Category: Class 10

Subject Matter Experts at SamacheerKalvi.Guide have created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th Social Science Model Question Papers 2020-2021 Pdf Free Download in English Medium and Tamil Medium of TN SSLC Class 10th Social Science Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2021 Tamilnadu.

Let us look at these Government of TN State Board Samacheer Kalvi 10th Std Social Science Model Question Papers with Answers 2020-21 Tamil Medium Pdf. Students can view or download the Samacheer Kalvi Class 10th Social Science New Model Question Papers 2021 Tamil Nadu English Medium Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samacheer Kalvi 10th Social Science Guide.

Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Papers 2020 2021 Tamil English Medium

Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Papers English Medium

• Samacheer Kalvi 10th Social Science Model Question Paper 1
• Samacheer Kalvi 10th Social Science Model Question Paper 2
• Samacheer Kalvi 10th Social Science Model Question Paper 3
• Samacheer Kalvi 10th Social Science Model Question Paper 4
• Samacheer Kalvi 10th Social Science Model Question Paper 5

Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Papers Tamil Medium

• Samacheer Kalvi 10th Social Science Model Question Paper 1
• Samacheer Kalvi 10th Social Science Model Question Paper 2
• Samacheer Kalvi 10th Social Science Model Question Paper 3
• Samacheer Kalvi 10th Social Science Model Question Paper 4
• Samacheer Kalvi 10th Social Science Model Question Paper 5

Samacheer Kalvi 10th Social Science Model Question Paper Design 2020-2021 Tamil Nadu

Types of Questions	Marks	No. of Questions to be Answered	Total Marks
Part-I Multiple Choice Questions	1	14	14
Part-II (Totally 14 questions will be given. Answer any Ten. Any one question should be answered compulsorily)	2	10	20
Part-Ill (Totally 14 questions will be given. Answer any Ten. Any one question should be answered compulsorily)	5	10	50
Part-IV	8	2	16
Total Marks			100

Samacheer Kalvi 10th Social Science Model Question Paper Weightage of Marks

Purpose	Weightage
1. Knowledge	30%
2. Understanding	40%
3. Application	20%
4. Skill/Creativity	10%

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th Social Science Model Question Papers 2021 according to the latest exam pattern. These State Board 10th Standard Social Science Public Exam Model Question Papers 2020-21 Tamil Nadu in Tamil Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th Social Science Model Question Papers 2020 2021 Pdf Free Download in English Medium and Tamil Medium will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard Social Science Model Question Papers with Answers 2020 21, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you as soon as possible.

Subject Matter Experts at SamacheerKalvi.Guide have created Tamil Nadu State Board New Syllabus Samacheer Kalvi 10th English Model Question Papers 2020-2021 Pdf Free Download of TN SSLC Class 10th English Model Question Papers, Previous Year Question Papers, Sample Papers are part of Samacheer Kalvi 10th Model Question Papers 2021 Tamilnadu.

Let us look at these Government of TN State Board Samacheer Kalvi 10th Std English Model Question Papers with Answers 2020-21 Pdf. Students can view or download the Samacheer Kalvi Class 10th English New Model Question Papers 2021 Tamil Nadu Pdf for their upcoming TN SSLC board examinations. Students can also read Tamilnadu Samacheer Kalvi 10th English Guide.

Tamil Nadu Samacheer Kalvi 10th English Model Question Papers 2020 2021

Samacheer Kalvi 10th English Model Question Papers with Answers 2021 Tamil Nadu

• Samacheer Kalvi 10th English Model Question Paper 1
• Samacheer Kalvi 10th English Model Question Paper 2
• Samacheer Kalvi 10th English Model Question Paper 3
• Samacheer Kalvi 10th English Model Question Paper 4
• Samacheer Kalvi 10th English Model Question Paper 5

Samacheer Kalvi 10th English Model Question Paper Design 2020-2021 Tamil Nadu

It is necessary that students will understand the new pattern and style of Tamil Nadu Samacheer Kalvi 10th English Model Question Papers 2021 Answer Keys according to the latest exam pattern. These State Board 10th Standard English Public Exam Model Question Papers 2020-21 Tamil Nadu are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for TN SSLC Board Exams and Score More marks.

We hope the given Tamil Nadu State Board New Syllabus Samacheer Kalvi Class 10th English Model Question Papers 2020 2021 Pdf Free Download will help you get through your subjective questions in the exam.

Let us know if you have any concerns regarding the Government of TN State Board Samacheer Kalvi SSLC 10th Standard English Model Question Papers with Answers 2020 21, Sample Papers, Previous Year Question Papers, drop a comment below and we will get back to you as soon as possible.

Students can download 10th Social Science History Chapter 4 The World after World War II Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Social Science Solutions History Chapter 4 The World after World War II

Samacheer Kalvi 10th Social Science The World after World War II Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Who was the first director of Whampoa Military Academy?
(a) Sun Yat-Sen
(b) Chiang Kai-Shek
(c) Michael Borodin
(d) Chou En Lai
Answer:
(b) Chiang Kai-Shek

Question 2.
Which American President followed the policy of containment of Communism?
(a) Woodrow Wilson
(b) Truman
(c) Theodore Roosevelt
(d) Franklin Roosevelt
Answer:
(b) Truman

Question 3.
When was People’s Political Consultative Conference held in China?
(a) September 1959
(b) September 1948
(c) September 1954
(d) September 1949
Answer:
(d) September 1949

Question 4.
The United States and European allies formed to resist any Soviet aggression in Europe.
(a) SEATO
(b) NATO
(c) SENTO
(d) Warsaw Pact
Answer:
(b) NATO

Question 5.
Who became the Chairman of the PLO’s Executive Committee in 1969?
(a) Hafez al-Assad
(b) Yasser Arafat
(c) Nasser
(d) Saddam Hussein
Answer:
(b) Yasser Arafat

Question 6.
When was North and South Vietnam united?
(a) 1975
(b) 1976
(c) 1973
(d) 1974
Answer:
(b) 1976

Question 7.
Where was Arab League formed?
(a) Cairo
(b) Jordan
(c) Lebanon
(d) Syria
Answer:
(a) Cairo

Question 8.
When was the Warsaw Pact dissolved?
(a) 1979
(b) 1989
(c) 1990
(d) 1991
Answer:
(d) 1991

II. Fill in the blanks

1. ………………. was known as the “Father of modern China”.
2. in 1918, the society for the study of Marxism was formed in ………………. University.
3. After the death of Dr. Sun Yat-Sen, the leader of the Kuomintang party was ……………….
4. ……………….. treaty is open to any Arab nation desiring peace and security in the region.
5. The treaty of ………………. provided for mandates in Turkish -Arab Empire.
6. Germany joined the NATO in ……………….
7. ………………. was the Headquarters of the Council of Europe.
8. ………………. treaty signed on February 7,1992 created the European Union.

Answers:

1. Dr. Sun Yat-Sen
2. Peking
3. Chiang-Kai-Sheik
4. Central Treaty organisation
5. Versailles
6. 1955
7. Strasbourg
8. Maastricht

III. Choose the correct statement / statements

Question 1.
(i) In China (1898) the young emperor, under the influence of the educated
minority, initiated a series of reforms known as the 100 days of reforms.
(ii) The Kuomintang Party represented the interests of the workers and peasants.
(iii) Yuan Shih-Kai had lost prestige in the eyes of Nationalists, when he agreed to the demand of Japan to have economic control of Manchuria and Shantung.
(iv) Soviet Union refused to recognize the People’s Republic of China for more than two decades.
(a) (i) and (ii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iii) are correct
(d) (i) and (iv) are correct
Answer:
(c) (i) and (iii) are correct

Question 2.
(i) In 1948, the Soviets had established left wing government in the countries of Eastern Europe that had been liberated by the Soviet Army.
(ii) The chief objective of NATO was to preserve peace and security in the North Atlantic region.
(iii) The member countries of SEATO were committed to prevent democracy from gaining ground in the region.
(iv) Britain used the atomic bomb against Japan to convey its destructive capability to the USSR.
(a) (ii), (iii) and (iv) are correct
(b) (i) and (ii) are correct
(c) (iii) and (iv) are correct
(d) (i), (ii) and (iii) are correct
Answer:
(b) (i) and (ii) are correct

Question 3.
Assertion (A): America’s Marshall Plan was for reconstruction of the war¬. ravaged Europe.
Reason (R): The US conceived the Marshal Plan to bring the countries in the Western Europe under its influence.
(a) Both (A) and (R) are correct, but R is not the correct explanation of A
(b) Both (A)and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct and R is the correct explanation of A

IV. Match the following

Answer:
A. (ii)
B. (iii)
C. (iv)
D. (v)
E. (i)

V. Answer briefly

Question 1.
Write any three causes for the Chinese Revolution of 1911.
Answer:
Three causes of the Chinese Revolution of 1911 are

1. The government of Manchu dynasty began to disintegrate with the death of the Empress Dowager Cixi in 1908. The new emperor was two-years old and the provincial governors began to assert their independence. In October 1911 the local army mutinied and the revolt spread.
2. There were a few middle class leaders. Dr. Sun Yat-sen was one among them. He took part in the rising against the Manchus in 1895. The rising failed and Sun Yat-Sen was sent in the prison. But he continued to spread nationalist ideas.
3. Yuan Shih-Kai, who had earlier served as a minister in the Manchu administration, persuaded those responsible for the ascension of the young Emperor to prevant on him to abdicate.

Question 2.
Explain how in 1928 Kuomintang and Chiang-Kai Shek established Central Government in China.
Answer:
Chiang Kai-shek started conquering China. Starting from Canton, in 1928, he captured Peking and also removed all communists in the Kuomintang party. Thus, he established Central government in China.

Question 3.
Write a note on Mao’s Long March.
Answer:
Mao was an active leader who had gained full control of the Chinese Communist Party, by 1933. In 1934, he organised communist army of about 100,000 and set out on a long March. The marchers were continually harassed by Kuominatang forces, by local war lords and by unfriendly tribesmen of the 100,000 who set out, only 20,000 finally arrived in northern Shemi in late 1935, after crossing nearly 6000 miles. They were soon poined by other communist armies. By 1937, Mao had become the leader of over 10 million people. Mao’s Long March, as it is called so, has become legendary in the history of China.

Question 4.
What do you know of Baghdad Pact?
Answer:
In 1955, Turkey, Iraq, Great Britain, Iran and Pakistan signed a pact known as Baghdad pact. In 1958, when United States joined, then it was called as Central Treaty organization. The treaty was open to any Arab nation desiring peace and security in their region. It was dissolved in 1979.

Question 5.
What was the Marshall Plan?
Answer:
The Marshall Plan was an American initiative passed in 1948 to and Western Europe, in which the United States gave over $12 Billion in economic assistance to help rebuild Western European economies after the end of World War II. It operated for four years beginning in April 1948. The goals of the United States were to rebuild war-tom regions, modernise industry and improve European prosperity.

Question 6.
The Suez Canal crisis confirmed that Israel had been created to serve the cause of western interests-Elaborate.
Answer:
In 1956, Egypt invaded Suez Canal under Colonel Nasser and nationalized it. With the failure of diplomacy, Britain and France decided to use force. They bombed Egyptian air fields as well as Suez Canal area. However, United States and United Nations pressure, all the three invaders withdrew from Suez Canal.

Question 7.
Write a note on Third World Countries.
Answer:
The capitalist countries led by the US were politically designated as the First Worlds, while the communist states led by the Soviet Union came to be known as the Second World states, outside these two were called third World. During the Cold War, third World consisted of the developing world the former colonies of Africa, Asia, and Latin America. With the break up of the Soviet Union in 1991, and the process of globalisation, the term Third World has lost its relevance.

Question 8.
How was the Cuban missile crisis defused?
Answer:
In April 1961, on the island of Bay of Pigs, U.S bombed Cuban airfields and surrounded Cuba, with their warships. At the same time, USSR was secretly installing nuclear missiles in Cuba. Finally when Soviet President Khrushchev agreed to withdraw the missiles, the Cuban missile-crisis defused.

VI. Answer all the questions under each caption

Question 1.
Cold War

(a) Name the two military blocs that emerged in the Post-World War II.
Answer:
United States and the Soviet Union were the two military blocs that emerged in the post World War-II.

(b) Who coined the term “Cold War” and who used it first?
Answer:
The term Cold war was first coined by the English writer George Orwell in 1945 and it was used for the first time by Bernard Baruch, a multimillionaire from USA.

(c) What was the response of Soviet Russia to the formation of NATO?
Answer:
Warsaw Pact was the response of Soviet Russia to the formation of NATO.

(d) What was the context in which Warsaw Pact was dissolved?
Answer:
With the Break-up of USSR in 1991, the Warsaw Pact was dissolved.

Question 2.
Korean War

(a) Who was the President of North Korea during the Korean War?
Answer:
Kim II was the President of North Korea during the Korean War.

(b) Name the southern rival to the President of North Korea.
Answer:
Syngman Rhee

(c) How long did the Korean War last?
Answer:
The Korean War lasted three years

(d) What was the human cost of the War?
Answer:
The human cost was enormous, there were 500,000 western casualties and three times that number on the other side. Approximately two million Korean civilian died.

Question 3.
Non-Aligned Movement (NAM)

(a) When and where was the first conference on Non-Aligned Movement held?
Answer:
At Belgrade, in 1961, the first conference on NAM was held.

(b) Who were the prominent personalities present in the first conference?
Answer:
The prominent personalities present in the first conference were Tito (Yugoslavia), Nasser (Egypt), Nehru (India), Nikrumah (Ghana), Sukrano (Indonesia),.

