: Asia has the largest deposits of iron ore in the world. China and India are the important iron ore deposit countries of Asia.
3. Coal: Coal is a fossil fuel. China and India are the largest producers of coal in Asia.
4. Petroleum: Petroleum is a mineral oil. The largest petroleum reserves are found in South West Asia. The important petroleum-producing countries are Saudi Arabia, Kuwait, Iran, Bahrain, Qatar and UAE.
5. Bauxite is found in India and Indonesia.
6. Mica: India is the largest producer in the world.
7. Tin is found in Myanmar, Thailand, Malaysia, and Indonesia.

Question 3.
What are fjords? How do they protect harbours from bad weather conditions?
Answer:

1. A fjord is a narrow and deep-sea inlet between steep clifts.
2. Fjords reduce the speed of wind irrespective of their direction.
3. The force of sea waves is also controlled.
4. Fjord coast was created by glaciations in the past.
5. Areas with fjords are best suited for natural harbours. Example : Norway.

Question 4.
Describe the climatic divisions of Europe.
Answer:

1. The climate of Europe varies from the subtropical to the polar climate.
2. The Mediterranean climate of the south has warm summer and rainy winter.
3. The western and northwestern parts have a mild, generally humid climate, influenced by the North Atlantic Drift.
4. In central and eastern Europe, the climate is humid continental-type.
5. In the northeast, subarctic and tundra climates are found.
6. The whole of Europe is subject to the moderating influence of prevailing westerly winds from the Atlantic Ocean.
7. North Atlantic Drift is a warm ocean current which brings warmth to Western Europe.
8. The westerly wind further transports warmth across Europe.

X. Activity

Question 1.
Complete the following
I belong to ………… district. My district is famous for the following : …………, ………… and ………… . The boundaries of my districts are in the north, in the east, in the south and in the west. It spreads for an area of km². There are ………… taluks and ………… villages in my district. …………, …………, ………… are the important mountain / plain / plateaus (If all, mention all features). The rivers …………, ………… ………… criss – cross my district. …………, …………, ………… are common trees and wildlife such as, are found here. …………, …………, ………… are important minerals available in my district. Based on this …………, ………… industries are located here. The major crops are …………, …………, ………… . (Coastal districts may write the variety of fish). The total population is …………. We celebrate …………, …………, ………… festivals.
Answer:
Tirunelveli, Nellaiyappar Temple, Courttalam, Halwa, Virudhunagar, Tuticorin, Kanyakumari, Western Ghats, 6823sq.kms, 16 taluks, 559 Villages, Thamirabarani, Chittar, Manimuthar, Palm, Neem, Cocount, Monkeys, Tigers, Elephants, Bluemetal, Lime stone, Thorium, Cement, Ginning, Vessels, Paddy, Cotton, Sugarcane, 33, 22, 644, Pongal, Deepavali, Christmas.

Question 2.
If you get a chance to settle in Europe, which country would you choose? List out the reasons why?

Question 3.
Choose any region is Asia. In the map of Asia, mark its distribution of natural vegetation and wildlife. Paste related pictures.

Samacheer Kalvi 6th Social Science Asia and Europe Additional Important Questions and Answers

I. Choose the Correct Answer

Question 1.
……………. separates Asia from Africa
(a) Suez Canal
(b) Bering Strait
(c) Mediterranean Sea
(d) Palk Striat
Answer:
(a) Suez Canal

Question 2.
Shan plateau is located in ______
(a) Saudi Arabia
(b) Myanmar
(c) India
(d) China

Question 3.
The South Asian rivers
(i) remain frozen during winter
(ii) flow towards the north
(iii) are perennial
Of the statements given above
(a) i alone is correct
(b) iii alone is correct
(c) All the three are correct
(d) All the three are wrong
Answer:
(b) iii alone is correct

Question 4.
Find out the wrong pair.
(a) Coal – China
(b) Iron ore – India
(c) Bauxite – Iran
(d) Tin – Myanmar
Answer:
(c) Bauxite – Iran

Question 5.
River yangtze flows in ……………
(a) Inida
(b) Japan
(c) Myanmar
(d) China
Answer:
(d) China

Question 6.
The light house of the Mediterranean is ______
(a) Mt. Stromboli
(b) Mt. Etana
(c) Mt. Vesuvius
(d) None of the above

Question 7.
Choose the incorrect pair:
(a) Siberian Plain – Ob, Yenisey
(b) Manchurian Plain – Amur
(c) Greet plain of China – Yangtze, Sikiang
(d) Mesopotamian Plain – Irrawaddy
Answer:
(d) Mesopotamian Plain – Irrawaddy

II. Match the following

Answer:
1. – c
2. – e
3. – d
4. – b
5. – a

III. True/False

1. There are two knots found in Asia.
2. Ob, Yensie, Lena remain frozen during winter
3. Teak, Sandal wood are coniferous trees
4. The fjord region has a lot of lakes
5. Wheat is the dominant crop throughout Europe.

Answer:

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

IV. Answer in Brief

Question 1.
What are the cradles of ancient civilization? Why?
Answer:

1. The river valleys are the cradles of civilizations.
2. Because the ancient civilizations Indus valley, Mesopotamian and Chinese civilizations were born in Asian river valleys.

Question 2.
Mention the Physical divisions of Asia.
Answer:
The physical divisions of Asia are

1. The Northern lowlands
2. The Central High Mountains
3. The Southern Plateaus
4. The Great Plains and
5. The Island Groups.

Question 3.
What are the rare species found in Asia?
Answer:

1. Orang – Utan
2. Komodo Dragon
3. Giant Panda

Question 4.
Mention the Great Plains of Asia.
Answer:
The great plains of Asia are

1. West Siberian plain (Ob and Yenisey)
2. Manchurian Plain (Amur)
3. Great Plain of China (Yangtze and Sikiang)
4. Indo-Gangetic Plain (Indus and Ganga)
5. Mesopotamian plain (Tigris and Euphrates) and
6. The Irrawaddy plain (Irrawaddy)

Question 5.
Name the countries where fishing is a large industry.
Answer:

1. Norway
2. Iceland
3. Russia
4. Denmark
5. The United Kingdom
6. The Netherlands

V. Give Reasons

Question 1.
Europe is a modern and economically developed continent.
Answer:
Avilability of sources, efficient educated work force, research, contact with other nations and innovations are the factors that transformed like this.

Question 2.
Varied patterns of agricultural activities are in use in Europe.
Answer:

1. Europe is an industrially developed continent. It has great diversity in its topography, climate and soil.
2. These interact to produce varied patterns of agriculture activities.

[The varied patterns of agriculture activities in use: Mediterranean agriculture, Dairy farming, mixed livestock, crop farming, horticulture]

Question 3.
There is no winter in the equatorial region.
Answer:
The areas found in and around the equator of Asia have uniform climate throughout the year. There is no winter. Average temperature 27°c. The mean rainfall 1270 mm.

VI. Answer in Paragraph

Question 1.
Describe Industries in Europe.
Answer:

1. Large scale Industries : Steel and Iron are, Ship building, Motor vehicle, Aircraft construction, Pharmaceutical drugs.
2. Small scale Industries that produce nondurable goods are found throughout Europe.
3. Some countries have a reputation for speciality goods.
   • Bicycles – English, Italian and Dutch
   • Glass – Swedish and Finnish
   • Perfumes and fashion goods – Parisian
   • Precision instruments – Swiss

VII. Mind map

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Tamilnadu Samacheer Kalvi 6th Social Science History Solutions Term 3 Chapter 4 South Indian Kingdoms

Samacheer Kalvi 6th Social Science South Indian Kingdoms Text Book Back Questions and Answers

I. Choose the correct Answer

Question 1.
Who among the following built the Vaikundaperumal temple?
(a) Narasimhavarma II
(b) Nandivarma II
(c) Dantivarman
(d) Parameshvaravarma
Answer:
(b) Nandivarma II

Question 2.
Which of the following titles were the titles of Mahendra Varma I?
(a) Mattavilasa
(b) Vichitra Chitta
(c) Gunabara
(d) all the three
Answer:
(d) all the three

Question 3.
Which of the following inscriptions describes the victories of Pulakesin II?
(a) Aihole
(b) Saranath
(c) Sanchi
(d) Junagath
Answer:
(a) Aihole

II. Read the statement and tick the appropriate answer

Question 1.
Statement I : Pallava art shows transition from rock – cut – monolithic structure to stone built temple.
Statement II : Kailasanatha temple at Kanchipuram is an example of Pallava art and architecture.
(a) Statement I is wrong
(b) Statement II is wrong
(c) Both the statements are correct
(d) Both the statements are wrong.
Answer:
(c) Both the statements are correct

Question 2.
Consider the following statement(s) about Pallava Kingdom.
Statement I : Tamil literature flourished under Pallava rule, with the rise in popularity of Thevaram composed by Appar.
Statement II : Pallava King Mahendravarman was the author of the play MattavilasaPrahasana.
(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II
Answer:
(c) Both I and II

Question 3.
Consider the following statements about the Rashtrakuta dynasty and find out which of the following statements are correct.
(1) It was founded by Dantidurga
(2) Amogavarsha wrote Kavirajmarga
(3) Krishna I built the Kailasanatha temple at Ellora.
(a) 1 only
(b) 2 and 3
(c) 1 and 3
(d) all the three
Answer:
(d) all the three

Question 4.
Which of the following is not a correct pair?
Answer:
(a) Ellora caves – Rashtrakutas
(b) Mamallapuram – Narasimhavarma I
(c) Elephanta caves – Ashoka
(d) Pattadakal – Chalukyas
Answer:
(c) Elephanta caves – Ashoka

Question 5.
Find out the wrong pair.
(a) Dandin – Dasakumara Charitam
(b) Vatsyaya – Bharathavenba
(c) Bharavi – Kiratarjuneeyam
(d) Amogavarsha – Kavirajamarga
Answer:
(b) Vatsyaya – Bharathavenba

III. Fill in the blanks

1. ……………….. defeated Harsha Vardhana on the banks of the river Narmada.
2. ……………….. destroyed Vatapi and assumed the title VatapiKondan.
3. ……………….. was the author of Aihole Inscription.
4. ……………….. was the army general of Narasimhavarma I
5. The music inscriptions in ……………….. and ……………….. show Pallavas ‘interest in music.

Answer:

1. Pulakesin II
2. Narasimhavarma
3. Ravikirti
4. Paranjothi
5. Kudumianmalai, Thiruamayam

IV. Match the following

Answer:
1. – c
2. – d
3. – a
4. – b

V. State True or False

1. The famous musician Rudracharya lived during Mahendravarma I.
2. The greatest king of the Rashtrakuta dynasty was Pulakesin II.
3. Mamallapuram is one of the UNESCO World Heritage Sites.
4. Thevaram was composed by Azhwars.
5. The Virupaksha temple was built on the model of Kanchi Kailasanatha Temple.

Answer:

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

VI. Answer in one or two sentences

Question 1.
Name the three gems of Kannada literature.
Answer:
The three gems of Kannada literature were Pampa, Sri Ponna, and Raima.

Question 2.
How can we classify the Pallava architecture?
Answer:
Pallava architecture can be classified as

1. Rock-cut temples – Mahendravarman style.
2. Monolithic Rathas and Sculptural Mandapas Mamallan style.
3. Structural Temples – Rajasimhan style and Nandivarman style.

Question 3.
What do you know of Gatika?
Answer:

1. Gatika means monastery or Centre of learning.
2. It was popular during the Pallava times at Kanchi.
3. It attracted students from all parts of India and abroad.
4. Vatsyayana who wrote Nyaya Bhashya was a teacher at Kanchi (Gatika). Panchapandavar rathas are monolithic rathas. Explain.