(c) What were the objectives of NAM?
Answer:
Peaceful co-existence, commitment to peace and security.

(d) List out any two basic principles of Non-Alignment Movement enunciated in the Belgrade Conference.
Answer:

1. Non-Alignment with any of the two super powers (USA/USSR).
2. Fight all forms of colonialism and Imperialism.

VII. Answer in detail

Question 1.
Estimate the role of Mao Tse tung in making China a communist country.
Answer:

1. Mao was greatly influenced by the ideas of Max and Lenin. He wanted to make China a communist country. So, he became active in the political activities of Hunan from the year 1912.
2. After the death of Sun Yat-Sen in 1925, The Kuomintang was organised under the leadership of Chang Kai-Shek. Being an avowed critic of communists, Chiang removed all the important position holders in the Communist Party including Mao Tse Tung to weaken the party. However, the communists continued to influence the workers and peasants the Kuomintang represented the interests of landlords and capitalists.
3. Mao had understood that the Kuomintang grip on the towns was too strong. So, he started organising the peasantry. When the relationships between Kuomintang and Communist Party broke, a few hundred Communist-led by Mao retreated into the wild mountains on the border between the provinces of Kiangsi and Hunan. The Kuomintang could not penetrate the mountains.
4. Meanwhile, the campaign against the communists got distracted as Chiang Kai-Shek had to deal with the constant threat from Japan.
5. By 1933 Mao had gained full control of the Chinese Communist Party. In 1934, he set out on a long march with the help of about 100,000 communist army. He also got support of other communist armies.
6. By 1937, Mao had become the leader of over 10 million people. He organised workers and peasants councils in villages of Shensi and Kansu and finally got success in making China a communist country.

Question 2.
Attempt an essay on the Arab-lsraeii wars of 1967 and 1973.
Answer:

1. Before the creation of the State of Israel in 1948, all Arabs and their . descendants lived in the Palestine.
2. Ever since the formation, the Palestinian Liberation organisation (PLO). Israel came to be attacked by the Palestinian Guerrilla groups based in Syria, Lebanon and Jordan.
3. Israel resorted to violent military force .
4. In November 1966, an Israeli strike on the village of Al-Samu in the west bank of Jordan killed 18 people and wounded 54 people.
5. Israel’s air battle with Syria in April 1967 ended in the shooting down of six Syrian MIG fighter jets.
6. After air battle with Syria, when Arab States were mobilized by Nasser on June 5th Israel staged a sudden air strike that destroyed more than, 90% of Egypt’s air force on the tarmac.
7. Later, Egypt and Syria under Presidents Anwar Sadat and Hafez-al- Assad respectively concluded a secret agreement in January 1973 to bring their armies under one command.
8. When peace deals did not work out with Israel, Egypt and Syria launched a sudden and surprise attack on the Yom Kippur religious holiday on 6^th October 1973.
9. Though Israel suffered heavy casualities, it finally pushed back the Arab forces.
10. Due to U.N intervention, Israel was forced to return and Arabs gained nothing.
11. U.S succeeded in showing their control on their region and oil resource, led to U. S led war against Iraq in 1991.

Question 3.
Narrate the history of transformation of Council of Europe into an European Union.
Answer:
(i) After World War II, it was decided to integrate the states of western Europe. One of the chief objectives was to prevent further European wars by ending the rivalry between France and Germany. In May 1949, ten countries met in London and signed a form called Council of Europe

(ii) Since the Council of Europe had no real power, a proposal to set up two European organisations were made. Accordingly, the European Defence Community (EDC) and the European Coal and Steel Community (ECSC) were established. Six countries belonging to ECSC signed the treaty of Rome which established the European Economic Community (EEC) or the European Common Market, with headquarters at Brussels.

(iii) The EEC facilitated the elimination of barriers to the movement of goods and services, capital, and labour. It also prohibited public policies or private agreements that restricted market comptetion. Throughout the 1970s and 80s the EEC kept expanding its membership.

(iv) The single European Act came into force on July 1, 1987. It significantly expanded the EEC’s scope giving the meetings of the EPC a legal basis. It also called for more intensive coordination of foreign policy among member countries. According to the SEA each member was given multiple votes, depending on the countries population.

(v) The Maastricht (Netherlands) Treaty signed on February 7, 1992, created the European Union (EU). Today the European Union has 28 member states, and is functioning from its headquarters at Brussels, Belgium.

VIII. Activity

Question 1.
Divide the class into two groups. Let one group act as supporters of USA and the other group act as supporters of Soviet Union, Organise a debate.
Answer:
One group of students act as the supports of USA and the other group act as supporters of Soviet Union on their role in the cold war era.
(Students can have debate based on the following aspects)

During the time of cold war, the two super powers that emerged was USA and USSR.

Strengths of USA:

1. Strongest navy both the Pacific and Atlantic.
2. Continued and strong economic position that the USA held in the beginning till the end.
3. Foreign policy of the USA. The actions of USA during the cold war era, very powerful and could guarantee others the strong power of the nation is to prevent communism spreading elsewhere.
4. The country was able to generate the vast amount of wealth necessary to sustain the investment in weapons, technology and other operations.
5. USA was also technologically powerful in target finding, tracking, sensors etc.
6. The Truman’s doctrine was used throughout the war time. The doctrine helped USA to negotiate with other states to adopt capitalism.

Strengths of USSR:

1. Strongest land based military, especially tanks.
2. USSR developed air defense equipment and networks.
3. Spread of Communism joined together backed USSR.
4. As against NATO, USSR also formed military block, the Warsaw pact.
5. With Western European countries USSR facilitated trade relations.
6. Highly strengthened space exploration.

In the end of cold war, the Soviet Union fell, and the Communism expired.

Question 2.
Involving the entire class, an album may be prepared with pictures relating to Korean, Arab-lsraeli and Vietnam Wars to highlight the human sufferings in terms of death and devastation.
Answer:
Students should go to the search engine in google and type “The War’s impact on the Korean peninsula and trace the article of the journal of America – East Asia Relations” for a detailed Report.

Korean War: To give a sample answer: According to a U.S. Department of state publication, the number of killed, wounded and missing from the Armed forces of the Republic of Korea exceeded 4,00.000. From the U.S. side, 1,42,000. In addition, the heavy toll in death and injuries to the civilian population as well as wide property damage.

Arab-Israel War: Students should trace the answer from the google search engine as: The U.N. Report on Israel’s Gaza War.

To give a sample answer: From the 183 page report of the U.N. report, it is very clear that many civilians died. More than 6,000 Israeli airstrikes, 14,500 tank shells and 35,000 artillery shells led to the destruction of about 18,000 dwellings in Gaza. Nearly 1,462 Palestinian civilians, 299 women and 551 children were killed (According to U.N. Investigation). In Israel six civilians and 67 soldiers were killed.

Vietnam War: Students should trace the answer in the google search engine as “The War’s effect on the Vietnamese Land and People” for a detailed report.

To give a sample answer: About 58,000 American soldiers were killed and another 3,04,000 were wounded. Since most of the fighting took place in Vietnam, an estimated 4 million Vietnamese were killed on both sides including as many as 1 -3 million civilians. Many Vietnamese in country side turned into homeless refuges. Many forms and forests were destroyed.

Along with the detailed report, students should collect pictures showing Korea, Arab-Israel and Vietnam War, highlighting human sufferings. Students should click to google search engine and go to → Images → then type Korean War pictures / Arab Israel War pictures / Vietnam War pictures, download it take a print out of its and paste it in a A4 sheet paper or an a chart paper neatly and create an album.

Timeline:

Samacheer Kalvi 10th Social Science The World after World War II Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The emergence of ……………. and ……………. as super powers resulted in the division of the World into two block after World War II.
(a) Japan, USA
(b) USA, USSR
(c) China, Japan
(d) USA, Germany
Answer:
(b) USA, USSR

Question 2.
Dr.Sun yat-sen was born in a poor family near ………
(a) Canton
(b) France
(c) Spain
Answer:
(a) Canton

Question 3
……………. was called the father of modern China.
(a) Chiang-Kai-Shek
(b) Yuan-Shi-Kai
(c) Mao-Tse-Tung
(d) Dr. Sun Yat-Sen
Answer:
(d) Dr. Sun Yat-Sen

Question 4.
Mao was born in ……… in South – East China.
(a) Cambodia
(b) Vietnam
(c) Hunan
Answer:
(c) Hunan

Question 5.
Yuan-Shi-Kai died in the year …………….
(a) 1912
(b) 1913
(c) 1915
(d) 1916
Answer:
(d) 1916

Question 6.
Which of the following is not a part of Indo – China?
(a) Cambodia
(b) China
(c) Vietnam
Answer:
(b) China

Question 7.
The Historical Long March set out in China in …………….
(a) 1935
(b) 1937
(c) 1934
(d) 1936
Answer:
(c) 1934

Question 8.
In which of the following was indentured Vietnamese labour widely used?
(a) Rice cultivation
(b) Rubber plantation
(c) Industry
Answer:
(b) Rubber plantation

Question 9.
……………. was the leader of the people’s Republic of China.
(a) Marshall
(b) Truman
(c) Mao-Tse-Tung
(d) Chiang-Kai-shek
Answer:
(c) Mao-Tse-Tung

Question 10.
Ho Chi Minh means ………
(a) He, Who enlightens
(b) Enlightenment
(c) The Enlightened one
Answer:
(a) He, Who enlightens

Question 11.
The Idea of European self-help programme financed by the United States was called as …………….
(a) NATO
(b) SEATO
(c) ECA
(d) Marshall plan
Answer:
(d) Marshall plan

Question 12.
EURATOM was established by the ………
(a) Treaty of Nanking
(b) Treaty of Rome
(c) Treaty of London
Answer:
(b) Treaty of Rome

Question 13.
The term ‘cold war’ was first coined by the English writer …………….
(a) Shakespeare
(b) George Orwell
(c) William Dexter
(d) Peter Alphonse
Answer:
(b) George Orwell

Question 14.
As a part of Marshall plan financing, European nations received nearly ……………. billion in aid.
(a) $ 12
(b) $ 11
(c) $ 15
(d) $ 13
Answer:
(d) $ 13

Question 15.
Greece and Turkey joined NATO in the year …………….
(a) 1945
(b) 1947
(c) 1952
(d) 1955
Answer:
(c) 1952

Question 16.
……………. was otherwise called as pact.
(a) NATO
(b) CENTO
(c) SEATO
(d) EC SC
Answer:
(d) EC SC

Question 17
……………. are included as member countries in NATO.
(a) Canada, Belgium
(b) U.K., Portugal
(c) Both (a) and (b)
(d) None of the above
Answer:
(c) Both (a) and (b)

Question 18.
NATO had ……………. members by the year 2017.
(a) 30
(b) 51
(c) 29
(d) 24
Answer:
(c) 29

Question 19.
In December 1954, a conference of eight European nations took place in Moscow. This was called as …………….
(a) Warsaw pact
(b) Counter to NATO
(c) SEATO
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 20.
The headquarters of the Warsaw pact was …………….
(a) Belgium
(b) Poland
(c) Moscow
(d) Romania
Answer:
(c) Moscow

Question 21.
The Korea was partitioned into North and South Korea in the year …………….
(a) 1944
(b) 1945
(c) 1946
(d) 1947
Answer:
(b) 1945

Question 22.
With the collapse of the ……………. the idea of non-alignment lost relevance.
(a) Berlin
(b) CIA
(c) ECA
(d) Soviet Union
Answer:
(d) Soviet Union

Question 23.
The NAM held its first conference at ……………. in 1961.
(a) Bandung
(b) Belgrade
(c) Thailand
(d) Philippine
Answer:
(b) Belgrade

Question 24.
The World Zionist Organisation was founded in the year …………….
(a) 1857
(b) 1887
(c) 1897
(d) 1867
Answer:
(c) 1897

Question 25.
Castro nationalised the U.S. owned ……………. companies.
(a) cotton
(b) sugar
(c) oil
(d) petrol
Answer:
(b) sugar

Question 26.
The Cuban Missile crisis was defused by ……………. as the agreed to withdraw the missiles.
(a) Khrushchev
(b) Fidel Castro
(c) Leumi
(d) Stem Gang
Answer:
(a) Khrushchev

Question 27.
In the Arab World, it is treated as the ……………. when large number of Arabs become refugees.
(a) Catastrophe
(b) Nakbah
(c) Negev and (b)
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 28.
For resolving the Suez Canal Crisis, ……………. from Indian played a crucial role.
(a) Gandhi
(b) Nehru
(c) Bose
(d) Tilak
Answer:
(b) Nehru

Question 29.
Arab-lsrael war took place in the years …………….
(a) 1967,69
(b) 1969,74
(c) 1967,73
(d) 1972,73
Answer:
(c) 1967,73

Question 30.
……………. became the first president of the state of Palestine in 1989.
(a) Anwar Sadat
(b) Yasser Arafat
(c) Nasser Arafat
(d) Nasser Hussain
Answer:
(b) Yasser Arafat

Question 31.
By the end of second world war, ……………. controlled the northern half of Vietnam.
(a) Vietminh
(b) Ho-Chi-Minh
(c) Ngo Dinh Diem
(d) None
Answer:
(a) Vietminh

Question 32.
……………. controlled the southern half of the Vietnam.
(a) Viet Minh
(b) Bao Dai
(c) Ngo Dinh Diam
(d) None
Answer:
(b) Bao Dai

Question 33.
The North and the South Vietnam were formally united as one country in …………….
(a) 1973
(b) 1976
(c) 1975
(d) 1974
Answer:
(b) 1976