Question 4.
Panchapandavar rathas are monolithic rathas. Explain
Answer:

1. Under Mamalla style, the five rathas (chariots), popularly called Panchapandavar rathas, signify the different style of temple architecture.
2. Each ratha has been carved out of a single rock.
3. So they are called monolithic.
4. The popular mandapams here are Mahishasuramardhini mandapam, Thirumoorthi mandapam and Varaha mandapam.

Question 5.
Make a note on Battle of Takkolam.
Answer:

1. Krishna III was the last able ruler of the Rashtrakuta dynasty,
2. He defeated the Cholas in the battle of Takkolam (presently in Vellore Dt) and captured Thanjavur.

VII. Answer the following

Question 1.
Examine Pallavas’ contributions to architecture.
Answer:
Pallava period is known for architectural splendour. In 1984 Mamallapuram was added to the list of UNESCO World Heritage Sites. The illustrious examples of Pallava architecture are the Shore Temple, other temples and the Varaha cave.

Mahendra style : (Rock – cut temples)
Mahendravarma style of cave temples are seen at Mandagapattu, Mahendra vadi, Mamandur, Dalavanur, Trichirapalli, Vallam, Tirukazhukkundram and Siyamangalam.

Mamalla Style : (Monolithic Rathas and Sculptural Mandapas)

1. The last example for Mamalla style monolithic is Panchapandavar rathas. Mahishasuramardhini mandapam, Thirumoorthi mandapam and Varaha mandapam are the popular mandapams.
2. The most important is the open art gallery. The fall of the Ganges from the head of Lord Shiva and the Arjun’s penance are notable sculptures.

Rajasimha Style : (Structural Temples)
Rajasimha (Narasimhavarma II) Style was a structural temple. The best example is the Kailasanatha temple at Kanchipuram. This temple is called Rajasimheswaram.

Nandivarma Style : (Structural Temples)
Nandivarma style was the last stage of pallava architecture. The best example is Vaikunda Perumal temple at Kanchipuram.

Question 2.
Write a note on Elephanta island and Kailasanatha temple at Ellora.
Answer:
Elephanta island:

1. Elephanta is an island near Mumbai. It is originally known as Sripuri and the local people called Gharapuri.
2. The Portuguese named it as Elephanta after seeing the image.
3. The Trimurthi Siva icon and the images of dwarapalakas are seen in the cave temple.

Kailasanatha temple at Ellora:

1. Krishna I built the Kailasanatha temple. It was one of the 30 temples carved out at Ellora.
2. The temple covers an area of over 60,000 sq. feet and vimanam rises to a height of 90 feet.
3. It portrays typical Dravidian features and has a resemblance to the shore temple at Mamallapuram.

VIII. HOTS

Question 1.
Give an account of the Western Chalukyas of Kalyani.
Answer:
Western Chalukyas of Kalyani:

1. They were the descendants of Badami Chalukyas ruled from Kalyani (modem day Basavakalyan).
2. In 973, Tailapa II, a feudatory of the Rashtrakuta mling from the Bijapur region, defeated Parmara of Malwa.
3. Tailapa II occupied Kalyani and his dynasty quickly grew into an empire under Someswara I.
4. Someswara I moved the capital from Manyakheta to Kalyani.
5. For over a century, both the Chalukyas and the Cholas fought many fierce battles to control the fertile region of Vengi.
6. In the late 11th century, under Vikramaditya VI, vast areas between the Narmada river in the north and Kaveri river in the South came under Chalukya control.
7. The Kasi Vishwesvara Temple at Lakkundi, the Mallikaijuna temple at Kuruvatti, the Kalleshwara temple at Bagali and the Mahadeva temple at Itagi represent well-known examples of the architecture of Western Chalukyas of Kalyani.

IX. Life Skills (For Students)

1. Collect temple architecture pictures of Pallavas, Chalukyas and Rashtrakutas and identify the distinguishing features of each period.
2. Field Trip : Plan a trip to any place of historical importance.

X. Activity

Question 1.
(a) Sketch the biography of Mahendravarma I and Pulakesin II.
(b) See the picture and write a few sentences on it.
Answer:
Arjuna is fasting in one leg At the centre there is a naga Siva is attended by devaganas Hunters, animals, birds and trees are seen. There is also a herd of elephants.

XI. Answer Grid

Question 1.
Give examples for the structural temples of Pallava period.
Answer:
Kailasanatha temple, Vaikunda Perumal temple

Question 2.
Name the new style of architecture developed during Chalukya period.
Answer:
Vesara

Question 3.
What does Aihole inscription mention?
Answer:
Defeat of Harsha Vardhana by Pulakesan II

Question 4.
Who built the Kailasanatha temple at Ellora?
Answer:
Krishna I

Question 5.
Name the sculptural mandapas of Mamallan style of architecture.
Answer:
Monolithic

Question 6.
Where do structural temples of Chalukya exist?
Answer:
Aihole, Badami, Pattadakal

Question 7.
Name two Saivite saints and Vaishnavite saints who practised bhakticult during Pallava period?
Answer:
Appar, Manikkavasakar. Nammazhvar, Andal

Question 8.
Who was the founder of Rashtrakuta dynasty?
Answer:
Dantidurga

Question 9.
What were the titles adopted by Narasimhavarma I?
Answer:
Mamallan, Vatapi kondan

Samacheer Kalvi 6th Social Science South Indian Kingdoms Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
Who was Avanisimha?
(a) Simhavishnu
(b) Pulakesin II
(c) Mahendravarman
(d) Aparajita
Answer:
(a) Simhavishnu

Question 2.
Cave temples of the Pallavas are found at ________
(a) Vallam
(b) Vengi
(c) Badami
(d) Pattadakal
Answer:
(a) Vallam

Question 3.
Mangalesa belonged to Dynasty.
(a) Chalukya
(b) Pallava
(c) Rashtrakuta
(d) Gupta
Answer:
(a) Chalukya

Question 4.
The new style of architecture Vesara was introduced by
(a) Chalukyas
(b) Pallavas
(c) Rastrakutadas Vakataka
(d) Cave
Answer:
(a) Chalukyas

II. Read the statement and tick the appropriate answer

Question 1.
Statement I : The life of Rishabadeva, the first Jain Tirthankara is depicted in Adipurana
Statement II : The Rashtrakutas followed Jainism
(a) Statement I is wrong
(b) Statement II is wrong
(c) Both the statements are correct
(d) Both the statements are wrong.
Answer:
(b) Statement II is wrong

Question 2.
Statement I : Appar and Manikkavasakar were Vaishnavites.
Statement II : Nammazhvar and Andal were saivites
(a) Statement I is correct
(b) Statement II is correct
(c) Both the statements are wrong
(d) Both the statements are correct.
Answer:
(c) Both the statements are wrong

III. Fill in the blanks

1. The last Pallava ruler …………..
2. The poet who was patronized by Nandivarman II was …………..
3. The style adopted by Chalukyas in paintings is ……………
4. The last ruler to hold the empire intact was ……………

Answer:

1. Aprajita
2. Perundevanar
3. Vakataka
4. Govinda III

IV. State True or False

1. Rudracharya was a famous musician during Pallava Period
2. Varaha, Narasimha, Vamana are the Avatar of Siva
3. The Rashtrakutas were of Telugu origin
4. The Chalukyas perfected the art of stone building without mortar.

Answer:

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

V. Answer in one or two sentences

Question 1.
Write about ‘Siruthondar’
Answer:

1. The army general of Narasimhavarman I, Paranjothi, led the army during the invasion of Vatapi.
2. After the victory devoted himself to Lord Siva, he was known as Siruthondar (one of the 63 Nayanmars)

Question 2.
Write a note on Simhavishnu.
Answer:

1. Simhavishnu, son of Simhavarman-II, created a strong Pallava kingdom after destroying the Kalabhras.
2. He defeated many kings in the south including the Cholas and Pandyas.
3. His able son was Mahendravarman I.

Question 3.
Name the distinct but closely related and independent Chalukya dynasties.
Answer:

1. Chalukyas of Badami
2. Chalukyas of Vengi (Eastern chalukyas)
3. Chalukyas of Kalyani (Western chalukyas)

Question 4.
Who were Chalukyas?
Answer:

1. There were three distinct but closely related and independent Chalukya dynasties,
2. They were:
   • Chalukyas of Badami
   • Chalukyas of Vengi (Eastern Chalukyas)
   • Chalukyas of Kalyani (Western Chalukyas).
3. These Chalukyas held Harsha in the north, the Pallavas in the south and Kalinga (Odisha) in the east.

VII. Answer the following

Question 1.
Explain – ’Pattatakal’
Answer:

1. Pattatakal is a small village in Bagalkot district of Karnataka.
2. Here are 10 temples. Out of them 4 were in Nagara style and 6 were in Dravida style.
3. Dravida style – Virupaksha Temple, Sangameshwara Temple.
4. Nagara style – Papanatha Temple.
5. Pattatakal is one of the UNESCO World Heritage sites.

Question 2.
What do you know about ‘Rashtrakutas?
Answer:

1. Rashtrakutas were of Kannada origin. They ruled not only the Deccan but parts of the far south and the Ganges plain during 8 – 10 centuries.
2. Dantidurga was the founder of the Rashtrakuta dynasty. He was an official of high rank under the Chalukyas of Badami.
3. Krishna I was the successor of Dantidurga and he built the Kailasanatha temple at Ellora.
4. The greatest king of Amogavarsha. Krishna III was the last able ruler of the Rashtrakuta dynasty.
5. Capital: Malkhed (Manyakheta). Port: Broach.

VIII. Mind map

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TN State Board 11th Maths Model Question Paper 5 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 90

PART – I

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

Question 1.
If n((A × B) ∩ (A × C)) = 8 and n(B ∩ C) = 2 then n(A) = …………………
(a) 6
(b) 4
(c) 8
(d) 16
Answer:
(b) 4

Question 2.
The value of log₃ \(\frac{1}{81}\) is ……………….
(a) -2
(b) -8
(c) -4
(d) -9
Answer:
(c) -4

Question 3.
The value of log₃ 11 log₁₁ 13 log₁₃ 15 log₁₅ 27 log₂₇ 81 is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 4.
The value of sin(45° + θ) – cos (45° – θ) is …………………..
(a) 2 cos θ
(b) 1
(c) 0
(d) 2 sin θ
Answer:
(c) 0

Question 5.
If tanα and tan β are the roots of x² + ax + b = 0 then \(\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}\) is equal to ……………………..
(a) \(\frac{b}{a}\)
(b) \(\frac{a}{b}\)
(c) –\(\frac{a}{b}\)
(d) –\(\frac{b}{a}\)
Answer:
(c) –\(\frac{a}{b}\)

Question 6.
If a² – aC₂ = a² – aC₄ then the value of a is …………………….
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 7.
If ⁿP_r = 840, ⁿC_r= 35 then n = …………………..
(a) 1
(b) 6
(c) 5
(d) 4
Answer:
(a) 1

Question 8.
If 2x² + 3xy – cy² = 0 represents a pair of perpendicular lines then c = …………………….
(a) -2
(b) \(\frac{1}{2}\)
(c) – \(\frac{1}{2}\)
(d) 2
Answer:
(d) 2

Question 9.
The number of terms in the expansion of [(a + b)²]¹⁸ = …………………..
(a) 19
(b) 18
(c) 36
(d) 37
Answer:
(d) 37

Question 10.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …………………..
(a) (7, 3)
(b) (4, 1)
(c) (1,-1)
(d) (3, 4)
Answer:
(b) (4, 1)

Question 11.
Let A and B be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) A + B is a symmetric matrix
(b) AB is a symmetric matrix
(c) (AB) = (BA)^T
(d) A^TB = AB^T
Answer:
(d) A^TB = AB^T