Question 34.
In May 1949, ten countries met in ……………. and signed to form a council of Europe.
(a) Japan
(b) Syria
(c) London
(d) France
Answer:
(c) London

Question 35.
To prevent further European War ……………. was founded.
(a) Council of Europe
(b) Council of Trent
(c) ECSC
(d) EU
Answer:
(a) Council of Europe

Question 36.
……………. did not join the EEC, when it was formed.
(a) U.K
(b) U.S.A
(c) USSR
(d) Italy
Answer:
(a) U.K

Question 37.
Britain, voted to exit the EU in the year …………….
(a) 2015
(b) 2014
(c) 2017
(d) 2019
Answer:
(c) 2017

Question 38.
West Berlin was supported by …………….
(a) USA
(b) UK
(c) USSR
(d) Germany
Answer:
(a) USA

Question 39.
East Berlin was supported by …………….
(a) USA
(b) UK
(c) USSR
(d) Germany
Answer:
(c) USSR

Question 40.
Germany was officially united on …………….
(a) 1987
(b) 1989
(c) 1990
(d) 2003
Answer:
(c) 1990

Question 41.
in 1985, ……………. became the head of the USSR.
(a) Mikhai Gorbachev
(b) Reagen
(c) Stalin
(d) Kohl
Answer:
(a) Mikhai Gorbachev

Question 42.
The Chernobyl disaster took place in the year …………….
(a) 1987
(b) 1984
(c) 1986
(d) 2006
Answer:
(c) 1986

Question 43.
Gorbachev announced his resignation on …………….
(a) 21st November
(b) 25th December
(c) 2nd October
(d) 15th August
Answer:
(b) 25th December

Question 44.
USSR dissolved formally In the year …………….
(a) 31st Dec. 1990
(b) 30th Dec. 1991
(c) 31st Dec. 1991
(d) 28,h Feb. 1991
Answer:
(c) 31st Dec. 1991

Question 45.
……………. was an ally of Gorbachev.
(a) Yeltsin
(b) Brezhnev
(c) Khrushchev
(d) None
Answer:
(a) Yeltsin

II. Fill in the blanks

1. ………………… and ………………… were the Superpowers after the World War II.
2. The cold war period ended with the fall of …………………
3. Empress Dowager died in ………………… led to the disintegration of the Manchu dynasty.
4. Kuomintang party of China was otherwise called as ………………… party.
5. ………………… was the ultimate aim of Dr. Sun Yat-sen’s party.
6. The Chinese revolution broke out in the year …………………
7. The first director of the Whampoa Military academy was …………………
8. The campaign against the communists led by Chiang-Kai-Shek was distracted by ………………… and warlords.
9. By ………………… Mao became the leader of the Chinese people.
10. Over ………………… delegates from various countries attended the people’s political consultative conference.
11. When Japanese surrendered in 1945, the Japanese areas were occupied by the …………………
12. By the year …………………, communist control has been established over most parts of China.
13. The two mighty communist powers in the world were ………………… and …………………
14. The term cold war was first coined by …………………
15. ………………… nations became a part of Marshall’s plan of self-help programme.
16. The Marshall plan funding ended in …………………
17. ………………… was created to resist Soviet aggression in Europe.
18. NATO means …………………
19. NATO was formed in the year …………………
20. The members of NATO agreed that, any armed attack on any one of them would be considered attack on …………………
21. For the collective security of the South-east Asia, ………………… was formed in 1954.
22. Member of SEATO were committed to prevent …………………
23. SEATO was formed in 1954, after signing of the …………………
24. ………………… was formed by Soviet Union as a counter to NATO.
25. ………………… European nations attended the Warsaw pact in December 1954.
26. The treaty on Warsaw pact was concluded on …………………
27. The Warsaw pact was dissolved in …………………
28. The Warsaw pact dissolved because of the break-up of …………………
29. The Baghdad pact was otherwise called as …………………
30. Turkey, Iran, Iraq, Pakistan and Great Britain signed a treaty in 1955, known as …………………
31. When U.S.A joined the Baghdad pact, it was called by name …………………
32. USA joined the Baghdad pact in the year …………………
33. CENTO was dissolved in the year …………………
34. Korea was partitioned as North Korea and South Korea in the year …………………
35. The president of North Korea was …………………
36. The party of Kim II was called as …………………
37. The president of the South Korea was …………………
38. The party of Syngman Rhee was …………………
39. Korean War lasted for ………………… years.
40. NAM refers to the …………………
41. NAM was signed in ………………… at the ………………… conference.
42. Bandung is a city in …………………
43. With the collapse of ………………… the idea of NAM lost importance.
44. CIA is a Central Intelligence Agency of …………………
45. The Act of ………………… in Cuba, threatened American economic interests.
46. USA bombed Cuban ………………… with the aim of overthrowing Castro’s regime.
47. ………………… was the president of USSR, when the Cuban Missile crisis took place.
48. Cuba was helped by …………………
49. When USSR remove missile from Cuba, USA had to agree to remove missile from ………………… and …………………
50. ………………… was a Zionist Terrorist Organisation.
51. The Zionist Para-military organization was called as …………………
52. The World Zionist organization was founded in the year …………………
53. Jews living outside their ancient home scattered around North America and Europe was called as …………………
54. ………………… Nationalised Suez Canal.
55. India represented by ………………… played a crucial role in resolving the crisis.
56. The Arab-lsraeli War took place in the years ………………… and …………………
57. PLO refers to …………………
58. PLO was formed in …………………
59. ………………… was the prominent leader of Palestine.
60. By the end of the Second World War, ………………… Controlled the northern half of Vietnam.
61. South Vietnam was ruled by …………………
62. America supported troops of ………………… Vietnam.
63. The city of ………………… was renamed as Ho-Chi-Minh city.
64. To create a United Europe to resist any threat from Soviet Russia, ………………… was formed.
65. European Economic Community was otherwise called as …………………
66. SEA refers to …………………
67. According to the SEA, each member was given ………………… votes.
68. The ………………… Treaty created the European Union.
69. Maastricht is in …………………
70. EU was created on Maastricht treaty signed on …………………
71. The headquarters of the EU is at …………………
72. Brussels is at …………………
73. ………………… Germany was prosperous.
74. ………………… Germany was suffering from lack of democracy and freedom.
75. Germany was officially reunited on …………………
76. Glasnost means …………………
77. Perestroika means …………………
78. ………………… was introduced by Gorbachev to restructure Soviet economic and political system.
79. After Gorbachev, power fell into the hands of …………………
80. For ………………… days, Soviet Union, continued to exist only in name.
81. Soviet Union dissolved formally on 31st December …………………
82. USSR split into ………………… countries.
83. ………………… was the president of the newly independent Russian State.
84. U.S.A. troops used ………………… weapons in their war against Vietnam.
85. Napalm and Agent Orange are the names of incendiary …………………

Answers:

1. USA and USSR
2. Berlin Wall
3. 1908
4. National People’s Party
5. Socialism
6. 1911
7. Chiang-Kai-Shek
8. Japan
9. 1937
10. 650
11. Kuomintang
12. 1948
13. Soviet Union, People’s Republic of China
14. George Orwell
15. Sixty
16. 1951
17. NATO
18. North Atlantic Treaty Organisation
19. 1949
20. NATO
21. SEATO
22. Communism
23. Manila Pact
24. Warsaw Pact
25. 8 (eight)
26. May 14, 1955
27. 1991
28. USSR
29. Central Treaty Organisation
30. Baghdad Pact
31. Central Treaty Organisation
32. 1958
33. 1979
34. 1945
35. Kim II
36. People’s Republic of Korea
37. Syngman Rhee
38. The Republic of Korea
39. three
40. Non-Aligned Movement
41. 1955, Bandung
42. Indonesia
43. Soviet Union
44. America
45. Castro
46. Airfields
47. Khrushchev
48. USSR
49. Turkey, Italy
50. Stem Gang
51. Irgun Zvai Leumi
52. 1897
53. Diaspora
54. Colonel Nasser
55. Nehru
56. 1967, 1973
57. Palestine Liberation Organisation
58. 1948
59. Yasser Arafat
60. Viet Minh
61. Ngo Dinh Diem
62. South
63. Saigon
64. Council of Europe
65. European Common Market
66. Single European Act
67. multiple
68. Maastricht
69. Netherlands
70. Feb 7, 1992
71. Brussels
72. Belgium
73. West Berlin’s
74. East Berlin
75. 3rd October 1990
76. openness
77. restructuring
78. Perestroika
79. Boris Yeltsin
80. Six 81.1991
81. 15
82. Boris Yeltsin
83. Bacteriological
84. bombs

III. Choose the correct statement / statements

Question 1.
(i) The cold war period ended with the fall of Berlin Wall
(ii) Mao concentrated mainly on organizing the peasantry.
(iii) In 1937, the communist army of about 100,000 set out on the Long march.
(iv) Marshall plan funded nearly $ 15 billion.
(a) (ii and (ii) are correct
(b) (i) and (iii) are correct
(c) (ii and (iv) are correct
(d) (iii) and (iv) are correct.
Answer:
(a) (ii and (ii) are correct

Question 2.
(i) The USSR was much concerned about the destruction caused by the Second World War.
(ii) The South East Asia Treaty organization was organised for the collective security of countries in South East Asia.
(iii) The Communist States led by the Soviet Union came to be known First World Countries.
(iv) The Capitalist countries led by the U.S.A. were politically designated as Second World Countries.
(a) (i) and (ii) are correct
(b) (i), (iii), (iv) are wrong
(c) (i) and (iv) are correct
(d) (ii) and (iii) are correct.
Answer:
(b) (i), (iii), (iv) are wrong

Question 3.
(i) NAM refers to the Non-Aligned Movement.
(ii) The Single European Act of the EU as called as SEA.
(iii) Anwar Sadat was the president of the Palestine in 1989.
(iv) The fighters of South Vietnam were trained in guerrilla warfare.
(a) (i) and (ii) are correct
(b) (i), (iii), (iv) are wrong
(c) (i) and (iv) are correct
(d) (ii) and (iii) are correct.
Answer:
(a) (i) and (ii) are correct

Question 4.
(i) The Third World principally consist of the developing World.
(ii) With the breakup of the Soviet Union in 1991, and the process of globalization, the term has lost its relevance.
(iii) The former colonies of Asia, Africa, and Latin America were called as Third World Countries.
(iv) The division of Germany into West and East led to glaring differences in living standards.
(a) (i) and (ii) are correct
(b) (i), (ii), (iii) are correct
(c) (i) and (iii) are correct
(d) All the four are correct.
Answer:
(d) All the four are correct.

Question 5.
(i) Yeltsin worked as a Mayor of Moscow.
(ii) Yeltsin was returned to power with overwhelming support of a Moscow in 1899.
(iii) For twelve days, the Soviet Union continued to exist only in name.
(iv) On 28th February 1991, USSR was formally dissolved.
(a) (i) and (ii) are correct
(b) (i), (ii), (iii) are correct
(c) (i) and (iii) are correct
(d) All the four are correct.
Answer:
(a) (i) and (ii) are correct

Question 6.
(i) CENTO was otherwise called as Manila pact.
(ii) As a counter to SEATO, NATO was formed.
(iii) The Korean war helped to bring down the intensity of the Cold war.
(iv) The EEC eliminated barriers to the movement of goods, Capital and labour.
(a) (i) and (ii) are correct
(b) (i), (ii), (iii) are wrong
(c) (i) and (iv) are correct
(d) All the four are correct.
Answer:
(b) (i), (ii), (iii) are wrong

Question 7.
(i) The Berlin Wall was just a physical barrier.
(ii) Berlin Wall divided East Germany and West Germany.
(iii) USA supported East Berlin.
(iv) USSR supported West Berlin.
(a) (i) and (ii) are correct
(b) (i), (ii), (iii) are wrong
(c) (i) and (iv) are correct
(d) All the four are wrong.
Answer:
(a) (i) and (ii) are correct

IV. Assertion and Reason

Question 1.
Assertion (A): Yuan Shih-Kai of China lost prestige in his country.
Reason (R): He agreed to the demand of Japan to have economic control of Manchuria and Shantung.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct, and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct, and R is the correct explanation of A

Question 2.
Assertion (A): The rivalry that developed after World War 11 is referred to as “Cold War”.
Reason (R): This war did not take recourse to weapons.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct, and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct, and R is the correct explanation of A

Question 3.
Assertion (A): There was High military expenditure on both sides of USA and USSR.
Reason (R): Soviet Union tested the nuclear bomb and America used the nuclear bomb against Japan.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct, and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct, and R is the correct explanation of A

Question 4.
Assertion (A): The U.S. and its European allies formed the NATO to wage war against Vietnam.
Reason (R): As a counter to the NATO, Soviet Union organised the Warsaw pact.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R.) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(d) (A) is wrong and (R) is correct.

Question 5.
Assertion (A): A small country had succeeded in winning Independence and the greatest power of the World-The country Vietnam.
Reason (R): The help given to Vietnam by the Socialist Countries, the political support given by Asia and Africa is evident.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct and R is the correct explanation of A

Question 6.
Assertion (A): Sun Yat-sen sent Chiang Kai-shek to Moscow, in Russia. The Russians in turn sent Michael Borodin to China.
Reason {R): Chiang Kai-shek started conquering China from Canton.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(a) Both (A) and (R) are correct, R is not the correct explanation of A

Question 7.
Assertion (A): U.S.A. supported Diem government in South Vietnam. Reason (R): U.S. wanted to establish a strong government in South Vietnam.
(a) Both (A) and (R) are correct, R is not the correct explanation of A
(b) Both (A) and (R) are wrong
(c) Both (A) and (R) are correct and R is the correct explanation of A
(d) (A) is wrong and (R) is correct.
Answer:
(c) Both (A) and (R) are correct and R is the correct explanation of A

V. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (v)
B. (iii)
C. (ii)
D. (i)
E. (vi)

Question 2.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (i)
D. (ii)
E. (vi)

VI. Answer briefly

Question 1.
Describe about Zionist movement.
Answer:
(i) In Palestine, the ancient home of Jews, only a few thousand Jews were living in 1900.
(ii) Some 15 million were scattered around Europe and North America.
(iii) These Jews had been subjected to systematic persecution for centuries.
(iv) But in the late nineteenth century, the persecution in Russia (Where two-thirds of the world’s jews lived), France and Germany was intense.
(v) Some Jews emigrated to Palestine, while many more went to the United States and Britain.
(vi) In 1896, Thodore Herzel, a Viennese journalist, published a pamphlet called the Jewish state in which he called for the creation of a Jewish national home. In 1897 the world zionist organisation was founded.

Question 2.
What was Truman’s policy?
Answer:
Truman, the president of USA announced a policy of containment of communism. This implies U.S. would support those countries which were threatened by USSR to spread communism.

Question 3.
Mention the initial member countries of the EU.
Answer:

1. Belgium
2. France
3. Italy
4. Luxemburg
5. Netherlands
6. West Germany

Question 4.
What is meant by SEA?
Answer:
SEA refers to the single European Act which came into force on July 1,1987. According to the SEA, each member was given multiple votes depending on the country’s population. Its main aim is establishing a single market.

Question 5.
Name the organs of the EU.
Answer:

1. European parliament
2. Council of the European Union
3. European commission
4. Court of Justice
5. Court of Auditors

Question 6.
Write a note on European Union.
Answer:
On February 7, 1992, the Maastricht Treaty that signed in Netherlands created the European Union. All the member countries of the EU will use common currency Euro, a single market and common Act. EU at present has 28 members with head quarters at Brussels, Belgium.

Question 7.
How was the European union formed?
Answer:
According to the Merger Treaty of 1967, the three communities namely the European coal and steel community, the European Economic community and the European Atomic Energy community were merged together to form the European union.

VII. Answer all the questions under each caption

Question 1.
People’s Republic of China

(a) Who was the leader of the People’s Republic of China?
Answer:
Mao Tse-Tung was the leader of the People’s Republic of China.

(b) Name the two mighty Communist powers of the world?
Answer:
The Soviet-Union and the people’s Republic of China.

(c) Who did not recognise People’s Republic of China and for how long?
Answer:
The UNO refused to recognise people’s Republic of China for more than twenty years.

(d) How did the government of Taiwan got recognition?
Answer:
The government of Chiang Kai-shek in Taiwan was given recognition ‘ due to the pressure from USA.

Question 2.
Achievements of EU

(a) What is the symbol of the Euro?
Answer:
The symbol of the Euro is €.

(b) What did the Euro eliminate?
Answer:
The Euro eliminated foreign exchange hurdles encountered by companies doing business across European border.

(c) How many members are there in the EU at present?
Answer:
At present, there are 28 members in the EU.

(d) Who allocates funds to European research projects?
Answer:
The European Research council.

Question 3.
Disintegration of The Soviet Union.

(a) Who became the head of USSR in 1985?
Answer:
Mikhail Gorbachev became the head of the USSR in 1985.

(b) What is meant by “thaw”?
Answer:
The “Thaw” refers to the period from the early 1950’s and 1960’s when repression and censorship in the Soviet Union was relaxed and millions of political prisioners were released. It was the period of Khrushchev’s reign.

(c) What is meant by Glasnost?
Answer:
Glasnost means openness. It was the policy of more transperancy and openness in the government policy of former Soviet Russia introduced by Mikhail Gorbachev.

(d) What is meant by Perestroika?
Answer:
Perestroika means restructuring. It refers to the programme introduced by Mikhail Gorbhachev to restructure Soviet economic and political system.

Question 4.
Berlin Wall

(a) Who constructed a wall which virtually cut off West Berlin and East Berlin? and when?
Answer:
East German began to construct a wall in 1961 which virtually cut off West Berlin and East Berlin.

(b) How was it guarded?
Answer:
It was guarded with watchtowers and other lethal impediments to stop people from the east.

(c) What does the Berlin wall symbolise?
Answer:
It was symbolic boundary between Communism and Capitalism.

(d) What happened with the fall of the Berlin wall?
Answer:
With the fall of the Berlin wall , followed by the collapse of the Soviet Union, the cold war came to an end.

VIII. Answer in detail

Question 1.
Write a brief account of the life and achievements of Ho Chi Minh.
Answer:
He was born in a small town in Central Vietnam. He studied in French school that produced great leaders. After his studies he worked on a French liner operating between Saigon and Marseilles. He was greatly inspired by European communist parties, became member of commintem and was instrumental in bringing together competing nationalists groups to form the Vietnamese communist party in 1930. It was later renamed the Indo-Chinese communist party. After spending 30 years abroad in Europe, China Thailand, he returned to Vietnam in 1941. He became President of the Vietnam Democratic Republic in 1943.

After the split of Vietnam, he and the communists took control of North Vietnam. With the help of his government, National Liberation Front in the south fought for unification of the country. He fought hard to maintain the autonomy of Vietnam and till the end proved true to his name. Ho Chi Minh meaning He, Who Enlightens.

Question 2.
Illustrate the cold war developments in case of the Vietnam war. Narrate how North and South Vietnam unified as Independent Nation.
Solution:

1. By 1945, the end of the second world war, Viet Minh controlled the northern half of Vietnam, led by Ho-chi- Minh.
2. Viet Minh and french reached an agreement by which North Vietnam would be a free state.
3. While French was helped by America, Viet Minh was helped by the new Chinese communist government.
4. War broke out between them. Eventually , France troops were defeated.
5. The Geneva conference that met on Korea and Indo China in 1954, decided that of Laos. Combodia and Vietnam. The independent states would be Laos and Cambodia. Vietnam, temporarily divided.
6. While North Vietnam controlled by Viet Minh with leader Ho-Chi-Minh and south Vietnam would be under the leadership of Bao Dai.
7. At the same time, South Vietnam was ruled by Ngo Dinh Diem.
8. When U.S wanted to establish a strong Non-communist government in South Vietnam. In 1965, marines landed on Danang naval base and namely 2,10,000 traps in the country, j (ix) The U.S bombed both North and South.
9. The fighters of North Vietnam trained in Guerrilla warfare sustained.
10. America suffered heavy casualties , vast devastated and many were killed.
11. The youth rebelled against the horrors of the war.
12. The protest against the war spread all over the world.
13. By 1975, the armies of the North and the only one party of South Vietnam called National Liberation front of South Vietnam attacked America.
14. By 30th April 1975, all the American troops had withdrawn and capital of South Vietnam Saigon was liberated.
15. North Vietnam and South Vietnam formally united as one country in 1976.
16. The city of Saigon was renamed as the Ho-Chi-Minh city .
17. Thus, the emergence of Vietnam as a united and Independent nation was an historic event.

Question 3.
Explain the breakup of the Soviet Union.
Answer:

1. In the middle of 1980’s Soviet Union economy was suffering.
2. In 1985, Mikhail Gorbachev took over as the president of USSR.
3. Gorbachev spoke about the need for openness (Glasnost) and Perestroika (restructuring).
4. But his ideas of reform did not work out for him because, to compete with U.S, USSR need to allocate more funds to the military.
5. The economic stagnation of the Soviet Union aggrevated tension and promoted nationalist feelings.
6. In the year 1988, Mass protest broke out in Armenia and in the Baltic states.
7. Gorbachev made attempts to stabilize his position by relying on conservative forces in 1989, 1991.
8. But the massive miner’s strike interrupted. The series of worker’s strike under mined the communist regimes first in Poland, then in Hungary.
9. The fall of Berlin wall in Germany, encouraged people to be united.
10. Gorbachev made a last attempt to take a hard line against miner’s strike and huge demonstrations in Moscow in 1991.
11. In response, the conservative forces in his government used troops in Moscow and held Gorbachev under house arrest.
12. Power fell into the hands of Boris Yeltsin.
13. In November 1991, eleven republics announced that they would establish a common wealth of independent states.
14. On 25th Dec 1991 , Gorbachev resigned.
15. For six days, the Soviet Union continued to remain only in name and at midnight on 31st December 1991 it was formally dissolved.
16. The USSR was no more.

Students can download 10th Science Chapter 2 Optics Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Science Solutions Chapter 2 Optics

Samacheer Kalvi 10th Science Optics Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
The refractive index of four substances A, B, C and D are 1.31,1.43,1.33, 2.4 respectively. The speed of light is maximum in:
(a) A
(b) B
(c) C
(d) D
Answer:
(a) A

Question 2.
Where should an object be placed so that a real and inverted image of same size is obtained by a convex lens:
(a) f
(b) 2f
(c) infinity
(d) between f and 2f
Answer:
(b) 2f

Question 3.
Where should an object be placed so that a real and inverted image of the same size is obtained by a convex lens ______.
(a) f
(b) 2f
(c) infinity
(d) between f and 2f.
Answer:
(b) 2f

Question 4.
Magnification of a convex lens is _____.
(a) positive
(b) negative
(c) either positive or negative
(d) zero.
Answer:
(b) negative

Question 5.
A convex lens forms a real, diminished point sized image at focus. Then the position of the object is at:
(a) focus
(b) infinity
(c) at 2f
(d) between f and 2f
Answer:
(b) infinity

Question 6.
Power of a lens is -4D, then its focal length is:
(a) 4 m
(b) -40 m
(c) -0.25 m
(d) -2.5 m
Answer:
(d) -2.5 m

Question 7.
In a myopic eye, the image of the object is formed _____.
(a) behind the retina
(b) on the retina
(c) in front of the retina
(d) on the blind spot.
Answer:
(c) in front of the retina

Question 8.
The eye defect ‘presbyopia’ can be corrected by:
(a) convex lens
(b) concave lens
(c) convex mirror
(d) Bi focal lenses
Answer:
(d) Bi focal lenses

Question 9.
Which of the following lens would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 5 cm
(b) A concave lens of focal length 5 cm
(c) A convex lens of focal length 10 cm
(d) A concave lens of focal length 10 cm
Answer:
(d) A concave lens of focal length 10 cm

Question 10.
If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?
(a) V_B = V_G = V_R
(b) V_B > V_G > V_R
(c) V_B < V_G < V_R
(d) V_B < V_G > V_R
Answer:
(c) V_B < V_G < V_R

II. Fill in the blanks.

1. The path of the light is called as ………
2. The refractive index of a transparent medium is always greater than ……….
3. If the energy of incident beam and the scattered beam are same, then the ………. scattering of light is called as scattering ……….
4. According to Rayleigh’s scattering law, the amount of scattering of light is inversely proportional to the fourth power of its ……….
5. Amount of light entering into the eye is controlled by ……….

Answer:

1. ray
2. unity
3. elastic
4. wavelength
5. iris

III. True or False. If false correct it.

1. Velocity of light is greater in denser medium than in rarer medium.
2. The power of lens depends on the focal length of the lens.
3. Increase in the converging power of eye lens cause ‘hypermetropia’
4. The convex lens always gives small virtual image.

Answer:

1. False – Velocity of light is greater in rarer medium than in denser medium.
2. True
3. True
4. False – The convex lens does not give small virtual image always.

IV. Match the following.

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

V. Assertion and reasoning type.

Mark the correct choice as-
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
1. Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small
Reason: Refractive index of the medium is inversely proportional to the velocity of the light.

2. Assertion: Myopia is due to the increase in the converging power of eye lens.
Reason: Myopia can be corrected with the help of concave lens.
Answer:
1. (a)
2. (a)

VI. Answer Briefly.

Question 1.
What is refractive index?
Answer:
Refractive index gives us an idea of how fast or how slow light travels in a medium.

Question 2.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer:

Question 4.
Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

Question 5.
State Rayleigh’s law of scattering.
Answer:
Rayleigh’s scattering law states that, “The amount of scattering of light is inversely proportional to the fourth power of its wavelength”.
Amount of scattering ‘S’ ∝\(\frac{1}{λ^4}\)

Question 6.
Differentiate convex lens and concave lens.
Answer:

Question 7.
What is the power of accommodation of the eye?
Answer:

• The ability of the eye lens to focus nearby as well as the distant objects is called the power of accommodation of the eye.
• This is achieved by changing the focal length of the eye lens with the help of ciliary muscles.

Question 8.
What are the causes of ‘Myopia’?
Answer:

1. The lengthening of eye ball.
2. The focal length of eye lens is reduced.
3. The distance between eye lens and retina increases.
4. The far point will not be at infinity.
5. The far point comes closer.

Question 9.
Why does the sky appear blue in colour?
Answer:
When sunlight passes through the atmosphere, the blue colour (shorter wavelength) is scattered to a greater extent than the red colour (longer wavelength). This scattering causes the sky to appear blue in colour.

Question 10.
Why are traffic signals red in colour?
Answer:

• Red light has the highest wavelength.
• It is scattered by atmospheric particles.
• So red light is able to travel the longest distance through a fog, rain etc.