Question 12.
If [3 -1 2] B = [5, 6] then the order of B is ………………….
(a) 3 × 2
(b) 2 × 3
(c) 3 × 1
(d) 1 × 1
Answer:
(a) 3 × 2

Question 13.
1f \(\underset { x\rightarrow 0 }{ lim } \) \(\frac{sin px}{tan 3x}\) = 4 then the value of p is …………………….
(a) 6
(b) 9
(c) 12
(d) 4
Answer:
(c) 12

Question 14.
For \(\vec { a } \) = \(\vec { i } \) + \(\vec { j } \) – 2\(\vec { k } \), \(\vec { b } \) = \(\vec { i } \) + 2\(\vec { j } \) + \(\vec { k } \) and \(\vec { c } \) = \(\vec { i } \) – 2\(\vec { j } \) + 2\(\vec { k } \) the unit vector parallal to is \(\vec { a } \)
+ \(\vec { b } \) + \(\vec { c } \) is ……………………….
(a) \(\frac{\vec{i}+\vec{j}-\vec{k}}{\sqrt{3}}\)
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)
(c) \(\frac{\vec{i}+\vec{j}+\vec{k}}{3}\)
(d) \(\frac{\vec{i}-\vec{j}+\vec{k}}{\sqrt{6}}\)
Answer:
(b) \(\frac{\vec{i}+\vec{j}+\vec{k}}{\sqrt{3}}\)

Question 15.
The differential co-efficient of log₁₀ x with respect to log_x 10 is …………………….
(a) 1
(b) -(log₁₀x)²
(c) (log_x10)²
(d) \(\frac { x^{ 2 } }{ 100 } \)
Answer:
(b) -(log₁₀x)²

Question 16.
\(\frac{d}{dx}\)(e^x+5logx) is …………………..
(a) e^xx⁴(x + 5)
(b) e^xx(x + 5)
(c) e^x + \(\frac{5}{x}\)
(d) e^x – \(\frac{5}{x}\)
Answer:
(a) e^xx⁴(x + 5)

Question 17.
If f(x) = x tan^-1x thenf'(1) = ……………………..
(a) 1 + \(\frac { \pi }{ 4 } \)
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)
(c) \(\frac{1}{2}\) – \(\frac { \pi }{ 4 } \)
(d) 2
Answer:
(b) \(\frac{1}{2}\) + \(\frac { \pi }{ 4 } \)

Question 18.
∫cosec xdx = …………………….
(a) log tan \(\frac{x}{2}\) + c
(b) -log (cosec x + cot x) + c
(c) log (cosecx – cot x) + c
(d) all of them
Answer:
(d) all of them

Question 19.
Ten coins are tossed; The probability of getting atleast 8 heads is …………………..
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{7}{128}\)
(d) \(\frac{7}{16}\)
Answer:
(c) \(\frac{7}{128}\)

Question 20.
Two items are chosen from a lot containing twelve items of which four are defective. Then the probability that atleast one of the item is defective is …………………..
(a) \(\frac{19}{33}\)
(b) \(\frac{17}{33}\)
(c) \(\frac{23}{33}\)
(d) \(\frac{13}{34}\)
Answer:
(a) \(\frac{19}{33}\)

PART – II

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

Question 21.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A?
Answer:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 22.
Prove that \(\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}\) = \(\frac{1+sinθ}{cosθ}\)
Answer:

Question 23.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
No. of non-collincar points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.
∴ No. of Tnangle formed ¹⁵C₃
= \(\frac{15 \times 14 \times 13}{3 \times 2 \times 1}\) = 455

Question 24.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1?
Answer:
Givc sum of the intercepts = 1
⇒ when x intercept a then y intercept = 1 – a
Equation of the line is \(\frac{x}{a}\) + \(\frac{y}{1-a}\) = 1
The line passes through (8, 3) ⇒ \(\frac{8}{a}\) + \(\frac{3}{1-a}\) = 1
(i.e) 8 (1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 6a + 8 =0
(a – 2)(a – 4) = 0 ⇒a = 2 or 4
1. When a = 2 equation of the line is \(\frac{x}{2}\) + \(\frac{y}{-2}\) = 1 (i.e) \(\frac{x}{2}\) – y = 1 ⇒ x – 2y = 2
2.When a = 4 equation of the line is \(\frac{x}{4}\) + \(\frac{y}{1-4}\) = 1 (i.e) \(\frac{x}{4}\) – \(\frac{y}{3}\) = 1 ⇒ 3x – 4y = 12

Question 25.
Find the values of p, q, r, & s if \(\left[\begin{array}{ccc}
p^{2}-1 & 0 & -31-q^{3} \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]\)
Answer:
When two matrices (of some order) are equal then their correspondings entries are equal.

Question 26.
Find |\(\vec { a } \) × \(\vec { b } \)| where \(\vec { a } \) = 3\(\vec { i } \) + 4\(\vec { j } \) and \(\vec { b } \) = \(\vec { i } \) + \(\vec { j } \) + \(\vec { k } \)
Answer:

Question 27.
At the given point x₀ discover whether the given function is continous or discontinous citing the reasons for your answer?
Answer:

Question 28.
Evaluate y = xe^-x2
Answer:

Question 29.
Evaluate ∫\(\frac{1}{x logx}\) dx?
Answer:

Question 30.
Evaluate [((256)^-1/2)^-1/4]³
Answer:

PART – III

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

Question 31.
Find the largest possible domain for the real valued functionsf defined by f(x) = \(\sqrt { x^{ 2 }-5x+6 } \)

Question 32.
Show that tan 75° + cot 75° = 4?

Question 33.
There are 10 bulbs in a room. Each one of them can be operated independently. Find the number of ways in which the room can be illuminated?

Question 34.
Find the \(\sqrt [ 3 ]{ 126 } \) approximately to two decimal places?