VII. Give the answer in detail.

Question 1.
List any five properties of light?
Answer:

• Light is a form of energy.
• Light always travels along a straight line.
• Light does not need any medium for its propagation. It can even travel through a vacuum.
• The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1
• Since light is in the form of waves, it is characterized by a wavelength (λ) and a frequency (v), which are related by the following equation: c = vλ (c = velocity of light).
• Different coloured light has a different wavelength and frequency.

Question 2.
Explain the rules for obtaining images formed by a convex lens with the help of ray diagram.
Answer:
Rule-1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation.

Rule-2: When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus.

Rule-3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis.

Question 3.
Differentiate the eye defects: Myopia and Hypermetropia.
Answer:

Question 4.
Explain the construction and working of a ‘Compound Microscope’.
Answer:
Construction : A compound microscope consists of two convex lenses. The lens with the shorter focal length is placed near the object, and is called as ‘objective lens’ or ‘objective piece’. The lens with larger focal length and larger aperture placed near the observer’s eye is called as ‘eye lens’ or ‘eye piece’. Both the lenses are fixed in a narrow tube with adjustable provision.

Working : The object (AB) is placed at a distance slightly greater than the focal length of objective lens (u > F₀). A real, inverted and magnified image (A’B’) is formed at the other side of the objective lens. This image behaves as the object for the eye lens. The position of the eye lens is adjusted in such a way, that the image (B’B’) falls within the principal focus of the eye piece. This eye piece forms a virtual, enlarged and erect image (A”B”) on the same side of the object.

Compound microscope has 50 to 200 times more magnification power than simple microscope.

VIII. Numerical Problems.

Question 1.
An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.
Answer:
Distance of an object u = 20 cm
Focal length of a convex lens f = 10 cm
Let the image distance be v
We know

v = 20 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{20}{20}\) = 1
Hence a real image of same size is formed at 20 cm.
Image distance = 20 cm

Question 2.
An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.
Answer:
Object distance u = 10 cm
Focal length of a concave lens f= -15 cm
Let v be the image distance,

Distance of image v = 6 cm
Magnification m = \(\frac{v}{u}\) = \(\frac{6}{10}\) = 0.6
And Magnification m = \(\frac{h’}{h}\)
Where h’ – height of image
h – height of object
0.6 = \(\frac{h’}{3}\)
∴ h’ = 3 × 0.6 = 1.8 cm
∴ Height of image = 1.8 cm

IX. Higher order thinking (HOT) questions.

Question 1.
While doing an experiment for the determination of focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens,
(a) can he get the image?
(b) Is there any change in the focal length?
Answer:
(a) He can get the image.
(b) The focal length of the lens will be doubled.

Question 2.
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
Answer:

• The large pupil opens wider and allows the maximum amount of light to enter the eye in the dark.
• Their lens is large and situated near the retina. This also allows a lot of light to register on the retina. The retina contains 2 types of light-sensing cells rods and cones.
• Cones are responsible for the coloured vision and require bright, focused light.
• Rods are extremely sensitive to light and have a photosensitive pigment called rhodopsin which plays a vital role in night vision.

Samacheer Kalvi 10th Science Optics Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The scattering of sun light by the atoms or molecules of the gases in the Earth’s atmosphere is known as:
(a) Mie scattering
(b) Tyndall scattering
(c) Rayleigh scattering
(d) Raman scattering
Answer:
(c) Rayleigh scattering

Question 2.
Mie scattering is responsible for the _____ appearance of the clouds.
(a) red
(b) blue
(c) colourless
(d) white.
Answer:
(d) white

Question 3.
In an inelastic scattering the energy of the incident beam of light is ……….. that of scattering beam.
(a) greater than
(b) less than
(c) equal to
(d) different from
Answer:
(d) different from

Question 4.
As per Rayleigh’s scattering law, amount of scattering is:
(a) directly proportioanl to fourth power of wavelength
(b) inversely proportioanl to fourth power of wavelength
(c) inversely proportioanl to square of wavelength
(d) directly proportional to square of wavelength
Answer:
(b) inversely proportioanl to fourth power of wavelength

Question 5.
The refractive index of a medium is dependent on the _____ of the light.
(a) wavelength
(b) strength
(c) density
(d) refraction.
Answer:
(a) wavelength

Question 6.
The scattering of light by colloidal particles in the colloidal solution is called:
(a) Raman scattering
(b) Tyndall scattering
(c) Mie scattering
(d) Elastic scattering
Answer:
(b) Tyndall scattering

Question 7.
A piece of transparent material bounded by curved surfaces is called:
(a) mirror
(b) prism
(c) slab
(d) lens
Answer:
(d) lens

Question 8.
If the energy of the incident and the scattered beam of light are not the same, then it is called as _____.
(a) Elastic
(b) Raman
(c) Inelastic
(d) Mie.
Answer:
(c) Inelastic

Question 9.
A convex lens does not produce:
(a) real magnified image
(b) virtual magnified image
(c) virtual diminished image
(d) real diminished image
Answer:
(c) virtual diminished image

Question 10.
A lens which is thicker in the middle than at the edges is known as:
(a) concave lens
(b) convex lens
(c) bifocal lens
(d) cylindrical lens
Answer:
(b) convex lens

Question 11.
The object is always placed on the _____ side of the lens.
(a) left
(b) right
(c) top
(d) bottom.
Answer:
(a) left

Question 12.
The parallel rays from the outer edge are deviated towards the middle in a:
(a) convex mirror
(b) concave lens
(c) concave mirror
(d) convex lens
Answer:
(d) convex lens

Question 13.
The light rays passing through the optic centre will:
(a) diverged
(b) scattered
(c) converged
(d) emerge undeviated
Answer:
(d) emerge undeviated

Question 14.
All the distances are measured from the ______ of the lense.
(a) centre of curvature
(b) optical centre
(c) principal focus
(d) infinity.
Answer:
(b) optical centre

Question 15.
A ray passing through the principal focus and incident on the lens will:
(a) converge
(b) diverge
(c) emerge parallel to the principal axis
(d) not emerge out
Answer:
(c) emerge parallel to the principal axis

Question 16.
When the object is placed at infinity from the convex lens, the image is formed at:
(a) F
(b) C
(c) infinity
(d) between F and 2F
Answer:
(a) F

Question 17.
The human eye is ____ in nature.
(a) convex
(b) concave
(c) transparent glass
(d) Plano – concave.
Answer:
(a) convex

Question 18.
The image formed by a concave lens is:
(a) virtual
(b) diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer:
(c) virtual and diminished

Question 19.
To get a real image using convex lens, the object must be placed at:
(a) infinity
(b) principal focus
(c) beyond principal focus and infinity
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 20.
_____ is the centre part of the iris.
(a) cornea
(b) retina
(c) pupil
(d) eye lens.
Answer:
(c) pupil

Question 21.
For a convex lens the point at which the parallel rays converge is called of the lens.
(a) pole
(b) centre of curvature
(c) principal focus
(d) none
Answer:
(c) principal focus

Question 22.
A real image formed by a convex lens is always:
(a) erect
(b) magnified
(c) inverted
(d) diminished
Answer:
(c) inverted

Question 23.
The law of distances is given by:

Answer:
(b) \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)

Question 24.
The unit of focal length is:
(a) dioptre
(b) metre
(c) ohm
(d) ampere
Answer:
(b) metre

Question 25.
The sign of focal length of a convex lens is ………. sign.
(a) negative
(b) positive
(c) negative or positive
(d) none of the above
Answer:
(b) positive

Question 26.
The focal length of concave lens has ……… sign.
(a) positive
(b) negative
(c) positive or negative
(d) none of the above
Answer:
(b) negative

Question 27.
The magnification in terms of object distance u and image distance v is m :
(a) \(\frac{u}{v}\)
(b) u + v
(c) \(\frac{v}{u}\)
(d) uv
Answer:
(c) \(\frac{v}{u}\)

Question 28.
In terms of object distance u and focal length/, magnification is given by m =

Answer:
(b) \(\frac{f}{u-f}\)

Question 29.
The magnification in terms of v and/:
(a) f
(b) v – f
(c) \(\frac{f}{v-f}\)
(d) \(\frac{v-f}{f}\)
Answer:
(d) \(\frac{v-f}{f}\)

Question 30.
The unit of power is:
(a) m
(b) ohm
(C) dioptre
(d) ampere
Answer:
(C) dioptre

Question 31.
If the focal length of a convex lens is 1 m then its power is:
(a) 1 dioptre
(b) 0.1 dioptre
(c) 10 dioptre
(d) 0.01 dioptre
Answer:
(a) 1 dioptre

Question 32.
In a simple microscope, the magnification can be increased by:
(a) lens of long focal length
(b) lens
(c) lens of short focal length
(d) lens of infinite focal length
Answer:
(c) lens of short focal length

Question 33.
Convex lenses are used:
(a) as camera lenses
(b) as magnifying lenses
(c) to correct hypermetropia
(d) all the above
Answer:
(d) all the above

Question 34.
Which lens is used in wide angle spyhole in doors?
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) parabolic lens
Answer:
(b) concave lens

Question 35.
The mathematical form of lens maker’s formula is:

Answer:
(a)

Question 36.
If f is the focal length of the lens then its power is given by:
(a) P = \(\frac{2}{f}\)
(b) p = \(\frac{1}{f}\)
(c) p = f
(d) p = f₁
Answer:
(b) p = \(\frac{1}{f}\)

Question 37.
Which part of the human eye changes the focal length of the eye lens?
(a) pupil
(b) retina
(c) ciliary muscles
(d) cornea
Answer:
(c) ciliary muscles

Question 38.
On which part of human eye, image is formed?
(a) cornea
(b) iris
(c) retina
(d) pupil
Answer:
(c) retina

Question 39.
For normal human eye the value of near point is:
(a) 25 cm
(b) 25 m
(c) 2.5 m
(d) 25 mm
Answer:
(a) 25 cm

Question 40.
In hypermeteropia, the focal length of the eye lens is:
(a) decreased
(b) remains the same
(c) increased
(d) none of the above
Answer:
(c) increased

Question 41.
Presbyopia can be corrected by using:
(a) convex lens
(b) bifocal lens
(c) concave lens
(d) cylindrical lens
Answer:
(b) bifocal lens

Question 42.
Astigmatism can be corrected by using:
(a) bifocal lens
(b) cylindrical lens
(c) convex lens
(d) concave lens
Answer:
(b) cylindrical lens

Question 43.
The magnifying power of compound microscope is:
(a) 10
(b) 20
(c) 50
(d) 50 to 200
Answer:
(d) 50 to 200

Question 44.
The accuracy of travelling microscope is of the order of:
(a) 0.01 cm
(b) 0.01 mm
(c) 0.1 mm
(d) 0.1 cm
Answer:
(b) 0.01 mm

II. Fill in the blanks.

1. The velocity of light in vacuum is ……….
2. If v is the frequency and λ is the wavelength then velocity of the wave is c = ……….
3. Among colours of visible light ……… colour has the highest wavelength.
4. According to Snell’s law refractive index, µ₂ = ……….
5. In a medium having high value of refractive index then speed of light in that medium is ……….
6. Angle of refraction is the smallest for ……… and the highest for ……….
7. The refractive index depends on ………. of light.
8. Colours having shorter wavelength scattered more than longer wavelength colours according to ……….. law.
9. After passing through a convex lens ……….. rays ………. at the principal focus.
10. For a convex lens, as the object distance increases, the image distance ……….
11. A ray passing through the optic centre of a lens emerges …………
12. ……… is due to irregular curvature of the surface of the eye lens.
13. When a parallel beam of light passes through a convex lens, the rays from the outer edges are …………
14. A ray parallel to the principal axis of a convex lens after refraction passes through …………
15. When the object is placed between ………….. and ………… of a convex lens a virtual image will be formed.
16. For a convex lens, as the object approaches the lens the image becomes …………
17. In a phographic camera ………… lens is used.
18. The shorter the focal length, the ………. is the magnification.
19. The nature of the image formed by a simple microscope is ……….., ………… and …………
20. Real images are formed by a ……….. lens.
21. Concave lens produces ………… images.
22. The value of power of a lens having focal length one metre is ………..
23. For a normal eye the value of far point is …………
24. …………. is known as short sightedness.
25. Hyper metropia is known as ………..
26. The mathematical form of focal length of a concave lens used to correct myopia is f = ……….
27. ……….. lenses are used to correct astigmatism.
28. For a normal eye, the value of least distance of distinct vision is ………..
29. The objective of the compound microscope has …………. focal length.
30. The focal length of ………. is greater in a compound microscope.
31. ……… is an optical instrument to see the distant objects.
32. A terrestrial telescope produces ……… image.
33. Elaborate view of galaxies and planets is obtained by ………
Answer:
1. 3 × 10⁸ m/s
2. vλ
3. red
4. \(\frac{sin i}{sin r}\)
5. low
6. red, violet
7. wavelength
8. Rayleigh scattering
9. parallel, converge
10. will decrease
11. undeviated
12. Astigmatism
13. deviated towards the centre of the lens
14. the principal focus
15. principal focus, optical centre
16. bigger
17. biconvex
18. greater
19. virtual, erect, magnified
20. convex
21. virtual
22. One dioptre
23. infinity
24. Myopia
25. long sightedness
26. xy/x – y
27. Cylindrical
28. 25 cm
29. shorter
30. eye piece
31. Telescope
32. an erect
33. Telescope

III. True or False. If false correct it.

1. Light does not travel along a straight line.
2. All coloured light has same wavelength.
3. In refraction incident ray, refracted ray and normal lie in the same plane.
4. Velocity of light is greater in rarer medium is greater than that in denser medium.
5. For red colour angle of refraction is the least.
6. The refractive index of a medium ¡s independent of wavelength.
7. Tyndall scattering, is the scattering of light by colloids.
8. According to Rayleigh’s scattering law, red colour is scattered to a greater extent than blue colour.
9. Mie scattering takes place when the diameter is larger than the wavelength of the incident light.
10. The lines in Raman scattering having frequencies lower than the incident frequency are called Antistoke’s lines.
11. In front of a convex lens when the object is placed at infinity the formed image is smaller than that of the object.
12. When an object is placed at finite distance from the concave lens a virtual image is formed between optical centre and focus of the concave lens.
13. Pupil of human eye bends the incident light on to the lens.
14. For a normal human eye, the value of far point is 25 cm.
15. Astigmatism is corrected by cylindrical lenses.
Answer:
1. False – Light always travels along a straight lines.
2. False – Different coloured light has different wavelength.
3. True
4. True
5. True
6. False – The refractive index of a medium depends on wavelength.
7. True
8. False – According to Rayleigh’s scattering law, blue colour is scattered to a greater extent than red colour.
9. True
10. False – The lines in Raman scattering having frequencies higher than the incident frequency are called Antistoke’s lines.
11. True
12. True
13. False – Cornea of human eye bends the incident light on to the lens.
14. False – For a normal human eye, the value of near point is 25 cm.
15. True

IV. Match the following.

Question 1.
Match the column I with column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 2.
Position of the object placed infront of a convex lens are given in Column
I. Match them with the natures of the images formed by the convex lens given in column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 3.
Match the following:

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)

Question 4.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

Question 5.
Match the Column I with Column II.

Answer:
A. (iv)
B. (i)
C. (ii)
D. (iii)

Question 6.
Match the column I with column II.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)

Question 7.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (ii)
D. (i)

V. Assertion and Reasoning type.

Question 1.
Assertion : The sun looks bigger in size at sunrise and sunset than during day.
Reason : In detraction light rays bend around the edges of the obstacle.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 2.
Assertion: Colours can be scan in thin layers of oil on the surface water. Reason: White light is composed of several colours.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 3.
Assertion: Raman spectrum of a liquids contains lines whose frequencies are not equal to that of incident radiation.
Reason: If a photon strikes an atom in a liquid that is in existed state photon losses energy.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 4.
Assertion: The refractive index of a prism depends only on the material of the prism.
Reason: The refractive index of a prism depends upon the refracting angle and angel of minimum deviation.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 5.
Assertion: A single lens produces a coloured image of an object illuminated by white light.
Reason: The refractive index of material of lens is different for different wavelength of light.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 6.
Assertion: If a convex lens is placed in water, its convergence power decrease.
Reason: Focal length of lens is independent of refractive index of the medium.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(c) If Assertion is true but Reason is false.

Question 7.
Assertion: Light waves travel in straight lines.
Reason: Rectilinear propagation of light confirm the above mentioned properly.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 8.
Assertion: Raman scattering the scattering of monochromatic light by atoms and molecule of a liquid.
Reason: The wavelength of Raman lines is same.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Question 9.
Assertion: Power of a lens is the reciprocal of its focal length.
Reason: The unit of power is one dioptre when the unit of focal length is one metre.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 10.
Assertion: Presbyopia is due to ageing of human beings.
Reason: For those persons, ciliary muscles of the eye become weak.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) If Assertion is true but Reason is false.
(d) If Assertion is false but Reason is true.
Answer:
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

VI. Answer Briefly.

Question 1.
What is meant by refraction?
Answer:
When a ray of light travels from one transparent medium into another obliquely, the path of the light undergoes deviation. This deviation of ray of light is called refraction.

Question 2.
State laws of refraction.
Answer:
First law of refraction: The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.

Second law of refraction: The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{sin i}{sin r}\) = \(\frac{µ_2}{µ_1}\)

Question 3.
Define refractive index of a medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in a medium is defined as refractive index ‘p’ of that medium.

Question 4.
What is meant by monochromatic source?
Answer:
If a source of light produces a light of single colour, it is known as a monochromatic source.

Question 5.
When white light is refracted by a transparent medium what will you get? Why?
Answer:

1. When white light is refracted by a transparent medium, a spectrum is obtained.
2. This is because, different coloured lights are bent through different angles.

Question 6.
What is scattering of light?
Answer:
When sunlight enters the Earth’s atmosphere, the atoms and molecules of different gases present in the atmosphere refract the light in all possible directions. This is called as ‘Scattering of light’.

Question 7.
State the types of scattering.
Answer:

1. Elastic scattering
2. Inelastic scattering

Question 8.
What is elastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are same, then it is called as ‘elastic scattering’.

Question 9.
What is inelastic scattering?
Answer:
If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as ‘inelastic scattering’.

Question 10.
How are different types of scattering formed? Mention the types of scattering.
Answer:
The nature and size of the scatterer results in different types of scattering. They are

1. Rayleigh scattering
2. Mie scattering
3. Tyndall scattering
4. Raman scattering

Question 11.
What is Rayleigh scattering?
Answer:
The scattering of sunlight by the atoms or molecules of the gases in the earth’s atmosphere is known as Rayleigh scattering.

Question 12.
Why the colour of the Sun is red at sunrise and sunset?
Answer:
At sunrise and sunset, the light rays from the Sun have to travel a larger distance in the atmosphere than at noon. Hence, most of the blue lights are scattered away and only the red light which gets least scattered reaches us. Therefore, the colour of the Sun is red at sunrise and sunset.

Question 13.
When does Mie scattering take place?
Answer:
Mie scattering takes place when the diameter of the scatterer is similar to or larger than the wavelength of the incident light.

Question 14.
What are the causes of Mie scattering?
Answer:
Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.

Question 15.
Why the clouds have white appearance?
Answer:
Mie scattering is responsible for the white appearance of the clouds. When white light falls on the water drop, all the colours are equally scattered which together form the white light.

Question 16.
What is Tyndall Scattering?
Answer:
The scattering of light rays by the colloidal particles in the colloidal solution is called Tyndall Scattering or Tyndall Effect.

Question 17.
What is meant by colloid? State few examples.
Answer:
Colloid is a microscopically small substance that is equally dispersed throughout another material. Eg: Milk, Ice cream, muddy water, smoke.

Question 18.
What is meant by Raman Scattering?
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

Question 19.
Define Raman Scattering.
Answer:
Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

Question 20.
What is Rayleigh line?
Answer:
The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’.

Question 21.
What are Raman lines?
Answer:
The spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’.

Question 22.
What are stokes lines and Antistokes lines?
Answer:
The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 23.
What is a lens?
Answer:
A lens is an optically transparent medium bounded by two spherical refracting surfaces or one plane and one spherical surface.

Question 24.
How is lens classified?
Answer:
Lens is basically classified into two types. They are:

1. Convex Lens
2. Concave Lens.

Question 25.
What is biconvex lens?
Answer:
Convex or bi-convex lens: It is a lens bounded by two spherical surfaces such that it is thicker at the centre than at the edges. A beam of light passing through it, is converged to a point. So, a convex lens is also called as converging lens.

Question 26.
What is meant by biconcave lens?
Answer:
Concave or bi-concave Lens: It is a lens bounded by two spherical surfaces such that it is thinner at the centre than at the edges. A parallel beam of light passing through it, is diverged or spread out. So, a concave lens is also called as diverging lens.

Question 27.
What are
(i) Plano-convex lens?
(ii) Plano-concave lens?
Answer:
(i) If one of the faces of a bi-convex lens is plane, it is known as a plano-convex lens.
(ii) If one of the faces of a bi-concave lens is plane, it is known as a plano-concave lens.

Question 28.
State the applications of convex lenses.
Answer:

1. Convex lenses are used as camera lenses.
2. They are used as magnifying lenses.
3. They are used in making microscope, telescope and slide projectors.
4. They are used to correct the defect of vision called hypermetropia.

Question 29.
Draw diagrams of different types converging lenses.
Answer:

Question 30.
Represent different types of lenses by diagram.
Answer:

Question 31.
What is the nature of the image formed by an object is placed behind the centre of curvature of a convex leas. Draw a ray diagram.
Answer:
When an object is placed behind the center of curvature (beyond C), a real and inverted image is formed between the center of curvature and the principal focus. The size of the image is the same as that of the object.

Question 32.
Draw a ray diagram to indicate the nature of the image formed when an object is placed in between the centre of curvature and principal focus of a convex lens.
Answer:
When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

Question 33.
Draw a ray diagram for the formation of image (by the concave lens) when object is at infinity.
Answer:
When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Question 34.
What are the applications of concave lens?
Answer:

1. Concave lenses are used as eye lens of‘Galilean Telescope’.
2. They are used in wide angle spy hole in doors.
3. They are used to correct the defect of vision called ‘myopia’.

Question 35.
What do you know about lens formula?
Answer:
The lens formula gives the relationship among distance of the object (u), distance of the image (v) and the focal length (f) of the lens. It is expressed as
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)

Question 36.
Define magnification.
Answer:
It is defined as the ratio of the height of the image to the height of an object. Magnification is denoted by the letter ‘m’. If height of the object is h and height of the image is h’, the magnification produced by lens is,

Question 37.
What is lens formula?
Answer:
The lens formula relates the focal length of a lens with the distance of object and image.

where µ is the refractive index of the material of the lens; R₁ and R₂ are the radii of curvature of the two faces of the lens; f is the focal length of the lens.

Question 38.
What is meant by power of lens?
Answer:
The ability of a lens to converge (convex lens) or diverge (concave lens) light rays is called as its power.

Question 39.
Define power of a lens. State its unit.
Answer:
Power of a lens is numerically defined as the reciprocal of its focal length.
P = \(\frac{1}{f}\)
The SI unit of power of a lens is dioptre.

Question 40.
What is meant by dioptre.
Answer:
Dioptre is the power of a lens, whose focal length is 1 metre.
1 Dioptre = 1 m^-1.

Question 41.
Differentiate convex lens from concave lens.
Answer:

Question 42.
What are
(i) Pupil &
(ii) Retina?
Answer:
(i) Pupil: It is the centre part of the Iris. It is the pathway for the light to retina.
(ii) Retina: This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Question 43.
What is persistence of vision?
Answer:
If the time interval between two consecutive light pulses is less than 0.1 second, human eye cannot distinguish them separately. It is called persistence of vision.

Question 44.
What is least distance of distinct vision?
Answer:
The minimum distance required to see the objects distinctly without strain is called least distance of distinct vision. It is called as near point of eye. It is 25 cm for normal human eye.

Question 45.
What is far point?
Answer:
The maximum distance up to which the eye can see objects clearly is called as far point of the eye.

Question 46.
What is Presbyopia?
Answer:
Due to ageing, ciliary muscles become weak and the eye-lens become rigid (inflexible) and so the eye loses its power of accommodation. Because of this, an aged person cannot see the nearby objects clearly. So, it is also called as ‘old age hypermetropia’.

Question 47.
What is meant by astigmatism?
Answer:
In this defect, eye cannot see parallel and horizontal lines clearly. It may be inherited or acquired. It is due to the imperfect structure of eye lens because of the development of cataract on the lens, ulceration of cornea, injury to the . refracting surfaces, etc. Astigmatism can be corrected by using cylindrical lenses (Torrid lenses).

Question 48.
State the principle of microscope. How is it classified?
Answer:
It works under the principle of angular magnification of lenses. It is classified as

1. Simple microscope
2. Compound microscope

Question 49.
What are the uses of simple microscope?
Answer:
Simple microscopes are used

1. By watch repairers and jewellers.
2. To read small letters clearly.
3. To observe parts of flower, insects etc.
4. To observe finger prints in the field of forensic science.

Question 50.
How is telescope classified?
Answer:
According to optical property, it is classified into two groups:

1. refracting telescope
2. reflecting telescope

Question 51.
Mention the advantages of telescope.
Answer:

1. Elaborate view of the Galaxies, Planets, stars and other heavenly bodies is possible.
2. Camera can be attached for taking photograph for the celestial objects.
3. Telescope can be viewed even with the low intensity of light.

Question 52.
What are the disadvantages of telescope?
Answer:

1. Frequent maintenance is needed.
2. It is not easily portable.

VII. Give the answer in Detail.

Question 1.
State the Laws of Refraction.
Answer:
The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.
Second law of Refraction:

1. The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
   \(\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}\)
2. Refractive index gives us an idea of how fast or how slow light travels in a medium. The ratio of the speed of light in a vacuum to the speed of light in a medium is defined as the refractive index ‘µ’ of that medium.
3. The speed of light in a medium is low if the refractive index of the medium is high and vice versa.
4. When light travels from a denser medium into a rarer medium, the refracted ray is bent away from the normal drawn to the interface.
5. When light travels from a rarer medium into a denser medium, the refracted ray is bent towards the normal drawn to the interface.

Question 2.
Describe Raman Scattering.
Answer:
When a parallel beam of monochromatic (single coloured) light passes through a gas or liquid or transparent solid, a part of light rays are scattered.

The scattered light contains some additional frequencies (or wavelengths) other than that of incident frequency (or wavelength). This is known as Raman scattering or Raman Effect.

Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

The spectral lines having frequency equal to the incident ray frequency is called ‘Rayleigh line’ and the spectral lines which are having frequencies other than the incident ray frequency are called ‘Raman lines’. The lines having frequencies lower than the incident frequency is called stokes lines and the lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 3.
With the help of ray diagram, explain the nature, size and position of the image formed by a convex lens. When object is placed at
(i) infinity
(ii) beyond C
(iii) placed at C
(iv) Placed between F and C,
(v) placed at F
(vi) placed between F and optical centre O.
Answer:
(i) Object at infinity: When an object is placed at infinity, a real image is formed at the principal focus. The size of the image is much smaller than that of the object.