Question 35.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = O and parallel to the line 3x + 4y = 7?

Question 36.
If \(\left|\begin{array}{ccc}
a & b & a \alpha+b \\
b & c & b \alpha+c \\
a \alpha+b & b \alpha+c & 0
\end{array}\right|\) = 0 prove that a, b, c are in G.P. or a is a root of ax² + 2bx + c =0

Question 37.
Evaluate \(\underset { x\rightarrow 3 }{ lim } \)\(\frac { x^{ 2 }-9 }{ x-3 } \) ¡f it exists by finding f(3^–) and f(3⁺)

Question 38.
Find the derivative of tan^-1 (1 + x²) with respect x² + x + 1?

Question 39.
Evaluate: ∫x⁵e^{x2 dx}

Question 40.
Prove that the line segment joining the mid points of the adjacent sides of a quadrilateral from parllelogram?

PART – IV

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

Question 41.
(a) Graph the functions f(x) = x³ and g(x) = \(\sqrt [ 3 ]{ x } \) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.

[OR]

(b) 1f x = -2 is one root of x³ – x² – 17x = 22 then find the other roots of the equation?

Question 42.
(a) lf A + B + C= it prove that cosA + cos B + cosC = 1 + 4 sin (\(\frac{A}{2}\)) sin (\(\frac{B}{2}\) sin (\(\frac{C}{2}\))

[OR]

(b) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) show that A² – 4A – 5I = O

Question 43.
(a) If ⁿ⁺¹C₈ : ^(n-3)P₄ = 57 : 16, find the value of n?

[OR]

(b) If the letters of the word IITJEE arc permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE?

Question 44.
(a) The line \(\frac{x}{a}\) + \(\frac{y}{b}\)= 1 moves in such a way that \(\frac { 1 }{ a^{ 2 } } \) + \(\frac { 1 }{ b^{ 2 } } \) = \(\frac { 1 }{ c^{ 2 } } \) where c is a constant. Find the locus of the foot of the perpendicular from the origin on the given line?

[OR]

(b) Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them?

Question 45 (a).
Prove that \(\left|\begin{array}{lll}
1 & x^{2} & x^{3} \\
1 & y^{2} & y^{3} \\
1 & z^{2} & z^{3}
\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)

(b) Evaluate \(\underset { x\rightarrow \infty }{ lim } \) \(\frac{3}{x-2}\) – \(\frac{2 x+11}{x^{2}+x-6}\)

Question 46.
(a) Evaluate \(\frac{1}{6 x-7-x^{2}}\)

(b) Evaluate ∫e^tan-1x (\(\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \)) dx

Question 47 (a).
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

[OR]

(b) Firm manufactures PVC pipes in three plants viz. X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,

1. find the probability that the selected pipe is a defective one.
2. if the selected pipe ¡s a defective, then what is the probability that it was produced by plant Y?

Students can download 6th Social Science Term 3 History Chapter 3 The Age of Empires: Guptas and Vardhanas Questions and Answers, Notes, Samacheer Kalvi 6th Social Science Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Social Science History Solutions Term 3 Chapter 3 The Age of Empires: Guptas and Vardhanas

Samacheer Kalvi 6th Social Science The Age of Empires: Guptas and Vardhanas Text Book Back Questions and Answers

I. Choose the correct Answer

Question 1.
………………. was the founder of Gupta dynasty.
(a) Chandragupta I
(b) Sri Gupta
(c) Vishnu Gopa
(d) Vishnugupta
Answer:
(b) Sri Gupta

Question 2.
Prayog Prashasti was composed by ________
(a) Kalidasa
(b) Amarasimha
(c) Harisena
(d) Dhanvantri
Answer:
(c) Harisena

Question 3.
The monolithic iron pillar of Chandragupta is at ………………
(a) Mehrauli
(b) Bhitari
(c) Gadhva
(d) Mathura
Answer:
(a) Mehrauli

Question 4.
________ was the first Indian to explain the process Of surgery.
(a) Charaka
(b) Sushruta
(c) Dhanvantri
(d) Agnivasa
Answer:
(b) Sushrutal

Question 5.
……………… was the Gauda ruler of Bengal.
(a) Sasanka
(b) Maitraka
(c) Rajavardhana
(d) Pulikesin II
Answer:
(a) Sasanka

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : Chandragupta I crowned himself as a monarch of a large kingdom after eliminating various small states in Northern India.
Reason (R) : Chandragupta I married Kumaradevi of Lichchavi family.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A.
(c) A is correct but R is not correct.
(d) A is not correct but R is correct.
Answer:
(a) Both A and R are true and R is the correct explanation of A

Question 2.
Statement I : Chandragupta II did not have cordial relationship with the rules of South India.
Statement II : The divine theory of kingship was practised by the Gupta rulers.
(a) Statement I is wrong but statement II is correct.
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
(a) Statement I is wrong but statement II is correct.

Question 3.
Which of the following is arranged in chronological order?
(a) Srigupta – Chandragupta I – Samudragupta – Vikramaditya
(b) Chandragupta I – Vikramaditya – Srigupta – Samudragupta
(c) Srigupta – Samudragupta – Vikramaditya – Chandragupta I
(d) Vikramaditya – Srigupta – Samudragupta – Chandragupta I
Answer:
(a) Srigupta – Chandragupta I – Samudragupta -Vikramaditya

Question 4.
Consider the following statements and find out which of the following statements (s) is/are correct.
(1) Lending money at high rate of interest was practised.
(2) Pottery and mining were the most flourishing industries,
(a) 1. is correct
(b) 2. is correct
(c) Both 1 and 2 are correct
(d) Both 1 and 2 are wrong
Answer:
(a) 1. is correct

Question 5.
Circle the odd one

Answer:
Samudragupta.

Answer:
Harshacharita.

III. Fill in the blanks Answer

1. ……………., the king of Ceylon, was a contemporary of Samudragupta.
2. A Buddhist monk from China ……………., visited India during the reign of Chandragupta II.
3. ……………. invasion led to the downfall of the Gupta Empire.
4. ……………. was the main revenue to the Government.
5. The official language of the Guptas was …………….
6. ……………., the Pallava king was defeated by Samudragupta.
7. ……………. was the popular king of the Vardhana dynasty.
8. Harsha shifted his capital from ……………. to Kanauj.

Answer:

1. Reign of
2. Fahien
3. Huns
4. Land tax
5. Sanskrit
6. Vishnugopa
7. Harsha Vardhana
8. Thaneswar

IV. State whether True or False

1. Dhanvantri was a famous scholar in the field of medicine.
2. The structural temples built during the Gupta period resemble the Indo – Aryan style.
3. Sati was not in practice in the Gupta Empire.
4. Harsha belonged to the Hinayana school of thought.
5. Harsha was noted for his religious intolerance.

Answer:

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

V. Match the following

Question 1.

Answer:
b) 2, 4, 1, 3, 5

Question 2.

Answer:
c) 3, 5, 1, 2, 4

VI. Answer in one or two sentences

Question 1.
Who was given the title Kaviraja? Why?
Answer:

1. The title Kaviraja was given to Samudragupta.
2. He was a great lover of poetry and music.
3. In one of the gold coins, he is portrayed playing the harp (Veenai)

Question 2.
What were the subjects taught at Nalanda University?
Answer:

1. At Nalanda University Buddhism was the main subject of study.
2. Other subjects like Yoga, Vedic literature, and medicine were also taught.

Question 3.
Explain the Divine Theory of Kingship.
Answer:

1. The divine theory of Kingship meant that the king is the representative of God on earth.
2. He is answerable only to God and not to anyone else.

Question 4.
Highlight the achievements of Guptas in metallurgy.
Answer:

1. Mining and metallurgy were the most flourishing industries during the Gupta period.
2. The most important evidence of development in metallurgy was the Mehrauli Iron Pillar installed by King Chandragupta in Delhi.
3. This monolithic iron pillar has lasted through the centuries without rusting.

Question 5.
Who were the Huns?
Answer:

1. Huns were the nomadic tribe, who under their great Attila were terrorizing Rome and Constantinople.
2. They came to India through Central Asia, defeated Skandagupta, and spread across central India.
3. Their chief Toromana crowned himself asking.
4. After him, his son Mihirakula ruled and got finally defeated by Yasodharman, ruler of Malwa.

Question 6.
Name the three kinds of tax collected during Harsha’s reign.
Answer:
A Bhaga, Hiranya, and Bali were three kinds of tax collected during Harsha’s reign.

Question 7.
Name the books authored by Harsha.
Answer:
The books authored by Harsha were Ratnavali, Nagananda, and Priyadharshika.

VII. Answer the following briefly

Question 1.
Write a note on Prashasti.
Answer:

1. Prashasti is a Sanskrit word, meaning communication or in praise of.
2. Court poets flattered their kings listing out their achievements.
3. These accounts were later engraved on pillars so that the people could read them.

Question 2.
Give an account of Samudragupta’s military conquests.
Answer:

1. Samudragupta was a great general and he carried on a vigorous campaign all over the country.
2. He defeated the Pallava king Vishnugopa.
3. He conquered nine kingdoms in northern India.
4. He reduced 12 rulers of southern India to the status of feudatories and to pay tribute.
5. He received homage from the rulers of East Bengal, Assam, Nepal, the eastern part of Punjab, and various tribes of Rajasthan.

Question 3.
Describe the land classification during the Gupta period.
Answer:
Classification of land during the Gupta period.

1. Kshetra – Cultivable land
2. Khila – Wasteland
3. Aprahata – Jungle (or) Forest land
4. Vasti – Habitable land
5. Gapata saraha – Pastoral and

Question 4.
Write about Sresti and Sarthavaha traders.
Answer:

1. Sresti: Sresti traders were usually settled at a standard place.
2. Sarthavaha: Sarthavaha traders caravan traders who carried their goods to different places.

Question 5.
Highlight the contribution of Guptas to architecture.
Answer:

1. From the earlier tradition of rock – out shrines, the Guptas were the first to construct temples.
2. These temples, adorned with towers and elaborate carvings, were dedicated to all Hindu deities.
3. The most notable rock-cut caves are found at Ajanta and Ellora, Bagh, and Udaygiri.
4. The structural temples built during this period resemble the Dravidian style.

Question 6.
Name the works of Kalidasa.
Answer:

1. Kalidasa’s famous dramas were Sakunthala, Malavikagnimitra and Vikramaoorvashiyam.
2. Other significant works were Meghaduta, Raghuvamsa, Kumarasambava and Ritusamhara

Question 7.
Estimate Harshvardhana as a poet and a dramatist.
Answer:

1. Harsha himself was a poet and dramatist.
2. Around him gathered the best of poets and artists.
3. His popular works are Ratnavali, Nagananda and Priyadharshika
4. is royal court was adorned by Banabhatta, Mayura, Hardatta, and Jayasena.

VIII. HOTs

Question 1.
The gold coins issued by Gupta kings indicate ………………
Answer:
(a) the availability of gold mines in the kingdom
(b) the ability of the people to work with gold
(c) the prosperity of the kingdom
(d) the extravagant nature of kings.
Answer:
(c) the prosperity of the kingdom

Question 2.
The famous ancient paintings at Ajanta were painted on __________
a. walls of caves
b. ceilings of temples
c. Rocks
d. papyrus
Answer:
a. walls of caves

Question 3.
Gupta period is remembered for ……………….
(a) renaissance in literature and art
(b) expeditions to southern India
(c) invasion of Huns
(d) religious tolerance
Answer:
(a) renaissance in literature and art

Question 4.
What did Indian scientists achieve in astronomy and mathematics during the Gupta period?
Answer:

1. The invention of zero and the consequent evolution of the decimal system was the legacy of Guptas to the modem world.
2. Aryabhatta, Varahamihira, and Brahmagupta were the foremost astronomers and mathematicians of the time.
3. Aryabhatta, in his book ‘ Surya Siddhanta’, explained the true causes of solar and lunar eclipses.
4. He was the first Indian astronomer to declare that the earth revolves around its own axis.
5. Dhanvantri was a famous scholar in the field of medicine.
6. He was a specialist in Ayurveda.
7. Charaka was a medical scientist.
8. Susruta was the first Indian to explain the process of surgery.