(ii) Object placed beyond C (>2F): When an object is placed behind the center of curvature(beyond C), a real and inverted image is formed between the center of curvature and the principal focus. Th e size of the image is the same as that of the object.

(iii) Object placed at C: When an object is placed at the center of curvature, a real and inverted image is formed at the other center of curvature. The size of the image is the same as that of the object.

(iv) Object placed between F and C: When an object is placed in between the center of curvature and principal focus, a real and inverted image is formed behind the center of curvature. The size of the image is bigger than that of the object.

(v) Object placed at the principal focus F: When an object is placed at the focus, a real image is formed at infinity. The size of the image is much larger than that of the object.

(vi) Object placed between the principal focus F and optical centre O: When an object is placed in between principal focus and optical centre, a virtual image is formed. The size of the image is larger than that of the object.

Question 4.
Explain the formation of images formed by a concave lens.
Answer:
Object at Infinity: When an object is placed at infinity, a virtual image is formed at the focus. The size of the image is much smaller than that of the object.

Object anywhere on the principal axis at a finite distance: When an object is placed at a finite distance from the lens, a virtual image is formed between optical center and focus of the concave lens. The size of the image is smaller than that of the object.

But, as the distance between the object and the lens is decreased, the distance between the image and the lens also keeps decreasing. Further, the size of the image formed increases as the distance between the object and the lens is decreased.

Question 5.
Explain Mie Scattering.
Answer:
Mie scattering:

1. Mie scattering takes place when the diameter of the Scatterer is similar to or larger than the wavelength of the incident light. It is also an elastic scattering.
2. The amount of scattering is independent of wavelength.
3. Mie scattering is caused by pollen, dust, smoke, water droplets, and other particles in the lower portion of the atmosphere.
4. Mie scattering is responsible for the white appearance of the clouds.
5. When white light falls on the water drop, all the colours are equally scattered which, together form the white light.

Question 6.
With the help of a diagram, explain the structure and working of human eye.
Answer:
Structure of the eye:
The eye ball is approximately spherical in shape with a diameter of about 2.3 cm. It consists of a tough membrane called sclera, which protects the internal parts of the eye.

Cornea : This is the thin and transparent layer on the front surface of the eyeball as shown in figure. It is the main refracting surface. When light enters through the cornea, it refracts or bends the light on to the lens.

Iris : It is the coloured part of the eye. It may be blue, brown or green in colour. Every person has a unique colour, pattern and texture. Iris controls amount of light entering into the pupil like camera aperture.

Pupil : It is the centre part of the Iris. It is the pathway for the light to retina.

Retina : This is the back surface of the eye. It is the most sensitive part of human eye, on which real and inverted image of objects is formed.

Ciliary muscles : Eye lens is fixed between the ciliary muscles. It helps to change the focal length of the eye lens according to the position of the object.

Eye Lens : It is the important part of human eye. It is convex in nature.

Working of the eye : The transparent layer cornea bends the light rays through pupil located at the centre part of the Iris. The adjusted light passes through the eye lens. Eye lens is convex in nature. So, the light rays from the objects are converged and a real and inverted image is formed on retina. Then, retina passes the received real and inverted image to the brain through optical nerves. Finally, the brain senses it as erect image.

Question 7.
Describe simple microscope.
Answer:
Simple microscope: It has a convex lens of short focal length. It is held near the eye to get enlarged image of small objects.
Let an object (AB) is placed at a point within the principal focus (u < f) of the convex lens and the observer’s eye is placed just behind the lens. As per this position the convex lens produces an erect, virtual and enlarged image (A’B’), The image formed is in the same side of the object and the distance equal to the least distance of distinct vision (D) (For normal human eye D = 25 cm).

Question 8.
Write short notes on
(i) Astronomical telescope
(ii) Terrestrial telescope.
Answer:
(i) Astronomical Telescope: An astronomical telescope is used to view heavenly bodies like stars, planets, galaxies and satellites.

(ii) Terrestrial Telescope: The image in an astronomical telescope is inverted. So, it is not suitable for viewing objects on the surface of the Earth. Therefore, a terrestrial telescope is used. It provides an erect image. The major difference between astronomical and terrestrial telescope is erecting the final image with respect to the object.

VIII. Numerical Problems.

Question 1.
A needle of size 5 cm is placed 45 cm from a lens produced an image on a screen placed 90 cm away from the lens.
Answer:
(i) Identify the types of lens.
Calculate focal length of the lens.
Height of the object h₁ = 5 cm
Distance of the object u = -45 cm
Distance of the image v = 90 cm
We know that

Focal length of the lens = 30 cm
Since focal length is positive the lens is convex lens.

(ii) Identify the size of the image

∴ h₂ = -10 cm
The negative sign indicates that the image is real and inverted.

Question 2.
A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image 10 cm from the lens?
Answer:
v = -10 cm; f =-15 cm; u = ?
Lens formula:
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\)
Type of lens: Concave lens

u =-30 cm
Thus, the object distance is 30 cm.

Question 3.
An object is placed at a distance of 30 cm from a concave lens of focal length 15 cm. An erect and virtual image is formed at a distance of 10 cm from the lens. Calculate the magnification.
Answer:
Type of lens is Cancave lens.
Formula:
Magnification m = \(\frac{v}{u}\)
Object distance u = -30 cm
Image distance v = -10 cm
m = \(\frac{-10}{-30}\) = \(\frac{1}{3}\) = +0.33

Question 4.
The focal length of a concave lens is 2 cm. Calculate the power of the lens.
Answer:
Formula:
P = \(\frac{1}{f}\)
Type of lens is concave lens.
Focal length of concave lens,
f = -2 m power of the lens.
P = \(\frac{1}{-2 m}\)
P = -0.5 dioptre

Question 5.
A needle placed at 30 cm from the lens forms an image on a screen placed 60 cm on the other side of the lens. Identify the type of lens and determine the focal length.
Answer:
u = -30 cm
v = 60 cm
u is negative because image is formed on the on the other side of the lens.

It is a convex lens.

Question 6.
A 3 cm tall bulb is placed at a distance of 20 cm from a diverging lens having a focal length of 10.5 cm. Determine the distance of the image.
Answer:
u = -20 cm
f = -10.5 cm

The distance of the image is -6.88 cm

Question 7.
A ray from medium 1 is refracted below while passing through medium 2. Find the refractive index of the second medium with respect to medium 1.
Answer:
Refractive index µ

Refractive index = 0.707

Question 8.
The optical prescription of a pair of spectacle is
Right eye: -3.5 D, Left eye: -4.00 D.
(i) Name the defect of the eye.
Answer:
Shortsighted (Myopia)

(ii) Are these lenses thinner at the middle or at the edges?
Answer:
These lenses are thinner in the middle.

(iii) Which lens has a greater focal length?
Answer:
power = \(\frac{1}{focal length}\)
Right eye: power P = -3.5 D

Left eye: Power P = -4 D

Hence the lens having power of -3.5 D has greater focal length.

Question 9.
The radii of curvature of two surfaces of a double convex lens are 10 cm each. Calculate its focal length and power of the lens in air and liquid. Refractive indices of glass and liquid are 1.5 and 1.8 respectively.
Answer:
Radius of curvature of first surface R₁ = 10 cm
Radius of curvature of second surface R₂ = 10 cm
In air

p_l = -3.33 d

Question 10.
An object 2 cm tall is placed 10 cm in front of a convex lens of focal length 15 cm. Find the position, size and nature of the image formed.
Answer:
Focal length of a convex lens f = 15 × 10^-2 m
Weight of the object ho = 2 × 10^-2 m
Let weight of the image be h_v
Distance of the object u = 10 × 10^-2 m
Distance of the image v = 15 × 10^-2 m
We know

∴ v = -30 × 10^-2 m
Distance of the image = 30 × 10^-2 m

Hence a virtual image 6 × 10^-2 m height is formed at a distance of 30 × 10^-2 m from the lens on the same side of the lens.

IX. Higher order thinking (HOT) questions.

Question 1.
Ramu passes white light through a quartz prism. For which colour refractive index is greater?
Answer:
Refractive index is maximum for violet light when white light passes through a quartz prism.

Question 2.
Sita has kept a stud consists of diamond. What will she observe? Give reason.
Answer:
The diamond stud appears bright because of total internal reflection.

Question 3.
Guna passes a ray light through a glass slab. Which optical phenomenon will take place? What can he observe with reference to wavelength?
Answer:
When a ray of light enters a glass slab he can observe refraction of light. He observed that wavelength of light decreases.

Question 4.
A prism is placed in the minimum deviation position. Chari has passed a ray of light at an angle of 45°, then what is the value of angle of emergence? Why?
Answer:
The angle of emergence = 45°.
Since, in the minimum deviation positron, angle of incidence is equal to angle of emergence.

Question 5.
Mani is using a lens of power 2 dioptre. What is the focal length of the lens?
Answer:
Focal length = \(\frac{1}{power}\)
F= \(\frac{1}{2}\) = 0.5 m

Question 6.
Surya has placed a lens of power 1 D in side water. What will happen to power of the lens?
Answer:
The power of the lens will be more than original power.

Question 7.
Sonu has observed some lines in solar spectrum are absorbed by the elements present in the atmosphere. What are the lines?
Answer:
The lines are Fraunhofer lines.

Students can download Maths Chapter 6 Trigonometry Ex 6.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Multiple Choice Questions

Question 1.
The value of sin2 θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to ………………
(1) tan² θ
(2) 1
(3) cot² θ
(4) 0
Answer:
(2) 1
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1

Question 2.
tan θ cosec² θ – tan θ is equal to ………………
(1) sec θ
(2) cot² θ
(3) sin θ
(4) cot θ
Answer:
(4) cot θ
Hint:
tan θ cosec² θ – tan θ = tan θ (cosec² θ – 1)
= tan θ × cot² θ = \(\frac{1}{\cot \theta}\) × cot² θ = cot θ

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
(sin α + cos α)² + (cos α + sec α)²
= sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α
= 1 + cosec² α + 2 + sec² α + 2
= 1 + cot² α + 1 + 2 + tan² α + 1 + 2
= 7 + tan² α + cot² α
k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b (a2 – 1) is equal to ……………
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a
Hint:
b (a² – 1) = (sec θ + cosec θ) [(sin θ + cos θ)² – 1]
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\) [sin² θ + cos² θ + 2 sin θ cos θ – 1]

Question 5.
If 5x = sec θ and \(\frac { 5 }{ x } \) = tan θ, then x2 – \(\frac{1}{x^{2}}\) is equal to …………….
(1) 25
(2) \(\frac { 1 }{ 25 } \)
(3) 5
(4) 1
Answer:
(2) \(\frac { 1 }{ 25 } \)
Hint:



Question 6.
If sin θ = cos θ , then 2 tan² θ + sin² θ – 1 is equal to ………………
(1) \(\frac { -3 }{ 2 } \)
(2) \(\frac { 3 }{ 2 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { -2 }{ 3 } \)
Answer:
(2) \(\frac { 3 }{ 2 } \)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then …………..
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:
x = a tan θ
\(\frac { x }{ a } \) = tan θ
\(\frac{x^{2}}{a^{2}}\) = tan² θ
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = sec² θ – tan² θ = 1
y = b sec θ
\(\frac{y}{b}\) = sec θ
\(\frac{y^{2}}{b^{2}}\) = sec² θ

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to ……………
(1) 0
(2) 1
(3) 2
(4) -1
Answer:
(3) 2
Hint:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b-a
Solution:
(2) b² – a²
(a cot θ + b cosec θ)² = p²
(b cot θ + a cosec θ )² = q²
p² – q² = a² cost²θ + a² cot²θ + 2ab cot θ cosec θ – (b²cot²θ + a² cosec²θ + 2ab cot θ cosec θ) = (a² – b²) cot²θ + (b² – a²)cosec²θ = (a² – b²) (cosec²θ – 1) + (b² – a²) (cosec²θ)
= (a² – b²)cosec²θ – (a² – b²) – (a² – b²) cosec²θ
= b² – a²

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt { 3 }\) : 1, then the angle of elevation of the sun has a measure
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°
Hint:
Ratio of length of the tower : length of the shadow = \(\sqrt { 3 }\) : 1

Let the tower be \(\sqrt { 3 }\) x and the shadow be x
tan C = \(\frac { AB }{ BC } \) ⇒ tan C = \(\frac{\sqrt{3} x}{x}\) = \(\sqrt { 3 }\)
tan C = tan 60° ⇒ ∴ ∠C = 60°

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘6’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to ……………
(1) \(\sqrt { 3 }\) b
(2) \(\frac { b }{ 3 } \)
(3) \(\frac { b }{ 2 } \)
(4) \(\frac{b}{\sqrt{3}}\)
Answer:
(3) \(\frac { b }{ 2 } \)
Hint:
Let the height of the pole BC be h
AC = b + h
Let CD be x
In the right ∆ BCD, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ x } \)
x = \(\sqrt { 3 }\) h ………. (1)

In the right ∆ ACD, tan 60° = \(\frac { AC }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { b+h }{ x } \)
x = \(\frac{b+h}{\sqrt{3}}\) ………(2)
From (1) and (2) we get
\(\sqrt { 3 }\) h = \(\frac{b+h}{\sqrt{3}}\) ⇒ 3 h = b + h
2 h = b ⇒ h = \(\frac { b }{ 2 } \)

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to
(1) 41. 92 m
(2) 43. 92 m
(3) 43 m
(4) 45. 6 m
Answer:
(2) 43. 92 m
Hint:
In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \) = \(\frac { 60 }{ x+y } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 60 }{ x+y } \) ⇒ x + y = 60 \(\sqrt { 3 }\)

y = 60 \(\sqrt { 3 }\) – x …….(1)
In the right ∆ ABD, tan 45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 60 }{ y } \) ⇒ y = 60 ………..(2)
From (1) and (2) we get
60 = 60 \(\sqrt { 3 }\) – x
x = 60 \(\sqrt { 3 }\) – 60 = 60 (\(\sqrt { 3 }\) – 1) = 60 (1.732 – 1)
= 60 × 0.732
x = 43.92 m

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is …………….
(1) 20,10\(\sqrt { 3 }\)
(2) 30, 5 \(\sqrt { 3 }\)
(3) 20, 10
(4) 30, 10\(\sqrt { 3 }\)
Answer:
(4) 30, 10\(\sqrt { 3 }\)
Hint:
Let the height of the multistoried building AB be “h”
AE = h – 20
Let BC be x
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \) ⇒ \(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ………..(1)

In the right ∆ ABC, tan 30° = \(\frac { AE }{ ED } \) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
x = (h – 20) \(\sqrt { 3 }\) ………(2)
From (1) and (2) we get,
\(\frac{h}{\sqrt{3}}\) = (h – 20) \(\sqrt { 3 }\)
h = 3h – 60 ⇒ 60 = 2 h
h = \(\frac { 60 }{ 2 } \) = 30
Distance between the building (x) = \(\frac{h}{\sqrt{3}}=\frac{30}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}=10 \sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is ……………….
(1) \(\sqrt { 2 }\)x
(2) \(\frac{x}{2 \sqrt{2}}\)
(3) \(\frac{x}{\sqrt{2}}\)
(4) 2 x
Answer:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:
Consider the height of the 2^nd person ED be “h”
Height of the second person is 2 h
C is the mid point of BD
In the right ∆ ABC, tan θ = \(\frac { AB }{ BC } \)

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β . The angle of depression of its reflection in the lake is 45° . The height of the location of the cloud from the lake is ………….

Students can download Maths Chapter 6 Trigonometry Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 6 Trigonometry Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile

Students can download Maths Chapter 5 Coordinate Geometry Unit Exercise 5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 5 Coordinate Geometry Unit Exercise 5

Question 1.
PQRS is a rectangle formed by joining the points P(- 1, – 1), Q(- 1, 4) , R(5, 4) and S (5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Answer:

Mid point of a line = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
Mid point of PQ (A) = (\(\frac { -1-1 }{ 2 } \),\(\frac { -1+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (-1,\(\frac { 3 }{ 2 } \))
Mid point of QR (B) = (\(\frac { -1+5 }{ 2 } \),\(\frac { 4+4 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { 8 }{ 2 } \)) = (2,4)
Mid point of RS (C) = (\(\frac { 5+5 }{ 2 } \),\(\frac { 4-1 }{ 2 } \)) = (\(\frac { 10 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (5,\(\frac { 3 }{ 2 } \))
Mid point of PS (D) = (\(\frac { 5-1 }{ 2 } \),\(\frac { -1-1 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (2,-1)


img 355
AB = BC = CD = AD = \(\sqrt{\frac{61}{4}}\)
Since all the four sides are equal,
∴ ABCD is a rhombus.

Question 2.
The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.
Answer:
Let the vertices A(2,1), B(3, – 2) and C(x, y)
Area of a triangle = 5 sq. unit

\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 5
\(\frac { 1 }{ 2 } \) [-4 + 3y + x – (3 – 2x + 2y)] = 5
-4 + 3y + x – 3 + 2x – 2y = 10
3x + y – 7 = 10
3x + y = 17 ……(1)
Given y = x + 3
Substitute the value ofy = x + 3 in (1)
3x + x + 3 = 17
4x = 17 – 3
4x = 14
x = \(\frac { 14 }{ 4 } \) = \(\frac { 7 }{ 2 } \)
Substitute the value of x in y = x + 3
y = \(\frac { 7 }{ 2 } \) + 3 ⇒ y = \(\frac { 7+6 }{ 2 } \) = \(\frac { 13 }{ 2 } \)
∴ The coordinates of the third vertex is (\(\frac { 7 }{ 2 } \),\(\frac { 13 }{ 2 } \))

Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Answer:
3x + y = 2 ……..(1)
5x + 2y = 3 ………(2)
2x – y = 3 ……….(3)
Solve (1) and (2) to get the vertices B


Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = – 1
The point B is (1, – 1)
Solve (2) and (3) to get the vertices C

Substitute the value of x = 1 in (3)
2(1) – y = 3 ⇒ -y = 3 – 2
– y = 1 ⇒ y = – 1
The point C is (1, – 1)
Solve (1) and (3) to get the vertices A

Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = -1
The point A is (1, – 1)
The points A (1, – 1), B (1, -1), C(1, -1)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of the triangle = 0 sq. units.
Note: All the three vertices are equal, all the point lies in a same points.

Question 4.
If vertices of a quadrilateral are at A(- 5, 7), B(- 4, k), C(- 1, – 6) and D(4, 5) and its area is 72 sq.units. Find the value of k.
Answer:

Area of the quadrilateral ABCD = 72 sq. units.
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 72

-5k + 24 – 5 + 28 – (- 28 – K – 24 – 25) = 144
– 5k + 47 – k – 77 = 144
– 5k + 47 + k + 77 = 144
– 4k + 124 = 144
-4k = 144 – 124
– 4k = 20
k = -5
The value of k = – 5

Question 5.
Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
Answer:
The vertices A(-2, -1), B(4, 0), C(3, 3) and D(- 3, 2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 0+1 }{ 4+2 } \) = \(\frac { 1 }{ 6 } \)
Slope of BC = \(\frac { 3-0 }{ 3-4 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of CD = \(\frac { 2-3 }{ -3-3 } \) = \(\frac { -1 }{ -6 } \) = \(\frac { 1 }{ 6 } \)
Slope of AD = \(\frac { 2+1 }{ -3+2 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of AB = Slope of CD = \(\frac { 1 }{ 6 } \)
∴ AB || CD ……(1)
Slope of BC = Slope of AD = -3
∴ BC || AD …..(2)
From (1) and (2) we get ABCD is a parallelogram.

Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Answer:
Let the “x” intercept be “a”
y intercept = 1 – a (sum of the intercept is 1)
Product of the intercept = – 6
a (1 – a) = – 6 ⇒ a – a² = – 6
– a² + a + 6 = 0 ⇒ a² – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a – 3 = 0 (or) a + 2 = 0
a = 3 (or) a = -2
When a = 3
x – intercept = 3
y – intercept = 1 – 3 = – 2
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 3 } \) + \(\frac { y }{ -2 } \) = 1
\(\frac { x }{ 3 } \) – \(\frac { y }{ 2 } \) = 1
2x – 3y = 6
2x – 3y – 6 = 0

When a =-2
x – intercept = -2
y – intercept = 1 – (- 2) = 1 + 2 = 3
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ -2 } \) + \(\frac { y }{ 3 } \) = 1
– \(\frac { x }{ 2 } \) + \(\frac { y }{ 3 } \) = 1
– 3x + 2y = 6
3x – 2y + 6 = 0

Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?
Answer:
Let the selling price of a milk be “x”
Let the demand be “y”
We have to find the linear equation connecting them
Two points on the line are (14, 980) and (16,1220)
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 1220-980 }{ 16-14 } \) = \(\frac { 240 }{ 2 } \) = 120
Equation of the line is y – y₁ = m (x – x₁)
y – 980 = 120 (x – 14) ⇒ y – 980 = 120 x – 1680
-120 x + y = -1680 + 980 ⇒ -120 x + y = -700 ⇒ 120 x – y = 700
Given the value of x = 17
120(17) – y = 700
-y = 700 – 2040 ⇒ – y = – 1340
y = 1340
The demand is 1340 liters

Question 8.
Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer:
Let the image of P(3, 8) and P’ (a, b)
Let the point of intersection be O

Slope of x + 3y = 7 is – \(\frac { 1 }{ 3 } \)
Slope of PP’ = 3 (perpendicular)
Equation of PP’ is
y – y₁ = m(x – x₁)
y – 8 = 3 (x – 3)
y – 8 = 3x – 9
-8 + 9 = 3x – y
∴ 3x – y = 1 ………(1)
The two line meet at 0

Substitute the value of x = 1 in (1)
3 – y = 1
3 – 1 = y
2 = y
The point O is (1,2)
Mid point of pp’ = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(1,2) = (\(\frac { 3+a }{ 2 } \),\(\frac { 8+b }{ 2 } \))
∴ \(\frac { 3+a }{ 2 } \) = 1 ⇒ 3 + a = 2
a = 2 – 3 = -1
\(\frac { 8+b }{ 2 } \) = 2
8 + b = 4
b = 4 – 8 = – 4
The point P’ is (-1, -4)

Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer:
Given lines

Substitute the value of y = \(\frac { 5 }{ 13 } \) in (2)
2x – 3 × \(\frac { 5 }{ 13 } \) = -1
2x – \(\frac { 15 }{ 13 } \) = -1
26x – 15 = -13
26x = -13 + 15
26x = 2
x = \(\frac { 2 }{ 26 } \) = \(\frac { 1 }{ 13 } \)
The point of intersection is (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))

Let the x – intercept and y intercept be “a”
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (equal intercepts)
It passes through (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))
\(\frac { 1 }{ 13a } \) + \(\frac { 5 }{ 13a } \) = 1
\(\frac { 1+5 }{ 13a } \) = 1
13a = 6
a = \(\frac { 6 }{ 13 } \)
The equation of the line is

Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Answer:
Two straight path will intersect at one point.
Solving this equations

2x – 3y + 4 = 0

Substitute the value of x = \(\frac { -1 }{ 17 } \) in (2)

The point of intersection is (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
Any equation perpendicular to 6x – 7y + 8 = 0 is 7x + 6y + k = 0
It passes through (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
7(-\(\frac { 1 }{ 17 } \)) + 6 (\(\frac { 22 }{ 17 } \)) + k = 0
Multiply by 17
-7 + 6 (22) + 17k = 0
-7 + 132 + 17k = 0
17k = -125 ⇒ k = – \(\frac { 125 }{ 17 } \)
The equation of a line is 7x + 6y – \(\frac { 125 }{ 17 } \) = 0
119x + 102y – 125 = 0
∴ Equation of the path is 119x + 102y – 125 = 0

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
Answer:
(2) 25 sq.units Hint.
Hint:
Area of the ∆

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Answer:
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Answer:
(2) parallel to Y axis

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
Answer:
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
Answer:
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 } \). The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2
Hint:
Slope of a line = \(\frac { 1 }{ 8 } \)

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 } \)
(4) -8
Answer:
(2) 1
Hint:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-0 }{ -8-0 } \) = \(\frac { 8 }{ -8 } \) = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is …………..
(1) \(\sqrt { 3 }\)
(2) –\(\sqrt { 3 }\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Answer:
(2) –\(\sqrt { 3 }\)
Hint:
Slope of a line = \(\frac{1}{\sqrt{3}}\)
Slope of the ⊥r bisector = –\(\sqrt { 3 }\)

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Answer:
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= \(\frac { -7 }{ -3 } \) = \(\frac { 7 }{ 3 } \)
Slope of its ⊥^r = \(\frac { -3 }{ 7 } \)
The line passes through (0,0)
Equation of a line is
y – y₁ = m(x – x₁)
y – 0 = \(\frac { -3 }{ 7 } \) (x – 0)
y = \(\frac { -3 }{ 7 } \) x ⇒ 7y = -3x
3x + 7y = 0

Question 11.
Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂ : 4y = 3x – 1
(iii) l₃ : 4y + 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true?
(1) l₁ and l₂ are perpendicular
(2) l₂ and l₄ are parallel
(3) l₂ and l₄ are perpendicular
(4) l₂ and l₃ are parallel
Answer:
(3) l₂ and l₄ are perpendicular
Hint:
Slope of l₁ = \(\frac { 4 }{ 3 } \); Slope of l₂ = \(\frac { 3 }{ 4 } \)
Slope of l₃ = – \(\frac { 3 }{ 4 } \); Slope of l₄ = –\(\frac { 4 }{ 3 } \)
(1) l₁ × l₂ = \(\frac { 4 }{ 3 } \) × \(\frac { 3 }{ 4 } \) = 1 …….False
(2) l₁ = \(\frac { 4 }{ 3 } \); l₄ = – \(\frac { 4 }{ 3 } \) not parallel ………False
(3) l₂ × l₄ = \(\frac { 3 }{ 4 } \) × – \(\frac { 4 }{ 3 } \) = -1 …….True
(4) l₂ = \(\frac { 3 }{ 4 } \); l₃ = – \(\frac { 3 }{ 4 } \) not parallel ………False

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
Answer:
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = \(\frac { 4 }{ 8 } \) x + \(\frac { 21 }{ 8 } \)
= \(\frac { 1 }{ 2 } \) x + \(\frac { 21 }{ 8 } \)
\(\frac { 1 }{ 2 } \) = 0.5
\(\frac { 21 }{ 8 } \) = 2.625
Slope = \(\frac { 1 }{ 2 } \) = 0.5
y intercept = \(\frac { 21 }{ 8 } \) = 2.6

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Answer:
(2) The slopes of two pair of opposite sides

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths 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 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0