IX. Student activity (For Students)

1. Stage any one of the dramas of Kalidasa in the classroom.
2. Compare and contrast the society of Guptas with that of Mauryas.

X. Life Skills (For Students)

1. Collect information about the contribution of Aryabhatta, Varahamihira and Brahmagupta to astronomy.
2. Visit a nearby ISRO centre to know more about satellite launching.

XI. Answer Grid

Question 1.
Who was Toromana?
Answer:
Toromana was the chief of White Huns.

Question 2.
Name the high-ranking officials of the Gupta Empire.
Answer:
Dandanayakas and Maha dandanayakas.

Question 3.
Name the Gupta kings who performed AsVamedha yagna.
Answer:
Samudragupta and Kumaragupta I

Question 4.
Name the book which explained the causes for the lunar and solar eclipses,
Answer:
Surya Siddhanta

Question 5.
Name the first Gupta king to find a place on coins.
Answer:
Samudragupta

Question 6.
Which was the main source of information to know about the Samudragupta’s reign?
Answer:
Allahabad Pillar

Question 7.
Harsha was the worshipper in the beginning.
Answer:
Shiva

Question 8.
University reached its fame during the Harsha period.
Answer:
The Nalanda

Samacheer Kalvi 6th Social Science The Post-Mauryan India Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
The successor of Sri Gupta ……………
(a) Kumaragupta I
(b) Skandagupta
(c) Vishnugupta
(d) Ghatotkacha
Answer:
(d) Ghatotkacha

Question 2.
Sri Gupta was succeeded by _______
(a) Chandra Gupta
(b) Samundra Gupta
(c) Ghatotkacha
(d) Skanda Gupta
Answer:
(c) Ghatotkacha

Question 3.
The Huhs chief crowned himself as king.
(a) Yasodharman
(b) Attila
(c) Mihirakula
(d) Toromana
Answer:
(d) Toromana

Question 4.
Srimeghavarman was the ruler of _______
(a) Singapore
(b) Ceylon
(c) Malaysia
(b) Hansena
Answer:
(b) Ceylon

Question 5.
The place Harsha went to participate in the great Kumbhamela held.
(a) Allahabad
(b) Kasi
(c) Ayodhya
(d) Prayag
Answer:
(d) Prayag

II. Match the statement with the reason and tick the appropriate answer

Question 1.
Assertion (A) : The last of the great Guptas Narasimha Gupta I was paying tribute to Mihirakula.
Reason (R) : He stopped paying tribute to Mihirakula’s hostility towards Buddhism.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is not correct
(d) A is not correct but R is correct
Answer:
(b) Both A and R are correct but R is not the correct explanation of A

Question 2.
Statement I : Criminal law was not more severe than that of the Gupta age.
Statement II : Death punishment was the punishment for violation of the laws and for plotting against the king.
(a) Statement I is wrong but statement II is correct
(b) Statement II is wrong but statement I is correct.
(c) Both the statements are correct.
(d) Both the statements are wrong.
Answer:
d) Both the statements are wrong

III. Fill in the blanks

1. In the assembly at ……………. Harsha distributed his wealth.
2. The capital of China ……………. was a great center of art and learning.
3. ……………. was the wife of Chandragupta I.
4. The military campaigns of kings were financed through revenue ……………..
5. The peasants were required to pay various taxes and were reduced to the position of ……………..

Answer:

1. Prayag
2. Xian
3. Kumaradevi
4. Surpluses
5. serfs

IV. Match the following

Answer:
b) 4, 5, 2, 1, 3

V. Answer in one or two sentences

Question 1.
Write a note on ‘Lichchhavi’.
Answer:

1. Lichchhavi was an old gana – Sanga and its territory lay between the Ganges and the Nepal Terai.
2. Chandragupta, I married Kumaradevi of the famous and powerful lichchhavi family.

Question 2.
How did Chandragupta I crown himself the monarch of a larger kingdom?
Answer:

1. Chandragupta, I married Kumaradevi of the famous and powerful Lichchhavi family.
2. With the support of this family, Chandragupta eliminated various small states and| crowned himself the monarch of a larger kingdom.

Question 3.
What did the travel accounts of Fahien provide information about the condi¬tions of the people of Magadha?
Answer:

1. According to Fahien the people of Magadha were happy and prosperous.
2. Gaya was desolated. Kapilvasthu had become a jungle, but at Pataliputra people were rich and prosperous.

VII. Answer the following briefly

Question 1.
Name the officials employed by the Gupta rulers.
Answer:

1. High – ranking officials were called dandanayakas and mahadandnayakas.
2. The provinces known as deshas or bhuktis were administered by the governors designated as Uparikas. The districts such as vaishyas, were controlled by vishyapatis. At the village level, gramika and gramadhyaksha were the functionaries.
3. The military designations.
4. Baladhikrita (Commander of infantry)
5. Mahabaladhikrita (Commander of the cavalry)
6. Dutakas (spies)

Question 2.
Mention the importance of Fahien’s travel accounts.
Answer:

1. During the reign of Chandragupta II, the Buddhist monk Fahien visited India.
2. His travel accounts provided us information about the socio-economic, religious and moral conditions of the people of the Gupta age.
3. According to Fahien, the people of Magadha were happy and prosperous.
4. Justice was mildly administered and there was no death penalty.
5. Gaya was desolated, Kapilavasthu had become a jungle, but at Pataliputra, people were rich and prosperous.

VIII. Mind map